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

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A sequentially compact metric space is complete, with no choice principle used

Statement

Let $(X, d)$ be a sequentially compact metric space (Countably compact, sequentially compact and limit point compact metric spaces, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). Then $(X, d)$ is complete (Complete metric space: every Cauchy sequence converges in the space).

The proof is a theorem of ZF: it instantiates two existential statements and selects nothing.

Facts & Assumptions

Given: A sequentially compact metric space $(X, d)$ .

[L1]

$(X, d)$ is sequentially compact: every sequence in $X$ has a subsequence converging to a point of $X$ (Countably compact, sequentially compact and limit point compact metric spaces, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[L2]

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

[L3]

A Cauchy sequence with a subsequence converging to $p$ converges to $p$ itself (A Cauchy sequence in a metric space with a convergent subsequence converges to that subsequence’s limit).

Proof

technique · direct

1.1

Let $(x_{k})$ be a Cauchy sequence in $(X, d)$ .

2.1

By sequential compactness there is a strictly increasing index map $j \mapsto n_{j}$ and a point $p \in X$ with $x_{n_{j}} \to p$ in $(X, d)$ .

L1step 1.1

3.1

Since $(x_{k})$ is Cauchy and one of its subsequences converges to $p$ , the whole sequence converges to $p$ , and $p \in X$ .

L3step 2.1

4.1

So every Cauchy sequence in $(X, d)$ converges in $X$ , that is $(X, d)$ is complete.

L2step 3.1∎

Remarks

The converse fails. A complete metric space need not be sequentially compact: $R$ with its usual metric is complete, and the sequence $x_{k} = k$ has no convergent subsequence, every subsequence being unbounded. What has to be added to completeness is total boundedness, and that pair is equivalent to compactness (A complete, totally bounded metric space is compact, proved from countable choice used exactly once, For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice).

Why this direction is free while the companion is not. Here the sequence is handed to the proof and sequential compactness hands back a subsequence: one object is produced, once. In A sequentially compact metric space is totally bounded, proved from the axiom of dependent choice a point has to be produced at every stage, each in terms of the points already produced, and that is where a choice principle enters the page.

Depends on

Used by

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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