Proof of Contraction Mapping Theorem on a Nonempty Complete Metric Space

Let $\lambda\in[0,1)$ be the contraction constant from the contraction definition, so $d(T(x),T(y))\le\lambda\,d(x,y)\quad\text{for all }x,y\in X.$

Existence and convergence. Fix any $x_0\in X$ (possible since $X\ne\emptyset$), and define the sequence of iterates by $x_{m+1}=T(x_m)$ for each natural number $m$ and for $m=0$.

Step 1: consecutive-distance bound. By the contraction property applied $m$ times, $d(x_m,x_{m+1})\le\lambda^m d(x_0,x_1).$

Step 2: $(x_m)$ is Cauchy. If $d(x_0,x_1)=0$ then all $x_m=x_0$ and the sequence is trivially Cauchy. Otherwise, for $m,\ell\in\mathbb{N}$ with $m\le\ell$, the triangle inequality (axiom 4 of the metric $d$) gives $d(x_m,x_\ell)\le\sum_{k=m}^{\ell-1}d(x_k,x_{k+1})\le\frac{\lambda^m(1-\lambda^{\ell-m})}{1-\lambda}\,d(x_0,x_1)\le\frac{\lambda^m}{1-\lambda}\,d(x_0,x_1).$ Since $0\le\lambda<1$, we have $\lambda^m\to 0$ in $\mathbb{R}$ (a standard consequence of the least upper bound property). Hence there exists a natural number $N$ such that $\frac{\lambda^N}{1-\lambda}d(x_0,x_1)<\varepsilon$; then $d(x_m,x_\ell)<\varepsilon$ for all $m,\ell\ge N$, so $(x_m)$ is a Cauchy sequence in $(X,d)$.

Step 3: convergence to $x^*$. Since $(X,d)$ is complete, $(x_m)$ converges: there exists $x^*\in X$ with $\lim_{m\to\infty}x_m=x^*$.

$x^*$ is a fixed point. By the contraction property, $d(x_{m+1},T(x^*))=d(T(x_m),T(x^*))\le\lambda\,d(x_m,x^*).$ Since $d(x_m,x^*)\to 0$, we get $d(x_{m+1},T(x^*))\to 0$, so $(x_{m+1})$ converges to $T(x^*)$. But $(x_{m+1})$ is a subsequence of $(x_m)$ and therefore also converges to $x^*$. Limits in metric spaces are unique: if $(y_m)\to a$ and $(y_m)\to b$ in $(X,d)$, then for any $\varepsilon>0$ we can find $M$ large enough that $d(a,b)\le d(a,y_M)+d(y_M,b)<2\varepsilon$ by the triangle inequality; since $\varepsilon$ is arbitrary, $d(a,b)=0$ and hence $a=b$ by axiom 2 of the metric. Applying uniqueness of limits, $T(x^*)=x^*$.

Uniqueness of the fixed point. Suppose $y^*\in X$ also satisfies $T(y^*)=y^*$. Then $d(x^*,y^*)=d(T(x^*),T(y^*))\le\lambda\,d(x^*,y^*).$ Since $\lambda<1$, this gives $(1-\lambda)\,d(x^*,y^*)\le 0$, so $d(x^*,y^*)=0$, hence $x^*=y^*$ by axiom 2 of the metric.