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\ihead{Linear optimization}
\ohead{\today}
\begin{document}

\title{Elementary derivations of some results of linear optimization}
\author{Darij Grinberg}
\date{version 1.0, 7 August 2017}
\maketitle
\tableofcontents

\newpage

\begin{statement}
\textbf{Remark (2017):}

This note has been mostly written in 2012, when I was learning combinatorial
optimization from Schrijver's online notes \cite{Schrij-CCO} and felt a need
for elementary and constructive proofs of the main theorems of linear
optimization (as opposed to the standard short proofs using compactness and
the Hahn-Banach theorem). This note provides such proofs, although I cannot
vouch for their readability\footnote{They are overdetailed in many places, the
result of my attempts to ensure their correctness; unfortunately the trees
often obscure the forest.}.

Needless to say, this note breaks no new ground, and probably these proofs (or
easier ones) appear often enough in the literature (in fact, I suspect that
almost all of the material of this note is covered by each of the two
textbooks \cite{Laurit13} and \cite{Schrij98}). Writing them up was a learning
experience which, I fear, reading them will not be.

While this note often refers to \cite{Schrij-CCO}, it is actually
self-contained and can be read separately (except for the few places where it
modifies arguments from \cite{Schrij-CCO} to make them constructive; but these
are tangential to the note).
\end{statement}

The purpose of this note is to give elementary proofs for various results in
linear optimization theory. Here, ``elementary'' means that no analysis is
being used, and that the proofs are ``morally'' constructive.

Let me explain what \textquotedblleft morally constructive\textquotedblright%
\ means: The results and proofs given below are not valid in constructive
logic, but this is solely for the reason that $\mathbb{R}$ is not a discrete
field in constructive logic. If we would formulate the results and proofs
below for $\mathbb{Q}$ instead of $\mathbb{R}$, then they would become valid
in constructive logic. Thus, when I make any claim in constructive logic
below, I tacitly want it to be understood with all $\mathbb{R}$'s replaced by
$\mathbb{Q}$'s (and all \textquotedblleft real numbers\textquotedblright%
\ replaced by \textquotedblleft rational numbers\textquotedblright, and so on).

Note that the following proofs, being free of analysis, generalize to any
ordered field instead of $\mathbb{R}$ (for example, to $\mathbb{Q}$ or to
$\mathbb{Q}\left(  \sqrt{2}\right)  $). However, we are going to formulate
them for the field $\mathbb{R}$ only (trusting that the reader, if necessary,
can generalize them on his own). In particular, whenever we speak of
\textquotedblleft vector spaces\textquotedblright, \textquotedblleft
vectors\textquotedblright\ and \textquotedblleft matrices\textquotedblright%
\ below, we mean vector spaces, vectors and matrices over $\mathbb{R}$.

This note was originally intended as a supplement to Chapter 2 of Schrijver's
notes \cite{Schrij-CCO}\footnote{In particular, this explains the strange
numbering of results in this note: The numbers have been chosen so as not to
conflict with the labeling of \cite{Schrij-CCO}.}; but it is fully
self-contained. It proves separation theorems for convex hulls and cones (of
finite sets), the Farkas, Gordan, Stiemke and Motzkin theorems, and two
versions of the linear programming duality theorem. Also, it \textquotedblleft
patches\textquotedblright\ the proof of Theorem 2.3 in \cite{Schrij-CCO} (the
theorem saying that any bounded polyhedron is the convex hull of its vertices)
to make it constructive.

\section{\label{sect.basics-conv}Notations and basics 1: Convex sets}

Before we start formulating and proving substantial theorems, let us introduce
some notations. The notations that we introduce will usually be identical to
the ones introduced in \cite{Schrij-CCO}, save for a few exceptions.

In the following, $\mathbb{N}$ will always denote the set $\left\{
0,1,2,\ldots\right\}  $; thus, we regard $0$ as an element of $\mathbb{N}$
(unlike some other authors).

We will use the words ``vector'' and ``point'' as synonyms; both of them will
denote elements of a vector space.

\begin{statement}
\textbf{Definition.} Let $E$ be an $\mathbb{R}$-vector space. Let $C$ be a
subset of $E$. We say that $C$ is \textit{convex} if every two elements $x\in
C$ and $y\in C$ and every real number $\lambda\in\left[  0,1\right]  $ satisfy
$\lambda x+\left(  1-\lambda\right)  y\in C$.
\end{statement}

Geometrically, this definition is often put into words as follows: Let $E$ be
an $\mathbb{R}$-vector space. Let $C$ be a subset of $E$. We say that $C$ is
\textit{convex} if every point on a segment containing two points of $C$ must
also lie in $C$.

The following property of convex sets is easy to see:

\begin{statement}
\textbf{Proposition 2.0a.} Let $E$ be an $\mathbb{R}$-vector space. Let $C$ be
a convex subset of $E$. Let $I$ be any set, and let $\left(  x_{i}\right)
_{i\in I}$ be a family of elements of $C$ indexed by elements of $I$. Let
$\left(  \lambda_{i}\right)  _{i\in I}$ be a family of nonnegative reals
indexed by elements of $I$ such that all but finitely many $i\in I$ satisfy
$\lambda_{i}=0$. Assume also that $\sum\limits_{i\in I}\lambda_{i}=1$. Then,
$\sum\limits_{i\in I}\lambda_{i}x_{i}\in C$.
\end{statement}

For the sake of completeness, we shall give a proof of Proposition 2.0a in
Section \ref{sect.basics-proofs}.

\begin{statement}
\textbf{Definition.} Let $E$ be an $\mathbb{R}$-vector space. Let $I$ be any
set, and let $\left(  x_{i}\right)  _{i\in I}$ be a family of elements of $E$
indexed by elements of $I$. If $\left(  \lambda_{i}\right)  _{i\in I}$ is a
family of nonnegative reals indexed by elements of $I$ such that all but
finitely many $i\in I$ satisfy $\lambda_{i}=0$, and such that $\sum
\limits_{i\in I}\lambda_{i}=1$, then the vector $\sum\limits_{i\in I}%
\lambda_{i}x_{i}$ is said to be a \textit{convex combination} of the vectors
$x_{i}$ for $i\in I$.
\end{statement}

Using this definition, Proposition 2.0a rewrites as follows:

\begin{statement}
\textbf{Proposition 2.0b.} Let $E$ be an $\mathbb{R}$-vector space. Let $C$ be
a convex subset of $E$. Let $I$ be any set, and let $\left(  x_{i}\right)
_{i\in I}$ be a family of elements of $C$ indexed by elements of $I$. Then,
any convex combination of the vectors $x_{i}$ for $i\in I$ lies in $C$.
\end{statement}

This result is often stated in words as follows: A convex set is closed under
convex combinations.

The notion of convex sets gives rise to another notion: that of a convex hull.
The notion of a convex hull can be defined in several ways; here are three:

\begin{statement}
\textbf{Definition 2.0c.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Then, the \textit{convex hull} of $S$ will denote the
intersection of all convex subsets of $E$ which contain $S$ as a subset. We
denote the convex hull of $S$ by $\operatorname*{conv}.\operatorname*{hull}S$.
\end{statement}

\begin{statement}
\textbf{Definition 2.0d.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Then, the \textit{convex hull} of $S$ will denote the set of
all convex combinations of the vectors $s$ for $s\in S$. (This will often be
abbreviated as follows: ``The convex hull of $S$ will denote the set of all
convex combinations of the elements of $S$.'') We denote the convex hull of
$S$ by $\operatorname*{conv}.\operatorname*{hull}S$.
\end{statement}

\begin{statement}
\textbf{Definition 2.0e.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Then, the \textit{convex hull} of $S$ will denote the set%
\[
\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
We denote the convex hull of $S$ by $\operatorname*{conv}.\operatorname*{hull}%
S$.
\end{statement}

Definitions 2.0c, 2.0d and 2.0e are equivalent (according to Proposition 2.0f
\textbf{(a)} below).

The following result gathers some fundamental properties of convex hulls, most
of which are standardly used without explicitly calling them out:

\begin{statement}
\textbf{Proposition 2.0f.} \textbf{(a)} Definitions 2.0c, 2.0d and 2.0e are equivalent.

Let $E$ be an $\mathbb{R}$-vector space.

\textbf{(b)} We have $\operatorname*{conv}.\operatorname*{hull}\varnothing
=\varnothing$.

Let now $S$ be a subset of $E$.

\textbf{(c)} We have $S\subseteq\operatorname*{conv}.\operatorname*{hull}S$.

\textbf{(d)} The convex hull $\operatorname*{conv}.\operatorname*{hull}S$ is a
convex set.

\textbf{(e)} If $T$ is a subset of $S$, then $\operatorname*{conv}%
.\operatorname*{hull}T\subseteq\operatorname*{conv}.\operatorname*{hull}S$.

\textbf{(f)} Every convex subset of $E$ which contains $S$ as a subset also
contains $\operatorname*{conv}.\operatorname*{hull}S$ as a subset.

\textbf{(g)} If $T$ is a subset of $\operatorname*{conv}.\operatorname*{hull}%
S$, then $\operatorname*{conv}.\operatorname*{hull}T\subseteq
\operatorname*{conv}.\operatorname*{hull}S$.
\end{statement}

Again, we refer to Section \ref{sect.basics-proofs} for a proof of Proposition 2.0f.

\begin{statement}
\textbf{Remark.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be a subset
of $E$. Then, the convex hull $\operatorname*{conv}.\operatorname*{hull}S$ is
a convex set (by Proposition 2.0f \textbf{(d)}) and contains $S$ as a subset
(by Proposition 2.0f \textbf{(c)}). Moreover, every convex subset of $E$ which
contains $S$ as a subset also contains $\operatorname*{conv}%
.\operatorname*{hull}S$ as a subset (by Proposition 2.0f \textbf{(f)}). This
result is often put into words as follows: \textquotedblleft The convex hull
$\operatorname*{conv}.\operatorname*{hull}S$ is the smallest convex set
containing $S$ as a subset.\textquotedblright
\end{statement}

The following proposition (which is, again, fundamental and will be used
without explicit mention) is a simple consequence of Definition 2.0e:

\begin{statement}
\textbf{Proposition 2.0g.} Let $E$ be an $\mathbb{R}$-vector space.

\textbf{(a)} If $S$ is any subset of $E$, then%
\begin{align*}
&  \operatorname*{conv}.\operatorname*{hull}S\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  .
\end{align*}


\textbf{(b)} Let $F$ be a vector subspace of $E$. Let $S$ be a subset of $F$.
Then, the convex hull $\operatorname*{conv}.\operatorname*{hull}S$ does not
depend on whether we consider $S$ as a subset of $F$ or as a subset of $E$.
\end{statement}

Again, we refer to Section \ref{sect.basics-proofs} for a proof of Proposition 2.0g.

Here is yet another proposition that will be used tacitly:

\begin{statement}
\textbf{Proposition 2.0h.} Let $E$ be an $\mathbb{R}$-vector space. Let
$x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ be finitely many vectors in $E$.

\textbf{(a)} Then,%
\begin{align*}
&  \operatorname*{conv}.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots
,x_{n}\right\} \\
&  =\left(  \text{the set of all convex combinations of the vectors }%
x_{1}\text{, }x_{2}\text{, }\ldots\text{, }x_{n}\right)  .
\end{align*}
\textbf{(b)} Let $x\in E$. Then, we have $x\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  $ if and only
if $x$ is a convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$,
$x_{n}$.
\end{statement}

Again, we refer to Section \ref{sect.basics-proofs} for a proof of Proposition 2.0h.

Now, we can easily define polytopes:

\begin{statement}
\textbf{Definition.} Let $E$ be an $\mathbb{R}$-vector space. Let $P$ be a
subset of $E$. Then, $P$ is said to be a \textit{polytope} if $P$ is the
convex hull of a finite subset of $E$. In other words, $P$ is said to be a
\textit{polytope} if there are finitely many vectors $x_{1}$, $x_{2}$,
$\ldots$, $x_{n}$ in $E$ such that $P=\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  $.
\end{statement}

We are not going to define the notion of a polyhedron (nor will we use this
notion); we refer to \cite{Schrij-CCO} for that.

\section{\label{sect.basics-cone}Notations and basics 2: Cones}

Convex cones are a notion similar to that of convex sets.

\begin{statement}
\textbf{Definition.} Let $E$ be an $\mathbb{R}$-vector space. Let $C$ be a
subset of $E$. We say that $C$ is a \textit{convex cone in }$E$ if it
satisfies the following two conditions:

\begin{itemize}
\item We have $0\in C$.

\item Every two elements $x\in C$ and $y\in C$ and every nonnegative reals
$\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C$.
\end{itemize}

We will abbreviate \textquotedblleft convex cone in $E$\textquotedblright\ as
\textquotedblleft convex cone\textquotedblright\ when the value of $E$ is
clear from the context.
\end{statement}

Note that this definition is in slight conflict with the definition of a
convex cone in \cite{Schrij-CCO}. In fact, the definition of a convex cone in
\cite{Schrij-CCO} does not contain the condition that $0\in C$. This is
irrelevant for nonempty subsets $C$ of $E$, because if a nonempty subset $C$
of $E$ is a convex cone in the sense of \cite{Schrij-CCO}, then it must
automatically contain $0$ and therefore is also a convex cone in the sense of
our definition. So the only difference between our definition of a convex cone
and the definition given in \cite{Schrij-CCO} is that the empty set
$\varnothing$ is a convex cone in the sense of \cite{Schrij-CCO}, but not a
convex cone in the sense of our definition. This is not a particularly
significant difference.\footnote{One advantage of our definition is that it
satisfies the identity $\operatorname*{cone}\left(  A\cup B\right)
=\operatorname*{cone}A+\operatorname*{cone}B$ for any two subsets $A$ and $B$
of $E$ (whereas with Schrijver's definition, this holds only when $A$ and $B$
are both nonempty or both empty). We will not use this identity, however.}

The name \textquotedblleft convex cone\textquotedblright\ is somewhat
presumptuous: It seems to imply that any convex cone is a convex set, although
we have not proved this. Fortunately, this is true: Any convex cone is indeed
a convex set\footnote{\textit{Proof.} Let $E$ be an $\mathbb{R}$-vector space.
Let $C$ be a convex cone in $E$. We shall prove that $C$ is a convex set.
\par
Recall that $C$ is a convex cone. Hence, every two elements $x\in C$ and $y\in
C$ and every nonnegative reals $\lambda$ and $\mu$ satisfy%
\begin{equation}
\lambda x+\mu y\in C \label{eq.convcone.pf.1}%
\end{equation}
(by the definition of a convex cone).
\par
Now, let $x\in C$, $y\in C$ and $\lambda\in\left[  0,1\right]  $. Then, both
$\lambda$ and $1-\lambda$ are nonnegative reals (since $\lambda\in\left[
0,1\right]  $). Hence, $\lambda x+\left(  1-\lambda\right)  y\in C$ (by
(\ref{eq.convcone.pf.1}), applied to $\mu=1-\lambda$).
\par
Let us now forget that we fixed $x$, $y$ and $\lambda$. We thus have proven
that every two elements $x\in C$ and $y\in C$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(  1-\lambda\right)
y\in C$. In other words, $C$ is a convex set (by the definition of a
\textquotedblleft convex set\textquotedblright).
\par
Now, let us forget that we fixed $C$. We thus have shown that if $C$ is a
convex cone in $E$, then $C$ is a convex set. In other words, any convex cone
is a convex set. Qed.}.

\begin{statement}
\textbf{Remark.} Let $E$ be an $\mathbb{R}$-vector space. Let $C$ be a convex
cone in $E$. It is easy to see that every $x\in C$ and every nonnegative real
$\lambda$ satisfy $\lambda x\in C$. In words, this is often stated as follows:
``Convex cones are closed under multiplication by a nonnegative scalar''.
\end{statement}

Similarly to Proposition 2.0a, we have:

\begin{statement}
\textbf{Proposition 2.0i.} Let $E$ be an $\mathbb{R}$-vector space. Let $C$ be
a convex cone in $E$. Let $I$ be any set, and let $\left(  x_{i}\right)
_{i\in I}$ be a family of elements of $C$ indexed by elements of $I$. Let
$\left(  \lambda_{i}\right)  _{i\in I}$ be a family of nonnegative reals
indexed by elements of $I$ such that all but finitely many $i\in I$ satisfy
$\lambda_{i}=0$. Then, $\sum\limits_{i\in I}\lambda_{i}x_{i}\in C$.
\end{statement}

In words, Proposition 2.0i is often stated as follows: \textquotedblleft Any
linear combination of finitely many elements of a convex cone with nonnegative
coefficients must lie in this cone.\textquotedblright

The proof of Proposition 2.0i is similar to (and somewhat easier than) that of
Proposition 2.0a; for the sake of completeness, we shall give it in Section
\ref{sect.basics-proofs}.

Just as the notion of convex sets gave rise to the notion of convex hulls, the
notion of convex cones will give rise to the notion of convex conic hulls.
Here are three definitions for this notion:

\begin{statement}
\textbf{Definition 2.0j.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Then, the \textit{convex conic hull} of $S$ will denote the
intersection of all convex cones in $E$ which contain $S$ as a subset. We
denote the convex conic hull of $S$ by $\operatorname*{cone}S$.
\end{statement}

\begin{statement}
\textbf{Definition 2.0k.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Then, the \textit{convex conic hull} of $S$ will denote the
set of all linear combinations of the vectors $s$ for $s\in S$ with
nonnegative coefficients. (This will often be abbreviated as follows:
\textquotedblleft The convex conic hull of $S$ will denote the set of all
linear combinations of the elements of $S$ with nonnegative
coefficients.\textquotedblright) We denote the convex conic hull of $S$ by
$\operatorname*{cone}S$.
\end{statement}

\begin{statement}
\textbf{Definition 2.0l.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Then, the \textit{convex conic hull} of $S$ will denote the
set%
\[
\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
We denote the convex conic hull of $S$ by $\operatorname*{cone}S$.
\end{statement}

Definitions 2.0j, 2.0k and 2.0l are equivalent (according to Proposition 2.0m
\textbf{(a)} below).

The following result is analogous to Proposition 2.0f (except for its part
\textbf{(h)}, which is easy)\footnote{And its proof (apart from part
\textbf{(h)}) is analogous to the proof of Proposition 2.0f. For example,
Proposition 2.0m \textbf{(f)} follows from Definition 2.0k in the same way as
Proposition 2.0f \textbf{(f)} follows from Definition 2.0d.}:

\begin{statement}
\textbf{Proposition 2.0m.} \textbf{(a)} Definitions 2.0j, 2.0k and 2.0l are equivalent.

Let $E$ be an $\mathbb{R}$-vector space.

\textbf{(b)} We have\footnote{Here and in the following, we use the symbol
\textquotedblleft$0$\textquotedblright\ not just for the number zero and the
zero vector in a vector space, but also for the zero subspace of any vector
subspace. In other words, if $E$ is any $\mathbb{R}$-vector space, then the
subspace $\left\{  0\right\}  $ of $E$ is also denoted by $0$. Hopefully, this
abuse of notation will sow no confusion.} $\operatorname*{cone}\varnothing=0$
(not $\varnothing$).

Let now $S$ be a subset of $E$.

\textbf{(c)} We have $S\subseteq\operatorname*{cone}S$.

\textbf{(d)} The convex conic hull $\operatorname*{cone}S$ is a convex cone.

\textbf{(e)} If $T$ is a subset of $S$, then $\operatorname*{cone}%
T\subseteq\operatorname*{cone}S$.

\textbf{(f)} Every convex cone in $E$ which contains $S$ as a subset also
contains $\operatorname*{cone}S$ as a subset.

\textbf{(g)} If $T$ is a subset of $\operatorname*{cone}S$, then
$\operatorname*{cone}T\subseteq\operatorname*{cone}S$.

\textbf{(h)} We have $\operatorname*{conv}.\operatorname*{hull}S\subseteq
\operatorname*{cone}S$.
\end{statement}

We refer (again) to Section \ref{sect.basics-proofs} for a proof of
Proposition 2.0m.

\begin{statement}
\textbf{Remark.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be a subset
of $E$. Then, the convex conic hull $\operatorname*{cone}S$ is a convex cone
(by Proposition 2.0m \textbf{(d)}) and contains $S$ as a subset (by
Proposition 2.0m \textbf{(c)}). Moreover, every convex cone in $E$ which
contains $S$ as a subset also contains $\operatorname*{cone}S$ as a subset (by
Proposition 2.0m \textbf{(f)}). This result is often put into words as
follows: \textquotedblleft The convex conic hull $\operatorname*{cone}S$ is
the smallest convex cone containing $S$ as a subset.\textquotedblright
\end{statement}

The following proposition (which is, again, fundamental and will be used
without explicit mention) is a simple consequence of Definition 2.0l:

\begin{statement}
\textbf{Proposition 2.0n.} Let $E$ be an $\mathbb{R}$-vector space.

\textbf{(a)} If $S$ is any subset of $E$, then%
\begin{align*}
&  \operatorname*{cone}S\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\}  .
\end{align*}


\textbf{(b)} Let $F$ be a vector subspace of $E$. Let $S$ be a subset of $F$.
Then, the convex conic hull $\operatorname*{cone}S$ does not depend on whether
we consider $S$ as a subset of $F$ or as a subset of $E$.
\end{statement}

See Section \ref{sect.basics-proofs} for a proof of Proposition 2.0n.

Here is yet another proposition (an analogue of Proposition 2.0h) that will be
used tacitly:

\begin{statement}
\textbf{Proposition 2.0o.} Let $E$ be an $\mathbb{R}$-vector space. Let
$x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ be finitely many vectors in $E$.

\textbf{(a)} Then,%
\begin{align*}
&  \operatorname*{cone}\left\{  x_{1},x_{2},\ldots,x_{n}\right\} \\
&  =\left(  \text{the set of all linear combinations of the}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{vectors }x_{1}\text{, }x_{2}\text{,
}\ldots\text{, }x_{n}\text{ with nonnegative coefficients}\right)  .
\end{align*}
\textbf{(b)} Let $x\in E$. Then, we have $x\in\operatorname*{cone}\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  $ if and only if $x$ is a linear combination
of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients.
\end{statement}

A proof of Proposition 2.0o can be found in Section \ref{sect.basics-proofs}.

The following proposition is nearly trivial; we state it merely for convenience:

\begin{statement}
\textbf{Proposition 2.0p.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a finite subset of $E$.

\textbf{(a)} Then,%
\[
\operatorname*{cone}S=\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(
\nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative reals}\right\}  .
\]
In particular:

\textbf{(b)} If $\left(  \nu_{s}\right)  _{s\in S}$ is a family of nonnegative
reals indexed by elements of $S$, then $\sum\limits_{s\in S}\nu_{s}%
s\in\operatorname*{cone}S$.

\textbf{(c)} Conversely, if $p$ is an element of $\operatorname*{cone}S$, then
there exists a family $\left(  \nu_{s}\right)  _{s\in S}$ of nonnegative reals
indexed by elements of $S$ such that $p=\sum\limits_{s\in S}\nu_{s}s$.
\end{statement}

For the sake of completeness, the proof of Proposition 2.0p will be given in
Section \ref{sect.basics-proofs}.

\section{\label{sect.basics-dual}Notations and basics 3: On the dual space}

Here are two further notations that we will be using:

\begin{itemize}
\item If $E$ is any $\mathbb{R}$-vector space, then $E^{\ast}$ will mean the
$\mathbb{R}$-vector space $\operatorname*{Hom}\nolimits_{\mathbb{R}}\left(
E,\mathbb{R}\right)  $ of all $\mathbb{R}$-linear maps from $E$ to
$\mathbb{R}$. This $E^{\ast}$ is called the \textit{dual space} of $E$.

\item If $\ell\in\mathbb{N}$, then we regard the vectors in $\mathbb{R}^{\ell
}$ as column vectors of length $\ell$, and we regard the vectors in $\left(
\mathbb{R}^{\ell}\right)  ^{\ast}$ as row vectors of length $\ell$. More
precisely, we identify every row vector $\left(  a_{1},a_{2},\ldots,a_{\ell
}\right)  \in\mathbb{R}^{\ell}$ with the $\mathbb{R}$-linear map $f\in\left(
\mathbb{R}^{\ell}\right)  ^{\ast}=\operatorname*{Hom}\nolimits_{\mathbb{R}%
}\left(  \mathbb{R}^{\ell},\mathbb{R}\right)  $ which sends every column
vector $\left(
\begin{array}
[c]{c}%
b_{1}\\
b_{2}\\
\vdots\\
b_{\ell}%
\end{array}
\right)  \in\mathbb{R}^{\ell}$ to the scalar $\left(  a_{1},a_{2}%
,\ldots,a_{\ell}\right)  \left(
\begin{array}
[c]{c}%
b_{1}\\
b_{2}\\
\vdots\\
b_{\ell}%
\end{array}
\right)  =\sum\limits_{i=1}^{\ell}a_{i}b_{i}\in\mathbb{R}$. Thus, for every
$f\in\left(  \mathbb{R}^{\ell}\right)  ^{\ast}$ and $b\in\mathbb{R}^{\ell}$,
we have $f\left(  b\right)  =fb$ (where $f$ is regarded as an element of
$\left(  \mathbb{R}^{\ell}\right)  ^{\ast}=\operatorname*{Hom}%
\nolimits_{\mathbb{R}}\left(  \mathbb{R}^{\ell},\mathbb{R}\right)  $ on the
left hand side, and regarded as a row vector on the right hand side).
\end{itemize}

Let us state a simple property of convex hulls that we will be using several times:

\begin{statement}
\textbf{Proposition 2.0r.} Let $E$ be an $\mathbb{R}$-vector space. Let
$t\in\mathbb{N}$. Let $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$ be $t$ vectors in
$E$. Let $f\in E^{\ast}$ and $\delta\in\mathbb{R}$. Assume that%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,t\right\}  \text{ satisfies }f\left(
x_{i}\right)  <\delta. \label{2.0r.ass}%
\end{equation}


Let $C=\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1},x_{2}%
,\ldots,x_{t}\right\}  $. Then, every $x\in C$ satisfies $f\left(  x\right)
<\delta$.
\end{statement}

The (rather easy) proof of Proposition 2.0r can be found in Section
\ref{sect.basics-proofs} below.

Here is an analogous property of convex conic hulls:

\begin{statement}
\textbf{Proposition 2.0s.} Let $E$ be an $\mathbb{R}$-vector space. Let $S$ be
a subset of $E$. Let $f\in E^{\ast}$. Assume that%
\begin{equation}
\text{every }s\in S\text{ satisfies }f\left(  s\right)  \leq0.
\label{2.0s.ass}%
\end{equation}


Then, every $x\in\operatorname*{cone}S$ satisfies $f\left(  x\right)  \leq0$.
\end{statement}

Again, we refer to Section \ref{sect.basics-proofs} for a proof of Proposition 2.0s.

Next, let us define the notion of a hyperplane:

\begin{statement}
\textbf{Definition.} Let $E$ be an $\mathbb{R}$-vector space. If $f$ is a
nonzero element of $E^{\ast}$, and $\delta$ is an element of $\mathbb{R}$,
then the subset $f^{-1}\left(  \delta\right)  $ of $E$ will be called a
\textit{hyperplane} in $E$.
\end{statement}

Note that this definition of a hyperplane is equivalent to the one given in
\cite{Schrij-CCO} when $E=\mathbb{R}^{n}$ for some $n\in\mathbb{N}$. In fact,
when $E=\mathbb{R}^{n}$, then any element $f\in E^{\ast}$ has the form
\begin{align*}
E  &  \rightarrow\mathbb{R},\\
x  &  \mapsto c^{T}x
\end{align*}
for some $c\in\mathbb{R}^{n}$, and this $c$ is nonzero if and only if $f$ is nonzero.

\begin{statement}
\textbf{Definition.} Let $E$ be an $\mathbb{R}$-vector space. Let $H$ be a
hyperplane in $E$. Let $z\in E$ and $C\subseteq E$. We say that the hyperplane
$H$ \textit{separates} $z$ and $C$ if there exist a nonzero $f\in E^{\ast}$
and a $\delta\in\mathbb{R}$ such that $H=f^{-1}\left(  \delta\right)  $,
$f\left(  z\right)  >\delta$ and $\left(  \text{every }x\in C\text{ satisfies
}f\left(  x\right)  <\delta\right)  $.
\end{statement}

Note that this definition of ``separate'' is equivalent to the one given by
Schrijver in \cite{Schrij-CCO} when $E=\mathbb{R}^{n}$ for some $n\in
\mathbb{N}$. Schrijver's definition, however, has the disadvantage of using a
topological notion (that of a ``component''), which makes it difficult to
generalize to other ordered fields instead of $\mathbb{R}$. It should be
remarked that Schrijver never really uses his definition of ``separate'' in
\cite{Schrij-CCO}; instead, he more or less uses my definition.

Furthermore, we introduce the relations $\geq$ and $\leq$ on vectors:

\begin{statement}
\textbf{Definition.} Let $\ell\in\mathbb{N}$. Let $u$ and $v$ be two column
vectors in $\mathbb{R}^{\ell}$. Then, we write $u\leq v$ if and only if every
$i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }u\right)  \leq\left(  \text{the
}i\text{-th coordinate of }v\right)  .
\]
Also, we write $u\geq v$ if and only if every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }u\right)  \geq\left(  \text{the
}i\text{-th coordinate of }v\right)  .
\]
The same notations apply if $u$ and $v$ are row vectors in $\left(
\mathbb{R}^{\ell}\right)  ^{\ast}$ rather than column vectors in
$\mathbb{R}^{\ell}$.
\end{statement}

Of course, it is true that if $u$ and $v$ are two column vectors in
$\mathbb{R}^{\ell}$ (or two row vectors in $\left(  \mathbb{R}^{\ell}\right)
^{\ast}$), then $u\geq v$ holds if and only if $v\leq u$. The relation $\leq$
is the smaller-or-equal relation of a partial order on $\mathbb{R}^{\ell}$
(namely, the so-called \textit{componentwise partial order}), and the relation
$\geq$ is the greater-or-equal relation of this order. However, this order is
not a total order (unless $\ell\leq1$): for example, if $u=\left(
\begin{array}
[c]{c}%
0\\
1
\end{array}
\right)  $ and $v=\left(
\begin{array}
[c]{c}%
1\\
0
\end{array}
\right)  $, then we have neither $u\geq v$ nor $v\geq u$.

Clearly, the relation $\leq$ respects the addition of vectors: If
$u\in\mathbb{R}^{\ell}$, $v\in\mathbb{R}^{\ell}$ and $w\in\mathbb{R}^{\ell}$
are such that $u\leq v$, then $u+w\leq v+w$. Moreover, if $u\in\mathbb{R}%
^{\ell}$, $v\in\mathbb{R}^{\ell}$, $x\in\mathbb{R}^{\ell}$ and $y\in
\mathbb{R}^{\ell}$ are such that $u\leq v$ and $x\leq y$, then $u+x\leq v+y$.
As a consequence, two vectors $u\in\mathbb{R}^{\ell}$ and $v\in\mathbb{R}%
^{\ell}$ satisfy $u\leq v$ if and only if $u-v\leq0$. Furthermore, if
$x\in\mathbb{R}^{\ell}$ and $y\in\mathbb{R}^{\ell}$ are such that $x\leq y$,
and if $\lambda$ is a nonnegative real, then $\lambda x\leq\lambda y$. Similar
rules hold for the relation $\geq$.

\begin{statement}
\textbf{Definition.} Let $\ell\in\mathbb{N}$. A vector $v$ lying either in
$\mathbb{R}^{\ell}$ or in $\left(  \mathbb{R}^{\ell}\right)  ^{\ast}$ is said
to be \textit{nonnegative} if it satisfies $v\geq0$.
\end{statement}

It is clear that a sum of nonnegative vectors in $\mathbb{R}^{\ell}$ is again
nonnegative. More generally, any linear combination of nonnegative vectors in
$\mathbb{R}^{\ell}$ with nonnegative coefficients is again nonnegative.

Let us state a trivial but important fact:

\begin{statement}
\textbf{Lemma 2.0t.} Let $n\in\mathbb{N}$.

\textbf{(a)} If a column vector $v\in\mathbb{R}^{n}$ satisfies $v\geq0$, then
all coordinates of the column vector $v$ are nonnegative.

\textbf{(b)} If a row vector $v\in\left(  \mathbb{R}^{n}\right)  ^{\ast}$
satisfies $v\geq0$, then all coordinates of the row vector $v$ are nonnegative.
\end{statement}

A proof of Lemma 2.0t can be found in Section \ref{sect.basics-proofs}. Of
course, the converse of Lemma 2.0t also holds; we just won't use it often
enough to have a reason to state it.

Here is a simple property of nonnegative vectors, which will not be directly
used in these notes but which provides some context for them:

\begin{statement}
\textbf{Lemma 2.0u.} Let $n\in\mathbb{N}$. Let $x\in\mathbb{R}^{n}$. Then,
there exist two vectors $y$ and $z$ in $\mathbb{R}^{n}$ such that $y\geq0$,
$z\geq0$ and $x=y-z$.
\end{statement}

A proof of Lemma 2.0u can be found in Section \ref{sect.basics-proofs}.

Let us record another simple fact, which we will use many times:

\begin{statement}
\textbf{Lemma 2.0v.} Let $n\in\mathbb{N}$. Let $x\in\left(  \mathbb{R}%
^{n}\right)  ^{\ast}$ be a row vector such that $x\geq0$. Let $y\in
\mathbb{R}^{n}$ be a column vector such that $y\geq0$.\textbf{ }Then,
$xy\geq0$.
\end{statement}

Again, we refer to Section \ref{sect.basics-proofs} for the proof of this fact.

\section{\label{sect.closed-int}Closed intervals and \cite[Theorem
2.3]{Schrij-CCO}}

In this brief section, we shall state three simple lemmas about closed
intervals and use them to \textquotedblleft patch\textquotedblright\ the proof
of \cite[Theorem 2.3]{Schrij-CCO} to make this latter proof constructive. This
section can be freely skipped, since none of what is done here will be used afterwards.

In the following, a \textit{closed interval} will mean a set which has the
form $\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some
elements $a$ and $b$ of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $
(where, of course, $-\infty$ is supposed to be smaller than each element of
$\mathbb{R}\cup\left\{  \infty\right\}  $, and $\infty$ is supposed to be
larger than each element of $\mathbb{R}\cup\left\{  -\infty\right\}  $). In
particular, $\varnothing$ is a closed interval (since $\varnothing=\left\{
x\in\mathbb{R}\ \mid\ 1\leq x\leq0\right\}  $), and $\mathbb{R}$ is a closed
interval (since $\mathbb{R}=\left\{  x\in\mathbb{R}\ \mid\ -\infty\leq
x\leq\infty\right\}  $).

We will now state three simple lemmas, the proofs of which can all be found in
Section \ref{sect.basics-proofs}.

\begin{statement}
\textbf{Lemma 2.0x.} The intersection of finitely many closed intervals always
is a closed interval.

\textbf{Lemma 2.0y.} Let $\alpha$ and $\beta$ be two reals. Then, the set
$\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a closed interval.

\textbf{Lemma 2.0z.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$b\in\mathbb{R}^{m}$. Let $A\in\mathbb{R}^{m\times n}$ be an $m\times
n$-matrix. Let $P=\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $.

Let $z\in\mathbb{R}^{n}$ and $c\in\mathbb{R}^{n}$. Then, the set $\left\{
\mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ is a closed interval.
\end{statement}

With the help of Lemma 2.0z, we can modify the proof of Theorem 2.3 in
\cite{Schrij-CCO} in such a way that it no longer uses analysis:

\begin{proof}
[Modifications to the proof of Theorem 2.3 in \cite{Schrij-CCO}.]In the proof
of Theorem 2.3 in \cite{Schrij-CCO}, it is claimed that\footnote{We are using
the notations of \cite{Schrij-CCO} here.} the numbers
\[
\mu_{0}:=\max\left\{  \mu\ \mid\ z+\mu c\in P\right\}
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \nu_{0}:=\max\left\{
\nu\ \mid\ z-\nu c\in P\right\}
\]
``exist since $P$ is compact''. We want to avoid this use of compactness.
Instead, we will prove the existence of these numbers $\mu_{0}$ and $\nu_{0}$
as follows:

\textit{Alternative proof of the existence of the numbers $\mu_{0}%
:=\max\left\{  \mu\ \mid\ z+\mu c\in P\right\}  $ and $\nu_{0}:=\max\left\{
\nu\ \mid\ z-\nu c\in P\right\}  $ in the proof of Theorem 2.3 in
\cite{Schrij-CCO}:}

Lemma 2.0z shows that $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}
$ is a closed interval.

We also notice that $z+\underbrace{0c}_{=0}=z\in P$, so that $0\in\left\{
\mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $. Hence, the set $\left\{
\mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ is nonempty.

Finally, we notice that the set $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in
P\right\}  $ is bounded (from both sides)\footnote{\textit{Proof.} Since $P$
is bounded, there exists an $M\in\mathbb{R}$ such that every $w\in P$
satisfies $\left\vert w\right\vert \leq M$. Consider this $M$.
\par
Now, let $\xi$ be an element of $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in
P\right\}  $. Then, $\xi\in\mathbb{R}$ and $z+\xi c\in P$. Recall that every
$w\in P$ satisfies $\left\vert w\right\vert \leq M$. Applied to $w=z+\xi c$,
this yields $\left\vert z+\xi c\right\vert \leq M$. But the triangle
inequality yields $\left\vert z+\xi c\right\vert \geq\underbrace{\left\vert
\xi c\right\vert }_{=\left\vert \xi\right\vert \left\vert c\right\vert
}-\left\vert z\right\vert =\left\vert \xi\right\vert \left\vert c\right\vert
-\left\vert z\right\vert $. Thus, $M\geq\left\vert z+\xi c\right\vert
\geq\left\vert \xi\right\vert \left\vert c\right\vert -\left\vert z\right\vert
$, so that $\left\vert \xi\right\vert \left\vert c\right\vert \leq M+z$. Since
$\left\vert c\right\vert >0$ (because $c\neq0$), we can divide this inequality
by $\left\vert c\right\vert $ and obtain $\left\vert \xi\right\vert \leq
\dfrac{M+z}{\left\vert c\right\vert }$.
\par
Now, forget that we fixed $\xi$. We thus have proven that every $\xi
\in\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ satisfies
$\left\vert \xi\right\vert \leq\dfrac{M+z}{\left\vert c\right\vert }$. This
yields that the set $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $
is bounded (from both sides), qed.}.

Now, a nonempty bounded closed interval always has a maximum. Applied to the
closed interval $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $,
this yields that $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ has
a maximum (since we know that $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in
P\right\}  $ is a nonempty bounded closed interval). In other words,
$\max\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ exists. Since we
defined $\mu_{0}$ as $\max\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in
P\right\}  $, this means that $\mu_{0}$ exists.

We have thus shown that $\mu_{0}$ exists. Similarly, $\nu_{0}$ exists.

Also note that $0$ is an element of the set $\left\{  \mu\in\mathbb{R}%
\ \mid\ z+\mu c\in P\right\}  $, whereas $\mu_{0}$ is the maximum of this set
(since $\mu_{0}=\max\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $).
Since any element of a set is $\leq$ to the maximum of this set (if the
maximum exists), this yields $0\leq\mu_{0}$. Similarly, $0\geq\nu_{0}$. This
shows that $z$ is a convex combination of the points $x:=z+\mu_{0}c$ and
$y:=z-\nu_{0}c$. (This is used further below in the proof of Theorem 2.3.)

This completes the modifications necessary to make the proof of Theorem 2.3 in
\cite{Schrij-CCO} independent of analysis.
\end{proof}

\section{The separation theorem for polytopes}

We now come to the topic of separation theorems. In this section, we are only
going to state them; for their proofs, we refer to Section
\ref{sect.sep-proofs} below.

Theorem 2.1 in \cite{Schrij-CCO} is a non-elementary fact; its proof cannot be
ridden of analysis. This is not surprising: Theorem 2.1 is too general (it
speaks of arbitrary convex sets). However, the following weaker version of
Theorem 2.1 (which will still be enough for most of what we want) can be
proven elementarily:

\begin{statement}
\textbf{Theorem 2.1a.} Let $C$ be a polytope in $\mathbb{R}^{n}$, and let
$z\in\mathbb{R}^{n}$ be such that $z\notin C$. Then, there exists a hyperplane
separating $z$ and $C$.
\end{statement}

Let us first give a basis-free version of this theorem:

\begin{statement}
\textbf{Theorem 2.1b.} Let $E$ be a finite-dimensional $\mathbb{R}$-vector
space. Let $C$ be a polytope in $E$, and let $z\in E$ be such that $z\notin
C$. Then, there exist an $f\in E^{\ast}$ and a $\delta\in\mathbb{R}$ such
that
\[
\left(  f\left(  z\right)  >\delta\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{every }x\in C\text{ satisfies }f\left(
x\right)  <\delta\right)  .
\]

\end{statement}

Here is another way to rewrite this:

\begin{statement}
\textbf{Theorem 2.1c.} Let $E$ be a finite-dimensional $\mathbb{R}$-vector
space. Let $C$ be a polytope in $E$. Then, exactly one of the following two
assertions holds:

\textit{Assertion C1:} We have $0\in C$.

\textit{Assertion C2:} There exists an $f\in E^{\ast}$ such that every $x\in
C$ satisfies $f\left(  x\right)  <0$.
\end{statement}

We shall prove Theorem 2.1c, Theorem 2.1b and Theorem 2.1a (in this order) in
Section \ref{sect.sep-proofs}.

Theorem 2.1a does not have the full strength of Theorem 2.1, but it is enough
to replace many applications of Theorem 2.1. For example, in the proof of
Theorem 2.4 in \cite{Schrij-CCO}, Schrijver writes: ``Suppose $x\notin P$.
Then there exists a hyperplane separating $x$ and $P$.'' This is (tacitly)
being derived from Theorem 2.1, but it also follows from Theorem 2.1a (applied
to $C=P$ and $z=x$).

\section{The separation theorem for finitely generated cones}

Next, we will show a separation theorem for finitely generated cones (more
precisely, the ``cone version'' of Theorem 2.1c):

\begin{statement}
\textbf{Theorem 2.5c.} Let $E$ be a finite-dimensional $\mathbb{R}$-vector
space. Let $S$ be a finite subset of $E$. Let $b\in E$. Then, exactly one of
the following two assertions holds:

\textit{Assertion D1:} We have $b\in\operatorname*{cone}S$.

\textit{Assertion D2:} There exists an $f\in E^{\ast}$ such that $f\left(
b\right)  >0$ and $\left(  \text{every }x\in\operatorname*{cone}S\text{
satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

Again, we refer to Section \ref{sect.sep-proofs} for the proof of this theorem.

\section{\label{sect.sep-proofs}Proofs of the separation theorems}

We owe the reader four proofs now: the proofs of Theorems 2.1a, 2.1b, 2.1c and
2.5c. Now is the time to pay back this debt. We begin with a proof of Theorem
2.5c; the following neat proof I have learnt from \cite{Bartl12} (which shows
a more general result):\footnote{A different (and much uglier) proof can be
found in Section \ref{sect.sep-proofs-old} below.}

\begin{proof}
[First proof of Theorem 2.5c.]We shall prove Theorem 2.5c by strong induction
over $\left\vert S\right\vert $:

\textit{Induction step:} Let $n$ be a nonnegative integer. Assume that Theorem
2.5c holds whenever $\left\vert S\right\vert <n$. We will now prove that
Theorem 2.5c holds whenever $\left\vert S\right\vert =n$.

So, let $E$ be a finite-dimensional $\mathbb{R}$-vector space. Let $S$ be a
finite subset of $E$ such that $\left\vert S\right\vert =n$. Let $b\in E$.
Then, we must prove that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{1}$\textit{1:} We have $b\in\operatorname*{cone}S$.

\textit{Assertion D}$_{1}$\textit{2:} There exists an $f\in E^{\ast}$ such
that $f\left(  b\right)  >0$ and \newline$\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

\begin{vershort}
Since $\operatorname*{cone}S$ is a convex cone, it is clear that
$\operatorname*{cone}S$ is closed under multiplication by a nonnegative
scalar, and that any linear combination of finitely many elements of
$\operatorname*{cone}S$ with nonnegative coefficients must lie in
$\operatorname*{cone}S$. (These facts follow from general properties of convex cones.)
\end{vershort}

\begin{verlong}
Since $\operatorname*{cone}S$ is a convex cone, it is clear that
$\operatorname*{cone}S$ is closed under multiplication by a nonnegative scalar
(because convex cones are closed under multiplication by a nonnegative
scalar). It is also clear that any linear combination of finitely many
elements of $\operatorname*{cone}S$ with nonnegative coefficients must lie in
$\operatorname*{cone}S$ (because $\operatorname*{cone}S$ is a convex cone, and
because any linear combination of finitely many elements of a convex cone with
nonnegative coefficients must lie in this cone). In particular, the sum of any
two elements of $\operatorname*{cone}S$ must lie in $\operatorname*{cone}S$.
\end{verlong}

The Assertions D$_{1}$1 and D$_{1}$2 cannot hold at the same
time\footnote{\textit{Proof.} Assume the opposite. Then, the Assertions
D$_{1}$1 and D$_{1}$2 hold at the same time. Since Assertion D$_{1}$2 holds,
there exists an $f\in E^{\ast}$ such that $f\left(  b\right)  >0$ and $\left(
\text{every }x\in\operatorname*{cone}S\text{ satisfies }f\left(  x\right)
\leq0\right)  $. Consider this $f$. We know that every $x\in
\operatorname*{cone}S$ satisfies $f\left(  x\right)  \leq0$. Since
$b\in\operatorname*{cone}S$ (because Assertion D$_{1}$1 holds), we can apply
this to $x=b$, and thus obtain $f\left(  b\right)  \leq0$. But this
contradicts $f\left(  b\right)  >0$. This contradiction shows that our
assumption was wrong, qed.}. Therefore, \textbf{at most} one of the two
Assertions D$_{1}$1 and D$_{1}$2 holds. We will now show that at least one of
these assertions holds.

\begin{vershort}
If $n=0$, then this is easy to prove\footnote{\textit{Proof.} Assume that
$n=0$. We must show that at least one of the two Assertions D$_{1}$1 and
D$_{1}$2 holds.
\par
We have $\left\vert S\right\vert =n=0$, and thus $S=\varnothing$. Hence,
$\operatorname*{cone}S=\operatorname*{cone}\varnothing=0$. Therefore,
$0\in0=\operatorname*{cone}S$.
\par
If $b=0$, then Assertion D$_{1}$1 holds (since $b=0\in\operatorname*{cone}S$).
Hence, if $b=0$, then at least one of the two Assertions D$_{1}$1 and D$_{1}$2
holds. Thus, for the rest of this proof, we WLOG assume that we don't have
$b=0$. Thus, $b\neq0$.
\par
Recall the following well-known fact from linear algebra: If $v$ is a vector
in a finite-dimensional $\mathbb{R}$-vector space $V$, and if $v\neq0$, then
there exists some $g\in V^{\ast}$ such that $g\left(  v\right)  \neq0$.
\par
Applying this fact to $V=E$ and $v=b$, we conclude that there exists a $g\in
E^{\ast}$ such that $g\left(  b\right)  \neq0$ (since $b\neq0$). Consider this
$g$. Define an $h\in E^{\ast}$ by $h=g\left(  b\right)  \cdot g$. Then,
$h\left(  b\right)  =\left(  g\left(  b\right)  \cdot g\right)  \left(
b\right)  =g\left(  b\right)  g\left(  b\right)  =\left(  g\left(  b\right)
\right)  ^{2}>0$ (since $g\left(  b\right)  \neq0$). On the other hand, every
$x\in\operatorname*{cone}S$ satisfies $h\left(  x\right)  \leq0$ (since
$x\in\operatorname*{cone}S=0$, so that $x=0$, so that $h\left(  x\right)
=h\left(  0\right)  =0$ (since $h$ is linear)). Altogether, we have thus shown
that $h\left(  b\right)  >0$ and $\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }h\left(  x\right)  \leq0\right)  $.
\par
Thus, we have proven that there exists an $f\in E^{\ast}$ such that $f\left(
b\right)  >0$ and $\left(  \text{every }x\in\operatorname*{cone}S\text{
satisfies }f\left(  x\right)  \leq0\right)  $ (namely, $f=h$). In other words,
Assertion D$_{1}$2 holds. Thus, at least one of the two Assertions D$_{1}$1
and D$_{1}$2 holds. Qed.}. Hence, for the rest of this proof, we WLOG assume
that $n\neq0$.

The set $S$ is nonempty (since $\left\vert S\right\vert =n\neq0$). In other
words, there exists some $t\in S$. Consider such a $t$.
\end{vershort}

\begin{verlong}
If $n=0$, then this is easy to prove\footnote{\textit{Proof.} Assume that
$n=0$. We must show that at least one of the two Assertions D$_{1}$1 and
D$_{1}$2 holds.
\par
We have $\left\vert S\right\vert =n=0$, and thus $S=\varnothing$. Hence,
$\operatorname*{cone}S=\operatorname*{cone}\varnothing=0$. Therefore,
$0\in0=\operatorname*{cone}S$.
\par
If $b=0$, then Assertion D$_{1}$1 holds (since $b=0\in\operatorname*{cone}S$).
Hence, if $b=0$, then at least one of the two Assertions D$_{1}$1 and D$_{1}$2
holds. Thus, for the rest of this proof, we can WLOG assume that we don't have
$b=0$. Assume this.
\par
We have $b\neq0$ (since we don't have $b=0$).
\par
Recall the following well-known fact from linear algebra: If $v$ is a vector
in a finite-dimensional $\mathbb{R}$-vector space $V$, and if $v\neq0$, then
there exists some $g\in V^{\ast}$ such that $g\left(  v\right)  \neq0$.
\par
Applying this fact to $V=E$ and $v=b$, we conclude that there exists a $g\in
E^{\ast}$ such that $g\left(  b\right)  \neq0$ (since $b\neq0$). Consider this
$g$. Define an $h\in E^{\ast}$ by $h=g\left(  b\right)  \cdot g$. Then,
$h\left(  b\right)  =\left(  g\left(  b\right)  \cdot g\right)  \left(
b\right)  =g\left(  b\right)  g\left(  b\right)  =\left(  g\left(  b\right)
\right)  ^{2}>0$ (since $g\left(  b\right)  \neq0$). On the other hand, every
$x\in\operatorname*{cone}S$ satisfies $h\left(  x\right)  \leq0$ (since
$x\in\operatorname*{cone}S=0$, so that $x=0$, so that $h\left(  x\right)
=h\left(  0\right)  =0$ (since $h$ is linear)). Altogether, we have thus shown
that $h\left(  b\right)  >0$ and $\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }h\left(  x\right)  \leq0\right)  $.
\par
Thus, we have proven that there exists an $f\in E^{\ast}$ such that $f\left(
b\right)  >0$ and $\left(  \text{every }x\in\operatorname*{cone}S\text{
satisfies }f\left(  x\right)  \leq0\right)  $ (namely, $f=h$). In other words,
Assertion D$_{1}$2 holds. Thus, at least one of the two Assertions D$_{1}$1
and D$_{1}$2 holds. Qed.}. Hence, for the rest of this proof, we can WLOG
assume that we don't have $n=0$. Assume this.

We have $n\in\mathbb{N}$ and $n\neq0$ (since we don't have $n=0$). Thus, $n>0$.

We have $\left\vert S\right\vert =n>0$ (since $n$ is positive). Hence, the set
$S$ is nonempty. In other words, there exists some $t\in S$. Consider such a
$t$.
\end{verlong}

From $t\in S$, we obtain $\left\vert S\setminus\left\{  t\right\}  \right\vert
=\underbrace{\left\vert S\right\vert }_{=n}-1=n-1<n$. Hence, we can apply
Theorem 2.5c to $E$, $S\setminus\left\{  t\right\}  $ and $b$ instead of $E$,
$S$ and $b$ (since we assumed that Theorem 2.5c holds whenever $\left\vert
S\right\vert <n$), and conclude that\textbf{ }exactly one of the following two
assertions holds:

\begin{statement}
\textit{Assertion D}$_{2}$\textit{1:} We have $b\in\operatorname*{cone}\left(
S\setminus\left\{  t\right\}  \right)  $.

\textit{Assertion D}$_{2}$\textit{2:} There exists an $f\in E^{\ast}$ such
that $f\left(  b\right)  >0$ and $\left(  \text{every }x\in
\operatorname*{cone}\left(  S\setminus\left\{  t\right\}  \right)  \text{
satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

Thus, we must be in one of the following two cases:

\textit{Case 1:} Assertion D$_{2}$1 holds.

\textit{Case 2:} Assertion D$_{2}$2 holds.

First, let us consider Case 1. In this case, Assertion D$_{2}$1 holds. In
other words, we have $b\in\operatorname*{cone}\left(  S\setminus\left\{
t\right\}  \right)  $.

But $S\setminus\left\{  t\right\}  \subseteq S$ and therefore
$\operatorname*{cone}\left(  S\setminus\left\{  t\right\}  \right)
\subseteq\operatorname*{cone}S$ (by Proposition 2.0m \textbf{(e)}, applied to
$S\setminus\left\{  t\right\}  $ instead of $T$). Hence, $b\in
\operatorname*{cone}\left(  S\setminus\left\{  t\right\}  \right)
\subseteq\operatorname*{cone}S$. In other words, Assertion D$_{1}$1 holds.
Hence, at least one of the two Assertions D$_{1}$1 and D$_{1}$2 holds. We have
thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2 holds in
Case 1.

Let us now consider Case 2. In this case, Assertion D$_{2}$2 holds. In other
words, there exists an $f\in E^{\ast}$ such that $f\left(  b\right)  >0$ and
$\left(  \text{every }x\in\operatorname*{cone}\left(  S\setminus\left\{
t\right\}  \right)  \text{ satisfies }f\left(  x\right)  \leq0\right)  $.
Consider this $f$, and denote it by $g$. Thus, $g$ is an element of $E^{\ast}$
satisfying $g\left(  b\right)  >0$ and%
\begin{equation}
\left(  \text{every }x\in\operatorname*{cone}\left(  S\setminus\left\{
t\right\}  \right)  \text{ satisfies }g\left(  x\right)  \leq0\right)
\label{2.5c.pf.first.3}%
\end{equation}


We must be in one of the following two subcases:

\textit{Subcase 2.1:} We have $g\left(  t\right)  \leq0$.

\textit{Subcase 2.2:} We have $g\left(  t\right)  >0$.

Let us first consider Subcase 2.1. In this subcase, we have $g\left(
t\right)  \leq0$. Thus, every $x\in S$ satisfies $g\left(  x\right)  \leq
0$\ \ \ \ \footnote{\textit{Proof.} Let $x\in S$. We must show that $g\left(
x\right)  \leq0$.
\par
If $x=t$, then this follows immediately from $g\left(  \underbrace{x}%
_{=t}\right)  =g\left(  t\right)  \leq0$. Hence, for the rest of this proof,
we can WLOG assume that we don't have $x=t$. Assume this.
\par
We have $x\neq t$ (since we don't have $x=t$). Combining $x\in S$ with $x\neq
t$, we find $x\in S\setminus\left\{  t\right\}  \subseteq\operatorname*{cone}%
\left(  S\setminus\left\{  t\right\}  \right)  $ (by Proposition 2.0m
\textbf{(c)}, applied to $S\setminus\left\{  t\right\}  $ instead of $S$).
Hence, (\ref{2.5c.pf.first.3}) shows that $g\left(  x\right)  \leq0$. Qed.}.
Hence, every $x\in\operatorname*{cone}S$ satisfies $g\left(  x\right)  \leq0$
(by Proposition 2.0s, applied to $f=g$). Now, we know that $g$ is an element
of $E^{\ast}$ satisfying $g\left(  b\right)  >0$ and $\left(  \text{every
}x\in\operatorname*{cone}S\text{ satisfies }g\left(  x\right)  \leq0\right)
$. Hence, there exists an $f\in E^{\ast}$ such that $f\left(  b\right)  >0$
and $\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies }f\left(
x\right)  \leq0\right)  $ (namely, $f=g$). In other words, Assertion D$_{1}$2
holds. Hence, at least one of the two Assertions D$_{1}$1 and D$_{1}$2 holds.
We have thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds in Subcase 2.1.

Let us now consider Subcase 2.2. In this subcase, we have $g\left(  t\right)
>0$. Thus, the real number $\dfrac{1}{g\left(  t\right)  }$ is well-defined
and satisfies $\dfrac{1}{g\left(  t\right)  }>0$. In particular, $\dfrac
{1}{g\left(  t\right)  }$ is a nonnegative real.

Define a vector $q\in E$ by $q=\dfrac{1}{g\left(  t\right)  }t$. Then,
$g\left(  q\right)  =1$\ \ \ \ \footnote{\textit{Proof.} Applying the map $g$
to the equality $q=\dfrac{1}{g\left(  t\right)  }t$, we obtain%
\begin{align*}
g\left(  q\right)   &  =g\left(  \dfrac{1}{g\left(  t\right)  }t\right)
=\dfrac{1}{g\left(  t\right)  }g\left(  t\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }g\text{ is }\mathbb{R}\text{-linear}\right) \\
&  =1.
\end{align*}
}. Also, $q\in\operatorname*{cone}S$\ \ \ \ \footnote{\textit{Proof.} We have
$t\in S\subseteq\operatorname*{cone}S$ (by Proposition 2.0m \textbf{(c)}).
\par
Recall that $\operatorname*{cone}S$ is closed under multiplication by a
nonnegative scalar. In other words, every nonnegative real $\omega$ and every
$y\in\operatorname*{cone}S$ satisfy $\omega y\in\operatorname*{cone}S$.
Applying this to $\omega=\dfrac{1}{g\left(  t\right)  }$ and $y=t$, we obtain
$\dfrac{1}{g\left(  t\right)  }t\in\operatorname*{cone}S$ (since $\dfrac
{1}{g\left(  t\right)  }$ is a nonnegative real). Thus, $q=\dfrac{1}{g\left(
t\right)  }t\in\operatorname*{cone}S$.}.

Define a subset $S^{\prime}$ of $E$ by%
\[
S^{\prime}=\left\{  s-g\left(  s\right)  q\ \mid\ s\in S\setminus\left\{
t\right\}  \right\}  .
\]
Thus, $\left\vert S^{\prime}\right\vert \leq\left\vert S\setminus\left\{
t\right\}  \right\vert <n$. Therefore, we can apply Theorem 2.5c to $E$,
$S^{\prime}$ and $b-g\left(  b\right)  q$ instead of $E$, $S$ and $b$ (since
we assumed that Theorem 2.5c holds whenever $\left\vert S\right\vert <n$), and
conclude that\textbf{ }exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{3}$\textit{1:} We have $b-g\left(  b\right)
q\in\operatorname*{cone}\left(  S^{\prime}\right)  $.

\textit{Assertion D}$_{3}$\textit{2:} There exists an $f\in E^{\ast}$ such
that $f\left(  b-g\left(  b\right)  q\right)  >0$ and $\left(  \text{every
}x\in\operatorname*{cone}\left(  S^{\prime}\right)  \text{ satisfies }f\left(
x\right)  \leq0\right)  $.
\end{statement}

Thus, we must be in one of the following two subsubcases:

\textit{Subsubcase 2.2.1:} Assertion D$_{3}$1 holds.

\textit{Subsubcase 2.2.2:} Assertion D$_{3}$2 holds.

First, let us consider Subsubcase 2.2.1. In this subsubcase, Assertion D$_{3}%
$1 holds. In other words, we have $b-g\left(  b\right)  q\in
\operatorname*{cone}\left(  S^{\prime}\right)  $.

But $S^{\prime}\subseteq\operatorname*{cone}S$%
\ \ \ \ \footnote{\textit{Proof.} Let $s\in S\setminus\left\{  t\right\}  $.
We shall show that $s-g\left(  s\right)  q\in\operatorname*{cone}S$.
\par
Indeed, $s\in S\setminus\left\{  t\right\}  \subseteq\operatorname*{cone}%
\left(  S\setminus\left\{  t\right\}  \right)  $ (by Proposition 2.0m
\textbf{(c)}, applied to $S\setminus\left\{  t\right\}  $ instead of $S$).
Hence, (\ref{2.5c.pf.first.3}) (applied to $x=s$) yields $g\left(  s\right)
\leq0$. Thus, $-g\left(  s\right)  \geq0$. In other words, $-g\left(
s\right)  $ is nonnegative.
\par
We have $s\in S\setminus\left\{  t\right\}  \subseteq S\subseteq
\operatorname*{cone}S$ (by Proposition 2.0m \textbf{(c)}) and $q\in
\operatorname*{cone}S$. Hence, $1s+\left(  -g\left(  s\right)  \right)  q$ is
a linear combination of finitely many elements of $\operatorname*{cone}S$ with
nonnegative coefficients (since $s\in\operatorname*{cone}S$ and $q\in
\operatorname*{cone}S$, and since $1$ and $-g\left(  s\right)  $ are
nonnegative). Therefore, $1s+\left(  -g\left(  s\right)  \right)  q$ lies in
$\operatorname*{cone}S$ (since any linear combination of finitely many
elements of $\operatorname*{cone}S$ with nonnegative coefficients must lie in
$\operatorname*{cone}S$). In other words, $1s+\left(  -g\left(  s\right)
\right)  q\in\operatorname*{cone}S$. Since $1s+\left(  -g\left(  s\right)
\right)  q=s-g\left(  s\right)  q$, this rewrites as $s-g\left(  s\right)
q\in\operatorname*{cone}S$.
\par
Now, forget that we fixed $s$. We thus have shown that $s-g\left(  s\right)
q\in\operatorname*{cone}S$ for each $s\in S\setminus\left\{  t\right\}  $. In
other words, $\left\{  s-g\left(  s\right)  q\ \mid\ s\in S\setminus\left\{
t\right\}  \right\}  \subseteq\operatorname*{cone}S$. Thus, $S^{\prime
}=\left\{  s-g\left(  s\right)  q\ \mid\ s\in S\setminus\left\{  t\right\}
\right\}  \subseteq\operatorname*{cone}S$.}. Hence, Proposition 2.0m
\textbf{(g)} (applied to $T=S^{\prime}$) yields $\operatorname*{cone}\left(
S^{\prime}\right)  \subseteq\operatorname*{cone}S$. Hence, $b-g\left(
b\right)  q\in\operatorname*{cone}\left(  S^{\prime}\right)  \subseteq
\operatorname*{cone}S$.

Define $c\in E$ by $c=b-g\left(  b\right)  q$. Thus, $c=b-g\left(  b\right)
q\in\operatorname*{cone}S$.

Now, $g\left(  b\right)  $ is a nonnegative real (since $g\left(  b\right)
>0$). Hence, $1$ and $g\left(  b\right)  $ are nonnegative reals. Also, $c$
and $q$ are elements of $\operatorname*{cone}S$ (since $c\in
\operatorname*{cone}S$ and $q\in\operatorname*{cone}S$). Therefore,
$1c+g\left(  b\right)  q$ is a linear combination of finitely many elements of
$\operatorname*{cone}S$ with nonnegative coefficients. Hence, $1c+g\left(
b\right)  q$ must lie in $\operatorname*{cone}S$ (since any linear combination
of finitely many elements of $\operatorname*{cone}S$ with nonnegative
coefficients must lie in $\operatorname*{cone}S$). In other words,
$1c+g\left(  b\right)  q\in\operatorname*{cone}S$. Since $\underbrace{1c}%
_{=c=b-g\left(  b\right)  q}+g\left(  b\right)  q=\left(  b-g\left(  b\right)
q\right)  +g\left(  b\right)  q=b$, this rewrites as $b\in\operatorname*{cone}%
S$. In other words, Assertion D$_{1}$1 holds. Hence, at least one of the two
Assertions D$_{1}$1 and D$_{1}$2 holds. We have thus proven that at least one
of Assertions D$_{1}$1 and D$_{1}$2 holds in Subsubcase 2.2.1.

Next, let us consider Subsubcase 2.2.2. In this subsubcase, Assertion D$_{3}$2
holds. In other words, there exists an $f\in E^{\ast}$ such that $f\left(
b-g\left(  b\right)  q\right)  >0$ and $\left(  \text{every }x\in
\operatorname*{cone}\left(  S^{\prime}\right)  \text{ satisfies }f\left(
x\right)  \leq0\right)  $. Consider this $f$, and denote it by $h$. Thus, $h$
is an element of $E^{\ast}$ satisfying $h\left(  b-g\left(  b\right)
q\right)  >0$ and%
\begin{equation}
\left(  \text{every }x\in\operatorname*{cone}\left(  S^{\prime}\right)  \text{
satisfies }h\left(  x\right)  \leq0\right)  . \label{2.5c.pf.first.h2}%
\end{equation}


Now, define a $k\in E^{\ast}$ by $k=h-h\left(  q\right)  \cdot g$. Then, each
$x\in E$ satisfies%
\begin{align}
\underbrace{k}_{=h-h\left(  q\right)  \cdot g}\left(  x\right)   &  =\left(
h-h\left(  q\right)  \cdot g\right)  \left(  x\right)  =h\left(  x\right)
-h\left(  q\right)  \cdot g\left(  x\right) \nonumber\\
&  =h\left(  x\right)  -g\left(  x\right)  \cdot h\left(  q\right)  =h\left(
x-g\left(  x\right)  q\right)  \label{2.5c.pf.first.k}%
\end{align}
(since the map $h$ is $\mathbb{R}$-linear). Applying this to $x=b$, we obtain
$k\left(  b\right)  =h\left(  b-g\left(  b\right)  q\right)  >0$. Moreover,
each $x\in S$ satisfies $k\left(  x\right)  \leq0$%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in S$. We must prove that $k\left(
x\right)  \leq0$.
\par
From $q=\dfrac{1}{g\left(  t\right)  }t$, we obtain $t=g\left(  t\right)  q$.
But applying (\ref{2.5c.pf.first.k}) to $x=t$, we obtain%
\[
k\left(  t\right)  =h\left(  \underbrace{t}_{=g\left(  t\right)  q}-g\left(
t\right)  q\right)  =h\left(  \underbrace{g\left(  t\right)  q-g\left(
t\right)  q}_{=0}\right)  =h\left(  0\right)  =0
\]
(since the map $h$ is $\mathbb{R}$-linear). Therefore, $k\left(  t\right)
\leq0$. Thus, if $x=t$, then $k\left(  \underbrace{x}_{=t}\right)  =k\left(
t\right)  \leq0$. Hence, for the rest of our proof of $k\left(  x\right)
\leq0$, we can WLOG assume that we don't have $x=t$. Assume this.
\par
We have $x\neq t$ (since we don't have $x=t$). Combining $x\in S$ with $x\neq
t$, we obtain $x\in S\setminus\left\{  t\right\}  $. Thus, $x-g\left(
x\right)  q\in\left\{  s-g\left(  s\right)  q\ \mid\ s\in S\setminus\left\{
t\right\}  \right\}  $ (since $x-g\left(  x\right)  q=s-g\left(  s\right)  q$
for some $s\in S\setminus\left\{  t\right\}  $ (namely, for $s=x$)). Hence,%
\[
x-g\left(  x\right)  q\in\left\{  s-g\left(  s\right)  q\ \mid\ s\in
S\setminus\left\{  t\right\}  \right\}  =S^{\prime}\subseteq
\operatorname*{cone}\left(  S^{\prime}\right)
\]
(by Proposition 2.0m \textbf{(c)}, applied to $S^{\prime}$ instead of $S$).
Thus, (\ref{2.5c.pf.first.h2}) (applied to $x-g\left(  x\right)  q$ instead of
$x$) yields $h\left(  x-g\left(  x\right)  q\right)  \leq0$.
\par
Now, (\ref{2.5c.pf.first.k}) yields $k\left(  x\right)  =h\left(  x-g\left(
x\right)  q\right)  \leq0$. Qed.}. Hence, each $x\in\operatorname*{cone}S$
satisfies $k\left(  x\right)  \leq0$ (by Proposition 2.0s, applied to $f=k$).

Thus, there exists an $f\in E^{\ast}$ such that $f\left(  b\right)  >0$ and
$\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies }f\left(
x\right)  \leq0\right)  $ (namely, $f=k$). In other words, Assertion D$_{1}$2
holds. Hence, at least one of Assertions D$_{1}$1 and D$_{1}$2 holds.

We have thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds in Subsubcase 2.2.2.

\begin{vershort}
Recall that our goal was to prove that at least one of Assertions D$_{1}$1 and
D$_{1}$2 holds. We have now proven this in Case 1, in Subcase 2.1, and in
Subsubcases 2.2.1 and 2.2.2. Thus, we have proven this in all possible
situations. So we know that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds. Thus, exactly one of Assertions D$_{1}$1 and D$_{1}$2 holds (since we
know that \textbf{at most} one of the two Assertions D$_{1}$1 and D$_{1}$2 holds).
\end{vershort}

\begin{verlong}
Hence, in each of the Subsubcases 2.2.1 and 2.2.2, at least one of Assertions
D$_{1}$1 and D$_{1}$2 holds. Since these Subsubcases 2.2.1 and 2.2.2 cover the
whole Subcase 2.2, this yields that at least one of Assertions D$_{1}$1 and
D$_{1}$2 holds in Subcase 2.2.

Hence, in each of the Subcases 2.1 and 2.2, at least one of Assertions D$_{1}%
$1 and D$_{1}$2 holds. Since these Subcases 2.1 and 2.2 cover the whole Case
2, this yields that at least one of Assertions D$_{1}$1 and D$_{1}$2 holds in
Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions D$_{1}$1 and
D$_{1}$2 holds. Since these Cases 1 and 2 cover all possibilities, this yields
that, in every situation, at least one of Assertions D$_{1}$1 and D$_{1}$2
holds. Thus, exactly one of Assertions D$_{1}$1 and D$_{1}$2 holds (since we
know that \textbf{at most} one of the two Assertions D$_{1}$1 and D$_{1}$2 holds).
\end{verlong}

Now, forget that we fixed $E$, $S$ and $b$. We have thus proven that if $E$ is
a finite-dimensional $\mathbb{R}$-vector space, $S$ is a finite subset of $E$
such that $\left\vert S\right\vert =n$, and $b$ is an element of $E$, then
exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{1}$\textit{1:} We have $b\in\operatorname*{cone}S$.

\textit{Assertion D}$_{1}$\textit{2:} There exists an $f\in E^{\ast}$ such
that $f\left(  b\right)  >0$ and \newline$\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

In other words, we have proven that Theorem 2.5c holds in the case when
$\left\vert S\right\vert =n$. This completes the induction step, and thus the
induction proof of Theorem 2.5c is complete.
\end{proof}

Before we step to the proof of Theorem 2.1c, let us state a basic result that
connects convex conic hulls with convex hulls:

\begin{statement}
\textbf{Proposition 2.4a.} Let $E$ be a finite-dimensional $\mathbb{R}$-vector
space. Let $T$ be a finite subset of $E$. Let $x\in E$.

Consider the direct sum $E\oplus\mathbb{R}$ (which is also a
finite-dimensional $\mathbb{R}$-vector space). Consider the vector $\left(
x,1\right)  \in E\oplus\mathbb{R}$. Let $S$ be the subset $\left\{  \left(
y,1\right)  \ \mid\ y\in T\right\}  $ of $E\oplus\mathbb{R}$.

Then, $x\in\operatorname*{conv}.\operatorname*{hull}T$ holds if and only if
$\left(  x,1\right)  \in\operatorname*{cone}S$.
\end{statement}

\begin{proof}
[Proof of Proposition 2.4a.]The definition of $S$ shows that $S=\left\{
\left(  y,1\right)  \ \mid\ y\in T\right\}  $. Thus, the set $S$ is finite
(since the set $T$ is finite).

Let us first show that%
\begin{equation}
\text{if }x\in\operatorname*{conv}.\operatorname*{hull}T\text{, then }\left(
x,1\right)  \in\operatorname*{cone}S. \label{2.4a.pf.dir1}%
\end{equation}


[\textit{Proof of (\ref{2.4a.pf.dir1}):} Assume that $x\in\operatorname*{conv}%
.\operatorname*{hull}T$. We want to show that $\left(  x,1\right)
\in\operatorname*{cone}S$.

We have
\begin{align*}
x  &  \in\operatorname*{conv}.\operatorname*{hull}T\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }T\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}
\end{align*}
(by Proposition 2.0g \textbf{(a)}, applied to $T$ instead of $S$). In other
words, there exist some element $t$ of $\mathbb{N}$, some $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $T$, and some $t$-tuple
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative
reals such that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $x=\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}$. Consider this $t$, this $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{t}\right)  $. Clearly, the family $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,t\right\}  }$ is a family of
nonnegative reals (since $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  $ is a $t$-tuple of nonnegative reals) and has the property that
all but finitely many $i\in\left\{  1,2,\ldots,t\right\}  $ satisfy
$\lambda_{i}=0$ (since there are only finitely many $i\in\left\{
1,2,\ldots,t\right\}  $).

Each $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $\left(  x_{i},1\right)
\in S$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots
,t\right\}  $. Then, $x_{i}\in T$ (since $\left(  x_{1},x_{2},\ldots
,x_{t}\right)  $ is a $t$-tuple of elements of $T$). Thus, $\left(
x_{i},1\right)  \in\left\{  \left(  y,1\right)  \ \mid\ y\in T\right\}  $
(since $\left(  x_{i},1\right)  =\left(  y,1\right)  $ for some $y\in T$
(namely, for $y=x_{i}$)). Hence, $\left(  x_{i},1\right)  \in\left\{  \left(
y,1\right)  \ \mid\ y\in T\right\}  =S$. Qed.} and therefore $\left(
x_{i},1\right)  \in\operatorname*{cone}S$ (since $\left(  x_{i},1\right)  \in
S\subseteq\operatorname*{cone}S$ (by Proposition 2.0m \textbf{(c)}, applied to
$E\oplus\mathbb{R}$ instead of $E$)). Thus, $\left(  \left(  x_{i},1\right)
\right)  _{i\in\left\{  1,2,\ldots,t\right\}  }$ is a family of elements of
$\operatorname*{cone}S$. Also, Proposition 2.0m \textbf{(d)} (applied to
$E\oplus\mathbb{R}$ instead of $E$) shows that the convex conic hull
$\operatorname*{cone}S$ is a convex cone. Hence, Proposition 2.0i (applied to
$E\oplus\mathbb{R}$, $\operatorname*{cone}S$, $\left\{  1,2,\ldots,t\right\}
$ and $\left(  x_{i},1\right)  $ instead of $E$, $C$, $I$ and $x_{i}$) shows
that $\sum\limits_{i\in\left\{  1,2,\ldots,t\right\}  }\lambda_{i}\left(
x_{i},1\right)  \in\operatorname*{cone}S$. Since
\begin{align*}
\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,t\right\}  }}_{=\sum
\limits_{i=1}^{t}}\lambda_{i}\left(  x_{i},1\right)   &  =\sum\limits_{i=1}%
^{t}\lambda_{i}\left(  x_{i},1\right)  =\left(  \sum\limits_{i=1}^{t}%
\lambda_{i}x_{i},\sum\limits_{i=1}^{t}\underbrace{\lambda_{i}1}_{=\lambda_{i}%
}\right) \\
&  =\left(  \underbrace{\sum\limits_{i=1}^{t}\lambda_{i}x_{i}}_{=x}%
,\underbrace{\sum\limits_{i=1}^{t}\lambda_{i}}_{=1}\right)  =\left(
x,1\right)  ,
\end{align*}
this rewrites as $\left(  x,1\right)  \in\operatorname*{cone}S$. Thus,
(\ref{2.4a.pf.dir1}) is proven.]

Now, let us show that%
\begin{equation}
\text{if }\left(  x,1\right)  \in\operatorname*{cone}S\text{, then }%
x\in\operatorname*{conv}.\operatorname*{hull}T. \label{2.4a.pf.dir2}%
\end{equation}


[\textit{Proof of (\ref{2.4a.pf.dir2}):} Assume that $\left(  x,1\right)
\in\operatorname*{cone}S$. We want to show that $x\in\operatorname*{conv}%
.\operatorname*{hull}T$.

Write the set $T$ in the form $T=\left\{  t_{1},t_{2},\ldots,t_{n}\right\}  $
for some $n\in\mathbb{N}$. (This is possible, since the set $T$ is finite.)

We have%
\begin{align*}
S  &  =\left\{  \left(  y,1\right)  \ \mid\ y\in\underbrace{T}_{=\left\{
t_{1},t_{2},\ldots,t_{n}\right\}  }\right\}  =\left\{  \left(  y,1\right)
\ \mid\ y\in\left\{  t_{1},t_{2},\ldots,t_{n}\right\}  \right\} \\
&  =\left\{  \left(  t_{1},1\right)  ,\left(  t_{2},1\right)  ,\ldots,\left(
t_{n},1\right)  \right\}  .
\end{align*}


We have
\[
\left(  x,1\right)  \in\operatorname*{cone}\underbrace{S}_{=\left\{  \left(
t_{1},1\right)  ,\left(  t_{2},1\right)  ,\ldots,\left(  t_{n},1\right)
\right\}  }=\operatorname*{cone}\left\{  \left(  t_{1},1\right)  ,\left(
t_{2},1\right)  ,\ldots,\left(  t_{n},1\right)  \right\}  .
\]
But Proposition 2.0o \textbf{(b)} (applied to $E\oplus\mathbb{R}$, $\left(
t_{i},1\right)  $ and $\left(  x,1\right)  $ instead of $E$, $x_{i}$ and $x$)
shows that we have $\left(  x,1\right)  \in\operatorname*{cone}\left\{
\left(  t_{1},1\right)  ,\left(  t_{2},1\right)  ,\ldots,\left(
t_{n},1\right)  \right\}  $ if and only if $\left(  x,1\right)  $ is a linear
combination of the vectors $\left(  t_{1},1\right)  ,\left(  t_{2},1\right)
,\ldots,\left(  t_{n},1\right)  $ with nonnegative coefficients. Therefore,
$\left(  x,1\right)  $ is a linear combination of the vectors $\left(
t_{1},1\right)  ,\left(  t_{2},1\right)  ,\ldots,\left(  t_{n},1\right)  $
with nonnegative coefficients (since\newline$\left(  x,1\right)
\in\operatorname*{cone}\left\{  \left(  t_{1},1\right)  ,\left(
t_{2},1\right)  ,\ldots,\left(  t_{n},1\right)  \right\}  $). In other words,
there exists an $n$-tuple $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{n}\right)  $ of nonnegative reals such that%
\[
\left(  x,1\right)  =\sum_{i=1}^{n}\lambda_{i}\left(  t_{i},1\right)  .
\]
Consider this $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $.

We have%
\[
\left(  x,1\right)  =\sum_{i=1}^{n}\lambda_{i}\left(  t_{i},1\right)  =\left(
\sum_{i=1}^{n}\lambda_{i}t_{i},\sum_{i=1}^{n}\lambda_{i}1\right)  .
\]
In other words,%
\[
x=\sum_{i=1}^{n}\lambda_{i}t_{i}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ 1=\sum_{i=1}^{n}\lambda_{i}1.
\]
Now, $1=\underbrace{\sum_{i=1}^{n}}_{=\sum_{i\in\left\{  1,2,\ldots,n\right\}
}}\underbrace{\lambda_{i}1}_{=\lambda_{i}}=\sum_{i\in\left\{  1,2,\ldots
,n\right\}  }\lambda_{i}$, so that $\sum_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}=1$.

Clearly, the family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$ is a family of nonnegative reals (since $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $ is a $n$-tuple of nonnegative
reals) and has the property that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$ (since there are only finitely
many $i\in\left\{  1,2,\ldots,n\right\}  $).

Each $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies $t_{i}\in T$ (since
$t_{i}\in\left\{  t_{1},t_{2},\ldots,t_{n}\right\}  =T$) and therefore
$t_{i}\in\operatorname*{conv}.\operatorname*{hull}T$ (since $t_{i}\in
T\subseteq\operatorname*{conv}.\operatorname*{hull}T$ (by Proposition 2.0f
\textbf{(c)}, applied to $T$ instead of $S$)). Thus, $\left(  t_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family of elements of
$\operatorname*{conv}.\operatorname*{hull}T$. Also, Proposition 2.0f
\textbf{(d)} (applied to $T$ instead of $S$) shows that the convex hull
$\operatorname*{conv}.\operatorname*{hull}T$ is a convex set. Hence,
Proposition 2.0a (applied to $\operatorname*{conv}.\operatorname*{hull}T$,
$\left\{  1,2,\ldots,n\right\}  $ and $t_{i}$ instead of $C$, $I$ and $x_{i}$)
shows that $\sum_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}t_{i}%
\in\operatorname*{conv}.\operatorname*{hull}T$. Since
\[
\underbrace{\sum_{i\in\left\{  1,2,\ldots,n\right\}  }}_{=\sum_{i=1}^{n}%
}\lambda_{i}t_{i}=\sum_{i=1}^{n}\lambda_{i}t_{i}=x,
\]
this rewrites as $x\in\operatorname*{conv}.\operatorname*{hull}T$. Thus,
(\ref{2.4a.pf.dir2}) is proven.]

