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A Cauchy sequence in a metric space satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. Suppose a subsequence $x_{n_k}$ converges to $x\in X$. Choose $N$ so that the Cauchy bound is $\epsilon /2$, and then choose $k$ with $n_k\ge N$ and $d(x_{n_k},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole Cauchy sequence converges to the subsequential limit.

Given $\epsilon >0$, the Cauchy sequence property gives $N$ with $d(x_m,x_n)<\epsilon /2$ for all $m,n\ge N$. Fixing $m\ge N$ and letting $n\to \infty$ yields $0\le d_m\le \epsilon /2<\epsilon$. Therefore

For fixed $m$, the reverse triangle inequality givesThe right side tends uniformly to zero as $n,k\to \infty$ because $(x_n)$ is a Cauchy sequence. Hence $(d(x_m,x_n){)}_n$ is a real Cauchy sequence. By completeness of $\mathbb{R}$, it has a finite, nonnegative limit:This conclusion uses completeness only of the real line, without assuming that $X$ is complete.

We argue by contradiction using the previous subsequence construction. If $X$ is not complete, choose a Cauchy sequence with no limit in $X$. None of its $d_m$ can be zero: if $d_m=0$, then $d(x_m,x_n)\to 0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ withLet $Y=\{y_1,y_2,\dots \}$. This subset is closed in $X$. Indeed, if a point $x\notin Y$ lay in its closure, there would be points of $Y$ tending to $x$. Their indices must be unbounded, because the distance from $x$ to any finite subset of $Y$ is positive. Passing to increasing indices gives a convergent subsequence of the original Cauchy sequence. Part (a) would then make the whole sequence converge to $x$, a contradiction.

Now define $F:Y\to Y$ by $F(y_j)=y_{j+1}$. Distinctness makes this well-defined, and the displayed inequality makes it a contraction mapping with constant at most $1/2$. It has no fixed point, contradicting the assumed property of closed subsets. ConsequentlyThis is the converse direction of contraction property of closed subsets characterizes completeness.

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.

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

For each $x\in X$, density of $Y$ allows a sequence $y_n\in Y$ with $d(y_n,x)<1/n$. If $f_1,f_2$ are two continuous extensions, continuity givesThus a continuous extension is unique if it exists.

Now assume $g$ is uniformly continuous. Each chosen $y_n\to x$ is a Cauchy sequence, so part (a), applied to $Y$, shows that $g(y_n)$ converges in $\mathbb{R}$. DefineThis is independent of the choice: for another $z_n\to x$ in $Y$, $d(y_n,z_n)\to 0$, and uniform continuity gives $g(y_n)-g(z_n)\to 0$. At $x\in Y$, comparison with the constant sequence $x$ proves $f(x)=g(x)$.

To prove uniform continuity of the extension, choose $\delta >0$ so that $|g(y)-g(z)|<\epsilon /2$ whenever $y,z\in Y$ and $d(y,z)<\delta$. If $d(x,x^{\prime })<\delta /3$, choose sequences $y_n\to x$ and $z_n\to x^{\prime }$ in $Y$. For large $n$, both approximation distances are less than $\delta /3$, so $d(y_n,z_n)<\delta$. Passing to the limit gives $|f(x)-f(x^{\prime })|\le \epsilon /2<\epsilon$. The same $\delta /3$ works for all $x,x^{\prime }$. Therefore the extension exists and is uniformly continuous. No completeness of $X$ is needed; completeness of the target $\mathbb{R}$ is the relevant requirement.

The contraction mapping theorem states that a contraction mapping of a nonempty complete metric space into itself has a unique fixed point $x_{\ast }$. For any starting point $x_0$, the iterates $x_{n+1}=f(x_n)$ converge to $x_{\ast }$, with the quantitative estimateCompleteness and the self-map condition are essential: the geometric estimate makes the iterates a Cauchy sequence, completeness supplies their limit in $X$, and continuity of the contraction mapping makes that limit a fixed point. If two fixed points existed, their distance would be at most $q$ times itself, forcing equality of the points.

