Added the inverse function theorem.

src/topology/metric/metric.tex

@@ -41,3 +41,33 @@
     (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{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
+    \[
+    d(f(x), f(y)) \le Cd(x, y) \quad \forall x, y \in X
+    \]
+
+    then:
+    \begin{enumerate}
+        \item There exists a unique $x \in X$ such that $f(x) = x$. 
+        \item For any $y \in X$, $\limv{n}f^n(y) = x$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Let $x_0 \in X$ be arbitrary, and $x_n = f^n(x_0)$, then for ecah $n \in \natp$,
+    \[
+    d(x_n, x_{n+1}) \le C d(x_{n-1}, x_n) \le C^n d(x_0, x_1)
+    \]
+
+    Thus $\seq{x_n} \subset X$ is Cauchy, and converges to a point $x \in X$. 
+
+    (2): For any $y_0 \in X$, let $y_n = f^n(y_0)$, then $d(x_n, y_n) \to 0$ as $n \to \infty$, so $\limv{n}f^n(y_0) = x$. 
+
+    (1): Since $f$ is Lipschitz continuous, 
+    \[
+    f(x) = f\braks{\limv{n}f^n(x)} = \limv{n}f^{n+1}(x) = x
+    \]
+\end{proof}
+