Completeness, Completion, and Uniform Continuity

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_k)$ be a sequence in $X$, that is a function $\mathbb{N}\to X$ written $x_k:=x(k)$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$, Sequences of reals: bounded, eventually, frequently, tails, subsequences). As everywhere in this library $\mathbb{N}$ contains $0$, so a sequence is indexed from $0$.

$(x_k)$ is a Cauchy sequence in $(X,d)$ if for every rational $\epsilon >0$ there is $K\in \mathbb{N}$ such that

$d(x_m,x_n)<\epsilon \text{for all }m,n\ge K\mathrm{.}$

Rational and real $\epsilon$ agree here. The test is written with a rational $\epsilon$ to match Limits and Cauchy sequences of reals and Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$, and nothing is lost by using a real one: below any real $\eta >0$ lies a positive rational (The rationals embed densely in the reals), and the index belonging to that rational serves for $\eta$. So a proof may establish Cauchyness by producing an index for every real $\epsilon >0$, and may use a Cauchy hypothesis at a real $\epsilon$ by first passing to a rational below it. Both moves are used on this page and are always cited.

The condition is exactly $d(x_m,x_n)\to 0$ as $m$ and $n$ grow independently. The distances $d(x_m,x_n)$ are nonnegative reals (Nonnegativity of a metric is a consequence of the other axioms, not an axiom), and the displayed condition asks them to be uniformly small on a tail of the doubly indexed family. It is not the same as $d(x_{k+1},x_k)\to 0$, which is a strictly weaker condition and is a standing source of error. The partial sums $H_n$ of the harmonic series separate the two: consecutive ones differ by $1\mathrm{/}(n+1)$, which tends to $0$, while the sequence is unbounded, and an unbounded sequence of reals is not Cauchy (Every Cauchy sequence in a metric space is bounded).

Consistency with the real line. For $X=\mathbb{R}$ with the usual metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded) the condition above reads $\mathrm{\mid }{x}_m-x_n\mathrm{\mid }<\epsilon$ for $m,n\ge K$, which is verbatim the definition of a Cauchy sequence of reals (Limits and Cauchy sequences of reals). So the notion introduced here restricts on $\mathbb{R}$ to the one already in use, and no ambiguity is created.

• A Cauchy sequence need not converge. The definition mentions no candidate limit, and that is the whole point of it: it is a condition on the sequence alone, testable without knowing where the sequence is going. Whether every Cauchy sequence converges is a property of the space, namely completeness (Complete metric space: every Cauchy sequence converges in the space), and it genuinely fails in some spaces (FALSE: every Cauchy sequence in a metric space converges).
• Cauchyness is a property of the metric, not of the topology. Two metrics on the same set may have exactly the same open sets and different Cauchy sequences (FALSE: two metrics inducing the same topology have the same Cauchy sequences). What does preserve Cauchy sequences is uniform equivalence (Topologically, uniformly and Lipschitz equivalent metrics on a set), and the reason is A uniformly continuous map sends Cauchy sequences to Cauchy sequences.
• Every subsequence of a Cauchy sequence is Cauchy, since a strictly increasing index map satisfies $n_j\ge j$ (A strictly increasing index map satisfies $n_k\ge k$), so the same $K$ works for the subsequence. Conversely a Cauchy sequence with one convergent subsequence already converges (A Cauchy sequence in a metric space with a convergent subsequence converges to that subsequence’s limit).

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_k)$ be a sequence in $X$ converging to $p\in X$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$). Then $(x_k)$ is Cauchy in $(X,d)$ (Cauchy sequence in a metric space).

The converse fails, and that failure is the subject of this page (FALSE: every Cauchy sequence in a metric space converges).

• The proof spends the triangle inequality and symmetry, but not separation. Symmetry rewrites $d(p,x_n)$ as the bounded quantity $d(x_n,p)$ in step 2.1. The separation axiom (M1) is not used, so the same argument shows that a sequence converging in a pseudometric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) is Cauchy for the pseudometric.
• Halving is the whole idea. The Cauchy condition compares two terms of the sequence, and a limit compares one term with the limit; routing $m$ and $n$ through $p$ costs two applications of the convergence hypothesis, so each is run at half the target. Every proof on this page that produces a Cauchy sequence out of a convergent one repeats this step.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_k)$ be a Cauchy sequence in $X$ (Cauchy sequence in a metric space). Then its range $R:=\{x_k:k\in \mathbb{N}\}$ is a bounded subset of $X$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): there are a point $x_0\in X$ and a real $r>0$ with $R\subseteq B(x_0,r)$ (Open ball, closed ball and sphere in a metric space).

