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For a function $F:(0,1)\to Z$ into a complete metric space, the displayed pairwise condition is equivalent to existence of a limit as $h\to 0^{+}$. Necessity follows from the triangle inequality. For sufficiency, the values $F(1/(n+1))$ form a Cauchy sequence, hence converge to some $z\in Z$. Given $\epsilon >0$, use the pairwise condition with $\epsilon /2$ and then send $n\to \infty$ in $d(F(h),F(1/(n+1)))<\epsilon /2$. This gives $d(F(h),z)\le \epsilon /2<\epsilon$ for every sufficiently small positive $h$.

If every factor is a complete metric space, the countable product metric is complete. For a product-Cauchy sequence, the inequality $2^{-i}\varphi (d_i(x_i^{(m)},x_i^{(n)}))\le D(x^{(m)},x^{(n)})$, with $\varphi (t)=t/(1+t)$, makes each coordinate sequence Cauchy. Its coordinate limits define a product point. Coordinatewise convergence and the uniformly small summable tail imply convergence to that point in $D$. It matters that the original distances are bounded before they are summed: summing arbitrary unbounded metrics directly need not be finite.

A Cauchy sequence in the uniform metric has a pointwise limit in the complete target, and the common Cauchy bound makes convergence uniform. The triangle inequality with one fixed continuous approximant shows that the limit is continuous. Thus the space of all continuous maps into a bounded complete metric space is complete, with no compactness assumption on the domain.

For any nonempty complete metric space $X$, the family of its nonempty closed bounded subsets is a complete metric space under the Hausdorff distance. Boundedness of the entire ambient space is unnecessary in this general form.

For a Cauchy sequence $(F_n)$ of these sets, define $A_n=\overline{{\bigcup }_{m\ge n}{F}_m}$ and ${\delta }_n={\sup }_{p,q\ge n}{d}_H(F_p,F_q)$. A Cauchy sequence in any metric space is bounded, so its tail sets are contained in a common finite-radius neighbourhood of one bounded $F_N$; hence every $A_n$ is bounded. The distance to a set is one-Lipschitz, so estimates pass to the closure, giving $d_H(F_n,A_n)\le {\delta }_n$ and $d_H(A_n,A_m)\le {\delta }_n$ for $m\ge n$. Choose $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. The Summably nested Hausdorff intersection theorem gives $A_{n_j}\to G\ne \varnothing$. The triangle inequality now gives $F_n\to G$, proving completeness.

A contraction mapping $T:X\to X$ on a nonempty complete metric space has exactly one fixed point $x_{\ast }$, and the sequence $x_{n+1}=T(x_n)$ converges to $x_{\ast }$ from every $x_0\in X$. The same constant $0\le L<1$ must satisfy $d(Tx,Ty)\le Ld(x,y)$ for all $x,y\in X$. The statement applies to metric spaces, including closed subsets of a Banach space; neither linearity nor a vector space structure is required.^[ref]

The force of the result is that existence, uniqueness, and a practical approximation procedure arrive together. Repeated application produces corrections bounded by a geometric series, hence a Cauchy sequence; completeness supplies its limit. The Lipschitz continuity of $T$ makes that limit a fixed point. Two different fixed points would have their positive distance multiplied by at most $L<1$ while remaining unchanged, which is impossible.

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]

For Picard iteration $y_{n+1}(x)=b+{\int }_a^{x}f(s,y_n(s))ds$, $y_0=b$, assume $|f|\le M$ and a Lipschitz constant $L$ on a rectangle containing every iterate. The first difference is at most $M|x-a|$. Integrating the preceding bound multiplied by $L$ proves the displayed estimate by mathematical induction. Its uniform bounds sum to a convergent exponential series for every finite $Lh$, so the Weierstrass M-test gives uniform convergence without requiring $Lh<1$.

For two arbitrary initial functions $y,z$ in a function set preserved by the Picard iteration operator $T$, the same induction givesThus some iterate is a contraction mapping on any fixed finite interval for which every iterate stays in that set. The iterated contraction argument gives a unique fixed point of $T$ on a nonempty complete metric space, without requiring $Lh<1$. For a rectangle with field bound $M$ and vertical radius $k$, $Mh\le k$ supplies the required self-mapping property.

For two contraction mappings of a nonempty complete metric space, their fixed points obey $d(x_f,x_g)\le q_{f}d(x_f,x_g)+d_{\infty }(f,g)$. Rearranging proves the displayed bound whenever the uniform metric is finite. It establishes continuity of the fixed-point map at each $f$ without assuming a contraction constant shared by all maps near $f$.

