Banach fixed point theorem

Theorem
Let $$(M,d)$$ be a complete metric space and $$f$$ a contraction on $$M$$ with Lipschitz constant $$k$$. Then there exists a unique $$x\in M$$ such that $$f(x)=x$$.

Proof
Let us define a sequence $$(x_n)$$ of elements in $$M$$ in the following way and show that this sequence is a Cauchy sequence. Since $$f$$ is a contraction, we have that, for $$n\in \mathbb N_0$$ which results in the formula As such, we obtain, for any $$n,m\in \mathbb N_0$$ ($$n\leq m$$) by basic results on geometric series. Since $$k\in [0,1)$$, $$(d(x_0,x_1)\frac{k^n}{1-k})$$ is a sequence converging to $$0$$, and hence which results in or, in other words, $$(x_n)$$ is a Cauchy sequence. $$M$$ being complete, $$(x_n)$$ converges to a point which we will call $$x$$. Finally, since $$f$$ is a contraction, it is continuous, and hence, the following holds which states that $$x$$ is a fixed point of $$f$$. To prove uniqueness, suppose $$y$$ $$(\not =x)$$ is also a fixed point of $$f$$. Then and we would also obtain resulting in a contradiction.
 * $$\begin{cases} x_0\in M\\x_{n+1}=f(x_n), & n\in \mathbb N\end{cases}$$,
 * $$d(x_{n+1},x_{n+2})=d(f(x_n),f(x_{n+1}))\leq kd(x_n,x_{n+1})$$
 * $$d(x_n,x_{n+1})\leq k^n d(x_0,x_1)$$, for any $$n\in \mathbb N$$.
 * $$d(x_n,x_m)\leq \sum_{i=n}^{m-1} d(x_i,x_{i+1})\leq d(x_0,x_1)\sum_{i=n}^{m-1}k^i\leq d(x_0,x_1)\frac{k^n}{1-k}$$
 * $$\underset{\epsilon >0}{\forall} \; \underset{n\in \mathbb N}{\exists} \; \underset{m>n}{\forall} \; d(x_0,x_1)\frac{k^m}{1-k}<\epsilon$$
 * $$\underset{\epsilon >0}{\forall} \; \underset{p\in \mathbb N}{\exists}\; \underset{m,n>p}{\forall} \; d(x_n,x_m)<\epsilon$$,
 * $$x=\lim_{n\to \infty} x_n=\lim_{n\to \infty} f(x_n)=f(\lim_{n\to \infty} x_n)=f(x)$$
 * $$d(x,y)>0$$,
 * $$d(x,y)=d(f(x),f(y))\leq kd(x,y)<d(x,y)$$,