Added a few points about Cauchy completeness of metric spaces.

src/topology/metric/metric.tex

@@ -55,6 +55,20 @@
 \end{proof}
 
 
+\begin{proposition}
+\label{proposition:complete-metric-space}
+    Let $(X, d)$ be a metric space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any Cauchy sequence $\seq{x_n} \subset X$, there exists $x \in X$ such that $x = \limv{n}x_n$. 
+        \item For any Cauchy sequence $\seq{x_n} \subset X$, there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{k}x_{n_k}$. 
+        \item For every Cauchy filter $\fF \subset 2^X$, there exists $x \in X$ such that $\fF \to x$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (3): Since $\fF$ is Cauchy, there exists $\seq{E_n} \subset \fF$ such that for each $n \in \natp$, $E_n \supset E_{n+1}$ and $\sup_{y, z \in E_n}d(y, z) \le 1/n$. For each $n \in \natp$, let $x_n \in E_n$, then there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{n}x_n$. In which case, $x \in \bigcap_{n \in \natp}\overline{E_n}$. For each $n \in \natp$, $\sup_{y, z\in E_n}d(y, z) \le 1/n$, so $B_X(x, 2/n) \supset E_n$. Therefore $\fF \to x$. 
+\end{proof}
+
+
 \begin{theorem}[Banach's Fixed Point Theorem]
 \label{theorem:banach-fixed-point}
     Let $(X, d)$ be a complete metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that