TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-09-10 (gpt-6-astra)

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Closed subspaces of complete metric spaces are complete; the converse under countable choice

Statement

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:

Under countable choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), 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.
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.

Facts & Assumptions

Given: A metric space $(X, d)$ and a subset $A \subseteq X$ with the subspace metric $d_{A} = d ↾ (A \times A)$ .

[A1]

Completeness of $(A, d_{A})$ : every $d_{A}$ -Cauchy sequence in $A$ converges in $(A, d_{A})$ to a point of $A$ (Complete metric space: every Cauchy sequence converges in the space, Cauchy sequence in a metric space).

[A2]

Completeness of $(X, d)$ : every $d$ -Cauchy sequence in $X$ converges in $(X, d)$ to a point of $X$ (Complete metric space: every Cauchy sequence converges in the space).

[L1]

Distances inside $A$ are computed in $X$ : $d_{A} (a, b) = d (a, b)$ for $a, b \in A$ (Isometry, isometric embedding, and the subspace metric on a subset). Hence a sequence in $A$ is $d_{A}$ -Cauchy exactly when it is $d$ -Cauchy, and for $p \in A$ it converges to $p$ in $(A, d_{A})$ exactly when it converges to $p$ in $(X, d)$ (Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[L2]

Under countable choice, each point of $\overline{A}$ is the limit of a sequence from $A$ (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, claim 1, sequence-manufacturing direction). In ZF, a closed set contains the limit of every ambient-convergent sequence of its points (the same theorem, proof step 2.2). We do not use the converse characterization of closed sets without its choice hypothesis.

[A3]

Countable choice is assumed only for claim 1: a sequence of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L3]

A convergent sequence in a metric space is Cauchy (Every convergent sequence in a metric space is Cauchy).

[L4]

Limits in a metric space are unique (A sequence in a metric space has at most one limit).

[L5]

By the ball definition of closure, $A \subseteq \overline{A}$ , and if $\overline{A} \subseteq A$ then every point outside $A$ has a ball disjoint from $A$ , so $A$ is closed (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, 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).

Proof

technique · direct

1.1

First work without choice: assume [A1] and let $(a_{k})$ be any sequence of points of $A$ which converges to $x \in X$ . We will prove $x \in A$ for this already given sequence.

A1given

1.2

For claim 2, assume [A2], assume $A$ closed, and let $(a_{k})$ be a $d_{A}$ -Cauchy sequence in $A$ ; by [L1] it is $d$ -Cauchy in $X$ , so by [A2] it converges in $(X, d)$ to some $x \in X$ .

A2L1

2.1

That sequence is $d$ -Cauchy by [L3], hence $d_{A}$ -Cauchy by [L1], since all its terms lie in $A$ .

step 1.1L1L3

2.2

The sequence lies in $A$ and converges in $X$ , and $A$ is closed, so $x \in A$ by [L2]; by [L1] the sequence then converges to $x$ in $(A, d_{A})$ , and $x \in A$ , so $(A, d_{A})$ is complete. This is claim 2.

step 1.2L1L2

3.1

By [A1] it therefore converges in $(A, d_{A})$ to some $a \in A$ , and by [L1] it converges to $a$ in $(X, d)$ as well.

step 2.1A1L1

4.1

The sequence converges in $(X, d)$ both to $x$ and to $a$ , so $x = a \in A$ by [L4]. Thus a complete subspace contains all ambient limits of sequences of its points, in ZF. In particular, if each $x \in \overline{A}$ is already known to admit such a sequence, applying this argument to one fixed $x$ at a time gives $\overline{A} \subseteq A$ , and [L5] makes $A$ closed without choosing a family of sequences.

step 1.1step 3.1L4L5

5.1

Now assume [A3] as well as [A1], and fix $x \in \overline{A}$ . The proof of [L2] applies [A3] to the nonempty sets $A \cap B (x, 1 / (k + 1))$ , producing a sequence from $A$ that converges to $x$ . Step 4.1 gives $x \in A$ , so [L5] gives closedness. This proves claim 1 under countable choice. Claim 2 was proved in step 2.2 without [A3]; combining these directions gives the stated equivalence under countable choice. If $A$ is empty, it is closed and has no Cauchy sequences, so all relevant conclusions hold vacuously as well.

A1A3L2L5step 4.1step 2.2∎

Remarks

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 ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) 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 $R^{n}$ is a complete metric space, because $R^{n}$ is ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ). Every appeal to Banach's fixed point theorem on a closed subset of $R^{n}$ passes through this remark.

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Sources

Keremedis and Wajch, On densely complete metric spaces and extensions of uniformly continuous functions in ZF, arXiv:1901.08709v1, Theorem 4.1 (standard reference, not scraped)
Complete metric space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)