In the l2 sequence space, let $e_n$ have value one in coordinate $n$ and zero elsewhere. This orthonormal sequence is bounded: $\Vert e_n{\Vert }_2=1$. For distinct indices $n,m$, its terms obey $\Vert e_n-e_m{\Vert }_2=\sqrt{2}$. No subsequence can therefore be a Cauchy sequence, so none converges in the norm. Thus even a complete metric space can have a bounded sequence with no convergent subsequence. Finite dimension is essential to the usual Euclidean space extension, and in a general metric space the relevant replacement is compactness or total boundedness together with completeness.

If a self-map $T$ of a nonempty complete metric space has an iterate $T^n$ that is a contraction mapping, the contraction mapping theorem gives a unique fixed point $a$ of $T^n$. The maps commute, so $T^n(Ta)=T(T^{n}a)=Ta$, and uniqueness forces $Ta=a$. Conversely every fixed point of $T$ is fixed by $T^n$. Thus $a$ is the unique fixed point of $T$.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

For a prime number $p$, a p-adic number is an element of the field ${\mathbb{Q}}_p$ obtained by completing the rational numbers for the metric $d_p(x,y)=|x-y{|}_p$. Here $|\cdot {|}_p$ is the p-adic absolute value. The Cauchy-sequence construction of a metric completion supplies limits for every Cauchy sequence, and addition, multiplication, and the p-adic absolute value extend to this complete metric space. The rational numbers form a dense subset. ^[ref]

For a concrete example with $p=2$, consider the partial sums of a geometric series:The first values $1,3,7,15,31$ have distances $1/2,1/4,1/8,1/16,1/32$ from $-1$ in the 2-adic absolute value. ConsequentlyThis is a statement about convergence in a metric space with the 2-adic absolute value. In the ordinary real number absolute value, the same partial sums grow without bound. Choosing a metric changes which approximation errors shrink.

More generally every nonzero P-adic number has a unique digit expansionOnly finitely many negative powers occur, while positive powers can continue indefinitely. Written with the highest powers on the left, the digits extend infinitely to the left; for example the P-adic number $-1$ at $p=2$ is written $\dots 111_2$. The p-adic integers are exactly the elements with $|x{|}_p\le 1$, whose digit expansions use only nonnegative powers. ^[ref]

The construction can be understood through modular arithmetic. Successively choose the low-order digits to agree modulo $p$, $p^2$, and higher powers. After fixing digits below $p^n$, the remaining error has p-adic absolute value at most $p^{-n}$; the partial sums form a Cauchy sequence, so completeness supplies a limit. If two expansions first differ at position $j$, their difference has P-adic valuation $j$, proving uniqueness. This turns compatible finite congruence classes into one exact number. In particular, Hensel's lemma upgrades suitable approximate roots of a polynomial modulo powers of $p$ into roots in the p-adic integers.

For $0<\eta \le h$, work on the closed subsetIt is a nonempty complete metric space because it is closed in the Banach space of continuous functions with the supremum norm. Define the Picard iteration operatorChoose $\eta$ so that $M\eta \le k$ and $L\eta <1$, omitting a bound whose constant is zero. Such a positive $\eta$ exists. The first inequality makes $T$ a self-map, since $|(Ty)(x)-b|\le M|x-a|\le k$. The Lipschitz condition givesso the contraction mapping theorem yields a unique fixed point. The fundamental theorem of calculus makes it a differentiable solution of $y^{\prime }=f(x,y)$ with $y(a)=b$; conversely every solution staying in the rectangle satisfies the integral equation and is a fixed point. This is the local Picard-Lindelöf theorem construction. The interval-restricted space $C_{\eta }$ avoids treating functions defined on a shorter interval as elements of $C_h$.

The contraction mapping theorem states that a contraction mapping $T:S\to S$ on a nonempty complete metric space has a unique fixed point. Explicitly, if $d(Tu,Tv)\le qd(u,v)$ for all $u,v\in S$ and one fixed $0\le q<1$, then the iterates $u_{n+1}=T{u}_n$ converge to that fixed point from every $u_0\in S$. Both self-mapping and completeness are needed; the set need not be a vector space.

For the direct contraction mapping theorem proof with the usual supremum norm, take the complete metric spaceand the Picard iteration map $Tz(x)=b+{\int }_a^{x}f(s,z(s))ds$. The estimates areThus this proof requiresThe first inequality makes $T$ preserve $\mathcal{B}$; the strict second inequality makes it a contraction mapping. Its unique fixed point is the desired solution of a differential equation.

This is more restrictive than the preceding Weierstrass M-test argument: a single application of $T$ must shrink every difference in the supremum norm, whereas repeated integration in the other proof produces the factorial $(n+1)!$ and convergence even when $Lh\ge 1$. If $L=0$, the additional condition is automatic. These are conditions for the two stated direct proofs; they are not necessary bounds on the actual existence interval.