Now, combining the two logical implications (\ref{2.4a.pf.dir1}) and
(\ref{2.4a.pf.dir2}), we conclude that $x\in\operatorname*{conv}%
.\operatorname*{hull}T$ holds if and only if $\left(  x,1\right)
\in\operatorname*{cone}S$. This proves Proposition 2.4a.
\end{proof}

Next, we can prove Theorem 2.1c:\footnote{A different (and much uglier) proof
can be found in Section \ref{sect.sep-proofs-old} below.}

\begin{proof}
[First proof of Theorem 2.1c.]The Assertions C1 and C2 cannot hold at the same
time\footnote{\textit{Proof.} Assume the contrary. Thus, the Assertions C1 and
C2 hold at the same time. Since Assertion C2 holds, there exists an $f\in
E^{\ast}$ such that every $x\in C$ satisfies $f\left(  x\right)  <0$. Consider
this $f$. We know that every $x\in C$ satisfies $f\left(  x\right)  <0$. Since
$0\in C$ (because Assertion C1 holds), we can apply this to $x=0$, and thus
obtain $f\left(  0\right)  <0$. But this contradicts $f\left(  0\right)  =0$
(which is because $f$ is linear). This contradiction shows that our assumption
was wrong, qed.}. We will now show that at least one of these assertions holds.

The set $C$ is a polytope in $E$. In other words, $C$ is the convex hull of a
finite subset of $E$ (by the definition of a \textquotedblleft
polytope\textquotedblright). In other words, there exists a finite subset $T$
of $E$ such that $C=\operatorname*{conv}.\operatorname*{hull}T$. Consider this
$T$.

Let $\iota:E\rightarrow E\oplus\mathbb{R}$ be the canonical inclusion of the
$\mathbb{R}$-vector space $E$ into the direct sum $E\oplus\mathbb{R}$. This
inclusion $\iota$ maps each $x\in E$ to $\left(  x,0\right)  \in
E\oplus\mathbb{R}$; it is an $\mathbb{R}$-linear map.

Let $S$ be the subset $\left\{  \left(  y,1\right)  \ \mid\ y\in T\right\}  $
of $E\oplus\mathbb{R}$. Then, $S$ is a finite subset of $E\oplus\mathbb{R}$
(since $T$ is a finite set).

Let $b$ be the vector $\left(  0,1\right)  \in E\oplus\mathbb{R}$. Theorem
2.5c (applied to $E\oplus\mathbb{R}$ instead of $E$) then shows that exactly
one of the following two assertions holds:

\begin{statement}
\textit{Assertion C}$_{1}$\textit{1:} We have $b\in\operatorname*{cone}S$.
\end{statement}

\begin{statement}
\textit{Assertion C}$_{1}$\textit{2:} There exists an $f\in\left(
E\oplus\mathbb{R}\right)  ^{\ast}$ such that $f\left(  b\right)  >0$ and
$\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies }f\left(
x\right)  \leq0\right)  $.
\end{statement}

Thus, we are in one of the following two cases:

\textit{Case 1:} Assertion C$_{1}$1 holds.

\textit{Case 2:} Assertion C$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion C$_{1}$1 holds. In other
words, we have $b\in\operatorname*{cone}S$. Thus, $\left(  0,1\right)
=b\in\operatorname*{cone}S$. Proposition 2.4a (applied to $x=0$) shows that
$0\in\operatorname*{conv}.\operatorname*{hull}T$ holds if and only if $\left(
0,1\right)  \in\operatorname*{cone}S$. Thus, $0\in\operatorname*{conv}%
.\operatorname*{hull}T$ holds (since $\left(  0,1\right)  \in
\operatorname*{cone}S$). Hence, $0\in\operatorname*{conv}.\operatorname*{hull}%
T=C$. In other words, Assertion C1 holds. Thus, at least one of the two
assertions C1 and C2 holds.

We thus have shown that in Case 1, at least one of the two assertions C1 and
C2 holds.

Now, let us consider Case 2. In this case, Assertion C$_{1}$2 holds. In other
words, there exists an $f\in\left(  E\oplus\mathbb{R}\right)  ^{\ast}$ such
that $f\left(  b\right)  >0$ and $\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }f\left(  x\right)  \leq0\right)  $.
Consider such an $f$, and denote it by $h$. Thus, $h$ is an element of
$\left(  E\oplus\mathbb{R}\right)  ^{\ast}$ satisfying $h\left(  b\right)  >0$
and%
\begin{equation}
\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies }h\left(
x\right)  \leq0\right)  . \label{2.1c.pf.first.4}%
\end{equation}


The map $h\circ\iota$ is a well-defined $\mathbb{R}$-linear map $E\rightarrow
\mathbb{R}$ (since $h$ is an $\mathbb{R}$-linear map $E\oplus\mathbb{R}%
\rightarrow\mathbb{R}$, and since $\iota$ is an $\mathbb{R}$-linear map
$E\rightarrow E\oplus\mathbb{R}$). In other words, $h\circ\iota\in E^{\ast}$.

Now, let $x\in C$ be arbitrary. Thus, $x\in C=\operatorname*{conv}%
.\operatorname*{hull}T$. But Proposition 2.4a shows that $x\in
\operatorname*{conv}.\operatorname*{hull}T$ holds if and only if $\left(
x,1\right)  \in\operatorname*{cone}S$. Hence, $\left(  x,1\right)
\in\operatorname*{cone}S$ (since $x\in\operatorname*{conv}%
.\operatorname*{hull}T$). Therefore, (\ref{2.1c.pf.first.4}) (applied to
$\left(  x,1\right)  $ instead of $x$) shows that $h\left(  \left(
x,1\right)  \right)  \leq0$.

The definition of $\iota$ yields $\iota\left(  x\right)  =\left(  x,0\right)
$. Hence, $\underbrace{\iota\left(  x\right)  }_{=\left(  x,0\right)
}+\underbrace{b}_{=\left(  0,1\right)  }=\left(  x,0\right)  +\left(
0,1\right)  =\left(  \underbrace{x+0}_{=x},\underbrace{0+1}_{=1}\right)
=\left(  x,1\right)  $. Applying the map $h$ to both sides of this equality,
we find $h\left(  \iota\left(  x\right)  +b\right)  =h\left(  \left(
x,1\right)  \right)  \leq0$.

But the map $h$ is $\mathbb{R}$-linear. Thus,%
\[
h\left(  \iota\left(  x\right)  +b\right)  =h\left(  \iota\left(  x\right)
\right)  +\underbrace{h\left(  b\right)  }_{>0}>h\left(  \iota\left(
x\right)  \right)  =\left(  h\circ\iota\right)  \left(  x\right)  .
\]
Hence, $\left(  h\circ\iota\right)  \left(  x\right)  <h\left(  \iota\left(
x\right)  +b\right)  \leq0$.

Now, forget that we fixed $x$. We thus have shown that every $x\in C$
satisfies $\left(  h\circ\iota\right)  \left(  x\right)  <0$. Hence, there
exists an $f\in E^{\ast}$ such that every $x\in C$ satisfies $f\left(
x\right)  <0$ (namely, $f=h\circ\iota$). In other words, Assertion C2 holds.
Thus, at least one of the two assertions C1 and C2 holds.

We thus have shown that in Case 2, at least one of the two assertions C1 and
C2 holds.

Thus, in each of the two Cases 1 and 2, we have proven that at least one of
the two assertions C1 and C2 holds. Since these two Cases cover all
possibilities, we therefore conclude that at least one of the two assertions
C1 and C2 holds. Therefore, exactly one of the two assertions C1 and C2 holds
(since we already know that the Assertions C1 and C2 cannot hold at the same
time). This proves Theorem 2.1c.
\end{proof}

Now, we are going to prove Theorem 2.1b and Theorem 2.1a:

\begin{proof}
[Proof of Theorem 2.1b.]We know that $C$ is a polytope. By the definition of a
polytope, this shows that $C$ is the convex hull of a finite set of vectors in
$E$. In other words, there exist some $t\in\mathbb{N}$ and some vectors
$x_{1}$, $x_{2}$, $\ldots$, $x_{t}$ in $E$ such that $C=\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots,x_{t}\right\}  $. Consider
this $t$ and these $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$.

Clearly, $\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}%
-z,x_{2}-z,\ldots,x_{t}-z\right\}  $ is a polytope. Moreover,%
\[
0\notin\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}-z,x_{2}%
-z,\ldots,x_{t}-z\right\}
\]
\footnote{\textit{Proof.} Assume the opposite. Then, $0\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $. But
Proposition 2.0h \textbf{(b)} (applied to $t$, $x_{i}-z$ and $0$ instead of
$n$, $x_{i}$ and $x$) shows that we have $0\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $ if and
only if $0$ is a convex combination of the vectors $x_{1}-z$, $x_{2}-z$,
$\ldots$, $x_{t}-z$. Hence, $0$ is a convex combination of the vectors
$x_{1}-z$, $x_{2}-z$, $\ldots$, $x_{t}-z$ (since we have $0\in
\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}-z,x_{2}-z,\ldots
,x_{t}-z\right\}  $). In other words, there exist nonnegative elements
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{t}$ of $\mathbb{R}$ such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}\left(  x_{i}-z\right)  =0$. Consider these $\lambda_{1}$,
$\lambda_{2}$, $\ldots$, $\lambda_{t}$.
\par
We have
\[
0=\sum\limits_{i=1}^{t}\lambda_{i}\left(  x_{i}-z\right)  =\sum\limits_{i=1}%
^{t}\lambda_{i}x_{i}-\underbrace{\sum\limits_{i=1}^{t}\lambda_{i}}_{=1}%
z=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}-z,
\]
so that $z=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}$. Since $\lambda_{1}$,
$\lambda_{2}$, $\ldots$, $\lambda_{t}$ are nonnegative elements of
$\mathbb{R}$ such that $\sum\limits_{i=1}^{t}\lambda_{i}=1$, this yields that
$z$ is a convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{t}%
$. In other words, $z\in\operatorname*{conv}.\operatorname*{hull}\left\{
x_{1},x_{2},\ldots,x_{t}\right\}  $. Hence, $z\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots,x_{t}\right\}  =C$, which
contradicts $z\notin C$. This contradiction shows that our assumption was
wrong, qed.}.

Now, applying Theorem 2.1c to $\operatorname*{conv}.\operatorname*{hull}%
\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $ instead of $C$, we obtain
that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion C}$_{z}$\textit{1:} We have $0\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $.

\textit{Assertion C}$_{z}$\textit{2:} There exists an $f\in E^{\ast}$ such
that every \newline$x\in\operatorname*{conv}.\operatorname*{hull}\left\{
x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $ satisfies $f\left(  x\right)  <0$.
\end{statement}

Since Assertion C$_{z}$1 cannot hold (because $0\notin\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $), this
yields that Assertion C$_{z}$2 must hold. In other words, there exists an
$f\in E^{\ast}$ such that every $x\in\operatorname*{conv}.\operatorname*{hull}%
\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $ satisfies $f\left(
x\right)  <0$. Consider this $f$.

Clearly, every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $f\left(
z\right)  -f\left(  x_{i}\right)  >0$\ \ \ \ \footnote{\textit{Proof.} Let
$i\in\left\{  1,2,\ldots,t\right\}  $. Then, $x_{i}-z\in\left\{  x_{1}%
-z,x_{2}-z,\ldots,x_{t}-z\right\}  \subseteq\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1}-z,x_{2}-z,\ldots,x_{t}-z\right\}  $ (by
Proposition 2.0f \textbf{(c)}, applied to $\left\{  x_{1}-z,x_{2}%
-z,\ldots,x_{t}-z\right\}  $ instead of $S$). But we know that every
$x\in\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}-z,x_{2}%
-z,\ldots,x_{t}-z\right\}  $ satisfies $f\left(  x\right)  <0$. Applying this
to $x=x_{i}-z$, we obtain $f\left(  x_{i}-z\right)  <0$ (since $x_{i}%
-z\in\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}-z,x_{2}%
-z,\ldots,x_{t}-z\right\}  $). Since $f\left(  x_{i}-z\right)  =f\left(
x_{i}\right)  -f\left(  z\right)  $ (since $f$ is linear), this rewrites as
$f\left(  x_{i}\right)  -f\left(  z\right)  <0$. In other words, $f\left(
x_{i}\right)  <f\left(  z\right)  $. Hence, $f\left(  z\right)  -f\left(
x_{i}\right)  >0$, qed.}. Hence,
\[
\left\{  f\left(  z\right)  -f\left(  x_{1}\right)  ,f\left(  z\right)
-f\left(  x_{2}\right)  ,\ldots,f\left(  z\right)  -f\left(  x_{t}\right)
\right\}
\]
is a set of positive reals. Since this set is finite, it must be bounded from
below by a positive real (because any finite set of positive reals is bounded
from below by a positive real). In other words, there exists a positive real
$\varepsilon$ such that every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies
$f\left(  z\right)  -f\left(  x_{i}\right)  \geq\varepsilon$. Consider this
$\varepsilon$.

Since $\varepsilon$ is positive, we have $\varepsilon>\dfrac{\varepsilon}{2}$.

Now, let $\delta=f\left(  z\right)  -\dfrac{\varepsilon}{2}$. Then,%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,t\right\}  \text{ satisfies }f\left(
x_{i}\right)  <\delta\label{2.1b.pf.1}%
\end{equation}
(since every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $f\left(
z\right)  -f\left(  x_{i}\right)  \geq\varepsilon$ and thus $f\left(
x_{i}\right)  \leq f\left(  z\right)  -\underbrace{\varepsilon}_{>\dfrac
{\varepsilon}{2}}<f\left(  z\right)  -\dfrac{\varepsilon}{2}=\delta$).
Therefore, Proposition 2.0r shows that%
\begin{equation}
\text{every }x\in C\text{ satisfies }f\left(  x\right)  <\delta.
\label{2.1b.pf.2}%
\end{equation}


Altogether, we thus know that $f\left(  z\right)  >\delta$ (since
$\varepsilon$ is positive, so that $\dfrac{\varepsilon}{2}>0$, and thus
$\delta=f\left(  z\right)  -\underbrace{\dfrac{\varepsilon}{2}}_{>0}<f\left(
z\right)  $) and that every $x\in C$ satisfies $f\left(  x\right)  <\delta$
(by (\ref{2.1b.pf.2})). We thus have shown that there exist an $f\in E^{\ast}$
and a $\delta\in\mathbb{R}$ such that
\[
\left(  f\left(  z\right)  >\delta\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{every }x\in C\text{ satisfies }f\left(
x\right)  <\delta\right)  .
\]
This proves Theorem 2.1b.
\end{proof}

\begin{proof}
[Proof of Theorem 2.1a.]Theorem 2.1b (applied to $E=\mathbb{R}^{n}$) shows
that there exist an $f\in\left(  \mathbb{R}^{n}\right)  ^{\ast}$ and a
$\delta\in\mathbb{R}$ such that
\[
\left(  f\left(  z\right)  >\delta\right)  \ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{every }x\in C\text{ satisfies }f\left(
x\right)  <\delta\right)  .
\]
Consider this $f$. Then, the hyperplane $f^{-1}\left(  \delta\right)
\subseteq\mathbb{R}^{n}$ clearly separates $z$ and $C$. Thus, there exists a
hyperplane separating $z$ and $C$. This proves Theorem 2.1a.
\end{proof}

\section{The Farkas lemma}

Next, we are going to give a proof of \cite[Theorem 2.5]{Schrij-CCO} without
recourse to \cite[Exercise 2.7]{Schrij-CCO}. This won't be very rich in
substance, since the main work has already been done proving Theorems 2.1c and 2.5c.

First, we reformulate \cite[Theorem 2.5]{Schrij-CCO}:

\begin{statement}
\textbf{Theorem 2.5d.} Let $m\in\mathbb{N}$ and $n\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix and let $b\in\mathbb{R}^{m}$. Then, exactly one of the
following two assertions holds:

\textit{Assertion F1:} The system $Ax=b$ has a nonnegative solution
$x\in\mathbb{R}^{n}$.

\textit{Assertion F2:} There exists a vector $y\in\mathbb{R}^{m}$ such that
$y^{T}A\geq0$ and $y^{T}b<0$.
\end{statement}

In classical logic, Theorem 2.5d is equivalent to \cite[Theorem 2.5]%
{Schrij-CCO}, but constructively Theorem 2.5d is stronger.

Notice that Theorem 2.5d appears in \cite[Teorema 1]{Jacim11} (in an
equivalent form) and in \cite[Lemma 1]{Bartl12} (in a generalized version).

\begin{proof}
[Proof of Theorem 2.5d.]The Assertions F1 and F2 cannot hold at the same
time\footnote{\textit{Proof.} Assume the opposite. Then, the Assertions F1 and
F2 hold at the same time. Since Assertion F2 holds, there exists a vector
$y\in\mathbb{R}^{m}$ such that $y^{T}A\geq0$ and $y^{T}b<0$. Consider this
$y$. Since Assertion F1 holds, the system $Ax=b$ has a nonnegative solution
$x\in\mathbb{R}^{n}$. Consider this solution $x$. We have $x \geq0$ (since $x$
is nonnegative).
\par
Lemma 2.0v (applied to $y^{T}A$ and $x$ instead of $x$ and $y$) shows that
$y^{T}Ax\geq0$ (since $y^{T}A\geq0$ and $x\geq0$). This contradicts
$y^{T}\underbrace{Ax}_{=b}=y^{T}b<0$. This contradiction shows that our
assumption was wrong, qed.}. We will now show that at least one of these
assertions holds.

Let $a_{1}$, $a_{2}$, $\ldots$, $a_{n}$ be the columns of $A$. Let $S=\left\{
a_{1},a_{2},\ldots,a_{n}\right\}  $. Then, $S$ is a finite subset of
$\mathbb{R}^{m}$. Hence, Theorem 2.5c (applied to $E=\mathbb{R}^{m}$) yields
that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion F}$_{1}$\textit{1:} We have $b\in\operatorname*{cone}S$.

\textit{Assertion F}$_{2}$\textit{2:} There exists an $f\in\left(
\mathbb{R}^{m}\right)  ^{\ast}$ such that $f\left(  b\right)  >0$ and
\newline$\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies
}f\left(  x\right)  \leq0\right)  $.
\end{statement}

We must thus be in one of the following two cases:

\textit{Case 1:} Assertion F$_{1}$1 holds.

\textit{Case 2:} Assertion F$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion F$_{1}$1 holds. In other
words, $b\in\operatorname*{cone}S$. Thus,%
\[
b\in\operatorname*{cone}S=\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(
\nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative reals}\right\}
\]
(by Proposition 2.0p \textbf{(a)}). Hence, there exists a family $\left(
\nu_{s}\right)  _{s\in S}$ of nonnegative reals such that $b=\sum\limits_{s\in
S}\nu_{s}s$. Consider this family $\left(  \nu_{s}\right)  _{s\in S}$.

Let $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ be the standard basis of
$\mathbb{R}^{n}$. In other words, for every $i\in\left\{  1,2,\ldots
,n\right\}  $, let $e_{i}$ be the vector in $\mathbb{R}^{n}$ whose $i$-th
coordinate is $1$ and whose other coordinates are all $0$. Then, for every
$i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{equation}
Ae_{i}=\left(  \text{the }i\text{-th column of }A\right)  =a_{i}
\label{2.5d.pf.0}%
\end{equation}
(since the columns of $A$ are $a_{1}$, $a_{2}$, $\ldots$, $a_{n}$). Also, for
every $i\in\left\{  1,2,\ldots,n\right\}  $, we have $e_{i}\geq0$ (since every
coordinate of $e_{i}$ is either $1$ or $0$, and thus nonnegative).

Now, it is pretty clear that%
\begin{equation}
\text{for every }s\in S\text{, there exists a vector }x_{s}\in\mathbb{R}%
^{n}\text{ such that }x_{s}\geq0\text{ and }Ax_{s}=s\text{.} \label{2.5d.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5d.pf.1}):} Let $s\in S$. Then, $s\in
S=\left\{  a_{1},a_{2},\ldots,a_{n}\right\}  $. Hence, there exists some
$i\in\left\{  1,2,\ldots,n\right\}  $ such that $s=a_{i}$. Consider this $i$.
Then, $s=a_{i}=Ae_{i}$ (by (\ref{2.5d.pf.0})). Thus, $e_{i}\geq0$ and
$Ae_{i}=s$. Thus, there exists a vector $x_{s}\in\mathbb{R}^{n}$ such that
$x_{s}\geq0$ and $Ax_{s}=s$ (namely, $x_{s}=e_{i}$). This proves
(\ref{2.5d.pf.1}).} Consider this $x_{s}$.

Now, $\sum\limits_{s\in S}\nu_{s}x_{s}$ is a linear combination of nonnegative
vectors (namely, the vectors $x_{s}$) with nonnegative coefficients (namely,
the coefficients $\nu_{s}$). Thus, $\sum\limits_{s\in S}\nu_{s}x_{s}$ must
itself be a nonnegative vector (since any linear combination of nonnegative
vectors with nonnegative coefficients must itself be a nonnegative vector).
Since%
\[
A\left(  \sum\limits_{s\in S}\nu_{s}x_{s}\right)  =\sum\limits_{s\in S}\nu
_{s}\underbrace{Ax_{s}}_{=s}=\sum\limits_{s\in S}\nu_{s}s=b,
\]
this yields that the system $Ax=b$ has a nonnegative solution $x\in
\mathbb{R}^{n}$ (namely, $x=\sum\limits_{s\in S}\nu_{s}x_{s}$). In other
words, Assertion F1 holds. Hence, at least one of Assertions F1 and F2 holds.

We thus have proven that at least one of Assertions F1 and F2 holds in Case 1.

Now, let us consider Case 2. In this case, Assertion F$_{1}$2 holds. In other
words, there exists an $f\in\left(  \mathbb{R}^{m}\right)  ^{\ast}$ such that
$f\left(  b\right)  >0$ and $\left(  \text{every }x\in\operatorname*{cone}%
S\text{ satisfies }f\left(  x\right)  \leq0\right)  $. Consider this $f$.

Let $T$ denote the map%
\begin{align*}
\mathbb{R}^{m}  &  \rightarrow\left(  \mathbb{R}^{m}\right)  ^{\ast},\\
v  &  \mapsto v^{T}.
\end{align*}
Then, it is known that $T$ is an isomorphism of $\mathbb{R}$-vector spaces.
Hence, $T^{-1}$ is well-defined. Let $z=T^{-1}\left(  f\right)  $. Then,
$f=T\left(  z\right)  =z^{T}$ (by the definition of $T$). Thus, $f\left(
b\right)  =z^{T}b$. Hence, $z^{T}b=f\left(  b\right)  >0$, so that $\left(
-z\right)  ^{T}b=-\underbrace{z^{T}b}_{>0}<-0=0$.

Also, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies $\left(
-z\right)  ^{T}a_{i}\geq0$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
1,2,\ldots,n\right\}  $. Then, $a_{i}\in\left\{  a_{1},a_{2},\ldots
,a_{n}\right\}  =S\subseteq\operatorname*{cone}S$ (by Proposition 2.0m
\textbf{(c)}). But we know that every $x\in\operatorname*{cone}S$ satisfies
$f\left(  x\right)  \leq0$. Applying this to $x=a_{i}$, we obtain $f\left(
a_{i}\right)  \leq0$ (since $a_{i}\in\operatorname*{cone}S$). Since $f=z^{T}$,
this rewrites as $z^{T}a_{i}\leq0$, so that $\left(  -z\right)  ^{T}%
a_{i}=-\underbrace{z^{T}a_{i}}_{\leq0}\geq-0=0$, qed.}. Now, for every
$i\in\left\{  1,2,\ldots,n\right\}  $, the rule for multiplying a row vector
by a matrix yields that%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }\left(  -z\right)
^{T}A\right) \\
&  =\left(  -z\right)  ^{T}\cdot\underbrace{\left(  \text{the }i\text{-th
column of }A\right)  }_{\substack{=a_{i}\\\text{(since the columns of }A\text{
are }a_{1}\text{, }a_{2}\text{, }\ldots\text{, }a_{n}\text{)}}}=\left(
-z\right)  ^{T}a_{i}\\
&  \geq0=\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $\left(  -z\right)  ^{T}A\geq0$.

We thus know that $\left(  -z\right)  ^{T}A\geq0$ and $\left(  -z\right)
^{T}b<0$. Hence, there exists a vector $y\in\mathbb{R}^{m}$ such that
$y^{T}A\geq0$ and $y^{T}b<0$ (namely, $y=-z$). In other words, Assertion F2
holds. Hence, at least one of Assertions F1 and F2 holds.

We thus have proven that at least one of Assertions F1 and F2 holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions F1 and F2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions F1 and F2 holds. Since we know
that the Assertions F1 and F2 cannot hold at the same time, this yields that
exactly one of Assertions F1 and F2 holds. This proves Theorem 2.5d.
\end{proof}

Thus, of course, \cite[Theorem 2.5]{Schrij-CCO} is also proven.

\section{\label{sect.lt-gt}The $<$ and $>$ relations for vectors}

Let us now take a break from proving theorems, and introduce another piece of notation:

\begin{statement}
\textbf{Definition.} Let $\ell\in\mathbb{N}$. Let $u$ and $v$ be two column
vectors in $\mathbb{R}^{\ell}$. Then, we write $u<v$ if and only if every
$i\in\left\{  1,2,\ldots,\ell\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }u\right)  <\left(  \text{the
}i\text{-th coordinate of }v\right)  .
\]
Also, we write $u>v$ if and only if every $i\in\left\{  1,2,\ldots
,\ell\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }u\right)  >\left(  \text{the
}i\text{-th coordinate of }v\right)  .
\]
The same notations apply if $u$ and $v$ are row vectors in $\left(
\mathbb{R}^{\ell}\right)  ^{\ast}$ rather than column vectors in
$\mathbb{R}^{\ell}$.
\end{statement}

Note that for a positive integer $\ell$ and two vectors $u\in\mathbb{R}^{\ell
}$ and $v\in\mathbb{R}^{\ell}$, the assertion $\left(  u>v\right)  $ is
\textbf{not} (in general) equivalent to the assertion $\left(  u\geq v\text{
and }u\neq v\right)  $, but is stronger. Similarly, the assertion $\left(
u<v\right)  $ is stronger than $\left(  u\leq v\text{ and }u\neq v\right)  $.
For example, if $u=\left(
\begin{array}
[c]{c}%
1\\
1
\end{array}
\right)  $ and $v=\left(
\begin{array}
[c]{c}%
1\\
0
\end{array}
\right)  $, then we do have $\left(  u\geq v\text{ and }u\neq v\right)  $, but
we don't have $u>v$.

Here is one further caveat: If $0_{0}$ denotes the zero vector in
$\mathbb{R}^{0}$, then $0_{0}>0_{0}$ and $0_{0}<0_{0}$. This is not
surprising, since $0_{0}$ has \textbf{no coordinates at all}.

The newly introduced notation has several properties which are similar to
properties proven before. For example, here is an analogue of Lemma 2.0t:

\begin{statement}
\textbf{Lemma 2.2a.} Let $n\in\mathbb{N}$.

\textbf{(a)} If a column vector $v\in\mathbb{R}^{n}$ satisfies $v>0$, then all
coordinates of the column vector $v$ are positive.

\textbf{(b)} If a row vector $v\in\left(  \mathbb{R}^{n}\right)  ^{\ast}$
satisfies $v>0$, then all coordinates of the row vector $v$ are positive.
\end{statement}

A proof of Lemma 2.2a can be found in Section \ref{sect.basics-proofs}. Of
course, the converse of Lemma 2.2a also holds; we just won't use it often
enough to have a reason to state it.

Let us state three further facts, which also are proven in Section
\ref{sect.basics-proofs} below.

\begin{statement}
\textbf{Lemma 2.2b.} Let $n\in\mathbb{N}$. Let $x\in\left(  \mathbb{R}%
^{n}\right)  ^{\ast}$ be a nonzero row vector such that $x\geq0$. Let
$y\in\mathbb{R}^{n}$ be a column vector such that $y>0$. Then, $xy>0$.

\textbf{Lemma 2.2c.} Let $n\in\mathbb{N}$. Let $x\in\left(  \mathbb{R}%
^{n}\right)  ^{\ast}$ be a row vector such that $x>0$. Let $y\in\mathbb{R}%
^{n}$ be a nonzero column vector such that $y\geq0$. Then, $xy>0$.

\textbf{Lemma 2.2d.} Let $n\in\mathbb{N}$. Let $x\in\left(  \mathbb{R}%
^{n}\right)  ^{\ast}$ be a row vector such that $x\geq0$. Let $y\in
\mathbb{R}^{n}$ be a column vector such that $y>0$. Assume that $xy=0$. Then,
$x=0$.
\end{statement}

\section{The Gordan and Stiemke theorems}

Next, we will state two further important results.

\begin{statement}
\textbf{Theorem 2.5e.} Let $A$ be an $m\times n$-matrix. Then, exactly one of
the following two assertions holds:

\textit{Assertion G1:} There exists a nonzero vector $x\in\mathbb{R}^{n}$ such
that $x\geq0$ and $Ax=0$.

\textit{Assertion G2:} There exists a vector $y\in\mathbb{R}^{m}$ such that
$y^{T}A>0$.
\end{statement}

Theorem 2.5e is known as \textit{Gordan's theorem} and is equivalent to
Exercise 2.17 in \cite{Schrij-CCO} in classical logic. In constructive logic,
Theorem 2.5e is somewhat stronger than Exercise 2.17 in \cite{Schrij-CCO}.

\begin{statement}
\textbf{Theorem 2.5f.} Let $A$ be an $m\times n$-matrix. Then, exactly one of
the following two assertions holds:

\textit{Assertion S1:} There exists a vector $x\in\mathbb{R}^{n}$ such that
$x>0$ and $Ax=0$.

\textit{Assertion S2:} There exists a vector $y\in\mathbb{R}^{m}$ such that
$y^{T}A\geq0$ and $y^{T}A\neq0$.
\end{statement}

Theorem 2.5f is known as \textit{Stiemke's theorem} and is equivalent to
Exercise 2.16 in \cite{Schrij-CCO} in classical logic. In constructive logic,
Theorem 2.5f is somewhat stronger than Exercise 2.16 in \cite{Schrij-CCO}.

\begin{proof}
[Proof of Theorem 2.5e.]The Assertions G1 and G2 cannot hold at the same
time\footnote{\textit{Proof.} Assume the opposite. Then, the Assertions G1 and
G2 hold at the same time. Since Assertion G2 holds, there exists a vector
$y\in\mathbb{R}^{m}$ such that $y^{T}A>0$. Consider this $y$. Since Assertion
G1 holds, there exists a nonzero vector $x\in\mathbb{R}^{n}$ such that
$x\geq0$ and $Ax=0$. Consider this $x$.
\par
Lemma 2.2c (applied to $y^{T}A$ and $x$ instead of $x$ and $y$) yields
$y^{T}Ax>0$ (since $y^{T}A>0$ and $x\geq0$ and since $x$ is nonzero). This
contradicts $y^{T}Ax=y^{T}\underbrace{Ax}_{=0}=0$. This contradiction shows
that our assumption was wrong, qed.}. We will now show that at least one of
these assertions holds.

Let $a_{1}$, $a_{2}$, $\ldots$, $a_{n}$ be the columns of $A$. Let
$C=\operatorname*{conv}.\operatorname*{hull}\left\{  a_{1},a_{2},\ldots
,a_{n}\right\}  $. Then, $C$ is a polytope in $\mathbb{R}^{m}$. Hence, Theorem
2.1c (applied to $E=\mathbb{R}^{m}$) yields that exactly one of the following
two assertions holds:

\begin{statement}
\textit{Assertion G}$_{1}$\textit{1:} We have $0\in C$.

\textit{Assertion G}$_{1}$\textit{2:} There exists an $f\in\left(
\mathbb{R}^{m}\right)  ^{\ast}$ such that every $x\in C$ satisfies $f\left(
x\right)  <0$.
\end{statement}

Thus, we must be in one of the following two cases:

\textit{Case 1:} Assertion G$_{1}$1 holds.

\textit{Case 2:} Assertion G$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion G$_{1}$1 holds. In other
words, $0\in C$. Thus, $0\in C=\operatorname*{conv}.\operatorname*{hull}%
\left\{  a_{1},a_{2},\ldots,a_{n}\right\}  $.

Proposition 2.0h \textbf{(b)} (applied to $\mathbb{R}^{m}$, $a_{i}$ and $0$
instead of $E$, $x_{i}$ and $x$) shows that we have $0\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  a_{1},a_{2},\ldots,a_{n}\right\}  $ if and only
if $0$ is a convex combination of the vectors $a_{1}$, $a_{2}$, $\ldots$,
$a_{n}$. Thus, $0$ is a convex combination of the vectors $a_{1}$, $a_{2}$,
$\ldots$, $a_{n}$ (since we have $0\in\operatorname*{conv}%
.\operatorname*{hull}\left\{  a_{1},a_{2},\ldots,a_{n}\right\}  $). In other
words, there exist $n$ nonnegative elements $\lambda_{1}$, $\lambda_{2}$,
$\ldots$, $\lambda_{n}$ of $\mathbb{R}$ such that $\sum\limits_{i=1}%
^{n}\lambda_{i}=1$ and $\sum\limits_{i=1}^{n}\lambda_{i}a_{i}=0$. Consider
these $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$.

Let $u$ be the vector $\left(
\begin{array}
[c]{c}%
\lambda_{1}\\
\lambda_{2}\\
\vdots\\
\lambda_{n}%
\end{array}
\right)  \in\mathbb{R}^{n}$. Then, the sum of the coordinates of the vector
$u$ is $\sum\limits_{i=1}^{n}\lambda_{i}=1\neq0$, while the sum of the
coordinates of the vector $0$ is $0$. Thus, $u\neq0$. Also, $u\geq0$ (because
$u=\left(
\begin{array}
[c]{c}%
\lambda_{1}\\
\lambda_{2}\\
\vdots\\
\lambda_{n}%
\end{array}
\right)  $, but every $\lambda_{i}$ is nonnegative). Moreover, by the
definition of the product of a matrix with a column vector, we have%
\begin{align*}
Au  &  =\sum\limits_{i=1}^{n}\underbrace{\left(  \text{the }i\text{-th column
of }A\right)  }_{\substack{=a_{i}\\\text{(since the columns of }A\\\text{are
}a_{1}\text{, }a_{2}\text{, }\ldots\text{, }a_{n}\text{)}}}\cdot
\underbrace{\left(  \text{the }i\text{-th coordinate of }u\right)
}_{\substack{=\lambda_{i}\\\text{(since }u=\left(
\begin{array}
[c]{c}%
\lambda_{1}\\
\lambda_{2}\\
\vdots\\
\lambda_{n}%
\end{array}
\right)  \text{)}}}=\sum\limits_{i=1}^{n}a_{i}\lambda_{i}\\
&  =\sum\limits_{i=1}^{n}\lambda_{i}a_{i}=0.
\end{align*}


Hence, $u$ is nonzero, and satisfies $u\geq0$ and $Au=0$. Thus, there exists a
nonzero vector $x\in\mathbb{R}^{n}$ such that $x\geq0$ and $Ax=0$ (namely,
$x=u$). In other words, Assertion G1 holds. Hence, at least one of Assertions
G1 and G2 holds.

We thus have proven that at least one of Assertions G1 and G2 holds in Case 1.

Next, let us consider Case 2. In this case, Assertion G$_{1}$2 holds. In other
words, there exists an $f\in\left(  \mathbb{R}^{m}\right)  ^{\ast}$ such that
\begin{equation}
\text{every }x\in C\text{ satisfies }f\left(  x\right)  <0. \label{2.5e.pf.1}%
\end{equation}
Consider this $f$.

Let $T$ denote the map%
\begin{align*}
\mathbb{R}^{m}  &  \rightarrow\left(  \mathbb{R}^{m}\right)  ^{\ast},\\
v  &  \mapsto v^{T}.
\end{align*}
Then, it is known that $T$ is an isomorphism of $\mathbb{R}$-vector spaces.
Hence, $T^{-1}$ is well-defined. Let $z=T^{-1}\left(  f\right)  $. Then,
$f=T\left(  z\right)  =z^{T}$ (by the definition of $T$).

Every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
a_{i}  &  \in\left\{  a_{1},a_{2},\ldots,a_{n}\right\}  \subseteq
\operatorname*{conv}.\operatorname*{hull}\left\{  a_{1},a_{2},\ldots
,a_{n}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition 2.0f \textbf{(c)}, applied
to }S=\left\{  a_{1},a_{2},\ldots,a_{n}\right\}  \right) \\
&  =C
\end{align*}
and thus $f\left(  a_{i}\right)  <0$ (by (\ref{2.5e.pf.1}), applied to
$x=a_{i}$). Now, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
-z\right)  ^{T}A\right) \\
&  =\underbrace{\left(  -z\right)  ^{T}}_{\substack{=-z^{T}=-f\\\text{(since
}f=z^{T}\text{)}}}\cdot\underbrace{\left(  \text{the }i\text{-th column of
}A\right)  }_{\substack{=a_{i}\\\text{(since the columns of }A\\\text{are
}a_{1}\text{, }a_{2}\text{, }\ldots\text{, }a_{n}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the product of a row
vector with a matrix}\right) \\
&  =\left(  -f\right)  \cdot a_{i}=-\underbrace{f\left(  a_{i}\right)  }%
_{<0}>-0=0=\left(  \text{the }i\text{-th coordinate of the row vector
}0\right)  .
\end{align*}
Thus, $\left(  -z\right)  ^{T}A>0$. Hence, there exists a vector
$y\in\mathbb{R}^{m}$ such that $y^{T}A>0$ (namely, $y=-z$). In other words,
Assertion G2 holds. Hence, at least one of Assertions G1 and G2 holds.

We thus have proven that at least one of Assertions G1 and G2 holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions G1 and G2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions G1 and G2 holds. Since we know
that the Assertions G1 and G2 cannot hold at the same time, this yields that
exactly one of Assertions G1 and G2 holds. This proves Theorem 2.5e.
\end{proof}

As we have seen, Theorem 2.5e is no more than a simple corollary of Theorem
2.1c. Theorem 2.5f is more interesting.

\begin{proof}
[Proof of Theorem 2.5f.]The Assertions S1 and S2 cannot hold at the same
time\footnote{\textit{Proof.} Assume the opposite. Then, the Assertions S1 and
S2 hold at the same time. Since Assertion S2 holds, there exists a vector
$y\in\mathbb{R}^{m}$ such that $y^{T}A\geq0$ and $y^{T}A\neq0$. Consider this
$y$. Since Assertion S1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such
that $x>0$ and $Ax=0$. Consider this $x$.
\par
The row vector $y^{T}A$ is nonzero (since $y^{T}A\neq0$) and satisfies
$y^{T}A\geq0$. Hence, Lemma 2.2b (applied to $y^{T}A$ and $x$ instead of $x$
and $y$) shows that $y^{T}Ax>0$ (since $x>0$). This contradicts $y^{T}%
Ax=y^{T}\underbrace{Ax}_{=0}=0$. This contradiction shows that our assumption
was wrong, qed.}. We will now show that at least one of these assertions holds.

Let $p$ be the vector $\left(
\begin{array}
[c]{c}%
1\\
1\\
\vdots\\
1
\end{array}
\right)  \in\mathbb{R}^{n}$. Clearly, $p>0$.

Applying Theorem 2.5d to $b=-Ap$, we see that exactly one of the following two
assertions holds:

\begin{statement}
\textit{Assertion S}$_{1}$\textit{1:} The system $Ax=-Ap$ has a nonnegative
solution $x\in\mathbb{R}^{n}$.

\textit{Assertion S}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{m}$ such that $y^{T}A\geq0$ and $y^{T}\left(  -Ap\right)  <0$.
\end{statement}

Therefore, we must be in one of the following two cases:

\textit{Case 1:} Assertion S$_{1}$1 holds.

\textit{Case 2:} Assertion S$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion S$_{1}$1 holds. In other
words, the system $Ax=-Ap$ has a nonnegative solution $x\in\mathbb{R}^{n}$.
Let $y$ be this solution. Then, $y\in\mathbb{R}^{n}$ is nonnegative and
satisfies $Ay=-Ap$. Since $y\geq0$ (because $y$ is nonnegative) and $p>0$, we
have $\underbrace{y}_{\geq0}+\underbrace{p}_{>0}>0$. Also, $A\left(
y+p\right)  =\underbrace{Ay}_{=-Ap}+Ap=-Ap+Ap=0$.

Hence, we know that $y+p>0$ and $A\left(  y+p\right)  =0$. Thus, there exists
a vector $x\in\mathbb{R}^{n}$ such that $x>0$ and $Ax=0$ (namely, $x=y+p$). In
other words, Assertion S1 holds. Hence, at least one of Assertions S1 and S2 holds.

We have thus proven that at least one of Assertions S1 and S2 holds in Case 1.

Now, let us consider Case 2. In this case, Assertion S$_{1}$2 holds. In other
words, there exists a vector $y\in\mathbb{R}^{m}$ such that $y^{T}A\geq0$ and
$y^{T}\left(  -Ap\right)  <0$. Denote this vector $y$ by $w$. Then,
$w\in\mathbb{R}^{m}$ satisfies $w^{T}A\geq0$ and $w^{T}\left(  -Ap\right)  <0$.

Since $-w^{T}Ap=w^{T}\left(  -Ap\right)  <0$, we have $-w^{T}Ap\neq0$, so that
$w^{T}A\neq0$.

We thus know that $w^{T}A\geq0$ and $w^{T}A\neq0$. Hence, there exists a
vector $y\in\mathbb{R}^{m}$ such that $y^{T}A\geq0$ and $y^{T}A\neq0$ (namely,
$y=w$). In other words, Assertion S2 holds. Hence, at least one of Assertions
S1 and S2 holds.

We have thus proven that at least one of Assertions S1 and S2 holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions S1 and S2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions S1 and S2 holds. Since we know
that the Assertions S1 and S2 cannot hold at the same time, this yields that
exactly one of Assertions S1 and S2 holds. This proves Theorem 2.5f.
\end{proof}

\section{\label{sect.block}Block matrices}

Let us recall the following convention:

\begin{statement}
\textbf{Convention 2.5g.} \textbf{(a)} Let $u\in\mathbb{N}$ and $v\in
\mathbb{N}$. Let $n_{1}$, $n_{2}$, $\ldots$, $n_{v}$ be nonnegative integers,
and let $m_{1}$, $m_{2}$, $\ldots$, $m_{u}$ be nonnegative integers. For every
$\left(  i,j\right)  \in\left\{  1,2,\ldots,u\right\}  \times\left\{
1,2,\ldots,v\right\}  $, let $A_{i,j}$ be an $m_{i}\times n_{j}$-matrix. Then,
$\left(
\begin{array}
[c]{cccc}%
A_{1,1} & A_{1,2} & \ldots & A_{1,v}\\
A_{2,1} & A_{2,2} & \ldots & A_{2,v}\\
\vdots & \vdots & \ddots & \vdots\\
A_{u,1} & A_{u,2} & \ldots & A_{u,v}%
\end{array}
\right)  $ will not mean a $u\times v$-matrix whose entries themselves are
matrices, but rather a \textbf{block matrix} of size $\left(  m_{1}%
+m_{2}+\cdots+m_{u}\right)  \times\left(  n_{1}+n_{2}+\cdots+n_{v}\right)  $
whose blocks are the matrices $A_{i,j}$.

\textbf{(b)} Let $n\in\mathbb{N}$. Column vectors in $\mathbb{R}^{n}$ are
considered as $n\times1$-matrices, and row vectors in $\left(  \mathbb{R}%
^{n}\right)  ^{\ast}$ are considered as $1\times n$-matrices. In particular,
Convention 2.5g \textbf{(a)} also applies when some of the $A_{i,j}$ are
column vectors or row vectors (as long as their sizes \textquotedblleft
fit\textquotedblright). For example, if $x$ is the vector $\left(
\begin{array}
[c]{c}%
2\\
1
\end{array}
\right)  $ and $y$ is the vector $\left(
\begin{array}
[c]{c}%
3\\
7\\
5
\end{array}
\right)  $, then $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $ denotes the vector $\left(
\begin{array}
[c]{c}%
2\\
1\\
3\\
7\\
5
\end{array}
\right)  $.
\end{statement}

Let us state yet another trivial property of nonnegative vectors using
Convention 2.5g:

\begin{statement}
\textbf{Lemma 2.5h.} Let $N\in\mathbb{N}$ and $M\in\mathbb{N}$. Let
$x\in\mathbb{R}^{N}$ and $y\in\mathbb{R}^{M}$. Consider the block matrix
$\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $; this is an $\left(  N+M\right)  \times1$-matrix, i. e., a vector
in $\mathbb{R}^{N+M}$ (since we identify matrices having only one column with
column vectors).

\textbf{(a)} If $x\geq0$ and $y\geq0$, then $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \geq0$.

\textbf{(b)} If $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \geq0$, then $x\geq0$ and $y\geq0$.
\end{statement}

See Section \ref{sect.basics-proofs} for the proof of Lemma 2.5h.

The following simple lemma further illustrates the use of Convention 2.5g:

\begin{statement}
\textbf{Lemma 2.5i.} Let $n\in\mathbb{N}$. Let $x\in\mathbb{R}^{n}$ and
$y\in\mathbb{R}^{n}$ be such that $\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \geq\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  $ (where $\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  $ and $\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  $ are to be understood according to Convention 2.5g). Then, $x=y$.
\end{statement}

We shall not use Lemma 2.5i, but a proof nevertheless is given in Section
\ref{sect.basics-proofs}.

\section{The Motzkin theorem}

We will now prove another of the basic theorems of linear optimization theory:

\begin{statement}
\textbf{Theorem 2.5k.} Let $n\in\mathbb{N}$, $m\in\mathbb{N}$ and $m^{\prime
}\in\mathbb{N}$. Let $A$ be an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$.
Let $A^{\prime}$ be an $m^{\prime}\times n$-matrix. Let $b^{\prime}%
\in\mathbb{R}^{m^{\prime}}$. Then, exactly one of the following two assertions holds:

\textit{Assertion M1:} There exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax<b$ and $A^{\prime}x\leq b^{\prime}$.

\textit{Assertion M2:} There exist two vectors $y\in\mathbb{R}^{m}$ and
$y^{\prime}\in\mathbb{R}^{m^{\prime}}$ such that $y\geq0$, $y^{\prime}\geq0$,
$y^{T}A+y^{\prime T}A^{\prime}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  \right)  .
\]

\end{statement}

Theorem 2.5k is known as \textit{Motzkin's theorem} and is equivalent to
Exercise 2.19 in \cite{Schrij-CCO} in classical logic. In constructive logic,
Theorem 2.5k is somewhat stronger than Exercise 2.19 in \cite{Schrij-CCO}.

To prove Theorem 2.5k, we will first show:

\begin{statement}
\textbf{Theorem 2.5l.} Let $n\in\mathbb{N}$, $n^{\prime}\in\mathbb{N}$ and
$m\in\mathbb{N}$. Let $A$ be an $m\times n$-matrix. Let $A^{\prime}$ be an
$m\times n^{\prime}$-matrix. Then, exactly one of the following two assertions holds:

\textit{Assertion L1:} There exist two vectors $x\in\mathbb{R}^{n}$ and
$x^{\prime}\in\mathbb{R}^{n^{\prime}}$ such that $x>0$, $x^{\prime}\geq0$ and
$Ax+A^{\prime}x^{\prime}=0$.

\textit{Assertion L2:} There exists a vector $y\in\mathbb{R}^{m}$ such that
$y^{T}A\geq0$, $y^{T}A^{\prime}\geq0$ and $y^{T}A\neq0$.
\end{statement}

\begin{proof}
[Proof of Theorem 2.5l.]The Assertions L1 and L2 cannot hold at the same
time\footnote{\textit{Proof.} Assume the opposite. Then, the Assertions L1 and
L2 hold at the same time. Since Assertion L2 holds, there exists a vector
$y\in\mathbb{R}^{m}$ such that $y^{T}A\geq0$, $y^{T}A^{\prime}\geq0$ and
$y^{T}A\neq0$. Consider this $y$.
\par
Since Assertion L1 holds, there exist two vectors $x\in\mathbb{R}^{n}$ and
$x^{\prime}\in\mathbb{R}^{n^{\prime}}$ such that $x>0$, $x^{\prime}\geq0$ and
$Ax+A^{\prime}x^{\prime}=0$. Consider these two vectors $x$ and $x^{\prime}$.
\par
From $x>0$, we conclude that $x\geq0$.
\par
Applying Lemma 2.0v to $y^{T}A$ and $x$ instead of $x$ and $y$, we obtain
$y^{T}Ax\geq0$. Applying Lemma 2.0v to $n^{\prime}$, $y^{T}A^{\prime}$ and
$x^{\prime}$ instead of $n$, $x$ and $y$, we obtain $y^{T}A^{\prime}x^{\prime
}\geq0$.
\par
If we had $y^{T}Ax=0$, then Lemma 2.2d (applied to $y^{T}A$ and $x$ instead of
$x$ and $y$) would yield $y^{T}A=0$, contradicting $y^{T}A\neq0$. Thus, we
cannot have $y^{T}Ax=0$. Hence, $y^{T}Ax\neq0$, so that $y^{T}Ax>0$ (since
$y^{T}Ax\geq0$).
\par
We have $y^{T}\underbrace{\left(  Ax+A^{\prime}x^{\prime}\right)  }_{=0}=0$,
thus%
\[
0=y^{T}\left(  Ax+A^{\prime}x^{\prime}\right)  =\underbrace{y^{T}Ax}%
_{>0}+\underbrace{y^{T}A^{\prime}x^{\prime}}_{\geq0}>0.
\]
This is absurd. This contradiction shows that our assumption was wrong, qed.}.
We will now show that at least one of these assertions holds.

Consider the block matrix $\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  $; this is an $m\times\left(  n+n^{\prime}\right)  $-matrix.

Let $p$ be the vector $\left(
\begin{array}
[c]{c}%
1\\
1\\
\vdots\\
1
\end{array}
\right)  \in\mathbb{R}^{n}$. Clearly, $p>0$.

Applying Theorem 2.5d to $n+n^{\prime}$, $\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  $ and $-Ap$ instead of $n$, $A$ and $b$, we conclude that exactly one
of the following two assertions holds:

\begin{statement}
\textit{Assertion L}$_{1}$\textit{1:} The system $\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  x=-Ap$ has a nonnegative solution $x\in\mathbb{R}^{n+n^{\prime}}$.

\textit{Assertion L}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{m}$ such that $y^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \geq0$ and $y^{T}\left(  -Ap\right)  <0$.
\end{statement}

Therefore, we must be in one of the following two cases:

\textit{Case 1:} Assertion L$_{1}$1 holds.

\textit{Case 2:} Assertion L$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion L$_{1}$1 holds. In other
words, the system $\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  x=-Ap$ has a nonnegative solution $x\in\mathbb{R}^{n+n^{\prime}}$.
Let $\xi$ be this solution. Then, $\xi\in\mathbb{R}^{n+n^{\prime}}$ is a
nonnegative vector satisfying $\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \xi=-Ap$.

Let us write the vector $\xi$ in the form $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, where $u\in\mathbb{R}^{n}$ and $v\in\mathbb{R}^{n^{\prime}}$.

We have $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  =\xi\geq0$ (since $\xi$ is a nonnegative vector). Hence, Lemma 2.5h
\textbf{(b)} (applied to $n$, $n^{\prime}$, $u$ and $v$ instead of $N$, $M$,
$x$ and $y$) shows that $u\geq0$ and $v\geq0$.

Since $\xi=\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, we have%
\begin{align*}
\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \xi &  =\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  =Au+A^{\prime}v\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right)  .
\end{align*}
Compared with $\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \xi=-Ap$, this yields $-Ap=Au+A^{\prime}v$. Thus,
$0=\underbrace{Ap+Au}_{=A\left(  p+u\right)  }+A^{\prime}v=A\left(
p+u\right)  +A^{\prime}v$. Also, $\underbrace{p}_{>0}+\underbrace{u}_{\geq
0}>0$.

Altogether, we know that $p+u>0$, $v\geq0$ and $A\left(  p+u\right)
+A^{\prime}v=0$. Thus, there exist two vectors $x\in\mathbb{R}^{n}$ and
$x^{\prime}\in\mathbb{R}^{n^{\prime}}$ such that $x>0$, $x^{\prime}\geq0$ and
$Ax+A^{\prime}x^{\prime}=0$ (namely, $x=p+u$ and $x^{\prime}=v$). In other
words, Assertion L1 holds. Hence, at least one of Assertions L1 and L2 holds.

We have thus proven that at least one of Assertions L1 and L2 holds in Case 1.

Now, let us consider Case 2. In this case, Assertion L$_{1}$2 holds. In other
words, there exists a vector $y\in\mathbb{R}^{m}$ such that $y^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \geq0$ and $y^{T}\left(  -Ap\right)  <0$. Denote this $y$ by $w$.
Then, $w\in\mathbb{R}^{m}$ satisfies $w^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \geq0$ and $w^{T}\left(  -Ap\right)  <0$.

Since $w^{T}\left(  -Ap\right)  <0$, we have $w^{T}\left(  -Ap\right)  \neq0$
and thus $w^{T}A\left(  -p\right)  =w^{T}\left(  -Ap\right)  \neq0$, hence
$w^{T}A\neq0$.

\begin{vershort}
We have $w^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \geq0$. Since $w^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  $ (by the multiplication rule for block matrices), this rewrites as
$\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \geq0$. Thus, $w^{T}A\geq0$ and $w^{T}A^{\prime}\geq0$ (by an
analogue of Lemma 2.5h \textbf{(b)} for block matrices of the form $\left(
\begin{array}
[c]{cc}%
x & y
\end{array}
\right)  $ instead of $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $).
\end{vershort}

\begin{verlong}
We have $w^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  \geq0$. Since $w^{T}\left(
\begin{array}
[c]{cc}%
A & A^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  $ (by the multiplication rule for block matrices), this rewrites as
$\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \geq0$. In other words, for every $i\in\left\{  1,2,\ldots
,n+n^{\prime}\right\}  $, we have%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \right) \nonumber\\
&  \geq\left(  \text{the }i\text{-th coordinate of the row vector }0\right)
=0. \label{2.5l.pf.1}%
\end{align}


But for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }w^{T}A\right)
.
\end{align*}
Thus, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }w^{T}A\right)
\\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \right)  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5l.pf.1}%
)}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }0\right)  .
\end{align*}
Thus, $w^{T}A\geq0$.

Also, for every $i\in\left\{  1,2,\ldots,n^{\prime}\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }\left(  n+i\right)  \text{-th coordinate of the row
vector }\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }w^{T}%
A^{\prime}\right)  .
\end{align*}
Thus, for every $i\in\left\{  1,2,\ldots,n^{\prime}\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }w^{T}A\right)
\\
&  =\left(  \text{the }\left(  n+i\right)  \text{-th coordinate of the row
vector }\left(
\begin{array}
[c]{cc}%
w^{T}A & w^{T}A^{\prime}%
\end{array}
\right)  \right)  \geq0\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5l.pf.1}), applied to
}n+i\text{ instead of }i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }0\right)  .
\end{align*}
Thus, $w^{T}A^{\prime}\geq0$.
\end{verlong}

So we have proven that $w^{T}A\geq0$, $w^{T}A^{\prime}\geq0$ and $w^{T}A\neq
0$. Thus, there exists a vector $y\in\mathbb{R}^{m}$ such that $y^{T}A\geq0$,
$y^{T}A^{\prime}\geq0$ and $y^{T}A\neq0$ (namely, $y=w$). In other words,
Assertion L2 holds. Hence, at least one of Assertions L1 and L2 holds.

We have thus proven that at least one of Assertions L1 and L2 holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions L1 and L2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions L1 and L2 holds. Since we know
that the Assertions L1 and L2 cannot hold at the same time, this yields that
exactly one of Assertions L1 and L2 holds. This proves Theorem 2.5l.
\end{proof}

\begin{proof}
[Proof of Theorem 2.5k.]%
\begin{vershort}
The Assertions M1 and M2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions M1 and M2 hold at the same time.
Since Assertion M2 holds, there exist two vectors $y\in\mathbb{R}^{m}$ and
$y^{\prime}\in\mathbb{R}^{m^{\prime}}$ such that $y\geq0$, $y^{\prime}\geq0$,
$y^{T}A+y^{\prime T}A^{\prime}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  \right)  .
\]
Consider these two vectors $y$ and $y^{\prime}$.
\par
Since Assertion M1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax<b$ and $A^{\prime}x\leq b^{\prime}$. Consider this $x$.
\par
From $Ax<b$, we obtain $b>Ax$, so that $b-Ax>0$. Thus, $b-Ax\geq0$. Also,
$y\geq0$, so that $y^{T}\geq0$ (since the transpose of any nonnegative vector
is nonnegative). Now, applying Lemma 2.0v to $m$, $y^{T}$ and $b-Ax$ instead
of $n$, $x$ and $y$, we obtain $y^{T}\left(  b-Ax\right)  \geq0$. Thus,
$y^{T}b-y^{T}Ax=y^{T}\left(  b-Ax\right)  \geq0$, so that $y^{T}b\geq y^{T}%
Ax$.
\par
From $A^{\prime}x\leq b^{\prime}$, we obtain $b^{\prime}\geq A^{\prime}x$, so
that $b^{\prime}-A^{\prime}x\geq0$. Also, $y^{\prime}\geq0$, so that
$y^{\prime T}\geq0$ (since the transpose of any nonnegative vector is
nonnegative). Now, applying Lemma 2.0v to $m^{\prime}$, $y^{\prime T}$ and
$b^{\prime}-A^{\prime}x$ instead of $n$, $x$ and $y$, we obtain $y^{\prime
T}\left(  b^{\prime}-A^{\prime}x\right)  \geq0$. Thus, $y^{\prime T}b^{\prime
}-y^{\prime T}A^{\prime}x=y^{\prime T}\left(  b^{\prime}-A^{\prime}x\right)
\geq0$, so that $y^{\prime T}b^{\prime}\geq y^{\prime T}A^{\prime}x$.
\par
Now, recall that we have $\left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  $ or
$\left(  y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  $.
Hence, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $y^{T}b+y^{\prime T}b^{\prime}<0$.
\par
\textit{Case 2:} We have $y\neq0$ and $y^{T}b+y^{\prime T}b^{\prime}\leq0$.
\par
Let us first consider Case 1. In this case, $y^{T}b+y^{\prime T}b^{\prime}<0$.
Thus,
\[
0>\underbrace{y^{T}b}_{\geq y^{T}Ax}+\underbrace{y^{\prime T}b^{\prime}}_{\geq
y^{\prime T}A^{\prime}x}\geq y^{T}Ax+y^{\prime T}A^{\prime}%
x=\underbrace{\left(  y^{T}A+y^{\prime T}A^{\prime}\right)  }_{=0}x=0.
\]
This is absurd. Thus, we have obtained a contradiction in Case 1.
\par
Let us now consider Case 2. In this case, $y\neq0$ and $y^{T}b+y^{\prime
T}b^{\prime}\leq0$. From $y\neq0$, we obtain $y^{T}\neq0$. If we had
$y^{T}\left(  b-Ax\right)  =0$, then Lemma 2.2d (applied to $m$, $y^{T}$ and
$b-Ax$ instead of $n$, $x$ and $y$) would yield $y^{T}=0$ (since $y^{T}\geq0$,
$y^{T}\left(  b-Ax\right)  =0$ and $b-Ax>0$), contradicting $y^{T}\neq0$.
Hence, we cannot have $y^{T}\left(  b-Ax\right)  =0$. In other words,
$y^{T}\left(  b-Ax\right)  \neq0$. Combined with $y^{T}\left(  b-Ax\right)
\geq0$, this yields $y^{T}\left(  b-Ax\right)  >0$. Hence, $y^{T}%
b-y^{T}Ax=y^{T}\left(  b-Ax\right)  >0$, so that $y^{T}b>y^{T}Ax$.
\par
Now, from $y^{T}b+y^{\prime T}b^{\prime}\leq0$, we obtain
\[
0\geq\underbrace{y^{T}b}_{>y^{T}Ax}+\underbrace{y^{\prime T}b^{\prime}}_{\geq
y^{\prime T}A^{\prime}x}>y^{T}Ax+y^{\prime T}A^{\prime}x=\underbrace{\left(
y^{T}A+y^{\prime T}A^{\prime}\right)  }_{=0}x=0.
\]
This is absurd. Thus, we have obtained a contradiction in Case 2.
\par
Hence, in each of the cases 1 and 2, we have obtained a contradiction. Since
cases 1 and 2 cover all possibilities, this yields that we have a
contradiction in any situation. Thus, our assumption was wrong, qed.}. We will
now show that at least one of these assertions holds.
\end{vershort}

\begin{verlong}
The Assertions M1 and M2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions M1 and M2 hold at the same time.
Since Assertion M2 holds, there exist two vectors $y\in\mathbb{R}^{m}$ and
$y^{\prime}\in\mathbb{R}^{m^{\prime}}$ such that $y\geq0$, $y^{\prime}\geq0$,
$y^{T}A+y^{\prime T}A^{\prime}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  \right)  .
\]
Consider these two vectors $y$ and $y^{\prime}$.
\par
Since Assertion M1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax<b$ and $A^{\prime}x\leq b^{\prime}$. Consider this $x$.
\par
For every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of }Ax\right)  <\left(  \text{the
}i\text{-th coordinate of }b\right)
\]
(since $Ax<b$). Thus, for every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Ax\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Ax\right)
}_{<\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  >\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(  \text{the
}i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b-Ax>0$. Thus, $b-Ax\geq0$. Also, $y\geq0$, so that $y^{T}\geq0$
(since the transpose of any nonnegative vector is nonnegative). Now, applying
Lemma 2.0v to $m$, $y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$, we obtain
$y^{T}\left(  b-Ax\right)  \geq0$. Thus, $y^{T}b-y^{T}Ax=y^{T}\left(
b-Ax\right)  \geq0$, so that $y^{T}b\geq y^{T}Ax$.
\par
On the other hand, for every $i\in\left\{  1,2,\ldots,m^{\prime}\right\}  $,
we have%
\[
\left(  \text{the }i\text{-th coordinate of }A^{\prime}x\right)  \leq\left(
\text{the }i\text{-th coordinate of }b^{\prime}\right)
\]
(since $A^{\prime}x\leq b^{\prime}$). Thus, for every $i\in\left\{
1,2,\ldots,m^{\prime}\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b^{\prime}-A^{\prime}x\right)
\\
&  =\left(  \text{the }i\text{-th coordinate of }b^{\prime}\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }A^{\prime}x\right)
}_{\leq\left(  \text{the }i\text{-th coordinate of }b^{\prime}\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b^{\prime}\right)
-\left(  \text{the }i\text{-th coordinate of }b^{\prime}\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b^{\prime}-A^{\prime}x\geq0$. Also, $y^{\prime}\geq0$, so that
$y^{\prime T}\geq0$ (since the transpose of any nonnegative vector is
nonnegative). Now, applying Lemma 2.0v to $m^{\prime}$, $y^{\prime T}$ and
$b^{\prime}-A^{\prime}x$ instead of $n$, $x$ and $y$, we obtain $y^{\prime
T}\left(  b^{\prime}-A^{\prime}x\right)  \geq0$. Thus, $y^{\prime T}b^{\prime
}-y^{\prime T}A^{\prime}x=y^{\prime T}\left(  b^{\prime}-A^{\prime}x\right)
\geq0$, so that $y^{\prime T}b^{\prime}\geq y^{\prime T}A^{\prime}x$.
\par
Now, recall that we have $\left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  $ or
$\left(  y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  $.
Hence, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $y^{T}b+y^{\prime T}b^{\prime}<0$.
\par
\textit{Case 2:} We have $y\neq0$ and $y^{T}b+y^{\prime T}b^{\prime}\leq0$.
\par
Let us first consider Case 1. In this case, $y^{T}b+y^{\prime T}b^{\prime}<0$.
Thus,
\[
0>\underbrace{y^{T}b}_{\geq y^{T}Ax}+\underbrace{y^{\prime T}b^{\prime}}_{\geq
y^{\prime T}A^{\prime}x}\geq y^{T}Ax+y^{\prime T}A^{\prime}%
x=\underbrace{\left(  y^{T}A+y^{\prime T}A^{\prime}\right)  }_{=0}x=0.
\]
This is absurd. Thus, we have obtained a contradiction in Case 1.
\par
Let us now consider Case 2. In this case, $y\neq0$ and $y^{T}b+y^{\prime
T}b^{\prime}\leq0$. If we had $y^{T}\left(  b-Ax\right)  =0$, then Lemma 2.2d
(applied to $m$, $y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$) would yield
$y^{T}=0$ (since $y^{T}\geq0$, $y^{T}\left(  b-Ax\right)  =0$ and $b-Ax>0$),
thus $y=\left(  \underbrace{y^{T}}_{=0}\right)  ^{T}=0^{T}=0$, contradicting
$y\neq0$. Hence, we cannot have $y^{T}\left(  b-Ax\right)  =0$. In other
words, $y^{T}\left(  b-Ax\right)  \neq0$. Combined with $y^{T}\left(
b-Ax\right)  \geq0$, this yields $y^{T}\left(  b-Ax\right)  >0$. Hence,
$y^{T}b-y^{T}Ax=y^{T}\left(  b-Ax\right)  >0$, so that $y^{T}b>y^{T}Ax$.
\par
Now, from $y^{T}b+y^{\prime T}b^{\prime}\leq0$, we obtain
\[
0\geq\underbrace{y^{T}b}_{>y^{T}Ax}+\underbrace{y^{\prime T}b^{\prime}}_{\geq
y^{\prime T}A^{\prime}x}>y^{T}Ax+y^{\prime T}A^{\prime}x=\underbrace{\left(
y^{T}A+y^{\prime T}A^{\prime}\right)  }_{=0}x=0.
\]
This is absurd. Thus, we have obtained a contradiction in Case 2.
\par
Hence, in each of the cases 1 and 2, we have obtained a contradiction. Since
cases 1 and 2 cover all possibilities, this yields that we have a
contradiction in any situation. Thus, our assumption was wrong, qed.}. We will
now show that at least one of these assertions holds.
\end{verlong}

For any nonnegative integers $\alpha$ and $\beta$, let $0_{\alpha,\beta}$
denote the $\alpha\times\beta$ zero matrix. For any nonnegative integer
$\gamma$, let $I_{\gamma}$ denote the $\gamma\times\gamma$ identity matrix.

Consider the block matrix $\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  $; this is a $\left(  m+m^{\prime}\right)  \times\left(  m+1\right)
$-matrix.

Also, consider the block matrix $\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  $; this is a $\left(  m+m^{\prime}\right)  \times\left(  m^{\prime
}+2n\right)  $-matrix.

Applying Theorem 2.5l to $m+m^{\prime}$, $m+1$, $m^{\prime}+2n$, $\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  $ and $\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  $ instead of $m$, $n$, $n^{\prime}$, $A$ and $A^{\prime}$, we
conclude that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion M}$_{1}$\textit{1:} There exist two vectors $x\in
\mathbb{R}^{m+1}$ and $x^{\prime}\in\mathbb{R}^{m^{\prime}+2n}$ such that
$x>0$, $x^{\prime}\geq0$ and $\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  x+\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  x^{\prime}=0$.

\textit{Assertion M}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{m+m^{\prime}}$ such that $y^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \geq0$, $y^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \geq0$ and $y^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \neq0$.
\end{statement}

Therefore, we must be in one of the following two cases:

\textit{Case 1:} Assertion M$_{1}$1 holds.

\textit{Case 2:} Assertion M$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion M$_{1}$1 holds. In other
words, there exist two vectors $x\in\mathbb{R}^{m+1}$ and $x^{\prime}%
\in\mathbb{R}^{m^{\prime}+2n}$ such that $x>0$, $x^{\prime}\geq0$ and $\left(
%
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  x+\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  x^{\prime}=0$. Denote these vectors $x$ and $x^{\prime}$ by $\xi$ and
$\xi^{\prime}$, respectively. Then, $\xi\in\mathbb{R}^{m+1}$ and $\xi^{\prime
}\in\mathbb{R}^{m^{\prime}+2n}$ satisfy $\xi>0$, $\xi^{\prime}\geq0$ and
$\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \xi+\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \xi^{\prime}=0$.

\begin{vershort}
Denote the $\left(  m+1\right)  $-th coordinate of the vector $\xi$ by $t$.
Then, $t>0$ (since $\xi>0$). Hence, $t\neq0$, so that $t$ is invertible, and
$\dfrac{1}{t}>0$ (since $t>0$).

Since $\xi$ is a vector in $\mathbb{R}^{m+1}$ whose $\left(  m+1\right)  $-th
coordinate is $t$, we can write $\xi$ in the form $\xi=\left(
\begin{array}
[c]{c}%
u\\
t
\end{array}
\right)  $ for some $u\in\mathbb{R}^{m}$. Consider this $u$. Then, $u>0$
(since $\xi>0$).
\end{vershort}

\begin{verlong}
Since $\xi>0$, every $i\in\left\{  1,2,\ldots,m+1\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\xi\right)  >\left(  \text{the
}i\text{-th coordinate of }0\right)  =0. \label{2.5k.pf.4}%
\end{equation}
Applied to $i=m+1$, this yields
\[
\left(  \text{the }\left(  m+1\right)  \text{-th coordinate of }\xi\right)
>0.
\]
Denote the $\left(  m+1\right)  $-th coordinate of $\xi$ by $t$. Then,%
\[
t=\left(  \text{the }\left(  m+1\right)  \text{-th coordinate of }\xi\right)
>0.
\]
Hence, $t\neq0$, so that $t$ is invertible, and $\dfrac{1}{t}>0$ (since $t>0$).

Since $\xi$ is a vector in $\mathbb{R}^{m+1}$ whose $\left(  m+1\right)  $-th
coordinate is $t$, we can write $\xi$ in the form $\xi=\left(
\begin{array}
[c]{c}%
u\\
t
\end{array}
\right)  $ for some $u\in\mathbb{R}^{m}$. Consider this $u$. For every
$i\in\left\{  1,2,\ldots,m\right\}  $, we have
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }\xi\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
u\\
t
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\xi=\left(
\begin{array}
[c]{c}%
u\\
t
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }u\right)
\end{align*}
and thus%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }u\right)   &  =\left(  \text{the
}i\text{-th coordinate of }\xi\right)  >0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{2.5k.pf.4})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $u>0$.
\end{verlong}

\begin{vershort}
Write the vector $\xi^{\prime}\in\mathbb{R}^{m^{\prime}+2n}$ in the form
$\left(
\begin{array}
[c]{c}%
v\\
w\\
z
\end{array}
\right)  $, where $v\in\mathbb{R}^{m^{\prime}}$, $w\in\mathbb{R}^{n}$ and
$z\in\mathbb{R}^{n}$. Then, from $\xi^{\prime}\geq0$, we obtain $v\geq0$,
$w\geq0$ and $z\geq0$.
\end{vershort}

\begin{verlong}
Since $\xi^{\prime}\geq0$, every $i\in\left\{  1,2,\ldots,m^{\prime
}+2n\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\xi^{\prime}\right)  \geq\left(
\text{the }i\text{-th coordinate of }0\right)  =0. \label{2.5k.pf.5}%
\end{equation}


Let us write the vector $\xi^{\prime}$ in the form $\left(
\begin{array}
[c]{c}%
v\\
w\\
z
\end{array}
\right)  $, where $v\in\mathbb{R}^{m^{\prime}}$, $w\in\mathbb{R}^{n}$ and
$z\in\mathbb{R}^{n}$. Then, for every $i\in\left\{  1,2,\ldots,m^{\prime
}\right\}  $, we have%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }\xi^{\prime}\right)   &  =\left(
\text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
v\\
w\\
z
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\xi^{\prime
}=\left(
\begin{array}
[c]{c}%
v\\
w\\
z
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }v\right)  ,
\end{align*}
so that%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }v\right)   &  =\left(  \text{the
}i\text{-th coordinate of }\xi^{\prime}\right)  \geq
0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5k.pf.5})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Thus, $v\geq0$.
\end{verlong}

But%
\begin{align*}
0  &  =\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \underbrace{\xi}_{=\left(
\begin{array}
[c]{c}%
u\\
t
\end{array}
\right)  }+\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \underbrace{\xi^{\prime}}_{=\left(
\begin{array}
[c]{c}%
v\\
w\\
z
\end{array}
\right)  }\\
&  =\underbrace{\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
u\\
t
\end{array}
\right)  }_{\substack{=\left(
\begin{array}
[c]{c}%
I_{m}u+\left(  -b\right)  t\\
0_{m^{\prime},m}u+\left(  -b^{\prime}\right)  t
\end{array}
\right)  \\\text{(by the multiplication rule for block matrices)}%
}}+\underbrace{\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
v\\
w\\
z
\end{array}
\right)  }_{\substack{=\left(
\begin{array}
[c]{c}%
0_{m,m^{\prime}}v+Aw+\left(  -A\right)  z\\
I_{m^{\prime}}v+A^{\prime}w+\left(  -A^{\prime}\right)  z
\end{array}
\right)  \\\text{(by the multiplication rule for block matrices)}}}\\
&  =\left(
\begin{array}
[c]{c}%
I_{m}u+\left(  -b\right)  t\\
0_{m^{\prime},m}u+\left(  -b^{\prime}\right)  t
\end{array}
\right)  +\left(
\begin{array}
[c]{c}%
0_{m,m^{\prime}}v+Aw+\left(  -A\right)  z\\
I_{m^{\prime}}v+A^{\prime}w+\left(  -A^{\prime}\right)  z
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{c}%
I_{m}u+\left(  -b\right)  t+0_{m,m^{\prime}}v+Aw+\left(  -A\right)  z\\
0_{m^{\prime},m}u+\left(  -b^{\prime}\right)  t+I_{m^{\prime}}v+A^{\prime
}w+\left(  -A^{\prime}\right)  z
\end{array}
\right)  .
\end{align*}
Hence, $\left(
\begin{array}
[c]{c}%
I_{m}u+\left(  -b\right)  t+0_{m,m^{\prime}}v+Aw+\left(  -A\right)  z\\
0_{m^{\prime},m}u+\left(  -b^{\prime}\right)  t+I_{m^{\prime}}v+A^{\prime
}w+\left(  -A^{\prime}\right)  z
\end{array}
\right)  =0$, so that $I_{m}u+\left(  -b\right)  t+0_{m,m^{\prime}%
}v+Aw+\left(  -A\right)  z=0$ and $0_{m^{\prime},m}u+\left(  -b^{\prime
}\right)  t+I_{m^{\prime}}v+A^{\prime}w+\left(  -A^{\prime}\right)  z=0$.

Thus,%
\[
0=\underbrace{I_{m}u}_{=u>0}+\underbrace{\left(  -b\right)  t}_{=-bt}%
+\underbrace{0_{m,m^{\prime}}v}_{=0}+Aw+\underbrace{\left(  -A\right)
z}_{=-Az}>-bt+\underbrace{Aw-Az}_{=A\left(  w-z\right)  }=-bt+A\left(
w-z\right)  .
\]
Since $t>0$, we can divide this inequality by $t$, and obtain $0>-b+\dfrac
{1}{t}A\left(  w-z\right)  =-b+A\cdot\dfrac{1}{t}\left(  w-z\right)  $. Thus,
$b>A\cdot\dfrac{1}{t}\left(  w-z\right)  $, so that $A\cdot\dfrac{1}{t}\left(
w-z\right)  <b$.

On the other hand,%
\[
0=\underbrace{0_{m^{\prime},m}u}_{=0}+\underbrace{\left(  -b^{\prime}\right)
t}_{=-b^{\prime}t}+\underbrace{I_{m^{\prime}}v}_{=v\geq0}+A^{\prime
}w+\underbrace{\left(  -A^{\prime}\right)  z}_{=-A^{\prime}z}\geq-b^{\prime
}t+\underbrace{A^{\prime}w-A^{\prime}z}_{=A^{\prime}\left(  w-z\right)
}=-b^{\prime}t+A^{\prime}\left(  w-z\right)  .
\]
Since $t>0$, we can divide this inequality by $t$, and obtain $0\geq
-b^{\prime}+\dfrac{1}{t}A^{\prime}\left(  w-z\right)  =-b^{\prime}+A^{\prime
}\cdot\dfrac{1}{t}\left(  w-z\right)  $. Thus, $b^{\prime}\geq A^{\prime}%
\cdot\dfrac{1}{t}\left(  w-z\right)  $, so that $A^{\prime}\cdot\dfrac{1}%
{t}\left(  w-z\right)  \leq b^{\prime}$.

So we know that $A\cdot\dfrac{1}{t}\left(  w-z\right)  <b$ and $A^{\prime
}\cdot\dfrac{1}{t}\left(  w-z\right)  \leq b^{\prime}$. Hence, there exists a
vector $x\in\mathbb{R}^{n}$ such that $Ax<b$ and $A^{\prime}x\leq b^{\prime}$
(namely, $x=\dfrac{1}{t}\left(  w-z\right)  $). In other words, Assertion M1
holds. Hence, at least one of Assertions M1 and M2 holds.

We have thus proven that at least one of Assertions M1 and M2 holds in Case 1.

Next, let us consider Case 2. In this case, Assertion M$_{1}$2 holds. In other
words, there exists a vector $y\in\mathbb{R}^{m+m^{\prime}}$ such that
$y^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \geq0$, $y^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \geq0$ and $y^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \neq0$. Denote this vector $y$ by $\eta$. Then, $\eta\in
\mathbb{R}^{m+m^{\prime}}$ satisfies $\eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \geq0$, $\eta^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \geq0$ and $\eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \neq0$.

\begin{vershort}
Let us write the vector $\eta$ in the form $\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  $, where $v\in\mathbb{R}^{m}$ and $w\in\mathbb{R}^{m^{\prime}}$.
Then, $\eta^{T}=\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  ^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  $. Hence,%
\begin{align*}
&  \eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
v^{T}I_{m}+w^{T}0_{m^{\prime},m} & v^{T}\left(  -b\right)  +w^{T}\left(
-b^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{v^{T}I_{m}}_{=v^{T}%
}+\underbrace{w^{T}0_{m^{\prime},m}}_{=0}=v^{T}\text{ and }v^{T}\left(
-b\right)  +w^{T}\left(  -b^{\prime}\right)  =-\left(  v^{T}b+w^{T}b^{\prime
}\right)  \right)  .
\end{align*}
Thus, $\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  =\eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \geq0$. Therefore, $v^{T}\geq0$ and $-\left(  v^{T}b+w^{T}b^{\prime
}\right)  \geq0$ (by an analogue of Lemma 2.5h \textbf{(b)} for block matrices
of the form $\left(
\begin{array}
[c]{cc}%
x & y
\end{array}
\right)  $ instead of $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $). From $-\left(  v^{T}b+w^{T}b^{\prime}\right)  \geq0$, we obtain
$v^{T}b+w^{T}b^{\prime}\leq0$. From $v^{T}\geq0$, we obtain $v\geq0$ (since a
column vector is nonnegative if and only if its transpose is nonnegative).
\end{vershort}

\begin{verlong}
Let us write the vector $\eta$ in the form $\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  $, where $v\in\mathbb{R}^{m}$ and $w\in\mathbb{R}^{m^{\prime}}$.
Then, $\eta=\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  $, so that $\eta^{T}=\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  ^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  $. Hence,%
\begin{align*}
&  \eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
v^{T}I_{m}+w^{T}0_{m^{\prime},m} & v^{T}\left(  -b\right)  +w^{T}\left(
-b^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{v^{T}I_{m}}_{=v^{T}%
}+\underbrace{w^{T}0_{m^{\prime},m}}_{=0}=v^{T}\text{ and }v^{T}\left(
-b\right)  +w^{T}\left(  -b^{\prime}\right)  =-\left(  v^{T}b+w^{T}b^{\prime
}\right)  \right)  .
\end{align*}
Thus, $\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  =\eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \geq0$. In other words, every $i\in\left\{  1,2,\ldots,m+1\right\}  $
satisfies%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \right) \nonumber\\
&  \geq\left(  \text{the }i\text{-th coordinate of the row vector }0\right)
=0. \label{2.5k.pf.7}%
\end{align}


But%
\begin{align*}
&  \left(  \text{the }\left(  m+1\right)  \text{-th coordinate of the row
vector }\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \right) \\
&  =-\left(  v^{T}b+w^{T}b^{\prime}\right)  ,
\end{align*}
so that%
\begin{align*}
&  -\left(  v^{T}b+w^{T}b^{\prime}\right) \\
&  =\left(  \text{the }\left(  m+1\right)  \text{-th coordinate of the row
vector }\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5k.pf.7}), applied to
}i=m+1\right)  .
\end{align*}
Hence, $v^{T}b+w^{T}b^{\prime}\leq0$.

