Let $(X,d)$ be a non-empty, complete metric space, and let $T$ be a contraction mapping on $(X,d)$ with constant $q$ . Pick an arbitrary $x_0 \in X$ , and define the sequence $(x_n)_{n=0}^{\infty}$ by $x_n:=T^nx_0$ . Let $a:=d(Tx_0,x_0)$ . We first show by induction that for any $n\ge 0$ , $$ d(T^nx_0,x_0)\le\frac{1-q^n}{1-q} a. $$ For $n=0$ , this is obvious. For any $n\ge 1$ , suppose that $d(T^{n-1}x_0,x_0)\le\frac{1-q^{n-1}}{1-q}a$ . Then \begin{eqnarray*} d(T^nx_0,x_0)&\le&d(T^nx_0,T^{n-1}x_0)+d(x_0,T^{n-1}x_0)\\ &\le&q^{n-1}d(Tx_0,x_0)+\frac{1-q^{n-1}}{1-q}a\\ &=&\frac{q^{n-1}-q^n}{1-q}a+\frac{1-q^{n-1}}{1-q}a\\ &=&\frac{1-q^n}{1-q}a \end{eqnarray*}by the triangle inequality and repeated application of the property $d(Tx,Ty)\le qd(x,y)$ of $T$ . By induction, the inequality holds for all $n \ge 0$ .

Given any $\epsilon>0$ , it is possible to choose a natural number $N$ such that $\frac{q^n}{1-q}a<\epsilon$ for all $n\ge N$ , because $\frac{q^n}{1-q}a\to 0$ as $n\to\infty$ . Now, for any $m,n\ge N$ (we may assume that $m\ge n$ ), \begin{eqnarray*} d(x_m,x_n)&=&d(T^mx_0,T^nx_0)\\ &\le&q^nd(T^{m-n}x_0,x_0)\\ &\le&q^n\frac{1-q^{m-n}}{1-q}a\\ &<&\frac{q^n}{1-q}a<\epsilon, \end{eqnarray*}so the sequence $(x_n)$ is a Cauchy sequence. Because $(X,d)$ is complete, this implies that the sequence has a limit in $(X,d)$ ; define $x^*$ to be this limit. We now prove that $x^*$ is a fixed point of $T$ . Suppose it is not, then $\delta:=d(Tx^*,x^*)>0$ . However, because $(x_n)$ converges to $x^*$ , there is a natural number $N$ such that $d(x_n,x^*)<\delta/2$ for all $n\ge N$ . Then \begin{eqnarray*} d(Tx^*,x^*)&\le&d(Tx^*,x_{N+1})+d(x^*,x_{N+1})\\ &\le&qd(x^*,x_N)+d(x^*,x_{N+1})\\ &<&\delta/2+\delta/2=\delta, \end{eqnarray*}contradiction. So $x^*$ is a fixed point of $T$ . It is also unique. Suppose there is another fixed point $x'$ of $T$ ; because $x'\neq x^*$ , $d(x',x^*)>0$ . But then $$ d(x',x^*)=d(Tx',Tx^*)\le qd(x',x^*)<d(x',x^*), $$ contradiction. Therefore, $x^*$ is the unique fixed point of $T$ .