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

Let $x_n=(\pi /2+2\pi n{)}^{-1}$. The points $p_n=(x_n,1)$ lie on the specified graph and converge in ${\mathbb{R}}^2$ to $(0,1)$. They form a Cauchy sequence in the induced metric, but their only possible limit is absent from the space. Adding $(0,0)$ does not supply that missing limit. Hence the specified space is not complete.

For the unheaded request on $\mathbb{R}$, use the arctangent pullback metricPositivity, symmetry and the triangle inequality follow from the ordinary distance, and injectivity of $\arctan$ gives $d(x,y)=0$ only if $x=y$. The sequence $n$ is Cauchy for this metric, because $\arctan n\to \pi /2$, but cannot converge to any finite real $x$: that would require $\arctan x=\pi /2$. Thus this metric makes $\mathbb{R}$ incomplete, even though it induces its usual topology.

The rational sequence $q_n=10^{-n}\lfloor 10^n\sqrt{2}\rfloor$ satisfies $0\le \sqrt{2}-q_n<10^{-n}$. It is therefore a Cauchy sequence in the induced metric on $\mathbb{Q}$. Any limit in $\mathbb{Q}$ would also be its real limit, which is $\sqrt{2}\notin \mathbb{Q}$. ThusCompleteness concerns limits lying in the space, so being a dense subset of $\mathbb{R}$ does not suffice.

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.

Put $M=\mathcal{N}(K{)}^{\perp }$ for a bounded linear operator between Hilbert spaces. Its restriction $K{|}_M:M\to \mathcal{R}(K)$ is bijective. Define the Moore–Penrose inverse of an operator byThe inverse vanishes on $\mathcal{R}(K{)}^{\perp }$. For each datum in this domain, $K^{†}f$ is the minimum-norm least-squares solution; all least-squares solutions are $K^{†}f+\mathcal{N}(K)$. On their natural domains,The equivalent continuity condition isIf the range is closed, the inverse restriction is bounded by the bounded inverse theorem. Conversely, if $\Vert K^{†}f\Vert \le C\Vert f\Vert$ on its domain and $K{u}_n$ converges, replace $u_n$ by $P_M{u}_n$. The inequality makes these preimages a Cauchy sequence, whose limit maps to the limiting datum; the range is therefore closed. Boundedness here uses the data norm, not a graph norm that would conceal instability.

The Baire category theorem says that in a nonempty complete metric space, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty open set $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline{B}(x_n,r_n)$ with $0<r_n\le 2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a Cauchy sequence; completeness supplies a limit lying in every closed ball, hence in $O\cap {\bigcap }_n{G}_n$. The proof also works at isolated points, where sufficiently small balls are singletons.

The incomplete metric space $\mathbb{Q}$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.

For the sequence in the problem, putEach $E_N$ is closed by continuity. Pointwise convergence makes each point's values a Cauchy sequence, so $[0,1]={\bigcup }_N{E}_N$. By the Baire category theorem, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.

If continuous $g_n$ converged pointwise to the rationality indicator, take $ϵ=1/4$ and let $m\to \infty$ in this bound. On $J$, $|g_n-g|\le 1/4$ for $n\ge N$. Hence a fixed continuous $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts continuity. Thus the rationality indicator is not a pointwise limit of continuous functions.

Nevertheless $h_n(x)=x^n$ are continuous and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This limit is discontinuous at $1$: a pointwise limit of continuous functions can be discontinuous.

A complete metric space is one in which every Cauchy sequence converges. On $I=(0,1]$ defineThe map $x\mapsto 1/x$ is an isometry from $(I,d)$ onto the closed subset $[1,\infty )$ of $\mathbb{R}$, so $d$ is a metric and is complete. That map and its inverse are continuous in the ordinary Euclidean metrics, so it is also a homeomorphism. Therefore $d$ induces exactly the usual subspace topology on $I$.