Every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}\right)
=\left(  \text{the }i\text{-th coordinate of }v\right)  ,
\end{align*}
so that%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }v\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5k.pf.7})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Thus, $v\geq0$.
\end{verlong}

\begin{vershort}
Since $\eta^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  $, we have%
\begin{align*}
&  \eta^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
v^{T}0_{m,m^{\prime}}+w^{T}I_{m^{\prime}} & v^{T}A+w^{T}A^{\prime} &
v^{T}\left(  -A\right)  +w^{T}\left(  -A^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{v^{T}0_{m,m^{\prime}}%
}_{=0}+\underbrace{w^{T}I_{m^{\prime}}}_{=w^{T}}=w^{T}\text{ and }v^{T}\left(
-A\right)  +w^{T}\left(  -A^{\prime}\right)  =-\left(  v^{T}A+w^{T}A^{\prime
}\right)  \right)  .
\end{align*}
Thus, $\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  =\eta^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \geq0$. Hence, we have $w^{T}\geq0$, $v^{T}A+w^{T}A^{\prime}\geq0$
and $-\left(  v^{T}A+w^{T}A^{\prime}\right)  \geq0$ (by an analogue of Lemma
2.5h \textbf{(b)} for block matrices of the form $\left(
\begin{array}
[c]{ccc}%
x & y & z
\end{array}
\right)  $ instead of $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $). From $-\left(  v^{T}A+w^{T}A^{\prime}\right)  \geq0$, we obtain
$v^{T}A+w^{T}A^{\prime}\leq0$. Combining this with $v^{T}A+w^{T}A^{\prime}%
\geq0$, we obtain $v^{T}A+w^{T}A^{\prime}=0$. From $w^{T}\geq0$, we obtain
$w\geq0$ (since a column vector is nonnegative if and only if its transpose is nonnegative).
\end{vershort}

\begin{verlong}
Since $\eta^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  $, we have%
\begin{align*}
&  \eta^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T} & w^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
v^{T}0_{m,m^{\prime}}+w^{T}I_{m^{\prime}} & v^{T}A+w^{T}A^{\prime} &
v^{T}\left(  -A\right)  +w^{T}\left(  -A^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{v^{T}0_{m,m^{\prime}}%
}_{=0}+\underbrace{w^{T}I_{m^{\prime}}}_{=w^{T}}=w^{T}\text{ and }v^{T}\left(
-A\right)  +w^{T}\left(  -A^{\prime}\right)  =-\left(  v^{T}A+w^{T}A^{\prime
}\right)  \right)  .
\end{align*}
Thus, $\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  =\eta^{T}\left(
\begin{array}
[c]{ccc}%
0_{m,m^{\prime}} & A & -A\\
I_{m^{\prime}} & A^{\prime} & -A^{\prime}%
\end{array}
\right)  \geq0$. Hence, for every $i\in\left\{  1,2,\ldots,m^{\prime
}+2n\right\}  $, we have%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \nonumber\\
&  \geq\left(  \text{the }i\text{-th coordinate of the row vector }0\right)
=0. \label{2.5k.pf.9}%
\end{align}


But every $i\in\left\{  1,2,\ldots,m^{\prime}\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }w^{T}\right)
=\left(  \text{the }i\text{-th coordinate of }w\right)  ,
\end{align*}
so that%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }w\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5k.pf.9})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Thus, $w\geq0$.

Also, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }\left(  m^{\prime}+i\right)  \text{-th coordinate of the
row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A+w^{T}A^{\prime}\right)  ,
\end{align*}
so that%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A+w^{T}A^{\prime}\right) \nonumber\\
&  =\left(  \text{the }\left(  m^{\prime}+i\right)  \text{-th coordinate of
the row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \nonumber\\
&  \geq0\label{2.5k.pf.10}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5k.pf.9}), applied to
}m^{\prime}+i\text{ instead of }i\right)  .\nonumber
\end{align}
Furthermore, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }\left(  m^{\prime}+n+i\right)  \text{-th coordinate of
the row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }-\left(
v^{T}A+w^{T}A^{\prime}\right)  \right) \\
&  =-\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A+w^{T}A^{\prime}\right)
\end{align*}
so that%
\begin{align*}
&  -\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A+w^{T}A^{\prime}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }-\left(
v^{T}A+w^{T}A^{\prime}\right)  \right) \\
&  =\left(  \text{the }\left(  m^{\prime}+n+i\right)  \text{-th coordinate of
the row vector }\left(
\begin{array}
[c]{ccc}%
w^{T} & v^{T}A+w^{T}A^{\prime} & -\left(  v^{T}A+w^{T}A^{\prime}\right)
\end{array}
\right)  \right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5k.pf.9}), applied to
}m^{\prime}+n+i\text{ instead of }i\right)  ,
\end{align*}
thus%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A+w^{T}A^{\prime}\right)  \leq0. \label{2.5k.pf.11}%
\end{equation}


Hence, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A+w^{T}A^{\prime}\right)  =0
\]
(this follows by combining (\ref{2.5k.pf.10}) with (\ref{2.5k.pf.11})). In
other words, $v^{T}A+w^{T}A^{\prime}=0$.
\end{verlong}

Altogether, we now know that $v\geq0$, $w\geq0$, $v^{T}A+w^{T}A^{\prime}=0$,
$v^{T}b+w^{T}b^{\prime}\leq0$ and $\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  =\eta^{T}\left(
\begin{array}
[c]{cc}%
I_{m} & -b\\
0_{m^{\prime},m} & -b^{\prime}%
\end{array}
\right)  \neq0$.

\begin{vershort}
Since $\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \neq0$, at least one of the relations $v^{T}\neq0$ and $-\left(
v^{T}b+w^{T}b^{\prime}\right)  \neq0$ must hold. Thus, we must be in one of
the following two subcases:
\end{vershort}

\begin{verlong}
Since $\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \neq0$, at least one of the relations $v^{T}\neq0$ and $-\left(
v^{T}b+w^{T}b^{\prime}\right)  \neq0$ must hold (because otherwise, we would
have $v^{T}=0$ and $-\left(  v^{T}b+w^{T}b^{\prime}\right)  =0$, so that
$\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
0 & 0
\end{array}
\right)  =0$, which would contradict $\left(
\begin{array}
[c]{cc}%
v^{T} & -\left(  v^{T}b+w^{T}b^{\prime}\right)
\end{array}
\right)  \neq0$). Thus, we must be in one of the following two subcases:
\end{verlong}

\textit{Subcase 2.1:} We have $v^{T}\neq0$.

\textit{Subcase 2.2:} We have $-\left(  v^{T}b+w^{T}b^{\prime}\right)  \neq0$.

Let us consider Subcase 2.1 first. In this subcase, $v^{T}\neq0$. Thus,
$v\neq0$. Combined with $v^{T}b+w^{T}b^{\prime}\leq0$, this yields $\left(
v\neq0\text{ and }v^{T}b+w^{T}b^{\prime}\leq0\right)  $, so that%
\[
\left(  \left(  v^{T}b+w^{T}b^{\prime}<0\right)  \text{ or }\left(
v\neq0\text{ and }v^{T}b+w^{T}b^{\prime}\leq0\right)  \right)  .
\]


Thus, we know that $v\geq0$, $w\geq0$, $v^{T}A+w^{T}A^{\prime}=0$ and%
\[
\left(  \left(  v^{T}b+w^{T}b^{\prime}<0\right)  \text{ or }\left(
v\neq0\text{ and }v^{T}b+w^{T}b^{\prime}\leq0\right)  \right)  .
\]
Hence, there exist two vectors $y\in\mathbb{R}^{m}$ and $y^{\prime}%
\in\mathbb{R}^{m^{\prime}}$ such that $y\geq0$, $y^{\prime}\geq0$,
$y^{T}A+y^{\prime T}A^{\prime}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  \right)
\]
(namely, $y=v$ and $y^{\prime}=w$). In other words, Assertion M2 holds. Thus,
Assertion M2 holds in Subcase 2.1.

Let us now consider Subcase 2.2. In this subcase, $-\left(  v^{T}%
b+w^{T}b^{\prime}\right)  \neq0$, so that $v^{T}b+w^{T}b^{\prime}\neq0$.
Combined with $v^{T}b+w^{T}b^{\prime}\leq0$, this yields $v^{T}b+w^{T}%
b^{\prime}<0$. Hence,
\[
\left(  \left(  v^{T}b+w^{T}b^{\prime}<0\right)  \text{ or }\left(
v\neq0\text{ and }v^{T}b+w^{T}b^{\prime}\leq0\right)  \right)  .
\]
Thus, we know that $v\geq0$, $w\geq0$, $v^{T}A+w^{T}A^{\prime}=0$ and%
\[
\left(  \left(  v^{T}b+w^{T}b^{\prime}<0\right)  \text{ or }\left(
v\neq0\text{ and }v^{T}b+w^{T}b^{\prime}\leq0\right)  \right)  .
\]
Hence, there exist two vectors $y\in\mathbb{R}^{m}$ and $y^{\prime}%
\in\mathbb{R}^{m^{\prime}}$ such that $y\geq0$, $y^{\prime}\geq0$,
$y^{T}A+y^{\prime T}A^{\prime}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}b^{\prime}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}b^{\prime}\leq0\right)  \right)
\]
(namely, $y=v$ and $y^{\prime}=w$). In other words, Assertion M2 holds. Thus,
Assertion M2 holds in Subcase 2.2.

We have thus proven that Assertion M2 holds in each of Subcases 2.1 and 2.2.
Thus, in Case 2, Assertion M2 always holds (because Subcases 2.1 and 2.2 cover
all of Case 2). Hence, in Case 2, at least one of Assertions M1 and M2 holds.

Hence, in each of the Cases 1 and 2, at least one of Assertions M1 and M2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions M1 and M2 holds. Since we know
that the Assertions M1 and M2 cannot hold at the same time, this yields that
exactly one of Assertions M1 and M2 holds. This proves Theorem 2.5k.
\end{proof}

Let us mention two direct corollaries of Theorem 2.5k:

\begin{statement}
\textbf{Corollary 2.5n.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Then, exactly one of the
following two assertions holds:

\textit{Assertion N1:} There exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax<b$.

\textit{Assertion N2:} There exists a nonzero vector $y\in\mathbb{R}^{m}$ such
that $y\geq0$, $y^{T}A=0$ and $y^{T}b\leq0$.
\end{statement}

Corollary 2.5n is classically equivalent to Exercise 2.18 in \cite{Schrij-CCO}%
, but constructively stronger.

Here is the second corollary:

\begin{statement}
\textbf{Corollary 2.5o.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Then, exactly one of the
following two assertions holds:

\textit{Assertion O1:} There exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$.

\textit{Assertion O2:} There exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=0$ and $y^{T}b<0$.
\end{statement}

Corollary 2.5o is classically equivalent to \cite[Corollary 2.5a]{Schrij-CCO},
but constructively stronger.

\begin{proof}
[Proof of Corollary 2.5n.]%
\begin{vershort}
The Assertions N1 and N2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions N1 and N2 hold at the same time.
Since Assertion N2 holds, there exists a nonzero vector $y\in\mathbb{R}^{m}$
such that $y\geq0$, $y^{T}A=0$ and $y^{T}b\leq0$. Consider this $y$.
\par
Since Assertion N1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax<b$. Consider this $x$.
\par
From $Ax<b$, we obtain $b>Ax$, hence $b-Ax>0$. Thus, $b-Ax\geq0$. Also,
$y\geq0$, so that $y^{T}\geq0$ (since the transpose of any nonnegative vector
is nonnegative). Now, applying Lemma 2.0v to $m$, $y^{T}$ and $b-Ax$ instead
of $n$, $x$ and $y$, we obtain $y^{T}\left(  b-Ax\right)  \geq0$. Combined
with $y^{T}\left(  b-Ax\right)  =y^{T}b-\underbrace{y^{T}A}_{=0}x=y^{T}b\leq
0$, this yields $y^{T}\left(  b-Ax\right)  =0$. Hence, applying Lemma 2.2d to
$m$, $y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$, we obtain $y^{T}=0$, thus
$y=0$, contradicting the fact that $y$ be nonzero. Thus, our assumption was
wrong, qed.}. We will now show that at least one of these assertions holds.
\end{vershort}

\begin{verlong}
The Assertions N1 and N2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions N1 and N2 hold at the same time.
Since Assertion N2 holds, there exists a nonzero vector $y\in\mathbb{R}^{m}$
such that $y\geq0$, $y^{T}A=0$ and $y^{T}b\leq0$. Consider this $y$.
\par
Since Assertion N1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax<b$. Consider this $x$.
\par
For every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of }Ax\right)  <\left(  \text{the
}i\text{-th coordinate of }b\right)
\]
(since $Ax<b$). Thus, for every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Ax\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Ax\right)
}_{<\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  >\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(  \text{the
}i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b-Ax>0$. Thus, $b-Ax\geq0$. Also, $y\geq0$, so that $y^{T}\geq0$
(since the transpose of any nonnegative vector is nonnegative). Now, applying
Lemma 2.0v to $m$, $y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$, we obtain
$y^{T}\left(  b-Ax\right)  \geq0$. Combined with $y^{T}\left(  b-Ax\right)
=y^{T}b-\underbrace{y^{T}A}_{=0}x=y^{T}b\leq0$, this yields $y^{T}\left(
b-Ax\right)  =0$. Hence, applying Lemma 2.2d to $m$, $y^{T}$ and $b-Ax$
instead of $n$, $x$ and $y$, we obtain $y^{T}=0$. Thus, $y=\left(
\underbrace{y^{T}}_{=0}\right)  ^{T}=0^{T}=0$, contradicting the fact that $y$
be nonzero. Thus, our assumption was wrong, qed.}. We will now show that at
least one of these assertions holds.
\end{verlong}

Let $0_{0,n}$ denote the zero $0\times n$-matrix, and let $0_{0}$ denote the
zero vector in $\mathbb{R}^{0}$. (Both $0_{0,n}$ and $0_{0}$ are matrices with
no entries at all.) Applying Theorem 2.5k to $m^{\prime}=0$, $A^{\prime
}=0_{0,n}$ and $b^{\prime}=0_{0}$, we conclude that exactly one of the
following two assertions holds:

\begin{statement}
\textit{Assertion N}$_{1}$\textit{1:} There exists a vector $x\in
\mathbb{R}^{n}$ such that $Ax<b$ and $0_{0,n}x\leq0_{0}$.

\textit{Assertion N}$_{1}$\textit{2:} There exist two vectors $y\in
\mathbb{R}^{m}$ and $y^{\prime}\in\mathbb{R}^{0}$ such that $y\geq0$,
$y^{\prime}\geq0$, $y^{T}A+y^{\prime T}0_{0,n}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}0_{0}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}0_{0}\leq0\right)  \right)  .
\]

\end{statement}

Hence, we must be in one of the following two cases:

\textit{Case 1:} Assertion N$_{1}$1 holds.

\textit{Case 2:} Assertion N$_{1}$2 holds.

\begin{vershort}
Let us first consider Case 1. In this case, Assertion N$_{1}$1 holds. In other
words, there exists a vector $x\in\mathbb{R}^{n}$ such that $Ax<b$ and
$0_{0,n}x\leq0_{0}$. Thus, Assertion N1 holds. Hence, at least one of
Assertions N1 and N2 holds.
\end{vershort}

\begin{verlong}
Let us first consider Case 1. In this case, Assertion N$_{1}$1 holds. In other
words, there exists a vector $x\in\mathbb{R}^{n}$ such that $Ax<b$ and
$0_{0,n}x\leq0_{0}$. Denote this $x$ by $\xi$. Then, $\xi\in\mathbb{R}^{n}$
satisfies $A\xi<b$ and $0_{0,n}\xi\leq0_{0}$. Thus, there exists a vector
$x\in\mathbb{R}^{n}$ such that $Ax<b$ (namely, $x=\xi$). In other words,
Assertion N1 holds. Hence, at least one of Assertions N1 and N2 holds.
\end{verlong}

We have thus proven that at least one of Assertions N1 and N2 holds in Case 1.

Next, let us consider Case 2. In this case, Assertion N$_{1}$2 holds. In other
words, there exist two vectors $y\in\mathbb{R}^{m}$ and $y^{\prime}%
\in\mathbb{R}^{0}$ such that $y\geq0$, $y^{\prime}\geq0$, $y^{T}A+y^{\prime
T}0_{0,n}=0$ and%
\[
\left(  \left(  y^{T}b+y^{\prime T}0_{0}<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}b+y^{\prime T}0_{0}\leq0\right)  \right)  .
\]
Denote these vectors $y$ and $y^{\prime}$ by $\eta$ and $\eta^{\prime}$. Then,
$\eta\in\mathbb{R}^{m}$ and $\eta^{\prime}\in\mathbb{R}^{0}$ satisfy $\eta
\geq0$, $\eta^{\prime}\geq0$, $\eta^{T}A+\eta^{\prime T}0_{0,n}=0$ and%
\begin{equation}
\left(  \left(  \eta^{T}b+\eta^{\prime T}0_{0}<0\right)  \text{ or }\left(
\eta\neq0\text{ and }\eta^{T}b+\eta^{\prime T}0_{0}\leq0\right)  \right)  .
\label{2.5n.pf.1}%
\end{equation}


\begin{vershort}
Since $\eta^{\prime T}0_{0}=0$, the relation (\ref{2.5n.pf.1}) simplifies to
$\left(  \left(  \eta^{T}b<0\right)  \text{ or }\left(  \eta\neq0\text{ and
}\eta^{T}b\leq0\right)  \right)  $. Thus, $\eta^{T}b\leq0$ (because otherwise,
neither $\eta^{T}b<0$ nor $\left(  \eta\neq0\text{ and }\eta^{T}b\leq0\right)
$ would be possible) and $\eta\neq0$ (for the same reason). Also, $\eta
^{T}A+\eta^{\prime T}0_{0,n}=0$ simplifies to $\eta^{T}A=0$.
\end{vershort}

\begin{verlong}
Now, it is easy to see that $\eta$ is nonzero\footnote{\textit{Proof.} Assume
the contrary. Then, $\eta$ is not nonzero. Hence, $\eta=0$, so that $\eta
^{T}=0^{T}=0$. Now, $\underbrace{\eta^{T}}_{=0}b+\underbrace{\eta^{\prime
T}0_{0}}_{=0}=0b+0=0$. Thus, the assertion $\eta^{T}b+\eta^{\prime T}0_{0}<0$
cannot hold. The assertion $\left(  \eta\neq0\text{ and }\eta^{T}%
b+\eta^{\prime T}0_{0}\leq0\right)  $ cannot hold either (since $\eta\neq0$ is
false (because $\eta=0$)).
\par
Thus, neither of the assertions $\left(  \eta^{T}b+\eta^{\prime T}%
0_{0}<0\right)  $ and $\left(  \eta\neq0\text{ and }\eta^{T}b+\eta^{\prime
T}0_{0}\leq0\right)  $ holds. But this contradicts (\ref{2.5n.pf.1}). This
contradiction shows that our assumption was wrong, qed.}. We have $\eta
^{T}b\leq0$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Then,
$\eta^{T}b\leq0$ does not hold. Hence, $\eta^{T}b>0$. Thus, the assertion
$\left(  \eta^{T}b+\eta^{\prime T}0_{0}<0\right)  $ is false (because it
contradicts $\eta^{T}b+\underbrace{\eta^{\prime T}0_{0}}_{=0}=\eta^{T}b>0$),
and the assertion $\left(  \eta\neq0\text{ and }\eta^{T}b+\eta^{\prime T}%
0_{0}\leq0\right)  $ is false (because it yields $\eta^{T}b+\eta^{\prime
T}0_{0}\leq0$, which contradicts $\eta^{T}b+\underbrace{\eta^{\prime T}0_{0}%
}_{=0}=\eta^{T}b>0$). So we have shown that both assertions $\left(  \eta
^{T}b+\eta^{\prime T}0_{0}<0\right)  $ and $\left(  \eta\neq0\text{ and }%
\eta^{T}b+\eta^{\prime T}0_{0}\leq0\right)  $ are false. This contradicts
(\ref{2.5n.pf.1}). This contradiction proves that our assumption was wrong,
qed.}. Also, comparing $\eta^{T}A+\underbrace{\eta^{\prime T}0_{0,n}}%
_{=0}=\eta^{T}A$ with $\eta^{T}A+\eta^{\prime T}0_{0,n}=0$, we obtain
$\eta^{T}A=0$.
\end{verlong}

Altogether, we know that $\eta$ is nonzero and satisfies $\eta\geq0$,
$\eta^{T}A=0$ and $\eta^{T}b\leq0$. Thus, there exists a nonzero vector
$y\in\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=0$ and $y^{T}b\leq0$ (namely,
$y=\eta$). In other words, Assertion N2 holds. Hence, at least one of
Assertions N1 and N2 holds.

We have thus proven that at least one of Assertions N1 and N2 holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions N1 and N2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions N1 and N2 holds. Since we know
that the Assertions N1 and N2 cannot hold at the same time, this yields that
exactly one of Assertions N1 and N2 holds. This proves Corollary 2.5n.
\end{proof}

\begin{proof}
[Proof of Corollary 2.5o.]%
\begin{vershort}
The Assertions O1 and O2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions O1 and O2 hold at the same time.
Since Assertion O2 holds, there exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=0$ and $y^{T}b<0$. Consider this $y$.
\par
Since Assertion O1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$. Consider this $x$.
\par
From $Ax\leq b$, we obtain $b\geq Ax$, thus $b-Ax\geq0$. Also, $y\geq0$, so
that $y^{T}\geq0$ (since the transpose of any nonnegative vector is
nonnegative). Now, applying Lemma 2.0v to $m$, $y^{T}$ and $b-Ax$ instead of
$n$, $x$ and $y$, we obtain $y^{T}\left(  b-Ax\right)  \geq0$. This
contradicts $y^{T}\left(  b-Ax\right)  =y^{T}b-\underbrace{y^{T}A}_{=0}%
x=y^{T}b<0$. This contradiction shows that our assumption was wrong, qed.}. We
will now show that at least one of these assertions holds.
\end{vershort}

\begin{verlong}
The Assertions O1 and O2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions O1 and O2 hold at the same time.
Since Assertion O2 holds, there exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=0$ and $y^{T}b<0$. Consider this $y$.
\par
Since Assertion O1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$. Consider this $x$.
\par
For every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of }Ax\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)
\]
(since $Ax\leq b$). Thus, for every $i\in\left\{  1,2,\ldots,m\right\}  $, we
have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Ax\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Ax\right)  }%
_{\leq\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b-Ax\geq0$. Also, $y\geq0$, so that $y^{T}\geq0$ (since the transpose
of any nonnegative vector is nonnegative). Now, applying Lemma 2.0v to $m$,
$y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$, we obtain $y^{T}\left(
b-Ax\right)  \geq0$. This contradicts $y^{T}\left(  b-Ax\right)
=y^{T}b-\underbrace{y^{T}A}_{=0}x=y^{T}b<0$. This contradiction shows that our
assumption was wrong, qed.}. We will now show that at least one of these
assertions holds.
\end{verlong}

Let $0_{0,n}$ denote the zero $0\times n$-matrix, and let $0_{0}$ denote the
zero vector in $\mathbb{R}^{0}$. (Both $0_{0,n}$ and $0_{0}$ are matrices with
no entries at all.) Applying Theorem 2.5k to $0$, $m$, $0_{0,n}$, $0_{0}$, $A$
and $b$ instead of $m$, $m^{\prime}$, $A$, $b$, $A^{\prime}$ and $b^{\prime}$,
we conclude that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion O}$_{1}$\textit{1:} There exists a vector $x\in
\mathbb{R}^{n}$ such that $0_{0,n}x<0_{0}$ and $Ax\leq b$.

\textit{Assertion O}$_{1}$\textit{2:} There exist two vectors $y\in
\mathbb{R}^{0}$ and $y^{\prime}\in\mathbb{R}^{m}$ such that $y\geq0$,
$y^{\prime}\geq0$, $y^{T}0_{0,n}+y^{\prime T}A=0$ and%
\[
\left(  \left(  y^{T}0_{0}+y^{\prime T}b<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}0_{0}+y^{\prime T}b\leq0\right)  \right)  .
\]

\end{statement}

Hence, we must be in one of the following two cases:

\textit{Case 1:} Assertion O$_{1}$1 holds.

\textit{Case 2:} Assertion O$_{1}$2 holds.

\begin{vershort}
Let us first consider Case 1. In this case, Assertion O$_{1}$1 holds. In other
words, there exists a vector $x\in\mathbb{R}^{n}$ such that $0_{0,n}x<0_{0}$
and $Ax\leq b$. Hence, Assertion O1 holds. Hence, at least one of Assertions
O1 and O2 holds.
\end{vershort}

\begin{verlong}
Let us first consider Case 1. In this case, Assertion O$_{1}$1 holds. In other
words, there exists a vector $x\in\mathbb{R}^{n}$ such that $0_{0,n}x<0_{0}$
and $Ax\leq b$. Denote this $x$ by $\xi$. Then, $\xi\in\mathbb{R}^{n}$
satisfies $0_{0,n}\xi<0_{0}$ and $Ax\leq b$. Thus, there exists a vector
$x\in\mathbb{R}^{n}$ such that $Ax\leq b$ (namely, $x=\xi$). In other words,
Assertion O1 holds. Hence, at least one of Assertions O1 and O2 holds.
\end{verlong}

We have thus proven that at least one of Assertions O1 and O2 holds in Case 1.

Next, let us consider Case 2. In this case, Assertion O$_{1}$2 holds. In other
words, there exist two vectors $y\in\mathbb{R}^{0}$ and $y^{\prime}%
\in\mathbb{R}^{m}$ such that $y\geq0$, $y^{\prime}\geq0$, $y^{T}%
0_{0,n}+y^{\prime T}A=0$ and%
\[
\left(  \left(  y^{T}0_{0}+y^{\prime T}b<0\right)  \text{ or }\left(
y\neq0\text{ and }y^{T}0_{0}+y^{\prime T}b\leq0\right)  \right)  .
\]
Denote these vectors $y$ and $y^{\prime}$ by $\eta$ and $\eta^{\prime}$. Then,
$\eta\in\mathbb{R}^{0}$ and $\eta^{\prime}\in\mathbb{R}^{m}$ satisfy $\eta
\geq0$, $\eta^{\prime}\geq0$, $\eta^{T}0_{0,n}+\eta^{\prime T}A=0$ and%
\begin{equation}
\left(  \left(  \eta^{T}0_{0}+\eta^{\prime T}b<0\right)  \text{ or }\left(
\eta\neq0\text{ and }\eta^{T}0_{0}+\eta^{\prime T}b\leq0\right)  \right)  .
\label{2.5o.pf.1}%
\end{equation}


Since $\eta\in\mathbb{R}^{0}$, we have $\eta=0$ (since the only element of
$\mathbb{R}^{0}$ is $0$). Hence, the assertion $\eta\neq0$ is false. Thus, the
assertion $\left(  \eta\neq0\text{ and }\eta^{T}0_{0}+\eta^{\prime T}%
b\leq0\right)  $ is also false. Hence, from (\ref{2.5o.pf.1}), we conclude
that we must have $\eta^{T}0_{0}+\eta^{\prime T}b<0$. Since $\underbrace{\eta
^{T}0_{0}}_{=0}+\eta^{\prime T}b=\eta^{\prime T}b$, this simplifies to
$\eta^{\prime T}b<0$.

Furthermore, comparing $\underbrace{\eta^{T}0_{0,n}}_{=0}+\eta^{\prime
T}A=\eta^{\prime T}A$ with $\eta^{T}0_{0,n}+\eta^{\prime T}A=0$, we obtain
$\eta^{\prime T}A=0$.

Altogether, we now know that $\eta^{\prime}\geq0$, $\eta^{\prime T}A=0$ and
$\eta^{\prime T}b<0$. Hence, there exists a vector $y\in\mathbb{R}^{m}$ such
that $y\geq0$, $y^{T}A=0$ and $y^{T}b<0$ (namely, $y = \eta^{\prime}$). In
other words, Assertion O2 holds. Hence, at least one of Assertions O1 and O2 holds.

We have thus proven that at least one of Assertions O1 and O2 holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions O1 and O2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions O1 and O2 holds. Since we know
that the Assertions O1 and O2 cannot hold at the same time, this yields that
exactly one of Assertions O1 and O2 holds. This proves Corollary 2.5o.
\end{proof}

\section{The weak Duality theorem}

The following theorem is classically equivalent to, but constructively
stronger than \cite[Corollary 2.5b]{Schrij-CCO}:

\begin{statement}
\textbf{Theorem 2.5p.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$ and
$\delta\in\mathbb{R}$. Assume that the set $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $ has at least one element. Then, exactly one of the
following two assertions holds:

\textit{Assertion P1:} There exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$ and $c^{T}x>\delta$.

\textit{Assertion P2:} There exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=c^{T}$ and $y^{T}b\leq\delta$.
\end{statement}

The following proof of Theorem 2.5p is being given only for the sake of
completeness. It is more or less a copy of that of \cite[Corollary
2.5b]{Schrij-CCO}, with the only difference that the reductio to absurdum part
has been replaced by a constructive case distinction:

\begin{proof}
[Proof of Theorem 2.5p.]%
\begin{vershort}
The Assertions P1 and P2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions P1 and P2 hold at the same time.
Since Assertion P2 holds, there exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=c^{T}$ and $y^{T}b\leq\delta$. Consider this $y$.
\par
Since Assertion P1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$ and $c^{T}x>\delta$. Consider this $x$.
\par
From $Ax\leq b$, we obtain $b\geq Ax$, so that $b-Ax\geq0$. Also, $y\geq0$, so
that $y^{T}\geq0$ (since the transpose of any nonnegative vector is
nonnegative). Now, applying Lemma 2.0v to $m$, $y^{T}$ and $b-Ax$ instead of
$n$, $x$ and $y$, we obtain $y^{T}\left(  b-Ax\right)  \geq0$. This
contradicts $y^{T}\left(  b-Ax\right)  =\underbrace{y^{T}b}_{\leq\delta
}-\underbrace{y^{T}A}_{=c^{T}}x\leq\delta-\underbrace{c^{T}x}_{>\delta}%
<\delta-\delta=0$. This contradiction shows that our assumption was wrong,
qed.}. We will now show that at least one of these assertions holds.
\end{vershort}

\begin{verlong}
The Assertions P1 and P2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions P1 and P2 hold at the same time.
Since Assertion P2 holds, there exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=c^{T}$ and $y^{T}b\leq\delta$. Consider this $y$.
\par
Since Assertion P1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$ and $c^{T}x>\delta$. Consider this $x$.
\par
For every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of }Ax\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)
\]
(since $Ax\leq b$). Thus, for every $i\in\left\{  1,2,\ldots,m\right\}  $, we
have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Ax\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Ax\right)  }%
_{\leq\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b-Ax\geq0$. Also, $y\geq0$, so that $y^{T}\geq0$ (since the transpose
of any nonnegative vector is nonnegative). Now, applying Lemma 2.0v to $m$,
$y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$, we obtain $y^{T}\left(
b-Ax\right)  \geq0$. This contradicts $y^{T}\left(  b-Ax\right)
=\underbrace{y^{T}b}_{\leq\delta}-\underbrace{y^{T}A}_{=c^{T}}x\leq
\delta-\underbrace{c^{T}x}_{>\delta}<\delta-\delta=0$. This contradiction
shows that our assumption was wrong, qed.}. We will now show that at least one
of these assertions holds.
\end{verlong}

Let $0_{n}$ denote the zero vector in $\mathbb{R}^{n}$.

Consider the block matrix $\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  $ (where the scalar $1$ is considered as a vector in $\mathbb{R}^{1}%
$, thus as an $1\times1$-matrix); this is an $\left(  n+1\right)
\times\left(  m+1\right)  $-matrix.

Also, consider the block matrix $\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  $. This block matrix is an $\left(  n+1\right)  \times1$-matrix, i.
e., a vector in $\mathbb{R}^{n+1}$.

Applying Theorem 2.5d to $n+1$, $m+1$, $\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  $ and $\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  $ instead of $m$, $n$, $A$ and $b$, we conclude that exactly one of
the following two assertions holds:

\begin{statement}
\textit{Assertion P}$_{1}$\textit{1:} The system $\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  x=\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  $ has a nonnegative solution $x\in\mathbb{R}^{m+1}$.

\textit{Assertion P}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{n+1}$ such that $y^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \geq0$ and $y^{T}\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  <0$.
\end{statement}

Hence, we must be in one of the following two cases:

\textit{Case 1:} Assertion P$_{1}$1 holds.

\textit{Case 2:} Assertion P$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion P$_{1}$1 holds. In other
words, the system $\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  x=\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  $ has a nonnegative solution $x\in\mathbb{R}^{m+1}$. Let $\xi$ be
this solution. Then, $\xi\in\mathbb{R}^{m+1}$ is a nonnegative vector such
that $\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \xi=\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  $.

Let us write the vector $\xi$ in the form $\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  $, where $v\in\mathbb{R}^{m}$ and $w\in\mathbb{R}^{1}$. Consider
$w\in\mathbb{R}^{1}$ as a scalar (because elements of $\mathbb{R}^{1}$ can be
identified with scalars). Then,%
\[
\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \underbrace{\xi}_{=\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  }=\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
A^{T}v+0_{n}w\\
b^{T}v+1w
\end{array}
\right)
\]
(by the rules for multiplying block matrices). Thus, $c=A^{T}v+0_{n}w$ and
$\delta=b^{T}v+1w$. Hence, $c=A^{T}v+\underbrace{0_{n}w}_{=0}=A^{T}v$ and
$\delta=b^{T}v+\underbrace{1w}_{=w}=b^{T}v+w$.

\begin{vershort}
Now, $\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  =\xi\geq0$ (since $\xi$ is nonnegative). Thus, $v\geq0$ and $w\geq0$
(by Lemma 2.5h \textbf{(b)}).
\end{vershort}

\begin{verlong}
Since $\xi$ is nonnegative, we have $\xi\geq0$. Hence, every $i\in\left\{
1,2,\ldots,m+1\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\xi\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0. \label{2.5p.pf.1}%
\end{equation}


Now, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }\xi\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\xi=\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }v\right)
\end{align*}
and thus%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }v\right)   &  =\left(  \text{the
}i\text{-th coordinate of }\xi\right)  \geq0\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{2.5p.pf.1})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $v\geq0$. Besides,
\begin{align*}
&  \left(  \text{the }\left(  m+1\right)  \text{-th coordinate of }\xi\right)
\\
&  =\left(  \text{the }\left(  m+1\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\xi=\left(
\begin{array}
[c]{c}%
v\\
w
\end{array}
\right)  \right) \\
&  =\left(  \text{the }1\text{-st coordinate of }w\right)
=w\ \ \ \ \ \ \ \ \ \ \left(  \text{since we identify }w\in\mathbb{R}%
^{1}\text{ with a scalar}\right)  ,
\end{align*}
so that%
\[
w=\left(  \text{the }\left(  m+1\right)  \text{-th coordinate of }\xi\right)
\geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5p.pf.1}), applied to
}i=m+1\right)  .
\]

\end{verlong}

Thus,%
\begin{align*}
\delta &  =b^{T}v+\underbrace{w}_{\geq0}\geq b^{T}v=\left(  b^{T}v\right)
^{T}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }b^{T}v\text{ is a scalar, i. e.,
a }1\times1\text{-matrix}\right) \\
&  =v^{T}\underbrace{\left(  b^{T}\right)  ^{T}}_{=b}=v^{T}b.
\end{align*}
In other words, $v^{T}b\leq\delta$.

Also, from $c=A^{T}v$, we obtain $c^{T}=\left(  A^{T}v\right)  ^{T}%
=v^{T}\underbrace{\left(  A^{T}\right)  ^{T}}_{=A}=v^{T}A$.

We thus know that $v\geq0$, $v^{T}A=c^{T}$ and $v^{T}b\leq\delta$. Hence,
there exists a vector $y\in\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=c^{T}$
and $y^{T}b\leq\delta$ (namely, $y=v$). In other words, Assertion P2 holds.
Hence, at least one of Assertions P1 and P2 holds.

We have thus proven that at least one of Assertions P1 and P2 holds in Case 1.

Let us now consider Case 2. In this case, Assertion P$_{1}$2 holds. In other
words, there exists a vector $y\in\mathbb{R}^{n+1}$ such that $y^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \geq0$ and $y^{T}\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  <0$. Denote this $y$ by $w$. Then, $w\in\mathbb{R}^{n+1}$ satisfies
$w^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \geq0$ and $w^{T}\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  <0$.

Let us write the vector $w$ in the form $\left(
\begin{array}
[c]{c}%
v\\
t
\end{array}
\right)  $, where $v\in\mathbb{R}^{n}$ and $t\in\mathbb{R}^{1}$. Consider
$t\in\mathbb{R}^{1}$ as a scalar (because elements of $\mathbb{R}^{1}$ can be
identified with scalars). Then, $t^{T}=t$.

We have $\left(  c^{T}v\right)  ^{T}=c^{T}v$ (since $c^{T}v$ is a scalar).
Thus, $c^{T}v=\left(  c^{T}v\right)  ^{T}=v^{T}\underbrace{\left(
c^{T}\right)  ^{T}}_{=c}=v^{T}c$, therefore $v^{T}c=c^{T}v$.

From $w=\left(
\begin{array}
[c]{c}%
v\\
t
\end{array}
\right)  $, we obtain $w^{T}=\left(
\begin{array}
[c]{c}%
v\\
t
\end{array}
\right)  ^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & t^{T}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
v^{T} & t
\end{array}
\right)  $ (since $t^{T}=t$). Thus,%
\begin{align*}
w^{T}\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
v^{T} & t
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  =\underbrace{v^{T}c}_{=c^{T}v}+t\delta\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =c^{T}v+t\delta.
\end{align*}
Hence, the relation $w^{T}\left(
\begin{array}
[c]{c}%
c\\
\delta
\end{array}
\right)  <0$ (which we know to be true) rewrites as $c^{T}v+t\delta<0$.

\begin{vershort}
On the other hand, from $w^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & t
\end{array}
\right)  $, we deduce that%
\[
w^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
v^{T} & t
\end{array}
\right)  \left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & v^{T}0_{n}+t\cdot1
\end{array}
\right)
\]
(by the multiplication rule for block matrices). Thus, the relation
$w^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \geq0$ (which we know to be true) rewrites as $\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & v^{T}0_{n}+t\cdot1
\end{array}
\right)  \geq0$. Thus, (using the analogue of Lemma 2.5h \textbf{(b)} for
block matrices of the form $\left(
\begin{array}
[c]{cc}%
x & y
\end{array}
\right)  $ instead of $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $) we conclude that $v^{T}A^{T}+tb^{T}\geq0$ and $v^{T}0_{n}%
+t\cdot1\geq0$. From $v^{T}A^{T}+tb^{T}\geq0$, we obtain $\left(  v^{T}%
A^{T}+tb^{T}\right)  ^{T}\geq0$ (since the transpose of a nonnegative vector
is always nonnegative). In light of $\left(  v^{T}A^{T}+tb^{T}\right)
^{T}=\underbrace{\left(  A^{T}\right)  ^{T}}_{=A}\underbrace{\left(
v^{T}\right)  ^{T}}_{=v}+\underbrace{\left(  b^{T}\right)  ^{T}}%
_{=b}\underbrace{t^{T}}_{=t}=Av+bt$, this rewrites as $Av+bt\geq0$. Since
$bt=tb$ (as we can regard $t$ as a scalar), this rewrites as $Av+tb\geq0$.

Of course, $v^{T}0_{n}+t\cdot1\geq0$ simplifies to $t\geq0$. Hence, we must be
in one of the following two subcases:
\end{vershort}

\begin{verlong}
On the other hand, from $w^{T}=\left(
\begin{array}
[c]{cc}%
v^{T} & t
\end{array}
\right)  $, we deduce that%
\begin{align*}
w^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
v^{T} & t
\end{array}
\right)  \left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & v^{T}0_{n}+t\cdot1
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{v^{T}0_{n}}%
_{=0}+\underbrace{t\cdot1}_{=t}=t\right)  .
\end{align*}
Thus, the relation $w^{T}\left(
\begin{array}
[c]{cc}%
A^{T} & 0_{n}\\
b^{T} & 1
\end{array}
\right)  \geq0$ (which we know to be true) rewrites as $\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right)  \geq0$. As a consequence, every $i\in\left\{  1,2,\ldots,m+1\right\}
$ satisfies%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right)  \right) \nonumber\\
&  \geq\left(  \text{the }i\text{-th coordinate of the row vector }0\right)
=0. \label{2.5p.pf.2}%
\end{align}


But every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }v^{T}%
A^{T}+tb^{T}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }\left(
Av+tb\right)  ^{T}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{v^{T}A^{T}}_{=\left(
Av\right)  ^{T}}+tb^{T}=\left(  Av\right)  ^{T}+tb^{T}=\left(  Av+tb\right)
^{T}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the vector }Av+tb\right)  ,
\end{align*}
so that%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the vector }Av+tb\right)
\nonumber\\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right)  \right)  \geq0 \label{2.5p.pf.3}%
\end{align}
(by (\ref{2.5p.pf.2})).

Applying (\ref{2.5p.pf.2}) to $i=m+1$, we obtain%
\[
\left(  \text{the }\left(  m+1\right)  \text{-th coordinate of the row vector
}\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right)  \right)  \geq0.
\]
Since
\begin{align*}
&  \left(  \text{the }\left(  m+1\right)  \text{-th coordinate of the row
vector }\left(
\begin{array}
[c]{cc}%
v^{T}A^{T}+tb^{T} & t
\end{array}
\right)  \right) \\
&  =\left(  \text{the }1\text{-st coordinate of the row vector }t\right)  =t\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since we identify }t\in\mathbb{R}%
^{1}\text{ with a scalar}\right)  ,
\end{align*}
this rewrites as $t\geq0$. Hence, we must be in one of the following two subcases:
\end{verlong}

\textit{Subcase 2.1:} We have $t=0$.

\textit{Subcase 2.2:} We have $t>0$.

Let us first consider Subcase 2.1. In this subcase, we have $t=0$. Thus,
$c^{T}v+\underbrace{t}_{=0}\delta=c^{T}v+0\delta=c^{T}v$. Hence, the relation
$c^{T}v+t\delta<0$ (which we know to be true) rewrites as $c^{T}v<0$. Thus,
$c^{T}v\neq0$, so that $c^{T}v$ is invertible.

We know that the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $
has at least one element. Let $x_{0}$ be such an element. Then, $x_{0}%
\in\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $, so that $x_{0}%
\in\mathbb{R}^{n}$ and $Ax_{0}\leq b$.

\begin{verlong}
Since $Ax_{0}\leq b$, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the vector }Ax_{0}\right)
\nonumber\\
&  \leq\left(  \text{the }i\text{-th coordinate of the vector }b\right)  .
\label{2.5p.pf.6}%
\end{align}

\end{verlong}

Let $\tau=\max\left\{  0,\dfrac{c^{T}x_{0}-\delta}{c^{T}v}+1\right\}
$\ \ \ \ \footnote{This is well-defined, since the real number $c^{T}v$ is
invertible.}. Then, $\tau\geq0$ and $\tau\geq\dfrac{c^{T}x_{0}-\delta}{c^{T}%
v}+1$.

We have $\tau\geq\dfrac{c^{T}x_{0}-\delta}{c^{T}v}+1>\dfrac{c^{T}x_{0}-\delta
}{c^{T}v}$. Multiplying this inequality with $c^{T}v$, we obtain
\begin{equation}
\tau c^{T}v<c^{T}x_{0}-\delta\label{2.5p.pf.8}%
\end{equation}
(since $c^{T}v<0$).

\begin{vershort}
On the other hand, $Av+\underbrace{t}_{=0}b=Av+0b=Av$, so that $Av=Av+tb\geq0$.
\end{vershort}

\begin{verlong}
Every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the vector }Av\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the vector }Av+tb\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{t}_{=0}b=0\text{ and
thus }Av+tb=Av\text{, so that }Av=Av+tb\right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5p.pf.3})}\right)  ,
\end{align*}
so that%
\begin{equation}
\tau\left(  \text{the }i\text{-th coordinate of the vector }Av\right)  \geq0
\label{2.5p.pf.7}%
\end{equation}
(since $\tau\geq0$).
\end{verlong}

\begin{vershort}
Now, using $\tau\geq0$, we obtain $\tau\underbrace{Av}_{\geq0}\geq\tau0=0$.
Hence, $A\left(  x_{0}-\tau v\right)  =Ax_{0}-\underbrace{\tau Av}_{\geq0}\leq
Ax_{0}\leq b$.
\end{vershort}

\begin{verlong}
Now, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the vector }%
\underbrace{A\left(  x_{0}-\tau v\right)  }_{=Ax_{0}-\tau Av}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the vector }Ax_{0}-\tau
Av\right) \\
&  =\underbrace{\left(  \text{the }i\text{-th coordinate of the vector }%
Ax_{0}\right)  }_{\substack{\leq\left(  \text{the }i\text{-th coordinate of
the vector }b\right)  \\\text{(by (\ref{2.5p.pf.6}))}}}-\underbrace{\tau
\left(  \text{the }i\text{-th coordinate of the vector }Av\right)
}_{\substack{\geq0\\\text{(by (\ref{2.5p.pf.7}))}}}\\
&  \leq\left(  \text{the }i\text{-th coordinate of the vector }b\right)  .
\end{align*}
Thus, $A\left(  x_{0}-\tau v\right)  \leq b$.
\end{verlong}

On the other hand,%
\[
c^{T}\left(  x_{0}-\tau v\right)  =c^{T}x_{0}-\underbrace{\tau c^{T}%
v}_{\substack{<c^{T}x_{0}-\delta\\\text{(by (\ref{2.5p.pf.8}))}}}>c^{T}%
x_{0}-\left(  c^{T}x_{0}-\delta\right)  =\delta.
\]


Altogether, we thus know that $A\left(  x_{0}-\tau v\right)  \leq b$ and
$c^{T}\left(  x_{0}-\tau v\right)  >\delta$. Hence, there exists a vector
$x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and $c^{T}x>\delta$ (namely,
$x=x_{0}-\tau v$). In other words, Assertion P1 holds. We have thus proven
that Assertion P1 holds in Subcase 2.1.

\begin{vershort}
Now, let us consider Subcase 2.2. In this subcase, $t>0$. Thus, $t\neq0$, so
that $t$ is invertible, and $\dfrac{1}{t}>0$ (since $t>0$). Hence, from
$Av+tb\geq0$, we obtain $Av\geq-tb$. Multiplying this inequality by $\dfrac
{1}{t}$ (which is allowed, since $\dfrac{1}{t}>0$), we obtain $\dfrac{1}%
{t}Av\geq-b$. Hence, $A\left(  -\dfrac{1}{t}v\right)  =-\underbrace{\dfrac
{1}{t}Av}_{\geq-b}\leq-\left(  -b\right)  =b$.
\end{vershort}

\begin{verlong}
Now, let us consider Subcase 2.2. In this subcase, $t>0$. Thus, $t\neq0$, so
that $t$ is invertible, and $\dfrac{1}{t}>0$ (since $t>0$). Now, every
$i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the vector }A\left(  -\dfrac
{1}{t}v\right)  \right) \\
&  \leq\left(  \text{the }i\text{-th coordinate of the vector }b\right)
\end{align*}
(since every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the vector }b\right)  -\left(
\text{the }i\text{-th coordinate of the vector }A\left(  -\dfrac{1}%
{t}v\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the vector }b-A\left(
-\dfrac{1}{t}v\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of the vector }\dfrac{1}%
{t}\left(  Av+tb\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }b-\underbrace{A\left(  -\dfrac
{1}{t}v\right)  }_{=-\dfrac{1}{t}Av}=b-\left(  -\dfrac{1}{t}Av\right)
=b+\dfrac{1}{t}Av=\dfrac{1}{t}\left(  tb+Av\right)  =\dfrac{1}{t}\left(
Av+tb\right)  \right) \\
&  =\underbrace{\dfrac{1}{t}}_{>0}\underbrace{\left(  \text{the }i\text{-th
coordinate of the vector }Av+tb\right)  }_{\substack{\geq0\\\text{(by
(\ref{2.5p.pf.3}))}}}\geq0
\end{align*}
). In other words, $A\left(  -\dfrac{1}{t}v\right)  \leq b$.
\end{verlong}

On the other hand,%
\[
\delta-\underbrace{c^{T}\left(  -\dfrac{1}{t}v\right)  }_{=-\dfrac{1}{t}%
c^{T}v}=\delta-\left(  -\dfrac{1}{t}c^{T}v\right)  =\delta+\dfrac{1}{t}%
c^{T}v=\dfrac{1}{t}\left(  t\delta+c^{T}v\right)  =\underbrace{\dfrac{1}{t}%
}_{>0}\underbrace{\left(  c^{T}v+t\delta\right)  }_{<0}<0,
\]
so that $c^{T}\left(  -\dfrac{1}{t}v\right)  >\delta$.

Altogether, we know that $A\left(  -\dfrac{1}{t}v\right)  \leq b$ and
$c^{T}\left(  -\dfrac{1}{t}v\right)  >\delta$. Thus, there exists a vector
$x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and $c^{T}x>\delta$ (namely,
$x=-\dfrac{1}{t}v$). In other words, Assertion P1 holds. We have thus proven
that Assertion P1 holds in Subcase 2.2.

Altogether, we know that Assertion P1 holds in each of Subcases 2.1 and 2.2.
Thus, in Case 2, Assertion P1 always holds (since Subcases 2.1 and 2.2 cover
all of Case 2). Thus, in Case 2, at least one of Assertions P1 and P2 holds.

Hence, in each of the Cases 1 and 2, at least one of Assertions P1 and P2
holds. Since these Cases 1 and 2 cover all possibilities, this yields that, in
every situation, at least one of Assertions P1 and P2 holds. Since we know
that the Assertions P1 and P2 cannot hold at the same time, this yields that
exactly one of Assertions P1 and P2 holds. This proves Theorem 2.5p.
\end{proof}

Theorem 2.5p yields a weak version of linear programming duality:

\begin{statement}
\textbf{Corollary 2.5q.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$.
Assume that the number $\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $ exists. Then, the number\newline$\min\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exists and
satisfies
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]

\end{statement}

I am calling this version weak because it requires the existence of
\newline$\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $,
while the same conclusion can be proven under the (easier to verify)
assumption that the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $ is nonempty and bounded from below. (This will follow from
Theorem 2.6c below.) Here is yet another weak version of linear programming duality:

\begin{statement}
\textbf{Corollary 2.5r.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$.
Assume that the number \newline$\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exists. Then, the number
$\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ exists
and satisfies%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]

\end{statement}

Both Corollary 2.5q and Corollary 2.5r are easy corollaries of Theorem 2.6c
below. We will, however, give an alternative proof of Corollary 2.5q using
Theorem 2.5p first. In a similar vein, Corollary 2.5r could be shown using an
analogue of Theorem 2.5p.

\begin{proof}
[Proof of Corollary 2.5q.]The number $\max\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $ exists (by assumption). Denote this
number by $\delta$. Then, every element of $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is $\leq\delta$. In other words,
\begin{equation}
c^{T}x\leq\delta\text{ for every }x\in\mathbb{R}^{n}\text{ satisfying }Ax\leq
b. \label{2.5q.pf.1}%
\end{equation}


Also, the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $
has a maximum (since \newline$\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ exists) and thus has at least one element (since a
set that has a maximum must always have at least one element). In other words,
there exists some $x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. In other words,
the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least
one element. Hence, Theorem 2.5p yields that exactly one of the following two
assertions holds:

\begin{statement}
\textit{Assertion Q}$_{1}$\textit{1:} There exists a vector $x\in
\mathbb{R}^{n}$ such that $Ax\leq b$ and $c^{T}x>\delta$.

\textit{Assertion Q}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=c^{T}$ and $y^{T}b\leq\delta$.
\end{statement}

Since Assertion Q$_{1}$1 cannot hold\footnote{\textit{Proof.} Assume the
contrary. Then, Assertion Q$_{1}$1 holds. In other words, there exists a
vector $x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and $c^{T}x>\delta$. Consider
this $x$. Then, $Ax\leq b$, so that $c^{T}x\leq\delta$ (by (\ref{2.5q.pf.1})),
contradicting $c^{T}x>\delta$. This contradiction shows that our assumption
was wrong, qed.}, this yields that Assertion Q$_{1}$2 must hold. In other
words, there exists a vector $y\in\mathbb{R}^{m}$ such that $y\geq0$,
$y^{T}A=c^{T}$ and $y^{T}b\leq\delta$. Denote this $y$ by $w$. Thus,
$w\in\mathbb{R}^{m}$ satisfies $w\geq0$, $w^{T}A=c^{T}$ and $w^{T}b\leq\delta
$. Since $w\in\mathbb{R}^{m}$ satisfies $w\geq0$ and $w^{T}A=c^{T}$, we have
$w\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}%
\right\}  $ and thus%
\begin{equation}
w^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  . \label{2.5q.pf.2}%
\end{equation}


Now, we will prove that%
\begin{equation}
\lambda\geq\delta\text{ for every }\lambda\in\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  .
\label{2.5q.pf.3}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{2.5q.pf.3}):} Let $\lambda\in\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Thus,
there exists a $z\in\mathbb{R}^{m}$ such that $z\geq0$, $z^{T}A=c^{T}$ and
$z^{T}b=\lambda$. Consider this $z$.

By the definition of $\delta$, we have%
\[
\delta=\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  .
\]
Hence, there exists some $x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and
$c^{T}x=\delta$. Denote this $x$ by $q$. Then, $q\in\mathbb{R}^{n}$ satisfies
$Aq\leq b$ and $c^{T}q=\delta$. From $Aq\leq b$, we obtain $b\geq Aq$, thus
$b-Aq\geq0$. Also, $z\geq0$, so that $z^{T}\geq0$ (since the transpose of a
nonnegative vector must always be nonnegative). Thus, applying Lemma 2.0v to
$m$, $z^{T}$ and $b-Aq$ instead of $n$, $x$ and $y$, we obtain $z^{T}\left(
b-Aq\right)  \geq0$ (since $b-Aq\geq0$). Since $z^{T}\left(  b-Aq\right)
=\underbrace{z^{T}b}_{=\lambda}-\underbrace{z^{T}A}_{=c^{T}}q=\lambda
-\underbrace{c^{T}q}_{=\delta}=\lambda-\delta$, this rewrites as
$\lambda-\delta\geq0$. In other words, $\lambda\geq\delta$. This proves
(\ref{2.5q.pf.3}).]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{2.5q.pf.3}):} Let $\lambda\in\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Thus,
there exists a $z\in\mathbb{R}^{m}$ such that $z\geq0$, $z^{T}A=c^{T}$ and
$z^{T}b=\lambda$. Consider this $z$.

By the definition of $\delta$, we have%
\[
\delta=\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  .
\]
Hence, there exists some $x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and
$c^{T}x=\delta$. Denote this $x$ by $q$. Then, $q\in\mathbb{R}^{n}$ satisfies
$Aq\leq b$ and $c^{T}q=\delta$. In other words, for every $i\in\left\{
1,2,\ldots,m\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of }Aq\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
Thus, for every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Aq\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Aq\right)  }%
_{\leq\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0=\left(  \text{the
}i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b-Aq\geq0$. Also, $z\geq0$, so that $z^{T}\geq0$ (since the transpose
of a nonnegative vector must always be nonnegative). Thus, applying Lemma 2.0v
to $m$, $z^{T}$ and $b-Aq$ instead of $n$, $x$ and $y$, we obtain
$z^{T}\left(  b-Aq\right)  \geq0$ (since $b-Aq\geq0$). Since $z^{T}\left(
b-Aq\right)  =\underbrace{z^{T}b}_{=\lambda}-\underbrace{z^{T}A}_{=c^{T}%
}q=\lambda-\underbrace{c^{T}q}_{=\delta}=\lambda-\delta$, this rewrites as
$\lambda-\delta\geq0$. In other words, $\lambda\geq\delta$. This proves
(\ref{2.5q.pf.3}).]
\end{verlong}

From (\ref{2.5q.pf.2}), we know that $w^{T}b\in\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Hence,
applying (\ref{2.5q.pf.3}) to $\lambda=w^{T}b$, we obtain $w^{T}b\geq\delta$.
Combined with $w^{T}b\leq\delta$, this yields $w^{T}b=\delta$. Hence,
(\ref{2.5q.pf.2}) rewrites as%
\[
\delta\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]
Thus, $\delta$ is an element of the set $\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Combined with the
fact that every element of the set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is $\geq\delta$ (by
(\ref{2.5q.pf.3})), this yields that $\delta$ is the minimum of the set
\newline$\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $. In other words, the number \newline$\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
exists and satisfies%
\[
\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  =\delta.
\]
Since $\delta=\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $, this rewrites as follows: The number \newline$\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
exists and satisfies
\[
\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  =\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  .
\]
This proves Corollary 2.5q.
\end{proof}

\section{The Duality theorem}

We are almost ready to formulate the linear programming duality theorem in one
of its strongest forms. First, let us define two basic notions:

\begin{statement}
\textbf{Definition 2.6b. (i)} A subset $S$ of $\mathbb{R}$ is said to be
\textit{unbounded from above} if for every $\delta\in\mathbb{R}$, there exists
a $t\in S$ such that $t\geq\delta$.

\textbf{(ii)} A subset $S$ of $\mathbb{R}$ is said to be \textit{unbounded
from below} if for every $\delta\in\mathbb{R}$, there exists a $t\in S$ such
that $t\leq\delta$.
\end{statement}

In classical logic, a subset $S$ of $\mathbb{R}$ is unbounded from above if
and only if it is not bounded from above (i. e., there does not exist any
$m\in\mathbb{R}$ such that every $t\in S$ satisfies $t\leq m$). In
constructive logic, this is not generally true, and the assertion that $S$ be
unbounded from above is stronger than the assertion that $S$ not be bounded
from above.

We now state a theorem that is somewhat stronger (both classically and
constructively) than \cite[Theorem 2.6]{Schrij-CCO}:

\begin{statement}
\textbf{Theorem 2.6c.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$.
Then, exactly one of the following four assertions holds:

\textit{Assertion I1:} The sets $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ and $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ are empty.

\textit{Assertion I2:} The set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ is unbounded from above, and the set $\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is empty.

\textit{Assertion I3:} The set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is unbounded from below, and
the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty.

\textit{Assertion I4:} The numbers $\max\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $ and \newline$\min\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exist
and satisfy%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]

\end{statement}

Theorem 2.6c is often referred to as the \textquotedblleft asymmetric version
of the strong duality theorem of linear programming\textquotedblright.

Let me reiterate that Theorem 2.6c still holds if we replace $\mathbb{R}$ by
any other ordered field, such as $\mathbb{Q}$. This cannot be said of the
proof of Theorem 2.6 in \cite{Schrij-CCO}, since it uses the fact that any set
of real numbers has a supremum or is unbounded from above (and this fact does
not hold for $\mathbb{Q}$). So we are going to give a different proof.

First, we prove a slightly weaker result:

\begin{statement}
\textbf{Lemma 2.6d.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be an
$m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$. Assume
that the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at
least one element. Then, exactly one of the following two assertions holds:

\textit{Assertion J1:} The set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ is unbounded from above, and the set $\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is empty.

\textit{Assertion J2:} The numbers $\max\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $ and \newline$\min\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exist
and satisfy%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]

\end{statement}

It should be said that Lemma 2.6d is weaker than Theorem 2.6c, but still
stronger than \cite[Theorem 2.6]{Schrij-CCO}.

\begin{proof}
[Proof of Lemma 2.6d.]

\begin{vershort}
The Assertions J1 and J2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions J1 and J2 hold at the same time.
\par
The set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is
unbounded from above (by Assertion J1), and thus has no maximum. In other
words, the number $\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $ does not exist. But this contradicts Assertion J2. This
contradiction shows that our assumption was wrong, qed.}. We will now show
that at least one of these assertions holds.
\end{vershort}

\begin{verlong}
The Assertions J1 and J2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions J1 and J2 hold at the same time. We
know that Assertion J2 holds. In other words, the numbers $\max\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ and $\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
exist and satisfy%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]
\par
We know that Assertion J1 holds. In other words, the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is unbounded from
above, and the set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ is empty.
\par
A set which is unbounded from above can have no maximum. Hence, the set
$\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ has no
maximum (since $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
$ is unbounded from above). In other words, the number $\max\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ does not exist. This
contradicts the fact that the number $\max\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $ exists. This contradiction shows that our
assumption was wrong, qed.}. We will now show that at least one of these
assertions holds.
\end{verlong}

For any nonnegative integers $\alpha$ and $\beta$, let $0_{\alpha,\beta}$
denote the $\alpha\times\beta$ zero matrix. For any nonnegative integer
$\gamma$, let $I_{\gamma}$ denote the $\gamma\times\gamma$ identity matrix.

Consider the block matrix $\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  $; this is an $\left(  2m+2n+1\right)  \times\left(  n+m\right)  $-matrix.

Consider the block matrix $\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  $; this is an $\left(  2m+2n+1\right)  \times1$-matrix, i. e., a
vector in $\mathbb{R}^{2m+2n+1}$. We identify the $1\times1$-matrix $0_{1,1}$
with the scalar $0$; thus, the block matrix $\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  $ rewrites as $\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0
\end{array}
\right)  $.

Applying Corollary 2.5o to $2m+2n+1$, $n+m$, $\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  $ and $\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  $ instead of $m$, $n$, $A$ and $b$, we see that exactly one of the
following two assertions holds:

\begin{statement}
\textit{Assertion J}$_{1}$\textit{1:} There exists a vector $x\in
\mathbb{R}^{n+m}$ such that $\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  $.

\textit{Assertion J}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{2m+2n+1}$ such that $y\geq0$, $y^{T}\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  =0$ and $y^{T}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  <0$.
\end{statement}

Hence, we must be in one of the following two cases:

\textit{Case 1:} Assertion J$_{1}$1 holds.

\textit{Case 2:} Assertion J$_{1}$2 holds.

Let us first consider Case 1. In this case, Assertion J$_{1}$1 holds. In other
words, there exists a vector $x\in\mathbb{R}^{n+m}$ such that $\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  $. Denote this vector $x$ by $\xi$. Then, $\xi\in\mathbb{R}^{n+m}$
satisfies $\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  \xi\leq\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  $.

Let us write the vector $\xi$ in the form $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, where $u\in\mathbb{R}^{n}$ and $v\in\mathbb{R}^{m}$. Since
$\xi=\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, we have%
\begin{align*}
&  \left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  \xi\\
&  =\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
0_{m,n}u+\left(  -I_{m}\right)  v\\
Au+0_{m,m}v\\
0_{n,n}u+A^{T}v\\
0_{n,n}u+\left(  -A^{T}\right)  v\\
-c^{T}u+b^{T}v
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\underbrace{0_{m,n}u}_{=0}+\underbrace{\left(  -I_{m}\right)
v}_{=-v}=-v\text{, }Au+\underbrace{0_{m,m}v}_{=0}=Au\text{,}\\
\underbrace{0_{n,n}u}_{=0}+A^{T}v=A^{T}v\text{, and }\underbrace{0_{n,n}%
u}_{=0}+\underbrace{\left(  -A^{T}\right)  v}_{=-A^{T}v}=-A^{T}v
\end{array}
\right)  .
\end{align*}
Hence,%
\[
\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  \xi\leq\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  .
\]


\begin{vershort}
Hence, we have the five inequalities\footnote{Here, we are using the following
simple fact: Let $n_{1}$, $n_{2}$, $n_{3}$, $n_{4}$, and $n_{5}$ be five
nonnegative integers. Let $\alpha_{1}\in\mathbb{R}^{n_{1}}$, $\alpha_{2}%
\in\mathbb{R}^{n_{2}}$, $\alpha_{3}\in\mathbb{R}^{n_{3}}$, $\alpha_{4}%
\in\mathbb{R}^{n_{4}}$, and $\alpha_{5}\in\mathbb{R}^{n_{5}}$ be five column
vectors. Let $\beta_{1}\in\mathbb{R}^{n_{1}}$, $\beta_{2}\in\mathbb{R}^{n_{2}%
}$, $\beta_{3}\in\mathbb{R}^{n_{3}}$, $\beta_{4}\in\mathbb{R}^{n_{4}}$, and
$\beta_{5}\in\mathbb{R}^{n_{5}}$ be five column vectors. Assume that $\left(
\begin{array}
[c]{c}%
\alpha_{1}\\
\alpha_{2}\\
\alpha_{3}\\
\alpha_{4}\\
\alpha_{5}%
\end{array}
\right)  \geq\left(
\begin{array}
[c]{c}%
\beta_{1}\\
\beta_{2}\\
\beta_{3}\\
\beta_{4}\\
\beta_{5}%
\end{array}
\right)  $. Then, we have the five inequalities%
\[
\alpha_{1}\geq\beta_{1},\ \ \ \ \ \ \ \ \ \ \alpha_{2}\geq\beta_{2}%
,\ \ \ \ \ \ \ \ \ \ \alpha_{3}\geq\beta_{3},\ \ \ \ \ \ \ \ \ \ \alpha
_{4}\geq\beta_{4},\ \ \ \ \ \ \ \ \ \ \text{and }\alpha_{5}\geq\beta_{5}.
\]
}
\begin{align*}
-v  &  \leq0_{m,1},\ \ \ \ \ \ \ \ \ \ Au\leq b,\ \ \ \ \ \ \ \ \ \ A^{T}v\leq
c,\\
-A^{T}v  &  \leq-c,\ \ \ \ \ \ \ \ \ \ \text{and }-c^{T}u+b^{T}v\leq0_{1,1}.
\end{align*}
The inequality $-v\leq0_{m,1}$ leads to $v\geq-0_{m,1}=0_{m,1}=0$. Thus,
$v^{T}\geq0$ (since the transpose of any nonnegative vector is nonnegative).
The inequality $-A^{T}v\leq-c$ rewrites as $A^{T}v\geq c$; combining this with
$A^{T}v\leq c$, we obtain $A^{T}v=c$. The inequality $-c^{T}u+b^{T}%
v\leq0_{1,1}$ becomes $-c^{T}u+b^{T}v\leq0_{1,1}=0$, so that $b^{T}v\leq
c^{T}u$.
\end{vershort}

\begin{verlong}
Thus, for every $i\in\left\{  1,2,\ldots,2m+2n+1\right\}  $, we have%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)  \leq\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)  . \label{2.6d.pf.1}%
\end{equation}


This inequality has many consequences:

First, it is easy to see that $v\geq0$\ \ \ \ \footnote{\textit{Proof.} Let
$i\in\left\{  1,2,\ldots,m\right\}  $. Then,%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)   &  =\left(  \text{the }i\text{-th coordinate of }-v\right)
\\
&  =-\left(  \text{the }i\text{-th coordinate of }v\right)
\end{align*}
and%
\[
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }%
0_{m,1}\right)  =0.
\]
Thus,%
\begin{align*}
-\left(  \text{the }i\text{-th coordinate of }v\right)   &  =\left(  \text{the
}i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right) \\
&  \leq\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.1}%
)}\right) \\
&  =0,
\end{align*}
so that $\left(  \text{the }i\text{-th coordinate of }v\right)  \geq0=\left(
\text{the }i\text{-th coordinate of }0\right)  $.
\par
Now, forget that we fixed $i$. We thus have shown that $\left(  \text{the
}i\text{-th coordinate of }v\right)  \geq\left(  \text{the }i\text{-th
coordinate of }0\right)  $ for every $i\in\left\{  1,2,\ldots,m\right\}  $. In
other words, $v\geq0$, qed.}. Thus, $v^{T}\geq0$ (since the transpose of any
nonnegative vector is nonnegative).

Next, we have $Au\leq b$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
1,2,\ldots,m\right\}  $. Then,%
\[
\left(  \text{the }\left(  m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }Au\right)
\]
and%
\[
\left(  \text{the }\left(  m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }b\right)  .
\]
Thus,%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }Au\right) \\
&  =\left(  \text{the }\left(  m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)  \leq\left(  \text{the }\left(  m+i\right)  \text{-th
coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.1}), applied to
}m+i\text{ instead of }i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)  .
\end{align*}
\par
Now, forget that we fixed $i$. We thus have shown that $\left(  \text{the
}i\text{-th coordinate of }Au\right)  \leq\left(  \text{the }i\text{-th
coordinate of }b\right)  $ for every $i\in\left\{  1,2,\ldots,m\right\}  $. In
other words, $Au\leq b$, qed.}.

Furthermore, we have $A^{T}v=c$\ \ \ \ \footnote{\textit{Proof.} Let
$i\in\left\{  1,2,\ldots,n\right\}  $. Then,%
\[
\left(  \text{the }\left(  2m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }A^{T}v\right)
\]
and%
\[
\left(  \text{the }\left(  2m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }c\right)  .
\]
Thus,%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of }A^{T}v\right) \nonumber\\
&  =\left(  \text{the }\left(  2m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right) \nonumber\\
&  \leq\left(  \text{the }\left(  2m+i\right)  \text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.1}), applied to
}2m+i\text{ instead of }i\right) \nonumber\\
&  =\left(  \text{the }i\text{-th coordinate of }c\right)  . \label{2.6d.pf.2}%
\end{align}
\par
On the other hand,
\begin{align*}
\left(  \text{the }\left(  2m+n+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)   &  =\left(  \text{the }i\text{-th coordinate of }%
-A^{T}v\right) \\
&  =-\left(  \text{the }i\text{-th coordinate of }A^{T}v\right)
\end{align*}
and%
\begin{align*}
\left(  \text{the }\left(  2m+n+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)   &  =\left(  \text{the }i\text{-th coordinate of }-c\right)
\\
&  =-\left(  \text{the }i\text{-th coordinate of }c\right)  .
\end{align*}
Thus,%
\begin{align*}
&  -\left(  \text{the }i\text{-th coordinate of }A^{T}v\right) \\
&  =\left(  \text{the }\left(  2m+n+i\right)  \text{-th coordinate of }\left(
%
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right) \\
&  \leq\left(  \text{the }\left(  2m+n+i\right)  \text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.1}), applied to
}2m+n+i\text{ instead of }i\right) \\
&  =-\left(  \text{the }i\text{-th coordinate of }c\right)  ,
\end{align*}
so that%
\[
\left(  \text{the }i\text{-th coordinate of }A^{T}v\right)  \geq\left(
\text{the }i\text{-th coordinate of }c\right)  .
\]
Combined with (\ref{2.6d.pf.2}), this yields%
\[
\left(  \text{the }i\text{-th coordinate of }A^{T}v\right)  =\left(  \text{the
}i\text{-th coordinate of }c\right)  .
\]
\par
Now, forget that we fixed $i$. We thus have shown that $\left(  \text{the
}i\text{-th coordinate of }A^{T}v\right)  =\left(  \text{the }i\text{-th
coordinate of }c\right)  $ for every $i\in\left\{  1,2,\ldots,m\right\}  $. In
other words, $A^{T}v=c$, qed.}. Thus, $\left(  A^{T}v\right)  ^{T}=c^{T}$.
Since $\left(  A^{T}v\right)  ^{T}=v^{T}\underbrace{\left(  A^{T}\right)
^{T}}_{=A}=v^{T}A$, this rewrites as $v^{T}A=c^{T}$.

Finally we have $-c^{T}u+b^{T}v\leq0$\ \ \ \ \footnote{\textit{Proof.} We have%
\begin{align*}
\left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right)   &  =\left(  \text{the }1\text{-st coordinate of }%
-c^{T}u+b^{T}v\right) \\
&  =-c^{T}u+b^{T}v
\end{align*}
(since $-c^{T}u+b^{T}v$ is a scalar) and%
\[
\left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right)  =\left(  \text{the }1\text{-st coordinate of }%
0_{1,1}\right)  =0.
\]
Thus,%
\begin{align*}
&  -c^{T}u+b^{T}v\\
&  =\left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
-v\\
Au\\
A^{T}v\\
-A^{T}v\\
-c^{T}u+b^{T}v
\end{array}
\right)  \right) \\
&  \leq\left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.1}), applied to
}2m+2n+1\text{ instead of }i\right) \\
&  =0,
\end{align*}
qed.}. In other words, $b^{T}v\leq c^{T}u$.
\end{verlong}

Since $b^{T}v$ is a scalar, we have $\left(  b^{T}v\right)  ^{T}=b^{T}v$.
Thus, $b^{T}v=\left(  b^{T}v\right)  ^{T}=v^{T}\underbrace{\left(
b^{T}\right)  ^{T}}_{=b}=v^{T}b$. Hence, $v^{T}b=b^{T}v\leq c^{T}u$.

But $u\in\mathbb{R}^{n}$ satisfies $Au\leq b$. Thus, $u\in\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ and therefore $c^{T}u\in\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $.

\begin{vershort}
On the other hand,
\begin{equation}
\text{every }w\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
\text{ satisfies }w\leq v^{T}b. \label{2.6d.pf.short.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.6d.pf.short.4}):} Let $w\in\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $. Then, there exists
some $x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and $w=c^{T}x$. Consider this
$x$.
\par
We have $Ax\leq b$, hence $b\geq Ax$. Thus, $b-Ax\geq0$. Thus, Lemma 2.0v
(applied to $m$, $v^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$) yields
$v^{T}\left(  b-Ax\right)  \geq0$ (since $v^{T}\geq0$). Thus, $0\leq
v^{T}\left(  b-Ax\right)  =v^{T}b-\underbrace{v^{T}A}_{=c^{T}}x=v^{T}%
b-\underbrace{c^{T}x}_{=w}=v^{T}b-w$, so that $w\leq v^{T}b$. This proves
(\ref{2.6d.pf.short.4}).} Applying this to $w=c^{T}u$, we obtain $c^{T}u\leq
v^{T}b$ (since $c^{T}u\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $). Combined with $v^{T}b\leq c^{T}u$, this yields
\begin{equation}
v^{T}b=c^{T}u. \label{2.6d.pf.short.vtb=ctu}%
\end{equation}
Thus, $v^{T}b=c^{T}u\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $. In other words, $v^{T}b$ is an element of the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $. Combined with the fact
that every element of the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ is $\leq v^{T}b$ (due to (\ref{2.6d.pf.short.4})),
this yields that $v^{T}b$ is the maximum of the set $\left\{  c^{T}%
x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $. In other words, the number
$\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ exists
and satisfies%
\begin{equation}
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  =v^{T}b.
\label{2.6d.pf.short.max}%
\end{equation}

\end{vershort}

\begin{verlong}
On the other hand,
\begin{equation}
\text{every }w\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
\text{ satisfies }w\leq v^{T}b. \label{2.6d.pf.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.6d.pf.4}):} Let $w\in\left\{  c^{T}%
x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $. Then, there exists some
$x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and $w=c^{T}x$. Consider this $x$.
\par
We have $Ax\leq b$. Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }Ax\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
Hence, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }b-Ax\right)   &  =\left(
\text{the }i\text{-th coordinate of }b\right)  -\underbrace{\left(  \text{the
}i\text{-th coordinate of }Ax\right)  }_{\leq\left(  \text{the }i\text{-th
coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $b-Ax\geq0$. Thus, Lemma 2.0v (applied to $m$, $v^{T}$ and
$b-Ax$ instead of $n$, $x$ and $y$) yields $v^{T}\left(  b-Ax\right)  \geq0$
(since $v^{T}\geq0$). Thus, $0\leq v^{T}\left(  b-Ax\right)  =v^{T}%
b-\underbrace{v^{T}A}_{=c^{T}}x=v^{T}b-\underbrace{c^{T}x}_{=w}=v^{T}b-w$, so
that $w\leq v^{T}b$. This proves (\ref{2.6d.pf.4}).} Applying (\ref{2.6d.pf.4}%
) to $w=c^{T}u$ (this is allowed since $c^{T}u\in\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $), we obtain $c^{T}u\leq v^{T}b$.
Combined with $v^{T}b\leq c^{T}u$, this yields
\begin{equation}
v^{T}b=c^{T}u. \label{2.6d.pf.vtb=ctu}%
\end{equation}
Thus, $v^{T}b=c^{T}u\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $. In other words, $v^{T}b$ is an element of the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $. Combined with the fact
that every element of the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ is $\leq v^{T}b$ (due to (\ref{2.6d.pf.4})), this
yields that $v^{T}b$ is the maximum of the set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $. In other words, the number
$\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ exists
and satisfies%
\begin{equation}
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  =v^{T}b.
\label{2.6d.pf.max}%
\end{equation}

\end{verlong}

\begin{vershort}
Now,
\begin{equation}
\text{every }w\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A=c^{T}\right\}  \text{ satisfies }w\geq v^{T}b. \label{2.6d.pf.short.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.6d.pf.short.5}):} Let $w\in\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $.
Then, there exists an $y\in\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=c^{T}$
and $w=y^{T}b$. Consider this $y$. Since $y\geq0$, we have $y^{T}\geq0$ (since
the transpose of any nonnegative vector is nonnegative).
\par
We have $Au\leq b$. Thus, $b\geq Au$, so that $b-Au\geq0$. Thus, Lemma 2.0v
(applied to $m$, $y^{T}$ and $b-Au$ instead of $n$, $x$ and $y$) yields
$y^{T}\left(  b-Au\right)  \geq0$ (since $y^{T}\geq0$). Thus, $0\leq
y^{T}\left(  b-Au\right)  =y^{T}b-\underbrace{y^{T}A}_{=c^{T}}%
u=\underbrace{y^{T}b}_{=w}-\underbrace{c^{T}u}_{\substack{=v^{T}b\\\text{(by
(\ref{2.6d.pf.short.vtb=ctu}))}}}=w-v^{T}b$, so that $w\geq v^{T}b$. This
proves (\ref{2.6d.pf.short.5}).} Also, from $v\geq0$ and $v^{T}A=c^{T}$, we
conclude that $v\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $. Hence, $v^{T}b\in\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Combined with the
fact that every element of the set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is $\geq v^{T}b$ (by
(\ref{2.6d.pf.short.5})), this yields that $v^{T}b$ is the minimum of the set
\newline$\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $. In other words, the number \newline$\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
exists and satisfies%
\[
\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  =v^{T}b.
\]
Compared with (\ref{2.6d.pf.short.max}), this yields%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]

\end{vershort}

\begin{verlong}
Now,
\begin{equation}
\text{every }w\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A=c^{T}\right\}  \text{ satisfies }w\geq v^{T}b. \label{2.6d.pf.5}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.6d.pf.5}):} Let $w\in\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Then,
there exists an $y\in\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=c^{T}$ and
$w=y^{T}b$. Consider this $y$. Since $y\geq0$, we have $y^{T}\geq0$ (since the
transpose of any nonnegative vector is nonnegative).
\par
We have $Au\leq b$. Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }Au\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
Hence, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }b-Au\right)   &  =\left(
\text{the }i\text{-th coordinate of }b\right)  -\underbrace{\left(  \text{the
}i\text{-th coordinate of }Au\right)  }_{\leq\left(  \text{the }i\text{-th
coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $b-Au\geq0$. Thus, Lemma 2.0v (applied to $m$, $y^{T}$ and
$b-Au$ instead of $n$, $x$ and $y$) yields $y^{T}\left(  b-Au\right)  \geq0$
(since $y^{T}\geq0$). Thus, $0\leq y^{T}\left(  b-Au\right)  =y^{T}%
b-\underbrace{y^{T}A}_{=c^{T}}u=\underbrace{y^{T}b}_{=w}-\underbrace{c^{T}%
u}_{\substack{=v^{T}b\\\text{(by (\ref{2.6d.pf.vtb=ctu}))}}}=w-v^{T}b$, so
that $w\geq v^{T}b$. This proves (\ref{2.6d.pf.5}).} Also, from $v\geq0$ and
$v^{T}A=c^{T}$, we conclude that $v\in\left\{  y\in\mathbb{R}^{m}\ \mid
\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Hence, $v^{T}b\in\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $.
In other words, $v^{T}b$ is an element of the set $\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. Combined
with the fact that every element of the set $\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is $\geq
v^{T}b$ (by (\ref{2.6d.pf.5})), this yields that $v^{T}b$ is the minimum of
the set \newline$\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A=c^{T}\right\}  $. In other words, the number \newline$\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
exists and satisfies%
\[
\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  =v^{T}b.
\]
Compared with (\ref{2.6d.pf.max}), this yields%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]

\end{verlong}

Altogether, we have shown that the numbers $\max\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ and $\min\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exist
and satisfy%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]
In other words, Assertion J2 holds. Thus, at least one of Assertions J1 and J2 holds.

We have thus proven that in Case 1, at least one of Assertions J1 and J2 holds.