Consequently $R$ is nonempty and bounded, so $\mathrm{diam}(R)$ exists (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

• The maximum is taken over $j\le N$ and not over $j<N$, and the extra element $1$ is in the set as well. Both are deliberate. Indices run from $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so $N=0$ is possible and $\{d(x_j,x_N):j<N\}$ would then be empty, which has no maximum (Maximum and minimum of a set); adjoining $1$ makes the set nonempty in every case and simultaneously covers the tail bound.
• Boundedness of the range is strictly weaker than Cauchyness. The sequence $x_k=(-1{)}^k$ in $\mathbb{R}$ has bounded range and is not Cauchy, since two consecutive terms are always at distance $2$. So this lemma cannot be reversed, and nothing on this page reverses it.
• What the lemma is for. It is what makes $\mathrm{diam}$ available for the tails of a Cauchy sequence, which is the form in which Cauchyness enters the converse half of In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), let $(x_k)$ be a Cauchy sequence in $X$ (Cauchy sequence in a metric space), and suppose some subsequence $(x_{n_j})$ converges to $p\in X$, that is $p$ is a subsequential limit of $(x_k)$ in $(X,d)$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$, Subsequential limit of a real sequence, and the subsequential limit set). Then the whole sequence converges to $p$.

In particular a Cauchy sequence has at most one subsequential limit, and it has one exactly when it converges.

• Where $n_j\ge j$ is used. In step 2.1, and quoted again in step 3.1, and both places matter: it puts the single chosen subsequence term $x_{n_K}$ beyond the Cauchy index $K_1$, and it is what lets one index $K$ control the subsequence and the sequence at once. Without it the subsequence could in principle look only at small indices, and the Cauchy hypothesis would never be reached.
• Only one term of the subsequence is used. The proof does not pass to the limit along the subsequence; it fixes the single index $n_K$ and routes every later term of the whole sequence through it. That is why no further hypothesis on the subsequence is needed.
• This is the lemma that turns compactness into completeness. A space in which every sequence has a convergent subsequence is complete (Complete metric space: every Cauchy sequence converges in the space) by this lemma alone, applied to a Cauchy sequence. The converse is false, not merely harder: $\mathbb{R}$ is complete ($\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$) and the sequence $x_k=k$ has no convergent subsequence, its terms being at pairwise distance at least $1$ so that no subsequence is Cauchy. What does hold is that a complete space in which every sequence has a convergent subsequence is exactly a complete and totally bounded one, and total boundedness is a later page of this library.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric).

$(X,d)$ is complete if every Cauchy sequence in $(X,d)$ (Cauchy sequence in a metric space) converges to a point of $X$ (Convergence of a sequence in a metric space: $x_k\to x$ iff $d(x_k,x)\to 0$ in $\mathbb{R}$).

A subset $A\subseteq X$ is called complete when the metric subspace $(A,d_A)$ is complete (Isometry, isometric embedding, and the subspace metric on a subset); as always, the metric is part of the data, and $d_A$ is the restriction of $d$ to $A\times A$.

The limit is unique when it exists, since limits in a metric space are unique (A sequence in a metric space has at most one limit), so a complete space assigns to each of its Cauchy sequences one point and not a set of points.

Completeness is a property of the pair $(X,d)$, not of $X$ and not of the topology of $d$. Both quantifiers in the definition are about the metric: the Cauchy condition is stated with distances, and so is convergence. Two metrics on the same set can have the same open sets while exactly one of them is complete, which is the content of FALSE: completeness of a metric space is determined by its topology and its witness. Read the word complete as an abbreviation for complete with respect to this metric, always.

• Do not confuse this with Dedekind completeness. The least-upper-bound property of Complete ordered field (least-upper-bound property) is an order condition on an ordered field and is what defines $\mathbb{R}$; the condition here is a metric condition and makes sense in any metric space, with no order in sight. On $\mathbb{R}$ the first implies the second (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, $\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$) and the two are not the same statement: the rationals with the usual metric are an ordered field that fails both, while there are complete metric spaces with no field structure at all.
• Every convergent sequence is Cauchy (Every convergent sequence in a metric space is Cauchy), so completeness is exactly the assertion that the two classes of sequences coincide. It is the converse inclusion that carries all the content.
• Three sources of completeness are proved on this page. The real line and ${\mathbb{R}}^n$ are complete ($\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$); a closed subset of a complete space is complete (Closed subspaces of complete metric spaces are complete; the converse under countable choice); and, under Countable Choice, every metric space sits densely and isometrically inside a complete one (Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences).

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $A\subseteq X$ carry the subspace metric $d_A$ (Isometry, isometric embedding, and the subspace metric on a subset). Then:

1. Under countable choice ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), if $(A,d_A)$ is complete (Complete metric space: every Cauchy sequence converges in the space), then $A$ is closed in $(X,d)$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). No completeness hypothesis on $X$ is needed.
2. In ZF, without a choice axiom, if $(X,d)$ is complete and $A$ is closed in $(X,d)$, then $(A,d_A)$ is complete.

Consequently, under countable choice, a subspace of a complete metric space is complete if and only if it is closed.

The following form of the first direction is also choice-free: a complete subspace contains the ambient limit of every convergent sequence of its points. Hence it is closed whenever every point of its ambient closure is already known to be the limit of a sequence from that subspace. This last condition is pointwise existence, not a chosen family of sequences.

