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Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

The Banach fixed-point theorem states that a contraction mapping $T:X\to X$ of a nonempty complete metric space $(X,d)$ has a unique fixed point, and iterations of a map $x_{n+1}=T(x_n)$ converge to it from every $x_0\in X$. Precisely, assume a single constant $q\in [0,1)$ satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y\in X$.

Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. The triangle inequality and the geometric series therefore give, for $m>n$,Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x_{\ast }\in X$. The contraction mapping is a continuous map, so $T{x}_{\ast }={\lim }_{n}T{x}_n={\lim }_n{x}_{n+1}=x_{\ast }$. If $y_{\ast }$ is another fixed point, then $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing $x_{\ast }=y_{\ast }$. Letting $m\to \infty$ in the estimate also gives the contraction iteration error bound:If $q=0$, $T$ is constant and the sequence reaches its unique fixed point after the first step; this also removes any ambiguity about $0^0$ in the displayed estimate at $n=0$.