TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed

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:

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 hypothesis on $X$ is needed.
If $(X,d)$ is complete and $A$ is closed in $(X,d)$ , then $(A,d_A)$ is complete.

Consequently, for a complete $(X,d)$ a subset $A \subseteq X$ is complete if and only if it is closed.

Facts & Assumptions

Given: A metric space $(X,d)$ and a subset $A \subseteq X$ with the subspace metric $d_A = d \restriction (A \times A)$ .

[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 $\mathbb{R}$ ).

[L2]

A point lies in $\overline{A}$ if and only if some sequence in $A$ converges to it in $(X,d)$ ; and a subset $F \subseteq X$ is closed if and only if every sequence in $F$ converging in $X$ has its limit in $F$ (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, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space). The first claim, in the direction that manufactures a sequence, spends the Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )).

[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]

$A$ is closed in $(X,d)$ if and only if $\overline{A} = A$ , and $A \subseteq \overline{A}$ always (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, 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

For claim 1, assume [A1] and let $x \in \overline{A}$ ; by [L2] there is a sequence $(a_k)$ with $a_k \in A$ for every $k$ and $a_k \to x$ in $(X,d)$ .

A1L2

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]; as $x \in \overline{A}$ was arbitrary, $\overline{A} \subseteq A$ , hence $\overline{A} = A$ and $A$ is closed. This is claim 1.

step 1.1step 3.1L4L5

5.1

Claims 1 and 2 hold, by steps 4.1 and 2.2; for a complete $(X,d)$ they combine into the stated equivalence.

step 4.1step 2.2∎

Remarks

Claim 1 needs nothing about $X$ . A complete subspace is closed in whatever ambient metric space it sits in, complete or not, because the argument only compares a limit that exists in $X$ with a limit that exists in $A$ and uses uniqueness. This is what makes completeness so useful as a hypothesis: it is inherited downward by closed subsets and it forces closedness upward.
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 $\mathrm{AC}_\omega$ (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )) 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.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Complete metric space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)