Now, let us consider Case 2. In this case, Assertion J$_{1}$2 holds. In other
words, there exists a vector $y\in\mathbb{R}^{2m+2n+1}$ such that $y\geq0$,
$y^{T}\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  =0$ and $y^{T}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  <0$. Denote this vector $y$ by $\eta$. Then, $\eta\in\mathbb{R}%
^{2m+2n+1}$ satisfies $\eta\geq0$,%
\begin{equation}
\eta^{T}\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  =0 \label{2.6d.pf.eta.first}%
\end{equation}
and%
\begin{equation}
\eta^{T}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)  <0. \label{2.6d.pf.eta.second}%
\end{equation}


Let us write the vector $\eta$ in the form $\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  $, where $u\in\mathbb{R}^{m}$, $v\in\mathbb{R}^{m}$, $q\in
\mathbb{R}^{n}$, $r\in\mathbb{R}^{n}$ and $w\in\mathbb{R}^{1}$. The vector
$w\in\mathbb{R}^{1}$ will be regarded as a scalar (since we can identify
vectors of length $1$ with scalars). Thus, $w^{T}=w$.

Define a vector $s\in\mathbb{R}^{n}$ by $s=q-r$.

Since $\eta=\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  $, we have%
\begin{align}
\eta^{T}  &  =\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  ^{T}=\left(
\begin{array}
[c]{ccccc}%
u^{T} & v^{T} & q^{T} & r^{T} & w^{T}%
\end{array}
\right) \nonumber\\
&  =\left(
\begin{array}
[c]{ccccc}%
u^{T} & v^{T} & q^{T} & r^{T} & w
\end{array}
\right)  \label{2.6d.pf.etaT}%
\end{align}
(because $w^{T}=w$).

Now, (\ref{2.6d.pf.eta.first}) yields%
\[
\left(  \eta^{T}\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  \right)  ^{T}=0^{T}=0.
\]
Hence,%
\begin{align*}
0  &  =\left(  \eta^{T}\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  \right)  ^{T}\\
&  =\underbrace{\left(
\begin{array}
[c]{cc}%
0_{m,n} & -I_{m}\\
A & 0_{m,m}\\
0_{n,n} & A^{T}\\
0_{n,n} & -A^{T}\\
-c^{T} & b^{T}%
\end{array}
\right)  ^{T}}_{=\left(
\begin{array}
[c]{ccccc}%
\left(  0_{m,n}\right)  ^{T} & A^{T} & \left(  0_{n,n}\right)  ^{T} & \left(
0_{n,n}\right)  ^{T} & \left(  -c^{T}\right)  ^{T}\\
\left(  -I_{m}\right)  ^{T} & \left(  0_{m,m}\right)  ^{T} & \left(
A^{T}\right)  ^{T} & \left(  -A^{T}\right)  ^{T} & \left(  b^{T}\right)  ^{T}%
\end{array}
\right)  }\underbrace{\left(  \eta^{T}\right)  ^{T}}_{=\eta}\\
&  =\left(
\begin{array}
[c]{ccccc}%
\left(  0_{m,n}\right)  ^{T} & A^{T} & \left(  0_{n,n}\right)  ^{T} & \left(
0_{n,n}\right)  ^{T} & \left(  -c^{T}\right)  ^{T}\\
\left(  -I_{m}\right)  ^{T} & \left(  0_{m,m}\right)  ^{T} & \left(
A^{T}\right)  ^{T} & \left(  -A^{T}\right)  ^{T} & \left(  b^{T}\right)  ^{T}%
\end{array}
\right)  \eta\\
&  =\left(
\begin{array}
[c]{ccccc}%
0_{n,m} & A^{T} & 0_{n,n} & 0_{n,n} & -c\\
-I_{m} & 0_{m,m} & A & -A & b
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  0_{m,n}\right)  ^{T}=0_{n,m}\text{ and }\left(
-I_{m}\right)  ^{T}=-\underbrace{\left(  I_{m}\right)  ^{T}}_{=I_{m}}%
=-I_{m}\text{ and }\left(  0_{m,m}\right)  ^{T}=0_{m,m}\\
\text{and }\left(  0_{n,n}\right)  ^{T}=0_{n,n}\text{ and }\left(
A^{T}\right)  ^{T}=A\text{ and }\left(  -A^{T}\right)  ^{T}%
=-\underbrace{\left(  A^{T}\right)  ^{T}}_{=A}=-A\\
\text{and }\left(  -c^{T}\right)  ^{T}=-\underbrace{\left(  c^{T}\right)
^{T}}_{=c}=-c\text{ and }\left(  b^{T}\right)  ^{T}=b\text{ and }\eta=\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)
\end{array}
\right)
\end{align*}%
\begin{align*}
&  =\left(
\begin{array}
[c]{c}%
0_{n,m}u+A^{T}v+0_{n,n}q+0_{n,n}r+\left(  -c\right)  w\\
\left(  -I_{m}\right)  u+0_{m,m}v+Aq+\left(  -A\right)  r+bw
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right)  .
\end{align*}
Thus,%
\[
\left(
\begin{array}
[c]{c}%
0_{n,m}u+A^{T}v+0_{n,n}q+0_{n,n}r+\left(  -c\right)  w\\
\left(  -I_{m}\right)  u+0_{m,m}v+Aq+\left(  -A\right)  r+bw
\end{array}
\right)  =0=\left(
\begin{array}
[c]{c}%
0\\
0
\end{array}
\right)  .
\]
Thus,
\begin{equation}
0_{n,m}u+A^{T}v+0_{n,n}q+0_{n,n}r+\left(  -c\right)  w=0
\label{2.6d.pf.eta.first.1}%
\end{equation}
and%
\begin{equation}
\left(  -I_{m}\right)  u+0_{m,m}v+Aq+\left(  -A\right)  r+bw=0.
\label{2.6d.pf.eta.first.2}%
\end{equation}
Now, (\ref{2.6d.pf.eta.first.1}) yields%
\[
0=\underbrace{0_{n,m}u}_{=0}+A^{T}v+\underbrace{0_{n,n}q}_{=0}%
+\underbrace{0_{n,n}r}_{=0}+\underbrace{\left(  -c\right)  w}_{=-cw}%
=A^{T}v-\underbrace{cw}_{\substack{=wc\\\text{(since }w\text{ is a scalar)}%
}}=A^{T}v-wc,
\]
so that $A^{T}v=wc$. Thus, $wc=A^{T}v$, so that $\left(  \underbrace{wc}%
_{=A^{T}v}\right)  ^{T}=\left(  A^{T}v\right)  ^{T}=v^{T}\underbrace{\left(
A^{T}\right)  ^{T}}_{=A}=v^{T}A$. Hence,%
\begin{equation}
v^{T}A=\left(  wc\right)  ^{T}=wc^{T}. \label{2.6d.pf.eta.first.1a}%
\end{equation}
Also, (\ref{2.6d.pf.eta.first.2}) yields%
\begin{align*}
0  &  =\underbrace{\left(  -I_{m}\right)  u}_{=-u}+\underbrace{0_{m,m}v}%
_{=0}+Aq+\underbrace{\left(  -A\right)  r}_{=-Ar}+bw=-u+\underbrace{Aq-Ar}%
_{=A\left(  q-r\right)  }+\underbrace{bw}_{\substack{=wb\\\text{(since
}w\text{ is a scalar)}}}\\
&  =-u+A\underbrace{\left(  q-r\right)  }_{=s}+wb=-u+As+wb,
\end{align*}
so that%
\begin{equation}
u=As+wb. \label{2.6d.pf.eta.first.2a}%
\end{equation}


On the other hand, from (\ref{2.6d.pf.etaT}), we obtain%
\begin{align*}
\eta^{T}\left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{ccccc}%
u^{T} & v^{T} & q^{T} & r^{T} & w
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
0_{m,1}\\
b\\
c\\
-c\\
0_{1,1}%
\end{array}
\right) \\
&  =\underbrace{u^{T}0_{m,1}}_{=0}+v^{T}b+q^{T}c+\underbrace{r^{T}\left(
-c\right)  }_{=-r^{T}c}+\underbrace{w0_{1,1}}_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =v^{T}b+\underbrace{q^{T}c-r^{T}c}_{=\left(  q^{T}-r^{T}\right)  c}%
=v^{T}b+\underbrace{\left(  q^{T}-r^{T}\right)  }_{=\left(  q-r\right)  ^{T}%
}c=v^{T}b+\left(  \underbrace{q-r}_{=s}\right)  ^{T}c=v^{T}b+s^{T}c.
\end{align*}
Hence, (\ref{2.6d.pf.eta.second}) rewrites as
\begin{equation}
v^{T}b+s^{T}c<0. \label{2.6d.pf.eta.second.a}%
\end{equation}
Since $s^{T}c$ is a scalar, we have $\left(  s^{T}c\right)  ^{T}=s^{T}c$, so
that $s^{T}c=\left(  s^{T}c\right)  ^{T}=c^{T}\underbrace{\left(
s^{T}\right)  ^{T}}_{=s}=c^{T}s$. Hence, (\ref{2.6d.pf.eta.second.a}) rewrites
as%
\begin{equation}
v^{T}b+c^{T}s<0. \label{2.6d.pf.eta.second.b}%
\end{equation}


\begin{vershort}
We have $\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  =\eta\geq0$. Hence, we have the five inequalities\footnote{Here, we
are using the following simple fact: Let $n_{1}$, $n_{2}$, $n_{3}$, $n_{4}$,
and $n_{5}$ be five nonnegative integers. Let $\alpha_{1}\in\mathbb{R}^{n_{1}%
}$, $\alpha_{2}\in\mathbb{R}^{n_{2}}$, $\alpha_{3}\in\mathbb{R}^{n_{3}}$,
$\alpha_{4}\in\mathbb{R}^{n_{4}}$, and $\alpha_{5}\in\mathbb{R}^{n_{5}}$ be
five column vectors. Let $\beta_{1}\in\mathbb{R}^{n_{1}}$, $\beta_{2}%
\in\mathbb{R}^{n_{2}}$, $\beta_{3}\in\mathbb{R}^{n_{3}}$, $\beta_{4}%
\in\mathbb{R}^{n_{4}}$, and $\beta_{5}\in\mathbb{R}^{n_{5}}$ be five column
vectors. Assume that $\left(
\begin{array}
[c]{c}%
\alpha_{1}\\
\alpha_{2}\\
\alpha_{3}\\
\alpha_{4}\\
\alpha_{5}%
\end{array}
\right)  \geq\left(
\begin{array}
[c]{c}%
\beta_{1}\\
\beta_{2}\\
\beta_{3}\\
\beta_{4}\\
\beta_{5}%
\end{array}
\right)  $. Then, we have the five inequalities%
\[
\alpha_{1}\geq\beta_{1},\ \ \ \ \ \ \ \ \ \ \alpha_{2}\geq\beta_{2}%
,\ \ \ \ \ \ \ \ \ \ \alpha_{3}\geq\beta_{3},\ \ \ \ \ \ \ \ \ \ \alpha
_{4}\geq\beta_{4},\ \ \ \ \ \ \ \ \ \ \text{and }\alpha_{5}\geq\beta_{5}.
\]
}
\[
u\geq0,\ \ \ \ \ \ \ \ \ \ v\geq0,\ \ \ \ \ \ \ \ \ \ q\geq
0,\ \ \ \ \ \ \ \ \ \ r\geq0,\ \ \ \ \ \ \ \ \ \ \text{and }w\geq0.
\]

\end{vershort}

\begin{verlong}
Moreover, recall that $\eta\geq0$. Thus, every $i\in\left\{  1,2,\ldots
,2m+2n+1\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\eta\right)  \geq\left(
\text{the }i\text{-th coordinate of }0\right)  =0. \label{2.6d.pf.eta.0}%
\end{equation}
This easily yields that $u\geq0$\ \ \ \ \footnote{\textit{Proof.} Every
$i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }\eta\right)   &  =\left(
\text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta=\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }u\right)  .
\end{align*}
Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }u\right)   &  =\left(  \text{the
}i\text{-th coordinate of }\eta\right)  \geq0\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{2.6d.pf.eta.0})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $u\geq0$, qed.}, $v\geq0$\ \ \ \ \footnote{\textit{Proof.}
Every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }\left(  m+i\right)  \text{-th coordinate of }\eta\right)
\\
&  =\left(  \text{the }\left(  m+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta=\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }v\right)  .
\end{align*}
Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }v\right)   &  =\left(  \text{the
}\left(  m+i\right)  \text{-th coordinate of }\eta\right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.eta.0}), applied to
}m+i\text{ instead of }i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $v\geq0$, qed.} and $w\geq0$\ \ \ \ \footnote{\textit{Proof.}
We have%
\begin{align*}
&  \left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of }%
\eta\right) \\
&  =\left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta=\left(
\begin{array}
[c]{c}%
u\\
v\\
q\\
r\\
w
\end{array}
\right)  \right) \\
&  =\left(  \text{the }1\text{-st coordinate of }w\right)
=w\ \ \ \ \ \ \ \ \ \ \left(  \text{since }w\text{ is a scalar}\right)  .
\end{align*}
Thus,%
\begin{align*}
w  &  =\left(  \text{the }\left(  2m+2n+1\right)  \text{-th coordinate of
}\eta\right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.6d.pf.eta.0}), applied to
}i=2m+2n+1\right)  ,
\end{align*}
qed.}.
\end{verlong}

Since $w\geq0$, we can multiply the inequality (\ref{2.6d.pf.eta.second.b})
with $w$, and obtain%
\begin{equation}
w\left(  v^{T}b+c^{T}s\right)  \leq0. \label{2.6d.pf.eta.1}%
\end{equation}
But from $v\geq0$, we obtain $v^{T}\geq0$ (since the transpose of any
nonnegative vector is nonnegative). Thus, Lemma 2.0v (applied to $m$, $v^{T}$
and $u$ instead of $n$, $x$ and $y$) yields $v^{T}u\geq0$ (because $u\geq0$).
Hence,%
\begin{align}
0  &  \leq v^{T}\underbrace{u}_{\substack{=As+wb\\\text{(by
(\ref{2.6d.pf.eta.first.2a}))}}}=v^{T}\left(  As+wb\right)  =\underbrace{v^{T}%
A}_{\substack{=wc^{T}\\\text{(by (\ref{2.6d.pf.eta.first.1a}))}}%
}s+wv^{T}b\nonumber\\
&  =wc^{T}s+wv^{T}b=w\left(  v^{T}b+c^{T}s\right)  .
\label{2.6d.pf.eta.proofw0}%
\end{align}
Combined with (\ref{2.6d.pf.eta.second.b}), this easily yields that
$w=0$\ \ \ \ \footnote{\textit{Proof.} Assume that $w>0$. Then, we can divide
the inequality (\ref{2.6d.pf.eta.proofw0}) by $w$, and obtain $0\leq
v^{T}b+c^{T}s$. This contradicts (\ref{2.6d.pf.eta.second.b}). This
contradiction shows that our assumption was wrong. Thus, we don't have $w>0$.
Hence, $w\leq0$. Combined with $w\geq0$, this yields $w=0$, qed.}. As a
consequence, (\ref{2.6d.pf.eta.first.1a}) rewrites as
\begin{equation}
v^{T}A=0c^{T}=0. \label{2.6d.pf.eta.first.1b}%
\end{equation}
Also, (\ref{2.6d.pf.eta.first.2a}) becomes%
\begin{equation}
u=As+\underbrace{w}_{=0}b=As. \label{2.6d.pf.eta.first.2b}%
\end{equation}


Let us summarize what we have found so far: We have found vectors
$u\in\mathbb{R}^{m}$, $v\in\mathbb{R}^{m}$ and $s\in\mathbb{R}^{n}$ satisfying
$u\geq0$ and $v\geq0$ and the equations (\ref{2.6d.pf.eta.second.b}),
(\ref{2.6d.pf.eta.first.1b}) and (\ref{2.6d.pf.eta.first.2b}). These equations
(along with $u\geq0$ and $v\geq0$) are all that we are going to need from now
on; we can forget about $q$, $r$, $\eta$ and $w$.

Our goal now is to prove that Assertion J1 holds.

\begin{vershort}
By the assumptions of Lemma 2.6d, the set $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $ has at least one element. Let $z$ be this element.
Then, $z\in\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $. In other
words, $z\in\mathbb{R}^{n}$ and $Az\leq b$. From $Az\leq b$, we obtain $b\geq
Az$, thus $b-Az\geq0$.
\end{vershort}

\begin{verlong}
By the assumptions of Lemma 2.6d, the set $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $ has at least one element. Let $z$ be this element.
Then, $z\in\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $. In other
words, $z\in\mathbb{R}^{n}$ and $Az\leq b$. This easily yields $b-Az\geq
0$\ \ \ \ \footnote{\textit{Proof.} We have $Az\leq b$. In other words, every
$i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }Az\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }b-Az\right)   &  =\left(
\text{the }i\text{-th coordinate of }b\right)  -\underbrace{\left(  \text{the
}i\text{-th coordinate of }Az\right)  }_{\leq\left(  \text{the }i\text{-th
coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right) \\
&  =0=\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $b-Az\geq0$, qed.}.
\end{verlong}

Also, $v^{T}\geq0$ (since $v\geq0$, and since the transpose of any nonnegative
vector is nonnegative). Thus, Lemma 2.0v (applied to $m$, $v^{T}$ and $b-Az$
instead of $n$, $x$ and $y$) yields $v^{T}\left(  b-Az\right)  \geq0$ (since
$b-Az\geq0$). Thus, $v^{T}b-v^{T}Az=v^{T}\left(  b-Az\right)  \geq0$, so that
$v^{T}b\geq\underbrace{v^{T}A}_{\substack{=0\\\text{(by
(\ref{2.6d.pf.eta.first.1b}))}}}z=0$. But from (\ref{2.6d.pf.eta.second.b}),
it follows that%
\begin{equation}
c^{T}s<-\underbrace{v^{T}b}_{\geq0}\leq-0=0. \label{2.6d.pf.eta.cTs}%
\end{equation}
Hence, $c^{T}s\neq0$, so that $c^{T}s$ is invertible.

Now, let $\delta\in\mathbb{R}$. We will show that there exists a $t\in\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ such that $t\geq\delta$.

Let $\lambda=\max\left\{  0,\dfrac{c^{T}z-\delta}{c^{T}s}\right\}  $ (this is
well-defined since $c^{T}s$ is invertible). Then, $\lambda\geq0$ and
$\lambda\geq\dfrac{c^{T}z-\delta}{c^{T}s}$.

Multiplying the inequality $\lambda\geq\dfrac{c^{T}z-\delta}{c^{T}s}$ with
$c^{T}s$, we obtain%
\begin{equation}
\lambda\cdot c^{T}s\leq c^{T}z-\delta\label{2.6d.pf.eta.lambda}%
\end{equation}
(the sign has flipped since $c^{T}s<0$).

\begin{vershort}
From (\ref{2.6d.pf.eta.first.2b}), we have $As=u\geq0$. Hence, $\lambda
As\geq0$ (since $\lambda\geq0$). The vector $z-\lambda s\in\mathbb{R}^{n}$
satisfies $A\left(  z-\lambda s\right)  =Az-\underbrace{\lambda As}_{\geq
0}\leq Az\leq b$. Hence, $z-\lambda s\in\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $, so that $c^{T}\left(  z-\lambda s\right)
\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $.
\end{vershort}

\begin{verlong}
It is easy to see (using (\ref{2.6d.pf.eta.first.2b}) and the fact that
$Az\leq b$) that the vector $z-\lambda s\in\mathbb{R}^{n}$ satisfies $A\left(
z-\lambda s\right)  \leq b$\ \ \ \ \footnote{\textit{Proof.} We have $Az\leq
b$. In other words, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }Az\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
\par
On the other hand, $u\geq0$. In other words, every $i\in\left\{
1,2,\ldots,m\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }u\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0
\]
and thus $\lambda\left(  \text{the }i\text{-th coordinate of }u\right)  \geq0$
(since $\lambda\geq0$).
\par
Now, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }\underbrace{A\left(  z-\lambda
s\right)  }_{\substack{=Az-\lambda As=Az-\lambda u\\\text{(because
(\ref{2.6d.pf.eta.first.2b}) yields }As=u\text{)}}}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }Az-\lambda u\right)
=\underbrace{\left(  \text{the }i\text{-th coordinate of }Az\right)  }%
_{\leq\left(  \text{the }i\text{-th coordinate of }b\right)  }%
-\underbrace{\lambda\left(  \text{the }i\text{-th coordinate of }u\right)
}_{\geq0}\\
&  \leq\left(  \text{the }i\text{-th coordinate of }b\right)  -0=\left(
\text{the }i\text{-th coordinate of }b\right)  .
\end{align*}
In other words, $A\left(  z-\lambda s\right)  \leq b$, qed.}. Hence,
$z-\lambda s\in\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $, so that
$c^{T}\left(  z-\lambda s\right)  \in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $.
\end{verlong}

But%
\[
c^{T}\left(  z-\lambda s\right)  =c^{T}z-\underbrace{\lambda\cdot c^{T}%
s}_{\substack{\leq c^{T}z-\delta\\\text{(by (\ref{2.6d.pf.eta.lambda}))}}}\geq
c^{T}z-\left(  c^{T}z-\delta\right)  =\delta.
\]
Since $c^{T}\left(  z-\lambda s\right)  \in\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $, this yields that there exists a
$t\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ such that
$t\geq\delta$ (namely, $t=c^{T}\left(  z-\lambda s\right)  $).

Now, forget that we fixed $\delta$. We have thus shown that for every
$\delta\in\mathbb{R}$, there exists a $t\in\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ such that $t\geq\delta$. In other
words, the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
$ is unbounded from above.

Finally, let us prove that the set $\left\{  y\in\mathbb{R}^{m}\ \mid
\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is empty. In fact, let $\psi
\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}
$ be arbitrary. We will derive a contradiction (thus showing that $\psi$
cannot exist).

Since $\psi\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $, we have $\psi\in\mathbb{R}^{m}$, $\psi\geq0$ and
$\psi^{T}A=c^{T}$. From $\psi^{T}A=c^{T}$, it follows that $\psi^{T}As=c^{T}%
s$. Since $As=u$ (by (\ref{2.6d.pf.eta.first.2b})), this rewrites as $\psi
^{T}u=c^{T}s$. Combined with (\ref{2.6d.pf.eta.cTs}), this yields $\psi
^{T}u<0$. But since $\psi\geq0$, we have $\psi^{T}\geq0$ (since the transpose
of a nonnegative vector is always nonnegative), and thus Lemma 2.0v (applied
to $m$, $\psi^{T}$ and $u$ instead of $n$, $x$ and $y$) yields $\psi^{T}%
u\geq0$ (since $u\geq0$). This contradicts $\psi^{T}u<0$.

Now, forget that we fixed $\psi$. Thus, for every $\psi\in\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $, we have
derived a contradiction. Hence, there exists no \newline$\psi\in\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $. In other
words, the set \newline$\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and
}y^{T}A=c^{T}\right\}  $ is empty.

Altogether, we have proven that the set $\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $ is unbounded from above, and the set
$\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
is empty. In other words, Assertion J1 holds. Thus, at least one of Assertions
J1 and J2 holds.

We have thus proven that in Case 2, at least one of Assertions J1 and J2 holds.

We have thus proven that, in each of the Cases 1 and 2, at least one of
Assertions J1 and J2 holds. Since these cases cover all possibilities, this
yields that, in every situation, at least one of Assertions J1 and J2 holds.
Combined with the fact that the Assertions J1 and J2 cannot hold at the same
time, this yields that exactly one of the Assertions J1 and J2 holds. This
proves Lemma 2.6d.
\end{proof}

Lemma 2.6d was ``one piece'' of Theorem 2.6c; here is another ``piece'':

\begin{statement}
\textbf{Lemma 2.6e.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be an
$m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$. Assume
that the set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{, }y^{T}A=0\text{
and }y^{T}b<0\right\}  $ has at least one element. Then:

\textbf{(a)} The set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $
is empty.

\textbf{(b)} Exactly one of the following two assertions holds:

\textit{Assertion K1:} The sets $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ and $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ are empty.

\textit{Assertion K2:} The set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is unbounded from below, and
the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty.
\end{statement}

\begin{proof}
[Proof of Lemma 2.6e.]

\begin{vershort}
The Assertions K1 and K2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions K1 and K2 hold at the same time. The
set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ is unbounded from below (by Assertion K2), and thus
nonempty. In other words, there exists some $y\in\mathbb{R}^{m}$ satisfying
$y\geq0$ and $y^{T}A=c^{T}$. In other words, the set $\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is nonempty.
But this contradicts Assertion K1. This contradiction shows that our
assumption was wrong, qed.}. We will now show that at least one of these
assertions holds.
\end{vershort}

\begin{verlong}
The Assertions K1 and K2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions K1 and K2 hold at the same time. We
know that Assertion K2 holds. In other words, the set $\left\{  y^{T}%
b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is
unbounded from below, and the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ is empty.
\par
We know that Assertion K1 holds. In other words, the sets $\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ and $\left\{  y\in\mathbb{R}%
^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ are empty.
\par
The set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ is empty, and thus cannot be unbounded from below
(since the empty set is not unbounded from below). This contradicts the fact
that the set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A=c^{T}\right\}  $ is unbounded from below. This contradiction shows
that our assumption was wrong, qed.}. We will now show that at least one of
these assertions holds.
\end{verlong}

The set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{, }y^{T}A=0\text{ and
}y^{T}b<0\right\}  $ has at least one element (by assumption). Denote this
element by $z$. Then, \newline$z\in\left\{  y\in\mathbb{R}^{m}\ \mid
\ y\geq0\text{, }y^{T}A=0\text{ and }y^{T}b<0\right\}  $. In other words,
$z\in\mathbb{R}^{m}$, $z\geq0$, $z^{T}A=0$ and $z^{T}b<0$. From $z\geq0$, it
follows that $z^{T}\geq0$ (since the transpose of any nonnegative vector is nonnegative).

\begin{vershort}
Now, it is easy to see that the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ is empty\footnote{\textit{Proof.} Let $w\in\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $. Then, $w\in\mathbb{R}^{n}$ and
$Aw\leq b$.
\par
From $Aw\leq b$, we obtain $b\geq Aw$, so that $b-Aw\geq0$. Hence, Lemma 2.0v
(applied to $m$, $z^{T}$ and $b-Aw$ instead of $n$, $x$ and $y$) yields
$z^{T}\left(  b-Aw\right)  \geq0$ (since $z^{T}\geq0$). This contradicts
$z^{T}\left(  b-Aw\right)  =z^{T}b-\underbrace{z^{T}A}_{=0}w=z^{T}b<0$.
\par
Now, forget that we fixed $w$. We have thus shown that any $w\in\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ satisfies a contradiction. In
other words, there exists no $w\in\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $. In other words, the set $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $ is empty, qed.}. This proves Lemma 2.6e
\textbf{(a)}.
\end{vershort}

\begin{verlong}
Now, it is easy to see that the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ is empty\footnote{\textit{Proof.} Let $w\in\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $. Then, $w\in\mathbb{R}^{n}$ and
$Aw\leq b$.
\par
We have $Aw\leq b$. In other words, every $i\in\left\{  1,2,\ldots,m\right\}
$ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }Aw\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }b-Aw\right)   &  =\left(
\text{the }i\text{-th coordinate of }b\right)  -\underbrace{\left(  \text{the
}i\text{-th coordinate of }Aw\right)  }_{\leq\left(  \text{the }i\text{-th
coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right) \\
&  =0=\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $b-Aw\geq0$. Hence, Lemma 2.0v (applied to $m$, $z^{T}$ and
$b-Aw$ instead of $n$, $x$ and $y$) yields $z^{T}\left(  b-Aw\right)  \geq0$
(since $z^{T}\geq0$). This contradicts $z^{T}\left(  b-Aw\right)
=z^{T}b-\underbrace{z^{T}A}_{=0}w=z^{T}b<0$.
\par
Now, forget that we fixed $w$. We have thus shown that any $w\in\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ satisfies a contradiction. In
other words, there exists no $w\in\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $. In other words, the set $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $ is empty, qed.}. This proves Lemma 2.6e
\textbf{(a)}.
\end{verlong}

Applying Theorem 2.5d to $n$, $m$, $A^{T}$ and $c$ instead of $m$, $n$, $A$
and $b$, we conclude that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion K}$_{1}$\textit{1:} The system $A^{T}x=c$ has a nonnegative
solution $x\in\mathbb{R}^{m}$.

\textit{Assertion K}$_{1}$\textit{2:} There exists a vector $y\in
\mathbb{R}^{n}$ such that $y^{T}A^{T}\geq0$ and $y^{T}c<0$.
\end{statement}

Thus, we must be in one of the following two cases:

\textit{Case 1:} Assertion K$_{1}$1 holds.

\textit{Case 2:} Assertion K$_{1}$2 holds.

Let us consider Case 1 first. In this case, Assertion K$_{1}$1 holds. In other
words, the system $A^{T}x=c$ has a nonnegative solution $x\in\mathbb{R}^{m}$.
Denote this $x$ by $u$. Thus, $u\in\mathbb{R}^{m}$ is nonnegative and
satisfies $A^{T}u=c$.

Comparing $\left(  A^{T}u\right)  ^{T}=u^{T}\underbrace{\left(  A^{T}\right)
^{T}}_{=A}=u^{T}A$ with $\left(  \underbrace{A^{T}u}_{=c}\right)  ^{T}=c^{T}$,
we obtain $u^{T}A=c^{T}$. Also, $u$ is nonnegative, i. e., we have $u\geq0$.
Since $u\in\mathbb{R}^{m}$, $u\geq0$ and $u^{T}A=c^{T}$, we have $u\in\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ and thus%
\begin{equation}
u^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  . \label{2.6e.pf.1}%
\end{equation}


Now, let $\delta\in\mathbb{R}$. We will show that there exists a \newline%
$t\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ such that $t\leq\delta$.

In fact, since $z^{T}b<0$, we have $z^{T}b\neq0$, so that $z^{T}b$ is
invertible. Now, let $\lambda=\max\left\{  0,\dfrac{\delta-u^{T}b}{z^{T}%
b}\right\}  $ (this is well-defined since $z^{T}b$ is invertible). Then,
$\lambda\geq0$ and $\lambda\geq\dfrac{\delta-u^{T}b}{z^{T}b}$.

Multiplying the inequality $\lambda\geq\dfrac{\delta-u^{T}b}{z^{T}b}$ with
$z^{T}b$, we obtain
\begin{equation}
\lambda\cdot z^{T}b\leq\delta-u^{T}b \label{2.6e.pf.2}%
\end{equation}
(the sign flipped since $z^{T}b<0$).

\begin{vershort}
From $u\geq0$, $\lambda\geq0$ and $z\geq0$, we obtain $u+\lambda z\geq0$.
\end{vershort}

\begin{verlong}
It is easy to see (using $z\geq0$ and $u\geq0$) that $u+\lambda z\geq
0$\ \ \ \ \footnote{\textit{Proof.} We have $z\geq0$. Thus, every
$i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }z\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0.
\]
\par
We have $u\geq0$. Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }u\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0.
\]
\par
Now, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }u+\lambda z\right)   &
=\underbrace{\left(  \text{the }i\text{-th coordinate of }u\right)  }_{\geq
0}+\underbrace{\lambda}_{\geq0}\underbrace{\left(  \text{the }i\text{-th
coordinate of }z\right)  }_{\geq0}\\
&  \geq0=\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $u+\lambda z\geq0$, qed.}.
\end{verlong}

Also,%
\[
\underbrace{\left(  u+\lambda z\right)  ^{T}}_{=u^{T}+\lambda z^{T}}A=\left(
u^{T}+\lambda z^{T}\right)  A=\underbrace{u^{T}A}_{=c^{T}}+\lambda
\underbrace{z^{T}A}_{=0}=c^{T}.
\]


Since $u+\lambda z\in\mathbb{R}^{m}$, $u+\lambda z\geq0$ and $\left(
u+\lambda z\right)  ^{T}A=c^{T}$, we have $u+\lambda z\in\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ and thus%
\[
\left(  u+\lambda z\right)  ^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  .
\]
Since%
\[
\left(  u+\lambda z\right)  ^{T}b=u^{T}b+\underbrace{\lambda\cdot z^{T}%
b}_{\substack{\leq\delta-u^{T}b\\\text{(by (\ref{2.6e.pf.2}))}}}\leq
u^{T}b+\delta-u^{T}b=\delta,
\]
this yields that there exists a $t\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ such that $t\leq\delta$
(namely, $t=\left(  u+\lambda z\right)  ^{T}b$).

Now, forget that we fixed $\delta$. We thus have proven that for every
$\delta\in\mathbb{R}$, there exists a $t\in\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ such that
$t\leq\delta$. In other words, the set $\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is unbounded from below.

We now know that the set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m}%
;\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is unbounded from below, and the
set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In other
words, Assertion K2 holds. Thus, at least one of Assertions K1 and K2 holds.

Hence, we know that in Case 1, at least one of Assertions K1 and K2 holds.

Let us now consider Case 2. In this case, Assertion K$_{1}$2 holds. In other
words, there exists a vector $y\in\mathbb{R}^{n}$ such that $y^{T}A^{T}\geq0$
and $y^{T}c<0$. Denote this vector $y$ by $w$. Then, $w\in\mathbb{R}^{n}$ is a
vector such that $w^{T}A^{T}=0$ and $w^{T}c<0$.

Since $w^{T}A^{T}=0$, we have $\left(  w^{T}A^{T}\right)  ^{T}=0^{T}=0$. This
rewrites as $Aw=0$ (because $\left(  w^{T}A^{T}\right)  ^{T}%
=\underbrace{\left(  A^{T}\right)  ^{T}}_{=A}\underbrace{\left(  w^{T}\right)
^{T}}_{=w}=Aw$).

Since $w^{T}c$ is a scalar, we have $\left(  w^{T}c\right)  ^{T}=w^{T}c<0$.
Since $\left(  w^{T}c\right)  ^{T}=c^{T}\underbrace{\left(  w^{T}\right)
^{T}}_{=w}=c^{T}w$, this rewrites as $c^{T}w<0$.

Now, it is easy to see that the set $\left\{  y\in\mathbb{R}^{m}\ \mid
\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is empty\footnote{\textit{Proof.}
Let $s\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  $. Then, $s\in\mathbb{R}^{m}$ is a vector such that $s\geq0$
and $s^{T}A=c^{T}$. Thus, $\underbrace{s^{T}A}_{=c^{T}}w=c^{T}w<0$,
contradicting $s^{T}\underbrace{Aw}_{=0}=0$.
\par
Now, forget that we fixed $s$. We thus have shown that every $s\in\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ satisfies
a contradiction. In other words, no $s\in\left\{  y\in\mathbb{R}^{m}%
\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ can exist. In other words,
the set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  $ is empty, qed.}.

Altogether, we now know that the sets $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ Ax\leq b\right\}  $ and \newline$\left\{  y\in\mathbb{R}^{m}%
\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ are empty. In other words,
Assertion K1 holds. Thus, at least one of Assertions K1 and K2 holds.

Hence, we know that in Case 2, at least one of Assertions K1 and K2 holds.

We have thus proven that, in each of the Cases 1 and 2, at least one of
Assertions K1 and K2 holds. Since these cases cover all possibilities, this
yields that, in every situation, at least one of Assertions K1 and K2 holds.
Combined with the fact that the Assertions K1 and K2 cannot hold at the same
time, this yields that exactly one of the Assertions K1 and K2 holds. This
proves Lemma 2.6e \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem 2.6c.]Corollary 2.5o yields that exactly one of the two
assertions O1 and O2 holds.\footnote{See the statement of Corollary 2.5o for
these assertions.}

Thus, we must be in one of the following two cases:

\textit{Case 1:} Assertion O1 holds.

\textit{Case 2:} Assertion O2 holds.

Let us consider Case 1 first. In this case, Assertion O1 holds. In other
words, there exists a vector $x\in\mathbb{R}^{n}$ such that $Ax\leq b$. In
other words, the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $
has at least one element. Hence, Lemma 2.6d yields that exactly one of the two
assertions J1 and J2 holds.\footnote{See the statement of Lemma 2.6d for these
assertions.} Since Assertion J1 is identical to Assertion I2, and Assertion J2
is identical to Assertion I4, this rewrites as follows: Exactly one of the
assertions I2 and I4 holds.

\begin{vershort}
Assertion I1 is false\footnote{\textit{Proof.} Assume the contrary. Thus,
Assertion I1 holds. Hence, the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ is empty. This contradicts the fact that the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least one element. This
contradiction shows that our assumption was false, qed.}. Assertion I3 is also
false.\footnote{\textit{Proof.} Assume the contrary. Thus, Assertion I3 holds.
Thus, in particular, the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ is empty. This contradicts the fact that the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least one element. This
contradiction shows that our assumption was false, qed.}
\end{vershort}

\begin{verlong}
Assertion I1 is false\footnote{\textit{Proof.} Assume the contrary. Thus,
Assertion I1 holds. In other words, the sets $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ Ax\leq b\right\}  $ and $\left\{  y\in\mathbb{R}^{m}\ \mid
\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ are empty. In particular, the set
$\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty, i. e.,
contains no element. This contradicts the fact that the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least one element. This
contradiction shows that our assumption was false, qed.}. Assertion I3 is also
false.\footnote{\textit{Proof.} Assume the contrary. Thus, Assertion I3 holds.
In other words, the set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m}%
;\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is unbounded from below, and the
set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In
particular, the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is
empty, i. e., contains no element. This contradicts the fact that the set
$\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least one
element. This contradiction shows that our assumption was false, qed.}
\end{verlong}

So we know that Assertions I1 and I3 are both false. Combining this with the
fact that exactly one of the assertions I2 and I4 holds, we conclude that
exactly one of the four assertions I1, I2, I3 and I4 holds. In other words,
Theorem 2.6c is proven in Case 1.

Now, let us consider Case 2. In this case, Assertion O2 holds. In other words,
there exists a vector $y\in\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=0$ and
$y^{T}b<0$. In other words, the set $\left\{  y\in\mathbb{R}^{m}\ \mid
\ y\geq0\text{, }y^{T}A=0\text{ and }y^{T}b<0\right\}  $ has at least one
element. Thus, Lemma 2.6e \textbf{(b)} yields that exactly one of the two
assertions K1 and K2 holds.\footnote{See the statement of Lemma 2.6e for these
assertions.} Since Assertion K1 is identical to Assertion I1, and Assertion K2
is identical to Assertion I3, this rewrites as follows: Exactly one of the
assertions I1 and I3 holds.

\begin{vershort}
Assertion I2 is false\footnote{\textit{Proof.} Assume the contrary. Thus,
Assertion I2 holds. Thus, in particular, the set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is unbounded from above, and
therefore nonempty. In other words, there exists some $x\in\mathbb{R}^{n}$
satisfying $Ax\leq b$.
\par
But Lemma 2.6e \textbf{(a)} shows that the set $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In other words, there exists no
$x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This contradicts the fact that
there exists some $x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This
contradiction shows that our assumption was false, qed.}. Assertion I4 is also
false.\footnote{\textit{Proof.} Assume the contrary. Thus, Assertion I4 holds.
Hence, in particular, the number $\max\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $ exists. In other words, the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ has a maximum. Hence,
this set is nonempty. In other words, there exists some $x\in\mathbb{R}^{n}$
such that $Ax\leq b$.
\par
But Lemma 2.6e \textbf{(a)} shows that the set $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In other words, there exists no
$x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This contradicts the fact that
there exists some $x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This
contradiction shows that our assumption was false, qed.}
\end{vershort}

\begin{verlong}
Assertion I2 is false\footnote{\textit{Proof.} Assume the contrary. Thus,
Assertion I2 holds. In other words, the set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is unbounded from above, and the
set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}%
\right\}  $ is empty. In particular, the set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is unbounded from above, and
therefore nonempty. In other words, there exists some $x\in\mathbb{R}^{n}$
satisfying $Ax\leq b$.
\par
But Lemma 2.6e \textbf{(a)} shows that the set $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In other words, there exists no
$x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This contradicts the fact that
there exists some $x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This
contradiction shows that our assumption was false, qed.}. Assertion I4 is also
false.\footnote{\textit{Proof.} Assume the contrary. Thus, Assertion I4 holds.
In other words, the numbers $\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ and \newline$\min\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exist and
satisfy%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]
In particular, the number $\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ Ax\leq b\right\}  $ exists. In other words, the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ has a maximum. Hence,
this set is nonempty. In other words, there exists some $x\in\mathbb{R}^{n}$
such that $Ax\leq b$.
\par
But Lemma 2.6e \textbf{(a)} shows that the set $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In other words, there exists no
$x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This contradicts the fact that
there exists some $x\in\mathbb{R}^{n}$ satisfying $Ax\leq b$. This
contradiction shows that our assumption was false, qed.}
\end{verlong}

So we know that Assertions I2 and I4 are both false. Combining this with the
fact that exactly one of the assertions I1 and I3 holds, we conclude that
exactly one of the four assertions I1, I2, I3 and I4 holds. In other words,
Theorem 2.6c is proven in Case 2.

So we have proven Theorem 2.6c in each of the cases 1 and 2. Since these cases
1 and 2 cover all possibilities, this yields that Theorem 2.6c is proven in
all situations. In other words, the proof of Theorem 2.6c is complete.
\end{proof}

Note that Theorem 2.5p is an easy corollary of Theorem 2.6c:

\begin{proof}
[Alternative proof of Theorem 2.5p.]%
\begin{vershort}
The Assertions P1 and P2 cannot hold at the same time\footnote{This can be
proven just as in our first proof of Theorem 2.5p above.}. We will now show
that at least one of these assertions holds.
\end{vershort}

\begin{verlong}
The Assertions P1 and P2 cannot hold at the same time\footnote{\textit{Proof.}
Assume the opposite. Then, the Assertions P1 and P2 hold at the same time.
Since Assertion P2 holds, there exists a vector $y\in\mathbb{R}^{m}$ such that
$y\geq0$, $y^{T}A=c^{T}$ and $y^{T}b\leq\delta$. Consider this $y$.
\par
Since Assertion P1 holds, there exists a vector $x\in\mathbb{R}^{n}$ such that
$Ax\leq b$ and $c^{T}x>\delta$. Consider this $x$.
\par
For every $i\in\left\{  1,2,\ldots,m\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of }Ax\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)
\]
(since $Ax\leq b$). Thus, for every $i\in\left\{  1,2,\ldots,m\right\}  $, we
have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Ax\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Ax\right)  }%
_{\leq\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0\\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, $b-Ax\geq0$. Also, $y\geq0$, so that $y^{T}\geq0$ (since the transpose
of any nonnegative vector is nonnegative). Now, applying Lemma 2.0v to $m$,
$y^{T}$ and $b-Ax$ instead of $n$, $x$ and $y$, we obtain $y^{T}\left(
b-Ax\right)  \geq0$. This contradicts $y^{T}\left(  b-Ax\right)
=\underbrace{y^{T}b}_{\leq\delta}-\underbrace{y^{T}A}_{=c^{T}}x\leq
\delta-\underbrace{c^{T}x}_{>\delta}<\delta-\delta=0$. This contradiction
shows that our assumption was wrong, qed.}. We will now show that at least one
of these assertions holds.
\end{verlong}

Theorem 2.6c yields that exactly one of the four assertions I1, I2, I3 and I4
holds.\footnote{See the statement of Theorem 2.6c for these assertions.} Thus,
we must be in one of the following four cases:

\textit{Case 1:} Assertion I1 holds.

\textit{Case 2:} Assertion I2 holds.

\textit{Case 3:} Assertion I3 holds.

\textit{Case 4:} Assertion I4 holds.

\begin{vershort}
Let us consider Case 1 first. In this case, Assertion I1 holds. Thus, in
particular, the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is
empty. This contradicts the assumption that the set $\left\{  x\in
\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least one element. Thus, we
have obtained a contradiction. Hence, Theorem 2.5p holds in Case 1 (because ex
falso quod libet).
\end{vershort}

\begin{verlong}
Let us consider Case 1 first. In this case, Assertion I1 holds. In other
words, the sets $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ and
$\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $
are empty. In particular, the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq
b\right\}  $ is empty, i. e., contains no elements. This contradicts the
assumption that the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}
$ has at least one element. Thus, we have obtained a contradiction. Hence,
Theorem 2.5p holds in Case 1 (because ex falso quod libet).
\end{verlong}

\begin{vershort}
Let us now consider Case 2. In this case, Assertion I2 holds. Hence, in
particular, the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $ is unbounded from above. Thus, there exists a $t\in\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ such that $t\geq
\delta+1$. Consider this $t$. Then, $t\in\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ Ax\leq b\right\}  $, so that there exists some $z\in
\mathbb{R}^{n}$ such that $Az\leq b$ and $t=c^{T}z$. Consider this $z$. We
have $Az\leq b$ and $c^{T}z=t\geq\delta+1>\delta$. Thus, there exists a vector
$x\in\mathbb{R}^{n}$ such that $Ax\leq b$ and $c^{T}x>\delta$ (namely, $x=z$).
In other words, Assertion P1 holds. Hence, at least one of Assertions P1 and
P2 holds. Combined with the fact that the Assertions P1 and P2 cannot hold at
the same time, this yields that exactly one of Assertions P1 and P2 holds. In
other words, Theorem 2.5p holds in Case 2.
\end{vershort}

\begin{verlong}
Let us now consider Case 2. In this case, Assertion I2 holds. In other words,
the set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ is
unbounded from above, and the set \newline$\left\{  y\in\mathbb{R}^{m}%
\ \mid\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ is empty. In particular, the
set \newline$\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $
is unbounded from above. In other words, for every $\varepsilon\in\mathbb{R}$,
there exists a $t\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $ such that $t\geq\varepsilon$. Applying this to $\varepsilon
=\delta+1$, we conclude that there exists a $t\in\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ such that $t\geq\delta+1$. Consider
this $t$. Then, $t\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $, so that there exists some $z\in\mathbb{R}^{n}$ such that $Az\leq
b$ and $t=c^{T}z$. Consider this $z$. We have $Az\leq b$ and $c^{T}%
z=t\geq\delta+1>\delta$. Thus, there exists a vector $x\in\mathbb{R}^{n}$ such
that $Ax\leq b$ and $c^{T}x>\delta$ (namely, $x=z$). In other words, Assertion
P1 holds. Hence, at least one of Assertions P1 and P2 holds. Combined with the
fact that the Assertions P1 and P2 cannot hold at the same time, this yields
that exactly one of Assertions P1 and P2 holds. In other words, Theorem 2.5p
holds in Case 2.
\end{verlong}

\begin{vershort}
In Case 3, we obtain the very same contradiction as in Case 1. Hence, Theorem
2.5p holds in Case 3 (because ex falso quod libet).
\end{vershort}

\begin{verlong}
Let us now consider Case 3. In this case, Assertion I3 holds. In other words,
the set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  $ is unbounded from below, and the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty. In particular, the set
$\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ is empty, i. e.,
contains no elements. This contradicts the assumption that the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\}  $ has at least one element. Thus,
we have obtained a contradiction. Hence, Theorem 2.5p holds in Case 3 (because
ex falso quod libet).
\end{verlong}

Let us finally consider Case 4. In this case, Assertion I4 holds. In other
words, the numbers $\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq
b\right\}  $ and \newline$\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}  $ exist and satisfy%
\[
\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]
Hence, we can define an $\varepsilon\in\mathbb{R}$ by%
\[
\varepsilon=\max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}
=\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A=c^{T}\right\}  .
\]


Now, we must be in one of the following two subcases:

\textit{Subcase 4.1:} We have $\delta<\varepsilon$.

\textit{Subcase 4.2:} We have $\delta\geq\varepsilon$.

Let us first consider Subcase 4.1. In this subcase, $\delta<\varepsilon$.
Thus, $\varepsilon>\delta$. Since \newline$\varepsilon=\max\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  \in\left\{  c^{T}%
x\ \mid\ x\in\mathbb{R}^{n};\ Ax\leq b\right\}  $ (because the maximum of a
set always lies in this set), there exists a $z\in\mathbb{R}^{n}$ such that
$Az\leq b$ and $\varepsilon=c^{T}z$. Consider this $z$. Then, $Az\leq b$ and
$c^{T}z=\varepsilon>\delta$. Hence, there exists a vector $x\in\mathbb{R}^{n}$
such that $Ax\leq b$ and $c^{T}x>\delta$ (namely, $x=z$). In other words,
Assertion P1 holds. Hence, at least one of Assertions P1 and P2 holds.
Combined with the fact that the Assertions P1 and P2 cannot hold at the same
time, this yields that exactly one of Assertions P1 and P2 holds. In other
words, Theorem 2.5p holds in Subcase 4.1.

Let us now consider Subcase 4.2. In this subcase, $\delta\geq\varepsilon$.
Thus, $\varepsilon\leq\delta$. Since \newline$\varepsilon=\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A=c^{T}\right\}
\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}%
A=c^{T}\right\}  $ (because the minimum of a set always lies in this set),
there exists a $z\in\mathbb{R}^{m}$ such that $z\geq0$, $z^{T}A=c^{T}$ and
$\varepsilon=z^{T}b$. Consider this $z$. Then, $z\geq0$, $z^{T}A=c^{T}$ and
$z^{T}b=\varepsilon\leq\delta$. Hence, there exists a vector $y\in
\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A=c^{T}$ and $y^{T}b\leq\delta$
(namely, $y=z$). In other words, Assertion P2 holds. Hence, at least one of
Assertions P1 and P2 holds. Combined with the fact that the Assertions P1 and
P2 cannot hold at the same time, this yields that exactly one of Assertions P1
and P2 holds. In other words, Theorem 2.5p holds in Subcase 4.2.

We now know that Theorem 2.5p holds in each of Subcases 4.1 and 4.2. Since
these Subcases 4.1 and 4.2 cover all of Case 4, this yields that Theorem 2.5p
holds in Case 4.

We thus know that Theorem 2.5p holds in each of Cases 1, 2, 3 and 4. Since
these Cases 1, 2, 3 and 4 cover all possibilities, this yields that Theorem
2.5p holds in every situation. Theorem 2.5p is thus proven again.
\end{proof}

\section{The symmetric version of the Duality theorem}

We are now going to prove a variant of the duality theorem which is
``symmetric'' in the sense that the set we take the minimum of and the set we
take the maximum of are defined similarly (in particular, both are
parametrized by nonnegative vectors, which was not the case in Theorem 2.6c):

\begin{statement}
\textbf{Theorem 2.6f.} Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be
an $m\times n$-matrix. Let $b\in\mathbb{R}^{m}$. Let $c\in\mathbb{R}^{n}$.
Then, exactly one of the following four assertions holds:

\textit{Assertion Q1:} The sets $\left\{  x\in\mathbb{R}^{n}\ \mid
\ x\geq0\text{ and }Ax\leq b\right\}  $ and \newline$\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ are empty.

\textit{Assertion Q2:} The set $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $ is unbounded from above, and the
set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  $ is empty.

\textit{Assertion Q3:} The set $\left\{  y^{T}b\ \mid\ y\in\mathbb{R}%
^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ is unbounded from below,
and the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq
b\right\}  $ is empty.

\textit{Assertion Q4:} The numbers $\max\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $ and $\min\left\{
y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}
$ exist and satisfy%
\begin{align*}
&  \max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\} \\
&  =\min\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A\geq c^{T}\right\}  .
\end{align*}

\end{statement}

Theorem 2.6f is stronger than Exercise 2.23 in \cite{Schrij-CCO} (both
classically and constructively).

It should be noticed that most of the times when Schrijver applies linear
programming duality in \cite{Schrij-CCO}, it is being applied not in the form
of Theorem 2.6 in \cite{Schrij-CCO} (or our Theorem 2.6c), but in the form of
Exercise 2.23 in \cite{Schrij-CCO} (or our Theorem 2.6f). In particular, in
the proofs of Corollary 3.7b and Corollary 8.3a in \cite{Schrij-CCO},
Schrijver is applying Exercise 2.23 when he says that he is using linear
programming duality.

\begin{vershort}
\begin{proof}
[Proof of Theorem 2.6f.]For any nonnegative integer $\gamma$, let $I_{\gamma}$
denote the $\gamma\times\gamma$ identity matrix.

Let $0_{n}$ be the zero vector in $\mathbb{R}^{n}$.

Consider the block matrix $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  $; this is an $\left(  m+n\right)  \times n$-matrix.

Also, consider the block matrix $\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $; this is an $\left(  m+n\right)  \times1$-matrix, i. e., a vector
in $\mathbb{R}^{m+n}$ (since we identify matrices having only one column with
column vectors).

Applying Theorem 2.6c to $m+n$, $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  $ and $\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $ instead of $m$, $A$ and $b$, we see that exactly one of the
following four assertions holds:

\begin{statement}
\textit{Assertion Q}$_{1}$\textit{1:} The sets $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and \newline$\left\{  y\in\mathbb{R}^{m+n}\ \mid
\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ are empty.

\textit{Assertion Q}$_{1}$\textit{2:} The set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is unbounded from above, and the set $\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is empty.

\textit{Assertion Q}$_{1}$\textit{3:} The set $\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is unbounded from below, and the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is empty.

\textit{Assertion Q}$_{1}$\textit{4:} The numbers $\max\left\{  c^{T}%
x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and $\min\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ exist and satisfy%
\begin{align*}
&  \max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\} \\
&  =\min\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\end{align*}

\end{statement}

We will now prove that these assertions Q$_{1}$1, Q$_{1}$2, Q$_{1}$3, and
Q$_{1}$4 are equivalent to the assertions Q1, Q2, Q3, and Q4, respectively.

First, we notice that%
\begin{align}
&  \left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\} \nonumber\\
&  =\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  . \label{2.6f.pf.short.1}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.short.1}):} For each $x\in
\mathbb{R}^{n}$, we have%
\begin{align}
\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\underbrace{\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x}_{\substack{=\left(
\begin{array}
[c]{c}%
Ax\\
-x
\end{array}
\right)  \\\text{(by the multiplication rule}\\\text{for block matrices)}}}
&  =\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
Ax\\
-x
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
b-Ax\\
0_{n}-\left(  -x\right)
\end{array}
\right) \nonumber\\
&  =\left(
\begin{array}
[c]{c}%
b-Ax\\
x
\end{array}
\right)  . \label{2.6f.pf.short.1.1}%
\end{align}
Now, for each $x\in\mathbb{R}^{n}$, we have the following chain of
equivalences:%
\begin{align*}
&  \ \left(  \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right) \\
&  \Longleftrightarrow\ \left(  \left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \geq\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\right)  \ \Longleftrightarrow\ \left(  \left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
Ax\\
-x
\end{array}
\right)  \geq0\right) \\
&  \Longleftrightarrow\ \left(  \left(
\begin{array}
[c]{c}%
b-Ax\\
x
\end{array}
\right)  \geq0\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{2.6f.pf.short.1.1})}\right) \\
&  \Longleftrightarrow\ \left(  b-Ax\geq0\text{ and }x\geq0\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 2.5h}\right) \\
&  \Longleftrightarrow\ \left(  x\geq0\text{ and }\underbrace{b-Ax\geq
0}_{\Longleftrightarrow\ \left(  b\geq Ax\right)  \ \Longleftrightarrow
\ \left(  Ax\leq b\right)  }\right) \\
&  \Longleftrightarrow\ \left(  x\geq0\text{ and }Ax\leq b\right)  .
\end{align*}
Hence,%
\[
\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  =\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{
and }Ax\leq b\right\}  .
\]
This proves (\ref{2.6f.pf.short.1}).}

Next, we notice that%
\begin{align}
&  \left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\} \nonumber\\
&  =\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  . \label{2.6f.pf.short.2}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.short.2}):} Let $\lambda\in\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ be arbitrary. Then, there exists an $y\in
\mathbb{R}^{m+n}$ such that $y\geq0$, $y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$ and $\lambda=y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $. Denote this $y$ by $\xi$. Thus, $\xi\in\mathbb{R}^{m+n}$ satisfies
$\xi\geq0$, $\xi^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$ and $\lambda=\xi^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $.
\par
Let us write the vector $\xi$ in the form $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, where $u\in\mathbb{R}^{m}$ and $v\in\mathbb{R}^{n}$.
\par
We have $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  =\xi\geq0$. Thus, Lemma 2.5h \textbf{(b)} (applied to $N=m$, $M=n$,
$x=u$ and $y=v$) yields that $u\geq0$ and $v\geq0$. Thus, $u^{T}\geq0$ and
$v^{T}\geq0$.
\par
Since $\xi=\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, we have $\xi^{T}=\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  $, and thus%
\begin{align*}
\xi^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right) \\
&  =u^{T}A+\underbrace{v^{T}\left(  -I_{n}\right)  }_{=-v^{T}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =u^{T}A-v^{T}.
\end{align*}
Compared with $\xi^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$, this yields $c^{T}=u^{T}A-v^{T}$. In other words,%
\[
u^{T}A=c^{T}+v^{T}.
\]
\par
Hence, $u^{T}A=c^{T}+\underbrace{v^{T}}_{\geq0}\geq c^{T}$.
\par
So we now know that $u\in\mathbb{R}^{m}$, $u\geq0$ and $u^{T}A\geq c^{T}$. In
other words, $u\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A\geq c^{T}\right\}  $, so that%
\begin{equation}
u^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A\geq c^{T}\right\}  . \label{2.6f.pf.short.2.pf.2}%
\end{equation}
Now,
\begin{align*}
\lambda &  =\underbrace{\xi^{T}}_{=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right) \\
&  =u^{T}b+\underbrace{v^{T}0_{n}}_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the multiplication rule for block matrices}\right) \\
&  =u^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{2.6f.pf.short.2.pf.2})}\right)  .
\end{align*}
\par
Now, forget that we fixed $\lambda$. We thus have proven that every
$\lambda\in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ satisfies $\lambda\in\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $. In other
words,%
\begin{align}
&  \left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\} \nonumber\\
&  \subseteq\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  . \label{2.6f.pf.short.2.pf.3}%
\end{align}
\par
Now, let $\mu\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  $ be arbitrary. Then, there exists a $y\in
\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A\geq c^{T}$ and $\mu=y^{T}b$.
Denote this $y$ by $u$. Thus, $u\in\mathbb{R}^{m}$ satisfies $u\geq0$,
$u^{T}A\geq c^{T}$ and $\mu=u^{T}b$.
\par
We have $u^{T}A\geq c^{T}$. In other words, $u^{T}A-c^{T}\geq0$.
\par
Let $v$ be the vector $\left(  u^{T}A-c^{T}\right)  ^{T}$. Then,
$v^{T}=\left(  \left(  u^{T}A-c^{T}\right)  ^{T}\right)  ^{T}=u^{T}A-c^{T}%
\geq0$; therefore, $v\geq0$.
\par
Lemma 2.5h \textbf{(a)} (applied to $N=m$, $M=n$, $x=u$ and $y=v$) yields that
$\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \geq0$ (since $u\geq0$ and $v\geq0$). But on the other hand,%
\begin{align*}
\underbrace{\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}}_{=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  }\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =u^{T}A+\underbrace{v^{T}\left(  -I_{n}\right)  }_{=-v^{T}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =u^{T}A-\underbrace{v^{T}}_{=u^{T}A-c^{T}}=u^{T}A-\left(  u^{T}%
A-c^{T}\right)  =c^{T}.
\end{align*}
\par
Now, we know that our $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \in\mathbb{R}^{m+n}$ satisfies $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \geq0$ and $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$. In other words,%
\[
\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \in\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }%
y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  ,
\]
and thus%
\[
\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\]
Since%
\begin{align*}
\underbrace{\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}}_{=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  =\underbrace{u^{T}b}_{=\mu}+\underbrace{v^{T}0_{n}}_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\mu,
\end{align*}
this rewrites as%
\[
\mu\in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\]
\par
Now, forget that we fixed $\mu$. We thus have proven that every $\mu
\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  $ satisfies \newline$\mu\in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $. In other words,%
\begin{align*}
&  \left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\} \\
&  \subseteq\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\end{align*}
Combined with (\ref{2.6f.pf.short.2.pf.3}), this yields%
\begin{align*}
&  \left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\} \\
&  =\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  .
\end{align*}
This proves (\ref{2.6f.pf.short.2}).}

Next, we notice that we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right)  .
\label{2.6f.pf.short.3}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.short.3}):} We have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }x\in\mathbb{R}%
^{n}\text{ satisfying }\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  }_{\substack{=\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \\\text{(by (\ref{2.6f.pf.short.1}%
))}}}\text{ is empty }\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }x\in\mathbb{R}%
^{n}\text{ satisfying }x\geq0\text{ and }Ax\leq b\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right)  .
\end{align*}
This proves (\ref{2.6f.pf.short.3}).}

Also, we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \text{the set }\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{
and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y\in\mathbb{R}%
^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  . \label{2.6f.pf.short.4}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.short.4}):} We have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{the set }\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{
and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }y\in\mathbb{R}%
^{m+n}\text{ satisfying }y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  }_{\substack{=\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \\\text{(by
(\ref{2.6f.pf.short.2}))}}}\text{ is empty }\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }y\in\mathbb{R}%
^{m}\text{ satisfying }y\geq0\text{ and }y^{T}A\geq c^{T}\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y\in\mathbb{R}%
^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  .
\end{align*}
This proves (\ref{2.6f.pf.short.4}).}

Now, we make the following four observations:

\begin{itemize}
\item Assertion Q$_{1}$1 is equivalent to Assertion
Q1.\footnote{\textit{Proof.} Assertion Q$_{1}$1 says that the sets $\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and \newline$\left\{  y\in\mathbb{R}^{m+n}\ \mid
\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ are empty. Thus, we have the following equivalence
of assertions:%
\begin{align*}
&  \ \left(  \text{Assertion Q}_{1}\text{1 holds}\right) \\
&  \Longleftrightarrow\ \underbrace{\left(  \text{the set }\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right)  }_{\substack{\Longleftrightarrow
\ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and
}Ax\leq b\right\}  \text{ is empty}\right)  \\\text{(by (\ref{2.6f.pf.short.3}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \underbrace{\left(  \text{the set }\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right)  }%
_{\substack{\Longleftrightarrow\ \left(  \text{the set }\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{
is empty}\right)  \\\text{(by (\ref{2.6f.pf.short.4}))}}}\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{Assertion Q1 holds}\right)
\end{align*}
(because Assertion Q1 says that the sets $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  $ and $\left\{  y\in\mathbb{R}%
^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ are empty). In other
words, Assertion Q$_{1}$1 is equivalent to Assertion Q1.}

\item Assertion Q$_{1}$2 is equivalent to Assertion
Q2.\footnote{\textit{Proof.} Assertion Q$_{1}$2 says that the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is unbounded from above, and that the set $\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is empty. Thus, we have the following equivalence
of assertions:%
\begin{align*}
&  \ \left(  \text{Assertion Q}_{1}\text{2 holds}\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  }_{\substack{=\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \\\text{(by (\ref{2.6f.pf.short.1}%
))}}}\text{ is unbounded from above}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \underbrace{\left(  \text{the set }\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right)  }%
_{\substack{\Longleftrightarrow\ \left(  \text{the set }\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{
is empty}\right)  \\\text{(by (\ref{2.6f.pf.short.4}))}}}\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is unbounded
from above}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{Assertion Q2 holds}\right)
\end{align*}
(since Assertion Q2 says that the set $\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $ is unbounded from
above, and that the set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  $ is empty). In other words, Assertion Q$_{1}$2 is
equivalent to Assertion Q2.}

\item Assertion Q$_{1}$3 is equivalent to Assertion
Q3.\footnote{\textit{Proof.} Assertion Q$_{1}$3 says that the set $\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is unbounded from below, and that the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is empty. Thus, we have the following equivalence of
assertions:%
\begin{align*}
&  \ \left(  \text{Assertion Q}_{1}\text{3 holds}\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  }_{\substack{=\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \\\text{(by
(\ref{2.6f.pf.short.2}))}}}\text{ is unbounded from below}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \underbrace{\left(  \text{the set }\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right)  }_{\substack{\Longleftrightarrow
\ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and
}Ax\leq b\right\}  \text{ is empty}\right)  \\\text{(by (\ref{2.6f.pf.short.3}%
))}}}\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
unbounded from below}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  x\in
\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{Assertion Q3 holds}\right)
\end{align*}
(since Assertion Q3 says that the set $\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ is unbounded
from below, and that the set $\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{
and }Ax\leq b\right\}  $ is empty). In other words, Assertion Q$_{1}$3 is
equivalent to Assertion Q3.}

\item Assertion Q$_{1}$4 is equivalent to Assertion
Q4.\footnote{\textit{Proof.} If we use (\ref{2.6f.pf.short.1}) to replace
every $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  $ by $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $, and if we use (\ref{2.6f.pf.short.2}) to replace every
$\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  $ by $\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $, then Assertion Q4 turns into Assertion Q$_{1}$4.
Hence, Assertion Q$_{1}$4 is equivalent to Assertion Q4.}
\end{itemize}

Combining these observations, we see that Assertions Q$_{1}$1, Q$_{1}$2,
Q$_{1}$3, Q$_{1}$4 are equivalent to Assertions Q1, Q2, Q3, Q4, respectively.
Hence, we have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{exactly one of Assertions Q}_{1}\text{1, Q}_{1}\text{2,
Q}_{1}\text{3, Q}_{1}\text{4 holds}\right) \\
&  \Longleftrightarrow\ \left(  \text{exactly one of Assertions Q1, Q2, Q3, Q4
holds}\right)  .
\end{align*}
Since we know that exactly one of Assertions Q$_{1}$1, Q$_{1}$2, Q$_{1}$3,
Q$_{1}$4 holds, this yields that exactly one of Assertions Q1, Q2, Q3, Q4
holds. In other words, Theorem 2.6f is proven.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem 2.6f.]For any nonnegative integer $\gamma$, let $I_{\gamma}$
denote the $\gamma\times\gamma$ identity matrix.

Let $0_{n}$ be the zero vector in $\mathbb{R}^{n}$.

Consider the block matrix $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  $; this is an $\left(  m+n\right)  \times n$-matrix.

Also, consider the block matrix $\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $; this is an $\left(  m+n\right)  \times1$-matrix, i. e., a vector
in $\mathbb{R}^{m+n}$ (since we identify matrices having only one column with
column vectors).

Applying Theorem 2.6c to $m+n$, $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  $ and $\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $ instead of $m$, $A$ and $b$, we see that exactly one of the
following four assertions holds:

\begin{statement}
\textit{Assertion Q}$_{1}$\textit{1:} The sets $\left\{  x\in\mathbb{R}%
^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and \newline$\left\{  y\in\mathbb{R}^{m+n}\ \mid
\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ are empty.

\textit{Assertion Q}$_{1}$\textit{2:} The set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is unbounded from above, and the set $\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is empty.

\textit{Assertion Q}$_{1}$\textit{3:} The set $\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is unbounded from below, and the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is empty.

\textit{Assertion Q}$_{1}$\textit{4:} The numbers $\max\left\{  c^{T}%
x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and $\min\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ exist and satisfy%
\begin{align*}
&  \max\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\} \\
&  =\min\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\end{align*}

\end{statement}

We will now prove that these assertions Q$_{1}$1, Q$_{1}$2, Q$_{1}$3, and
Q$_{1}$4 are equivalent to the assertions Q1, Q2, Q3, and Q4, respectively.

First, we notice that%
\begin{align}
&  \left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\} \nonumber\\
&  =\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  . \label{2.6f.pf.1}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.1}):} Let $\lambda\in\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ be arbitrary. Then, there exists an $x\in\mathbb{R}^{n}$
such that $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $ and $\lambda=c^{T}x$. Denote this $x$ by $u$. Thus, $u\in
\mathbb{R}^{n}$ satisfies $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  u\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $ and $\lambda=c^{T}u$.
\par
We have $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  u\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $. Since%
\begin{align*}
\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  u  &  =\left(
\begin{array}
[c]{c}%
Au\\
-I_{n}u
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for
block matrices}\right) \\
&  =\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }I_{n}u=u\right)  ,
\end{align*}
this rewrites as $\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  \leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $. In other words, every $i\in\left\{  1,2,\ldots,m+n\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  \right)  \leq\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right)  .
\]
Now, every $i\in\left\{  1,2,\ldots,m+n\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right)  -\underbrace{\left(  \text{the }i\text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  \right)  }_{\leq\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right)  -\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right) \\
&  =0=\left(  \text{the }i\text{-th coordinate of }0\right)
\end{align*}
(where the last $0$ denotes the zero vector in $\mathbb{R}^{m+n}$). Thus,
$\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  \geq0$. Since $\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
Au\\
-u
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
b-Au\\
0_{n}-\left(  -u\right)
\end{array}
\right)  $, this rewrites as $\left(
\begin{array}
[c]{c}%
b-Au\\
0_{n}-\left(  -u\right)
\end{array}
\right)  \geq0$. Thus, Lemma 2.5h \textbf{(b)} (applied to $N=m$, $M=n$,
$x=b-Au$ and $y=0_{n}-\left(  -u\right)  $) yields that $b-Au\geq0$ and
$0_{n}-\left(  -u\right)  \geq0$. The inequality $0_{n}-\left(  -u\right)
\geq0$ simplifies to $u\geq0$ (since $0_{n}-\left(  -u\right)  =u$).
\par
We have $b-Au\geq0$. Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }b-Au\right)  \geq0.
\]
Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
0  &  \leq\left(  \text{the }i\text{-th coordinate of }b-Au\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(  \text{the
}i\text{-th coordinate of }Au\right)  ,
\end{align*}
so that $\left(  \text{the }i\text{-th coordinate of }Au\right)  \leq\left(
\text{the }i\text{-th coordinate of }b\right)  $.
\par
We thus have shown that every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies
$\left(  \text{the }i\text{-th coordinate of }Au\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  $. In other words, $Au\leq b$.
\par
So we know that $u\in\mathbb{R}^{n}$, $u\geq0$ and $Au\leq b$. In other words,
$u\in\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  $,
so that%
\[
c^{T}u\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  .
\]
Since $c^{T}u=\lambda$, this rewrites as
\[
\lambda\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  .
\]
\par
Now, forget that we fixed $\lambda$. We thus have proven that every
$\lambda\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ satisfies $\lambda\in\left\{  c^{T}x\ \mid\ x\in
\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $. In other words,%
\begin{equation}
\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \subseteq\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n}%
;\ x\geq0\text{ and }Ax\leq b\right\}  . \label{2.6f.pf.1.pf.1}%
\end{equation}
\par
Now, let $\mu\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and
}Ax\leq b\right\}  $ be arbitrary. Then, there exists an $x\in\mathbb{R}^{n}$
such that $x\geq0$, $Ax\leq b$ and $\mu=c^{T}x$. Denote this $x$ by $v$. Thus,
$v\in\mathbb{R}^{n}$ satisfies $v\geq0$, $Av\leq b$ and $\mu=c^{T}v$.
\par
We have $Av\leq b$. Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }Av\right)  \leq\left(  \text{the
}i\text{-th coordinate of }b\right)  .
\]
Thus, every $i\in\left\{  1,2,\ldots,m\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }b-Av\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }b\right)
-\underbrace{\left(  \text{the }i\text{-th coordinate of }Av\right)  }%
_{\leq\left(  \text{the }i\text{-th coordinate of }b\right)  }\\
&  \geq\left(  \text{the }i\text{-th coordinate of }b\right)  -\left(
\text{the }i\text{-th coordinate of }b\right)  =0=\left(  \text{the
}i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $b-Av\geq0$.
\par
Lemma 2.5h \textbf{(a)} (applied to $N=m$, $M=n$, $x=b-Av$ and $y=v$) yields
that $\left(
\begin{array}
[c]{c}%
b-Av\\
v
\end{array}
\right)  \geq0$ (since $b-Av\geq0$ and $v\geq0$). In other words, every
$i\in\left\{  1,2,\ldots,m+n\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b-Av\\
v
\end{array}
\right)  \right)  \geq\left(  \text{the }i\text{-th coordinate of }0\right)
=0. \label{2.6f.pf.1.pf.2}%
\end{equation}
But on the other hand,
\begin{align*}
\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  v  &  =\left(
\begin{array}
[c]{c}%
Av\\
-I_{n}v
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for
block matrices}\right) \\
&  =\left(
\begin{array}
[c]{c}%
b-\left(  b-Av\right) \\
0_{n}-v
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }Av=b-\left(  b-Av\right)
\text{ and }-\underbrace{I_{n}v}_{=v}=-v=0_{n}-v\right) \\
&  =\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
b-Av\\
v
\end{array}
\right)  .
\end{align*}
Hence, for every $i\in\left\{  1,2,\ldots,m+n\right\}  $, we have%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }\underbrace{\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  v}_{=\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
b-Av\\
v
\end{array}
\right)  }\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  -\left(
\begin{array}
[c]{c}%
b-Av\\
v
\end{array}
\right)  \right) \\
&  =\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right)  -\underbrace{\left(  \text{the }i\text{-th coordinate of
}\left(
\begin{array}
[c]{c}%
b-Av\\
v
\end{array}
\right)  \right)  }_{\substack{\geq0\\\text{(by (\ref{2.6f.pf.1.pf.2}))}}}\\
&  \leq\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right)  .
\end{align*}
In other words, $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  v\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $.
\par
So we know that our $v\in\mathbb{R}^{n}$ satisfies $\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  v\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $. In other words, $v\in\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $, and thus
\[
c^{T}v\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  .
\]
Since $c^{T}v=\mu$, this rewrites as
\[
\mu\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  .
\]
\par
Now, forget that we fixed $\mu$. We thus have proven that every $\mu
\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  $ satisfies $\mu\in\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $. In other words,%
\[
\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\}  \subseteq\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  .
\]
Combined with (\ref{2.6f.pf.1.pf.1}), this yields%
\begin{equation}
\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  =\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{
and }Ax\leq b\right\}  .\nonumber
\end{equation}
This proves (\ref{2.6f.pf.1}).}

Next, we notice that%
\begin{align}
&  \left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\} \nonumber\\
&  =\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  . \label{2.6f.pf.2}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.2}):} Let $\lambda\in\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ be arbitrary. Then, there exists an $y\in
\mathbb{R}^{m+n}$ such that $y\geq0$, $y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$ and $\lambda=y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $. Denote this $y$ by $\xi$. Thus, $\xi\in\mathbb{R}^{m+n}$ satisfies
$\xi\geq0$, $\xi^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$ and $\lambda=\xi^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  $.
\par
Let us write the vector $\xi$ in the form $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, where $u\in\mathbb{R}^{m}$ and $v\in\mathbb{R}^{n}$.
\par
We have $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  =\xi\geq0$. Thus, Lemma 2.5h \textbf{(b)} (applied to $N=m$, $M=n$,
$x=u$ and $y=v$) yields that $u\geq0$ and $v\geq0$.
\par
Since $\xi=\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  $, we have $\xi^{T}=\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  $, and thus%
\begin{align*}
\xi^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right) \\
&  =u^{T}A+\underbrace{v^{T}\left(  -I_{n}\right)  }_{=-v^{T}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =u^{T}A-v^{T}.
\end{align*}
Compared with $\xi^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$, this yields $c^{T}=u^{T}A-v^{T}$. In other words,%
\[
u^{T}A=c^{T}+v^{T}.
\]
\par
But $v\geq0$. In other words, every $i\in\left\{  1,2,\ldots,n\right\}  $
satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }v\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0. \label{2.6f.pf.2.pf.1}%
\end{equation}
\par
Now, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }%
\underbrace{u^{T}A}_{=c^{T}+v^{T}}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }c^{T}%
+v^{T}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the row vector }c^{T}\right)
+\underbrace{\left(  \text{the }i\text{-th coordinate of the row vector }%
v^{T}\right)  }_{\substack{=\left(  \text{the }i\text{-th coordinate of
}v\right)  \geq0\\\text{(by (\ref{2.6f.pf.2.pf.1}))}}}\\
&  \geq\left(  \text{the }i\text{-th coordinate of the row vector }%
c^{T}\right)  .
\end{align*}
In other words, $u^{T}A\geq c^{T}$.
\par
So we now know that $u\in\mathbb{R}^{m}$, $u\geq0$ and $u^{T}A\geq c^{T}$. In
other words, $u\in\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A\geq c^{T}\right\}  $, so that%
\begin{equation}
u^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }%
y^{T}A\geq c^{T}\right\}  . \label{2.6f.pf.2.pf.2}%
\end{equation}
Now,
\begin{align*}
\lambda &  =\underbrace{\xi^{T}}_{=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right) \\
&  =u^{T}b+\underbrace{v^{T}0_{n}}_{=0}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the multiplication rule for block matrices}\right) \\
&  =u^{T}b\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{2.6f.pf.2.pf.2})}\right)  .
\end{align*}
\par
Now, forget that we fixed $\lambda$. We thus have proven that every
$\lambda\in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ satisfies $\lambda\in\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $. In other
words,%
\begin{align}
&  \left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\} \nonumber\\
&  \subseteq\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  . \label{2.6f.pf.2.pf.3}%
\end{align}
\par
Now, let $\mu\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  $ be arbitrary. Then, there exists a $y\in
\mathbb{R}^{m}$ such that $y\geq0$, $y^{T}A\geq c^{T}$ and $\mu=y^{T}b$.
Denote this $y$ by $u$. Thus, $u\in\mathbb{R}^{m}$ satisfies $u\geq0$,
$u^{T}A\geq c^{T}$ and $\mu=u^{T}b$.
\par
We have $u^{T}A\geq c^{T}$. Thus, every $i\in\left\{  1,2,\ldots,n\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of the row vector }u^{T}A\right)
\geq\left(  \text{the }i\text{-th coordinate of the row vector }c^{T}\right)
.
\]
Thus, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of the row vector }u^{T}%
A-c^{T}\right) \\
&  =\underbrace{\left(  \text{the }i\text{-th coordinate of the row vector
}u^{T}A\right)  }_{\geq\left(  \text{the }i\text{-th coordinate of the row
vector }c^{T}\right)  }-\left(  \text{the }i\text{-th coordinate of the row
vector }c^{T}\right) \\
&  \geq\left(  \text{the }i\text{-th coordinate of the row vector }%
c^{T}\right)  -\left(  \text{the }i\text{-th coordinate of the row vector
}c^{T}\right) \\
&  =0=\left(  \text{the }i\text{-th coordinate of the row vector }0\right)  .
\end{align*}
In other words, $u^{T}A-c^{T}\geq0$. Thus, $\left(  u^{T}A-c^{T}\right)
^{T}\geq0$ (since the transpose of any nonnegative vector is nonnegative).
\par
Let $v$ be the vector $\left(  u^{T}A-c^{T}\right)  ^{T}$. Then, $v=\left(
u^{T}A-c^{T}\right)  ^{T}\geq0$ and $v^{T}=\left(  \left(  u^{T}%
A-c^{T}\right)  ^{T}\right)  ^{T}=u^{T}A-c^{T}$.
\par
Lemma 2.5h \textbf{(a)} (applied to $N=m$, $M=n$, $x=u$ and $y=v$) yields that
$\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \geq0$ (since $u\geq0$ and $v\geq0$). But on the other hand,%
\begin{align*}
\underbrace{\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}}_{=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  }\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =u^{T}A+\underbrace{v^{T}\left(  -I_{n}\right)  }_{=-v^{T}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =u^{T}A-\underbrace{v^{T}}_{=u^{T}A-c^{T}}=u^{T}A-\left(  u^{T}%
A-c^{T}\right)  =c^{T}.
\end{align*}
\par
Now, we know that our $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \in\mathbb{R}^{m+n}$ satisfies $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \geq0$ and $\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}$. In other words,%
\[
\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  \in\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }%
y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  ,
\]
and thus%
\[
\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\]
Since%
\begin{align*}
\underbrace{\left(
\begin{array}
[c]{c}%
u\\
v
\end{array}
\right)  ^{T}}_{=\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  }\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
u^{T} & v^{T}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  =\underbrace{u^{T}b}_{=\mu}+\underbrace{v^{T}0_{n}}_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the multiplication rule for block
matrices}\right) \\
&  =\mu,
\end{align*}
this rewrites as%
\[
\mu\in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\]
\par
Now, forget that we fixed $\mu$. We thus have proven that every $\mu
\in\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  $ satisfies \newline$\mu\in\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $. In other words,%
\begin{align*}
&  \left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\} \\
&  \subseteq\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  .
\end{align*}
Combined with (\ref{2.6f.pf.2.pf.3}), this yields%
\begin{align*}
&  \left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\} \\
&  =\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  .
\end{align*}
This proves (\ref{2.6f.pf.2}).}

Next, we notice that we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right)  .
\label{2.6f.pf.3}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.3}):} We have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }x\in\mathbb{R}%
^{n}\text{ satisfying }\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  }_{\substack{=\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \\\text{(by (\ref{2.6f.pf.1}))}%
}}\text{ is empty }\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }x\in\mathbb{R}%
^{n}\text{ satisfying }x\geq0\text{ and }Ax\leq b\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right)  .
\end{align*}
This proves (\ref{2.6f.pf.3}).}

Also, we have the following equivalence of assertions:%
\begin{align}
&  \ \left(  \text{the set }\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{
and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y\in\mathbb{R}%
^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  . \label{2.6f.pf.4}%
\end{align}
\footnote{\textit{Proof of (\ref{2.6f.pf.4}):} We have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{the set }\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{
and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }y\in\mathbb{R}%
^{m+n}\text{ satisfying }y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  }_{\substack{=\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \\\text{(by
(\ref{2.6f.pf.2}))}}}\text{ is empty }\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists no }y\in\mathbb{R}%
^{m}\text{ satisfying }y\geq0\text{ and }y^{T}A\geq c^{T}\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y\in\mathbb{R}%
^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  .
\end{align*}
This proves (\ref{2.6f.pf.4}).}

Now, we make the following four observations:

\begin{itemize}
\item Assertion Q$_{1}$1 is equivalent to Assertion
Q1.\footnote{\textit{Proof.} Assertion Q$_{1}$1 says that the sets $\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and \newline$\left\{  y\in\mathbb{R}^{m+n}\ \mid
\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ are empty. Thus, we have the following equivalence
of assertions:%
\begin{align}
&  \ \left(  \text{Assertion Q}_{1}\text{1 holds}\right) \nonumber\\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{c}%
\text{the sets }\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ and}\\
\text{ }\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ are empty}%
\end{array}
\right) \nonumber\\
&  \Longleftrightarrow\ \underbrace{\left(  \text{the set }\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right)  }_{\substack{\Longleftrightarrow
\ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and
}Ax\leq b\right\}  \text{ is empty}\right)  \\\text{(by (\ref{2.6f.pf.3}))}%
}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \underbrace{\left(  \text{the set }\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right)  }%
_{\substack{\Longleftrightarrow\ \left(  \text{the set }\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{
is empty}\right)  \\\text{(by (\ref{2.6f.pf.4}))}}}\nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  . \label{2.6f.pf.5}%
\end{align}
\par
On the other hand, Assertion Q1 says that the sets $\left\{  x\in
\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  $ and $\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ are
empty. In other words, we have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{Assertion Q1 holds}\right) \\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{c}%
\text{the sets }\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq
b\right\}  \text{ and}\\
\text{ }\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\}  \text{ are empty}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  x\in\mathbb{R}%
^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is empty}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  .
\end{align*}
Compared with (\ref{2.6f.pf.5}), this yields the following equivalence of
assertions:%
\[
\left(  \text{Assertion Q}_{1}\text{1 holds}\right)  \Longleftrightarrow
\left(  \text{Assertion Q1 holds}\right)  .
\]
In other words, Assertion Q$_{1}$1 is equivalent to Assertion Q1, qed.}

\item Assertion Q$_{1}$2 is equivalent to Assertion
Q2.\footnote{\textit{Proof.} Assertion Q$_{1}$2 says that the set $\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is unbounded from above, and that the set $\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is empty. Thus, we have the following equivalence
of assertions:%
\begin{align}
&  \ \left(  \text{Assertion Q}_{1}\text{2 holds}\right) \nonumber\\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{c}%
\text{the set }\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is unbounded from above,}\\
\text{and the set }\left\{  y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and
}y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}%
\end{array}
\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  }_{\substack{=\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \\\text{(by (\ref{2.6f.pf.1}))}%
}}\text{ is unbounded from above}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \underbrace{\left(  \text{the set }\left\{
y\in\mathbb{R}^{m+n}\ \mid\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is empty}\right)  }%
_{\substack{\Longleftrightarrow\ \left(  \text{the set }\left\{
y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{
is empty}\right)  \\\text{(by (\ref{2.6f.pf.4}))}}}\nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is unbounded
from above}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  . \label{2.6f.pf.6}%
\end{align}
\par
On the other hand, Assertion Q2 says that the set $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $ is unbounded from
above, and that the set $\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  $ is empty. In other words, we have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{Assertion Q2 holds}\right) \\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{c}%
\text{the set }\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and
}Ax\leq b\right\}  \text{ is unbounded from above,}\\
\text{and the set }\left\{  y\in\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }%
y^{T}A\geq c^{T}\right\}  \text{ is empty}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is unbounded
from above}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  y\in
\mathbb{R}^{m}\ \mid\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
empty}\right)  .
\end{align*}
Compared with (\ref{2.6f.pf.6}), this yields the following equivalence of
assertions:%
\[
\left(  \text{Assertion Q}_{1}\text{2 holds}\right)  \Longleftrightarrow
\left(  \text{Assertion Q2 holds}\right)  .
\]
In other words, Assertion Q$_{1}$2 is equivalent to Assertion Q2, qed.}

\item Assertion Q$_{1}$3 is equivalent to Assertion
Q3.\footnote{\textit{Proof.} Assertion Q$_{1}$3 says that the set $\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is unbounded from below, and that the set $\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ is empty. Thus, we have the following equivalence of
assertions:%
\begin{align}
&  \ \left(  \text{Assertion Q}_{1}\text{3 holds}\right) \nonumber\\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{c}%
\text{the set }\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  \text{ is unbounded from below,}\\
\text{and the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}%
\end{array}
\right) \nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\underbrace{\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  }_{\substack{=\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \\\text{(by
(\ref{2.6f.pf.2}))}}}\text{ is unbounded from below}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \underbrace{\left(  \text{the set }\left\{
x\in\mathbb{R}^{n}\ \mid\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  \text{ is empty}\right)  }_{\substack{\Longleftrightarrow
\ \left(  \text{the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and
}Ax\leq b\right\}  \text{ is empty}\right)  \\\text{(by (\ref{2.6f.pf.3}))}%
}}\nonumber\\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
unbounded from below}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  x\in
\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is
empty}\right)  . \label{2.6f.pf.7}%
\end{align}
\par
On the other hand, Assertion Q3 says that the set $\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ is
unbounded from below, and that the set $\left\{  x\in\mathbb{R}^{n}%
\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  $ is empty. In other words, we
have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{Assertion Q3 holds}\right) \\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{c}%
\text{the set }\left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and
}y^{T}A\geq c^{T}\right\}  \text{ is unbounded from below,}\\
\text{and the set }\left\{  x\in\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq
b\right\}  \text{ is empty}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \text{the set }\left\{  y^{T}b\ \mid
\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  \text{ is
unbounded from below}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \wedge\ \left(  \text{the set }\left\{  x\in
\mathbb{R}^{n}\ \mid\ x\geq0\text{ and }Ax\leq b\right\}  \text{ is
empty}\right)  .
\end{align*}
Compared with (\ref{2.6f.pf.7}), this yields the following equivalence of
assertions:%
\[
\left(  \text{Assertion Q}_{1}\text{3 holds}\right)  \Longleftrightarrow
\left(  \text{Assertion Q3 holds}\right)  .
\]
In other words, Assertion Q$_{1}$3 is equivalent to Assertion Q3, qed.}

\item Assertion Q$_{1}$4 is equivalent to Assertion
Q4.\footnote{\textit{Proof.} We have%
\begin{align*}
&  \left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq
b\right\} \\
&  =\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}
\end{align*}
(by (\ref{2.6f.pf.1})) and
\begin{align*}
&  \left\{  y^{T}b\ \mid\ y\in\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq
c^{T}\right\} \\
&  =\left\{  y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}
\end{align*}
(by (\ref{2.6f.pf.2})). Hence, Assertion Q4 is equivalent to the assertion
that we obtain from Assertion Q4 by replacing every $\left\{  c^{T}%
x\ \mid\ x\in\mathbb{R}^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $ by
$\left\{  c^{T}x\ \mid\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and replacing every $\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ by $\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $. But since the assertion that we obtain from
Assertion Q4 by replacing every $\left\{  c^{T}x\ \mid\ x\in\mathbb{R}%
^{n};\ x\geq0\text{ and }Ax\leq b\right\}  $ by $\left\{  c^{T}x\ \mid
\ x\in\mathbb{R}^{n};\ \left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  x\leq\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \right\}  $ and replacing every $\left\{  y^{T}b\ \mid\ y\in
\mathbb{R}^{m};\ y\geq0\text{ and }y^{T}A\geq c^{T}\right\}  $ by $\left\{
y^{T}\left(
\begin{array}
[c]{c}%
b\\
0_{n}%
\end{array}
\right)  \ \mid\ y\in\mathbb{R}^{m+n};\ y\geq0\text{ and }y^{T}\left(
\begin{array}
[c]{c}%
A\\
-I_{n}%
\end{array}
\right)  =c^{T}\right\}  $ is exactly Assertion Q$_{1}$4, this rewrites as
follows: Assertion Q4 is equivalent to Assertion Q$_{1}$4. In other words,
Assertion Q$_{1}$4 is equivalent to Assertion Q4, qed.}
\end{itemize}

Combining these observations, we see that Assertions Q$_{1}$1, Q$_{1}$2,
Q$_{1}$3, Q$_{1}$4 are equivalent to Assertions Q1, Q2, Q3, Q4, respectively.
Hence, we have the following equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{exactly one of Assertions Q}_{1}\text{1, Q}_{1}\text{2,
Q}_{1}\text{3, Q}_{1}\text{4 holds}\right) \\
&  \Longleftrightarrow\ \left(  \text{exactly one of Assertions Q1, Q2, Q3, Q4
holds}\right)  .
\end{align*}
Since we know that exactly one of Assertions Q$_{1}$1, Q$_{1}$2, Q$_{1}$3,
Q$_{1}$4 holds, this yields that exactly one of Assertions Q1, Q2, Q3, Q4
holds. In other words, Theorem 2.6f is proven.
\end{proof}
\end{verlong}

\section{\label{sect.basics-proofs}Appendix: Proofs omitted from the early
sections}

We have promised to give proofs for various statements made in Section
\ref{sect.basics-conv}, in Section \ref{sect.basics-cone}, in Section
\ref{sect.basics-dual}, in Section \ref{sect.closed-int}, and in Section
\ref{sect.lt-gt}. Let us now fulfill this promise.

