Added Urysohn Metrisation theorem and compactness theorems.

src/fa/lc/compact.tex

@@ -0,0 +1,85 @@
+\section{Compact Convex Sets}
+\label{section:compact-convex}
+
+\begin{definition}[Extreme Point]
+\label{definition:extreme-point}
+    Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$. 
+\end{definition}
+
+\begin{definition}[Extreme Subset]
+\label{definition:extreme-subset}
+    Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.  
+\end{definition}
+
+
+
+\begin{lemma}
+\label{lemma:extremal-face}
+    Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$. 
+\end{lemma}
+\begin{proof}
+    Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by \autoref{proposition:compact-extensions}. Let $x \in A$ and $y, z \in K$ such that $x \in (y, z)$. By definition of $\alpha$, $\dpn{y, \phi}{E} = \dpn{z, \phi}{E} = \alpha$. Thus $y, z \in A$ as well. 
+\end{proof}
+
+\begin{theorem}[Krein-Milman]
+\label{theorem:krein-milman}
+    Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.12.5]{Bogachev}}}. ]
+    Assume without loss of generality that $K \ne \emptyset$. 
+    
+    Let $K_0 \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. 
+    
+    By Zorn's lemma, there exists a minimal element $A$ of $\mathcal{E}(K_0)$. Let $x, y \in A$, $\phi \in E^*$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha} \cap K$ is a non-empty, closed, and extreme subset of $K$ by \autoref{lemma:extremal-face}, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_0$. 
+    
+    Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point. 
+    
+    Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. 
+\end{proof}
+
+\begin{theorem}[Markov-Kakutani Fixed Point Theorem]
+\label{theorem:markov-kakutani}
+    Let $E$ be a separated topological vector space over $\real$, $K \subset E$ be a compact convex set, and $\cf \subset C(K; K)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For any $f, g \in \cf$, $f \circ g = g \circ f$. 
+        \item For each $f \in \cf$, $x, y \in K$, and $t \in [0, 1]$, 
+        \[
+        f(tx + (1 - t)y) = tf(x) + (1 - t)f(y)
+        \]
+    \end{enumerate}
+
+    then there exists $x_0 \in K$ such that $f(x_0) = x_0$ for all $f \in \cf$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.12.10]{Bogachev}}}. ]
+    For each $f \in \cf$ and $n \in \natp$, define $f^{(n)} = \frac{1}{n}\sum_{k = 0}^{n - 1}f^k$, then $f^{(n)} \in C(K; K)$ as well. Via a closure operation, assume without loss of generality that:
+    \begin{enumerate}
+        \item For any $f \in \cf$ and $n \in \natp$, $f^{(n)} \in \cf$. 
+        \item For any $f, g \in \cf$, $f \circ g \in \cf$. 
+    \end{enumerate}
+
+    Let $\bracsn{f_j}_1^N \subset \cf$ and $\bracsn{n_j}_1^N \subset \natp$, then
+    \[
+        \bigcap_{j = 1}^n f_j^{(n_j)}(K) \supset \braks{\prod_{j = 1}^n f_j^{(n_j)}}(K) \ne \emptyset 
+    \]
+
+    thus any finite intersections of elements in
+    \[
+    \mathcal{K} = \bracsn{f^{(n)}(K)| f \in \cf, n \in \natp}
+    \]
+
+    is non-empty. Since $E$ is separated, by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}, $\mathcal{K}$ is a family of closed sets satisfying the finite intersection property. Hence $\bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K) \ne \emptyset$. 
+
+    Now, let $x \in \bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K)$ and $U \in \cn_E(0)$, then there exists $N \in \natp$ such that $NU \supset K$. In which case, there exists $y \in K$ such that
+    \[
+        x  = \frac{1}{N}\sum_{j = 0}^{N-1}f^j(y)  \quad f(x) = \frac{1}{N}\sum_{j = 1}^{N}f^j(y)
+    \]
+
+    In which case, 
+    \[
+    f(x) - x = \frac{1}{N}(f^N(y) - y) \in \frac{1}{N}(K - K) \subset U - U
+    \]
+
+    As this holds for all $U \in \cn_E(0)$ and $E$ is separated, $f(x) = x$.   
+    
+\end{proof}
+

src/fa/lc/index.tex

@@ -4,6 +4,7 @@
 
 \input{./convex.tex}
 \input{./continuous.tex}
+\input{./compact.tex}
 \input{./barrel.tex}
 \input{./bornologic.tex}
 \input{./quotient.tex}