The contraction mapping theorem states that a self-map $T$ of a nonempty complete metric space with a uniform contraction constant $q<1$ has a unique fixed point $p$. For every starting point, the iteration $x_{n+1}=T{x}_n$ converges to $p$, with $d(x_n,p)\le q^{n}d(x_1,x_0)/(1-q)$.

For the cosine fixed-point iteration, any real $x_0$ gives $x_1\in [-1,1]$. This closed interval is complete, and cosine maps it into $[\cos 1,1]\subseteq [-1,1]$. The mean value theorem giveswith $\sin (1)<1$. Apply the contraction mapping theorem to the tail starting at $x_1$. Therefore $\boxed{x_n\to p,p=\cos p}$ for every real initial value. Numerically $p\approx 0.739085$. The global real-line derivative bound is only one; entering the invariant interval supplies the strict uniform bound needed by the theorem.

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 $n=0$.

Take $S=T(B_X)$. A bounded linear operator is compact exactly when the closure of this set is compact. In a complete metric space this is equivalent to its being totally bounded.

If $T$ is compact, choose a finite $\epsilon$-net $z_1,\dots ,z_m$ for $T(B_X)$ and let $Y$ be their span. Then $\mathrm{dist}(Tx,Y)<\epsilon$ for $\Vert x\Vert \le 1$, giving $\Vert {\pi }_{Y}T\Vert \le \epsilon$.

Conversely, fix $\delta >0$ and take $Y$ with $\Vert {\pi }_{Y}T\Vert \le \delta /4$. For every $x\in B_X$ choose $y_x\in Y$ with $\Vert Tx-y_x\Vert <\delta /2$. These approximants satisfy $\Vert y_x\Vert \le \Vert T\Vert +\delta /2$, so form a bounded subset of a finite-dimensional space and have a finite $\delta /2$-net. The same centers give a finite $\delta$-net for $T(B_X)$. Thus it is totally bounded and $T$ is compact. In particular

The Baire category theorem says that a nonempty complete metric space is not a countable union of closed sets with empty interior; equivalently, every countable intersection of open dense sets is dense. To prove the latter, start with any nonempty open set $V$ and open dense sets $G_n$. Choose closed balls of positive radii withEach choice is possible by density and openness. Their centers form a Cauchy sequence; its limit exists by completeness and belongs to every closed ball, hence to $V\cap \bigcap G_n$. This proves density and the theorem.

The Uniform boundedness principle says that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Z$ has uniformly bounded operator norms. Put $E_m=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\Vert h\Vert <r$, the triangle inequality applied to $T(x_0+h)-T{x}_0$ gives $\Vert Th\Vert \le 2m$. Taking $h=(r/2)u$ for unit $u$ proves ${\sup }_T\Vert T\Vert \le 4m/r$.

For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_{k}x\to x$, the norms $\Vert P_{k}x\Vert$ form a bounded sequence with limit $\Vert x\Vert$. Also, with $P_0=0$,This proves both preliminary bounds. The supremum therefore defines a finite norm $\Vert \cdot {\Vert }^{\prime }$, with $\Vert x\Vert \le \Vert x{\Vert }^{\prime }$.

To prove (i) implies (ii), the identity map from $(X,\Vert \cdot {\Vert }^{\prime })$ to $(X,\Vert \cdot \Vert )$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the bounded inverse theorem makes its inverse bounded: $\Vert x{\Vert }^{\prime }\le C\Vert x\Vert$. This is precisely (ii).

The contraction mapping theorem states that a self-map $f$ of a nonempty complete metric space $(X,d)$ satisfying $d(fu,fv)\le qd(u,v)$ for a fixed $0\le q<1$ has a unique fixed point. Its iterates converge to that point from every starting point.

To prove it, set $u_{n+1}=f(u_n)$. Induction gives $d(u_{n+1},u_n)\le q^{n}d(u_1,u_0)$. Hence for $m>n$, the triangle inequality yieldsThus the iterates are a Cauchy sequence, with limit $u$ by completeness. The contraction inequality gives $d(fu,u)\le qd(u,u_n)+d(u_{n+1},u)\to 0$, so $f(u)=u$. For any two fixed points $u,v$, $d(u,v)\le qd(u,v)$ forces equality of the points. This proves existence, uniqueness and convergence; passing $m\to \infty$ also gives the displayed error bound.

For the three-point example, positivity and symmetry of $d^{\prime }$ are immediate. For three distinct points, the largest side has length two, no more than the sum of the other two; repeated-point triangle inequalities are trivial. Thus $d^{\prime }$ is a metric. The discrete metric satisfieswhich proves bilipschitz equivalence of the two metrics. DefineThe ratios of output to input distances for the three distinct pairs are $1/2,1/2,0$ in $d^{\prime }$, so it is a contraction with $q=1/2$. In $d$, the pair $(x,y)$ has input and output distance one, precluding any $q<1$. This illustrates the contraction property under equivalent metrics.