\subsection{Proofs for Section \ref{sect.basics-conv}}

\begin{proof}
[Proof of Proposition 2.0a.]We will first show that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }J\text{ is any finite subset of }I\text{, and if }\left(  \mu
_{i}\right)  _{i\in J}\text{ is a}\\
\text{family of nonnegative reals indexed by }J\text{ such}\\
\text{that }\sum\limits_{i\in J}\mu_{i}=1\text{, then }\sum\limits_{i\in J}%
\mu_{i}x_{i}\in C
\end{array}
\right)  . \label{pf.prop2.0a.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.prop2.0a.goal}):} We shall prove
(\ref{pf.prop2.0a.goal}) by strong induction over $\left\vert J\right\vert $:

\textit{Induction step:}\footnote{A strong induction needs no induction base.}
Fix $N\in\mathbb{N}$. Assume that (\ref{pf.prop2.0a.goal}) holds in the case
when $\left\vert J\right\vert <N$. We must prove that (\ref{pf.prop2.0a.goal})
holds in the case when $\left\vert J\right\vert =N$.

We have assumed that (\ref{pf.prop2.0a.goal}) holds in the case when
$\left\vert J\right\vert <N$. In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }J\text{ is any finite subset of }I\text{ satisfying }\left\vert
J\right\vert <N\text{, and}\\
\text{if }\left(  \mu_{i}\right)  _{i\in J}\text{ is a family of nonnegative
reals indexed by }J\\
\text{such that }\sum\limits_{i\in J}\mu_{i}=1\text{, then }\sum\limits_{i\in
J}\mu_{i}x_{i}\in C
\end{array}
\right)  . \label{pf.prop2.0a.indhyp}%
\end{equation}


Now, let us prove that (\ref{pf.prop2.0a.goal}) holds in the case when
$\left\vert J\right\vert =N$. Thus, let $J$ be any finite subset of $I$
satisfying $\left\vert J\right\vert =N$, and let $\left(  \mu_{i}\right)
_{i\in J}$ be a family of nonnegative reals indexed by $J$ such that
$\sum\limits_{i\in J}\mu_{i}=1$. We shall show that $\sum\limits_{i\in J}%
\mu_{i}x_{i}\in C$.

\begin{vershort}
If $\left\vert J\right\vert \leq1$, then $\sum\limits_{i\in J}\mu_{i}x_{i}\in
C$ holds\footnote{\textit{Proof.} Assume that $\left\vert J\right\vert \leq1$.
\par
The sum $\sum\limits_{i\in J}\mu_{i}$ cannot be empty (since $\sum
\limits_{i\in J}\mu_{i}=1\neq0$). Hence, $J$ cannot be the empty set. Thus,
$\left\vert J\right\vert \geq1$. Therefore, $\left\vert J\right\vert =1$
(since $\left\vert J\right\vert \leq1$). In other words, $J=\left\{
k\right\}  $ for some $k\in J$. Consider this $k$. We have $k\in J\subseteq I$
and thus $x_{k}\in C$. Now, from $J=\left\{  k\right\}  $, we obtain
$\sum\limits_{i\in J}\mu_{i}=\sum\limits_{i\in\left\{  k\right\}  }\mu_{i}%
=\mu_{k}$, so that $\mu_{k}=\sum\limits_{i\in J}\mu_{i}=1$. From $J=\left\{
k\right\}  $, we also obtain $\sum\limits_{i\in J}\mu_{i}x_{i}=\sum
\limits_{i\in\left\{  k\right\}  }\mu_{i}x_{i}=\underbrace{\mu_{k}}_{=1}%
x_{k}=x_{k}\in C$, qed.}. Hence, for the rest of our proof of $\sum
\limits_{i\in J}\mu_{i}x_{i}\in C$, we WLOG assume that we don't have
$\left\vert J\right\vert \leq1$.
\end{vershort}

\begin{verlong}
If $\left\vert J\right\vert \leq1$, then $\sum\limits_{i\in J}\mu_{i}x_{i}\in
C$ holds\footnote{\textit{Proof.} Assume that $\left\vert J\right\vert \leq1$.
\par
If we had $J=\varnothing$, then we would have $\sum\limits_{i\in J}\mu
_{i}=\left(  \text{empty sum}\right)  =0$, which would contradict
$\sum\limits_{i\in J}\mu_{i}=1\neq0$. Hence, we cannot have $J=\varnothing$.
In other words, we have $J\neq\varnothing$. Therefore, $\left\vert
J\right\vert \geq1$. Combined with $\left\vert J\right\vert \leq1$, this
yields $\left\vert J\right\vert =1$. Hence, $J$ is a one-element set. In other
words, $J=\left\{  k\right\}  $ for some $k\in J$. Consider this $k$. We have
$k\in J\subseteq I$ and thus $x_{k}\in C$ (since $\left(  x_{i}\right)  _{i\in
I}$ is a family of elements of $C$). Now, from $J=\left\{  k\right\}  $, we
obtain $\sum\limits_{i\in J}\mu_{i}=\sum\limits_{i\in\left\{  k\right\}  }%
\mu_{i}=\mu_{k}$, so that $\mu_{k}=\sum\limits_{i\in J}\mu_{i}=1$. From
$J=\left\{  k\right\}  $, we also obtain $\sum\limits_{i\in J}\mu_{i}%
x_{i}=\sum\limits_{i\in\left\{  k\right\}  }\mu_{i}x_{i}=\underbrace{\mu_{k}%
}_{=1}x_{k}=x_{k}\in C$, qed.}. Hence, for the rest of our proof of
$\sum\limits_{i\in J}\mu_{i}x_{i}\in C$, we can WLOG assume that we don't have
$\left\vert J\right\vert \leq1$. Assume this.
\end{verlong}

\begin{vershort}
There exists a $k\in J$ satisfying $\mu_{k}<1$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, every $k\in J$
satisfies $\mu_{k}\geq1$. In other words, every $i\in J$ satisfies $\mu
_{i}\geq1$. Now, $\sum\limits_{i\in J}\mu_{i}=1$, so that $1=\sum\limits_{i\in
J}\underbrace{\mu_{i}}_{\geq1}\geq\sum\limits_{k\in J}1=\left\vert
J\right\vert \cdot1=\left\vert J\right\vert $. In other words, $\left\vert
J\right\vert \leq1$. This contradicts the fact that we don't have $\left\vert
J\right\vert \leq1$. This contradiction proves that our assumption was wrong,
qed.}. Consider such a $k$. We have $\mu_{k}<1$, thus $1-\mu_{k}>0$.
\end{vershort}

\begin{verlong}
There exists a $k\in J$ satisfying $\mu_{k}<1$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, every $k\in J$
satisfies $\mu_{k}\geq1$. Now, $\sum\limits_{i\in J}\mu_{i}=1$, so that%
\begin{align*}
1  &  =\sum\limits_{i\in J}\mu_{i}=\sum\limits_{k\in J}\underbrace{\mu_{k}%
}_{\geq1}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the summation
index }i\text{ as }k\right) \\
&  \geq\sum\limits_{k\in J}1=\left\vert J\right\vert \cdot1=\left\vert
J\right\vert .
\end{align*}
In other words, $\left\vert J\right\vert \leq1$. This contradicts the fact
that we don't have $\left\vert J\right\vert \leq1$. This contradiction proves
that our assumption was wrong, qed.}. Consider such a $k$. We have $\mu_{k}%
<1$, thus $1-\mu_{k}>0$.
\end{verlong}

For every $i\in J\setminus\left\{  k\right\}  $, set $\mu_{i}^{\prime}%
=\dfrac{\mu_{i}}{1-\mu_{k}}$. This is a nonnegative real (since $\mu_{i}$ is a
nonnegative real, and since $1-\mu_{k}>0$). Hence, $\left(  \mu_{i}^{\prime
}\right)  _{i\in J\setminus\left\{  k\right\}  }$ is a family of nonnegative
reals indexed by $J\setminus\left\{  k\right\}  $. Moreover, $k\in J$ and thus
$\left\vert J\setminus\left\{  k\right\}  \right\vert =\underbrace{\left\vert
J\right\vert }_{=N}-1=N-1<N$. Furthermore,%
\[
1=\sum\limits_{i\in J}\mu_{i}=\underbrace{\sum\limits_{\substack{i\in
J;\\i\neq k}}}_{=\sum\limits_{i\in J\setminus\left\{  k\right\}  }}\mu
_{i}+\underbrace{\sum\limits_{\substack{i\in J;\\i=k}}\mu_{i}}_{\substack{=\mu
_{k}\\\text{(since }k\in J\text{)}}}=\sum_{i\in J\setminus\left\{  k\right\}
}\mu_{i}+\mu_{k}.
\]
Solving this equation for $\sum_{i\in J\setminus\left\{  k\right\}  }\mu_{i}$
gives us $\sum_{i\in J\setminus\left\{  k\right\}  }\mu_{i}=1-\mu_{k}$. Hence,
$\dfrac{\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}}{1-\mu_{k}%
}=1$. Now,%
\[
\sum\limits_{i\in J\setminus\left\{  k\right\}  }\underbrace{\mu_{i}^{\prime}%
}_{=\dfrac{\mu_{i}}{1-\mu_{k}}}=\sum\limits_{i\in J\setminus\left\{
k\right\}  }\dfrac{\mu_{i}}{1-\mu_{k}}=\dfrac{\sum\limits_{i\in J\setminus
\left\{  k\right\}  }\mu_{i}}{1-\mu_{k}}=1.
\]
Thus, we can apply (\ref{pf.prop2.0a.indhyp}) to $J\setminus\left\{
k\right\}  $ and $\left(  \mu_{i}^{\prime}\right)  _{i\in J\setminus\left\{
k\right\}  }$ instead of $J$ and $\left(  \mu_{i}\right)  _{i\in J}$. As a
result, we obtain $\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu
_{i}^{\prime}x_{i}\in C$.

On the other hand,%
\begin{equation}
\sum\limits_{i\in J}\mu_{i}x_{i}=\underbrace{\sum\limits_{\substack{i\in
J;\\i=k}}\mu_{i}}_{\substack{=\mu_{k}x_{k}\\\text{(since }k\in J\text{)}%
}}+\underbrace{\sum\limits_{\substack{i\in J;\\i\neq k}}}_{=\sum\limits_{i\in
J\setminus\left\{  k\right\}  }}\mu_{i}x_{i}=\mu_{k}x_{k}+\sum\limits_{i\in
J\setminus\left\{  k\right\}  }\mu_{i}x_{i}. \label{pf.prop2.0a.3}%
\end{equation}


\begin{vershort}
We have $k\in J\subseteq I$ and thus $x_{k}\in C$. Also, $\mu_{k}\in\left[
0,1\right]  $ (since $\mu_{k}$ is a nonnegative real and satisfies $\mu_{k}<1$).
\end{vershort}

\begin{verlong}
We have $k\in J\subseteq I$ and thus $x_{k}\in C$ (since $\left(
x_{i}\right)  _{i\in I}$ is a family of elements of $C$). Also, $\mu_{k}$ is a
nonnegative real (since $\left(  \mu_{i}\right)  _{i\in J}$ is a family of
nonnegative reals), so that we have $\mu_{k}\geq0$. On the other hand,
$\mu_{k}\leq1$ (since $\mu_{k}<1$). Combined with $\mu_{k}\geq0$, this leads
to $\mu_{k}\in\left[  0,1\right]  $.
\end{verlong}

Now, recall that the set $C$ is convex. In other words, every two elements
$x\in C$ and $y\in C$ and every real number $\lambda\in\left[  0,1\right]  $
satisfy $\lambda x+\left(  1-\lambda\right)  y\in C$ (because of the
definition of convexity). Applying this to $x=x_{k}$, $y=\sum\limits_{i\in
J\setminus\left\{  k\right\}  }\mu_{i}^{\prime}x_{i}$ and $\lambda=\mu_{k}$
gives us $\mu_{k}x_{k}+\left(  1-\mu_{k}\right)  \sum\limits_{i\in
J\setminus\left\{  k\right\}  }\mu_{i}^{\prime}x_{i}\in C$ (since $x_{k}\in
C$, $\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}^{\prime}x_{i}\in
C$ and $\mu_{k}\in\left[  0,1\right]  $). Since%
\begin{align*}
&  \mu_{k}x_{k}+\left(  1-\mu_{k}\right)  \sum\limits_{i\in J\setminus\left\{
k\right\}  }\underbrace{\mu_{i}^{\prime}}_{=\dfrac{\mu_{i}}{1-\mu_{k}}}x_{i}\\
&  =\mu_{k}x_{k}+\left(  1-\mu_{k}\right)  \underbrace{\sum\limits_{i\in
J\setminus\left\{  k\right\}  }\dfrac{\mu_{i}}{1-\mu_{k}}x_{i}}_{=\dfrac
{1}{1-\mu_{k}}\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}x_{i}%
}=\mu_{k}x_{k}+\underbrace{\left(  1-\mu_{k}\right)  \dfrac{1}{1-\mu_{k}}%
}_{=1}\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}x_{i}\\
&  =\mu_{k}x_{k}+\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}%
x_{i}=\sum\limits_{i\in J}\mu_{i}x_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop2.0a.3})}\right)  ,
\end{align*}
this rewrites as $\sum\limits_{i\in J}\mu_{i}x_{i}\in C$.

Now, let us forget that we fixed $J$ and $\left(  \mu_{i}\right)  _{i\in J}$.
We thus have proven that if $J$ is any finite subset of $I$ satisfying
$\left\vert J\right\vert =N$, and if $\left(  \mu_{i}\right)  _{i\in J}$ is a
family of nonnegative reals indexed by $J$ such that $\sum\limits_{i\in J}%
\mu_{i}=1$, then $\sum\limits_{i\in J}\mu_{i}x_{i}\in C$. In other words,
(\ref{pf.prop2.0a.goal}) holds in the case when $\left\vert J\right\vert =N$.
This completes the induction step.

Thus, (\ref{pf.prop2.0a.goal}) is proven by induction.]

Now, recall that all but finitely many $i\in I$ satisfy $\lambda_{i}=0$. In
other words, there exists a finite subset $J$ of $I$ such that%
\begin{equation}
\text{every }i\in I\setminus J\text{ satisfies }\lambda_{i}=0.
\label{pf.prop2.0a.cofin}%
\end{equation}
Consider this $J$. We have $\sum\limits_{i\in I}\lambda_{i}=1$. Since%
\begin{align*}
\sum\limits_{i\in I}\lambda_{i}  &  =\underbrace{\sum\limits_{\substack{i\in
I;\\i\in J}}}_{\substack{=\sum\limits_{i\in J}\\\text{(since }J\subseteq
I\text{)}}}\lambda_{i}+\underbrace{\sum\limits_{\substack{i\in I;\\i\notin
J}}}_{=\sum\limits_{i\in I\setminus J}}\lambda_{i}=\sum\limits_{i\in J}%
\lambda_{i}+\sum\limits_{i\in I\setminus J}\underbrace{\lambda_{i}%
}_{\substack{=0\\\text{(by (\ref{pf.prop2.0a.cofin}))}}}\\
&  =\sum\limits_{i\in J}\lambda_{i}+\underbrace{\sum\limits_{i\in I\setminus
J}0}_{=0}=\sum\limits_{i\in J}\lambda_{i},
\end{align*}
this rewrites as $\sum\limits_{i\in J}\lambda_{i}=1$. Thus,
(\ref{pf.prop2.0a.goal}) (applied to $\left(  \mu_{i}\right)  _{i\in
J}=\left(  \lambda_{i}\right)  _{i\in J}$) shows that $\sum\limits_{i\in
J}\lambda_{i}x_{i}\in C$. Now,%
\begin{align*}
\sum\limits_{i\in I}\lambda_{i}x_{i}  &  =\underbrace{\sum
\limits_{\substack{i\in I;\\i\in J}}}_{\substack{=\sum\limits_{i\in
J}\\\text{(since }J\subseteq I\text{)}}}\lambda_{i}x_{i}+\underbrace{\sum
\limits_{\substack{i\in I;\\i\notin J}}}_{=\sum\limits_{i\in I\setminus J}%
}\lambda_{i}x_{i}=\sum\limits_{i\in J}\lambda_{i}x_{i}+\sum\limits_{i\in
I\setminus J}\underbrace{\lambda_{i}}_{\substack{=0\\\text{(by
(\ref{pf.prop2.0a.cofin}))}}}\\
&  =\sum\limits_{i\in J}\lambda_{i}x_{i}+\underbrace{\sum\limits_{i\in
I\setminus J}0x_{i}}_{=0}=\sum\limits_{i\in J}\lambda_{i}x_{i}\in C.
\end{align*}
This proves Proposition 2.0a.
\end{proof}

\begin{proof}
[Proof of Proposition 2.0f.]\textbf{(a)} Let $E$ be an $\mathbb{R}$-vector
space. Let $S$ be a subset of $E$.

Let $C_{1}$ be the intersection of all convex subsets of $E$ which contain $S$
as a subset. Definition 2.0c defined the convex hull of $S$ to be this set
$C_{1}$. In other words,%
\begin{equation}
\left(  \text{the convex hull of }S\text{ defined according to Definition
2.0c}\right)  =C_{1}. \label{pf.prop2.0f.a.chull1}%
\end{equation}


Let $C_{2}$ be the set of all convex combinations of the vectors $s$ for $s\in
S$. Definition 2.0d defined the convex hull of $S$ to be this set $C_{2}$. In
other words,%
\begin{equation}
\left(  \text{the convex hull of }S\text{ defined according to Definition
2.0d}\right)  =C_{2}. \label{pf.prop2.0f.a.chull2}%
\end{equation}


Let $C_{3}$ be the set%
\[
\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
Definition 2.0e defined the convex hull of $S$ to be this set $C_{3}$. In
other words,%
\begin{equation}
\left(  \text{the convex hull of }S\text{ defined according to Definition
2.0e}\right)  =C_{3}. \label{pf.prop2.0f.a.chull3}%
\end{equation}


\begin{vershort}
The set $C_{3}$ is convex\footnote{\textit{Proof.} Let $y\in C_{3}$, $z\in
C_{3}$ and $\lambda\in\left[  0,1\right]  $. We shall show that $\lambda
y+\left(  1-\lambda\right)  z\in C_{3}$.
\par
Notice that $\lambda$ and $1-\lambda$ both are nonnegative reals (since
$\lambda\in\left[  0,1\right]  $).
\par
We have%
\[
y\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, we can write $y$ in the form $y=\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}$ for some $t\in\mathbb{N}$, some $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and some $t$-tuple
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative
reals such that $\sum\limits_{i=1}^{t}\lambda_{i}=1$. Let us denote this $t$,
this $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by $p$, $\left(  y_{1}%
,y_{2},\ldots,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{p}\right)  $, respectively. Thus, $p$ is an element of $\mathbb{N}$, and
$\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$,
and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{p}\mu_{i}=1$ and
$y=\sum\limits_{i=1}^{p}\mu_{i}y_{i}$.
\par
Similarly, use the assumption $z\in C_{3}$ to write $z$ in the form
$z=\sum\limits_{i=1}^{q}\nu_{i}z_{i}$, where $q$ is an element of $\mathbb{N}%
$, and where $\left(  z_{1},z_{2},\ldots,z_{q}\right)  $ is a $q$-tuple of
elements of $S$, and where $\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $
is a $q$-tuple of nonnegative reals such that $\sum\limits_{i=1}^{q}\nu_{i}%
=1$.
\par
Define a $\left(  p+q\right)  $-tuple $\left(  w_{1},w_{2},\ldots
,w_{p+q}\right)  $ of elements of $S$ by%
\[
\left(  w_{1},w_{2},\ldots,w_{p+q}\right)  =\left(  y_{1},y_{2},\ldots
,y_{p},z_{1},z_{2},\ldots,z_{q}\right)  .
\]
Then:
\par
\begin{itemize}
\item For every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{equation}
w_{i}=y_{i}. \label{pf.prop2.0f.short.wi.1}%
\end{equation}
\par
\item For every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{equation}
w_{i}=z_{i-p}. \label{pf.prop2.0f.short.wi.2}%
\end{equation}
\end{itemize}
\par
Furthermore, define a $\left(  p+q\right)  $-tuple $\left(  \eta_{1},\eta
_{2},\ldots,\eta_{p+q}\right)  $ of nonnegative reals by%
\[
\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  =\left(  \lambda\mu
_{1},\lambda\mu_{2},\ldots,\lambda\mu_{p},\left(  1-\lambda\right)  \nu
_{1},\left(  1-\lambda\right)  \nu_{2},\ldots,\left(  1-\lambda\right)
\nu_{q}\right)
\]
(this is well-defined because both $\lambda$ and $1-\lambda$ are nonnegative
reals and because $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ and
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $ are tuples of nonnegative
reals). Then:
\par
\begin{itemize}
\item For every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{equation}
\eta_{i}=\lambda\mu_{i}. \label{pf.prop2.0f.short.etai.1}%
\end{equation}
\par
\item For every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{equation}
\eta_{i}=\left(  1-\lambda\right)  \nu_{i-p}. \label{pf.prop2.0f.short.etai.2}%
\end{equation}
\end{itemize}
\par
Now,%
\begin{align*}
\sum_{i=1}^{p+q}\eta_{i}  &  =\sum_{i=1}^{p}\underbrace{\eta_{i}%
}_{\substack{=\lambda\mu_{i}\\\text{(by (\ref{pf.prop2.0f.short.etai.1}))}%
}}+\sum_{i=p+1}^{p+q}\underbrace{\eta_{i}}_{\substack{=\left(  1-\lambda
\right)  \nu_{i-p}\\\text{(by (\ref{pf.prop2.0f.short.etai.2}))}%
}}=\underbrace{\sum_{i=1}^{p}\lambda\mu_{i}}_{=\lambda\sum\limits_{i=1}^{p}%
\mu_{i}}+\underbrace{\sum_{i=p+1}^{p+q}\left(  1-\lambda\right)  \nu_{i-p}%
}_{=\left(  1-\lambda\right)  \sum\limits_{i=p+1}^{p+q}\nu_{i-p}}\\
&  =\lambda\sum\limits_{i=1}^{p}\mu_{i}+\left(  1-\lambda\right)
\sum\limits_{i=p+1}^{p+q}\nu_{i-p}=\lambda\underbrace{\sum\limits_{i=1}^{p}%
\mu_{i}}_{=1}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i=1}^{q}%
\nu_{i}}_{=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-p\text{ in the second sum}\right) \\
&  =\lambda+\left(  1-\lambda\right)  =1
\end{align*}
and%
\begin{align*}
\sum_{i=1}^{p+q}\eta_{i}w_{i}  &  =\sum_{i=1}^{p}\underbrace{\eta_{i}%
}_{\substack{=\lambda\mu_{i}\\\text{(by (\ref{pf.prop2.0f.short.etai.1}))}%
}}\underbrace{w_{i}}_{\substack{=y_{i}\\\text{(by
(\ref{pf.prop2.0f.short.wi.1}))}}}+\sum_{i=p+1}^{p+q}\underbrace{\eta_{i}%
}_{\substack{=\left(  1-\lambda\right)  \nu_{i-p}\\\text{(by
(\ref{pf.prop2.0f.short.etai.2}))}}}\underbrace{w_{i}}_{\substack{=z_{i-p}%
\\\text{(by (\ref{pf.prop2.0f.short.wi.2}))}}}\\
&  =\underbrace{\sum_{i=1}^{p}\lambda\mu_{i}y_{i}}_{=\lambda\sum
\limits_{i=1}^{p}\mu_{i}y_{i}}+\underbrace{\sum_{i=p+1}^{p+q}\left(
1-\lambda\right)  \nu_{i-p}z_{i-p}}_{=\left(  1-\lambda\right)  \sum
\limits_{i=p+1}^{p+q}\nu_{i-p}z_{i-p}}\\
&  =\lambda\sum\limits_{i=1}^{p}\mu_{i}y_{i}+\left(  1-\lambda\right)
\sum\limits_{i=p+1}^{p+q}\nu_{i-p}z_{i-p}=\lambda\underbrace{\sum
\limits_{i=1}^{p}\mu_{i}y_{i}}_{=y}+\left(  1-\lambda\right)  \underbrace{\sum
\limits_{i=1}^{q}\nu_{i}z_{i}}_{=z}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-p\text{ in the second sum}\right) \\
&  =\lambda y+\left(  1-\lambda\right)  z.
\end{align*}
Hence, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}=\lambda y+\left(  1-\lambda\right)  z$ (namely, $t=p+q$,
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w_{1},w_{2},\ldots
,w_{p+q}\right)  $ and $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  =\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  $). In other
words,%
\[
\lambda y+\left(  1-\lambda\right)  z\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Now, let us forget that we fixed $y$, $z$ and $\lambda$. We thus have proven
that every two elements $y\in C_{3}$ and $z\in C_{3}$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda y+\left(  1-\lambda\right)
z\in C_{3}$. If we rename $y$ and $z$ as $x$ and $y$ in this statement, we
obtain the following: Every two elements $x\in C_{3}$ and $y\in C_{3}$ and
every real number $\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(
1-\lambda\right)  y\in C_{3}$. In other words, the set $C_{3}$ is convex (by
the definition of convexity). Qed.}. Moreover, $S\subseteq C_{3}%
$\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Then, $\sum\limits_{i=1}%
^{1}1=1$ and $\sum\limits_{i=1}^{1}1w=1w=w$. Hence, there exist some
$t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ of
elements of $S$ and a $t$-tuple $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ of nonnegative reals such that $\sum
\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=w$
(namely, $t=1$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w\right)  $
and $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(
1\right)  $). In other words,%
\[
w\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{3}$ for
every $w\in S$. In other words, $S\subseteq C_{3}$, qed.}. Hence, $C_{3}$ is a
convex subset of $E$ which contains $S$ as a subset. Thus, the intersection of
all convex subsets of $E$ which contain $S$ as a subset is a subset of $C_{3}%
$. In other words, $C_{1}$ is a subset of $C_{3}$ (since $C_{1}$ is the
intersection of all convex subsets of $E$ which contain $S$ as a subset). In
other words, $C_{1}\subseteq C_{3}$.
\end{vershort}

\begin{verlong}
The set $C_{3}$ is convex\footnote{\textit{Proof.} Let $y\in C_{3}$, $z\in
C_{3}$ and $\lambda\in\left[  0,1\right]  $. We shall show that $\lambda
y+\left(  1-\lambda\right)  z\in C_{3}$.
\par
Notice that $\lambda$ and $1-\lambda$ both are nonnegative reals (since
$\lambda\in\left[  0,1\right]  $).
\par
We have%
\[
y\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}=y$. Let us denote this $t$, this $\left(  x_{1},x_{2}%
,\ldots,x_{t}\right)  $ and this $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ by $p$, $\left(  y_{1},y_{2},\ldots
,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $,
respectively. Thus, $p$ is an element of $\mathbb{N}$, and $\left(
y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$, and
$\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{p}\mu_{i}=1$ and
$\sum\limits_{i=1}^{p}\mu_{i}y_{i}=y$.
\par
Also,
\[
z\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}=z$. Let us denote this $t$, this $\left(  x_{1},x_{2}%
,\ldots,x_{t}\right)  $ and this $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ by $q$, $\left(  z_{1},z_{2},\ldots
,z_{q}\right)  $ and $\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $,
respectively. Thus, $q$ is an element of $\mathbb{N}$, and $\left(
z_{1},z_{2},\ldots,z_{q}\right)  $ is a $q$-tuple of elements of $S$, and
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $ is a $q$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{q}\nu_{i}=1$ and
$\sum\limits_{i=1}^{q}\nu_{i}z_{i}=z$.
\par
Define a $\left(  p+q\right)  $-tuple $\left(  w_{1},w_{2},\ldots
,w_{p+q}\right)  $ of elements of $S$ by%
\[
\left(  w_{1},w_{2},\ldots,w_{p+q}\right)  =\left(  y_{1},y_{2},\ldots
,y_{p},z_{1},z_{2},\ldots,z_{q}\right)  .
\]
Then,%
\[
w_{i}=%
\begin{cases}
y_{i}, & \text{if }i\leq p;\\
z_{i-p}, & \text{if }i>p
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,p+q\right\}  .
\]
Hence, for every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{align}
w_{i}  &  =%
\begin{cases}
y_{i}, & \text{if }i\leq p;\\
z_{i-p}, & \text{if }i>p
\end{cases}
=y_{i}\label{pf.prop2.0f.wi.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq p\text{ (since }i\in\left\{
1,2,\ldots,p\right\}  \text{)}\right)  .\nonumber
\end{align}
Also, for every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{align}
w_{i}  &  =%
\begin{cases}
y_{i}, & \text{if }i\leq p;\\
z_{i-p}, & \text{if }i>p
\end{cases}
=z_{i-p}\label{pf.prop2.0f.wi.2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>p\text{ (since }i\in\left\{
p+1,p+2,\ldots,p+q\right\}  \text{)}\right)  .\nonumber
\end{align}
\par
Furthermore, define a $\left(  p+q\right)  $-tuple $\left(  \eta_{1},\eta
_{2},\ldots,\eta_{p+q}\right)  $ of nonnegative reals by%
\[
\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  =\left(  \lambda\mu
_{1},\lambda\mu_{2},\ldots,\lambda\mu_{p},\left(  1-\lambda\right)  \nu
_{1},\left(  1-\lambda\right)  \nu_{2},\ldots,\left(  1-\lambda\right)
\nu_{q}\right)
\]
(this is well-defined because both $\lambda$ and $1-\lambda$ are nonnegative
reals and because $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ and
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $ are tuples of nonnegative
reals). Then,%
\[
\eta_{i}=%
\begin{cases}
\lambda\mu_{i}, & \text{if }i\leq p;\\
\left(  1-\lambda\right)  \nu_{i-p}, & \text{if }i>p
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,p+q\right\}  .
\]
Hence, for every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{align}
\eta_{i}  &  =%
\begin{cases}
\lambda\mu_{i}, & \text{if }i\leq p;\\
\left(  1-\lambda\right)  \nu_{i-p}, & \text{if }i>p
\end{cases}
=\lambda\mu_{i}\label{pf.prop2.0f.etai.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq p\text{ (since }i\in\left\{
1,2,\ldots,p\right\}  \text{)}\right)  .\nonumber
\end{align}
Also, for every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{align}
\eta_{i}  &  =%
\begin{cases}
\lambda\mu_{i}, & \text{if }i\leq p;\\
\left(  1-\lambda\right)  \nu_{i-p}, & \text{if }i>p
\end{cases}
=\left(  1-\lambda\right)  \nu_{i-p}\label{pf.prop2.0f.etai.2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>p\text{ (since }i\in\left\{
p+1,p+2,\ldots,p+q\right\}  \text{)}\right)  .\nonumber
\end{align}
\par
Now,%
\begin{align*}
\sum_{i=1}^{p+q}\eta_{i}  &  =\sum_{i=1}^{p}\underbrace{\eta_{i}%
}_{\substack{=\lambda\mu_{i}\\\text{(by (\ref{pf.prop2.0f.etai.1}))}}%
}+\sum_{i=p+1}^{p+q}\underbrace{\eta_{i}}_{\substack{=\left(  1-\lambda
\right)  \nu_{i-p}\\\text{(by (\ref{pf.prop2.0f.etai.2}))}}}=\underbrace{\sum
_{i=1}^{p}\lambda\mu_{i}}_{=\lambda\sum\limits_{i=1}^{p}\mu_{i}}%
+\underbrace{\sum_{i=p+1}^{p+q}\left(  1-\lambda\right)  \nu_{i-p}}_{=\left(
1-\lambda\right)  \sum\limits_{i=p+1}^{p+q}\nu_{i-p}}\\
&  =\lambda\sum\limits_{i=1}^{p}\mu_{i}+\left(  1-\lambda\right)
\sum\limits_{i=p+1}^{p+q}\nu_{i-p}=\lambda\underbrace{\sum\limits_{i=1}^{p}%
\mu_{i}}_{=1}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i=1}^{q}%
\nu_{i}}_{=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-p\text{ in the second sum}\right) \\
&  =\lambda+\left(  1-\lambda\right)  =1
\end{align*}
and%
\begin{align*}
\sum_{i=1}^{p+q}\eta_{i}w_{i}  &  =\sum_{i=1}^{p}\underbrace{\eta_{i}%
}_{\substack{=\lambda\mu_{i}\\\text{(by (\ref{pf.prop2.0f.etai.1}))}%
}}\underbrace{w_{i}}_{\substack{=y_{i}\\\text{(by (\ref{pf.prop2.0f.wi.1}))}%
}}+\sum_{i=p+1}^{p+q}\underbrace{\eta_{i}}_{\substack{=\left(  1-\lambda
\right)  \nu_{i-p}\\\text{(by (\ref{pf.prop2.0f.etai.2}))}}}\underbrace{w_{i}%
}_{\substack{=z_{i-p}\\\text{(by (\ref{pf.prop2.0f.wi.2}))}}}\\
&  =\underbrace{\sum_{i=1}^{p}\lambda\mu_{i}y_{i}}_{=\lambda\sum
\limits_{i=1}^{p}\mu_{i}y_{i}}+\underbrace{\sum_{i=p+1}^{p+q}\left(
1-\lambda\right)  \nu_{i-p}z_{i-p}}_{=\left(  1-\lambda\right)  \sum
\limits_{i=p+1}^{p+q}\nu_{i-p}z_{i-p}}\\
&  =\lambda\sum\limits_{i=1}^{p}\mu_{i}y_{i}+\left(  1-\lambda\right)
\sum\limits_{i=p+1}^{p+q}\nu_{i-p}z_{i-p}=\lambda\underbrace{\sum
\limits_{i=1}^{p}\mu_{i}y_{i}}_{=y}+\left(  1-\lambda\right)  \underbrace{\sum
\limits_{i=1}^{q}\nu_{i}z_{i}}_{=z}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-p\text{ in the second sum}\right) \\
&  =\lambda y+\left(  1-\lambda\right)  z.
\end{align*}
Hence, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}=\lambda y+\left(  1-\lambda\right)  z$ (namely, $t=p+q$,
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w_{1},w_{2},\ldots
,w_{p+q}\right)  $ and $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  =\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  $). In other
words,%
\[
\lambda y+\left(  1-\lambda\right)  z\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Now, let us forget that we fixed $y$, $z$ and $\lambda$. We thus have proven
that every two elements $y\in C_{3}$ and $z\in C_{3}$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda y+\left(  1-\lambda\right)
z\in C_{3}$. If we rename $y$ and $z$ as $x$ and $y$ in this statement, we
obtain the following: Every two elements $x\in C_{3}$ and $y\in C_{3}$ and
every real number $\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(
1-\lambda\right)  y\in C_{3}$. In other words, the set $C_{3}$ is convex (by
the definition of convexity). Qed.}. Moreover, $S\subseteq C_{3}%
$\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Then, $\sum\limits_{i=1}%
^{1}1=1$ and $\sum\limits_{i=1}^{1}1w=1w=w$. Hence, there exist some
$t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ of
elements of $S$ and a $t$-tuple $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ of nonnegative reals such that $\sum
\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=w$
(namely, $t=1$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w\right)  $
and $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(
1\right)  $). In other words,%
\[
w\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{3}$ for
every $w\in S$. In other words, $S\subseteq C_{3}$, qed.}. Hence, $C_{3}$ is a
convex subset of $E$ which contains $S$ as a subset. Thus, the intersection of
all convex subsets of $E$ which contain $S$ as a subset is a subset of $C_{3}%
$. In other words, $C_{1}$ is a subset of $C_{3}$ (since $C_{1}$ is the
intersection of all convex subsets of $E$ which contain $S$ as a subset). In
other words, $C_{1}\subseteq C_{3}$.
\end{verlong}

On the other hand, $C_{3}\subseteq C_{1}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in C_{3}$. We shall show that $w\in C_{1}$.
\par
Let $D$ be any convex subset of $E$ which contains $S$ as a subset. Thus, $D$
is a convex subset of $E$ and satisfies $S\subseteq D$.
\par
We have%
\[
w\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}=w$. Consider this $t$, this $\left(  x_{1},x_{2},\ldots
,x_{t}\right)  $ and this $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  $. For every $i\in\left\{  1,2,\ldots,t\right\}  $, we have
$x_{i}\in S\subseteq D$. Thus, $\left(  x_{i}\right)  _{i\in\left\{
1,2,\ldots,t\right\}  }$ is a family of elements of $D$. Also, clearly,
$\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,t\right\}  }$ is a
family of nonnegative reals (since $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ is a $t$-tuple of nonnegative reals).
Furthermore, all but finitely many $i\in\left\{  1,2,\ldots,t\right\}  $
satisfy $\lambda_{i}=0$ (since there are only finitely many $i\in\left\{
1,2,\ldots,t\right\}  $). Finally, $\sum\limits_{i\in\left\{  1,2,\ldots
,t\right\}  }\lambda_{i}=\sum\limits_{i=1}^{t}\lambda_{i}=1$. Hence,
Proposition 2.0a (applied to $D$ and $\left\{  1,2,\ldots,t\right\}  $ instead
of $C$ and $I$) shows that $\sum\limits_{i\in\left\{  1,2,\ldots,t\right\}
}\lambda_{i}x_{i}\in D$. Thus, $w=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}%
=\sum\limits_{i\in\left\{  1,2,\ldots,t\right\}  }\lambda_{i}x_{i}\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex subset of $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex subset of $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex subsets of $E$ which
contain $S$ as a subset. In other words, $w$ lies in $C_{1}$ (since $C_{1}$ is
the intersection of all convex subsets of $E$ which contain $S$ as a subset).
In other words, $w\in C_{1}$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{1}$ for
every $w\in C_{3}$. In other words, $C_{3}\subseteq C_{1}$, qed.}. Combined
with $C_{1}\subseteq C_{3}$, this shows that $C_{1}=C_{3}$.

\begin{vershort}
The set $C_{2}$ is convex\footnote{\textit{Proof.} Let $y\in C_{2}$, $z\in
C_{2}$ and $\lambda\in\left[  0,1\right]  $. We shall show that $\lambda
y+\left(  1-\lambda\right)  z\in C_{2}$.
\par
Notice that $\lambda$ and $1-\lambda$ both are nonnegative reals (since
$\lambda\in\left[  0,1\right]  $).
\par
We have $y\in C_{2}$. In other words, $y$ is a convex combination of the
vectors $s$ for $s\in S$ (since $C_{2}$ is the set of all convex combinations
of the vectors $s$ for $s\in S$). In other words, $y$ has the form
$y=\sum\limits_{i\in S}\mu_{i}i$ for some family $\left(  \mu_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\mu_{i}=0$, and the
property that $\sum\limits_{i\in S}\mu_{i}=1$. Similarly, $z$ has the form
$z=\sum\limits_{i\in S}\nu_{i}i$ for some family $\left(  \nu_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\nu_{i}=0$, and the
property that $\sum\limits_{i\in S}\nu_{i}=1$. Consider these two families
$\left(  \mu_{i}\right)  _{i\in S}$ and $\left(  \nu_{i}\right)  _{i\in S}$.
\par
For every $i\in S$, both $\lambda\mu_{i}$ and $\left(  1-\lambda\right)
\nu_{i}$ are nonnegative reals (since $\lambda$, $\mu_{i}$, $1-\lambda$ and
$\nu_{i}$ are nonnegative reals), and hence the sum $\lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}$ is a nonnegative real. Thus, we can define a family
$\left(  \eta_{i}\right)  _{i\in S}$ of nonnegative reals by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in S\right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in S}$. We know that all
but finitely many $i\in S$ satisfy $\mu_{i}=0$, and we also know that all but
finitely many $i\in S$ satisfy $\nu_{i}=0$. Using these two facts, we see that
all but finitely many $i\in S$ satisfy $\eta_{i}=\lambda\underbrace{\mu_{i}%
}_{=0}+\left(  1-\lambda\right)  \underbrace{\nu_{i}}_{=0}=0+0=0$. We have%
\[
\sum\limits_{i\in S}\underbrace{\eta_{i}}_{=\lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}}=\sum_{i\in S}\left(  \lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}\right)  =\lambda\underbrace{\sum\limits_{i\in S}%
\mu_{i}}_{=1}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in S}%
\nu_{i}}_{=1}=\lambda+\left(  1-\lambda\right)  =1
\]
and%
\[
\sum\limits_{i\in S}\underbrace{\eta_{i}}_{=\lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}}i=\sum_{i\in S}\left(  \lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}\right)  i=\lambda\underbrace{\sum\limits_{i\in S}%
\mu_{i}i}_{=y}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in S}%
\nu_{i}z}_{=z}=\lambda y+\left(  1-\lambda\right)  z.
\]
Hence, $\lambda y+\left(  1-\lambda\right)  z=\sum\limits_{i\in S}\eta_{i}i$.
Therefore, $\lambda y+\left(  1-\lambda\right)  z$ has the form $\sum
\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the
property that $\sum\limits_{i\in S}\lambda_{i}=1$ (namely, for the family
$\left(  \lambda_{i}\right)  _{i\in S}=\left(  \eta_{i}\right)  _{i\in S}$).
In other words, $\lambda y+\left(  1-\lambda\right)  z$ is a convex
combination of the vectors $s$ for $s\in S$. In other words, $\lambda
y+\left(  1-\lambda\right)  z\in C_{2}$ (since $C_{2}$ is the set of all
convex combinations of the vectors $s$ for $s\in S$).
\par
Now, let us forget that we fixed $y$, $z$ and $\lambda$. We thus have proven
that every two elements $y\in C_{2}$ and $z\in C_{2}$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda y+\left(  1-\lambda\right)
z\in C_{2}$. If we rename $y$ and $z$ as $x$ and $y$ in this statement, we
obtain the following: Every two elements $x\in C_{2}$ and $y\in C_{2}$ and
every real number $\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(
1-\lambda\right)  y\in C_{2}$. In other words, the set $C_{2}$ is convex (by
the definition of convexity).}. Moreover, $S\subseteq C_{2}$%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Define a family $\left(
\mu_{i}\right)  _{i\in S}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in S\right)  .
\]
Then, it is easy to see that:
\par
\begin{itemize}
\item All but finitely many $i\in S$ satisfy $\mu_{i}=0$.
\par
\item We have $\sum\limits_{i\in S}\mu_{i}=1$.
\par
\item We have $\sum\limits_{i\in S}\mu_{i}i=w$.
\end{itemize}
\par
Hence, $w=\sum\limits_{i\in S}\mu_{i}i$. Thus, $w$ has the form $\sum
\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the
property that $\sum\limits_{i\in S}\lambda_{i}=1$ (namely, for the family
$\left(  \lambda_{i}\right)  _{i\in S}=\left(  \mu_{i}\right)  _{i\in S}$). In
other words, $w$ is a convex combination of the vectors $s$ for $s\in S$. In
other words, $w\in C_{2}$ (since $C_{2}$ is the set of all convex combinations
of the vectors $s$ for $s\in S$).
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in C_{2}$
for every $w\in S$. In other words, $S\subseteq C_{2}$, qed.}. Hence, $C_{2}$
is a convex subset of $E$ which contains $S$ as a subset. Thus, the
intersection of all convex subsets of $E$ which contain $S$ as a subset is a
subset of $C_{2}$. In other words, $C_{1}$ is a subset of $C_{2}$ (since
$C_{1}$ is the intersection of all convex subsets of $E$ which contain $S$ as
a subset). In other words, $C_{1}\subseteq C_{2}$.
\end{vershort}

\begin{verlong}
The set $C_{2}$ is convex\footnote{\textit{Proof.} Let $y\in C_{2}$, $z\in
C_{2}$ and $\lambda\in\left[  0,1\right]  $. We shall show that $\lambda
y+\left(  1-\lambda\right)  z\in C_{2}$.
\par
Notice that $\lambda$ and $1-\lambda$ both are nonnegative reals (since
$\lambda\in\left[  0,1\right]  $).
\par
We have $y\in C_{2}$. In other words, $y$ belongs to the set $C_{2}$. In other
words, $y$ belongs to the set of all convex combinations of the vectors $s$
for $s\in S$ (since $C_{2}$ is the set of all convex combinations of the
vectors $s$ for $s\in S$). In other words, $y$ is a convex combination of the
vectors $s$ for $s\in S$. In other words, $y$ has the form $\sum\limits_{i\in
S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)  _{i\in S}$ of
nonnegative reals indexed by elements of $S$ and having the property that all
but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the property that
$\sum\limits_{i\in S}\lambda_{i}=1$ (by the definition of \textquotedblleft
convex combination\textquotedblright). Let us denote this family $\left(
\lambda_{i}\right)  _{i\in S}$ by $\left(  \mu_{i}\right)  _{i\in S}$. Thus,
$\left(  \mu_{i}\right)  _{i\in S}$ is a family of nonnegative reals indexed
by elements of $S$ and having the property that all but finitely many $i\in S$
satisfy $\mu_{i}=0$, and the property that $\sum\limits_{i\in S}\mu_{i}=1$,
and it furthermore satisfies $y=\sum\limits_{i\in S}\mu_{i}i$.
\par
We have $z\in C_{2}$. In other words, $z$ belongs to the set $C_{2}$. In other
words, $z$ belongs to the set of all convex combinations of the vectors $s$
for $s\in S$ (since $C_{2}$ is the set of all convex combinations of the
vectors $s$ for $s\in S$). In other words, $z$ is a convex combination of the
vectors $s$ for $s\in S$. In other words, $z$ has the form $\sum\limits_{i\in
S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)  _{i\in S}$ of
nonnegative reals indexed by elements of $S$ and having the property that all
but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the property that
$\sum\limits_{i\in S}\lambda_{i}=1$ (by the definition of \textquotedblleft
convex combination\textquotedblright). Let us denote this family $\left(
\lambda_{i}\right)  _{i\in S}$ by $\left(  \nu_{i}\right)  _{i\in S}$. Thus,
$\left(  \nu_{i}\right)  _{i\in S}$ is a family of nonnegative reals indexed
by elements of $S$ and having the property that all but finitely many $i\in S$
satisfy $\nu_{i}=0$, and the property that $\sum\limits_{i\in S}\nu_{i}=1$,
and it furthermore satisfies $z=\sum\limits_{i\in S}\nu_{i}i$.
\par
For every $i\in S$, both $\lambda\mu_{i}$ and $\left(  1-\lambda\right)
\nu_{i}$ are nonnegative reals (since $\lambda$, $\mu_{i}$, $1-\lambda$ and
$\nu_{i}$ are nonnegative reals), and hence the sum $\lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}$ is a nonnegative real. Thus, we can define a family
$\left(  \eta_{i}\right)  _{i\in S}$ of nonnegative reals by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in S\right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in S}$ of nonnegative
reals. We know that all but finitely many $i\in S$ satisfy $\mu_{i}=0$, and we
also know that all but finitely many $i\in S$ satisfy $\nu_{i}=0$. Using these
two facts, we see that all but finitely many $i\in S$ satisfy $\eta
_{i}=\lambda\underbrace{\mu_{i}}_{=0}+\left(  1-\lambda\right)
\underbrace{\nu_{i}}_{=0}=0+0=0$. We have%
\[
\sum\limits_{i\in S}\underbrace{\eta_{i}}_{=\lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}}=\sum_{i\in S}\left(  \lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}\right)  =\lambda\underbrace{\sum\limits_{i\in S}%
\mu_{i}}_{=1}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in S}%
\nu_{i}}_{=1}=\lambda+\left(  1-\lambda\right)  =1
\]
and%
\[
\sum\limits_{i\in S}\underbrace{\eta_{i}}_{=\lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}}i=\sum_{i\in S}\left(  \lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}\right)  i=\lambda\underbrace{\sum\limits_{i\in S}%
\mu_{i}i}_{=y}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in S}%
\nu_{i}z}_{=z}=\lambda y+\left(  1-\lambda\right)  z.
\]
Hence, $\lambda y+\left(  1-\lambda\right)  z=\sum\limits_{i\in S}\eta_{i}i$.
Therefore, $\lambda y+\left(  1-\lambda\right)  z$ has the form $\sum
\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the
property that $\sum\limits_{i\in S}\lambda_{i}=1$ (namely, for the family
$\left(  \lambda_{i}\right)  _{i\in S}=\left(  \eta_{i}\right)  _{i\in S}$).
In other words, $\lambda y+\left(  1-\lambda\right)  z$ is a convex
combination of the vectors $s$ for $s\in S$. In other words, $\lambda
y+\left(  1-\lambda\right)  z$ belongs to the set of all convex combinations
of the vectors $s$ for $s\in S$. In other words, $\lambda y+\left(
1-\lambda\right)  z$ belongs to $C_{2}$ (since $C_{2}$ is the set of all
convex combinations of the vectors $s$ for $s\in S$). In other words, $\lambda
y+\left(  1-\lambda\right)  z\in C_{2}$.
\par
Now, let us forget that we fixed $y$, $z$ and $\lambda$. We thus have proven
that every two elements $y\in C_{2}$ and $z\in C_{2}$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda y+\left(  1-\lambda\right)
z\in C_{2}$. If we rename $y$ and $z$ as $x$ and $y$ in this statement, we
obtain the following: Every two elements $x\in C_{2}$ and $y\in C_{2}$ and
every real number $\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(
1-\lambda\right)  y\in C_{2}$. In other words, the set $C_{2}$ is convex (by
the definition of convexity). Qed.}. Moreover, $S\subseteq C_{2}%
$\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Define a family $\left(
\mu_{i}\right)  _{i\in S}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in S\right)  .
\]
Then, every $i\in S\setminus\left\{  w\right\}  $ satisfies%
\[
\mu_{i}=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\neq w\text{ (since }i\in
S\setminus\left\{  w\right\}  \text{)}\right)  .
\]
Thus, there exists a finite subset $T$ of $S$ such that every $i\in S\setminus
T$ satisfies $\mu_{i}=0$ (namely, $T=\left\{  w\right\}  $). In other words,
all but finitely many $i\in S$ satisfy $\mu_{i}=0$. Also,%
\begin{align*}
\sum\limits_{i\in S}\mu_{i}  &  =\sum\limits_{\substack{i\in S;\\i=w}%
}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=1\\\text{(since }i=w\text{)}}}+\sum\limits_{\substack{i\in S;\\i\neq
w}}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=0\\\text{(since }i\neq w\text{)}}}=\sum\limits_{\substack{i\in S;\\i=w}%
}1+\underbrace{\sum\limits_{\substack{i\in S;\\i\neq w}}0}_{=0}\\
&  =\sum\limits_{\substack{i\in S;\\i=w}}1=1\ \ \ \ \ \ \ \ \ \ \left(
\text{since }w\in S\right)
\end{align*}
and%
\begin{align*}
\sum\limits_{i\in S}\mu_{i}i  &  =\sum\limits_{\substack{i\in S;\\i=w}%
}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=1\\\text{(since }i=w\text{)}}}i+\sum\limits_{\substack{i\in S;\\i\neq
w}}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=0\\\text{(since }i\neq w\text{)}}}i=\sum\limits_{\substack{i\in
S;\\i=w}}1i+\underbrace{\sum\limits_{\substack{i\in S;\\i\neq w}}0i}_{=0}\\
&  =\sum\limits_{\substack{i\in S;\\i=w}}\underbrace{1i}_{=i}=\sum
\limits_{\substack{i\in S;\\i=w}}i=w\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}w\in S\right)  .
\end{align*}
\par
Hence, $w=\sum\limits_{i\in S}\mu_{i}i$. Thus, $w$ has the form $\sum
\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the
property that $\sum\limits_{i\in S}\lambda_{i}=1$ (namely, for the family
$\left(  \lambda_{i}\right)  _{i\in S}=\left(  \mu_{i}\right)  _{i\in S}$). In
other words, $w$ is a convex combination of the vectors $s$ for $s\in S$. In
other words, $w$ belongs to the set of all convex combinations of the vectors
$s$ for $s\in S$. In other words, $w$ belongs to $C_{2}$ (since $C_{2}$ is the
set of all convex combinations of the vectors $s$ for $s\in S$). In other
words, $w\in C_{2}$.
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in C_{2}$
for every $w\in S$. In other words, $S\subseteq C_{2}$, qed.}. Hence, $C_{2}$
is a convex subset of $E$ which contains $S$ as a subset. Thus, the
intersection of all convex subsets of $E$ which contain $S$ as a subset is a
subset of $C_{2}$. In other words, $C_{1}$ is a subset of $C_{2}$ (since
$C_{1}$ is the intersection of all convex subsets of $E$ which contain $S$ as
a subset). In other words, $C_{1}\subseteq C_{2}$.
\end{verlong}

\begin{vershort}
On the other hand, $C_{2}\subseteq C_{1}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in C_{2}$. We shall show that $w\in C_{1}$.
\par
Let $D$ be any convex subset of $E$ which contains $S$ as a subset. Thus, $D$
is a convex subset of $E$ and satisfies $S\subseteq D$.
\par
We have $w\in C_{2}$. In other words, $w$ is a convex combination of the
vectors $s$ for $s\in S$ (since $C_{2}$ is the set of all convex combinations
of the vectors $s$ for $s\in S$). In other words, $w$ has the form
$w=\sum\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda
_{i}\right)  _{i\in S}$ of nonnegative reals indexed by elements of $S$ and
having the property that all but finitely many $i\in S$ satisfy $\lambda
_{i}=0$, and the property that $\sum\limits_{i\in S}\lambda_{i}=1$. Consider
this family $\left(  \lambda_{i}\right)  _{i\in S}$.
\par
For every $i\in S$, we have $i\in S\subseteq D$. Thus, $\left(  i\right)
_{i\in S}$ is a family of elements of $D$. Hence, Proposition 2.0a (applied to
$D$, $S$, $\left(  i\right)  _{i\in S}$ and $\left(  \lambda_{i}\right)
_{i\in S}$ instead of $C$, $I$, $\left(  x_{i}\right)  _{i\in I}$ and $\left(
\lambda_{i}\right)  _{i\in I}$) shows that $\sum\limits_{i\in S}\lambda
_{i}i\in D$. Thus, $w=\sum\limits_{i\in S}\lambda_{i}i\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex subset of $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex subset of $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex subsets of $E$ which
contain $S$ as a subset. In other words, $w$ lies in $C_{1}$ (since $C_{1}$ is
the intersection of all convex subsets of $E$ which contain $S$ as a subset).
In other words, $w\in C_{1}$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{1}$ for
every $w\in C_{2}$. In other words, $C_{2}\subseteq C_{1}$, qed.}. Combined
with $C_{1}\subseteq C_{2}$, this shows that $C_{1}=C_{2}$.
\end{vershort}

\begin{verlong}
On the other hand, $C_{2}\subseteq C_{1}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in C_{2}$. We shall show that $w\in C_{1}$.
\par
Let $D$ be any convex subset of $E$ which contains $S$ as a subset. Thus, $D$
is a convex subset of $E$ and satisfies $S\subseteq D$.
\par
We have $w\in C_{2}$. In other words, $w$ belongs to the set $C_{2}$. In other
words, $w$ belongs to the set of all convex combinations of the vectors $s$
for $s\in S$ (since $C_{2}$ is the set of all convex combinations of the
vectors $s$ for $s\in S$). In other words, $w$ is a convex combination of the
vectors $s$ for $s\in S$. In other words, $w$ has the form $\sum\limits_{i\in
S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)  _{i\in S}$ of
nonnegative reals indexed by elements of $S$ and having the property that all
but finitely many $i\in S$ satisfy $\lambda_{i}=0$, and the property that
$\sum\limits_{i\in S}\lambda_{i}=1$. Consider this family $\left(  \lambda
_{i}\right)  _{i\in S}$. Thus, $w=\sum\limits_{i\in S}\lambda_{i}i$.
\par
For every $i\in S$, we have $i\in S\subseteq D$. Thus, $\left(  i\right)
_{i\in S}$ is a family of elements of $D$. Hence, Proposition 2.0a (applied to
$D$, $S$, $\left(  i\right)  _{i\in S}$ and $\left(  \lambda_{i}\right)
_{i\in S}$ instead of $C$, $I$, $\left(  x_{i}\right)  _{i\in I}$ and $\left(
\lambda_{i}\right)  _{i\in I}$) shows that $\sum\limits_{i\in S}\lambda
_{i}i\in D$. Thus, $w=\sum\limits_{i\in S}\lambda_{i}i\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex subset of $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex subset of $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex subsets of $E$ which
contain $S$ as a subset. In other words, $w$ lies in $C_{1}$ (since $C_{1}$ is
the intersection of all convex subsets of $E$ which contain $S$ as a subset).
In other words, $w\in C_{1}$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{1}$ for
every $w\in C_{2}$. In other words, $C_{2}\subseteq C_{1}$, qed.}. Combined
with $C_{1}\subseteq C_{2}$, this shows that $C_{1}=C_{2}$.
\end{verlong}

Combining $C_{1}=C_{2}$ with $C_{1}=C_{3}$, we obtain $C_{1}=C_{2}=C_{3}$.
Now, (\ref{pf.prop2.0f.a.chull1}) becomes%
\begin{align}
&  \left(  \text{the convex hull of }S\text{ defined according to Definition
2.0c}\right) \nonumber\\
&  =C_{1}=C_{2}=\left(  \text{the convex hull of }S\text{ defined according to
Definition 2.0d}\right)  \label{pf.prop2.0f.a.chull12}%
\end{align}
(by (\ref{pf.prop2.0f.a.chull2})). Also, (\ref{pf.prop2.0f.a.chull2}) becomes%
\begin{align}
&  \left(  \text{the convex hull of }S\text{ defined according to Definition
2.0d}\right) \nonumber\\
&  =C_{2}=C_{3}=\left(  \text{the convex hull of }S\text{ defined according to
Definition 2.0e}\right)  \label{pf.prop2.0f.a.chull23}%
\end{align}
(by (\ref{pf.prop2.0f.a.chull3})).

Now, let us forget that we fixed $S$. We thus have shown that if $S$ is any
subset of $E$, then (\ref{pf.prop2.0f.a.chull12}) and
(\ref{pf.prop2.0f.a.chull23}) hold. In other words, if $S$ is any subset of
$E$, then
\begin{align*}
&  \left(  \text{the convex hull of }S\text{ defined according to Definition
2.0c}\right) \\
&  =\left(  \text{the convex hull of }S\text{ defined according to Definition
2.0d}\right) \\
&  =\left(  \text{the convex hull of }S\text{ defined according to Definition
2.0e}\right)  .
\end{align*}
In other words, Definitions 2.0c, 2.0d and 2.0e are equivalent. This proves
Proposition 2.0f \textbf{(a)}.

\textbf{(c)} We define the set $C_{3}$ as in our proof of Proposition 2.0f
\textbf{(a)}. Then, (\ref{pf.prop2.0f.a.chull3}) shows that%
\begin{align*}
C_{3}  &  =\left(  \text{the convex hull of }S\text{ defined according to
Definition 2.0e}\right) \\
&  =\left(  \text{the convex hull of }S\right)  =\operatorname*{conv}%
.\operatorname*{hull}S.
\end{align*}
But in our proof of Proposition 2.0f \textbf{(a)}, we have shown that
$S\subseteq C_{3}$. Thus, $S\subseteq C_{3}=\operatorname*{conv}%
.\operatorname*{hull}S$. This proves Proposition 2.0f \textbf{(c)}.

\textbf{(d)} We define the set $C_{3}$ as in our proof of Proposition 2.0f
\textbf{(a)}. Then, (\ref{pf.prop2.0f.a.chull3}) shows that%
\begin{align*}
C_{3}  &  =\left(  \text{the convex hull of }S\text{ defined according to
Definition 2.0e}\right) \\
&  =\left(  \text{the convex hull of }S\right)  =\operatorname*{conv}%
.\operatorname*{hull}S.
\end{align*}
But in our proof of Proposition 2.0f \textbf{(a)}, we have shown that the set
$C_{3}$ is convex. Since $C_{3}=\operatorname*{conv}.\operatorname*{hull}S$,
this rewrites as follows: The set $\operatorname*{conv}.\operatorname*{hull}S$
is convex. This proves Proposition 2.0f \textbf{(d)}.

\textbf{(f)} Let $D$ be a convex subset of $E$ which contains $S$ as a subset.
Thus, the intersection of all convex subsets of $E$ which contain $S$ as a
subset is a subset of $D$. In other words, $\operatorname*{conv}%
.\operatorname*{hull}S$ is a subset of $D$ (since $\operatorname*{conv}%
.\operatorname*{hull}S$ is the intersection of all convex subsets of $E$ which
contain $S$ as a subset (by Definition 2.0d)). In other words, $D$ contains
$\operatorname*{conv}.\operatorname*{hull}S$ as a subset.

Now, let us forget that we fixed $D$. We thus have shown that if $D$ is a
convex subset of $E$ which contains $S$ as a subset, then $D$ contains
$\operatorname*{conv}.\operatorname*{hull}S$ as a subset. In other words,
every convex subset of $E$ which contains $S$ as a subset also contains
$\operatorname*{conv}.\operatorname*{hull}S$ as a subset. This proves
Proposition 2.0f \textbf{(f)}.

\textbf{(b)} Every two elements $x\in\varnothing$ and $y\in\varnothing$ and
every real number $\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(
1-\lambda\right)  y\in\varnothing$\ \ \ \ \footnote{Indeed, this is vacuously
true (since there exist no $x\in\varnothing$).}. In other words, the set
$\varnothing$ is convex (by the definition of convexity). Thus, $\varnothing$
is a convex subset of $E$ which contains $\varnothing$ as a subset. Now,
Proposition 2.0f \textbf{(f)} (applied to $S=\varnothing$) shows that every
convex subset of $E$ which contains $\varnothing$ as a subset also contains
$\operatorname*{conv}.\operatorname*{hull}\varnothing$ as a subset. Thus, the
set $\varnothing$ contains $\operatorname*{conv}.\operatorname*{hull}%
\varnothing$ as a subset (since $\varnothing$ is a convex subset of $E$ which
contains $\varnothing$ as a subset). In other words, $\operatorname*{conv}%
.\operatorname*{hull}\varnothing\subseteq\varnothing$, so that
$\operatorname*{conv}.\operatorname*{hull}\varnothing=\varnothing$. This
proves Proposition 2.0f \textbf{(b)}.

\textbf{(g)} Let $T$ be a subset of $\operatorname*{conv}.\operatorname*{hull}%
S$. The set $\operatorname*{conv}.\operatorname*{hull}S$ is a convex set (by
Proposition 2.0f \textbf{(d)}) and contains $T$ as a subset (since
$T\subseteq\operatorname*{conv}.\operatorname*{hull}S$). In other words,
$\operatorname*{conv}.\operatorname*{hull}S$ is a convex subset of $E$ which
contains $T$ as a subset.

But Proposition 2.0f \textbf{(f)} (applied to $T$ instead of $S$) shows that
every convex subset of $E$ which contains $T$ as a subset also contains
$\operatorname*{conv}.\operatorname*{hull}T$ as a subset. Thus, the set
$\operatorname*{conv}.\operatorname*{hull}S$ contains $\operatorname*{conv}%
.\operatorname*{hull}T$ as a subset (since $\operatorname*{conv}%
.\operatorname*{hull}S$ is a convex subset of $E$ which contains $T$ as a
subset). In other words, $\operatorname*{conv}.\operatorname*{hull}%
T\subseteq\operatorname*{conv}.\operatorname*{hull}S$. This proves Proposition
2.0f \textbf{(g)}.

\textbf{(e)} Let $T$ be a subset of $S$. Thus, $T\subseteq S\subseteq
\operatorname*{conv}.\operatorname*{hull}S$ (by Proposition 2.0f
\textbf{(c)}). In other words, $T$ is a subset of $\operatorname*{conv}%
.\operatorname*{hull}S$. Hence, Proposition 2.0f \textbf{(g)} yields
$\operatorname*{conv}.\operatorname*{hull}T\subseteq\operatorname*{conv}%
.\operatorname*{hull}S$. This proves Proposition 2.0f \textbf{(e)}.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Proposition 2.0g.]\textbf{(a)} Let $S$ be a subset of $E$.
Definition 2.0e yields%
\begin{align*}
&  \operatorname*{conv}.\operatorname*{hull}S\\
&  =\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\} \\
&  =\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
x\text{ can be written in the form }x=\sum\limits_{i=1}^{t}\lambda_{i}x_{i},\\
\text{where }t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\} \\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  .
\end{align*}
This proves Proposition 2.0g \textbf{(a)}.

\textbf{(b)} Proposition 2.0g \textbf{(a)} provides an expression for
$\operatorname*{conv}.\operatorname*{hull}S$ which clearly does not depend on
whether we consider $S$ as a subset of $F$ or as a subset of $E$ (since it
only refers to the elements of $S$). Hence, $\operatorname*{conv}%
.\operatorname*{hull}S$ does not depend on whether we consider $S$ as a subset
of $F$ or as a subset of $E$. This proves Proposition 2.0g \textbf{(b)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition 2.0g.]\textbf{(a)} Let $S$ be a subset of $E$. Recall
that%
\begin{align}
&  \operatorname*{conv}.\operatorname*{hull}S\nonumber\\
&  =\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  \label{pf.prop2.0g.a.1}%
\end{align}
(by Definition 2.0e).

Let%
\[
Q=\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  .
\]


We have $Q\subseteq\operatorname*{conv}.\operatorname*{hull}S$%
\ \ \ \ \footnote{\textit{Proof.} Let $y\in Q$. Thus,%
\[
y\in Q=\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  .
\]
In other words, $y$ has the form $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}$ for
some element $t$ of $\mathbb{N}$, some $t$-tuple $\left(  x_{1},x_{2}%
,\ldots,x_{t}\right)  $ of elements of $S$, and some $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$. Let us denote this $t$, this
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by $p$, $\left(  y_{1}%
,y_{2},\ldots,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{p}\right)  $, respectively. Thus, $p$ is an element of $\mathbb{N}$, and
$\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$,
and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{p}\mu_{i}=1$ and
$y=\sum\limits_{i=1}^{p}\mu_{i}y_{i}$.
\par
Now, $\sum\limits_{i=1}^{p}\mu_{i}y_{i}=y$ (since $y=\sum\limits_{i=1}^{p}%
\mu_{i}y_{i}$). Also, $y=\sum\limits_{i=1}^{p}\mu_{i}\underbrace{y_{i}}_{\in
S\subseteq E}\in\sum\limits_{i=1}^{p}\mu_{i}E\subseteq E$ (since $E$ is an
$\mathbb{R}$-vector space). Hence, $y$ is an element of $E$ such that there
exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1},x_{2},\ldots
,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such that
$\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}\lambda
_{i}x_{i}=y$ (namely, $t=p$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)
=\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ and $\left(  \lambda_{1}%
,\lambda_{2},\ldots,\lambda_{t}\right)  =\left(  \mu_{1},\mu_{2},\ldots
,\mu_{p}\right)  $). In other words,%
\begin{align*}
y  &  \in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\} \\
&  =\operatorname*{conv}.\operatorname*{hull}S\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.prop2.0g.a.1})}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $y$. We thus have shown that $y\in
\operatorname*{conv}.\operatorname*{hull}S$ for every $y\in Q$. In other
words, $Q\subseteq\operatorname*{conv}.\operatorname*{hull}S$, qed.} and
$\operatorname*{conv}.\operatorname*{hull}S\subseteq Q$%
\ \ \ \ \footnote{\textit{Proof.} Let $y\in\operatorname*{conv}%
.\operatorname*{hull}S$. Thus,%
\begin{align*}
y  &  \in\operatorname*{conv}.\operatorname*{hull}S\\
&  =\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1\text{ and }\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}
\end{align*}
(by (\ref{pf.prop2.0g.a.1})). In other words, $y$ is an element of $E$ such
that there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  $ of elements of $S$, and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}=y$. Let us denote this $t$, this $\left(  x_{1},x_{2}%
,\ldots,x_{t}\right)  $ and this $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ by $p$, $\left(  y_{1},y_{2},\ldots
,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $,
respectively. Thus, $p$ is an element of $\mathbb{N}$, and $\left(
y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$, and
$\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{p}\mu_{i}=1$ and
$\sum\limits_{i=1}^{p}\mu_{i}y_{i}=y$.
\par
Now, $y=\sum\limits_{i=1}^{p}\mu_{i}y_{i}$ (since $\sum\limits_{i=1}^{p}%
\mu_{i}y_{i}=y$). Hence, $y$ has the form $\sum\limits_{i=1}^{t}\lambda
_{i}x_{i}$ for some element $t$ of $\mathbb{N}$, some $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$, and some $t$-tuple
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative
reals such that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ (namely, for $t=p$,
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  y_{1},y_{2},\ldots
,y_{p}\right)  $ and $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  =\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $). In other
words,%
\begin{align*}
y  &  \in\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\} \\
&  =Q.
\end{align*}
\par
Now, let us forget that we fixed $y$. We thus have shown that $y\in Q$ for
every $y\in\operatorname*{conv}.\operatorname*{hull}S$. In other words,
$\operatorname*{conv}.\operatorname*{hull}S\subseteq Q$, qed.}. Combining
these two relations, we obtain $\operatorname*{conv}.\operatorname*{hull}S=Q$.
Thus,%
\begin{align*}
&  \operatorname*{conv}.\operatorname*{hull}S\\
&  =Q=\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  .
\end{align*}
This proves Proposition 2.0g \textbf{(a)}.

\textbf{(b)} We have $S\subseteq F\subseteq E$. If we regard $S$ as a subset
of $F$, then we have%
\begin{align}
&  \operatorname*{conv}.\operatorname*{hull}S\nonumber\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  \label{pf.prop2.0g.b.1}%
\end{align}
(by Proposition 2.0g \textbf{(a)}, applied to $F$ instead of $E$). If we
regard $S$ as a subset of $E$, then we have%
\begin{align}
&  \operatorname*{conv}.\operatorname*{hull}S\nonumber\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}=1
\end{array}
\right)  \right\}  \label{pf.prop2.0g.b.2}%
\end{align}
(by Proposition 2.0g \textbf{(a)}). The right hand sides of the equalities
(\ref{pf.prop2.0g.b.1}) and (\ref{pf.prop2.0g.b.2}) are equal (because these
right hand sides do not involve $E$ or $F$ in their definition). Therefore,
the left hand sides of the equalities (\ref{pf.prop2.0g.b.1}) and
(\ref{pf.prop2.0g.b.2}) are equal as well. In other words, the value of
$\operatorname*{conv}.\operatorname*{hull}S$ when $S$ is regarded as a subset
of $F$ equals the value of $\operatorname*{conv}.\operatorname*{hull}S$ when
$S$ is regarded as a subset of $E$. In other words, $\operatorname*{conv}%
.\operatorname*{hull}S$ does not depend on whether we consider $S$ as a subset
of $F$ or as a subset of $E$. This proves Proposition 2.0g \textbf{(b)}.
\end{proof}
\end{verlong}

\begin{proof}
[Proof of Proposition 2.0h.]\textbf{(a)} Let $S=\left\{  x_{1},x_{2}%
,\ldots,x_{n}\right\}  $. Let $Q$ be the set of all convex combinations of the
vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$.

\begin{vershort}
Recall that $Q$ is the set of all convex combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$. In other words, $Q$ is the set of all convex
combinations of the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $.

The set $Q$ is convex\footnote{\textit{Proof.} Let $y\in Q$, $z\in Q$ and
$\lambda\in\left[  0,1\right]  $. We shall show that $\lambda y+\left(
1-\lambda\right)  z\in Q$.
\par
Notice that $\lambda$ and $1-\lambda$ both are nonnegative reals (since
$\lambda\in\left[  0,1\right]  $).
\par
We have $y\in Q$. In other words, $y$ is a convex combination of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $ (since $Q$ is the set of all
convex combinations of the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots
,n\right\}  $). In other words, $y$ has the form $y=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\mu_{i}x_{i}$ for some family $\left(  \mu_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by
elements of $\left\{  1,2,\ldots,n\right\}  $ and having the property that all
but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\mu_{i}=0$,
and the property that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}=1$. Similarly, $z$ has the form $z=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\nu_{i}x_{i}$ for some family $\left(  \nu_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by
elements of $\left\{  1,2,\ldots,n\right\}  $ and having the property that all
but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\nu_{i}=0$,
and the property that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\nu_{i}=1$. Consider these two families $\left(  \mu_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ and $\left(  \nu_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, both $\lambda\mu_{i}$ and
$\left(  1-\lambda\right)  \nu_{i}$ are nonnegative reals (since $\lambda$,
$\mu_{i}$, $1-\lambda$ and $\nu_{i}$ are nonnegative reals), and hence the sum
$\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}$ is a nonnegative real.
Thus, we can define a family $\left(  \eta_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}
\right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$. We know that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\eta_{i}=0$ (since there are only finitely
many $i\in\left\{  1,2,\ldots,n\right\}  $). We have%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\eta_{i}%
}_{=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}}  &  =\sum_{i\in\left\{
1,2,\ldots,n\right\}  }\left(  \lambda\mu_{i}+\left(  1-\lambda\right)
\nu_{i}\right) \\
&  =\lambda\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}}_{=1}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\nu_{i}}_{=1}=\lambda+\left(  1-\lambda\right)  =1
\end{align*}
and%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\eta_{i}%
}_{=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}}x_{i}  &  =\sum
_{i\in\left\{  1,2,\ldots,n\right\}  }\left(  \lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}\right)  x_{i}\\
&  =\lambda\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}x_{i}}_{=y}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in
\left\{  1,2,\ldots,n\right\}  }\nu_{i}x_{i}}_{=z}=\lambda y+\left(
1-\lambda\right)  z.
\end{align*}
Hence, $\lambda y+\left(  1-\lambda\right)  z=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\eta_{i}x_{i}$. Therefore, $\lambda y+\left(
1-\lambda\right)  z$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by
elements of $\left\{  1,2,\ldots,n\right\}  $ and having the property that all
but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda
_{i}=0$, and the property that $\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }\lambda_{i}=1$ (namely, for the family $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }=\left(  \eta_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$). In other words, $\lambda y+\left(
1-\lambda\right)  z$ is a convex combination of the vectors $x_{i}$ or
$i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $\lambda y+\left(
1-\lambda\right)  z\in Q$ (since $Q$ is the set of all convex combinations of
the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $).
\par
Now, let us forget that we fixed $y$, $z$ and $\lambda$. We thus have proven
that every two elements $y\in Q$ and $z\in Q$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda y+\left(  1-\lambda\right)
z\in Q$. If we rename $y$ and $z$ as $x$ and $y$ in this statement, we obtain
the following: Every two elements $x\in Q$ and $y\in Q$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(  1-\lambda\right)
y\in Q$. In other words, the set $Q$ is convex (by the definition of
convexity). Qed.}. Moreover, $S\subseteq Q$\ \ \ \ \footnote{\textit{Proof.}
Let $w\in S$. Then, $w\in S=\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  $.
Thus, there exists some $k\in\left\{  1,2,\ldots,n\right\}  $ such that
$w=x_{k}$. Consider this $k$.
\par
Define a family $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}
\right)  .
\]
Then, it is easy to see that:
\par
\begin{itemize}
\item All but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy
$\mu_{i}=0$.
\par
\item We have $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}=1$.
\par
\item We have $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}%
x_{i}=x_{k}$.
\end{itemize}
\par
Hence, $w=x_{k}=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}%
$. Thus, $w$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$, and the
property that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}=1$
(namely, for the family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }=\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$). In other words, $w$ is a convex combination of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $w\in Q$
(since $Q$ is the set of all convex combinations of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $).
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in Q$ for
every $w\in S$. In other words, $S\subseteq Q$, qed.}. Hence, $Q$ is a convex
subset of $E$ which contains $S$ as a subset. Thus, Proposition 2.0f
\textbf{(f)} (applied to the convex subset $Q$ of $E$) shows that $Q$ contains
$\operatorname*{conv}.\operatorname*{hull}S$ as a subset. In other words,
$\operatorname*{conv}.\operatorname*{hull}S\subseteq Q$.
\end{vershort}

\begin{verlong}
The set $Q$ is convex\footnote{\textit{Proof.} Let $y\in Q$, $z\in Q$ and
$\lambda\in\left[  0,1\right]  $. We shall show that $\lambda y+\left(
1-\lambda\right)  z\in Q$.
\par
Notice that $\lambda$ and $1-\lambda$ both are nonnegative reals (since
$\lambda\in\left[  0,1\right]  $).
\par
We have $y\in Q$. In other words, $y$ belongs to the set $Q$. In other words,
$y$ belongs to the set of all convex combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ (since $Q$ is the set of all convex combinations of
the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$). In other words, $y$ is a
convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$. In
other words, $y$ is a convex combination of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $y$ has the form
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$, and the property that $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}=1$. Let us denote
this family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ by $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$. Thus,
$\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family
of nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $
and having the property that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\mu_{i}=0$, and the property that
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}=1$, and it
furthermore satisfies $y=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}x_{i}$.
\par
We have $z\in Q$. In other words, $z$ belongs to the set $Q$. In other words,
$z$ belongs to the set of all convex combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ (since $Q$ is the set of all convex combinations of
the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$). In other words, $z$ is a
convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$. In
other words, $z$ is a convex combination of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $z$ has the form
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$, and the property that $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}=1$. Let us denote
this family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ by $\left(  \nu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$. Thus,
$\left(  \nu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family
of nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $
and having the property that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\nu_{i}=0$, and the property that
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\nu_{i}=1$, and it
furthermore satisfies $z=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\nu_{i}x_{i}$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, both $\lambda\mu_{i}$ and
$\left(  1-\lambda\right)  \nu_{i}$ are nonnegative reals (since $\lambda$,
$\mu_{i}$, $1-\lambda$ and $\nu_{i}$ are nonnegative reals), and hence the sum
$\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}$ is a nonnegative real.
Thus, we can define a family $\left(  \eta_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}%
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}
\right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$ of nonnegative reals. We know that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\eta_{i}=0$ (since there are
only finitely many $i\in\left\{  1,2,\ldots,n\right\}  $). We have%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\eta_{i}%
}_{=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}}  &  =\sum_{i\in\left\{
1,2,\ldots,n\right\}  }\left(  \lambda\mu_{i}+\left(  1-\lambda\right)
\nu_{i}\right) \\
&  =\lambda\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}}_{=1}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\nu_{i}}_{=1}=\lambda+\left(  1-\lambda\right)  =1
\end{align*}
and%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\eta_{i}%
}_{=\lambda\mu_{i}+\left(  1-\lambda\right)  \nu_{i}}x_{i}  &  =\sum
_{i\in\left\{  1,2,\ldots,n\right\}  }\left(  \lambda\mu_{i}+\left(
1-\lambda\right)  \nu_{i}\right)  x_{i}\\
&  =\lambda\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}x_{i}}_{=y}+\left(  1-\lambda\right)  \underbrace{\sum\limits_{i\in
\left\{  1,2,\ldots,n\right\}  }\nu_{i}x_{i}}_{=z}=\lambda y+\left(
1-\lambda\right)  z.
\end{align*}
Hence, $\lambda y+\left(  1-\lambda\right)  z=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\eta_{i}x_{i}$. Therefore, $\lambda y+\left(
1-\lambda\right)  z$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by
elements of $\left\{  1,2,\ldots,n\right\}  $ and having the property that all
but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda
_{i}=0$, and the property that $\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }\lambda_{i}=1$ (namely, for the family $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }=\left(  \eta_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$). In other words, $\lambda y+\left(
1-\lambda\right)  z$ is a convex combination of the vectors $x_{i}$ or
$i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $\lambda y+\left(
1-\lambda\right)  z$ is a convex combination of the vectors $x_{1}$, $x_{2}$,
$\ldots$, $x_{n}$. In other words, $\lambda y+\left(  1-\lambda\right)  z$
belongs to the set of all convex combinations of the vectors $x_{1}$, $x_{2}$,
$\ldots$, $x_{n}$. In other words, $\lambda y+\left(  1-\lambda\right)  z$
belongs to $Q$ (since $Q$ is the set of all convex combinations of the vectors
$x_{1}$, $x_{2}$, $\ldots$, $x_{n}$). In other words, $\lambda y+\left(
1-\lambda\right)  z\in Q$.
\par
Now, let us forget that we fixed $y$, $z$ and $\lambda$. We thus have proven
that every two elements $y\in Q$ and $z\in Q$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda y+\left(  1-\lambda\right)
z\in Q$. If we rename $y$ and $z$ as $x$ and $y$ in this statement, we obtain
the following: Every two elements $x\in Q$ and $y\in Q$ and every real number
$\lambda\in\left[  0,1\right]  $ satisfy $\lambda x+\left(  1-\lambda\right)
y\in Q$. In other words, the set $Q$ is convex (by the definition of
convexity). Qed.}. Moreover, $S\subseteq Q$\ \ \ \ \footnote{\textit{Proof.}
Let $w\in S$. Then, $w\in S=\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  $.
Thus, there exists some $k\in\left\{  1,2,\ldots,n\right\}  $ such that
$w=x_{k}$. Consider this $k$.
\par
Define a family $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}
\right)  .
\]
Then, all but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy
$\mu_{i}=0$ (since there are only finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $). Also,%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}  &  =\sum
\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}}\underbrace{\mu
_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
=1\\\text{(since }i=k\text{)}}}+\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
=0\\\text{(since }i\neq k\text{)}}}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}1+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i\neq k}}0}_{=0}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}1=1\ \ \ \ \ \ \ \ \ \ \left(  \text{since }k\in\left\{  1,2,\ldots
,n\right\}  \right)
\end{align*}
and%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}  &
=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
=1\\\text{(since }i=k\text{)}}}x_{i}+\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
=0\\\text{(since }i\neq k\text{)}}}x_{i}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}1x_{i}+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i\neq k}}0x_{i}}_{=0}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}\underbrace{1x_{i}}_{=x_{i}}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i=k}}x_{i}=x_{k}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }k\in\left\{  1,2,\ldots,n\right\}  \right) \\
&  =w.
\end{align*}
\par
Hence, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}$.
Thus, $w$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$, and the
property that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}=1$
(namely, for the family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }=\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$). In other words, $w$ is a convex combination of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $w$ is a
convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$. In
other words, $w$ belongs to the set of all convex combinations of the vectors
$x_{1}$, $x_{2}$, $\ldots$, $x_{n}$. In other words, $w$ belongs to $Q$ (since
$Q$ is the set of all convex combinations of the vectors $x_{1}$, $x_{2}$,
$\ldots$, $x_{n}$). In other words, $w\in Q$.
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in Q$ for
every $w\in S$. In other words, $S\subseteq Q$, qed.}. Hence, $Q$ is a convex
subset of $E$ which contains $S$ as a subset. Thus, the intersection of all
convex subsets of $E$ which contain $S$ as a subset is a subset of $Q$. In
other words, $\operatorname*{conv}.\operatorname*{hull}S$ is a subset of $Q$
(since $\operatorname*{conv}.\operatorname*{hull}S$ is the intersection of all
convex subsets of $E$ which contain $S$ as a subset (because of Definition
2.0c)). In other words, $\operatorname*{conv}.\operatorname*{hull}S\subseteq
Q$.
\end{verlong}

\begin{vershort}
On the other hand, $Q\subseteq\operatorname*{conv}.\operatorname*{hull}%
S$\ \ \ \ \footnote{\textit{Proof.} Let $w\in Q$. We shall show that
$w\in\operatorname*{conv}.\operatorname*{hull}S$.
\par
Let $D$ be any convex subset of $E$ which contains $S$ as a subset. Thus, $D$
is a convex subset of $E$ and satisfies $S\subseteq D$.
\par
We have $w\in Q$. In other words, $w$ is a convex combination of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $ (since $Q$ is the set of all
convex combinations of the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots
,n\right\}  $). In other words, $w$ has the form $w=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some family $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals
indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and having the
property that all but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $
satisfy $\lambda_{i}=0$, and the property that $\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\lambda_{i}=1$. Consider this family $\left(
\lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, we have $x_{i}\in\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  =S\subseteq D$. Thus, $\left(  x_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family of elements of $D$. Hence,
Proposition 2.0a (applied to $D$, $\left\{  1,2,\ldots,n\right\}  $, $\left(
x_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ and $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ instead of $C$, $I$,
$\left(  x_{i}\right)  _{i\in I}$ and $\left(  \lambda_{i}\right)  _{i\in I}$)
shows that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}%
x_{i}\in D$. Thus, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\lambda_{i}x_{i}\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex subset of $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex subset of $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex subsets of $E$ which
contain $S$ as a subset. In other words, $w$ lies in $\operatorname*{conv}%
.\operatorname*{hull}S$ (since $\operatorname*{conv}.\operatorname*{hull}S$ is
the intersection of all convex subsets of $E$ which contain $S$ as a subset
(because of Definition 2.0c)). In other words, $w\in\operatorname*{conv}%
.\operatorname*{hull}S$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in
\operatorname*{conv}.\operatorname*{hull}S$ for every $w\in Q$. In other
words, $Q\subseteq\operatorname*{conv}.\operatorname*{hull}S$, qed.}.
Combining this with $\operatorname*{conv}.\operatorname*{hull}S\subseteq Q$,
we obtain $\operatorname*{conv}.\operatorname*{hull}S=Q$. Thus,%
\begin{align*}
&  \operatorname*{conv}.\operatorname*{hull}\underbrace{\left\{  x_{1}%
,x_{2},\ldots,x_{n}\right\}  }_{=S}\\
&  =\operatorname*{conv}.\operatorname*{hull}S=Q\\
&  =\left(  \text{the set of all convex combinations of the vectors }%
x_{1}\text{, }x_{2}\text{, }\ldots\text{, }x_{n}\right)
\end{align*}
(by the definition of $Q$). This proves Proposition 2.0h \textbf{(a)}.
\end{vershort}

\begin{verlong}
On the other hand, $Q\subseteq\operatorname*{conv}.\operatorname*{hull}%
S$\ \ \ \ \footnote{\textit{Proof.} Let $w\in Q$. We shall show that
$w\in\operatorname*{conv}.\operatorname*{hull}S$.
\par
Let $D$ be any convex subset of $E$ which contains $S$ as a subset. Thus, $D$
is a convex subset of $E$ and satisfies $S\subseteq D$.
\par
We have $w\in Q$. In other words, $w$ belongs to the set $Q$. In other words,
$w$ belongs to the set of all convex combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ (since $Q$ is the set of all convex combinations of
the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$). In other words, $w$ is a
convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$. In
other words, $w$ is a convex combination of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $. In other words, $w$ has the form
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$, and the property that $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}=1$. Consider this
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$.
Thus, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, we have $x_{i}\in\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  =S\subseteq D$. Thus, $\left(  x_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family of elements of $D$. Hence,
Proposition 2.0a (applied to $D$, $\left\{  1,2,\ldots,n\right\}  $, $\left(
x_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ and $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ instead of $C$, $I$,
$\left(  x_{i}\right)  _{i\in I}$ and $\left(  \lambda_{i}\right)  _{i\in I}$)
shows that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}%
x_{i}\in D$. Thus, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\lambda_{i}x_{i}\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex subset of $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex subset of $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex subsets of $E$ which
contain $S$ as a subset. In other words, $w$ lies in $\operatorname*{conv}%
.\operatorname*{hull}S$ (since $\operatorname*{conv}.\operatorname*{hull}S$ is
the intersection of all convex subsets of $E$ which contain $S$ as a subset
(because of Definition 2.0c)). In other words, $w\in\operatorname*{conv}%
.\operatorname*{hull}S$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in
\operatorname*{conv}.\operatorname*{hull}S$ for every $w\in Q$. In other
words, $Q\subseteq\operatorname*{conv}.\operatorname*{hull}S$, qed.}.
Combining this with $\operatorname*{conv}.\operatorname*{hull}S\subseteq Q$,
we obtain $\operatorname*{conv}.\operatorname*{hull}S=Q$. Thus,%
\begin{align*}
&  \operatorname*{conv}.\operatorname*{hull}\underbrace{\left\{  x_{1}%
,x_{2},\ldots,x_{n}\right\}  }_{=S}\\
&  =\operatorname*{conv}.\operatorname*{hull}S=Q\\
&  =\left(  \text{the set of all convex combinations of the vectors }%
x_{1}\text{, }x_{2}\text{, }\ldots\text{, }x_{n}\right)
\end{align*}
(since $Q$ is the set of all convex combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$). This proves Proposition 2.0h \textbf{(a)}.
\end{verlong}

\textbf{(b)} We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  x\in\underbrace{\operatorname*{conv}.\operatorname*{hull}\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  }_{\substack{=\left(  \text{the set of all
convex combinations of the vectors }x_{1}\text{, }x_{2}\text{, }\ldots\text{,
}x_{n}\right)  \\\text{(by Proposition 2.0h \textbf{(a)})}}}\right) \\
&  \Longleftrightarrow\ \left(  x\in\left(  \text{the set of all convex
combinations of the vectors }x_{1}\text{, }x_{2}\text{, }\ldots\text{, }%
x_{n}\right)  \right) \\
&  \Longleftrightarrow\ \left(  x\text{ is a convex combination of the vectors
}x_{1}\text{, }x_{2}\text{, }\ldots\text{, }x_{n}\right)  .
\end{align*}
In other words, we have $x\in\operatorname*{conv}.\operatorname*{hull}\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  $ if and only if $x$ is a convex combination
of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$. This proves Proposition
2.0h \textbf{(b)}.
\end{proof}

\subsection{Proofs for Section \ref{sect.basics-cone}}

\begin{proof}
[Proof of Proposition 2.0i.]We will first show that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }J\text{ is any finite subset of }I\text{, and if }\left(  \mu
_{i}\right)  _{i\in J}\text{ is a}\\
\text{family of nonnegative reals indexed by }J\text{,}\\
\text{then }\sum\limits_{i\in J}\mu_{i}x_{i}\in C
\end{array}
\right)  . \label{pf.prop2.0i.goal}%
\end{equation}


[\textit{Proof of (\ref{pf.prop2.0i.goal}):} We shall prove
(\ref{pf.prop2.0i.goal}) by strong induction over $\left\vert J\right\vert $:

\textit{Induction step:}\footnote{A strong induction needs no induction base.}
Fix $N\in\mathbb{N}$. Assume that (\ref{pf.prop2.0i.goal}) holds in the case
when $\left\vert J\right\vert <N$. We must prove that (\ref{pf.prop2.0i.goal})
holds in the case when $\left\vert J\right\vert =N$.

We have assumed that (\ref{pf.prop2.0i.goal}) holds in the case when
$\left\vert J\right\vert <N$. In other words,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }J\text{ is any finite subset of }I\text{ satisfying }\left\vert
J\right\vert <N\text{, and}\\
\text{if }\left(  \mu_{i}\right)  _{i\in J}\text{ is a family of nonnegative
reals indexed}\\
\text{by }J\text{, then }\sum\limits_{i\in J}\mu_{i}x_{i}\in C
\end{array}
\right)  . \label{pf.prop2.0i.indhyp}%
\end{equation}


Now, let us prove that (\ref{pf.prop2.0i.goal}) holds in the case when
$\left\vert J\right\vert =N$. Thus, let $J$ be any finite subset of $I$
satisfying $\left\vert J\right\vert =N$, and let $\left(  \mu_{i}\right)
_{i\in J}$ be a family of nonnegative reals indexed by $J$. We shall show that
$\sum\limits_{i\in J}\mu_{i}x_{i}\in C$.

If $\left\vert J\right\vert \leq0$, then $\sum\limits_{i\in J}\mu_{i}x_{i}\in
C$ holds\footnote{\textit{Proof.} Assume that $\left\vert J\right\vert \leq0$.
We need to prove $\sum\limits_{i\in J}\mu_{i}x_{i}\in C$.
\par
From $\left\vert J\right\vert \leq0$, we obtain $\left\vert J\right\vert =0$.
Thus, $J=\varnothing$.
\par
Recall that $C$ is a convex cone. Hence, $0\in C$ (according to the definition
of a convex cone).
\par
But $J=\varnothing$, and thus $\sum\limits_{i\in J}\mu_{i}x_{i}=\left(
\text{empty sum}\right)  =0\in C$, qed.}. Hence, for the rest of our proof of
$\sum\limits_{i\in J}\mu_{i}x_{i}\in C$, we can WLOG assume that we don't have
$\left\vert J\right\vert \leq0$. Assume this.

We have $\left\vert J\right\vert >0$ (since we don't have $\left\vert
J\right\vert \leq0$). Thus, $J\neq\varnothing$. Hence, there exists a $k\in
J$. Consider such a $k$.

Now, $\left(  \mu_{i}\right)  _{i\in J\setminus\left\{  k\right\}  }$ is a
family of nonnegative reals indexed by $J\setminus\left\{  k\right\}  $.
Moreover, $k\in J$ and thus $\left\vert J\setminus\left\{  k\right\}
\right\vert =\underbrace{\left\vert J\right\vert }_{=N}-1=N-1<N$. Thus, we can
apply (\ref{pf.prop2.0i.indhyp}) to $J\setminus\left\{  k\right\}  $ and
$\left(  \mu_{i}\right)  _{i\in J\setminus\left\{  k\right\}  }$ instead of
$J$ and $\left(  \mu_{i}\right)  _{i\in J}$. As a result, we obtain
$\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}x_{i}\in C$. Also,
$x_{k}\in C$ (since $\left(  x_{i}\right)  _{i\in I}$ is a family of elements
of $C$).

Recall that $C$ is a convex cone. Thus, every two elements $x\in C$ and $y\in
C$ and every nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in
C$ (by the definition of a convex cone). Applying this to $x=x_{k}$,
$y=\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}x_{i}$,
$\lambda=\mu_{k}$ and $\mu=1$, we obtain $\mu_{k}x_{k}+1\sum\limits_{i\in
J\setminus\left\{  k\right\}  }\mu_{i}x_{i}\in C$. Now,%
\begin{align*}
\sum\limits_{i\in J}\mu_{i}x_{i}  &  =\underbrace{\sum\limits_{\substack{i\in
J;\\i=k}}\mu_{i}}_{\substack{=\mu_{k}x_{k}\\\text{(since }k\in J\text{)}%
}}+\underbrace{\sum\limits_{\substack{i\in J;\\i\neq k}}}_{=\sum\limits_{i\in
J\setminus\left\{  k\right\}  }}\mu_{i}x_{i}=\mu_{k}x_{k}+\underbrace{\sum
\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}x_{i}}_{=1\sum
\limits_{i\in J\setminus\left\{  k\right\}  }\mu_{i}x_{i}}\\
&  =\mu_{k}x_{k}+1\sum\limits_{i\in J\setminus\left\{  k\right\}  }\mu
_{i}x_{i}\in C.
\end{align*}


Now, let us forget that we fixed $J$ and $\left(  \mu_{i}\right)  _{i\in J}$.
We thus have proven that if $J$ is any finite subset of $I$ satisfying
$\left\vert J\right\vert =N$, and if $\left(  \mu_{i}\right)  _{i\in J}$ is a
family of nonnegative reals indexed by $J$, then $\sum\limits_{i\in J}\mu
_{i}x_{i}\in C$. In other words, (\ref{pf.prop2.0i.goal}) holds in the case
when $\left\vert J\right\vert =N$. This completes the induction step.

Thus, (\ref{pf.prop2.0i.goal}) is proven by induction.]

Now, recall that all but finitely many $i\in I$ satisfy $\lambda_{i}=0$. In
other words, there exists a finite subset $J$ of $I$ such that%
\begin{equation}
\text{every }i\in I\setminus J\text{ satisfies }\lambda_{i}=0.
\label{pf.prop2.0i.cofin}%
\end{equation}
Consider this $J$. Then, (\ref{pf.prop2.0i.goal}) (applied to $\left(  \mu
_{i}\right)  _{i\in J}=\left(  \lambda_{i}\right)  _{i\in J}$) shows that
$\sum\limits_{i\in J}\lambda_{i}x_{i}\in C$. Now,%
\begin{align*}
\sum\limits_{i\in I}\lambda_{i}x_{i}  &  =\underbrace{\sum
\limits_{\substack{i\in I;\\i\in J}}}_{\substack{=\sum\limits_{i\in
J}\\\text{(since }J\subseteq I\text{)}}}\lambda_{i}x_{i}+\underbrace{\sum
\limits_{\substack{i\in I;\\i\notin J}}}_{=\sum\limits_{i\in I\setminus J}%
}\lambda_{i}x_{i}=\sum\limits_{i\in J}\lambda_{i}x_{i}+\sum\limits_{i\in
I\setminus J}\underbrace{\lambda_{i}}_{\substack{=0\\\text{(by
(\ref{pf.prop2.0i.cofin}))}}}\\
&  =\sum\limits_{i\in J}\lambda_{i}x_{i}+\underbrace{\sum\limits_{i\in
I\setminus J}0x_{i}}_{=0}=\sum\limits_{i\in J}\lambda_{i}x_{i}\in C.
\end{align*}
This proves Proposition 2.0i.
\end{proof}

\begin{proof}
[Proof of Proposition 2.0m.]\textbf{(a)} Let $E$ be an $\mathbb{R}$-vector
space. Let $S$ be a subset of $E$.

Let $C_{1}$ be the intersection of all convex cones in $E$ which contain $S$
as a subset. Definition 2.0j defined the convex conic hull of $S$ to be this
set $C_{1}$. In other words,%
\begin{equation}
\left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0j}\right)  =C_{1}. \label{pf.prop2.0m.a.chull1}%
\end{equation}


Let $C_{2}$ be the set of all linear combinations of the vectors $s$ for $s\in
S$ with nonnegative coefficients. Definition 2.0k defined the convex conic
hull of $S$ to be this set $C_{2}$. In other words,%
\begin{equation}
\left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0k}\right)  =C_{2}. \label{pf.prop2.0m.a.chull2}%
\end{equation}


Let $C_{3}$ be the set%
\[
\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
Definition 2.0l defined the convex conic hull of $S$ to be this set $C_{3}$.
In other words,%
\begin{equation}
\left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0l}\right)  =C_{3}. \label{pf.prop2.0m.a.chull3}%
\end{equation}


\begin{vershort}
The set $C_{3}$ is a convex cone\footnote{\textit{Proof.} As usual, we let
$\left(  {}\right)  $ denote an empty $0$-tuple. If $\left(  x_{1}%
,x_{2},\ldots,x_{0}\right)  =\left(  {}\right)  $ and $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{0}\right)  =\left(  {}\right)  $, then
$\sum\limits_{i=1}^{0}\lambda_{i}x_{i}=\left(  \text{empty sum}\right)  =0$.
Hence, $0$ is an element of $E$ such that there exist some $t\in\mathbb{N}$, a
$t$-tuple $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a
$t$-tuple $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of
nonnegative reals such that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=0$ (namely,
$t=0$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  {}\right)  $ and
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(
{}\right)  $). In other words,%
\[
0\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Let $y\in C_{3}$ and $z\in C_{3}$. Let $\lambda$ and $\kappa$ be two
nonnegative reals. We shall show that $\lambda y+\kappa z\in C_{3}$.
\par
We have%
\[
y\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, we can write $y$ in the form $y=\sum\limits_{i=1}^{t}%
\lambda_{i}x_{i}$ for some $t\in\mathbb{N}$, some $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and some $t$-tuple
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative
reals. Let us denote this $t$, this $\left(  x_{1},x_{2},\ldots,x_{t}\right)
$ and this $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by
$p$, $\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ and $\left(  \mu_{1},\mu
_{2},\ldots,\mu_{p}\right)  $, respectively. Thus, $p$ is an element of
$\mathbb{N}$, and $\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple
of elements of $S$, and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a
$p$-tuple of nonnegative reals such that $y=\sum\limits_{i=1}^{p}\mu_{i}y_{i}%
$.
\par
Similarly, use the assumption $z\in C_{3}$ to write $z$ in the form
$z=\sum\limits_{i=1}^{q}\nu_{i}z_{i}$, where $q$ is an element of $\mathbb{N}%
$, and where $\left(  z_{1},z_{2},\ldots,z_{q}\right)  $ is a $q$-tuple of
elements of $S$, and where $\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $
is a $q$-tuple of nonnegative reals.
\par
Define a $\left(  p+q\right)  $-tuple $\left(  w_{1},w_{2},\ldots
,w_{p+q}\right)  $ of elements of $S$ by%
\[
\left(  w_{1},w_{2},\ldots,w_{p+q}\right)  =\left(  y_{1},y_{2},\ldots
,y_{p},z_{1},z_{2},\ldots,z_{q}\right)  .
\]
Then:
\par
\begin{itemize}
\item For every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{equation}
w_{i}=y_{i}. \label{pf.prop2.0m.short.wi.1}%
\end{equation}
\par
\item For every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{equation}
w_{i}=z_{i-p}. \label{pf.prop2.0m.short.wi.2}%
\end{equation}
\end{itemize}
\par
Furthermore, define a $\left(  p+q\right)  $-tuple $\left(  \eta_{1},\eta
_{2},\ldots,\eta_{p+q}\right)  $ of nonnegative reals by%
\[
\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  =\left(  \lambda\mu
_{1},\lambda\mu_{2},\ldots,\lambda\mu_{p},\kappa\nu_{1},\kappa\nu_{2}%
,\ldots,\kappa\nu_{q}\right)
\]
(this is well-defined because both $\lambda$ and $\kappa$ are nonnegative
reals and because $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ and
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $ are tuples of nonnegative
reals). Then:
\par
\begin{itemize}
\item For every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{equation}
\eta_{i}=\lambda\mu_{i}. \label{pf.prop2.0m.short.etai.1}%
\end{equation}
\par
\item For every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{equation}
\eta_{i}=\kappa\nu_{i-p}. \label{pf.prop2.0m.short.etai.2}%
\end{equation}
\end{itemize}
\par
Now,%
\begin{align*}
\sum_{i=1}^{p+q}\eta_{i}w_{i}  &  =\sum_{i=1}^{p}\underbrace{\eta_{i}%
}_{\substack{=\lambda\mu_{i}\\\text{(by (\ref{pf.prop2.0m.short.etai.1}))}%
}}\underbrace{w_{i}}_{\substack{=y_{i}\\\text{(by
(\ref{pf.prop2.0m.short.wi.1}))}}}+\sum_{i=p+1}^{p+q}\underbrace{\eta_{i}%
}_{\substack{=\kappa\nu_{i-p}\\\text{(by (\ref{pf.prop2.0m.short.etai.2}))}%
}}\underbrace{w_{i}}_{\substack{=z_{i-p}\\\text{(by
(\ref{pf.prop2.0m.short.wi.2}))}}}=\underbrace{\sum_{i=1}^{p}\lambda\mu
_{i}y_{i}}_{=\lambda\sum\limits_{i=1}^{p}\mu_{i}y_{i}}+\underbrace{\sum
_{i=p+1}^{p+q}\kappa\nu_{i-p}z_{i-p}}_{=\kappa\sum\limits_{i=p+1}^{p+q}%
\nu_{i-p}z_{i-p}}\\
&  =\lambda\sum\limits_{i=1}^{p}\mu_{i}y_{i}+\kappa\sum\limits_{i=p+1}%
^{p+q}\nu_{i-p}z_{i-p}=\lambda\underbrace{\sum\limits_{i=1}^{p}\mu_{i}y_{i}%
}_{=y}+\kappa\underbrace{\sum\limits_{i=1}^{q}\nu_{i}z_{i}}_{=z}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-p\text{ in the second sum}\right) \\
&  =\lambda y+\kappa z.
\end{align*}
Hence, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=\lambda y+\kappa z$ (namely,
$t=p+q$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w_{1},w_{2}%
,\ldots,w_{p+q}\right)  $ and $\left(  \lambda_{1},\lambda_{2},\ldots
,\lambda_{t}\right)  =\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  $).
In other words,%
\[
\lambda y+\kappa z\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Now, let us forget that we fixed $y$, $z$, $\lambda$ and $\kappa$. We thus
have proven that every two elements $y\in C_{3}$ and $z\in C_{3}$ and every
nonnegative reals $\lambda$ and $\kappa$ satisfy $\lambda y+\kappa z\in C_{3}%
$. If we rename $y$, $z$ and $\kappa$ as $x$, $y$ and $\mu$ in this statement,
we obtain the following: Every two elements $x\in C_{3}$ and $y\in C_{3}$ and
every nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{3}%
$.
\par
Thus, we have proven the following two statements:
\par
\begin{itemize}
\item We have $0\in C_{3}$.
\par
\item Every two elements $x\in C_{3}$ and $y\in C_{3}$ and every nonnegative
reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{3}$.
\end{itemize}
\par
In other words, the set $C_{3}$ is a convex cone (by the definition of a
convex cone).}. Moreover, $S\subseteq C_{3}$\ \ \ \ \footnote{\textit{Proof.}
Let $w\in S$. Then, $\sum\limits_{i=1}^{1}1w=1w=w$. Hence, there exist some
$t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ of
elements of $S$ and a $t$-tuple $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  $ of nonnegative reals such that $\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=w$ (namely, $t=1$, $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  =\left(  w\right)  $ and $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(  1\right)  $). In other
words,%
\[
w\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{3}$ for
every $w\in S$. In other words, $S\subseteq C_{3}$.}. Hence, $C_{3}$ is a
convex cone in $E$ which contains $S$ as a subset. Thus, the intersection of
all convex cones in $E$ which contain $S$ as a subset is a subset of $C_{3}$.
In other words, $C_{1}$ is a subset of $C_{3}$ (since $C_{1}$ is the
intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $C_{1}\subseteq C_{3}$.
\end{vershort}

\begin{verlong}
The set $C_{3}$ is a convex cone\footnote{\textit{Proof.} As usual, we let
$\left(  {}\right)  $ denote an empty $0$-tuple. If $\left(  x_{1}%
,x_{2},\ldots,x_{0}\right)  =\left(  {}\right)  $ and $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{0}\right)  =\left(  {}\right)  $, then
$\sum\limits_{i=1}^{0}\lambda_{i}x_{i}=\left(  \text{empty sum}\right)  =0$.
Hence, $0$ is an element of $E$ such that there exist some $t\in\mathbb{N}$, a
$t$-tuple $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a
$t$-tuple $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of
nonnegative reals such that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=0$ (namely,
$t=0$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  {}\right)  $ and
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(
{}\right)  $). In other words,%
\[
0\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Let $y\in C_{3}$ and $z\in C_{3}$. Let $\lambda$ and $\kappa$ be two
nonnegative reals. We shall show that $\lambda y+\kappa z\in C_{3}$.
\par
We have%
\[
y\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=y$. Let us denote this $t$, this
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by $p$, $\left(  y_{1}%
,y_{2},\ldots,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{p}\right)  $, respectively. Thus, $p$ is an element of $\mathbb{N}$, and
$\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$,
and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{p}\mu_{i}y_{i}=y$.
\par
Also,
\[
z\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=z$. Let us denote this $t$, this
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by $q$, $\left(  z_{1}%
,z_{2},\ldots,z_{q}\right)  $ and $\left(  \nu_{1},\nu_{2},\ldots,\nu
_{q}\right)  $, respectively. Thus, $q$ is an element of $\mathbb{N}$, and
$\left(  z_{1},z_{2},\ldots,z_{q}\right)  $ is a $q$-tuple of elements of $S$,
and $\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $ is a $q$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{q}\nu_{i}z_{i}=z$.
\par
Define a $\left(  p+q\right)  $-tuple $\left(  w_{1},w_{2},\ldots
,w_{p+q}\right)  $ of elements of $S$ by%
\[
\left(  w_{1},w_{2},\ldots,w_{p+q}\right)  =\left(  y_{1},y_{2},\ldots
,y_{p},z_{1},z_{2},\ldots,z_{q}\right)  .
\]
Then,%
\[
w_{i}=%
\begin{cases}
y_{i}, & \text{if }i\leq p;\\
z_{i-p}, & \text{if }i>p
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,p+q\right\}  .
\]
Hence, for every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{align}
w_{i}  &  =%
\begin{cases}
y_{i}, & \text{if }i\leq p;\\
z_{i-p}, & \text{if }i>p
\end{cases}
=y_{i}\label{pf.prop2.0m.wi.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq p\text{ (since }i\in\left\{
1,2,\ldots,p\right\}  \text{)}\right)  .\nonumber
\end{align}
Also, for every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{align}
w_{i}  &  =%
\begin{cases}
y_{i}, & \text{if }i\leq p;\\
z_{i-p}, & \text{if }i>p
\end{cases}
=z_{i-p}\label{pf.prop2.0m.wi.2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>p\text{ (since }i\in\left\{
p+1,p+2,\ldots,p+q\right\}  \text{)}\right)  .\nonumber
\end{align}
\par
Furthermore, define a $\left(  p+q\right)  $-tuple $\left(  \eta_{1},\eta
_{2},\ldots,\eta_{p+q}\right)  $ of nonnegative reals by%
\[
\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  =\left(  \lambda\mu
_{1},\lambda\mu_{2},\ldots,\lambda\mu_{p},\kappa\nu_{1},\kappa\nu_{2}%
,\ldots,\kappa\nu_{q}\right)
\]
(this is well-defined because both $\lambda$ and $\kappa$ are nonnegative
reals and because $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ and
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{q}\right)  $ are tuples of nonnegative
reals). Then,%
\[
\eta_{i}=%
\begin{cases}
\lambda\mu_{i}, & \text{if }i\leq p;\\
\kappa\nu_{i-p}, & \text{if }i>p
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,p+q\right\}  .
\]
Hence, for every $i\in\left\{  1,2,\ldots,p\right\}  $, we have%
\begin{align}
\eta_{i}  &  =%
\begin{cases}
\lambda\mu_{i}, & \text{if }i\leq p;\\
\kappa\nu_{i-p}, & \text{if }i>p
\end{cases}
=\lambda\mu_{i}\label{pf.prop2.0m.etai.1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq p\text{ (since }i\in\left\{
1,2,\ldots,p\right\}  \text{)}\right)  .\nonumber
\end{align}
Also, for every $i\in\left\{  p+1,p+2,\ldots,p+q\right\}  $, we have%
\begin{align}
\eta_{i}  &  =%
\begin{cases}
\lambda\mu_{i}, & \text{if }i\leq p;\\
\kappa\nu_{i-p}, & \text{if }i>p
\end{cases}
=\kappa\nu_{i-p}\label{pf.prop2.0m.etai.2}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }i>p\text{ (since }i\in\left\{
p+1,p+2,\ldots,p+q\right\}  \text{)}\right)  .\nonumber
\end{align}
\par
Now,%
\begin{align*}
\sum_{i=1}^{p+q}\eta_{i}w_{i}  &  =\sum_{i=1}^{p}\underbrace{\eta_{i}%
}_{\substack{=\lambda\mu_{i}\\\text{(by (\ref{pf.prop2.0m.etai.1}))}%
}}\underbrace{w_{i}}_{\substack{=y_{i}\\\text{(by (\ref{pf.prop2.0m.wi.1}))}%
}}+\sum_{i=p+1}^{p+q}\underbrace{\eta_{i}}_{\substack{=\kappa\nu
_{i-p}\\\text{(by (\ref{pf.prop2.0m.etai.2}))}}}\underbrace{w_{i}%
}_{\substack{=z_{i-p}\\\text{(by (\ref{pf.prop2.0m.wi.2}))}}}=\underbrace{\sum
_{i=1}^{p}\lambda\mu_{i}y_{i}}_{=\lambda\sum\limits_{i=1}^{p}\mu_{i}y_{i}%
}+\underbrace{\sum_{i=p+1}^{p+q}\kappa\nu_{i-p}z_{i-p}}_{=\kappa
\sum\limits_{i=p+1}^{p+q}\nu_{i-p}z_{i-p}}\\
&  =\lambda\sum\limits_{i=1}^{p}\mu_{i}y_{i}+\kappa\sum\limits_{i=p+1}%
^{p+q}\nu_{i-p}z_{i-p}=\lambda\underbrace{\sum\limits_{i=1}^{p}\mu_{i}y_{i}%
}_{=y}+\kappa\underbrace{\sum\limits_{i=1}^{q}\nu_{i}z_{i}}_{=z}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }i\text{ for
}i-p\text{ in the second sum}\right) \\
&  =\lambda y+\kappa z.
\end{align*}
Hence, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=\lambda y+\kappa z$ (namely,
$t=p+q$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w_{1},w_{2}%
,\ldots,w_{p+q}\right)  $ and $\left(  \lambda_{1},\lambda_{2},\ldots
,\lambda_{t}\right)  =\left(  \eta_{1},\eta_{2},\ldots,\eta_{p+q}\right)  $).
In other words,%
\[
\lambda y+\kappa z\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Now, let us forget that we fixed $y$, $z$, $\lambda$ and $\kappa$. We thus
have proven that every two elements $y\in C_{3}$ and $z\in C_{3}$ and every
nonnegative reals $\lambda$ and $\kappa$ satisfy $\lambda y+\kappa z\in C_{3}%
$. If we rename $y$, $z$ and $\kappa$ as $x$, $y$ and $\mu$ in this statement,
we obtain the following: Every two elements $x\in C_{3}$ and $y\in C_{3}$ and
every nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{3}%
$.
\par
Thus, we have proven the following two statements:
\par
\begin{itemize}
\item We have $0\in C_{3}$.
\par
\item Every two elements $x\in C_{3}$ and $y\in C_{3}$ and every nonnegative
reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{3}$.
\end{itemize}
\par
In other words, the set $C_{3}$ is a convex cone (by the definition of a
convex cone). Qed.}. Moreover, $S\subseteq C_{3}$%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Then, $\sum\limits_{i=1}%
^{1}1w=1w=w$. Hence, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=w$ (namely, $t=1$, $\left(
x_{1},x_{2},\ldots,x_{t}\right)  =\left(  w\right)  $ and $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(  1\right)  $). In other
words,%
\[
w\in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  =C_{3}.
\]
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{3}$ for
every $w\in S$. In other words, $S\subseteq C_{3}$, qed.}. Hence, $C_{3}$ is a
convex cone in $E$ which contains $S$ as a subset. Thus, the intersection of
all convex cones in $E$ which contain $S$ as a subset is a subset of $C_{3}$.
In other words, $C_{1}$ is a subset of $C_{3}$ (since $C_{1}$ is the
intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $C_{1}\subseteq C_{3}$.
\end{verlong}

On the other hand, $C_{3}\subseteq C_{1}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in C_{3}$. We shall show that $w\in C_{1}$.
\par
Let $D$ be any convex cone in $E$ which contains $S$ as a subset. Thus, $D$ is
a convex cone in $E$ and satisfies $S\subseteq D$.
\par
We have%
\[
w\in C_{3}=\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  .
\]
In other words, there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=w$. Consider this $t$, this
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $. For every $i\in\left\{
1,2,\ldots,t\right\}  $, we have $x_{i}\in S\subseteq D$. Thus, $\left(
x_{i}\right)  _{i\in\left\{  1,2,\ldots,t\right\}  }$ is a family of elements
of $D$. Also, clearly, $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,t\right\}  }$ is a family of nonnegative reals (since $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ is a $t$-tuple of
nonnegative reals). Clearly, all but finitely many $i\in\left\{
1,2,\ldots,t\right\}  $ satisfy $\lambda_{i}=0$ (since there are only finitely
many $i\in\left\{  1,2,\ldots,t\right\}  $). Hence, Proposition 2.0i (applied
to $D$ and $\left\{  1,2,\ldots,t\right\}  $ instead of $C$ and $I$) shows
that $\sum\limits_{i\in\left\{  1,2,\ldots,t\right\}  }\lambda_{i}x_{i}\in D$.
Thus, $w=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=\sum\limits_{i\in\left\{
1,2,\ldots,t\right\}  }\lambda_{i}x_{i}\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex cone in $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex cone in $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex cones in $E$ which
contain $S$ as a subset. In other words, $w$ lies in $C_{1}$ (since $C_{1}$ is
the intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $w\in C_{1}$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{1}$ for
every $w\in C_{3}$. In other words, $C_{3}\subseteq C_{1}$, qed.}. Combined
with $C_{1}\subseteq C_{3}$, this shows that $C_{1}=C_{3}$.

\begin{vershort}
The set $C_{2}$ is a convex cone\footnote{\textit{Proof.} We have
$\sum\limits_{i\in S}\underbrace{0i}_{=0}=\sum\limits_{i\in S}0=0$. Thus, $0$
has the form $\sum\limits_{i\in S}\lambda_{i}i$ for some family $\left(
\lambda_{i}\right)  _{i\in S}$ of nonnegative reals indexed by elements of $S$
and having the property that all but finitely many $i\in S$ satisfy
$\lambda_{i}=0$ (namely, for $\left(  \lambda_{i}\right)  _{i\in S}=\left(
0\right)  _{i\in S}$). In other words, $0$ is a linear combination of the
vectors $s$ for $s\in S$ with nonnegative coefficients. In other words, $0\in
C_{2}$ (since $C_{2}$ is the set of all linear combinations of the vectors $s$
for $s\in S$).
\par
Now, let $y\in C_{2}$ and $z\in C_{2}$. Let $\lambda$ and $\kappa$ be two
nonnegative reals. We shall show that $\lambda y+\kappa z\in C_{2}$.
\par
We have $y\in C_{2}$. In other words, $y$ is a linear combination of the
vectors $s$ for $s\in S$ with nonnegative coefficients (since $C_{2}$ is the
set of all linear combinations of the vectors $s$ for $s\in S$ with
nonnegative coefficients). In other words, $y$ has the form $y=\sum
\limits_{i\in S}\mu_{i}i$ for some family $\left(  \mu_{i}\right)  _{i\in S}$
of nonnegative reals indexed by elements of $S$ and having the property that
all but finitely many $i\in S$ satisfy $\mu_{i}=0$. Similarly, $z$ has the
form $z=\sum\limits_{i\in S}\nu_{i}i$ for some family $\left(  \nu_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\nu_{i}=0$. Consider
these two families $\left(  \mu_{i}\right)  _{i\in S}$ and $\left(  \nu
_{i}\right)  _{i\in S}$.
\par
For every $i\in S$, both $\lambda\mu_{i}$ and $\kappa\nu_{i}$ are nonnegative
reals (since $\lambda$, $\mu_{i}$, $\kappa$ and $\nu_{i}$ are nonnegative
reals), and hence the sum $\lambda\mu_{i}+\kappa\nu_{i}$ is a nonnegative
real. Thus, we can define a family $\left(  \eta_{i}\right)  _{i\in S}$ of
nonnegative reals by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\kappa\nu_{i}\ \ \ \ \ \ \ \ \ \ \text{for
every }i\in S\right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in S}$. We know that all
but finitely many $i\in S$ satisfy $\mu_{i}=0$, and we also know that all but
finitely many $i\in S$ satisfy $\nu_{i}=0$. Using these two facts, we see that
all but finitely many $i\in S$ satisfy $\eta_{i}=\lambda\underbrace{\mu_{i}%
}_{=0}+\kappa\underbrace{\nu_{i}}_{=0}=0+0=0$. We have%
\[
\sum\limits_{i\in S}\underbrace{\eta_{i}}_{=\lambda\mu_{i}+\kappa\nu_{i}%
}i=\sum_{i\in S}\left(  \lambda\mu_{i}+\kappa\nu_{i}\right)  i=\lambda
\underbrace{\sum\limits_{i\in S}\mu_{i}i}_{=y}+\kappa\underbrace{\sum
\limits_{i\in S}\nu_{i}z}_{=z}=\lambda y+\kappa z.
\]
Hence, $\lambda y+\kappa z=\sum\limits_{i\in S}\eta_{i}i$. Therefore, $\lambda
y+\kappa z$ has the form $\sum\limits_{i\in S}\lambda_{i}i$ for some family
$\left(  \lambda_{i}\right)  _{i\in S}$ of nonnegative reals indexed by
elements of $S$ and having the property that all but finitely many $i\in S$
satisfy $\lambda_{i}=0$ (namely, for the family $\left(  \lambda_{i}\right)
_{i\in S}=\left(  \eta_{i}\right)  _{i\in S}$). In other words, $\lambda
y+\kappa z$ is a linear combination of the vectors $s$ for $s\in S$ with
nonnegative coefficients. In other words, $\lambda y+\kappa z\in C_{2}$ (since
$C_{2}$ is the set of all linear combinations of the vectors $s$ for $s\in S$
with nonnegative coefficients).
\par
Now, let us forget that we fixed $y$, $z$, $\lambda$ and $\kappa$. We thus
have proven that every two elements $y\in C_{2}$ and $z\in C_{2}$ and every
nonnegative reals $\lambda$ and $\kappa$ satisfy $\lambda y+\kappa z\in C_{2}%
$. If we rename $y$, $z$ and $\kappa$ as $x$, $y$ and $\mu$ in this statement,
we obtain the following: Every two elements $x\in C_{2}$ and $y\in C_{2}$ and
every nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{2}%
$.
\par
Thus, we have proven the following two statements:
\par
\begin{itemize}
\item We have $0\in C_{2}$.
\par
\item Every two elements $x\in C_{2}$ and $y\in C_{2}$ and every nonnegative
reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{2}$.
\end{itemize}
\par
In other words, the set $C_{2}$ is a convex cone (by the definition of a
convex cone). Qed.}. Moreover, $S\subseteq C_{2}$%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Define a family $\left(
\mu_{i}\right)  _{i\in S}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in S\right)  .
\]
It is easy to see that:
\par
\begin{itemize}
\item All but finitely many $i\in S$ satisfy $\mu_{i}=0$.
\par
\item We have $\sum\limits_{i\in S}\mu_{i}i=w$.
\end{itemize}
\par
Hence, $w=\sum\limits_{i\in S}\mu_{i}i$. Thus, $w$ has the form $\sum
\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$ (namely,
for the family $\left(  \lambda_{i}\right)  _{i\in S}=\left(  \mu_{i}\right)
_{i\in S}$). In other words, $w$ is a linear combination of the vectors $s$
for $s\in S$ with nonnegative coefficients. In other words, $w\in C_{2}$
(since $C_{2}$ is the set of all linear combinations of the vectors $s$ for
$s\in S$ with nonnegative coefficients).
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in C_{2}$
for every $w\in S$. In other words, $S\subseteq C_{2}$.}. Hence, $C_{2}$ is a
convex cone in $E$ which contains $S$ as a subset. Thus, the intersection of
all convex cones in $E$ which contain $S$ as a subset is a subset of $C_{2}$.
In other words, $C_{1}$ is a subset of $C_{2}$ (since $C_{1}$ is the
intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $C_{1}\subseteq C_{2}$.
\end{vershort}

\begin{verlong}
The set $C_{2}$ is a convex cone\footnote{\textit{Proof.} We have
$\sum\limits_{i\in S}\underbrace{0i}_{=0}=\sum\limits_{i\in S}0=0$. Hence,
$0=\sum\limits_{i\in S}0i$. Thus, $0$ has the form $\sum\limits_{i\in
S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)  _{i\in S}$ of
nonnegative reals indexed by elements of $S$ and having the property that all
but finitely many $i\in S$ satisfy $\lambda_{i}=0$ (namely, for $\left(
\lambda_{i}\right)  _{i\in S}=\left(  0\right)  _{i\in S}$). In other words,
$0$ is a linear combination of the vectors $s$ for $s\in S$ with nonnegative
coefficients. In other words, $0$ belongs to the set of all linear
combinations of the vectors $s$ for $s\in S$ with nonnegative coefficients. In
other words, $0$ belongs to $C_{2}$ (since $C_{2}$ is the set of all linear
combinations of the vectors $s$ for $s\in S$). In other words, $0\in C_{2}$.
\par
Now, let $y\in C_{2}$ and $z\in C_{2}$. Let $\lambda$ and $\kappa$ be two
nonnegative reals. We shall show that $\lambda y+\kappa z\in C_{2}$.
\par
We have $y\in C_{2}$. In other words, $y$ belongs to the set $C_{2}$. In other
words, $y$ belongs to the set of all linear combinations of the vectors $s$
for $s\in S$ with nonnegative coefficients (since $C_{2}$ is the set of all
linear combinations of the vectors $s$ for $s\in S$ with nonnegative
coefficients). In other words, $y$ is a linear combination of the vectors $s$
for $s\in S$ with nonnegative coefficients. In other words, $y$ has the form
$\sum\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda
_{i}\right)  _{i\in S}$ of nonnegative reals indexed by elements of $S$ and
having the property that all but finitely many $i\in S$ satisfy $\lambda
_{i}=0$. Let us denote this family $\left(  \lambda_{i}\right)  _{i\in S}$ by
$\left(  \mu_{i}\right)  _{i\in S}$. Thus, $\left(  \mu_{i}\right)  _{i\in S}$
is a family of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\mu_{i}=0$, and it
furthermore satisfies $y=\sum\limits_{i\in S}\mu_{i}i$.
\par
We have $z\in C_{2}$. In other words, $z$ belongs to the set $C_{2}$. In other
words, $z$ belongs to the set of all linear combinations of the vectors $s$
for $s\in S$ with nonnegative coefficients (since $C_{2}$ is the set of all
linear combinations of the vectors $s$ for $s\in S$ with nonnegative
coefficients). In other words, $z$ is a linear combination of the vectors $s$
for $s\in S$ with nonnegative coefficients. In other words, $z$ has the form
$\sum\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda
_{i}\right)  _{i\in S}$ of nonnegative reals indexed by elements of $S$ and
having the property that all but finitely many $i\in S$ satisfy $\lambda
_{i}=0$. Let us denote this family $\left(  \lambda_{i}\right)  _{i\in S}$ by
$\left(  \nu_{i}\right)  _{i\in S}$. Thus, $\left(  \nu_{i}\right)  _{i\in S}$
is a family of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\nu_{i}=0$, and it
furthermore satisfies $z=\sum\limits_{i\in S}\nu_{i}i$.
\par
For every $i\in S$, both $\lambda\mu_{i}$ and $\kappa\nu_{i}$ are nonnegative
reals (since $\lambda$, $\mu_{i}$, $\kappa$ and $\nu_{i}$ are nonnegative
reals), and hence the sum $\lambda\mu_{i}+\kappa\nu_{i}$ is a nonnegative
real. Thus, we can define a family $\left(  \eta_{i}\right)  _{i\in S}$ of
nonnegative reals by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\kappa\nu_{i}\ \ \ \ \ \ \ \ \ \ \text{for
every }i\in S\right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in S}$ of nonnegative
reals. We know that all but finitely many $i\in S$ satisfy $\mu_{i}=0$, and we
also know that all but finitely many $i\in S$ satisfy $\nu_{i}=0$. Using these
two facts, we see that all but finitely many $i\in S$ satisfy $\eta
_{i}=\lambda\underbrace{\mu_{i}}_{=0}+\kappa\underbrace{\nu_{i}}_{=0}=0+0=0$.
We have%
\[
\sum\limits_{i\in S}\underbrace{\eta_{i}}_{=\lambda\mu_{i}+\kappa\nu_{i}%
}i=\sum_{i\in S}\left(  \lambda\mu_{i}+\kappa\nu_{i}\right)  i=\lambda
\underbrace{\sum\limits_{i\in S}\mu_{i}i}_{=y}+\kappa\underbrace{\sum
\limits_{i\in S}\nu_{i}z}_{=z}=\lambda y+\kappa z.
\]
Hence, $\lambda y+\kappa z=\sum\limits_{i\in S}\eta_{i}i$. Therefore, $\lambda
y+\kappa z$ has the form $\sum\limits_{i\in S}\lambda_{i}i$ for some family
$\left(  \lambda_{i}\right)  _{i\in S}$ of nonnegative reals indexed by
elements of $S$ and having the property that all but finitely many $i\in S$
satisfy $\lambda_{i}=0$ (namely, for the family $\left(  \lambda_{i}\right)
_{i\in S}=\left(  \eta_{i}\right)  _{i\in S}$). In other words, $\lambda
y+\kappa z$ is a linear combination of the vectors $s$ for $s\in S$ with
nonnegative coefficients. In other words, $\lambda y+\kappa z$ belongs to the
set of all linear combinations of the vectors $s$ for $s\in S$ with
nonnegative coefficients. In other words, $\lambda y+\kappa z$ belongs to
$C_{2}$ (since $C_{2}$ is the set of all linear combinations of the vectors
$s$ for $s\in S$ with nonnegative coefficients). In other words, $\lambda
y+\kappa z\in C_{2}$.
\par
Now, let us forget that we fixed $y$, $z$, $\lambda$ and $\kappa$. We thus
have proven that every two elements $y\in C_{2}$ and $z\in C_{2}$ and every
nonnegative reals $\lambda$ and $\kappa$ satisfy $\lambda y+\kappa z\in C_{2}%
$. If we rename $y$, $z$ and $\kappa$ as $x$, $y$ and $\mu$ in this statement,
we obtain the following: Every two elements $x\in C_{2}$ and $y\in C_{2}$ and
every nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{2}%
$.
\par
Thus, we have proven the following two statements:
\par
\begin{itemize}
\item We have $0\in C_{2}$.
\par
\item Every two elements $x\in C_{2}$ and $y\in C_{2}$ and every nonnegative
reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in C_{2}$.
\end{itemize}
\par
In other words, the set $C_{2}$ is a convex cone (by the definition of a
convex cone). Qed.}. Moreover, $S\subseteq C_{2}$%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in S$. Define a family $\left(
\mu_{i}\right)  _{i\in S}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in S\right)  .
\]
Then, every $i\in S\setminus\left\{  w\right\}  $ satisfies%
\[
\mu_{i}=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\neq w\text{ (since }i\in
S\setminus\left\{  w\right\}  \text{)}\right)  .
\]
Thus, there exists a finite subset $T$ of $S$ such that every $i\in S\setminus
T$ satisfies $\mu_{i}=0$ (namely, $T=\left\{  w\right\}  $). In other words,
all but finitely many $i\in S$ satisfy $\mu_{i}=0$. Also,%
\begin{align*}
\sum\limits_{i\in S}\mu_{i}i  &  =\sum\limits_{\substack{i\in S;\\i=w}%
}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=1\\\text{(since }i=w\text{)}}}i+\sum\limits_{\substack{i\in S;\\i\neq
w}}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=w;\\
0, & \text{if }i\neq w
\end{cases}
=0\\\text{(since }i\neq w\text{)}}}i=\sum\limits_{\substack{i\in
S;\\i=w}}1i+\underbrace{\sum\limits_{\substack{i\in S;\\i\neq w}}0i}_{=0}\\
&  =\sum\limits_{\substack{i\in S;\\i=w}}\underbrace{1i}_{=i}=\sum
\limits_{\substack{i\in S;\\i=w}}i=w\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}w\in S\right)  .
\end{align*}
\par
Hence, $w=\sum\limits_{i\in S}\mu_{i}i$. Thus, $w$ has the form $\sum
\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)
_{i\in S}$ of nonnegative reals indexed by elements of $S$ and having the
property that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$ (namely,
for the family $\left(  \lambda_{i}\right)  _{i\in S}=\left(  \mu_{i}\right)
_{i\in S}$). In other words, $w$ is a linear combination of the vectors $s$
for $s\in S$ with nonnegative coefficients. In other words, $w$ belongs to the
set of all linear combinations of the vectors $s$ for $s\in S$ with
nonnegative coefficients. In other words, $w$ belongs to $C_{2}$ (since
$C_{2}$ is the set of all linear combinations of the vectors $s$ for $s\in S$
with nonnegative coefficients). In other words, $w\in C_{2}$.
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in C_{2}$
for every $w\in S$. In other words, $S\subseteq C_{2}$, qed.}. Hence, $C_{2}$
is a convex cone in $E$ which contains $S$ as a subset. Thus, the intersection
of all convex cones in $E$ which contain $S$ as a subset is a subset of
$C_{2}$. In other words, $C_{1}$ is a subset of $C_{2}$ (since $C_{1}$ is the
intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $C_{1}\subseteq C_{2}$.
\end{verlong}

\begin{vershort}
On the other hand, $C_{2}\subseteq C_{1}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in C_{2}$. We shall show that $w\in C_{1}$.
\par
Let $D$ be any convex cone in $E$ which contains $S$ as a subset. Thus, $D$ is
a convex cone in $E$ and satisfies $S\subseteq D$.
\par
We have $w\in C_{2}$. In other words, $w$ belongs to the set $C_{2}$. In other
words, $w$ is a linear combination of the vectors $s$ for $s\in S$ with
nonnegative coefficients (since $C_{2}$ is the set of all linear combinations
of the vectors $s$ for $s\in S$ with nonnegative coefficients). In other
words, $w$ has the form $w=\sum\limits_{i\in S}\lambda_{i}i$ for some family
$\left(  \lambda_{i}\right)  _{i\in S}$ of nonnegative reals indexed by
elements of $S$ and having the property that all but finitely many $i\in S$
satisfy $\lambda_{i}=0$. Consider this family $\left(  \lambda_{i}\right)
_{i\in S}$.
\par
For every $i\in S$, we have $i\in S\subseteq D$. Thus, $\left(  i\right)
_{i\in S}$ is a family of elements of $D$. Hence, Proposition 2.0i (applied to
$D$, $S$, $\left(  i\right)  _{i\in S}$ and $\left(  \lambda_{i}\right)
_{i\in S}$ instead of $C$, $I$, $\left(  x_{i}\right)  _{i\in I}$ and $\left(
\lambda_{i}\right)  _{i\in I}$) shows that $\sum\limits_{i\in S}\lambda
_{i}i\in D$. Thus, $w=\sum\limits_{i\in S}\lambda_{i}i\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex cone in $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex cone in $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex cones in $E$ which
contain $S$ as a subset. In other words, $w$ lies in $C_{1}$ (since $C_{1}$ is
the intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $w\in C_{1}$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{1}$ for
every $w\in C_{2}$. In other words, $C_{2}\subseteq C_{1}$.}. Combined with
$C_{1}\subseteq C_{2}$, this shows that $C_{1}=C_{2}$.
\end{vershort}

\begin{verlong}
On the other hand, $C_{2}\subseteq C_{1}$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in C_{2}$. We shall show that $w\in C_{1}$.
\par
Let $D$ be any convex cone in $E$ which contains $S$ as a subset. Thus, $D$ is
a convex cone in $E$ and satisfies $S\subseteq D$.
\par
We have $w\in C_{2}$. In other words, $w$ belongs to the set $C_{2}$. In other
words, $w$ belongs to the set of all linear combinations of the vectors $s$
for $s\in S$ with nonnegative coefficients (since $C_{2}$ is the set of all
linear combinations of the vectors $s$ for $s\in S$ with nonnegative
coefficients). In other words, $w$ is a linear combination of the vectors $s$
for $s\in S$ with nonnegative coefficients. In other words, $w$ has the form
$\sum\limits_{i\in S}\lambda_{i}i$ for some family $\left(  \lambda
_{i}\right)  _{i\in S}$ of nonnegative reals indexed by elements of $S$ and
having the property that all but finitely many $i\in S$ satisfy $\lambda
_{i}=0$. Consider this family $\left(  \lambda_{i}\right)  _{i\in S}$. Thus,
$w=\sum\limits_{i\in S}\lambda_{i}i$.
\par
For every $i\in S$, we have $i\in S\subseteq D$. Thus, $\left(  i\right)
_{i\in S}$ is a family of elements of $D$. Hence, Proposition 2.0i (applied to
$D$, $S$, $\left(  i\right)  _{i\in S}$ and $\left(  \lambda_{i}\right)
_{i\in S}$ instead of $C$, $I$, $\left(  x_{i}\right)  _{i\in I}$ and $\left(
\lambda_{i}\right)  _{i\in I}$) shows that $\sum\limits_{i\in S}\lambda
_{i}i\in D$. Thus, $w=\sum\limits_{i\in S}\lambda_{i}i\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex cone in $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex cone in $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex cones in $E$ which
contain $S$ as a subset. In other words, $w$ lies in $C_{1}$ (since $C_{1}$ is
the intersection of all convex cones in $E$ which contain $S$ as a subset). In
other words, $w\in C_{1}$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in C_{1}$ for
every $w\in C_{2}$. In other words, $C_{2}\subseteq C_{1}$, qed.}. Combined
with $C_{1}\subseteq C_{2}$, this shows that $C_{1}=C_{2}$.
\end{verlong}

Combining $C_{1}=C_{2}$ with $C_{1}=C_{3}$, we obtain $C_{1}=C_{2}=C_{3}$.
Now, (\ref{pf.prop2.0m.a.chull1}) becomes%
\begin{align}
&  \left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0j}\right) \nonumber\\
&  =C_{1}=C_{2}=\left(  \text{the convex conic hull of }S\text{ defined
according to Definition 2.0k}\right)  \label{pf.prop2.0m.a.chull12}%
\end{align}
(by (\ref{pf.prop2.0m.a.chull2})). Also, (\ref{pf.prop2.0m.a.chull2}) becomes%
\begin{align}
&  \left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0k}\right) \nonumber\\
&  =C_{2}=C_{3}=\left(  \text{the convex conic hull of }S\text{ defined
according to Definition 2.0l}\right)  \label{pf.prop2.0m.a.chull23}%
\end{align}
(by (\ref{pf.prop2.0m.a.chull3})).

Now, let us forget that we fixed $S$. We thus have shown that if $S$ is any
subset of $E$, then (\ref{pf.prop2.0m.a.chull12}) and
(\ref{pf.prop2.0m.a.chull23}) hold. In other words, if $S$ is any subset of
$E$, then
\begin{align*}
&  \left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0j}\right) \\
&  =\left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0k}\right) \\
&  =\left(  \text{the convex conic hull of }S\text{ defined according to
Definition 2.0l}\right)  .
\end{align*}
In other words, Definitions 2.0j, 2.0k and 2.0l are equivalent. This proves
Proposition 2.0m \textbf{(a)}.

\textbf{(c)} We define the set $C_{3}$ as in our proof of Proposition 2.0m
\textbf{(a)}. Then, (\ref{pf.prop2.0m.a.chull3}) shows that%
\begin{align*}
C_{3}  &  =\left(  \text{the convex conic hull of }S\text{ defined according
to Definition 2.0l}\right) \\
&  =\left(  \text{the convex conic hull of }S\right)  =\operatorname*{cone}S.
\end{align*}
But in our proof of Proposition 2.0m \textbf{(a)}, we have shown that
$S\subseteq C_{3}$. Thus, $S\subseteq C_{3}=\operatorname*{cone}S$. This
proves Proposition 2.0m \textbf{(c)}.

\textbf{(d)} We define the set $C_{3}$ as in our proof of Proposition 2.0m
\textbf{(a)}. Then, (\ref{pf.prop2.0m.a.chull3}) shows that%
\begin{align*}
C_{3}  &  =\left(  \text{the convex conic hull of }S\text{ defined according
to Definition 2.0l}\right) \\
&  =\left(  \text{the convex conic hull of }S\right)  =\operatorname*{cone}S.
\end{align*}
But in our proof of Proposition 2.0m \textbf{(a)}, we have shown that the set
$C_{3}$ is a convex cone. Since $C_{3}=\operatorname*{cone}S$, this rewrites
as follows: The set $\operatorname*{cone}S$ is a convex cone. This proves
Proposition 2.0m \textbf{(d)}.

\textbf{(f)} Let $D$ be a convex cone in $E$ which contains $S$ as a subset.
Thus, the intersection of all convex cones in $E$ which contain $S$ as a
subset is a subset of $D$. In other words, $\operatorname*{cone}S$ is a subset
of $D$ (since $\operatorname*{cone}S$ is the intersection of all convex cones
in $E$ which contain $S$ as a subset (by Definition 2.0k)). In other words,
$D$ contains $\operatorname*{cone}S$ as a subset.

Now, let us forget that we fixed $D$. We thus have shown that if $D$ is a
convex cone in $E$ which contains $S$ as a subset, then $D$ contains
$\operatorname*{cone}S$ as a subset. In other words, every convex cone in $E$
which contains $S$ as a subset also contains $\operatorname*{cone}S$ as a
subset. This proves Proposition 2.0m \textbf{(f)}.

\textbf{(b)} Every two elements $x\in0$ and $y\in0$ and every nonnegative
reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in0$%
\ \ \ \ \footnote{\textit{Proof.} Let $x\in0$ and $y\in0$ be two elements. Let
$\lambda$ and $\mu$ be two nonnegative reals. Then, $x=0$ (since $x\in0$) and
$y=0$ (since $y\in0$), so that $\lambda\underbrace{x}_{=0}+\mu\underbrace{y}%
_{=0}=\lambda0+\mu0=0\in0$, qed.}. Furthermore, $0\in0$. Thus, we have proven
the following two statements:

\begin{itemize}
\item We have $0\in0$.

\item Every two elements $x\in0$ and $y\in0$ and every nonnegative reals
$\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in0$.
\end{itemize}

In other words, the set $0$ is a convex cone (by the definition of a convex cone).

Thus, $0$ is a convex cone in $E$ which contains $\varnothing$ as a subset.
Now, Proposition 2.0m \textbf{(f)} (applied to $S=\varnothing$) shows that
every convex cone in $E$ which contains $\varnothing$ as a subset also
contains $\operatorname*{cone}\varnothing$ as a subset. Thus, the set $0$
contains $\operatorname*{cone}\varnothing$ as a subset (since $0$ is a convex
cone in $E$ which contains $\varnothing$ as a subset). In other words,
$\operatorname*{cone}\varnothing\subseteq0$.

On the other hand, $0\subseteq\operatorname*{cone}\varnothing$%
\ \ \ \ \footnote{\textit{Proof.} Proposition 2.0m \textbf{(d)} (applied to
$S=\varnothing$) shows that the set $\operatorname*{cone}\varnothing$ is a
convex cone. In other words, it satisfies the following two statements:
\par
\begin{itemize}
\item We have $0\in\operatorname*{cone}\varnothing$.
\par
\item Every two elements $x\in\operatorname*{cone}\varnothing$ and
$y\in\operatorname*{cone}\varnothing$ and every nonnegative reals $\lambda$
and $\mu$ satisfy $\lambda x+\mu y\in\operatorname*{cone}\varnothing$.
\end{itemize}
\par
(This is because of our definition of a convex cone.)
\par
In particular, $0\in\operatorname*{cone}\varnothing$, so that $\left\{
0\right\}  \subseteq\operatorname*{cone}\varnothing$. Thus, $0=\left\{
0\right\}  \subseteq\operatorname*{cone}\varnothing$, qed.}. Combining this
with $\operatorname*{cone}\varnothing\subseteq0$, we obtain
$\operatorname*{cone}\varnothing=0$. This proves Proposition 2.0m \textbf{(b)}.

\textbf{(g)} Let $T$ be a subset of $\operatorname*{cone}S$. The set
$\operatorname*{cone}S$ is a convex cone (by Proposition 2.0m \textbf{(d)})
and contains $T$ as a subset (since $T\subseteq\operatorname*{cone}S$). In
other words, $\operatorname*{cone}S$ is a convex cone in $E$ which contains
$T$ as a subset.

But Proposition 2.0m \textbf{(f)} (applied to $T$ instead of $S$) shows that
every convex cone in $E$ which contains $T$ as a subset also contains
$\operatorname*{cone}T$ as a subset. Thus, the set $\operatorname*{cone}S$
contains $\operatorname*{cone}T$ as a subset (since $\operatorname*{cone}S$ is
a convex cone in $E$ which contains $T$ as a subset). In other words,
$\operatorname*{cone}T\subseteq\operatorname*{cone}S$. This proves Proposition
2.0m \textbf{(g)}.

\textbf{(e)} Let $T$ be a subset of $S$. Thus, $T\subseteq S\subseteq
\operatorname*{cone}S$ (by Proposition 2.0m \textbf{(c)}). In other words, $T$
is a subset of $\operatorname*{cone}S$. Hence, Proposition 2.0m \textbf{(g)}
yields $\operatorname*{cone}T\subseteq\operatorname*{cone}S$. This proves
Proposition 2.0m \textbf{(e)}.

\textbf{(h)} Proposition 2.0m \textbf{(d)} shows that $\operatorname*{cone}S$
is a convex cone.

Recall that any convex cone is a convex set. Applying this to the convex cone
$\operatorname*{cone}S$, we conclude that $\operatorname*{cone}S$ is a convex
set (since $\operatorname*{cone}S$ is a convex cone). In other words,
$\operatorname*{cone}S$ is a convex subset of $E$.

Proposition 2.0m \textbf{(c)} shows that $S\subseteq\operatorname*{cone}S$.
Thus, the set $\operatorname*{cone}S$ contains $S$ as a subset. Hence,
$\operatorname*{cone}S$ is a convex subset of $E$ which contains $S$ as a subset.

But Proposition 2.0f \textbf{(f)} shows that every convex subset of $E$ which
contains $S$ as a subset also contains $\operatorname*{conv}%
.\operatorname*{hull}S$ as a subset. Thus, the set $\operatorname*{cone}S$
contains $\operatorname*{conv}.\operatorname*{hull}S$ as a subset (since
$\operatorname*{cone}S$ is a convex subset of $E$ which contains $S$ as a
subset). In other words, $\operatorname*{conv}.\operatorname*{hull}%
S\subseteq\operatorname*{cone}S$. This proves Proposition 2.0m \textbf{(h)}.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Proposition 2.0n.]The proof of Proposition 2.0n is analogous to the
proof of Proposition 2.0g. (Of course, instead of using properties of convex
sets, we now need to use the corresponding properties of convex cones.)
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition 2.0n.]\textbf{(a)} Let $S$ be a subset of $E$. Recall
that%
\begin{align}
&  \operatorname*{cone}S\nonumber\\
&  =\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}  \label{pf.prop2.0n.a.1}%
\end{align}
(by Definition 2.0l).

Let%
\[
Q=\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\}  .
\]


We have $Q\subseteq\operatorname*{cone}S$\ \ \ \ \footnote{\textit{Proof.} Let
$y\in Q$. Thus,%
\[
y\in Q=\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\}  .
\]
In other words, $y$ has the form $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}$ for
some element $t$ of $\mathbb{N}$, some $t$-tuple $\left(  x_{1},x_{2}%
,\ldots,x_{t}\right)  $ of elements of $S$, and some $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals. Let
us denote this $t$, this $\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by $p$, $\left(
y_{1},y_{2},\ldots,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{p}\right)  $, respectively. Thus, $p$ is an element of $\mathbb{N}$, and
$\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$,
and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $y=\sum\limits_{i=1}^{p}\mu_{i}y_{i}$.
\par
Now, $\sum\limits_{i=1}^{p}\mu_{i}y_{i}=y$ (since $y=\sum\limits_{i=1}^{p}%
\mu_{i}y_{i}$). Also, $y=\sum\limits_{i=1}^{p}\mu_{i}\underbrace{y_{i}}_{\in
S\subseteq E}\in\sum\limits_{i=1}^{p}\mu_{i}E\subseteq E$ (since $E$ is an
$\mathbb{R}$-vector space). Hence, $y$ is an element of $E$ such that there
exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1},x_{2},\ldots
,x_{t}\right)  $ of elements of $S$ and a $t$-tuple $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such that
$\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=y$ (namely, $t=p$, $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  =\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ and
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  =\left(  \mu
_{1},\mu_{2},\ldots,\mu_{p}\right)  $). In other words,%
\begin{align*}
y  &  \in\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\} \\
&  =\operatorname*{cone}S\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.prop2.0n.a.1})}\right)  .
\end{align*}
\par
Now, let us forget that we fixed $y$. We thus have shown that $y\in
\operatorname*{cone}S$ for every $y\in Q$. In other words, $Q\subseteq
\operatorname*{cone}S$, qed.} and $\operatorname*{cone}S\subseteq
Q$\ \ \ \ \footnote{\textit{Proof.} Let $y\in\operatorname*{cone}S$. Thus,%
\begin{align*}
y  &  \in\operatorname*{cone}S\\
&  =\left\{  x\in E\ \mid\ \left(
\begin{array}
[c]{c}%
\text{there exist some }t\in\mathbb{N}\text{,}\\
\text{a }t\text{-tuple }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ of
elements of }S\\
\text{and a }t\text{-tuple }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{t}\right)  \text{ of nonnegative reals}\\
\text{such that }\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=x
\end{array}
\right)  \right\}
\end{align*}
(by (\ref{pf.prop2.0n.a.1})). In other words, $y$ is an element of $E$ such
that there exist some $t\in\mathbb{N}$, a $t$-tuple $\left(  x_{1}%
,x_{2},\ldots,x_{t}\right)  $ of elements of $S$, and a $t$-tuple $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative reals such
that $\sum\limits_{i=1}^{t}\lambda_{i}x_{i}=y$. Let us denote this $t$, this
$\left(  x_{1},x_{2},\ldots,x_{t}\right)  $ and this $\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ by $p$, $\left(  y_{1}%
,y_{2},\ldots,y_{p}\right)  $ and $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{p}\right)  $, respectively. Thus, $p$ is an element of $\mathbb{N}$, and
$\left(  y_{1},y_{2},\ldots,y_{p}\right)  $ is a $p$-tuple of elements of $S$,
and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)  $ is a $p$-tuple of
nonnegative reals such that $\sum\limits_{i=1}^{p}\mu_{i}y_{i}=y$.
\par
Now, $y=\sum\limits_{i=1}^{p}\mu_{i}y_{i}$ (since $\sum\limits_{i=1}^{p}%
\mu_{i}y_{i}=y$). Hence, $y$ has the form $\sum\limits_{i=1}^{t}\lambda
_{i}x_{i}$ for some element $t$ of $\mathbb{N}$, some $t$-tuple $\left(
x_{1},x_{2},\ldots,x_{t}\right)  $ of elements of $S$, and some $t$-tuple
$\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  $ of nonnegative
reals (namely, for $t=p$, $\left(  x_{1},x_{2},\ldots,x_{t}\right)  =\left(
y_{1},y_{2},\ldots,y_{p}\right)  $ and $\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{t}\right)  =\left(  \mu_{1},\mu_{2},\ldots,\mu_{p}\right)
$). In other words,%
\begin{align*}
y  &  \in\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\} \\
&  =Q.
\end{align*}
\par
Now, let us forget that we fixed $y$. We thus have shown that $y\in Q$ for
every $y\in\operatorname*{cone}S$. In other words, $\operatorname*{cone}%
S\subseteq Q$, qed.}. Combining these two relations, we obtain
$\operatorname*{cone}S=Q$. Thus,%
\begin{align*}
&  \operatorname*{cone}S\\
&  =Q=\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\}  .
\end{align*}
This proves Proposition 2.0n \textbf{(a)}.

\textbf{(b)} We have $S\subseteq F\subseteq E$. If we regard $S$ as a subset
of $F$, then we have%
\begin{align}
&  \operatorname*{cone}S\nonumber\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\}  \label{pf.prop2.0n.b.1}%
\end{align}
(by Proposition 2.0n \textbf{(a)}, applied to $F$ instead of $E$). If we
regard $S$ as a subset of $E$, then we have%
\begin{align}
&  \operatorname*{cone}S\nonumber\\
&  =\left\{  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\ \mid\ \left(
\begin{array}
[c]{c}%
t\text{ is an element of }\mathbb{N}\text{,}\\
\text{and }\left(  x_{1},x_{2},\ldots,x_{t}\right)  \text{ is a }t\text{-tuple
of elements of }S\text{,}\\
\text{and }\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{t}\right)  \text{
is a }t\text{-tuple of nonnegative reals}%
\end{array}
\right)  \right\}  \label{pf.prop2.0n.b.2}%
\end{align}
(by Proposition 2.0n \textbf{(a)}). The right hand sides of the equalities
(\ref{pf.prop2.0n.b.1}) and (\ref{pf.prop2.0n.b.2}) are equal (because these
right hand sides do not involve $E$ or $F$ in their definition). Therefore,
the left hand sides of the equalities (\ref{pf.prop2.0n.b.1}) and
(\ref{pf.prop2.0n.b.2}) are equal as well. In other words, the value of
$\operatorname*{cone}S$ when $S$ is regarded as a subset of $F$ equals the
value of $\operatorname*{cone}S$ when $S$ is regarded as a subset of $E$. In
other words, $\operatorname*{cone}S$ does not depend on whether we consider
$S$ as a subset of $F$ or as a subset of $E$. This proves Proposition 2.0n
\textbf{(b)}.
\end{proof}
\end{verlong}

\begin{proof}
[Proof of Proposition 2.0o.]\textbf{(a)} Let $S=\left\{  x_{1},x_{2}%
,\ldots,x_{n}\right\}  $. Let $Q$ be the set of all linear combinations of the
vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients.

\begin{vershort}
Recall that $Q$ is the set of all linear combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients. In other words, $Q$
is the set of all linear combinations of the vectors $x_{i}$ for $i\in\left\{
1,2,\ldots,n\right\}  $ with nonnegative coefficients.

