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The contraction mapping theorem states that a map $f$ of a nonempty complete metric space into itself, satisfying $d(fx,fy)\le qd(x,y)$ for some $0\le q<1$, has a unique fixed point. Iteration from any point converges to it; for example $d(f^{n}x,\tau (f))\le q^{n}d(x,fx)/(1-q)$.

Assume $X$ is nonempty. Its boundedness makes $\delta$ finite. Symmetry, positivity and separation follow pointwise from $d$, and taking the supremum of $d(fx,hx)\le d(fx,gx)+d(gx,hx)$ proves the triangle inequality. Hence $\delta$ is a uniform metric.

If $(f_n)$ is a Cauchy sequence in this uniform metric, each $(f_n(x))$ converges by completeness of $X$; denote its limit by $f(x)$. Given $ϵ>0$, choose $N$ so that $\delta (f_n,f_m)<ϵ$ for $n,m\ge N$. Letting $m\to \infty$ gives $d(f_n(x),f(x))\le ϵ$ for all $x$, so $f_n\to f$ uniformly. To prove continuity, fix $x$ and useChoose $N$ so the outside terms are small, then use continuity of $f_N$ for the middle term. Thus $f\in \mathrm{Maps}(X,X)$, proving completeness of the continuous-map space in the uniform metric.

The contraction subspace is not necessarily complete. On $X=[0,1]$, $f_n(x)=(1-1/n)x$, $n\ge 2$, are contraction mappings and converge to the identity by uniform convergence. Their only possible uniform limit is not a contraction, so this Cauchy sequence has no limit in $\mathcal{C}$.

Nevertheless the fixed point assignment is continuous. Fix $f\in \mathcal{C}$ with contraction constant $q_f<1$, and put $u=\tau (f)$, $v=\tau (g)$. ThenConsequently $d(\tau (f),\tau (g))\le \delta (f,g)/(1-q_f)$. This proves continuity of the fixed-point assignment at $f$ without requiring one contraction constant valid for every $g$.

A contraction mapping $T$ on a metric space satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. The contraction mapping theorem says that on a nonempty complete metric space such a map has a unique fixed point; successive iterates from any starting point converge to it. More precisely, if $x_n=T^n{x}_0$, then $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

The Volterra integration operator maps continuous functions to continuous functions. In the uniform norm,This constant cannot be decreased: for $f=1$ and $g=0$, $Af(x)=x$, and both uniform norms are one. Hence $A$ is not a contraction.

Changing the order of integration over the triangular region givesThus$A^2$ is a contraction, with sharp constant $1/2$. This is an example of an iterated contraction.

Take $X$ to be the set $\mathbb{Q}$, equipped with the discrete metricEvery Cauchy sequence is eventually constant: its terms must eventually have mutual distance in this metric less than $1/2$. Consequently $X$ is a complete metric space. Its topology is the discrete topology, since every singleton is a ball of radius $1/2$ and every subset is a union of such balls.

For the codomain use the usual subspace topology on $\mathbb{Q}$. The inverse image of any open set there is a subset of the discrete $X$, hence is open. Therefore $f$ is continuous and plainly surjective. A continuous image of a complete metric space need not be complete. The two occurrences of $\mathbb{Q}$ have different domain and codomain metrics.

Define $F_i(x)=a_i+\frac{1}{6}{\sum }_{j\ne i}\sin x_j$. Solving the system is equivalent to finding a fixed point of $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$. The mean value theorem implies $|\sin u-\sin v|\le |u-v|$. In the specified metric,Each coordinate difference occurs exactly twice. This metric makes ${\mathbb{R}}^3$ a nonempty complete metric space, because a Cauchy sequence in it is Cauchy in each real coordinate, and coordinate convergence implies convergence in the sum metric. Thus $F$ is a contraction mapping with constant $1/3$.

There is exactly one solution for every choice of $(a_1,a_2,a_3)$, by the contraction mapping theorem. Iteration of $F$ from any starting vector also constructs it.

A contraction mapping satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one fixed $0\le q<1$. The contraction mapping theorem states that such a map on a nonempty complete metric space has a unique fixed point and that every sequence of iterates converges to it.

Choose $x_0$ and put $x_{n+1}=T{x}_n$. Iteration gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$,Thus the iterates form a Cauchy sequence and converge by completeness to $x_{\ast }$. The Lipschitz bound gives continuity of $T$, hence $T{x}_{\ast }=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing $m\to \infty$ also gives the usual geometric error estimate. For $q=0$, convergence occurs after one application.

If $f^k$ is a contraction mapping, let $x_{\ast }$ be its unique fixed point. The commutation identity $f^{k}f=f{f}^k$ gives $f^k(f{x}_{\ast })=f{x}_{\ast }$, so uniqueness implies $f{x}_{\ast }=x_{\ast }$. Any fixed point of $f$ is a fixed point of $f^k$, giving uniqueness, without even assuming $f$ continuous. On the complete discrete metric space $\{0,1,2\}$, the map $f(0)=f(1)=0$, $f(2)=1$ has constant second iterate but is not a contraction: $d(f2,f0)=d(2,0)=1$.

For the specified function, $x+\sin x\ge 0$ on $[0,\infty )$ because it starts at zero and has derivative $1+\cos x\ge 0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfiesThe mean value theorem therefore makes it a contraction of constant at most $2/3$. Hence

A metric space is a complete metric space when every Cauchy sequence converges to a point of that space: for each $\epsilon >0$, the condition $d(x_m,x_n)<\epsilon$ for all sufficiently large $m,n$ must imply existence of $x$ with $d(x_n,x)\to 0$.

A Cauchy sequence in $[0,1]$ is also Cauchy in $\mathbb{R}$, so it converges there by the assumed completeness. Its limit remains between zero and one, since inequalities are preserved under limits. ThereforeThis is an instance of the fact that a closed subset of a complete metric space is complete with its induced metric.