Added measurability in separable metric spaces.

src/topology/index.tex

@@ -4,3 +4,4 @@
 \input{./src/topology/main/index.tex}
 \input{./src/topology/uniform/index.tex}
 \input{./src/topology/functions/index.tex}
+\input{./src/topology/metric/index.tex}

src/topology/main/interiorclosureboundary.tex

@@ -63,7 +63,7 @@
 
 \begin{definition}[Dense]
 \label{definition:dense}
-    Let $X$ be a topologicial space and $A \subset X$, then the following are equivalent:
+    Let $X$ be a topological space and $A \subset X$, then the following are equivalent:
     \begin{enumerate}
         \item $\ol{A} = X$.
         \item For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.

src/topology/metric/index.tex

@@ -0,0 +1,5 @@
+\chapter{Metric Spaces}
+\label{chap:metric-space}
+
+\input{./src/topology/metric/metric.tex}
+\input{./src/topology/metric/set-distance.tex}

src/topology/metric/metric.tex

@@ -0,0 +1,14 @@
+\section{Metrics}
+\label{section:metric}
+
+\begin{definition}[Metric Space]
+\label{definition:metric}
+    Let $X$ be a set and $d: X \times X$, 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 $x \ne y$, $d(x, y) > 0$.
+        \item[(PM2)] For any $x, y \in X$, $d(x, y) = d(y, x)$.
+        \item[(PM3)] For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
+    \end{enumerate}
+    The pair $(X, d)$ is a \textbf{metric space}, which comes with the metric uniformity induced by $d$, and the corresponding topology.
+\end{definition}

src/topology/metric/set-distance.tex

@@ -0,0 +1,55 @@
+\section{Distance Between Sets}
+\label{section:distance-between-sets}
+
+\begin{definition}[Distance Between Sets]
+\label{definition:distance-betwene-sets}
+    Let $X$ be a set, $d: X \times X \to [0, \infty)$ be a pseudometric, and $A, B \subset X$, then
+    \[
+    d(A, B) = \inf_{\substack{a \in A \\ b \in B}}d(a, b)
+    \]
+    is the \textbf{distance} between $A$ and $B$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:set-distance-continuous}
+    Let $(X, d)$ be a pseudometric space, $A \subset X$, and
+    \[
+    d_A: X \to [0, \infty) \quad x \mapsto d(x, A)
+    \]
+    then $d_A \in UC(X; [0, \infty))$, where for any $x, y \in X$, $|d_A(x) - d_A(y)| \le d(x, y)$.
+\end{proposition}
+\begin{proof}
+    Let $x, y \in X$, then
+    \[
+    d(y, A) = \inf_{a \in A}d(y, a) \le d(x, y) + \inf_{a \in A}d(x, a) = d(x, y) + d(x, A)
+    \]
+    As the argument is symmetric, $|d_A(x) - d_A(y)| \le d(x, y)$.
+\end{proof}
+
+\begin{definition}[Fattening]
+\label{definition:fattening}
+    Let $X$ be a set, $d: X \times X \to [0, \infty)$ be a pseudometric, $A \subset X$, and $\eps > 0$, then
+    \[
+    B(A, \eps) = \bracs{x \in X|d(x, A) < \eps} = \bigcup_{a \in A}B(a, \eps)
+    \]
+    is the \textbf{$\eps$-fattening} of $A$ with respect to $d$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:fattening-closure}
+    Let $(X, d)$ be a metric space and $A \subset X$, then
+    \[
+    \ol{A} = \bigcap_{\eps > 0}B(A, \eps)
+    \]
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:uniformclosure}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:distance-compact}
+    Let $(X, d)$ be a metric space, $C \subset X$ be closed, and $K \subset X$ be compact. If $C \cap K = \emptyset$, then $d(K, C) > 0$.
+\end{proposition}
+\begin{proof}
+    Suppose that $d(K, C) = 0$, then there exists $\seq{(x_n, y_n)} \subset K \times C$ such that $d(x_n, y_n) \to 0$ as $n \to \infty$. By \ref{definition:compact}, there exists a subsequence $\seq{n_k}$ and $x \in K$ such that $x_{n_k} \to x$ as $k \to \infty$. In which case, $d(x, C) = 0$, and $x \in \ol{K}$ by \ref{proposition:fattening-closure}, so $K \cap C \ne \emptyset$.
+\end{proof}

src/topology/uniform/metric.tex

@@ -13,7 +13,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
     \end{enumerate}
     If $d$ satisfies the above and
     \begin{enumerate}
-        \item[(M)] For any $x, y \in X$ with $x \ne y$, $d(x, y)$
+        \item[(M)] For any $x, y \in X$ with $x \ne y$, $d(x, y) > 0$.
     \end{enumerate}
     then $d$ is a \textbf{metric}.
 \end{definition}