TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); completions of it exist (Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences, A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace). Then:

Universal property. Let $((\hat{X}, \hat{d}), ι)$ be a completion of $(X, d)$ , let $(Z, d_{Z})$ be a complete metric space (Complete metric space: every Cauchy sequence converges in the space) and let $f : X \to Z$ be uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point). Then there is exactly one continuous $F : \hat{X} \to Z$ with $F \circ ι = f$ , and that $F$ is uniformly continuous.
Uniqueness of the completion. Let $(({\hat{X}}_{1}, {\hat{d}}_{1}), ι_{1})$ and $(({\hat{X}}_{2}, {\hat{d}}_{2}), ι_{2})$ be completions of $(X, d)$ . Then there is exactly one continuous $φ : {\hat{X}}_{1} \to {\hat{X}}_{2}$ with $φ \circ ι_{1} = ι_{2}$ , and that $φ$ is an isometry (Isometry, isometric embedding, and the subspace metric on a subset).

So a completion is determined by $(X, d)$ up to a unique isometry compatible with the embeddings, which is what licenses the phrase the completion from here on.

Facts & Assumptions

Given: The Axiom of Countable Choice; a metric space $(X, d)$ ; completions $((\hat{X}, \hat{d}), ι)$ , $(({\hat{X}}_{1}, {\hat{d}}_{1}), ι_{1})$ and $(({\hat{X}}_{2}, {\hat{d}}_{2}), ι_{2})$ of it; a complete metric space $(Z, d_{Z})$ ; a uniformly continuous $f : X \to Z$ ; a real $ε > 0$ .

[A1]

Completion: $\hat{X}$ is complete, $ι$ is an isometric embedding, and $ι [X]$ is dense in $\hat{X}$ (A completion of a metric space: a complete metric space together with an isometric embedding onto a dense subspace, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[A2]

Uniform continuity of $f$ : one $δ > 0$ per $ε > 0$ serving every pair (Uniform continuity of a map of metric spaces: one $δ$ serving every point).

[L1]

An isometric embedding is injective and is an isometry onto its image, whose subspace metric is the restriction (An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image, Isometry, isometric embedding, and the subspace metric on a subset, Injection, surjection, bijection).

[L2]

Extension from a dense subspace: a uniformly continuous map from a dense subspace of a metric space into a complete metric space has a uniformly continuous extension to the whole space, and it is the only continuous one (A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space).

[L3]

A point of the closure is a limit of a sequence from the set; this direction spends ${A C}_{ω}$ (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, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[L4]

A continuous map is sequentially continuous (Metric continuity characterisations, with countable choice for the sequential converse, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

[L5]

Quadrilateral estimate: $∣ d (u, v) - d (u^{'}, v^{'}) ∣ \leq d (u, u^{'}) + d (v, v^{'})$ , from the reverse triangle inequality and the triangle inequality for the absolute value (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space, Basic properties of the absolute value).

[L6]

Limits of real sequences are unique (A sequence has at most one limit, Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1A1L1

By [L1] the map $ι$ is an isometry of $X$ onto the subspace $ι [X]$ of $\hat{X}$ , so its inverse $ι^{- 1} : ι [X] \to X$ is an isometry and $\hat{d} (u, v) = d (ι^{- 1} (u), ι^{- 1} (v))$ for all $u, v \in ι [X]$ .

2.1step 1.1A2

Hence $f \circ ι^{- 1} : ι [X] \to Z$ is uniformly continuous: the $δ$ that [A2] supplies for $ε$ also serves here, since $\hat{d} (u, v) < δ$ gives $d (ι^{- 1} (u), ι^{- 1} (v)) < δ$ and hence $d_{Z} (f (ι^{- 1} (u)), f (ι^{- 1} (v))) < ε$ .

3.1step 2.1A1L2

$ι [X]$ is dense in $\hat{X}$ and $Z$ is complete, so [L2] gives a uniformly continuous $F : \hat{X} \to Z$ extending $f \circ ι^{- 1}$ , and $F$ is the only continuous map $\hat{X} \to Z$ that does so.

4.1step 3.1L1

$F \circ ι = f$ , since $F (ι (x)) = f (ι^{- 1} (ι (x))) = f (x)$ for every $x \in X$ ; and if $G : \hat{X} \to Z$ is continuous with $G \circ ι = f$ then $G$ agrees with $f \circ ι^{- 1}$ on $ι [X]$ , so $G = F$ by the uniqueness in step 3.1. This is claim 1.