The set $Q$ is a convex cone\footnote{\textit{Proof.} We have $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{0x_{i}}_{=0}%
=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }0=0$. Hence, $0=\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }0x_{i}$. Thus, $0$ has the form
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$ (namely, for $\left(  \lambda
_{i}\right)  _{i\in S}=\left(  0\right)  _{i\in S}$). In other words, $0$ is a
linear combination of the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots
,n\right\}  $ with nonnegative coefficients. In other words, $0\in Q$ (since
$Q$ is the set of all linear combinations of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients).
\par
Now, let $y\in Q$ and $z\in Q$. Let $\lambda$ and $\kappa$ be two nonnegative
reals. We shall show that $\lambda y+\kappa z\in Q$.
\par
We have $y\in Q$. In other words, $y$ is a linear combination of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative
coefficients (since $Q$ is the set of all linear combinations of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative
coefficients). In other words, $y$ has the form $y=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\mu_{i}x_{i}$ for some family $\left(  \mu_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by
elements of $\left\{  1,2,\ldots,n\right\}  $ and having the property that all
but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\mu_{i}=0$.
Similarly, $z$ has the form $z=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\nu_{i}x_{i}$ for some family $\left(  \nu_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\nu_{i}=0$. Consider these two
families $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ and
$\left(  \nu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, both $\lambda\mu_{i}$ and
$\kappa\nu_{i}$ are nonnegative reals (since $\lambda$, $\mu_{i}$, $\kappa$
and $\nu_{i}$ are nonnegative reals), and hence the sum $\lambda\mu_{i}%
+\kappa\nu_{i}$ is a nonnegative real. Thus, we can define a family $\left(
\eta_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals
by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\kappa\nu_{i}\ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$. We know that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\eta_{i}=0$ (since there are only finitely
many $i\in\left\{  1,2,\ldots,n\right\}  $). We have%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\eta_{i}%
}_{=\lambda\mu_{i}+\kappa\nu_{i}}x_{i}  &  =\sum_{i\in\left\{  1,2,\ldots
,n\right\}  }\left(  \lambda\mu_{i}+\kappa\nu_{i}\right)  x_{i}\\
&  =\lambda\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}x_{i}}_{=y}+\kappa\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }\nu_{i}x_{i}}_{=z}=\lambda y+\kappa z.
\end{align*}
Hence, $\lambda y+\kappa z=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\eta_{i}x_{i}$. Therefore, $\lambda y+\kappa z$ has the form $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some family
$\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$ (namely, for the family $\left(
\lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }=\left(  \eta
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$). In other words,
$\lambda y+\kappa z$ is a linear combination of the vectors $x_{i}$ or
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other
words, $\lambda y+\kappa z\in Q$ (since $Q$ is the set of all linear
combinations of the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $
with nonnegative coefficients).
\par
Now, let us forget that we fixed $y$, $z$, $\lambda$ and $\kappa$. We thus
have proven that every two elements $y\in Q$ and $z\in Q$ and every
nonnegative reals $\lambda$ and $\kappa$ satisfy $\lambda y+\kappa z\in Q$. If
we rename $y$, $z$ and $\kappa$ as $x$, $y$ and $\mu$ in this statement, we
obtain the following: Every two elements $x\in Q$ and $y\in Q$ and every
nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in Q$.
\par
Thus, we have proven the following two statements:
\par
\begin{itemize}
\item We have $0\in Q$.
\par
\item Every two elements $x\in Q$ and $y\in Q$ and every nonnegative reals
$\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in Q$.
\end{itemize}
\par
In other words, the set $Q$ is a convex cone (by the definition of a convex
cone). Qed.}. Moreover, $S\subseteq Q$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in S$. Then, $w\in S=\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  $. Thus,
there exists some $k\in\left\{  1,2,\ldots,n\right\}  $ such that $w=x_{k}$.
Consider this $k$.
\par
Define a family $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}
\right)  .
\]
Then, it is easy to see that:
\par
\begin{itemize}
\item All but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy
$\mu_{i}=0$.
\par
\item We have $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}%
x_{i}=x_{k}$.
\end{itemize}
\par
Hence, $w=x_{k}=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}%
$. Thus, $w$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$ (namely, for the
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}=\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$). In other
words, $w$ is a linear combination of the vectors $x_{i}$ for $i\in\left\{
1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other words, $w\in
Q$ (since $Q$ is the set of all linear combinations of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients).
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in Q$ for
every $w\in S$. In other words, $S\subseteq Q$.}. Hence, $Q$ is a convex cone
in $E$ which contains $S$ as a subset. Thus, the intersection of all convex
cones in $E$ which contain $S$ as a subset is a subset of $Q$. In other words,
$\operatorname*{cone}S$ is a subset of $Q$ (since $\operatorname*{cone}S$ is
the intersection of all convex cones in $E$ which contain $S$ as a subset
(because of Definition 2.0j)). In other words, $\operatorname*{cone}S\subseteq
Q$.
\end{vershort}

\begin{verlong}
The set $Q$ is a convex cone\footnote{\textit{Proof.} We have $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{0x_{i}}_{=0}%
=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }0=0$. Hence, $0=\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }0x_{i}$. Thus, $0$ has the form
$\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$ (namely, for $\left(  \lambda
_{i}\right)  _{i\in S}=\left(  0\right)  _{i\in S}$). In other words, $0$ is a
linear combination of the vectors $x_{i}$ for $i\in\left\{  1,2,\ldots
,n\right\}  $ with nonnegative coefficients. In other words, $0$ is a linear
combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with
nonnegative coefficients. In other words, $0$ belongs to the set of all linear
combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with
nonnegative coefficients. In other words, $0$ belongs to $Q$ (since $Q$ is the
set of all linear combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$,
$x_{n}$ with nonnegative coefficients). In other words, $0\in Q$.
\par
Now, let $y\in Q$ and $z\in Q$. Let $\lambda$ and $\kappa$ be two nonnegative
reals. We shall show that $\lambda y+\kappa z\in Q$.
\par
We have $y\in Q$. In other words, $y$ belongs to the set $Q$. In other words,
$y$ belongs to the set of all linear combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients (since $Q$ is the set
of all linear combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$
with nonnegative coefficients). In other words, $y$ is a linear combination of
the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients.
In other words, $y$ is a convex combination of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other
words, $y$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$. Let us denote
this family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ by $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$. Thus,
$\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family
of nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $
and having the property that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\mu_{i}=0$, and it furthermore satisfies
$y=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}$.
\par
We have $z\in Q$. In other words, $z$ belongs to the set $Q$. In other words,
$z$ belongs to the set of all linear combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients (since $Q$ is the set
of all linear combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$
with nonnegative coefficients). In other words, $z$ is a linear combination of
the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients.
In other words, $z$ is a convex combination of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other
words, $z$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$. Let us denote
this family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ by $\left(  \nu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$. Thus,
$\left(  \nu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family
of nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $
and having the property that all but finitely many $i\in\left\{
1,2,\ldots,n\right\}  $ satisfy $\nu_{i}=0$, and it furthermore satisfies
$z=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\nu_{i}x_{i}$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, both $\lambda\mu_{i}$ and
$\kappa\nu_{i}$ are nonnegative reals (since $\lambda$, $\mu_{i}$, $\kappa$
and $\nu_{i}$ are nonnegative reals), and hence the sum $\lambda\mu_{i}%
+\kappa\nu_{i}$ is a nonnegative real. Thus, we can define a family $\left(
\eta_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals
by setting%
\[
\left(  \eta_{i}=\lambda\mu_{i}+\kappa\nu_{i}\ \ \ \ \ \ \ \ \ \ \text{for
every }i\in\left\{  1,2,\ldots,n\right\}  \right)  .
\]
Consider this family $\left(  \eta_{i}\right)  _{i\in\left\{  1,2,\ldots
,n\right\}  }$ of nonnegative reals. We know that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\eta_{i}=0$ (since there are
only finitely many $i\in\left\{  1,2,\ldots,n\right\}  $). We have%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\eta_{i}%
}_{=\lambda\mu_{i}+\kappa\nu_{i}}x_{i}  &  =\sum_{i\in\left\{  1,2,\ldots
,n\right\}  }\left(  \lambda\mu_{i}+\kappa\nu_{i}\right)  x_{i}\\
&  =\lambda\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\mu_{i}x_{i}}_{=y}+\kappa\underbrace{\sum\limits_{i\in\left\{  1,2,\ldots
,n\right\}  }\nu_{i}x_{i}}_{=z}=\lambda y+\kappa z.
\end{align*}
Hence, $\lambda y+\kappa z=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\eta_{i}x_{i}$. Therefore, $\lambda y+\kappa z$ has the form $\sum
\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some family
$\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of
nonnegative reals indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and
having the property that all but finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfy $\lambda_{i}=0$ (namely, for the family $\left(
\lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }=\left(  \eta
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$). In other words,
$\lambda y+\kappa z$ is a linear combination of the vectors $x_{i}$ or
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other
words, $\lambda y+\kappa z$ is a linear combination of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients. In other words,
$\lambda y+\kappa z$ belongs to the set of all linear combinations of the
vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients. In
other words, $\lambda y+\kappa z$ belongs to $Q$ (since $Q$ is the set of all
linear combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with
nonnegative coefficients). In other words, $\lambda y+\kappa z\in Q$.
\par
Now, let us forget that we fixed $y$, $z$, $\lambda$ and $\kappa$. We thus
have proven that every two elements $y\in Q$ and $z\in Q$ and every
nonnegative reals $\lambda$ and $\kappa$ satisfy $\lambda y+\kappa z\in Q$. If
we rename $y$, $z$ and $\kappa$ as $x$, $y$ and $\mu$ in this statement, we
obtain the following: Every two elements $x\in Q$ and $y\in Q$ and every
nonnegative reals $\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in Q$.
\par
Thus, we have proven the following two statements:
\par
\begin{itemize}
\item We have $0\in Q$.
\par
\item Every two elements $x\in Q$ and $y\in Q$ and every nonnegative reals
$\lambda$ and $\mu$ satisfy $\lambda x+\mu y\in Q$.
\end{itemize}
\par
In other words, the set $Q$ is a convex cone (by the definition of a convex
cone). Qed.}. Moreover, $S\subseteq Q$\ \ \ \ \footnote{\textit{Proof.} Let
$w\in S$. Then, $w\in S=\left\{  x_{1},x_{2},\ldots,x_{n}\right\}  $. Thus,
there exists some $k\in\left\{  1,2,\ldots,n\right\}  $ such that $w=x_{k}$.
Consider this $k$.
\par
Define a family $\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}$ of nonnegative reals by setting%
\[
\left(  \mu_{i}=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,\ldots,n\right\}
\right)  .
\]
Then, all but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $ satisfy
$\mu_{i}=0$ (since there are only finitely many $i\in\left\{  1,2,\ldots
,n\right\}  $). Also,%
\begin{align*}
\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}  &
=\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
=1\\\text{(since }i=k\text{)}}}x_{i}+\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}\underbrace{\mu_{i}}_{\substack{=%
\begin{cases}
1, & \text{if }i=k;\\
0, & \text{if }i\neq k
\end{cases}
=0\\\text{(since }i\neq k\text{)}}}x_{i}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}1x_{i}+\underbrace{\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i\neq k}}0x_{i}}_{=0}\\
&  =\sum\limits_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i=k}%
}\underbrace{1x_{i}}_{=x_{i}}=\sum\limits_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i=k}}x_{i}=x_{k}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }k\in\left\{  1,2,\ldots,n\right\}  \right) \\
&  =w.
\end{align*}
\par
Hence, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\mu_{i}x_{i}$.
Thus, $w$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$ (namely, for the
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}
}=\left(  \mu_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$). In other
words, $w$ is a linear combination of the vectors $x_{i}$ for $i\in\left\{
1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other words, $w$ is
a linear combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with
nonnegative coefficients. In other words, $w$ belongs to the set of all linear
combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with
nonnegative coefficients. In other words, $w$ belongs to $Q$ (since $Q$ is the
set of all linear combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$,
$x_{n}$ with nonnegative coefficients). In other words, $w\in Q$.
\par
Now, let us forget that we fixed $w$. We thus have proven that $w\in Q$ for
every $w\in S$. In other words, $S\subseteq Q$, qed.}. Hence, $Q$ is a convex
cone in $E$ which contains $S$ as a subset. Thus, the intersection of all
convex cones in $E$ which contain $S$ as a subset is a subset of $Q$. In other
words, $\operatorname*{cone}S$ is a subset of $Q$ (since $\operatorname*{cone}%
S$ is the intersection of all convex cones in $E$ which contain $S$ as a
subset (because of Definition 2.0j)). In other words, $\operatorname*{cone}%
S\subseteq Q$.
\end{verlong}

\begin{vershort}
On the other hand, $Q\subseteq\operatorname*{cone}S$%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in Q$. We shall show that
$w\in\operatorname*{cone}S$.
\par
Let $D$ be any convex cone in $E$ which contains $S$ as a subset. Thus, $D$ is
a convex cone in $E$ and satisfies $S\subseteq D$.
\par
We have $w\in Q$. In other words, $w$ is a linear combination of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative
coefficients (since $Q$ is the set of all linear combinations of the vectors
$x_{i}$ for $i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative
coefficients). In other words, $w$ has the form $w=\sum\limits_{i\in\left\{
1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$ for some family $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ of nonnegative reals
indexed by elements of $\left\{  1,2,\ldots,n\right\}  $ and having the
property that all but finitely many $i\in\left\{  1,2,\ldots,n\right\}  $
satisfy $\lambda_{i}=0$. Consider this family $\left(  \lambda_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, we have $x_{i}\in\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  =S\subseteq D$. Thus, $\left(  x_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family of elements of $D$. Hence,
Proposition 2.0i (applied to $D$, $\left\{  1,2,\ldots,n\right\}  $, $\left(
x_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ and $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ instead of $C$, $I$,
$\left(  x_{i}\right)  _{i\in I}$ and $\left(  \lambda_{i}\right)  _{i\in I}$)
shows that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}%
x_{i}\in D$. Thus, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\lambda_{i}x_{i}\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex cone in $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex cone in $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex cones in $E$ which
contain $S$ as a subset. In other words, $w$ lies in $\operatorname*{cone}S$
(since $\operatorname*{cone}S$ is the intersection of all convex cones in $E$
which contain $S$ as a subset (because of Definition 2.0j)). In other words,
$w\in\operatorname*{cone}S$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in
\operatorname*{cone}S$ for every $w\in Q$. In other words, $Q\subseteq
\operatorname*{cone}S$, qed.}. Combining this with $\operatorname*{cone}%
S\subseteq Q$, we obtain $\operatorname*{cone}S=Q$. Thus,%
\begin{align*}
&  \operatorname*{cone}\underbrace{\left\{  x_{1},x_{2},\ldots,x_{n}\right\}
}_{=S}\\
&  =\operatorname*{cone}S=Q\\
&  =\left(  \text{the set of all linear combinations of the}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{vectors }x_{1}\text{, }x_{2}\text{,
}\ldots\text{, }x_{n}\text{ with nonnegative coefficients}\right)
\end{align*}
(by the definition of $Q$). This proves Proposition 2.0o \textbf{(a)}.
\end{vershort}

\begin{verlong}
On the other hand, $Q\subseteq\operatorname*{cone}S$%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in Q$. We shall show that
$w\in\operatorname*{cone}S$.
\par
Let $D$ be any convex cone in $E$ which contains $S$ as a subset. Thus, $D$ is
a convex cone in $E$ and satisfies $S\subseteq D$.
\par
We have $w\in Q$. In other words, $w$ belongs to the set $Q$. In other words,
$w$ belongs to the set of all linear combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients (since $Q$ is the set
of all linear combinations of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$
with nonnegative coefficients). In other words, $w$ is a linear combination of
the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients.
In other words, $w$ is a linear combination of the vectors $x_{i}$ for
$i\in\left\{  1,2,\ldots,n\right\}  $ with nonnegative coefficients. In other
words, $w$ has the form $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}
}\lambda_{i}x_{i}$ for some family $\left(  \lambda_{i}\right)  _{i\in\left\{
1,2,\ldots,n\right\}  }$ of nonnegative reals indexed by elements of $\left\{
1,2,\ldots,n\right\}  $ and having the property that all but finitely many
$i\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\lambda_{i}=0$. Consider this
family $\left(  \lambda_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$.
Thus, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}x_{i}$.
\par
For every $i\in\left\{  1,2,\ldots,n\right\}  $, we have $x_{i}\in\left\{
x_{1},x_{2},\ldots,x_{n}\right\}  =S\subseteq D$. Thus, $\left(  x_{i}\right)
_{i\in\left\{  1,2,\ldots,n\right\}  }$ is a family of elements of $D$. Hence,
Proposition 2.0i (applied to $D$, $\left\{  1,2,\ldots,n\right\}  $, $\left(
x_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ and $\left(  \lambda
_{i}\right)  _{i\in\left\{  1,2,\ldots,n\right\}  }$ instead of $C$, $I$,
$\left(  x_{i}\right)  _{i\in I}$ and $\left(  \lambda_{i}\right)  _{i\in I}$)
shows that $\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }\lambda_{i}%
x_{i}\in D$. Thus, $w=\sum\limits_{i\in\left\{  1,2,\ldots,n\right\}  }%
\lambda_{i}x_{i}\in D$.
\par
Let us now forget that we fixed $D$. We thus have proven that $w\in D$
whenever $D$ is any convex cone in $E$ which contains $S$ as a subset. In
other words, $w$ lies in every convex cone in $E$ which contains $S$ as a
subset. Hence, $w$ lies in the intersection of all convex cones in $E$ which
contain $S$ as a subset. In other words, $w$ lies in $\operatorname*{cone}S$
(since $\operatorname*{cone}S$ is the intersection of all convex cones in $E$
which contain $S$ as a subset (because of Definition 2.0j)). In other words,
$w\in\operatorname*{cone}S$.
\par
Let us now forget that we fixed $w$. We thus have proven that $w\in
\operatorname*{cone}S$ for every $w\in Q$. In other words, $Q\subseteq
\operatorname*{cone}S$, qed.}. Combining this with $\operatorname*{cone}%
S\subseteq Q$, we obtain $\operatorname*{cone}S=Q$. Thus,%
\begin{align*}
&  \operatorname*{cone}\underbrace{\left\{  x_{1},x_{2},\ldots,x_{n}\right\}
}_{=S}\\
&  =\operatorname*{cone}S=Q\\
&  =\left(  \text{the set of all linear combinations of the}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{vectors }x_{1}\text{, }x_{2}\text{,
}\ldots\text{, }x_{n}\text{ with nonnegative coefficients}\right)
\end{align*}
(since $Q$ is the set of all linear combinations of the vectors $x_{1}$,
$x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients). This proves
Proposition 2.0o \textbf{(a)}.
\end{verlong}

\textbf{(b)} We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  x\in\underbrace{\operatorname*{cone}\left\{  x_{1},x_{2}%
,\ldots,x_{n}\right\}  }_{\substack{=\left(  \text{the set of all linear
combinations of the}\right.  \\\left.  \text{vectors }x_{1}\text{, }%
x_{2}\text{, }\ldots\text{, }x_{n}\text{ with nonnegative coefficients}%
\right)  \\\text{(by Proposition 2.0o \textbf{(a)})}}}\right) \\
&  \Longleftrightarrow\ \left(
\begin{array}
[c]{l}%
x\in\left(  \text{the set of all linear combinations of the}\right. \\
\ \ \ \ \ \ \ \ \ \ \left.  \text{vectors }x_{1}\text{, }x_{2}\text{, }%
\ldots\text{, }x_{n}\text{ with nonnegative coefficients}\right)
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  x\text{ is a linear combination of the vectors
}x_{1}\text{, }x_{2}\text{, }\ldots\text{, }x_{n}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{with nonnegative coefficients}\right)  .
\end{align*}
In other words, we have $x\in\operatorname*{cone}\left\{  x_{1},x_{2}%
,\ldots,x_{n}\right\}  $ if and only if $x$ is a linear combination of the
vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ with nonnegative coefficients.
This proves Proposition 2.0o \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Proposition 2.0p.]\textbf{(b)} Let $\left(  \nu_{s}\right)  _{s\in
S}$ is a family of nonnegative reals indexed by elements of $S$.

Clearly, there are only finitely many $i\in S$ (since the set $S$ is finite).
Hence, all but finitely many $i\in S$ satisfy $\nu_{i}=0$.

We know that $\left(  \nu_{s}\right)  _{s\in S}$ is a family of nonnegative
reals. Renaming the index $s$ as $i$ in this statement, we conclude that
$\left(  \nu_{i}\right)  _{i\in S}$ is a family of nonnegative reals.

We have $\sum\limits_{s\in S}\nu_{s}s=\sum\limits_{i\in S}\nu_{i}i$ (here, we
renamed the summation index $s$ as $i$). Thus, $\sum\limits_{s\in S}\nu_{s}s$
has the form $\sum\limits_{i\in S}\lambda_{i}i$ for some family $\left(
\lambda_{i}\right)  _{i\in S}$ of nonnegative reals indexed by the elements of
$S$ such that all but finitely many $i\in S$ satisfy $\lambda_{i}=0$ (namely,
for the family $\left(  \lambda_{i}\right)  _{i\in S}=\left(  \nu_{i}\right)
_{i\in S}$). In other words, $\sum\limits_{s\in S}\nu_{s}s$ is a linear
combination of the vectors $s$ for $s\in S$ with nonnegative coefficients. In
other words, $\sum\limits_{s\in S}\nu_{s}s$ belongs to the set of all linear
combinations of the vectors $s$ for $s\in S$ with nonnegative coefficients. In
other words, $\sum\limits_{s\in S}\nu_{s}s$ belongs to $\operatorname*{cone}S$
(since $\operatorname*{cone}S$ is the set of all linear combinations of the
vectors $s$ for $s\in S$ with nonnegative coefficients (according to
Definition 2.0k)). In other words, $\sum\limits_{s\in S}\nu_{s}s\in
\operatorname*{cone}S$. This proves Proposition 2.0p \textbf{(b)}.

\begin{vershort}
\textbf{(c)} Let $p$ be an element of $\operatorname*{cone}S$. Thus,
$p\in\operatorname*{cone}S$. In other words, $p$ is a linear combination of
the vectors $s$ for $s\in S$ with nonnegative coefficients (since
$\operatorname*{cone}S$ is the set of all linear combinations of the vectors
$s$ for $s\in S$ with nonnegative coefficients (according to Definition
2.0k)). In other words, $p$ can be written in the form $\sum\limits_{i\in
S}\lambda_{i}i$ for some family $\left(  \lambda_{i}\right)  _{i\in S}$ of
nonnegative reals indexed by the elements of $S$ such that all but finitely
many $i\in S$ satisfy $\lambda_{i}=0$. Consider this $\left(  \lambda
_{i}\right)  _{i\in S}$. Thus, $p=\sum\limits_{i\in S}\lambda_{i}i$.
\end{vershort}

\begin{verlong}
\textbf{(c)} Let $p$ be an element of $\operatorname*{cone}S$. Thus, $p$
belongs to the set $\operatorname*{cone}S$. In other words, $p$ belongs to the
set of all linear combinations of the vectors $s$ for $s\in S$ with
nonnegative coefficients (since $\operatorname*{cone}S$ is the set of all
linear combinations of the vectors $s$ for $s\in S$ with nonnegative
coefficients (according to Definition 2.0k)). In other words, $p$ is a linear
combination of the vectors $s$ for $s\in S$ with nonnegative coefficients. In
other words, $p$ can be written in the form $\sum\limits_{i\in S}\lambda_{i}i$
for some family $\left(  \lambda_{i}\right)  _{i\in S}$ of nonnegative reals
indexed by the elements of $S$ such that all but finitely many $i\in S$
satisfy $\lambda_{i}=0$. Consider this $\left(  \lambda_{i}\right)  _{i\in S}%
$. Thus, $p=\sum\limits_{i\in S}\lambda_{i}i$.
\end{verlong}

We know that $\left(  \lambda_{i}\right)  _{i\in S}$ is a family of
nonnegative reals. Renaming the index $i$ as $s$ in this statement, we
conclude that $\left(  \lambda_{s}\right)  _{s\in S}$ is a family of
nonnegative reals. Now, $p=\sum\limits_{i\in S}\lambda_{i}i=\sum\limits_{s\in
S}\lambda_{s}s$ (here, we renamed the summation index $i$ as $s$). Thus, there
exists a family $\left(  \nu_{s}\right)  _{s\in S}$ of nonnegative reals
indexed by elements of $S$ such that $p=\sum\limits_{s\in S}\nu_{s}s$ (namely,
$\left(  \nu_{s}\right)  _{s\in S}=\left(  \lambda_{s}\right)  _{s\in S}$).
This proves Proposition 2.0p \textbf{(c)}.

\textbf{(a)} If $\left(  \nu_{s}\right)  _{s\in S}$ is a family of nonnegative
reals, then $\sum\limits_{s\in S}\nu_{s}s\in\operatorname*{cone}S$ (according
to Proposition 2.0p \textbf{(b)}). In other words,%
\begin{equation}
\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(  \nu_{s}\right)  _{s\in
S}\text{ is a family of nonnegative reals}\right\}  \subseteq
\operatorname*{cone}S. \label{pf.prop2.0p.a.1}%
\end{equation}


On the other hand,%
\[
\operatorname*{cone}S\subseteq\left\{  \sum\limits_{s\in S}\nu_{s}%
s\ \mid\ \left(  \nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative
reals}\right\}
\]
\footnote{\textit{Proof.} Let $p\in\operatorname*{cone}S$. Thus, there exists
a family $\left(  \nu_{s}\right)  _{s\in S}$ of nonnegative reals indexed by
elements of $S$ such that $p=\sum\limits_{s\in S}\nu_{s}s$ (according to
Proposition 2.0p \textbf{(c)}). In other words, $p$ has the form
$\sum\limits_{s\in S}\nu_{s}s$ for some family $\left(  \nu_{s}\right)  _{s\in
S}$ of nonnegative reals. In other words,%
\[
p\in\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(  \nu_{s}\right)
_{s\in S}\text{ is a family of nonnegative reals}\right\}  .
\]
\par
Now, let us forget that we fixed $p$. We thus have proven that every
$p\in\operatorname*{cone}S$ satisfies $p\in\left\{  \sum\limits_{s\in S}%
\nu_{s}s\ \mid\ \left(  \nu_{s}\right)  _{s\in S}\text{ is a family of
nonnegative reals}\right\}  $. In other words,%
\[
\operatorname*{cone}S\subseteq\left\{  \sum\limits_{s\in S}\nu_{s}%
s\ \mid\ \left(  \nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative
reals}\right\}  ,
\]
qed.}. Combining this with (\ref{pf.prop2.0p.a.1}), we obtain%
\[
\operatorname*{cone}S=\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(
\nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative reals}\right\}  .
\]
This proves Proposition 2.0p \textbf{(a)}.
\end{proof}

\subsection{Proofs for Section \ref{sect.basics-dual}}

\begin{proof}
[Proof of Proposition 2.0r.]Let $x\in C$.

\begin{vershort}
We have $x\in C=\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}%
,x_{2},\ldots,x_{t}\right\}  $, so that $x$ is a convex combination of the
vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$. In other words, there exist $t$
nonnegative elements $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{t}$ of
$\mathbb{R}$ such that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum
\limits_{i=1}^{t}\lambda_{i}x_{i}=x$. Consider these $\lambda_{1}$,
$\lambda_{2}$, $\ldots$, $\lambda_{t}$.
\end{vershort}

\begin{verlong}
We have $x\in C=\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}%
,x_{2},\ldots,x_{t}\right\}  $. But Proposition 2.0h \textbf{(b)} (applied to
$n=t$) shows that we have $x\in\operatorname*{conv}.\operatorname*{hull}%
\left\{  x_{1},x_{2},\ldots,x_{t}\right\}  $ if and only if $x$ is a convex
combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$. Hence, $x$ is
a convex combination of the vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$ (since
we have $x\in\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}%
,x_{2},\ldots,x_{t}\right\}  $). In other words, there exist $t$ nonnegative
elements $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{t}$ of $\mathbb{R}$
such that $\sum\limits_{i=1}^{t}\lambda_{i}=1$ and $\sum\limits_{i=1}%
^{t}\lambda_{i}x_{i}=x$ (by the definition of a \textquotedblleft convex
combination'). Consider these $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{t}$.
\end{verlong}

\begin{vershort}
There exists at least one $j\in\left\{  1,2,\ldots,t\right\}  $ such that
$\lambda_{j}\neq0$ (since otherwise, we would have $\lambda_{i}=0$ for every
$i\in\left\{  1,2,\ldots,t\right\}  $, so that $\sum\limits_{i=1}%
^{t}\underbrace{\lambda_{i}}_{=0}=\sum\limits_{i=1}^{t}0=0$, which would
contradict $\sum\limits_{i=1}^{t}\lambda_{i}=1\neq0$). Consider this $j$.
Then, $\lambda_{j}$ is nonnegative and satisfies $\lambda_{j}\neq0$. Thus,
$\lambda_{j}>0$.
\end{vershort}

\begin{verlong}
There exists at least one $j\in\left\{  1,2,\ldots,t\right\}  $ such that
$\lambda_{j}\neq0$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus,
there exists no $j\in\left\{  1,2,\ldots,t\right\}  $ such that $\lambda
_{j}\neq0$. In other words, each $j\in\left\{  1,2,\ldots,t\right\}  $
satisfies $\lambda_{j}=0$. Thus, $\sum_{j=1}^{t}\underbrace{\lambda_{j}}%
_{=0}=\sum_{j=1}^{t}0=0$. This contradicts%
\begin{align*}
\sum_{j=1}^{t}\lambda_{j}  &  =\sum_{i=1}^{t}\lambda_{i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}j\text{ as }i\right) \\
&  =1\neq0.
\end{align*}
This contradiction proves that our assumption was false. Qed.}. Consider this
$j$. We have $\lambda_{j}>0$ (since $\lambda_{j}$ is a nonnegative real
satisfying $\lambda_{j}\neq0$).
\end{verlong}

Combining $f\left(  x_{j}\right)  <\delta$ (by (\ref{2.0r.ass}), applied to
$i=j$) with $\lambda_{j}>0$, we obtain $\lambda_{j}f\left(  x_{j}\right)
<\lambda_{j}\delta$. In other words,%
\begin{equation}
\lambda_{j}f\left(  x_{j}\right)  -\lambda_{j}\delta<0. \label{2.1b.pf.2.pf.1}%
\end{equation}


Since $x=\sum\limits_{i=1}^{t}\lambda_{i}x_{i}$, we have%
\begin{align}
f\left(  x\right)   &  =f\left(  \sum\limits_{i=1}^{t}\lambda_{i}x_{i}\right)
=\sum\limits_{i=1}^{t}\lambda_{i}f\left(  x_{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{ is }\mathbb{R}%
\text{-linear}\right) \nonumber\\
&  =\underbrace{\sum\limits_{i=1}^{t}\lambda_{i}f\left(  x_{i}\right)
-\sum\limits_{i=1}^{t}\lambda_{i}\delta}_{=\sum\limits_{i=1}^{t}\left(
\lambda_{i}f\left(  x_{i}\right)  -\lambda_{i}\delta\right)  }%
+\underbrace{\sum\limits_{i=1}^{t}\lambda_{i}}_{=1}\delta\nonumber\\
&  =\sum\limits_{i=1}^{t}\left(  \lambda_{i}f\left(  x_{i}\right)
-\lambda_{i}\delta\right)  +\delta. \label{2.1b.pf.2.pf.2}%
\end{align}


\begin{vershort}
We know that every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies
$\lambda_{i}f\left(  x_{i}\right)  -\lambda_{i}\delta\leq0$ (since
(\ref{2.0r.ass}) yields $f\left(  x_{i}\right)  <\delta$, so that $\lambda
_{i}f\left(  x_{i}\right)  \leq\lambda_{i}\delta$ (since $\lambda_{i}\geq0$),
so that $\lambda_{i}f\left(  x_{i}\right)  -\lambda_{i}\delta\leq0$). Hence,
every addend of the sum $\sum\limits_{i=1}^{t}\left(  \lambda_{i}f\left(
x_{i}\right)  -\lambda_{i}\delta\right)  $ is nonpositive. Since we know that
at least one addend of this sum is actually negative (namely, the addend for
$i=j$, because of (\ref{2.1b.pf.2.pf.1})), this sum must thus be $<0$. Now,
(\ref{2.1b.pf.2.pf.2}) becomes
\[
f\left(  x\right)  =\underbrace{\sum\limits_{i=1}^{t}\left(  \lambda
_{i}f\left(  x_{i}\right)  -\lambda_{i}\delta\right)  }_{<0}+\delta<\delta.
\]

\end{vershort}

\begin{verlong}
We know that every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies%
\begin{equation}
\lambda_{i}f\left(  x_{i}\right)  -\lambda_{i}\delta\leq0
\label{2.1b.pf.2.pf.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.1b.pf.2.pf.4}):} Let $i\in\left\{
1,2,\ldots,t\right\}  $. From (\ref{2.0r.ass}), we obtain $f\left(
x_{i}\right)  <\delta$. Hence, $\lambda_{i}f\left(  x_{i}\right)  \leq
\lambda_{i}\delta$ (since $\lambda_{i}\geq0$), so that $\lambda_{i}f\left(
x_{i}\right)  -\lambda_{i}\delta\leq0$. This proves (\ref{2.1b.pf.2.pf.4}).}.
Now,%
\begin{align*}
&  \underbrace{\sum\limits_{i=1}^{t}}_{=\sum_{i\in\left\{  1,2,\ldots
,t\right\}  }}\left(  \lambda_{i}f\left(  x_{i}\right)  -\lambda_{i}%
\delta\right) \\
&  =\sum_{i\in\left\{  1,2,\ldots,t\right\}  }\left(  \lambda_{i}f\left(
x_{i}\right)  -\lambda_{i}\delta\right)  =\underbrace{\sum_{\substack{i\in
\left\{  1,2,\ldots,t\right\}  ;\\i=j}}\left(  \lambda_{i}f\left(
x_{i}\right)  -\lambda_{i}\delta\right)  }_{\substack{=\lambda_{j}f\left(
x_{j}\right)  -\lambda_{j}\delta\\\text{(since }j\in\left\{  1,2,\ldots
,t\right\}  \text{)}}}+\sum_{\substack{i\in\left\{  1,2,\ldots,t\right\}
;\\i\neq j}}\underbrace{\left(  \lambda_{i}f\left(  x_{i}\right)  -\lambda
_{i}\delta\right)  }_{\substack{\leq0\\\text{(by (\ref{2.1b.pf.2.pf.4}))}}}\\
&  \leq\left(  \lambda_{j}f\left(  x_{j}\right)  -\lambda_{j}\delta\right)
+0=\lambda_{j}f\left(  x_{j}\right)  -\lambda_{j}\delta
<0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.1b.pf.2.pf.1})}\right)  .
\end{align*}
Now, (\ref{2.1b.pf.2.pf.2}) becomes
\[
f\left(  x\right)  =\underbrace{\sum\limits_{i=1}^{t}\left(  \lambda
_{i}f\left(  x_{i}\right)  -\lambda_{i}\delta\right)  }_{<0}+\delta<\delta.
\]

\end{verlong}

This proves Proposition 2.0r.
\end{proof}

\begin{proof}
[Proof of Proposition 2.0s.]Let $x\in\operatorname*{cone}S$. Then,%
\[
x\in\operatorname*{cone}S=\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(
\nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative reals}\right\}
\]
(by the definition of $\operatorname*{cone}S$). Hence, there exists a family
$\left(  \nu_{s}\right)  _{s\in S}$ of nonnegative reals such that
$x=\sum\limits_{s\in S}\nu_{s}s$. Consider this $\left(  \nu_{s}\right)
_{s\in S}$.

Each $s\in S$ satisfies%
\begin{equation}
\nu_{s}f\left(  s\right)  \leq0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\nu_{s}\geq0\text{ and }f\left(  s\right)  \leq0\text{ (by (\ref{2.0s.ass}%
))}\right)  . \label{2.0s.pf.1}%
\end{equation}


Since $x=\sum\limits_{s\in S}\nu_{s}s$, we have%
\begin{align*}
f\left(  x\right)   &  =f\left(  \sum\limits_{s\in S}\nu_{s}s\right)
=\sum\limits_{s\in S}\underbrace{\nu_{s}f\left(  s\right)  }_{\substack{\leq
0\\\text{(by (\ref{2.0s.ass}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }f\text{ is }\mathbb{R}\text{-linear}\right) \\
&  \leq\sum\limits_{s\in S}0=0.
\end{align*}
This proves Proposition 2.0s.
\end{proof}

\begin{proof}
[Proof of Lemma 2.0t.]\textbf{(a)} Let $v\in\mathbb{R}^{n}$ be a column vector
satisfying $v\geq0$. We must prove that all coordinates of the column vector
$v$ are nonnegative.

We have $v\geq0$. In other words, every $i\in\left\{  1,2,\ldots,n\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }v\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)
\]
(by the definition of $v\geq0$). Hence, every $i\in\left\{  1,2,\ldots
,n\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }v\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0.
\]
In other words, for every $i\in\left\{  1,2,\ldots,n\right\}  $, the $i$-th
coordinate of $v$ is nonnegative. In other words, the coordinates of the
column vector $v$ are nonnegative. This proves Lemma 2.0t \textbf{(a)}.

\textbf{(b)} The proof of Lemma 2.0t \textbf{(b)} is completely identical to
the proof of Lemma 2.0t \textbf{(a)} given above (except that
\textquotedblleft$\mathbb{R}^{n}$\textquotedblright\ must be replaced by
\textquotedblleft$\left(  \mathbb{R}^{n}\right)  ^{\ast}$\textquotedblright,
and the word \textquotedblleft column\textquotedblright\ must be replaced by
\textquotedblleft row\textquotedblright).
\end{proof}

\begin{proof}
[Proof of Lemma 2.0u.]Write the vector $x$ in the form $\left(
\begin{array}
[c]{c}%
x_{1}\\
x_{2}\\
\vdots\\
x_{n}%
\end{array}
\right)  $ with $x_{1}$, $x_{2}$, $\ldots$, $x_{n}$ being real numbers. Then,
for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }x\right)  =x_{i}.
\label{2.5g.pf.1}%
\end{equation}


But now, let $u$ be the vector $\left(
\begin{array}
[c]{c}%
\max\left\{  x_{1},0\right\} \\
\max\left\{  x_{2},0\right\} \\
\vdots\\
\max\left\{  x_{n},0\right\}
\end{array}
\right)  $. Then, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align}
\left(  \text{the }i\text{-th coordinate of }u\right)   &  =\max\left\{
x_{i},0\right\} \label{2.5g.pf.2}\\
&  \geq0=\left(  \text{the }i\text{-th coordinate of }0\right)  .\nonumber
\end{align}
Thus, $u\geq0$.

Furthermore, let $v$ be the vector $\left(
\begin{array}
[c]{c}%
-\min\left\{  x_{1},0\right\} \\
-\min\left\{  x_{2},0\right\} \\
\vdots\\
-\min\left\{  x_{n},0\right\}
\end{array}
\right)  $. Then, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align}
\left(  \text{the }i\text{-th coordinate of }v\right)   &  =-\underbrace{\min
\left\{  x_{i},0\right\}  }_{\leq0}\label{2.5g.pf.3}\\
&  \geq-0=0=\left(  \text{the }i\text{-th coordinate of }0\right)  .\nonumber
\end{align}
Thus, $v\geq0$.

Now, every $\lambda\in\mathbb{R}$ satisfies
\begin{equation}
\max\left\{  \lambda,0\right\}  +\min\left\{  \lambda,0\right\}  =\lambda.
\label{2.5g.pf.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5g.pf.4}):} Let $\lambda\in\mathbb{R}$.
Then, we must be in one of the following two cases:
\par
\textit{Case 1:} We have $\lambda\geq0$.
\par
\textit{Case 2:} We have $\lambda<0$.
\par
Let us first consider Case 1. In this case, $\lambda\geq0$, so that
$\max\left\{  \lambda,0\right\}  =\lambda$ and $\min\left\{  \lambda
,0\right\}  =0$, and thus $\underbrace{\max\left\{  \lambda,0\right\}
}_{=\lambda}+\underbrace{\min\left\{  \lambda,0\right\}  }_{=0}=\lambda$.
Thus, (\ref{2.5g.pf.4}) is proven in Case 1.
\par
Let us next consider Case 2. In this case, $\lambda<0$, so that $\max\left\{
\lambda,0\right\}  =0$ and $\min\left\{  \lambda,0\right\}  =\lambda$, and
thus $\underbrace{\max\left\{  \lambda,0\right\}  }_{=0}+\underbrace{\min
\left\{  \lambda,0\right\}  }_{=\lambda}=\lambda$. Thus, (\ref{2.5g.pf.4}) is
proven in Case 2.
\par
Hence, we have proven (\ref{2.5g.pf.4}) in each of the cases 1 and 2. Thus,
(\ref{2.5g.pf.4}) always holds (since cases 1 and 2 cover all possibilities).
Qed.}

Now, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \left(  \text{the }i\text{-th coordinate of }u-v\right) \\
&  =\underbrace{\left(  \text{the }i\text{-th coordinate of }u\right)
}_{\substack{=\max\left\{  x_{i},0\right\}  \\\text{(by (\ref{2.5g.pf.2}))}%
}}-\underbrace{\left(  \text{the }i\text{-th coordinate of }v\right)
}_{\substack{=-\min\left\{  x_{i},0\right\}  \\\text{(by (\ref{2.5g.pf.3}))}%
}}\\
&  =\max\left\{  x_{i},0\right\}  -\left(  -\min\left\{  x_{i},0\right\}
\right)  =\max\left\{  x_{i},0\right\}  +\min\left\{  x_{i},0\right\}
=x_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5g.pf.4}), applied to }%
\lambda=x_{i}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5g.pf.1})}\right)  .
\end{align*}
Thus, $u-v=x$.

So we know that $u\geq0$, $v\geq0$ and $x=u-v$. Hence, there exist two vectors
$y$ and $z$ in $\mathbb{R}^{n}$ such that $y\geq0$, $z\geq0$ and $x=y-z$
(namely, $y=u$ and $z=v$). This proves Lemma 2.0u.
\end{proof}

\begin{proof}
[Proof of Lemma 2.0v.]The row vector $x\in\left(  \mathbb{R}^{n}\right)
^{\ast}$ satisfies $x\geq0$. Thus, Lemma 2.0t \textbf{(b)} (applied to $v=x$)
shows that the coordinates of the row vector $x$ are nonnegative. Let
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$ be the coordinates of
the row vector $x$. Then, $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda
_{n}$ are nonnegative (since the coordinates of the row vector $x$ are nonnegative).

The column vector $y\in\mathbb{R}^{n}$ satisfies $y\geq0$. Thus, Lemma 2.0t
\textbf{(a)} (applied to $v=y$) shows that the coordinates of the column
vector $y$ are nonnegative. Let $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$ be
the coordinates of the column vector $y$. Then, $\mu_{1}$, $\mu_{2}$, $\ldots
$, $\mu_{n}$ are nonnegative (since the coordinates of the column vector $y$
are nonnegative).

By the definition of the product of a row vector with a column vector, we have
$xy=\sum\limits_{i=1}^{n}\lambda_{i}\mu_{i}$ (since the coordinates of the row
vector $x$ are $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$, while
the coordinates of the column vector $y$ are $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{n}$).

But for every $i\in\left\{  1,2,\ldots,n\right\}  $, the real $\lambda_{i}%
\mu_{i}$ is nonnegative (since $\lambda_{i}$ is nonnegative and $\mu_{i}$ is
nonnegative), i. e., we have $\lambda_{i}\mu_{i}\geq0$. Thus, $xy=\sum
\limits_{i=1}^{n}\underbrace{\lambda_{i}\mu_{i}}_{\geq0}\geq\sum
\limits_{i=1}^{n}0=0$. This proves Lemma 2.0v.
\end{proof}

\subsection{Proofs for Section \ref{sect.closed-int}}

\begin{proof}
[Proof of Lemma 2.0x.]Let $n\in\mathbb{N}$, and let $I_{1}$, $I_{2}$, $\ldots
$, $I_{n}$ be $n$ closed intervals. We are going to prove that $I_{1}\cap
I_{2}\cap\cdots\cap I_{n}$ is a closed interval.

In order to show that, we will prove that%
\begin{equation}
\text{for every }k\in\left\{  0,1,\ldots,n\right\}  \text{, the set }I_{1}\cap
I_{2}\cap\cdots\cap I_{k}\text{ is a closed interval.} \label{2.0x.pf.1}%
\end{equation}


\textit{Proof of (\ref{2.0x.pf.1}):} We will prove (\ref{2.0x.pf.1}) by
induction over $k$:

\textit{Induction base:} For $k=0$, the set $I_{1}\cap I_{2}\cap\cdots\cap
I_{k}$ is a closed interval (because for $k=0$, the set $I_{1}\cap I_{2}%
\cap\cdots\cap I_{k}$ equals $I_{1}\cap I_{2}\cap\cdots\cap I_{0}=\mathbb{R}$,
and we know that $\mathbb{R}$ is a closed interval). In other words,
(\ref{2.0x.pf.1}) holds for $k=0$. This completes the induction base.

\textit{Induction step:} Let $i\in\left\{  1,2,\ldots,n\right\}  $. Assume
that (\ref{2.0x.pf.1}) holds for $k=i-1$. We now must show that
(\ref{2.0x.pf.1}) holds for $k=i$.

Since (\ref{2.0x.pf.1}) holds for $k=i-1$, we know that the set $I_{1}\cap
I_{2}\cap\cdots\cap I_{i-1}$ is a closed interval. Thus, the set $I_{1}\cap
I_{2}\cap\cdots\cap I_{i-1}$ has the form $\left\{  x\in\mathbb{R}%
\ \mid\ a\leq x\leq b\right\}  $ for some elements $a$ and $b$ of
$\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $\ \ \ \ \footnote{This is
because we have defined a closed interval to mean a set which has the form
$\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some elements $a$
and $b$ of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $.}. In other
words, there exist some elements $a$ and $b$ of $\mathbb{R}\cup\left\{
-\infty,\infty\right\}  $ such that $I_{1}\cap I_{2}\cap\cdots\cap
I_{i-1}=\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $. Denote this
$a$ and $b$ as $c$ and $d$, respectively. Then, $I_{1}\cap I_{2}\cap\cdots\cap
I_{i-1}=\left\{  x\in\mathbb{R}\ \mid\ c\leq x\leq d\right\}  $.

The set $I_{i}$ is a closed interval. Thus, $I_{i}$ has the form $\left\{
x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some elements $a$ and $b$
of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $\ \ \ \ \footnote{This is
because we have defined a closed interval to mean a set which has the form
$\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some elements $a$
and $b$ of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $.}. In other
words, there exist some elements $a$ and $b$ of $\mathbb{R}\cup\left\{
-\infty,\infty\right\}  $ such that $I_{i}=\left\{  x\in\mathbb{R}%
\ \mid\ a\leq x\leq b\right\}  $. Denote this $a$ and $b$ as $u$ and $v$,
respectively. Then, $I_{i}=\left\{  x\in\mathbb{R}\ \mid\ u\leq x\leq
v\right\}  $.

Now, combining the relations
\[
\left\{  x\in\mathbb{R}\ \mid\ \max\left\{  c,u\right\}  \leq x\leq
\min\left\{  d,v\right\}  \right\}  \subseteq I_{1}\cap I_{2}\cap\cdots\cap
I_{i}%
\]
\footnote{\textit{Proof.} Let $y\in\left\{  x\in\mathbb{R}\ \mid\ \max\left\{
c,u\right\}  \leq x\leq\min\left\{  d,v\right\}  \right\}  $. Then,
$y\in\mathbb{R}$ and $\max\left\{  c,u\right\}  \leq y\leq\min\left\{
d,v\right\}  $.
\par
Since $c\leq\max\left\{  c,u\right\}  \leq y$ and $y\leq\min\left\{
d,v\right\}  \leq d$, we have $c\leq y\leq d$, so that $y\in\left\{
x\in\mathbb{R}\ \mid\ c\leq x\leq d\right\}  =I_{1}\cap I_{2}\cap\cdots\cap
I_{i-1}$.
\par
Since $u\leq\max\left\{  c,u\right\}  \leq y$ and $y\leq\min\left\{
d,v\right\}  \leq v$, we have $u\leq y\leq v$, so that $y\in\left\{
x\in\mathbb{R}\ \mid\ u\leq x\leq v\right\}  =I_{i}$.
\par
Combining $y\in I_{1}\cap I_{2}\cap\cdots\cap I_{i-1}$ and $y\in I_{i}$, we
obtain $y\in\left(  I_{1}\cap I_{2}\cap\cdots\cap I_{i-1}\right)  \cap
I_{i}=I_{1}\cap I_{2}\cap\cdots\cap I_{i}$.
\par
Now, forget that we fixed $y$. We thus have proven that every $y\in\left\{
x\in\mathbb{R}\ \mid\ \max\left\{  c,u\right\}  \leq x\leq\min\left\{
d,v\right\}  \right\}  $ satisfies $y\in I_{1}\cap I_{2}\cap\cdots\cap I_{i}$.
In other words, $\left\{  x\in\mathbb{R}\ \mid\ \max\left\{  c,u\right\}  \leq
x\leq\min\left\{  d,v\right\}  \right\}  \subseteq I_{1}\cap I_{2}\cap
\cdots\cap I_{i}$, qed.} and%
\[
I_{1}\cap I_{2}\cap\cdots\cap I_{i}\subseteq\left\{  x\in\mathbb{R}%
\ \mid\ \max\left\{  c,u\right\}  \leq x\leq\min\left\{  d,v\right\}
\right\}
\]
\footnote{\textit{Proof.} Let $y\in I_{1}\cap I_{2}\cap\cdots\cap I_{i}$.
Then,
\[
y\in I_{1}\cap I_{2}\cap\cdots\cap I_{i}\subseteq I_{1}\cap I_{2}\cap
\cdots\cap I_{i-1}=\left\{  x\in\mathbb{R}\ \mid\ c\leq x\leq d\right\}  .
\]
Hence, $y\in\mathbb{R}$ and $c\leq y\leq d$. Also,%
\[
y\in I_{1}\cap I_{2}\cap\cdots\cap I_{i}\subseteq I_{i}=\left\{
x\in\mathbb{R}\ \mid\ u\leq x\leq v\right\}  ,
\]
so that $u\leq y\leq v$.
\par
Whenever $\alpha$, $\beta$, $\gamma$ are three reals satisfying $\alpha
\leq\gamma$ and $\beta\leq\gamma$, we have $\max\left\{  \alpha,\beta\right\}
\leq\gamma$. Applied to $\alpha=c$, $\beta=u$ and $\gamma=y$, this yields
$\max\left\{  c,u\right\}  \leq y$ (since $c\leq y$ and $u\leq y$).
\par
Whenever $\alpha$, $\beta$, $\gamma$ are three reals satisfying $\alpha
\leq\beta$ and $\alpha\leq\gamma$, we have $\alpha\leq\min\left\{
\beta,\gamma\right\}  $. Applied to $\alpha=y$, $\beta=d$ and $\gamma=v$, this
yields $y\leq\min\left\{  d,v\right\}  $ (since $y\leq d$ and $y\leq v$).
\par
Since $\max\left\{  c,u\right\}  \leq y\leq\min\left\{  d,v\right\}  $, we
have $y\in\left\{  x\in\mathbb{R}\ \mid\ \max\left\{  c,u\right\}  \leq
x\leq\min\left\{  d,v\right\}  \right\}  $.
\par
Now, forget that we fixed $y$. We thus have proven that every $y\in I_{1}\cap
I_{2}\cap\cdots\cap I_{i}$ satisfies $y\in\left\{  x\in\mathbb{R}\ \mid
\ \max\left\{  c,u\right\}  \leq x\leq\min\left\{  d,v\right\}  \right\}  $.
In other words, $I_{1}\cap I_{2}\cap\cdots\cap I_{i}\subseteq\left\{
x\in\mathbb{R}\ \mid\ \max\left\{  c,u\right\}  \leq x\leq\min\left\{
d,v\right\}  \right\}  $, qed.}, we obtain%
\[
I_{1}\cap I_{2}\cap\cdots\cap I_{i}=\left\{  x\in\mathbb{R}\ \mid
\ \max\left\{  c,u\right\}  \leq x\leq\min\left\{  d,v\right\}  \right\}  .
\]
Thus, $I_{1}\cap I_{2}\cap\cdots\cap I_{i}$ is a set of the form $\left\{
x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some elements $a$ and $b$
of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $ (namely, for
$a=\max\left\{  c,u\right\}  $ and $b=\min\left\{  d,v\right\}  $). In other
words, $I_{1}\cap I_{2}\cap\cdots\cap I_{i}$ is a closed
interval\footnote{This is because we have defined a closed interval to mean a
set which has the form $\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}
$ for some elements $a$ and $b$ of $\mathbb{R}\cup\left\{  -\infty
,\infty\right\}  $.}. In other words, (\ref{2.0x.pf.1}) holds for $k=i$. This
completes the induction step. The induction proof of (\ref{2.0x.pf.1}) is thus complete.

Now, applying (\ref{2.0x.pf.1}) to $k=n$, we conclude that $I_{1}\cap
I_{2}\cap\cdots\cap I_{n}$ is a closed interval.

Now forget that we fixed $n$ and $I_{1}$, $I_{2}$, $\ldots$, $I_{n}$. We have
thus shown that whenever $n\in\mathbb{N}$ and whenever $I_{1}$, $I_{2}$,
$\ldots$, $I_{n}$ are $n$ closed intervals, the set $I_{1}\cap I_{2}\cap
\cdots\cap I_{n}$ is a closed interval. In other words, we have proven that
the intersection of finitely many closed intervals always is a closed
interval. This proves Lemma 2.0x.
\end{proof}

\begin{proof}
[Proof of Lemma 2.0y.]We distinguish between three cases:

\textit{Case 1:} We have $\alpha<0$.

\textit{Case 2:} We have $\alpha=0$.

\textit{Case 3:} We have $\alpha>0$.

Let us first consider Case 1. In this case, $\alpha<0$. Hence, for every
$x\in\mathbb{R}$, the assertion $\alpha x\leq\beta$ is equivalent to
$x\geq\dfrac{\beta}{\alpha}$ (because dividing an inequality by a negative
real number reverses the sign of this inequality). Thus, for every
$x\in\mathbb{R}$, we have the following equivalence of assertions:%
\[
\left(  \alpha x\leq\beta\right)  \Longleftrightarrow\left(  x\geq\dfrac
{\beta}{\alpha}\right)  \Longleftrightarrow\left(  \dfrac{\beta}{\alpha}\leq
x\right)  \Longleftrightarrow\left(  \dfrac{\beta}{\alpha}\leq x\leq
\infty\right)  .
\]
Hence,
\[
\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  =\left\{
x\in\mathbb{R}\ \mid\ \dfrac{\beta}{\alpha}\leq x\leq\infty\right\}  .
\]
Thus, $\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a set of
the form $\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some
elements $a$ and $b$ of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $
(namely, for $a=\dfrac{\beta}{\alpha}$ and $b=\infty$). In other words,
$\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a closed
interval\footnote{This is because we have defined a closed interval to mean a
set which has the form $\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}
$ for some elements $a$ and $b$ of $\mathbb{R}\cup\left\{  -\infty
,\infty\right\}  $.}. This proves Lemma 2.0y in Case 1.

Let us now consider Case 2. In this case, $\alpha=0$. We must be in one of the
following two subcases:

\textit{Subcase 2.1:} We have $\beta\geq0$.

\textit{Subcase 2.2:} We have $\beta<0$.

First, let us consider Subcase 2.1. In this subcase, $\beta\geq0$. Thus, every
$x\in\mathbb{R}$ satisfies $\alpha x\leq\beta$ (since every $x\in\mathbb{R}$
satisfies $\underbrace{\alpha}_{=0}x=0\leq\beta$). Hence, $\left\{
x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  =\mathbb{R}$. Thus, $\left\{
x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a closed interval (since
$\mathbb{R}$ is a closed interval). This proves Lemma 2.0y in Subcase 2.1.

Next, let us consider Subcase 2.2. In this subcase, $\beta<0$. Thus, every
$x\in\mathbb{R}$ satisfies $\underbrace{\alpha}_{=0}x=0>\beta$. Hence, no
$x\in\mathbb{R}$ satisfies $\alpha x\leq\beta$. Therefore, $\left\{
x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  =\varnothing$. Thus,
$\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a closed
interval (since $\varnothing$ is a closed interval). This proves Lemma 2.0y in
Subcase 2.2.

Since Lemma 2.0y is proven in both Subcases 2.1 and 2.2, it follows that Lemma
2.0y always holds in Case 2 (because Subcases 2.1 and 2.2 cover all
possibilities within Case 2).

Finally, let us consider Case 3. In this case, $\alpha>0$. Hence, for every
$x\in\mathbb{R}$, the assertion $\alpha x\leq\beta$ is equivalent to
$x\leq\dfrac{\beta}{\alpha}$ (because dividing an inequality by a positive
real number leaves the sign of this inequality invariant). Thus, for every
$x\in\mathbb{R}$, we have the following equivalence of assertions:%
\[
\left(  \alpha x\leq\beta\right)  \Longleftrightarrow\left(  x\leq\dfrac
{\beta}{\alpha}\right)  \Longleftrightarrow\left(  -\infty\leq x\leq
\dfrac{\beta}{\alpha}\right)  .
\]
Hence,
\[
\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  =\left\{
x\in\mathbb{R}\ \mid\ -\infty\leq x\leq\dfrac{\beta}{\alpha}\right\}  .
\]
Thus, $\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a set of
the form $\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}  $ for some
elements $a$ and $b$ of $\mathbb{R}\cup\left\{  -\infty,\infty\right\}  $
(namely, for $a=-\infty$ and $b=\dfrac{\beta}{\alpha}$). In other words,
$\left\{  x\in\mathbb{R}\ \mid\ \alpha x\leq\beta\right\}  $ is a closed
interval\footnote{This is because we have defined a closed interval to mean a
set which has the form $\left\{  x\in\mathbb{R}\ \mid\ a\leq x\leq b\right\}
$ for some elements $a$ and $b$ of $\mathbb{R}\cup\left\{  -\infty
,\infty\right\}  $.}. This proves Lemma 2.0y in Case 3.

Thus, Lemma 2.0y is proven in each of Cases 1, 2 and 3. Since these cases
cover all possibilities, this yields that Lemma 2.0y is always proven.
\end{proof}

\begin{proof}
[Proof of Lemma 2.0z.]Let $b_{1},b_{2},\ldots,b_{m}$ be the $m$ coordinates of
the column vector $b$. Thus, we have%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }b\right)  =b_{i}
\label{2.0z.pf.1}%
\end{equation}
for each $i\in\left\{  1,2,\ldots,m\right\}  $.

Let $a_{1},a_{2},\ldots,a_{m}$ be the $m$ rows of the matrix $A$. Then, for
each column vector $x\in\mathbb{R}^{n}$, and for each $i\in\left\{
1,2,\ldots,m\right\}  $, we have%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }Ax\right)  =a_{i}x
\label{2.0z.pf.2}%
\end{equation}
(by the definition of the product of a matrix with a column vector).

For each $x\in\mathbb{R}^{n}$, we have the following chain of equivalences:%
\begin{align*}
&  \ \left(  Ax\leq b\right) \\
&  \Longleftrightarrow\ \left(  \text{every }i\in\left\{  1,2,\ldots
,m\right\}  \text{ satisfies}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  \underbrace{\left(
\text{the }i\text{-th coordinate of }Ax\right)  }_{\substack{=a_{i}%
x\\\text{(by (\ref{2.0z.pf.2}))}}}\leq\underbrace{\left(  \text{the
}i\text{-th coordinate of }b\right)  }_{\substack{=b_{i}\\\text{(by
(\ref{2.0z.pf.1}))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }Ax\leq b\right) \\
&  \Longleftrightarrow\ \left(  \text{every }i\in\left\{  1,2,\ldots
,m\right\}  \text{ satisfies }a_{i}x\leq b_{i}\right) \\
&  \Longleftrightarrow\ \left(  a_{i}x\leq b_{i}\text{ for every }i\in\left\{
1,2,\ldots,m\right\}  \right)  .
\end{align*}
Hence,%
\begin{align*}
&  \left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\} \\
&  =\left\{  x\in\mathbb{R}^{n}\ \mid\ a_{i}x\leq b_{i}\text{ for every }%
i\in\left\{  1,2,\ldots,m\right\}  \right\}  .
\end{align*}
Thus,%
\begin{align*}
P  &  =\left\{  x\in\mathbb{R}^{n}\ \mid\ Ax\leq b\right\} \\
&  =\left\{  x\in\mathbb{R}^{n}\ \mid\ a_{i}x\leq b_{i}\text{ for every }%
i\in\left\{  1,2,\ldots,m\right\}  \right\}  .
\end{align*}


For every $i\in\left\{  1,2,\ldots,m\right\}  $, the set $\left\{  \mu
\in\mathbb{R}\ \mid\ a_{i}c\mu\leq b_{i}-a_{i}z\right\}  $ is a closed
interval (by Lemma 2.0y, applied to $\alpha=a_{i}c$ and $\beta=b_{i}-a_{i}z$).

For any element $\mu\in\mathbb{R}$, we have the following equivalence of
assertions:%
\begin{align*}
\left(  z+\mu c\in P\right)   &  \Longleftrightarrow\left(  \underbrace{a_{i}%
\left(  z+\mu c\right)  }_{=a_{i}z+a_{i}c\mu}\leq b_{i}\text{ for every }%
i\in\left\{  1,2,\ldots,m\right\}  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }P=\left\{  x\in\mathbb{R}%
^{n}\ \mid\ a_{i}x\leq b_{i}\text{ for every }i\in\left\{  1,2,\ldots
,m\right\}  \right\}  \right) \\
&  \Longleftrightarrow\left(  \underbrace{a_{i}z+a_{i}c\mu\leq b_{i}%
}_{\substack{\text{this is equivalent to}\\a_{i}c\mu\leq b_{i}-a_{i}z}}\text{
for every }i\in\left\{  1,2,\ldots,m\right\}  \right) \\
&  \Longleftrightarrow\left(  a_{i}c\mu\leq b_{i}-a_{i}z\text{ for every }%
i\in\left\{  1,2,\ldots,m\right\}  \right)  .
\end{align*}
Thus,%
\begin{align*}
&  \left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\} \\
&  =\left\{  \mu\in\mathbb{R}\ \mid\ a_{i}c\mu\leq b_{i}-a_{i}z\text{ for
every }i\in\left\{  1,2,\ldots,m\right\}  \right\} \\
&  =\bigcap\limits_{i\in\left\{  1,2,\ldots,m\right\}  }\left\{  \mu
\in\mathbb{R}\ \mid\ a_{i}c\mu\leq b_{i}-a_{i}z\right\}  .
\end{align*}
This shows that $\left\{  \mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ is
an intersection of finitely many closed intervals (since $\left\{  \mu
\in\mathbb{R}\ \mid\ a_{i}c\mu\leq b_{i}-a_{i}z\right\}  $ is a closed
interval for every $i\in\left\{  1,2,\ldots,m\right\}  $). Thus, $\left\{
\mu\in\mathbb{R}\ \mid\ z+\mu c\in P\right\}  $ is a closed interval (since
Lemma 2.0x says that the intersection of finitely many closed intervals always
is a closed interval). This proves Lemma 2.0z.
\end{proof}

\subsection{Proofs for Section \ref{sect.lt-gt}}

\begin{proof}
[Proof of Lemma 2.2a.]\textbf{(a)} Let $v\in\mathbb{R}^{n}$ be a column vector
satisfying $v>0$. We must prove that all coordinates of the column vector $v$
are positive.

We have $v>0$. In other words, every $i\in\left\{  1,2,\ldots,n\right\}  $
satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }v\right)  >\left(  \text{the
}i\text{-th coordinate of }0\right)
\]
(by the definition of $v>0$). Hence, every $i\in\left\{  1,2,\ldots,n\right\}
$ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }v\right)  >\left(  \text{the
}i\text{-th coordinate of }0\right)  =0.
\]
In other words, for every $i\in\left\{  1,2,\ldots,n\right\}  $, the $i$-th
coordinate of $v$ is positive. In other words, the coordinates of the column
vector $v$ are positive. This proves Lemma 2.2a \textbf{(a)}.

\textbf{(b)} The proof of Lemma 2.2a \textbf{(b)} is completely identical to
the proof of Lemma 2.2a \textbf{(a)} given above (except that
\textquotedblleft$\mathbb{R}^{n}$\textquotedblright\ must be replaced by
\textquotedblleft$\left(  \mathbb{R}^{n}\right)  ^{\ast}$\textquotedblright,
and the word \textquotedblleft column\textquotedblright\ must be replaced by
\textquotedblleft row\textquotedblright).
\end{proof}

\begin{proof}
[Proof of Lemma 2.2b.]The row vector $x\in\left(  \mathbb{R}^{n}\right)
^{\ast}$ satisfies $x\geq0$. Thus, Lemma 2.0t \textbf{(b)} (applied to $v=x$)
shows that the coordinates of the row vector $x$ are nonnegative. Let
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$ be the coordinates of
the row vector $x$. Then, $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda
_{n}$ are nonnegative (since the coordinates of the row vector $x$ are nonnegative).

The column vector $y\in\mathbb{R}^{n}$ satisfies $y>0$. Thus, Lemma 2.2a
\textbf{(a)} (applied to $v=y$) shows that the coordinates of the column
vector $y$ are positive. Let $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$ be the
coordinates of the column vector $y$. Then, $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{n}$ are positive (since the coordinates of the column vector $y$ are positive).

By the definition of the product of a row vector with a column vector, we have
$xy=\sum\limits_{i=1}^{n}\lambda_{i}\mu_{i}$ (since the coordinates of the row
vector $x$ are $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$, while
the coordinates of the column vector $y$ are $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{n}$). But for every $i\in\left\{  1,2,\ldots,n\right\}  $, the real
$\lambda_{i}\mu_{i}$ is nonnegative (since $\lambda_{i}$ is
nonnegative\footnote{because $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{n}$ are nonnegative} and $\mu_{i}$ is positive\footnote{because
$\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$ are positive}). In other words, for
every $i\in\left\{  1,2,\ldots,n\right\}  $, we have $\lambda_{i}\mu_{i}\geq
0$. Hence, $\sum\limits_{i=1}^{n}\underbrace{\lambda_{i}\mu_{i}}_{\geq0}%
\geq\sum\limits_{i=1}^{n}0=0$. Hence, $xy=\sum\limits_{i=1}^{n}\lambda_{i}%
\mu_{i}\geq0$.

Assume (for the sake of contradiction) that $xy=0$.

Recall that, for every $i\in\left\{  1,2,\ldots,n\right\}  $, the real
$\lambda_{i}\mu_{i}$ is nonnegative. Thus, the sum $\sum\limits_{i=1}%
^{n}\lambda_{i}\mu_{i}$ is a sum of nonnegative reals. Since this sum is $0$
(because $\sum\limits_{i=1}^{n}\lambda_{i}\mu_{i}=0$), this yields that
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,n\right\}  \text{ satisfies }\lambda
_{i}\mu_{i}=0 \label{2.2b.pf.1}%
\end{equation}
(because if a sum of nonnegative reals is $0$, then each of these reals must
equal $0$). Thus, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
$\lambda_{i}=0$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{
1,2,\ldots,n\right\}  $. Recall that $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$
are positive; thus, $\mu_{i}$ is positive. Hence, $\mu_{i}\neq0$. But
(\ref{2.2b.pf.1}) yields $\lambda_{i}\mu_{i}=0$. We can divide this equality
by $\mu_{i}$ (since $\mu_{i}\neq0$), and thus obtain $\lambda_{i}=0$. Qed.}.
In other words, all the numbers $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{n}$ equal zero. Since $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{n}$ are the coordinates of the row vector $x$, this yields that all
the coordinates of the row vector $x$ equal zero. Hence, $x=0$. This
contradicts the fact that $x\neq0$ (since $x$ is nonzero). This contradiction
shows that our assumption (that $xy=0$) was false. Hence, we cannot have
$xy=0$. Thus, we have $xy\neq0$. Combining this with $xy\geq0$, we obtain
$xy>0$. This proves Lemma 2.2b.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Lemma 2.2c.]Lemma 2.2c is proven in the same way as Lemma 2.2b,
except that the roles of the vectors $x$ and $y$ are partly reversed. (The
details are left to the reader.)
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma 2.2c.]The row vector $x\in\left(  \mathbb{R}^{n}\right)
^{\ast}$ satisfies $x>0$. Thus, Lemma 2.2a \textbf{(b)} (applied to $v=x$)
shows that the coordinates of the row vector $x$ are positive. Let
$\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$ be the coordinates of
the row vector $x$. Then, $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda
_{n}$ are positive (since the coordinates of the row vector $x$ are positive).

The column vector $y\in\mathbb{R}^{n}$ satisfies $y\geq0$. Thus, Lemma 2.0t
\textbf{(a)} (applied to $v=y$) shows that the coordinates of the column
vector $y$ are nonnegative. Let $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$ be
the coordinates of the column vector $y$. Then, $\mu_{1}$, $\mu_{2}$, $\ldots
$, $\mu_{n}$ are nonnegative (since the coordinates of the column vector $y$
are nonnegative).

By the definition of the product of a row vector with a column vector, we have
$xy=\sum\limits_{i=1}^{n}\lambda_{i}\mu_{i}$ (since the coordinates of the row
vector $x$ are $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda_{n}$, while
the coordinates of the column vector $y$ are $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{n}$). But for every $i\in\left\{  1,2,\ldots,n\right\}  $, the real
$\lambda_{i}\mu_{i}$ is nonnegative (since $\lambda_{i}$ is
positive\footnote{because $\lambda_{1}$, $\lambda_{2}$, $\ldots$, $\lambda
_{n}$ are positive} and $\mu_{i}$ is nonnegative\footnote{because $\mu_{1}$,
$\mu_{2}$, $\ldots$, $\mu_{n}$ are nonnegative}). In other words, for every
$i\in\left\{  1,2,\ldots,n\right\}  $, we have $\lambda_{i}\mu_{i}\geq0$.
Hence, $\sum\limits_{i=1}^{n}\underbrace{\lambda_{i}\mu_{i}}_{\geq0}\geq
\sum\limits_{i=1}^{n}0=0$. Hence, $xy=\sum\limits_{i=1}^{n}\lambda_{i}\mu
_{i}\geq0$.

Assume (for the sake of contradiction) that $xy=0$.

Recall that, for every $i\in\left\{  1,2,\ldots,n\right\}  $, the real
$\lambda_{i}\mu_{i}$ is nonnegative. Hence, the sum $\sum\limits_{i=1}%
^{n}\lambda_{i}\mu_{i}$ is a sum of nonnegative reals. Since this sum is $0$
(because $\sum\limits_{i=1}^{n}\lambda_{i}\mu_{i}=0$), this yields that
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,n\right\}  \text{ satisfies }\lambda
_{i}\mu_{i}=0 \label{2.2c.pf.1}%
\end{equation}
(because if a sum of nonnegative reals is $0$, then each of these reals must
equal $0$). Thus, every $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
$\mu_{i}=0$\ \ \ \ \footnote{\textit{Proof.} Let $i\in\left\{  1,2,\ldots
,n\right\}  $. Recall that $\lambda_{1}$, $\lambda_{2}$, $\ldots$,
$\lambda_{n}$ are positive; thus, $\lambda_{i}$ is positive. Hence,
$\lambda_{i}\neq0$. But (\ref{2.2c.pf.1}) yields $\lambda_{i}\mu_{i}=0$. We
can divide this equality by $\lambda_{i}$ (since $\lambda_{i}\neq0$), and thus
obtain $\mu_{i}=0$. Qed.}. In other words, all the numbers $\mu_{1}$, $\mu
_{2}$, $\ldots$, $\mu_{n}$ equal zero. Since $\mu_{1}$, $\mu_{2}$, $\ldots$,
$\mu_{n}$ are the coordinates of the column vector $y$, this yields that all
the coordinates of the column vector $y$ equal zero. Hence, $y=0$. This
contradicts the fact that $y\neq0$ (since $y$ is nonzero). This contradiction
shows that our assumption (that $xy=0$) was false. Hence, we cannot have
$xy=0$. Thus, we have $xy\neq0$. Combining this with $xy\geq0$, we obtain
$xy>0$. This proves Lemma 2.2c.
\end{proof}
\end{verlong}

\begin{proof}
[Proof of Lemma 2.2d.]Assume the contrary. Thus, $x\neq0$. Hence, $x$ is
nonzero. Lemma 2.2b thus yields $xy>0$. This contradicts $xy=0$. This
contradiction shows that our assumption was false. This proves Lemma 2.2d.
\end{proof}

\subsection{Proofs for Section \ref{sect.block}}

\begin{vershort}
\begin{proof}
[Proof of Lemma 2.5h.]We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \geq0\right) \\
&  \Longleftrightarrow\ \left(  \text{all coordinates of the vector }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \text{ are }\geq0\right) \\
&  \Longleftrightarrow\ \left(  \text{all coordinates of the vectors }x\text{
and }y\text{ are }\geq0\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the coordinates of the vector }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \text{ are}\\
\text{precisely the coordinates of the vectors }x\text{ and }y
\end{array}
\right) \\
&  \Longleftrightarrow\ \underbrace{\left(  \text{all coordinates of the
vector }x\text{ are }\geq0\right)  }_{\Longleftrightarrow\ \left(
x\geq0\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \wedge\underbrace{\left(  \text{all coordinates of the
vector }y\text{ are }\geq0\right)  }_{\Longleftrightarrow\ \left(
y\geq0\right)  }\\
&  \Longleftrightarrow\ \left(  x\geq0\right)  \wedge\left(  y\geq0\right)  .
\end{align*}
In other words, $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \geq0$ holds if and only if $\left(  x\geq0\text{ and }y\geq0\right)
$. This yields both parts \textbf{(a)} and \textbf{(b)} of Lemma 2.5h.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma 2.5h.]\textbf{(a)} Assume that $x\geq0$ and $y\geq0$.

We know that $x\geq0$. In other words, every $i\in\left\{  1,2,\ldots
,N\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }x\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0. \label{2.5h.pf.1}%
\end{equation}


We know that $y\geq0$. In other words, every $i\in\left\{  1,2,\ldots
,M\right\}  $ satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }y\right)  \geq\left(  \text{the
}i\text{-th coordinate of }0\right)  =0. \label{2.5h.pf.2}%
\end{equation}


It is now easy to see that every $i\in\left\{  1,2,\ldots,N+M\right\}  $
satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right)  \geq\left(  \text{the }i\text{-th coordinate of }0\right)
\label{2.5h.pf.3}%
\end{equation}
(where $0$ means the zero vector in $\mathbb{R}^{N+M}$).
\footnote{\textit{Proof of (\ref{2.5h.pf.3}):} Let $i\in\left\{
1,2,\ldots,N+M\right\}  $. Then, either $i\in\left\{  1,2,\ldots,N\right\}  $
and $i\in\left\{  N+1,N+2,\ldots,N+M\right\}  $. In other words, we must be in
one of the following two cases:
\par
\textit{Case 1:} We have $i\in\left\{  1,2,\ldots,N\right\}  $.
\par
\textit{Case 2:} We have $i\in\left\{  N+1,N+2,\ldots,N+M\right\}  $.
\par
Let us consider Case 1 first. In this case, $i\in\left\{  1,2,\ldots
,N\right\}  $. Thus,%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right)   &  =\left(  \text{the }i\text{-th coordinate of }x\right)
\geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5h.pf.1})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, (\ref{2.5h.pf.3}) holds in Case 1.
\par
Now, let us consider Case 2. In this case, $i\in\left\{  N+1,N+2,\ldots
,N+M\right\}  $. Thus,%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right)   &  =\left(  \text{the }\left(  i-N\right)  \text{-th
coordinate of }y\right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5h.pf.2}), applied to
}i-N\text{ instead of }i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
Hence, (\ref{2.5h.pf.3}) holds in Case 2.
\par
We have thus proven that (\ref{2.5h.pf.3}) holds in each of Cases 1 and 2.
Since these Cases 1 and 2 cover all possibilities, this yields that
(\ref{2.5h.pf.3}) always holds. This proves (\ref{2.5h.pf.3}).} In other
words, $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \geq0$. This proves Lemma 2.5h \textbf{(a)}.

\textbf{(b)} Assume that $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \geq0$. In other words, every $i\in\left\{  1,2,\ldots,N+M\right\}  $
satisfies%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right)  \geq\left(  \text{the }i\text{-th coordinate of }0\right)
=0. \label{2.5h.pf.4}%
\end{equation}


Now, every $i\in\left\{  1,2,\ldots,N\right\}  $ satisfies%
\[
\left(  \text{the }i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }x\right)  .
\]
Thus, every $i\in\left\{  1,2,\ldots,N\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }x\right)   &  =\left(  \text{the
}i\text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5h.pf.4})}\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $x\geq0$.

Also, every $i\in\left\{  1,2,\ldots,M\right\}  $ satisfies%
\[
\left(  \text{the }\left(  N+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of }y\right)  .
\]
Thus, every $i\in\left\{  1,2,\ldots,M\right\}  $ satisfies%
\begin{align*}
\left(  \text{the }i\text{-th coordinate of }y\right)   &  =\left(  \text{the
}\left(  N+i\right)  \text{-th coordinate of }\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right) \\
&  \geq0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5h.pf.4}), applied to
}N+i\text{ instead of }i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of }0\right)  .
\end{align*}
In other words, $y\geq0$.

Thus, we have proven that $x\geq0$ and $y\geq0$. Lemma 2.5h \textbf{(b)} is proven.
\end{proof}
\end{verlong}

\begin{vershort}
\begin{proof}
[Proof of Lemma 2.5i.]Let $x_{1},x_{2},\ldots,x_{n}$ be the $n$ coordinates of
the vector $x$. Let $y_{1},y_{2},\ldots,y_{n}$ be the $n$ coordinates of the
vector $y$. Then, the $2n$ coordinates of the vector $\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  $ are $x_{1},x_{2},\ldots,x_{n},-x_{1},-x_{2},\ldots,-x_{n}$, whereas
the $2n$ coordinates of the vector $\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  $ are $y_{1},y_{2},\ldots,y_{n},-y_{1},-y_{2},\ldots,-y_{n}$. Hence,
the inequality $\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \geq\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  $ says that each of the $2n$ coordinates $x_{1},x_{2},\ldots
,x_{n},-x_{1},-x_{2},\ldots,-x_{n}$ is $\geq$ to the corresponding coordinate
among $y_{1},y_{2},\ldots,y_{n},-y_{1},-y_{2},\ldots,-y_{n}$. In other words,%
\begin{equation}
x_{i}\geq y_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots
,n\right\}  , \label{2.5i.pf.short.1}%
\end{equation}
and%
\begin{equation}
-x_{i}\geq-y_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{
1,2,\ldots,n\right\}  . \label{2.5i.pf.short.2}%
\end{equation}
But (\ref{2.5i.pf.short.2}) rewrites as%
\[
x_{i}\leq y_{i}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  1,2,\ldots
,n\right\}  .
\]
Combining this with (\ref{2.5i.pf.short.1}), we obtain $x_{i}=y_{i}$ for each
$i\in\left\{  1,2,\ldots,n\right\}  $. Thus, $x=y$. This proves Lemma 2.5i.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma 2.5i.]For every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align}
&  \left(  \text{the }i\text{-th coordinate of the vector }\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \right) \nonumber\\
&  \geq\left(  \text{the }i\text{-th coordinate of the vector }\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  \right)  \label{2.5i.pf.1}%
\end{align}
(since $\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \geq\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  $).

For every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of the vector }\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \right)  =\left(  \text{the }i\text{-th coordinate of the vector
}x\right)
\]
and thus%
\begin{align}
\left(  \text{the }i\text{-th coordinate of the vector }x\right)   &  =\left(
\text{the }i\text{-th coordinate of the vector }\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \right) \nonumber\\
&  \geq\left(  \text{the }i\text{-th coordinate of the vector }\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5i.pf.1})}\right) \nonumber\\
&  =\left(  \text{the }i\text{-th coordinate of the vector }y\right)  .
\label{2.5i.pf.2}%
\end{align}


On the other hand, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
\left(  \text{the }\left(  n+i\right)  \text{-th coordinate of the vector
}\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \right)   &  =\left(  \text{the }i\text{-th coordinate of the vector
}-x\right) \\
&  =-\left(  \text{the }i\text{-th coordinate of the vector }x\right)
\end{align*}
and thus%
\begin{align*}
&  -\left(  \text{the }i\text{-th coordinate of the vector }x\right) \\
&  =\left(  \text{the }\left(  n+i\right)  \text{-th coordinate of the vector
}\left(
\begin{array}
[c]{c}%
x\\
-x
\end{array}
\right)  \right) \\
&  \geq\left(  \text{the }\left(  n+i\right)  \text{-th coordinate of the
vector }\left(
\begin{array}
[c]{c}%
y\\
-y
\end{array}
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{2.5i.pf.1}), applied to
}n+i\text{ instead of }i\right) \\
&  =\left(  \text{the }i\text{-th coordinate of the vector }-y\right) \\
&  =-\left(  \text{the }i\text{-th coordinate of the vector }y\right)  ,
\end{align*}
so that%
\begin{equation}
\left(  \text{the }i\text{-th coordinate of the vector }x\right)  \leq\left(
\text{the }i\text{-th coordinate of the vector }y\right)  . \label{2.5i.pf.3}%
\end{equation}


Hence, for every $i\in\left\{  1,2,\ldots,n\right\}  $, we have%
\[
\left(  \text{the }i\text{-th coordinate of the vector }x\right)  =\left(
\text{the }i\text{-th coordinate of the vector }y\right)
\]
(by (\ref{2.5i.pf.2}), combined with (\ref{2.5i.pf.3})). Thus, $x=y$. Lemma
2.5i is proven.
\end{proof}
\end{verlong}

\section{\label{sect.sep-proofs-old}Appendix: Old (and ugly) proofs of
Theorems 2.1c and 2.5c}

This section gives alternative proofs for Theorem 2.1c and 2.5c. These are the
proofs I have found myself, before I became aware of the neat proofs given in
Section \ref{sect.sep-proofs}; they are clumsy and long-winded, but maybe
there is something of interest in them (although I don't know what that would be).

