Added Urysohn Metrisation theorem and compactness theorems.

.vscode/tasks.json

@@ -1,57 +1,57 @@
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+              "args": ["/d", "/c"]
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-    ]
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+  ]
+}

src/dg/complex/derivative.tex

@@ -212,7 +212,7 @@
     \]
 \end{corollary}
 \begin{proof}
-    By \autoref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, 
+    By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, 
     \begin{align*}
         D^kf(z_0) &= \frac{k!}{2\pi i}\int_{\omega_{z_0, r}} \frac{f(z)}{(z - z_0)^{k+1}}dz \\
         [D^kf(z_0)]_E &\le \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{|z - z_0|^{k+1}}dz \\

src/dg/complex/space.tex

@@ -11,7 +11,7 @@
     Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is complete. 
 \end{proposition}
 \begin{proof}
-    By \hyperref[corollary:cauchy-estimate]{Cauchy's estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}.
+    By \hyperref[Cauchy's estimate]{corollary:cauchy-estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}.
 \end{proof}
 
 \begin{theorem}[Montel]

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}

src/topology/main/cube.tex

@@ -59,5 +59,23 @@
     (2) $\Rightarrow$ (1): By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{C(X; [0, 1])}$, which is a uniform space. The subspace uniformity on $X$ then induces the topology on $X$. 
 \end{proof}
 
+\begin{theorem}[Uryson Metrisation Theorem]
+\label{theorem:urysohn-metrisation}
+    Let $X$ be a second countable regular space, then $X$ is metrisable. 
+\end{theorem}
+\begin{proof}
+    By \autoref{proposition:second-countable-regular}, $X$ is normal. Let $\cb \subset 2^X$ be a countable base for $X$, and let
+    \[
+    \mathcal{S} = \bracsn{(E, F) \in \mathcal{B}^2 | \ol{E} \subset F}
+    \]
+
+    By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f|_E = 1$ and $f|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore
+    \[
+    \cf = \bracsn{f_{EF}|(E, F) \in \mathcal{S}} \subset C(X; [0, 1])
+    \]
+    
+    is a countable family of continuous functions that separate points and closed sets. By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{\cf}$, which is metrisable by \autoref{proposition:countable-metric}. 
+\end{proof}
+
 
 

src/topology/main/definition.tex

@@ -77,6 +77,18 @@
     By definition, $\topo(\cb)$ satisfies (O3).
 \end{proof}
 
+\begin{definition}[First Countable]
+\label{definition:first-countable}
+    Let $X$ be a topological space, then $X$ is \textbf{first countable} if for every $x \in X$, there exists a countable fundamental system of neighbourhoods at $x$. 
+\end{definition}
+
+\begin{definition}[Second Countable]
+\label{definition:second-countable}
+    Let $X$ be a topological space, then $X$ is \textbf{second countable} if it admits a countable base. 
+\end{definition}
+
+
+
 \begin{definition}[Generated Topology]
 \label{definition:generated-topology}
     Let $X$ be a set and $\ce \subset 2^X$ such that $\bigcup_{U \in \ce}U = X$, then the smallest topology on $X$ containing $\ce$ is given by

src/topology/main/regular.tex

@@ -37,3 +37,25 @@
 
     as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:second-countable-regular}
+    Let $X$ be a second countable regular space, then $X$ is normal. 
+\end{proposition}
+\begin{proof}
+    Let $\cb \subset 2^X$ be a base for $X$ and $A, B \subset X$ be disjoint closed sets. For each $x \in A$ and $y \in B$, there exists $U_x, V_y \in \cb$ such that $x \in \ol{U_x} \subset B^c$ and $y \in \ol{V_y} \subset B^c$. Since $\cb$ is countable, let $\seq{U_n}$ and $\seq{V_n}$ be enumerations of $\bracs{U_x|x \in A}$ and $\bracs{V_y|y \in B}$, respectively. 
+
+    For each $n \in \natp$, let
+    \[
+    U_n' = U_n \setminus \bigcup_{j = 1}^n \ol{V_j} \quad V_n' = V_n \setminus \bigcup_{j = 1}^n \ol{V_j}
+    \]
+
+    then $U_n'$ and $V_n'$ are both open. Let
+    \[
+    U = \bigcup_{n \in \natp}U_n' \quad V = \bigcup_{n \in \natp}V_n'
+    \]
+
+    then $U \in \cn_X(A)$ and $V \in \cn_X(B)$. For each $m, n \in \natp$ with $m \le n$, $V_n \cap \bigcup_{j = 1}^n U_j = \emptyset$, so $U_m \cap V_n = \emptyset$. Likewise, if $m \ge n$, then $U_m \cap V_n = \emptyset$ as well. Therefore $U \cap V = \emptyset$. 
+\end{proof}
+
+

src/topology/metric/metric.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Metric Space]
 \label{definition:metric}
-    Let $X$ be a set and $d: X \times X$, then $d$ is a \textbf{metric} if:
+    Let $X$ be a set and $d: X \times X \to [0, \infty]$, then $d$ is a \textbf{metric} if:
     \begin{enumerate}
         \item[(PM1)] For any $x \in X$, $d(x, x) = 0$.
         \item[(M)] For any $x, y \in X$ with $x \ne y$, $d(x, y) > 0$.
@@ -41,10 +41,23 @@
     (3) $\Rightarrow$ (1): Let $\seq{x_n} \subset X$ be a countable dense subset. Let $x \in X$ and $k \in \natp$, then there exists $x_n \in \natp$ such that $d(x, x_n) < 1/(2k)$. In which case, $x \in B(x_n, 1/(2k)) \subset B(x_n, 1/k)$. Therefore $\bracs{B(x_n, 1/k)|n, k \in \natp}$ forms a countable basis for $X$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:countable-metric}
+    Let $\seq{(X_n, d_n)}$ be metrisable spaces, then $\prod_{n \in \natp}X_n$ is also metrisable. 
+\end{proposition}
+\begin{proof}
+    For each $n \in \natp$, let
+    \[
+    d_n': \braks{\prod_{n \in \natp}X_n}^2 \to [0, \infty] \quad (x, y) \mapsto d_n(\pi_n(x), \pi_n(y))
+    \]
+
+    then $d_n'$ is a pseudometric on $X$, and $\bracsn{d_n'}_1^\infty$ induces the product uniformity on $\prod_{n \in \natp}X_n$. By \autoref{theorem:uniform-metrisable}, $\prod_{n \in \natp}X_n$ is also metrisable. 
+\end{proof}
+
 
 \begin{theorem}[Banach's Fixed Point Theorem]
 \label{theorem:banach-fixed-point}
-    Let $(X, d)$ be a metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that
+    Let $(X, d)$ be a complete metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that
     \[
     d(f(x), f(y)) \le Cd(x, y) \quad \forall x, y \in X
     \]