• Claim 1 does not need ambient completeness. Under countable choice, it applies in any ambient metric space. The comparison of two limits is choice-free; producing an approximating sequence from arbitrary adherence is the step requiring the stated assumption.
• Both directions are genuinely about the metric. $A$ closed and $X$ complete are hypotheses about $(X,d)$; replacing $d$ by a topologically equivalent metric preserves closedness and can destroy completeness (FALSE: completeness of a metric space is determined by its topology), so no reading of this theorem survives the passage to the bare topology.
• Where choice enters. Only in claim 1, and only through A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, whose forward direction spends ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) to manufacture a sequence out of adherence. Claim 2 uses the choice-free direction of that theorem.
• The standard application. A closed interval, a closed ball, or any closed subset of ${\mathbb{R}}^n$ is a complete metric space, because ${\mathbb{R}}^n$ is ($\mathbb{R}$ and ${\mathbb{R}}^n$ for $n\ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$). Every appeal to Banach's fixed point theorem on a closed subset of ${\mathbb{R}}^n$ passes through this remark.

1. $(\mathbb{R},d_{\mathbb{R}})$ with the usual metric $d_{\mathbb{R}}(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded) is a complete metric space (Complete metric space: every Cauchy sequence converges in the space).
2. Let $n\in \mathbb{N}$ with $n\ge 1$ and let $d_2$ be the Euclidean metric on ${\mathbb{R}}^n$ (${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it). Then $({\mathbb{R}}^n,d_2)$ is complete.

The hypothesis $n\ge 1$ is inherited and is not decoration. ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it defines ${\mathbb{R}}^n$ and its three metrics only for $n\ge 1$, because at $n=0$ the metric $d_{\mathrm{\infty }}$ would be a maximum over the empty index set. Every statement about ${\mathbb{R}}^n$ in this library carries the hypothesis, and this one does too.

• The proof is the Cauchy criterion plus two inequalities. Step 1.2 says a coordinate difference is at most the Euclidean distance, which turns a Cauchy sequence of points into $n$ Cauchy sequences of reals; step 1.3 says that coordinates uniformly below $\epsilon \mathrm{/}(2n)$ force the Euclidean distance below $\epsilon$, which turns $n$ convergent coordinate sequences back into one convergent sequence of points. Nothing else about ${\mathbb{R}}^n$ is used, and in particular the Cauchy-Schwarz inequality is not needed here.
• The same two inequalities hold for $d_1$ and $d_{\mathrm{\infty }}$, with the same proof of completeness. For $d_1(x,y)={\sum }_{k<n}\mathrm{\mid }{x}_k-y_k\mathrm{\mid }$: each term is at most the sum (Laws of finite sums and finite products), so $\mathrm{\mid }{x}_j-y_j\mathrm{\mid }\le d_1(x,y)$; and $\mathrm{\mid }{x}_j-y_j\mathrm{\mid }<\epsilon \mathrm{/}(2n)$ for all $j<n$ gives $d_1(x,y)\le n\epsilon \mathrm{/}(2n)=\epsilon \mathrm{/}2<\epsilon$. For $d_{\mathrm{\infty }}(x,y)=\max \{\mathrm{\mid }{x}_k-y_k\mathrm{\mid }:k<n\}$: the maximum dominates each entry and is one of them (Every nonempty finite set of reals has a maximum and a minimum), so $\mathrm{\mid }{x}_j-y_j\mathrm{\mid }\le d_{\mathrm{\infty }}(x,y)$, and entries all below $\epsilon \mathrm{/}2$ make the maximum at most $\epsilon \mathrm{/}2$. Substituting either pair of inequalities for steps 1.2 and 1.3 leaves the rest of the proof unchanged, so $({\mathbb{R}}^n,d_1)$ and $({\mathbb{R}}^n,d_{\mathrm{\infty }})$ are complete as well. Nothing later on this page uses that.
• No choice is spent. The limit point $y$ is assembled coordinatewise in step 3.1 from limits that are unique, and the finitely many indices $K_j$ of step 3.2 are made canonical by taking the least one. This matters because completeness proofs elsewhere on this page do spend ${\mathrm{A}\mathrm{C}}_{\omega }$, and the contrast is worth keeping visible.
• Where the least-upper-bound property is. Entirely inside The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges. This theorem is a transfer result: it moves completeness from $\mathbb{R}$ to ${\mathbb{R}}^n$ and adds no new content about the reals.

Assume the Axiom of Countable Choice for assertion 1. Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric). Call a sequence $(F_k{)}_{k\in \mathbb{N}}$ of subsets of $X$ a Cantor chain if every $F_k$ is nonempty, closed (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and bounded, $F_{k+1}\subseteq F_k$ for every $k$, and $\mathrm{diam}(F_k)\to 0$ in $\mathbb{R}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Limits and Cauchy sequences of reals). Then:

1. If $(X,d)$ is complete (Complete metric space: every Cauchy sequence converges in the space), every Cantor chain in $X$ has an intersection ${\bigcap }_{k\in \mathbb{N}}{F}_k$ with exactly one element.
2. Conversely, if every Cantor chain in $X$ has nonempty intersection, then $(X,d)$ is complete.

Boundedness of each $F_k$ is part of the definition of a Cantor chain because $\mathrm{diam}$ is defined for nonempty bounded sets only in this library (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space); it is not an extra hypothesis but the precondition for writing the diameter condition down.