Added Urysohn Metrisation theorem and compactness theorems.

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
     \[
     d(f(x), f(y)) \le Cd(x, y) \quad \forall x, y \in X
     \]