We shall prove Theorem 2.1c first, and then derive Theorem 2.5c from it. But
first of all, we will need a basic fact (which is often used as an exercise in
linear algebra courses):

\begin{statement}
\textbf{Lemma 2.1d.} Let $k$ be an infinite field, and let $E$ be a $k$-vector
space. Let $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$ be vectors in $E$. Assume that
every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $x_{i}\neq0$. Then,
there exists an $f\in E^{\ast}$ such that every $i\in\left\{  1,2,\ldots
,t\right\}  $ satisfies $f\left(  x_{i}\right)  \neq0$.
\end{statement}

\begin{proof}
[Proof of Lemma 2.1d.]Let us prove that for every $j\in\left\{  0,1,\ldots
,t\right\}  $,%
\begin{equation}
\text{there exists some }f\in E^{\ast}\text{ such that every }i\in\left\{
1,2,\ldots,j\right\}  \text{ satisfies }f\left(  x_{i}\right)  \neq0.
\label{2.1c.pf.1}%
\end{equation}


\textit{Proof of (\ref{2.1c.pf.1}):} We will prove (\ref{2.1c.pf.1}) by
induction over $j$:

\textit{Induction base:} There exists some $f\in E^{\ast}$ such that every
$i\in\left\{  1,2,\ldots,0\right\}  $ satisfies $f\left(  x_{i}\right)  \neq0$
(namely, $f=0$ (because the assertion that every $i\in\left\{  1,2,\ldots
,0\right\}  $ satisfies $f\left(  x_{i}\right)  \neq0$ is vacuously true)).

\textit{Induction step:} Let $J\in\left\{  1,2,\ldots,t\right\}  $ be such
that (\ref{2.1c.pf.1}) holds for $j=J-1$. We then have to prove that
(\ref{2.1c.pf.1}) holds for $j=J$.

We have $x_{J}\neq0$ (because every $i\in\left\{  1,2,\ldots,t\right\}  $
satisfies $x_{i}\neq0$).

There is a known linear-algebraic fact that if $v$ is a vector in a $k$-vector
space $F$ such that $v\neq0$, then there exists an $h\in F^{\ast}$ such that
$h\left(  v\right)  \neq0$. Applying this fact to $F=E$ and $v=x_{J}$, we see
that there exists an $h\in E^{\ast}$ such that $h\left(  x_{J}\right)  \neq0$
(since $x_{J}\neq0$). Consider this $h$.

Since (\ref{2.1c.pf.1}) holds for $j=J-1$, there exists some $f\in E^{\ast}$
such that every $i\in\left\{  1,2,\ldots,J-1\right\}  $ satisfies $f\left(
x_{i}\right)  \neq0$. Denote this $f$ by $g$. Then, every $i\in\left\{
1,2,\ldots,J-1\right\}  $ satisfies $g\left(  x_{i}\right)  \neq0$.

Let $r$ be the element $\left\{
\begin{array}
[c]{c}%
\dfrac{-h\left(  x_{J}\right)  }{g\left(  x_{J}\right)  }%
,\ \ \ \ \ \ \ \ \ \ \text{if }g\left(  x_{J}\right)  \neq0;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }g\left(  x_{J}\right)  =0
\end{array}
\right.  $ of $k$. Let $M$ be the subset $\left\{  \dfrac{-h\left(
x_{1}\right)  }{g\left(  x_{1}\right)  },\dfrac{-h\left(  x_{2}\right)
}{g\left(  x_{2}\right)  },\ldots,\dfrac{-h\left(  x_{J-1}\right)  }{g\left(
x_{J-1}\right)  }\right\}  \cup\left\{  r\right\}  $ of $k$. Then,%
\begin{align*}
\left\vert M\right\vert  &  =\left\vert \left\{  \dfrac{-h\left(
x_{1}\right)  }{g\left(  x_{1}\right)  },\dfrac{-h\left(  x_{2}\right)
}{g\left(  x_{2}\right)  },\ldots,\dfrac{-h\left(  x_{J-1}\right)  }{g\left(
x_{J-1}\right)  }\right\}  \cup\left\{  r\right\}  \right\vert \\
&  \leq\underbrace{\left\vert \left\{  \dfrac{-h\left(  x_{1}\right)
}{g\left(  x_{1}\right)  },\dfrac{-h\left(  x_{2}\right)  }{g\left(
x_{2}\right)  },\ldots,\dfrac{-h\left(  x_{J-1}\right)  }{g\left(
x_{J-1}\right)  }\right\}  \right\vert }_{\leq J-1}+\underbrace{\left\vert
\left\{  r\right\}  \right\vert }_{=1}\leq J-1+1=J,
\end{align*}
so that we cannot have $k\subseteq M$ (because if we had $k\subseteq M$, then
we would have $\left\vert k\right\vert \leq\left\vert M\right\vert \leq J$,
which contradicts the fact that $k$ is infinite). As a consequence, there
exists an $s\in k$ such that $s\notin M$. Consider this $s$.

Now, every $i\in\left\{  1,2,\ldots,J\right\}  $ satisfies $\left(
sg+h\right)  \left(  x_{i}\right)  \neq0$.\ \ \ \ \footnote{\textit{Proof.}
Assume the opposite. Then, there exists some $i\in\left\{  1,2,\ldots
,J\right\}  $ such that $\left(  sg+h\right)  \left(  x_{i}\right)  =0$.
Consider this $i$. Since $i\in\left\{  1,2,\ldots,J\right\}  $, we must be in
one of the following two cases:
\par
\textit{Case 1:} We have $i\in\left\{  1,2,\ldots,J-1\right\}  $.
\par
\textit{Case 2:} We have $i=J$.
\par
Let us first consider Case 1. In this case, $i\in\left\{  1,2,\ldots
,J-1\right\}  $.
\par
Our $i$ satisfies $\left(  sg+h\right)  \left(  x_{i}\right)  =0$, so that
$0=\left(  sg+h\right)  \left(  x_{i}\right)  =sg\left(  x_{i}\right)
+h\left(  x_{i}\right)  $, thus $sg\left(  x_{i}\right)  =-h\left(
x_{i}\right)  $. Since $g\left(  x_{i}\right)  \neq0$, we can divide this by
$g\left(  x_{i}\right)  $ and obtain%
\begin{align*}
s  &  \in\dfrac{-h\left(  x_{i}\right)  }{g\left(  x_{i}\right)  }\in\left\{
\dfrac{-h\left(  x_{1}\right)  }{g\left(  x_{1}\right)  },\dfrac{-h\left(
x_{2}\right)  }{g\left(  x_{2}\right)  },\ldots,\dfrac{-h\left(
x_{J-1}\right)  }{g\left(  x_{J-1}\right)  }\right\} \\
&  \subseteq\left\{  \dfrac{-h\left(  x_{1}\right)  }{g\left(  x_{1}\right)
},\dfrac{-h\left(  x_{2}\right)  }{g\left(  x_{2}\right)  },\ldots
,\dfrac{-h\left(  x_{J-1}\right)  }{g\left(  x_{J-1}\right)  }\right\}
\cup\left\{  r\right\}  =M,
\end{align*}
contradicting $s\notin M$. Thus, we have obtained a contradiction in Case 1.
\par
Let us now consider Case 2. In this case, $i=J$. Thus, $\left(  sg+h\right)
\left(  x_{i}\right)  =\left(  sg+h\right)  \left(  x_{J}\right)  =sg\left(
x_{J}\right)  +h\left(  x_{J}\right)  $. Compared with $\left(  sg+h\right)
\left(  x_{i}\right)  =0$, this becomes $sg\left(  x_{J}\right)  +h\left(
x_{J}\right)  =0$. Thus, $sg\left(  x_{J}\right)  =-\underbrace{h\left(
x_{J}\right)  }_{\neq0}\neq-0=0$, so that $g\left(  x_{J}\right)  \neq0$.
Thus, $r=\left\{
\begin{array}
[c]{c}%
\dfrac{-h\left(  x_{J}\right)  }{g\left(  x_{J}\right)  }%
,\ \ \ \ \ \ \ \ \ \ \text{if }g\left(  x_{J}\right)  \neq0;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }g\left(  x_{J}\right)  =0
\end{array}
\right.  =\dfrac{-h\left(  x_{J}\right)  }{g\left(  x_{J}\right)  }$ (since
$g\left(  x_{J}\right)  \neq0$).
\par
But dividing $sg\left(  x_{J}\right)  =-h\left(  x_{J}\right)  $ by $g\left(
x_{J}\right)  $ (this is allowed since $g\left(  x_{J}\right)  \neq0$), we
obtain
\[
s=\dfrac{-h\left(  x_{J}\right)  }{g\left(  x_{J}\right)  }=r\in\left\{
\dfrac{-h\left(  x_{1}\right)  }{g\left(  x_{1}\right)  },\dfrac{-h\left(
x_{2}\right)  }{g\left(  x_{2}\right)  },\ldots,\dfrac{-h\left(
x_{J-1}\right)  }{g\left(  x_{J-1}\right)  }\right\}  \cup\left\{  r\right\}
=M,
\]
contradicting $s\notin M$. Thus, we have obtained a contradiction in Case 2.
\par
Hence, in each of the cases 1 and 2, we have obtained a contradiction. Thus,
we always have a contradiction. Hence, our assumption was wrong, qed.} Thus,
there exists some $f\in E^{\ast}$ such that every $i\in\left\{  1,2,\ldots
,J\right\}  $ satisfies $f\left(  x_{i}\right)  \neq0$ (namely, $f=sg+h$). In
other words, (\ref{2.1c.pf.1}) holds for $j=J$. This completes the induction
step. The induction proof of (\ref{2.1c.pf.1}) is thus complete.

Applying (\ref{2.1c.pf.1}) to $j=t$, we obtain that there exists an $f\in
E^{\ast}$ such that every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies
$f\left(  x_{i}\right)  \neq0$. This proves Lemma 2.1d.
\end{proof}

\begin{proof}
[Second proof of Theorem 2.1c.]We will prove Theorem 2.1c by induction over
$\dim E$.

\textit{Induction base:} In the case when $\dim E=0$, Theorem 2.1c is
obvious.\footnote{In fact, in the case when $\dim E=0$, the only polytopes in
$E$ are $0$ and $\varnothing$. If $C=0$, Assertion C1 holds and Assertion C2
does not; else, Assertion C2 holds and Assertion C1 does not. Thus, Theorem
2.1c holds in the case when $\dim E=0$.} This completes the induction base.

\textit{Induction step:} Let $n$ be a positive integer. Assume that Theorem
2.1c holds whenever $\dim E=n-1$. We will now prove that Theorem 2.1c holds
whenever $\dim E=n$.

So, let $G$ be an $\mathbb{R}$-vector space satisfying $\dim G=n$. Let $C$ be
a polytope in $G$. By the definition of a polytope, this shows that $C$ is the
convex hull of a finite set of vectors in $G$. In other words, there exist
some $t\in\mathbb{N}$ and some vectors $x_{1}$, $x_{2}$, $\ldots$, $x_{t}$ in
$G$ such that $C=\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}%
,x_{2},\ldots,x_{t}\right\}  $. Consider this $t$ and these $x_{1}$, $x_{2}$,
$\ldots$, $x_{t}$.

We will now prove that exactly one of the following assertions holds:

\begin{statement}
\textit{Assertion C}$_{1}$\textit{1:} We have $0\in C$.

\textit{Assertion C}$_{1}$\textit{2:} There exists an $f\in G^{\ast}$ such
that every $x\in C$ satisfies $f\left(  x\right)  <0$.
\end{statement}

First, it is clear that the Assertions C$_{1}$1 and C$_{1}$2 cannot hold at
the same time\footnote{\textit{Proof.} Assume the opposite. Then, the
Assertions C$_{1}$1 and C$_{1}$2 hold at the same time. Since Assertion
C$_{1}$2 holds, there exists an $f\in G^{\ast}$ such that every $x\in C$
satisfies $f\left(  x\right)  <0$. Consider this $f$. We know that every $x\in
C$ satisfies $f\left(  x\right)  <0$. Since $0\in C$ (because Assertion
C$_{1}$1 holds), we can apply this to $x=0$, and thus obtain $f\left(
0\right)  <0$. But this contradicts $f\left(  0\right)  =0$ (which is because
$f$ is linear). This contradiction shows that our assumption was wrong, qed.}.
We will now show that at least one of these assertions holds. Our proof will
proceed in several steps:\footnote{I have split this proof into several steps
for the reader's convenience. These steps, however, are not self-contained;
for example, Step 9 involves a case distinction, Step 10 handles its Case 1,
and Steps 11 until 18 handle its Case 2.}

\textit{Step 1:} If some $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies
$x_{i}=0$, then at least one of Assertions C$_{1}$1 and C$_{1}$2
holds\footnote{\textit{Proof.} Assume that some $i\in\left\{  1,2,\ldots
,t\right\}  $ satisfies $x_{i}=0$. Consider this $i$. Then, $0=x_{i}%
\in\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots
,x_{t}\right\}  =C$, so that Assertion C$_{1}$1 holds. Thus, at least one of
Assertions C$_{1}$1 and C$_{1}$2 holds, qed.}. Hence, for the rest of this
proof, we can WLOG assume that no $i\in\left\{  1,2,\ldots,t\right\}  $
satisfies $x_{i}=0$. Assume this.

\textit{Step 2:} If $t=0$, then at least one of Assertions C$_{1}$1 and
C$_{1}$2 holds\footnote{\textit{Proof.} Assume that $t=0$. Then,
$C=\operatorname*{conv}.\operatorname*{hull}\underbrace{\left\{  x_{1}%
,x_{2},\ldots,x_{t}\right\}  }_{\substack{=\varnothing\\\text{(since
}t=0\text{)}}}=\operatorname*{conv}.\operatorname*{hull}\varnothing
=\varnothing$, so that Assertion C$_{1}$2 holds (because it is vacuously true
for $f=0$). Thus, at least one of Assertions C$_{1}$1 and C$_{1}$2 holds.}.
Hence, for the rest of this proof, we can WLOG assume that $t\neq0$. Assume this.

\textit{Step 3:} Every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies
$x_{i}\neq0$ (since no $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies
$x_{i}=0$). Thus, Lemma 2.1d (applied to $k=\mathbb{R}$ and $E=G$) yields that
there exists an $f\in G^{\ast}$ such that every $i\in\left\{  1,2,\ldots
,t\right\}  $ satisfies $f\left(  x_{i}\right)  \neq0$. Denote this $f$ by
$g$. Thus, $g$ is an element of $G^{\ast}$ such that every $i\in\left\{
1,2,\ldots,t\right\}  $ satisfies $g\left(  x_{i}\right)  \neq0$. Now,
$\operatorname{Im}g=\mathbb{R}$\ \ \ \ \footnote{\textit{Proof.} Every
$i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $g\left(  x_{i}\right)  \neq
0$. Applied to $i=1$, this yields $g\left(  x_{1}\right)  \neq0$, so that
$g\neq0$, and thus $\operatorname{Im}g\neq0$. Hence, $\dim\left(
\operatorname{Im}g\right)  \geq1$. Since $1=\dim\mathbb{R}$, this rewrites as
$\dim\left(  \operatorname{Im}g\right)  \geq\dim\mathbb{R}$. Combined with
$\operatorname{Im}g\subseteq\mathbb{R}$, this yields $\operatorname{Im}%
g=\mathbb{R}$, qed.}.

\textit{Step 4:} Now, let $F=\operatorname*{Ker}g$. Then, $F$ is an
$\mathbb{R}$-vector subspace of $E$ and satisfies $\dim F=n-1$%
\ \ \ \ \footnote{\textit{Proof.} By the homomorphism theorem, $G\diagup
\left(  \operatorname*{Ker}g\right)  \cong\operatorname{Im}g$. Since
$F=\operatorname*{Ker}g$, we have $G\diagup F=G\diagup\left(
\operatorname*{Ker}g\right)  \cong\operatorname{Im}g=\mathbb{R}$, so that
$\dim\left(  G\diagup F\right)  =\dim\mathbb{R}=1$. Thus, $1=\dim\left(
G\diagup F\right)  =\underbrace{\dim G}_{=n}-\dim F=n-\dim F$. Hence, $\dim
F=n-1$, qed.}.

\textit{Step 5:} Let $I^{+}$ be the set $\left\{  i\in\left\{  1,2,\ldots
,t\right\}  \ \mid\ g\left(  x_{i}\right)  >0\right\}  $, and let $I^{-}$ be
the set $\left\{  i\in\left\{  1,2,\ldots,t\right\}  \ \mid\ g\left(
x_{i}\right)  <0\right\}  $. Clearly, these sets $I^{+}$ and $I^{-}$ are
disjoint (since no $i\in\left\{  1,2,\ldots,t\right\}  $ can satisfy $g\left(
x_{i}\right)  >0$ and $g\left(  x_{i}\right)  <0$ at the same time).

Since $I^{+}=\left\{  i\in\left\{  1,2,\ldots,t\right\}  \ \mid\ g\left(
x_{i}\right)  >0\right\}  $, it is clear that%
\begin{equation}
\text{every }i\in I^{+}\text{ satisfies }g\left(  x_{i}\right)  >0.
\label{2.1c.pf.4}%
\end{equation}
Since $I^{-}=\left\{  i\in\left\{  1,2,\ldots,t\right\}  \ \mid\ g\left(
x_{i}\right)  <0\right\}  $, it is clear that%
\begin{equation}
\text{every }j\in I^{-}\text{ satisfies }g\left(  x_{j}\right)  <0.
\label{2.1c.pf.5}%
\end{equation}


Every $i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $g\left(  x_{i}\right)
\neq0$. Thus,%
\begin{align*}
\left\{  1,2,\ldots,t\right\}   &  =\left\{  i\in\left\{  1,2,\ldots
,t\right\}  \ \mid\ \underbrace{g\left(  x_{i}\right)  \neq0}%
_{\substack{\text{this is equivalent to}\\\left(  g\left(  x_{i}\right)
>0\text{ or }g\left(  x_{i}\right)  <0\right)  }}\right\} \\
&  =\left\{  i\in\left\{  1,2,\ldots,t\right\}  \ \mid\ g\left(  x_{i}\right)
>0\text{ or }g\left(  x_{i}\right)  <0\right\} \\
&  =\underbrace{\left\{  i\in\left\{  1,2,\ldots,t\right\}  \ \mid\ g\left(
x_{i}\right)  >0\right\}  }_{=I^{+}}\cup\underbrace{\left\{  i\in\left\{
1,2,\ldots,t\right\}  \ \mid\ g\left(  x_{i}\right)  <0\right\}  }_{=I^{-}}\\
&  =I^{+}\cup I^{-}.
\end{align*}


\textit{Step 6:} If $I^{-}=\varnothing$, then at least one of Assertions
C$_{1}$1 and C$_{1}$2 holds\footnote{\textit{Proof.} Assume that
$I^{-}=\varnothing$.
\par
Let $i\in\left\{  1,2,\ldots,t\right\}  $ be arbitrary. Then, $i\in\left\{
1,2,\ldots,t\right\}  =I^{+}\cup\underbrace{I^{-}}_{=\varnothing}=I^{+}$, so
that $g\left(  x_{i}\right)  >0$ (by (\ref{2.1c.pf.4})).
\par
Now, forget that we fixed $i$. We thus have proven that every $i\in\left\{
1,2,\ldots,t\right\}  $ satisfies $g\left(  x_{i}\right)  >0$. Hence, every
$i\in\left\{  1,2,\ldots,t\right\}  $ satisfies $\left(  -g\right)  \left(
x_{i}\right)  =-\underbrace{g\left(  x_{i}\right)  }_{>0}<0$.
\par
Therefore, Proposition 2.0r (applied to $E=G$, $f=-g$ and $\delta=0$) shows
that every $x\in C$ satisfies $\left(  -g\right)  \left(  x\right)  <0$
(because $C=\operatorname*{conv}.\operatorname*{hull}\left\{  x_{1}%
,x_{2},\ldots,x_{t}\right\}  $). Thus, there exists an $f\in G^{\ast}$ such
that every $x\in C$ satisfies $f\left(  x\right)  <0$ (namely, $f=-g$). Hence,
Assertion C$_{1}$2 holds. Thus, at least one of Assertions C$_{1}$1 and
C$_{1}$2 holds, qed.}. Hence, for the rest of this proof, we can WLOG assume
that $I^{-}\neq\varnothing$. Assume this.

\textit{Step 7:} For every $i\in\left\{  1,2,\ldots,t\right\}  $, let
$\alpha_{i}$ be the element $\dfrac{1}{g\left(  x_{i}\right)  }\in\mathbb{R}$
(this is well-defined since $g\left(  x_{i}\right)  \neq0$).

For every $\left(  i,j\right)  \in I^{+}\times I^{-}$, we have $\alpha
_{i}-\alpha_{j}\neq0$\ \ \ \ \footnote{\textit{Proof.} Let $\left(
i,j\right)  \in I^{+}\times I^{-}$. Then, $i\in I^{+}$ and $j\in I^{-}$. Thus,
$g\left(  x_{i}\right)  >0$ (by (\ref{2.1c.pf.4})), so that $\dfrac
{1}{g\left(  x_{i}\right)  }>0$. Also, $g\left(  x_{j}\right)  <0$ (by
(\ref{2.1c.pf.5})), so that $\dfrac{1}{g\left(  x_{j}\right)  }<0$. Now, since
$\alpha_{i}=\dfrac{1}{g\left(  x_{i}\right)  }$ (by the definition of
$\alpha_{i}$) and $\alpha_{j}=\dfrac{1}{g\left(  x_{j}\right)  }$ (by the
definition of $\alpha_{j}$), we have $\alpha_{i}-\alpha_{j}=\underbrace{\dfrac
{1}{g\left(  x_{i}\right)  }}_{>0}-\underbrace{\dfrac{1}{g\left(
x_{j}\right)  }}_{<0}>0$, so that $\alpha_{i}-\alpha_{j}\neq0$, qed.}, so that
$\dfrac{1}{\alpha_{i}-\alpha_{j}}$ is well-defined.

Now, let $S$ denote the set
\[
\left\{  \dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha
_{j}x_{j}\right)  \ \mid\ \left(  i,j\right)  \in I^{+}\times I^{-}\right\}
.
\]
Since every $\left(  i,j\right)  \in I^{+}\times I^{-}$ satisfies $\dfrac
{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha_{j}x_{j}\right)
\in\operatorname*{Ker}g$\ \ \ \ \footnote{\textit{Proof.} Let $\left(
i,j\right)  \in I^{+}\times I^{-}$. Since $\alpha_{i}=\dfrac{1}{g\left(
x_{i}\right)  }$ (by the definition of $\alpha_{i}$) and $\alpha_{j}=\dfrac
{1}{g\left(  x_{j}\right)  }$ (by the definition of $\alpha_{j}$), we have%
\begin{align*}
g\left(  \alpha_{i}x_{i}-\alpha_{j}x_{j}\right)   &  =g\left(  \dfrac
{1}{g\left(  x_{i}\right)  }x_{i}-\dfrac{1}{g\left(  x_{j}\right)  }%
x_{j}\right)  =\underbrace{\dfrac{1}{g\left(  x_{i}\right)  }g\left(
x_{i}\right)  }_{=1}-\underbrace{\dfrac{1}{g\left(  x_{j}\right)  }g\left(
x_{j}\right)  }_{=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }g\text{ is }\mathbb{R}%
\text{-linear}\right) \\
&  =1-1=0.
\end{align*}
Thus, $\alpha_{i}x_{i}-\alpha_{j}x_{j}\in\operatorname*{Ker}g$. Since
$\operatorname*{Ker}g$ is a $k$-vector space, this yields $\dfrac{1}%
{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha_{j}x_{j}\right)
\in\operatorname*{Ker}g$, qed.}, we have $\left\{  \dfrac{1}{\alpha_{i}%
-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha_{j}x_{j}\right)  \ \mid\ \left(
i,j\right)  \in I^{+}\times I^{-}\right\}  \subseteq\operatorname*{Ker}g$.
Thus,%
\[
S=\left\{  \dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha
_{j}x_{j}\right)  \ \mid\ \left(  i,j\right)  \in I^{+}\times I^{-}\right\}
\subseteq\operatorname*{Ker}g=F.
\]
Thus, $\operatorname*{conv}.\operatorname*{hull}S$ is a polytope in $F$.

Since $S=\left\{  \dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}%
x_{i}-\alpha_{j}x_{j}\right)  \ \mid\ \left(  i,j\right)  \in I^{+}\times
I^{-}\right\}  $, we have%
\begin{equation}
\dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha_{j}%
x_{j}\right)  \in S\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in I^{+}\times I^{-}. \label{2.1c.pf.S}%
\end{equation}


\textit{Step 8:} Let us now see that $\operatorname*{conv}%
.\operatorname*{hull}S\subseteq C$:

Every $\left(  i,j\right)  \in I^{+}\times I^{-}$ satisfies $\dfrac{1}%
{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha_{j}x_{j}\right)  \in
C$\ \ \ \ \footnote{\textit{Proof.} Let $\left(  i,j\right)  \in I^{+}\times
I^{-}$. Then, $i\in I^{+}$ and $j\in I^{-}$. Thus, $g\left(  x_{i}\right)  >0$
(by (\ref{2.1c.pf.4})), so that $\dfrac{1}{g\left(  x_{i}\right)  }>0$. Also,
$g\left(  x_{j}\right)  <0$ (by (\ref{2.1c.pf.5})), so that $\dfrac
{1}{g\left(  x_{j}\right)  }<0$.
\par
Now, since $\alpha_{i}=\dfrac{1}{g\left(  x_{i}\right)  }$ (by the definition
of $\alpha_{i}$) and $\alpha_{j}=\dfrac{1}{g\left(  x_{j}\right)  }$ (by the
definition of $\alpha_{j}$), we have $\alpha_{i}=\dfrac{1}{g\left(
x_{i}\right)  }>0$ and $\alpha_{j}=\dfrac{1}{g\left(  x_{j}\right)  }<0$.
Thus, $\underbrace{\alpha_{i}}_{>0}-\underbrace{\alpha_{j}}_{<0}>0$.
\par
Since $\alpha_{i}>0$ and $\alpha_{i}-\alpha_{j}>0$, we have $\dfrac{\alpha
_{i}}{\alpha_{i}-\alpha_{j}}>0$. Since $-\alpha_{j}>0$ (this is because
$\alpha_{j}<0$) and $\alpha_{i}-\alpha_{j}>0$, we have $\dfrac{-\alpha_{j}%
}{\alpha_{i}-\alpha_{j}}>0$.
\par
But $C$ is convex. Thus, every point on a segment connecting two points of $C$
must lie in $C$.
\par
Since $\dfrac{\alpha_{i}}{\alpha_{i}-\alpha_{j}}>0$, $\dfrac{-\alpha_{j}%
}{\alpha_{i}-\alpha_{j}}>0$ and $\dfrac{\alpha_{i}}{\alpha_{i}-\alpha_{j}%
}+\dfrac{-\alpha_{j}}{\alpha_{i}-\alpha_{j}}=\dfrac{\alpha_{i}}{\alpha
_{i}-\alpha_{j}}-\dfrac{\alpha_{j}}{\alpha_{i}-\alpha_{j}}=\dfrac{\alpha
_{i}-\alpha_{j}}{\alpha_{i}-\alpha_{j}}=1$, it is clear that $\dfrac
{\alpha_{i}}{\alpha_{i}-\alpha_{j}}x_{i}+\dfrac{-\alpha_{j}}{\alpha_{i}%
-\alpha_{j}}x_{j}$ is a point on the segment connecting $x_{i}$ and $x_{j}$.
Since $x_{i}\in C$ (because $x_{i}\in\left\{  x_{1},x_{2},\ldots
,x_{t}\right\}  \subseteq\operatorname*{conv}.\operatorname*{hull}\left\{
x_{1},x_{2},\ldots,x_{t}\right\}  =C$) and $x_{j}\in C$ (because $x_{j}%
\in\left\{  x_{1},x_{2},\ldots,x_{t}\right\}  \subseteq\operatorname*{conv}%
.\operatorname*{hull}\left\{  x_{1},x_{2},\ldots,x_{t}\right\}  =C$), this
yields that $\dfrac{\alpha_{i}}{\alpha_{i}-\alpha_{j}}x_{i}+\dfrac{-\alpha
_{j}}{\alpha_{i}-\alpha_{j}}x_{j}$ is a point on a segment connecting two
points of $C$. Thus, $\dfrac{\alpha_{i}}{\alpha_{i}-\alpha_{j}}x_{i}%
+\dfrac{-\alpha_{j}}{\alpha_{i}-\alpha_{j}}x_{j}\in C$ (since every point on a
segment connecting two points of $C$ must lie in $C$). Since $\dfrac
{\alpha_{i}}{\alpha_{i}-\alpha_{j}}x_{i}+\dfrac{-\alpha_{j}}{\alpha_{i}%
-\alpha_{j}}x_{j}=\dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}%
x_{i}-\alpha_{j}x_{j}\right)  $, this rewrites as $\dfrac{1}{\alpha_{i}%
-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha_{j}x_{j}\right)  \in C$, qed.}.
Thus,
\[
\left\{  \dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha
_{j}x_{j}\right)  \ \mid\ \left(  i,j\right)  \in I^{+}\times I^{-}\right\}
\subseteq C.
\]
Thus,
\[
S=\left\{  \dfrac{1}{\alpha_{i}-\alpha_{j}}\left(  \alpha_{i}x_{i}-\alpha
_{j}x_{j}\right)  \ \mid\ \left(  i,j\right)  \in I^{+}\times I^{-}\right\}
\subseteq C.
\]
Hence, $C$ is a convex set containing $S$ as a subset. But since every convex
set containing $S$ as a subset must also contain $\operatorname*{conv}%
.\operatorname*{hull}S$ as a subset (because $\operatorname*{conv}%
.\operatorname*{hull}S$ is the smallest convex set containing $S$ as a
subset), this yields that $C$ contains $\operatorname*{conv}%
.\operatorname*{hull}S$ as a subset. In other words, $\operatorname*{conv}%
.\operatorname*{hull}S\subseteq C$.

\textit{Step 9:} Now, since $\dim F=n-1$, we can apply Theorem 2.1c to $F$ and
$\operatorname*{conv}.\operatorname*{hull}S$ instead of $E$ and $C$ (because
we assumed that Theorem 2.1c holds whenever $\dim E=n-1$). As a consequence,
we obtain that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion C}$_{2}$\textit{1:} We have $0\in\operatorname*{conv}%
.\operatorname*{hull}S$.

\textit{Assertion C}$_{2}$\textit{2:} There exists an $f\in F^{\ast}$ such
that every $x\in\operatorname*{conv}.\operatorname*{hull}S$ satisfies
$f\left(  x\right)  <0$.
\end{statement}

Thus, we must be in one of the following two cases:

\textit{Case 1:} Assertion C$_{2}$1 holds.

\textit{Case 2:} Assertion C$_{2}$2 holds.

\textit{Step 10:} Let us consider Case 1 first. In this case, Assertion
C$_{2}$1 holds. In other words, $0\in\operatorname*{conv}.\operatorname*{hull}%
S$. Since $\operatorname*{conv}.\operatorname*{hull}S\subseteq C$, this yields
$0\in C$, so that Assertion C$_{1}$1 holds. Thus, at least one of Assertions
C$_{1}$1 and C$_{1}$2 holds in Case 1.

\textit{Step 11:} Now, let us consider Case 2. In this case, Assertion C$_{2}%
$2 holds. In other words, there exists an $f\in F^{\ast}$ such that every
$x\in\operatorname*{conv}.\operatorname*{hull}S$ satisfies $f\left(  x\right)
<0$. Denote this $f$ by $h$. Then, every $x\in\operatorname*{conv}%
.\operatorname*{hull}S$ satisfies $h\left(  x\right)  <0$.

\textit{Step 12:}\footnote{This is no longer Step 11, but we are still in Case
2.} A known fact from linear algebra states that if $k$ is a field, $P$ and
$R$ are two $k$-vector spaces, $Q$ is a $k$-vector subspace of $P$, and
$\xi:Q\rightarrow R$ is a $k$-linear map, then there exists a $k$-linear map
$\eta:P\rightarrow R$ such that $\eta\mid_{Q}=\xi$.\ \ \ \ \footnote{Advanced
algebraists tend to state this fact in the following equivalent form: Every
$k$-vector space is an injective $k$-module.} Applying this fact to
$k=\mathbb{R}$, $P=G$, $Q=F$, $R=\mathbb{R}$ and $\xi=h$, we conclude that
there exists an $\mathbb{R}$-linear map $\eta:G\rightarrow\mathbb{R}$ such
that $\eta\mid_{F}=h$. Denote this $\eta$ by $e$. Thus, $e\mid_{F}=h$.

Now, $e$ is an $\mathbb{R}$-linear map $G\rightarrow\mathbb{R}$, and%
\begin{equation}
\text{every }x\in\operatorname*{conv}.\operatorname*{hull}S\text{ satisfies
}e\left(  x\right)  <0 \label{2.1c.pf.e}%
\end{equation}
(since every $x\in\operatorname*{conv}.\operatorname*{hull}S$ satisfies
$e\left(  x\right)  =\underbrace{\left(  e\mid_{F}\right)  }_{=h}\left(
x\right)  =h\left(  x\right)  <0$).

\textit{Step 13:} Let $k$ be the element $i$ of $I^{-}$ minimizing $\alpha
_{i}e\left(  x_{i}\right)  $ (this is well-defined since $I^{-}$ is a finite
set and $I^{-}\neq\varnothing$). Then, $k\in I^{-}$, and we have%
\begin{equation}
\alpha_{k}e\left(  x_{k}\right)  \leq\alpha_{i}e\left(  x_{i}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in I^{-}. \label{2.1c.pf.max}%
\end{equation}


\textit{Step 14:} Let us now prove that%
\begin{equation}
\alpha_{k}e\left(  x_{k}\right)  >\alpha_{i}e\left(  x_{i}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }i\in I^{+}. \label{2.1c.pf.otherside}%
\end{equation}


\textit{Proof of (\ref{2.1c.pf.otherside}):} Let $i\in I^{+}$. Since $i\in
I^{+}$ and $k\in I^{-}$, we have $\left(  i,k\right)  \in I^{+}\times I^{-}$.
Thus, (\ref{2.1c.pf.S}) (applied to $\left(  i,k\right)  $ instead of $\left(
i,j\right)  $) yields%
\[
\dfrac{1}{\alpha_{i}-\alpha_{k}}\left(  \alpha_{i}x_{i}-\alpha_{k}%
x_{k}\right)  \in S\subseteq\operatorname*{conv}.\operatorname*{hull}S.
\]
Hence, (\ref{2.1c.pf.e}) (applied to $x=\dfrac{1}{\alpha_{i}-\alpha_{k}%
}\left(  \alpha_{i}x_{i}-\alpha_{k}x_{k}\right)  $) yields $e\left(  \dfrac
{1}{\alpha_{i}-\alpha_{k}}\left(  \alpha_{i}x_{i}-\alpha_{k}x_{k}\right)
\right)  <0$. Since $e\left(  \dfrac{1}{\alpha_{i}-\alpha_{k}}\left(
\alpha_{i}x_{i}-\alpha_{k}x_{k}\right)  \right)  =\dfrac{1}{\alpha_{i}%
-\alpha_{k}}e\left(  \alpha_{i}x_{i}-\alpha_{k}x_{k}\right)  $ (because $e$ is
$\mathbb{R}$-linear), this rewrites as
\begin{equation}
\dfrac{1}{\alpha_{i}-\alpha_{k}}e\left(  \alpha_{i}x_{i}-\alpha_{k}%
x_{k}\right)  <0. \label{2.1c.pf.newfunc.pf.1}%
\end{equation}


Since $i\in I^{+}$, we have $g\left(  x_{i}\right)  >0$ (by (\ref{2.1c.pf.4}%
)). Since $k\in I^{-}$, we have $g\left(  x_{k}\right)  <0$ (by
(\ref{2.1c.pf.5}), applied to $j=k$). From $g\left(  x_{i}\right)  >0$ and
$g\left(  x_{k}\right)  <0$, it follows that $g\left(  x_{i}\right)  g\left(
x_{k}\right)  <0$.

By the definition of $\alpha_{i}$, we have $\alpha_{i}=\dfrac{1}{g\left(
x_{i}\right)  }>0$ (since $g\left(  x_{i}\right)  >0$). By the definition of
$\alpha_{k}$, we have $\alpha_{k}=\dfrac{1}{g\left(  x_{k}\right)  }<0$ (since
$g\left(  x_{k}\right)  <0$). From $\alpha_{i}>0$ and $\alpha_{k}<0$, it
follows that $\underbrace{\alpha_{i}}_{>0}-\underbrace{\alpha_{k}}_{<0}>0$.
Thus, we can multiply (\ref{2.1c.pf.newfunc}) with $\alpha_{i}-\alpha_{k}$,
and obtain $e\left(  \alpha_{i}x_{i}-\alpha_{k}x_{k}\right)  <0$. Since%
\[
e\left(  \alpha_{i}x_{i}-\alpha_{k}x_{k}\right)  =\alpha_{i}e\left(
x_{i}\right)  -\alpha_{k}e\left(  x_{k}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }e\text{ is }\mathbb{R}\text{-linear}\right)  ,
\]
this rewrites as $\alpha_{i}e\left(  x_{i}\right)  -\alpha_{k}e\left(
x_{k}\right)  <0$. In other words, $\alpha_{k}e\left(  x_{k}\right)
>\alpha_{i}e\left(  x_{i}\right)  $. This proves (\ref{2.1c.pf.otherside}).

\textit{Step 15:} Let us slightly improve (\ref{2.1c.pf.otherside}): It is
easy to see that there exists a positive real $\delta$ such that%
\begin{equation}
\text{every }i\in I^{+}\text{ satisfies }\alpha_{k}\left(  e\left(
x_{k}\right)  +\delta\right)  >\alpha_{i}e\left(  x_{i}\right)  .
\label{2.1c.pf.otherside.d}%
\end{equation}
\footnote{\textit{Proof.} Every $i\in I^{+}$ satisfies $\alpha_{k}e\left(
x_{k}\right)  >\alpha_{i}e\left(  x_{i}\right)  $ (by (\ref{2.1c.pf.otherside}%
)) and thus $\alpha_{k}e\left(  x_{k}\right)  -\alpha_{i}e\left(
x_{i}\right)  >0$. In other words, $\left\{  \alpha_{k}e\left(  x_{k}\right)
-\alpha_{i}e\left(  x_{i}\right)  \ \mid\ i\in I^{+}\right\}  $ is a set of
positive reals. Since this set is finite (because $I^{+}$ is finite), it is
thus bounded from below by a positive real. In other words, there exists a
positive real $\varepsilon$ such that%
\begin{equation}
\text{every }i\in I^{+}\text{ satisfies }\alpha_{k}e\left(  x_{k}\right)
-\alpha_{i}e\left(  x_{i}\right)  \geq\varepsilon.
\label{2.1c.pf.otherside.d.pf}%
\end{equation}
Consider this $\varepsilon$.
\par
Let $\delta=\dfrac{-\varepsilon}{2\alpha_{k}}$. Then, $\alpha_{k}\delta
=\dfrac{-\varepsilon}{2}$.
\par
Since $k\in I^{-}$, we have $g\left(  x_{k}\right)  <0$ (by (\ref{2.1c.pf.5}),
applied to $j=k$). By the definition of $\alpha_{k}$, we have $\alpha
_{k}=\dfrac{1}{g\left(  x_{k}\right)  }<0$ (since $g\left(  x_{k}\right)
<0$). Now, every $i\in I^{+}$ satisfies%
\begin{align*}
\underbrace{\alpha_{k}\left(  e\left(  x_{k}\right)  +\delta\right)
}_{=\alpha_{k}e\left(  x_{k}\right)  +\alpha_{k}\delta}-\alpha_{i}e\left(
x_{i}\right)   &  =\alpha_{k}e\left(  x_{k}\right)  +\alpha_{k}\delta
-\alpha_{i}e\left(  x_{i}\right) \\
&  =\underbrace{\alpha_{k}e\left(  x_{k}\right)  -\alpha_{i}e\left(
x_{i}\right)  }_{\substack{\geq\varepsilon\\\text{(by
(\ref{2.1c.pf.otherside.d.pf}))}}}+\underbrace{\alpha_{k}\delta}%
_{=\dfrac{-\varepsilon}{2}}\geq\varepsilon+\dfrac{-\varepsilon}{2}%
=\dfrac{\varepsilon}{2}>0
\end{align*}
(since $\varepsilon>0$). In other words, every $i\in I^{+}$ satisfies
$\alpha_{k}\left(  e\left(  x_{k}\right)  +\delta\right)  >\alpha_{i}e\left(
x_{i}\right)  $.
\par
We thus have proven that there exists a positive real $\delta$ such that every
$i\in I^{+}$ satisfies $\alpha_{k}\left(  e\left(  x_{k}\right)
+\delta\right)  >\alpha_{i}e\left(  x_{i}\right)  $. Qed.} Consider this
$\delta$.

\textit{Step 16:} We are now going to prove that%
\begin{equation}
\text{every }i\in\left\{  1,2,\ldots,t\right\}  \text{ satisfies }\left(
\left(  e\left(  x_{k}\right)  +\delta\right)  g-g\left(  x_{k}\right)
e\right)  \left(  x_{i}\right)  <0\text{.} \label{2.1c.pf.newfunc}%
\end{equation}


\textit{Proof of (\ref{2.1c.pf.newfunc}):} Let $i\in\left\{  1,2,\ldots
,t\right\}  $. Then, $i\in\left\{  1,2,\ldots,t\right\}  =I^{+}\cup I^{-}$.
Hence, either $i\in I^{+}$ or $i\in I^{-}$. We thus must be in one of the
following two subcases:

\textit{Subcase 2.1:} We have $i\in I^{+}$.

\textit{Subcase 2.2:} We have $i\in I^{-}$.

Let us first consider Subcase 2.1. In this subcase, $i\in I^{+}$. Thus,
$g\left(  x_{i}\right)  >0$ (by (\ref{2.1c.pf.4})). Since $k\in I^{-}$, we
have $g\left(  x_{k}\right)  <0$ (by (\ref{2.1c.pf.5}), applied to $j=k$).
From $g\left(  x_{i}\right)  >0$ and $g\left(  x_{k}\right)  <0$, it follows
that $g\left(  x_{i}\right)  g\left(  x_{k}\right)  <0$.

By the definition of $\alpha_{i}$, we have $\alpha_{i}=\dfrac{1}{g\left(
x_{i}\right)  }>0$ (since $g\left(  x_{i}\right)  >0$). By the definition of
$\alpha_{k}$, we have $\alpha_{k}=\dfrac{1}{g\left(  x_{k}\right)  }<0$ (since
$g\left(  x_{k}\right)  <0$). From (\ref{2.1c.pf.otherside.d}), we have
$\alpha_{k}\left(  e\left(  x_{k}\right)  +\delta\right)  >\alpha_{i}e\left(
x_{i}\right)  $. Since $\alpha_{i}=\dfrac{1}{g\left(  x_{i}\right)  }$ and
$\alpha_{k}=\dfrac{1}{g\left(  x_{k}\right)  }$, this rewrites as $\dfrac
{1}{g\left(  x_{k}\right)  }\left(  e\left(  x_{k}\right)  +\delta\right)
>\dfrac{1}{g\left(  x_{i}\right)  }e\left(  x_{i}\right)  $. Multiplied by
$g\left(  x_{i}\right)  g\left(  x_{k}\right)  $, this becomes $g\left(
x_{i}\right)  \left(  e\left(  x_{k}\right)  +\delta\right)  <g\left(
x_{k}\right)  e\left(  x_{i}\right)  $ (here, the sign has been switched since
$g\left(  x_{i}\right)  g\left(  x_{k}\right)  <0$). Thus,%
\begin{align*}
\left(  \left(  e\left(  x_{k}\right)  +\delta\right)  g-g\left(
x_{k}\right)  e\right)  \left(  x_{i}\right)   &  =\underbrace{\left(
e\left(  x_{k}\right)  +\delta\right)  g\left(  x_{i}\right)  }_{=g\left(
x_{i}\right)  \left(  e\left(  x_{k}\right)  +\delta\right)  <g\left(
x_{k}\right)  e\left(  x_{i}\right)  }-g\left(  x_{k}\right)  e\left(
x_{i}\right) \\
&  <g\left(  x_{k}\right)  e\left(  x_{i}\right)  -g\left(  x_{k}\right)
e\left(  x_{i}\right)  =0.
\end{align*}
Thus, (\ref{2.1c.pf.newfunc}) is proven in Subcase 2.1.

Let us now consider Subcase 2.2. In this subcase, $i\in I^{-}$. Hence,
(\ref{2.1c.pf.max}) yields $\alpha_{k}e\left(  x_{k}\right)  \leq\alpha
_{i}e\left(  x_{i}\right)  $.

Since $i\in I^{-}$, we have $g\left(  x_{i}\right)  <0$ (by (\ref{2.1c.pf.5}),
applied to $j=i$). Since $k\in I^{-}$, we have $g\left(  x_{k}\right)  <0$ (by
(\ref{2.1c.pf.5}), applied to $j=k$). From $g\left(  x_{i}\right)  <0$ and
$g\left(  x_{k}\right)  <0$, it follows that $g\left(  x_{i}\right)  g\left(
x_{k}\right)  >0$.

Since $\alpha_{i}=\dfrac{1}{g\left(  x_{i}\right)  }$ (by the definition of
$\alpha_{i}$) and $\alpha_{k}=\dfrac{1}{g\left(  x_{k}\right)  }$ (by the
definition of $\alpha_{k}$), the inequality $\alpha_{k}e\left(  x_{k}\right)
\leq\alpha_{i}e\left(  x_{i}\right)  $ (proven above) rewrites as $\dfrac
{1}{g\left(  x_{k}\right)  }e\left(  x_{k}\right)  \leq\dfrac{1}{g\left(
x_{i}\right)  }e\left(  x_{i}\right)  $. Multiplied by $g\left(  x_{i}\right)
g\left(  x_{k}\right)  $, this becomes $g\left(  x_{i}\right)  e\left(
x_{k}\right)  \leq g\left(  x_{k}\right)  e\left(  x_{i}\right)  $ (here, the
sign has not been switched since $g\left(  x_{i}\right)  g\left(
x_{k}\right)  >0$). Thus,%
\begin{align*}
&  \left(  \left(  e\left(  x_{k}\right)  +\delta\right)  g-g\left(
x_{k}\right)  e\right)  \left(  x_{i}\right)  =\underbrace{\left(  e\left(
x_{k}\right)  +\delta\right)  g\left(  x_{i}\right)  }_{=e\left(
x_{k}\right)  g\left(  x_{i}\right)  +\delta g\left(  x_{i}\right)  }-g\left(
x_{k}\right)  e\left(  x_{i}\right) \\
&  =\underbrace{e\left(  x_{k}\right)  g\left(  x_{i}\right)  }_{=g\left(
x_{i}\right)  e\left(  x_{k}\right)  \leq g\left(  x_{k}\right)  e\left(
x_{i}\right)  }+\underbrace{\delta g\left(  x_{i}\right)  }%
_{\substack{<0\\\text{(since }\delta>0\\\text{and }g\left(  x_{i}\right)
<0\text{)}}}-g\left(  x_{k}\right)  e\left(  x_{i}\right)  <g\left(
x_{k}\right)  e\left(  x_{i}\right)  -g\left(  x_{k}\right)  e\left(
x_{i}\right)  =0.
\end{align*}
Thus, (\ref{2.1c.pf.newfunc}) is proven in Subcase 2.2.

We have thus proven (\ref{2.1c.pf.newfunc}) in each of the Subcases 2.1 and
2.2. Since these subcases cover all possibilities, this yields that
(\ref{2.1c.pf.newfunc}) always holds. This completes the proof of
(\ref{2.1c.pf.newfunc}).

\textit{Step 17:} Now,%
\begin{equation}
\text{every }x\in C\text{ satisfies }\left(  \left(  e\left(  x_{k}\right)
+\delta\right)  g-g\left(  x_{k}\right)  e\right)  \left(  x\right)
<0\text{.} \label{2.1c.pf.newfunc.corr}%
\end{equation}


\textit{Proof of (\ref{2.1c.pf.newfunc.corr}):} We know that every
$i\in\left\{  1,2,\ldots,t\right\}  $ satisfies \newline$\left(  \left(
e\left(  x_{k}\right)  +\delta\right)  g-g\left(  x_{k}\right)  e\right)
\left(  x_{i}\right)  <0$ (because of (\ref{2.1c.pf.newfunc})). Thus,
Proposition 2.0r (applied to $G$, $\left(  e\left(  x_{k}\right)
+\delta\right)  g-g\left(  x_{k}\right)  e$ and $0$ instead of $E$, $f$ and
$\delta$) shows that every $x\in C$ satisfies $\left(  \left(  e\left(
x_{k}\right)  +\delta\right)  g-g\left(  x_{k}\right)  e\right)  \left(
x\right)  <0$. This proves (\ref{2.1c.pf.newfunc.corr}).

\textit{Step 18:} Due to (\ref{2.1c.pf.newfunc.corr}), there exists an $f\in
G^{\ast}$ such that every $x\in C$ satisfies $f\left(  x\right)  <0$ (namely,
$f=\left(  e\left(  x_{k}\right)  +\delta\right)  g-g\left(  x_{k}\right)
e$). In other words, Assertion C$_{1}$2 holds. Thus, at least one of
Assertions C$_{1}$1 and C$_{1}$2 holds in Case 2.

We have thus proven that, in each of the Cases 1 and 2, at least one of
Assertions C$_{1}$1 and C$_{1}$2 holds. Since these cases cover all
possibilities, this yields that, in every situation, at least one of
Assertions C$_{1}$1 and C$_{1}$2 holds.

Altogether, we have shown the following two claims:

\begin{itemize}
\item Assertions C$_{1}$1 and C$_{1}$2 cannot hold at the same time.

\item At least one of Assertions C$_{1}$1 and C$_{1}$2 holds.
\end{itemize}

Combining these claims, we conclude that exactly one of Assertions C$_{1}$1
and C$_{1}$2 holds.

Now, forget that we fixed $G$ and $C$. We have thus proven the following result:

\begin{statement}
\textbf{Result 1:} If $G$ is an $\mathbb{R}$-vector space satisfying $\dim
G=n$, and $C$ is a polytope in $G$, then exactly one of the following
assertions holds:

\textit{Assertion C}$_{1}$\textit{1:} We have $0\in C$.

\textit{Assertion C}$_{1}$\textit{2:} There exists an $f\in G^{\ast}$ such
that every $x\in C$ satisfies $f\left(  x\right)  <0$.
\end{statement}

If we rename $G$ as $E$ in Result 1, then this result takes the following form:

\begin{statement}
\textbf{Result 2:} If $E$ is an $\mathbb{R}$-vector space satisfying $\dim
E=n$, and $C$ is a polytope in $E$, then exactly one of the following
assertions holds:

\textit{Assertion C}$_{3}$\textit{1:} We have $0\in C$.

\textit{Assertion C}$_{3}$\textit{2:} There exists an $f\in E^{\ast}$ such
that every $x\in C$ satisfies $f\left(  x\right)  <0$.
\end{statement}

Clearly, Result 2 is exactly the statement of Theorem 2.1c in the case when
$\dim E=n$. Hence, we have proven that Theorem 2.1c holds in the case when
$\dim E=n$. This completes the induction step, and thus the induction proof of
Theorem 2.1c is complete.
\end{proof}

\begin{proof}
[Second proof of Theorem 2.5c.]We will prove Theorem 2.5c by strong induction
over $\left\vert S\right\vert $:

\textit{Induction step:} Let $n$ be a nonnegative integer. Assume that Theorem
2.5c holds whenever $\left\vert S\right\vert <n$. We will now prove that
Theorem 2.5c holds whenever $\left\vert S\right\vert =n$.

So, let $E$ be a finite-dimensional $\mathbb{R}$-vector space. Let $S$ be a
finite subset of $E$ such that $\left\vert S\right\vert =n$. Let $b\in E$.
Then, we must prove that exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{1}$\textit{1:} We have $b\in\operatorname*{cone}S$.

\textit{Assertion D}$_{1}$\textit{2:} There exists an $f\in E^{\ast}$ such
that $f\left(  b\right)  >0$ and \newline$\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

Since $\operatorname*{cone}S$ is a convex cone, it is clear that
$\operatorname*{cone}S$ is closed under multiplication by a nonnegative scalar
(because convex cones are closed under multiplication by a nonnegative
scalar). It is also clear that any linear combination of finitely many
elements of $\operatorname*{cone}S$ with nonnegative coefficients must lie in
$\operatorname*{cone}S$ (because $\operatorname*{cone}S$ is a convex cone, and
because any linear combination of finitely many elements of a convex cone with
nonnegative coefficients must lie in this cone). In particular, the sum of any
two elements of $\operatorname*{cone}S$ must lie in $\operatorname*{cone}S$.

First, it is clear that the Assertions D$_{1}$1 and D$_{1}$2 cannot hold at
the same time\footnote{\textit{Proof.} Assume the opposite. Then, the
Assertions D$_{1}$1 and D$_{1}$2 hold at the same time. Since Assertion
D$_{1}$2 holds, there exists an $f\in E^{\ast}$ such that $f\left(  b\right)
>0$ and $\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies
}f\left(  x\right)  \leq0\right)  $. Consider this $f$. We know that every
$x\in\operatorname*{cone}S$ satisfies $f\left(  x\right)  \leq0$. Since
$b\in\operatorname*{cone}S$ (because Assertion D$_{1}$1 holds), we can apply
this to $x=b$, and thus obtain $f\left(  b\right)  \leq0$. But this
contradicts $f\left(  b\right)  >0$. This contradiction shows that our
assumption was wrong, qed.}. We will now show that at least one of these
assertions holds.

Let $C$ be the polytope $\operatorname*{conv}.\operatorname*{hull}\left(
S\cup\left\{  -b\right\}  \right)  $ in $E$. According to Theorem 2.1c,
exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{2}$\textit{1:} We have $0\in C$.

\textit{Assertion D}$_{2}$\textit{2:} There exists an $f\in E^{\ast}$ such
that every $x\in C$ satisfies $f\left(  x\right)  <0$.
\end{statement}

Thus, we must be in one of the following two cases:

\textit{Case 1:} Assertion D$_{2}$1 holds.

\textit{Case 2:} Assertion D$_{2}$2 holds.

First, let us consider Case 1. In this case, Assertion D$_{2}$1 holds. In
other words, $0\in C$. Thus, $0\in C=\operatorname*{conv}.\operatorname*{hull}%
\left(  S\cup\left\{  -b\right\}  \right)  $. In other words, $0$ is a convex
combination of the elements of $S\cup\left\{  -b\right\}  $. In other words,
there exist a family $\left(  \lambda_{s}\right)  _{s\in S\cup\left\{
-b\right\}  }$ of nonnegative reals such that $\sum\limits_{s\in S\cup\left\{
-b\right\}  }\lambda_{s}=1$ and $\sum\limits_{s\in S\cup\left\{  -b\right\}
}\lambda_{s}s=0$. Consider this family $\left(  \lambda_{s}\right)  _{s\in
S\cup\left\{  -b\right\}  }$.

We must be in one of the following two subcases:

\textit{Subcase 1.1:} We have $\lambda_{-b}=0$.

\textit{Subcase 1.2:} We have $\lambda_{-b}\neq0$.

Let us first consider Subcase 1.1. In this subcase, we have $\lambda_{-b}=0$,
so that
\begin{align*}
1  &  =\sum\limits_{s\in S\cup\left\{  -b\right\}  }\lambda_{s}=\sum
\limits_{s\in S\setminus\left\{  -b\right\}  }\lambda_{s}+\underbrace{\sum
\limits_{s\in\left\{  -b\right\}  }\lambda_{s}}_{=\lambda_{-b}=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }S\cup\left\{  -b\right\}  \text{
is the union of the disjoint sets }S\setminus\left\{  -b\right\}  \text{ and
}\left\{  -b\right\}  \right) \\
&  =\sum\limits_{s\in S\setminus\left\{  -b\right\}  }\lambda_{s}.
\end{align*}
If every $s\in S\setminus\left\{  -b\right\}  $ would satisfy $\lambda_{s}=0$,
then we would have $\sum\limits_{s\in S\setminus\left\{  -b\right\}
}\underbrace{\lambda_{s}}_{=0}=\sum\limits_{s\in S\setminus\left\{
-b\right\}  }0=0$, contradicting $\sum\limits_{s\in S\setminus\left\{
-b\right\}  }\lambda_{s}=1$. Thus, not every $s\in S\setminus\left\{
-b\right\}  $ satisfies $\lambda_{s}=0$. In other words, there exists an $s\in
S\setminus\left\{  -b\right\}  $ such that $\lambda_{s}\neq0$. Let $t$ be such
an $s$. Then, $t\in S\setminus\left\{  -b\right\}  $ and $\lambda_{t}\neq0$.
Since $\lambda_{t}$ is nonnegative and $\lambda_{t}\neq0$, we have
$\lambda_{t}>0$.

We have%
\begin{align*}
0  &  =\sum\limits_{s\in S\cup\left\{  -b\right\}  }\lambda_{s}s=\sum
\limits_{s\in S\setminus\left\{  -b\right\}  }\lambda_{s}s+\underbrace{\sum
\limits_{s\in\left\{  -b\right\}  }\lambda_{s}s}_{\substack{=\lambda
_{-b}\left(  -b\right)  =0\\\text{(since }\lambda_{-b}=0\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }S\cup\left\{  -b\right\}  \text{
is the union of the disjoint sets }S\setminus\left\{  -b\right\}  \text{ and
}\left\{  -b\right\}  \right) \\
&  =\sum\limits_{s\in S\setminus\left\{  -b\right\}  }\lambda_{s}%
s=\sum\limits_{s\in\left(  S\setminus\left\{  -b\right\}  \right)
\setminus\left\{  t\right\}  }\lambda_{s}s+\lambda_{t}t,
\end{align*}
so that%
\begin{align}
-\lambda_{t}t  &  =\sum\limits_{s\in\left(  S\setminus\left\{  -b\right\}
\right)  \setminus\left\{  t\right\}  }\lambda_{s}s\in\operatorname*{cone}%
\underbrace{\left(  \left(  S\setminus\left\{  -b\right\}  \right)
\setminus\left\{  t\right\}  \right)  }_{\subseteq S}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda_{s}\text{ is nonnegative
for every }s\in\left(  S\setminus\left\{  -b\right\}  \right)  \setminus
\left\{  t\right\}  \right) \nonumber\\
&  \subseteq\operatorname*{cone}S. \label{2.5c.pf.1.aux}%
\end{align}


Now, let us notice that%
\begin{equation}
\text{every }\nu\in\mathbb{R}\text{ satisfies }\nu t\in\operatorname*{cone}S.
\label{2.5c.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5c.pf.1}):} Let $\nu\in\mathbb{R}$.
\par
We have $t\in S\subseteq\operatorname*{cone}S$. Thus, if $\nu\geq0$, then $\nu
t\in\operatorname*{cone}S$ (since $\operatorname*{cone}S$ is closed under
multiplication by a nonnegative scalar). Thus, if $\nu\geq0$, then
(\ref{2.5c.pf.1}) clearly holds. Therefore, we can WLOG assume that $\nu\geq0$
doesn't hold for the rest of this proof. Assume this.
\par
So, we know that $\nu\geq0$ doesn't hold. In other words, $\nu<0$. Thus,
$-\nu>0$. Combined with $\lambda_{t}>0$, this yields $\dfrac{-\nu}{\lambda
_{t}}>0$. Hence, $\dfrac{-\nu}{\lambda_{t}}\cdot\left(  -\lambda_{t}t\right)
\in\operatorname*{cone}S$ (since $-\lambda_{t}t\in\operatorname*{cone}S$ (by
(\ref{2.5c.pf.1.aux})), and since $\operatorname*{cone}S$ is closed under
multiplication by a nonnegative scalar). Since $\dfrac{-\nu}{\lambda_{t}}%
\cdot\left(  -\lambda_{t}t\right)  =\nu t$, this rewrites as $\nu
t\in\operatorname*{cone}S$. This proves (\ref{2.5c.pf.1}).} In other words,%
\begin{equation}
t\mathbb{R}\subseteq\operatorname*{cone}S. \label{2.5c.pf.1.rewr}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5c.pf.1.rewr}):} Let $x\in t\mathbb{R}$ be
arbitrary. Then, there exists a $\nu\in\mathbb{R}$ such that $x=\nu t$.
Consider this $\nu$. Then, $x=\nu t\in\operatorname*{cone}S$ (by
(\ref{2.5c.pf.1})). Now, forget that we fixed $x$. We have thus shown that
every $x\in t\mathbb{R}$ satisfies $x\in\operatorname*{cone}S$. In other
words, $t\mathbb{R}\subseteq\operatorname*{cone}S$. Thus,
(\ref{2.5c.pf.1.rewr}) is proven.}

Let $\pi$ be the canonical projection $E\rightarrow E\diagup\left(
t\mathbb{R}\right)  $. Clearly,
\begin{align*}
\left\vert \pi\left(  S\setminus\left\{  t\right\}  \right)  \right\vert  &
\leq\left\vert S\setminus\left\{  t\right\}  \right\vert
=\underbrace{\left\vert S\right\vert }_{=n}-1\ \ \ \ \ \ \ \ \ \ \left(
\text{since }t\in S\right) \\
&  =n-1<n.
\end{align*}
Hence, we can apply Theorem 2.5c to $E\diagup\left(  t\mathbb{R}\right)  $,
$\pi\left(  S\setminus\left\{  t\right\}  \right)  $ and $\pi\left(  b\right)
$ instead of $E$, $S$ and $b$ (since we assumed that Theorem 2.5c holds
whenever $\left\vert S\right\vert <n$), and conclude that\textbf{ }exactly one
of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{3}$\textit{1:} We have $\pi\left(  b\right)
\in\operatorname*{cone}\left(  \pi\left(  S\setminus\left\{  t\right\}
\right)  \right)  $.

\textit{Assertion D}$_{3}$\textit{2:} There exists an $f\in\left(
E\diagup\left(  t\mathbb{R}\right)  \right)  ^{\ast}$ such that $f\left(
\pi\left(  b\right)  \right)  >0$ and $\left(  \text{every }x\in
\operatorname*{cone}\left(  \pi\left(  S\setminus\left\{  t\right\}  \right)
\right)  \text{ satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

Thus, we must be in one of the following two subsubcases:

\textit{Subsubcase 1.1.1:} Assertion D$_{3}$1 holds.

\textit{Subsubcase 1.1.2:} Assertion D$_{3}$2 holds.

First, let us consider Subsubcase 1.1.1. In this subsubcase, Assertion D$_{3}%
$1 holds. In other words, we have $\pi\left(  b\right)  \in
\operatorname*{cone}\left(  \pi\left(  S\setminus\left\{  t\right\}  \right)
\right)  $. Thus, there exists a family $\left(  \mu_{w}\right)  _{w\in
\pi\left(  S\setminus\left\{  t\right\}  \right)  }$ of nonnegative reals such
that $\pi\left(  b\right)  =\sum\limits_{w\in\pi\left(  S\setminus\left\{
t\right\}  \right)  }\mu_{w}w$. Consider this family $\left(  \mu_{w}\right)
_{w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  }$.

For every $w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  $, let
$s_{w}$ be an (arbitrarily chosen) element of $S\setminus\left\{  t\right\}  $
satisfying $w=\pi\left(  s_{w}\right)  $. Such an $s_{w}$ exists, since
$w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  $. Now,%
\[
\pi\left(  b\right)  =\sum\limits_{w\in\pi\left(  S\setminus\left\{
t\right\}  \right)  }\mu_{w}\underbrace{w}_{=\pi\left(  s_{w}\right)  }%
=\sum\limits_{w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  }\mu
_{w}\pi\left(  s_{w}\right)  =\pi\left(  \sum\limits_{w\in\pi\left(
S\setminus\left\{  t\right\}  \right)  }\mu_{w}s_{w}\right)
\]
(since $\pi$ is linear). Since $\pi$ is linear, we have%
\[
\pi\left(  b-\sum\limits_{w\in\pi\left(  S\setminus\left\{  t\right\}
\right)  }\mu_{w}s_{w}\right)  =\pi\left(  b\right)  -\underbrace{\pi\left(
\sum\limits_{w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  }\mu
_{w}s_{w}\right)  }_{=\pi\left(  b\right)  }=\pi\left(  b\right)  -\pi\left(
b\right)  =0.
\]
Thus,%
\begin{align}
b-\sum\limits_{w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  }\mu
_{w}s_{w}  &  \in\operatorname*{Ker}\pi=t\mathbb{R}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is the canonical
projection }E\rightarrow E\diagup\left(  t\mathbb{R}\right)  \right)
\nonumber\\
&  \subseteq\operatorname*{cone}S\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{2.5c.pf.1.rewr})}\right)  . \label{2.5c.pf.2}%
\end{align}


But every $w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  $ satisfies
$s_{w}\in S\setminus\left\{  t\right\}  \subseteq S\subseteq
\operatorname*{cone}S$. Therefore, $\sum\limits_{w\in\pi\left(  S\setminus
\left\{  t\right\}  \right)  }\mu_{w}s_{w}\in\operatorname*{cone}S$ (since any
linear combination of finitely many elements of $\operatorname*{cone}S$ with
nonnegative coefficients must lie in $\operatorname*{cone}S$). Combined with
(\ref{2.5c.pf.2}), this shows that $\sum\limits_{w\in\pi\left(  S\setminus
\left\{  t\right\}  \right)  }\mu_{w}s_{w}$ and $b-\sum\limits_{w\in\pi\left(
S\setminus\left\{  t\right\}  \right)  }\mu_{w}s_{w}$ are two elements of
$\operatorname*{cone}S$. Thus, their sum $\left(  \sum\limits_{w\in\pi\left(
S\setminus\left\{  t\right\}  \right)  }\mu_{w}s_{w}\right)  +\left(
b-\sum\limits_{w\in\pi\left(  S\setminus\left\{  t\right\}  \right)  }\mu
_{w}s_{w}\right)  $ must also lie in $\operatorname*{cone}S$ (because the sum
of any two elements of $\operatorname*{cone}S$ must lie in
$\operatorname*{cone}S$). In other words, $\operatorname*{cone}S$ contains%
\[
\left(  \sum\limits_{w\in\pi\left(  S\setminus\left\{  t\right\}  \right)
}\mu_{w}s_{w}\right)  +\left(  b-\sum\limits_{w\in\pi\left(  S\setminus
\left\{  t\right\}  \right)  }\mu_{w}s_{w}\right)  =b.
\]
In other words, $b\in\operatorname*{cone}S$. Thus, Assertion D$_{1}$1 holds.
Hence, at least one of Assertions D$_{1}$1 and D$_{1}$2 holds.

We have thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds in Subsubcase 1.1.1.

Next, let us consider Subsubcase 1.1.2. In this subsubcase, Assertion D$_{3}$2
holds. In other words, there exists an $f\in\left(  E\diagup\left(
t\mathbb{R}\right)  \right)  ^{\ast}$ such that $f\left(  \pi\left(  b\right)
\right)  >0$ and \newline$\left(  \text{every }x\in\operatorname*{cone}\left(
\pi\left(  S\setminus\left\{  t\right\}  \right)  \right)  \text{ satisfies
}f\left(  x\right)  \leq0\right)  $. Denote this $f$ by $h$. Then,
$h\in\left(  E\diagup\left(  t\mathbb{R}\right)  \right)  ^{\ast}$ satisfies
$h\left(  \pi\left(  b\right)  \right)  >0$ and $\left(  \text{every }%
x\in\operatorname*{cone}\left(  \pi\left(  S\setminus\left\{  t\right\}
\right)  \right)  \text{ satisfies }h\left(  x\right)  \leq0\right)  $.

It is easy to see that%
\begin{equation}
\text{every }s\in S\text{ satisfies }\left(  h\circ\pi\right)  \left(
s\right)  \leq0. \label{2.5c.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5c.pf.3}):} Let $s\in S$.
\par
Since $\pi$ is the canonical projection $E\rightarrow E\diagup\left(
t\mathbb{R}\right)  $, we have $\pi\left(  t\mathbb{R}\right)  =0$, so that
$\pi\left(  t\right)  =0$ (since $t\in t\mathbb{R}$). Thus, $\left(  h\circ
\pi\right)  \left(  t\right)  =h\left(  \underbrace{\pi\left(  t\right)
}_{=0}\right)  =h\left(  0\right)  =0$ (since $h$ is linear). Hence, if $s=t$,
then (\ref{2.5c.pf.3}) is proven. Therefore, we can WLOG assume that $s\neq t$
for the rest of this proof. Assume this.
\par
Since $s\neq t$, we have $s\in S\setminus\left\{  t\right\}  $. Hence,
$\pi\left(  s\right)  \in\pi\left(  S\setminus\left\{  t\right\}  \right)
\subseteq\operatorname*{cone}\left(  \pi\left(  S\setminus\left\{  t\right\}
\right)  \right)  $.
\par
Recall that $\left(  \text{every }x\in\operatorname*{cone}\left(  \pi\left(
S\setminus\left\{  t\right\}  \right)  \right)  \text{ satisfies }h\left(
x\right)  \leq0\right)  $. Applying this to $x=\pi\left(  s\right)  $, we
obtain $h\left(  \pi\left(  s\right)  \right)  \leq0$. Hence, $\left(
h\circ\pi\right)  \left(  s\right)  =h\left(  \pi\left(  s\right)  \right)
\leq0$. This proves (\ref{2.5c.pf.3}).} From this, it is easy to see that%
\begin{equation}
\text{every }x\in\operatorname*{cone}S\text{ satisfies }\left(  h\circ
\pi\right)  \left(  x\right)  \leq0. \label{2.5c.pf.4}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5c.pf.4}):} Let $x\in\operatorname*{cone}%
S$. Then,%
\[
x\in\operatorname*{cone}S=\left\{  \sum\limits_{s\in S}\nu_{s}s\ \mid\ \left(
\nu_{s}\right)  _{s\in S}\text{ is a family of nonnegative reals}\right\}
\]
(by the definition of $\operatorname*{cone}S$). Hence, there exists a family
$\left(  \nu_{s}\right)  _{s\in S}$ of nonnegative reals such that
$x=\sum\limits_{s\in S}\nu_{s}s$. Consider this $\left(  \nu_{s}\right)
_{s\in S}$.
\par
Since $x=\sum\limits_{s\in S}\nu_{s}s$, we have%
\begin{align*}
\left(  h\circ\pi\right)  \left(  x\right)   &  =\left(  h\circ\pi\right)
\left(  \sum\limits_{s\in S}\nu_{s}s\right)  =\sum\limits_{s\in S}%
\underbrace{\nu_{s}\left(  h\circ\pi\right)  \left(  s\right)  }%
_{\substack{\leq0\\\text{(since }\nu_{s}\geq0\text{ and}\\\left(  h\circ
\pi\right)  \left(  s\right)  \leq0\text{ (by (\ref{2.5c.pf.3})))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h\circ\pi\text{ is }\mathbb{R}%
\text{-linear}\right) \\
&  \leq\sum\limits_{s\in S}0=0.
\end{align*}
This proves (\ref{2.5c.pf.4}).} Also, $\left(  h\circ\pi\right)  \left(
b\right)  =h\left(  \pi\left(  b\right)  \right)  >0$. Altogether, we have
thus shown that $\left(  h\circ\pi\right)  \left(  b\right)  >0$ and $\left(
\text{every }x\in\operatorname*{cone}S\text{ satisfies }\left(  h\circ
\pi\right)  \left(  x\right)  \leq0\right)  $.

Thus, there exists an $f\in E^{\ast}$ such that $f\left(  b\right)  >0$ and
$\left(  \text{every }x\in\operatorname*{cone}S\text{ satisfies }f\left(
x\right)  \leq0\right)  $ (namely, $f=h\circ\pi$). In other words, Assertion
D$_{1}$2 holds. Hence, at least one of Assertions D$_{1}$1 and D$_{1}$2 holds.

We have thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds in Subsubcase 1.1.2.

Hence, in each of the Subsubcases 1.1.1 and 1.1.2, at least one of Assertions
D$_{1}$1 and D$_{1}$2 holds. Since these Subsubcases 1.1.1 and 1.1.2 cover the
whole Subcase 1.1, this yields that at least one of Assertions D$_{1}$1 and
D$_{1}$2 holds in Subcase 1.1.

Next, let us consider Subcase 1.2. In this subcase, we have $\lambda_{-b}%
\neq0$. Combined with the fact that $\lambda_{-b}$ is nonnegative, this yields
$\lambda_{-b}>0$. Thus, $\dfrac{1}{\lambda_{-b}}$ exists and is $>0$. Now,%
\begin{align*}
0  &  =\sum\limits_{s\in S\cup\left\{  -b\right\}  }\lambda_{s}s=\sum
\limits_{s\in S\setminus\left\{  -b\right\}  }\lambda_{s}s+\underbrace{\sum
\limits_{s\in\left\{  -b\right\}  }\lambda_{s}s}_{=\lambda_{-b}\left(
-b\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }S\cup\left\{  -b\right\}  \text{
is the union of the disjoint sets }S\setminus\left\{  -b\right\}  \text{ and
}\left\{  -b\right\}  \right) \\
&  =\sum\limits_{s\in S\setminus\left\{  -b\right\}  }\lambda_{s}%
s+\lambda_{-b}\left(  -b\right)  =\sum\limits_{s\in S\setminus\left\{
-b\right\}  }\lambda_{s}s-\lambda_{-b}b,
\end{align*}
so that%
\begin{align*}
\lambda_{-b}b  &  =\sum\limits_{s\in S\setminus\left\{  -b\right\}  }%
\lambda_{s}s\in\operatorname*{cone}\underbrace{\left(  S\setminus\left\{
-b\right\}  \right)  }_{\subseteq S}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda_{s}\text{ is nonnegative
for every }s\in S\setminus\left\{  -b\right\}  \right) \\
&  \subseteq\operatorname*{cone}S.
\end{align*}
Since $\dfrac{1}{\lambda_{-b}}>0$, this yields $\dfrac{1}{\lambda_{-b}}%
\lambda_{-b}b\in\operatorname*{cone}S$ (because $\operatorname*{cone}S$ is
closed under multiplication by a nonnegative scalar). Since $\dfrac{1}%
{\lambda_{-b}}\lambda_{-b}b=b$, this rewrites as $b\in\operatorname*{cone}S$.
In other words, Assertion D$_{1}$1 holds. Hence, at least one of Assertions
D$_{1}$1 and D$_{1}$2 holds.

We have thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds in Subcase 1.2.

Hence, in each of the Subcases 1.1 and 1.2, at least one of Assertions D$_{1}%
$1 and D$_{1}$2 holds. Since these Subcases 1.1 and 1.2 cover the whole Case
1, this yields that at least one of Assertions D$_{1}$1 and D$_{1}$2 holds in
Case 1.

Finally, let us consider Case 2. In this case, Assertion D$_{2}$2 holds. In
other words, there exists an $f\in E^{\ast}$ such that every $x\in C$
satisfies $f\left(  x\right)  <0$. Denote this $f$ by $h$. Then,
\begin{equation}
\text{every }x\in C\text{ satisfies }h\left(  x\right)  <0\text{.}
\label{2.5c.pf.6}%
\end{equation}


Now, it is easy to see that
\begin{equation}
\text{every }x\in\operatorname*{cone}S\text{ satisfies }h\left(  x\right)
\leq0. \label{2.5c.pf.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{2.5c.pf.7}):} Let $x\in\operatorname*{cone}%
S$. Then, $x\in\operatorname*{cone}S=\left\{  \sum\limits_{s\in S}\nu
_{s}s\ \mid\ \left(  \nu_{s}\right)  _{s\in S}\text{ is a family of
nonnegative reals}\right\}  $. Thus, there exists a family $\left(  \nu
_{s}\right)  _{s\in S}$ of nonnegative reals such that $x=\sum\limits_{s\in
S}\nu_{s}s$. Consider this family $\left(  \nu_{s}\right)  _{s\in S}$.
\par
Every $s\in S$ satisfies $h\left(  s\right)  <0$ (by (\ref{2.5c.pf.6}),
applied to $x=s$ (since $s\in S\subseteq S\cup\left\{  -b\right\}
\subseteq\operatorname*{conv}.\operatorname*{hull}\left(  S\cup\left\{
-b\right\}  \right)  =C$)) and $\nu_{s}\geq0$, so that $\underbrace{\nu_{s}%
}_{\geq0}\underbrace{h\left(  s\right)  }_{<0}\leq0$. Now, since
$x=\sum\limits_{s\in S}\nu_{s}s$, we have%
\begin{align*}
h\left(  x\right)   &  =h\left(  \sum\limits_{s\in S}\nu_{s}s\right)
=\sum\limits_{s\in S}\underbrace{\nu_{s}h\left(  s\right)  }_{\leq
0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }h\text{ is }\mathbb{R}%
\text{-linear}\right) \\
&  \leq\sum\limits_{s\in S}0=0.
\end{align*}
This proves (\ref{2.5c.pf.7}).}

Also, $-b\in S\cup\left\{  -b\right\}  \subseteq\operatorname*{conv}%
.\operatorname*{hull}\left(  S\cup\left\{  -b\right\}  \right)  =C$, so that
$h\left(  -b\right)  <0$ (by (\ref{2.5c.pf.6})). Since $h$ is linear, we have
$-h\left(  b\right)  =h\left(  -b\right)  <0$, so that $h\left(  b\right)  >0$.

Altogether, we now know that $h\left(  b\right)  >0$ and $\left(  \text{every
}x\in\operatorname*{cone}S\text{ satisfies }h\left(  x\right)  \leq0\right)  $
(by (\ref{2.5c.pf.7})). Thus, there exists an $f\in E^{\ast}$ such that
$f\left(  b\right)  >0$ and \newline$\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }f\left(  x\right)  \leq0\right)  $
(namely, $f=h$). In other words, Assertion D$_{1}$2 holds. Hence, at least one
of Assertions D$_{1}$1 and D$_{1}$2 holds.

We have thus proven that at least one of Assertions D$_{1}$1 and D$_{1}$2
holds in Case 2.

Hence, in each of the Cases 1 and 2, at least one of Assertions D$_{1}$1 and
D$_{1}$2 holds. Since these Cases 1 and 2 cover all possibilities, this yields
that, in every situation, at least one of Assertions D$_{1}$1 and D$_{1}$2
holds. Since we know that the Assertions D$_{1}$1 and D$_{1}$2 cannot hold at
the same time, this yields that exactly one of Assertions D$_{1}$1 and D$_{1}%
$2 holds.

Now, forget that we fixed $E$, $S$ and $b$. We have thus proven that if $E$ is
a finite-dimensional $\mathbb{R}$-vector space, $S$ is a finite subset of $E$
such that $\left\vert S\right\vert =n$, and $b$ is an element of $E$, then
exactly one of the following two assertions holds:

\begin{statement}
\textit{Assertion D}$_{1}$\textit{1:} We have $b\in\operatorname*{cone}S$.

\textit{Assertion D}$_{1}$\textit{2:} There exists an $f\in E^{\ast}$ such
that $f\left(  b\right)  >0$ and \newline$\left(  \text{every }x\in
\operatorname*{cone}S\text{ satisfies }f\left(  x\right)  \leq0\right)  $.
\end{statement}

In other words, we have proven that Theorem 2.5c holds in the case when
$\left\vert S\right\vert =n$. This completes the induction step, and thus the
induction proof of Theorem 2.5c is complete.
\end{proof}

\begin{thebibliography}{99999999}                                                                                         %


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\end{document}