5.1step 4.1A1

For claim 2, note that $ι_{2} : X \to {\hat{X}}_{2}$ is an isometric embedding, hence uniformly continuous with $δ = ε$ , and ${\hat{X}}_{2}$ is complete. Claim 1, applied to the completion $(({\hat{X}}_{1}, {\hat{d}}_{1}), ι_{1})$ with $Z = {\hat{X}}_{2}$ and $f = ι_{2}$ , yields exactly one continuous $φ : {\hat{X}}_{1} \to {\hat{X}}_{2}$ with $φ \circ ι_{1} = ι_{2}$ , and $φ$ is uniformly continuous.

6.1step 5.1

Symmetrically there is exactly one continuous $ψ : {\hat{X}}_{2} \to {\hat{X}}_{1}$ with $ψ \circ ι_{2} = ι_{1}$ , and it is uniformly continuous.

6.2step 5.1A1L3L4

Let $u, v \in {\hat{X}}_{1}$ . Density of $ι_{1} [X]$ and [L3] supply sequences $(p_{k})$ and $(q_{k})$ in $X$ with $ι_{1} (p_{k}) \to u$ and $ι_{1} (q_{k}) \to v$ in ${\hat{X}}_{1}$ ; by continuity of $φ$ and $φ \circ ι_{1} = ι_{2}$ we get $ι_{2} (p_{k}) \to φ (u)$ and $ι_{2} (q_{k}) \to φ (v)$ in ${\hat{X}}_{2}$ .

7.1step 4.1step 5.1step 6.1

$ψ \circ φ : {\hat{X}}_{1} \to {\hat{X}}_{1}$ is continuous and satisfies $(ψ \circ φ) \circ ι_{1} = ψ \circ ι_{2} = ι_{1}$ ; the identity of ${\hat{X}}_{1}$ is continuous and satisfies the same identity; so by the uniqueness in claim 1, applied with $Z = {\hat{X}}_{1}$ and $f = ι_{1}$ , we get $ψ \circ φ = i d$ . Symmetrically $φ \circ ψ = i d$ , so $φ$ is a bijection with inverse $ψ$ .

7.2step 6.2A1L5L6

By [L5] the real sequence $({\hat{d}}_{1} (ι_{1} (p_{k}), ι_{1} (q_{k})))_{k}$ converges to ${\hat{d}}_{1} (u, v)$ and $({\hat{d}}_{2} (ι_{2} (p_{k}), ι_{2} (q_{k})))_{k}$ converges to ${\hat{d}}_{2} (φ (u), φ (v))$ ; but the two sequences are equal termwise, both being $d (p_{k}, q_{k})$ because $ι_{1}$ and $ι_{2}$ are isometric embeddings. Hence the limits agree and ${\hat{d}}_{2} (φ (u), φ (v)) = {\hat{d}}_{1} (u, v)$ .

8.1step 4.1step 7.1step 7.2∎

So $φ$ is a bijective isometric embedding, that is an isometry, and it is the only continuous map with $φ \circ ι_{1} = ι_{2}$ ; this is claim 2, and claim 1 is step 4.1.

Remarks

Claim 1 is the reason claim 2 is short. Once a completion is known to receive every uniformly continuous map into a complete space, in exactly one way, two completions each receive the other's embedding, and the two induced maps are mutually inverse for the same uniqueness reason. No property of the particular construction of Every metric space has a completion, constructed as the equivalence classes of its Cauchy sequences is used anywhere in this proof.
Uniform continuity is not decoration in claim 1. A merely continuous $f : X \to Z$ need not extend at all. Take $X = (0, 1)$ with the usual metric, which is dense in the complete space $[0, 1]$ , so that $[0, 1]$ with the inclusion is a completion of it; take $Z = R$ , which is complete ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ); and take $f (x) = 1 / x$ , which is continuous on $(0, 1)$ and not uniformly continuous ( $x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one ↗). No continuous $F : [0, 1] \to R$ extends it: the points $1 / (k + 2)$ converge to $0$ in $[0, 1]$ , so sequential continuity would force $F (1 / (k + 2)) = k + 2$ to converge to $F (0)$ , and the sequence $(k + 2)$ is unbounded, hence not convergent (Metric continuity characterisations, with countable choice for the sequential converse, Every complete ordered field is Archimedean).
Uniqueness is up to a unique isometry, not up to equality, and the compatibility condition $φ \circ ι_{1} = ι_{2}$ is what makes it unique. Dropping it leaves room for isometries of $\hat{X}$ onto itself that move $X$ around, and $R$ has many of those.
Where choice enters. Only at step 5.1, 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, which spends ${A C}_{ω}$ to produce one approximating sequence at one point at a time (The Axiom of Countable Choice ( ${A C}_{ω}$ )); and inside A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space, which is cited as [L2].

A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources