TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02‡ rests on unproved material (inherited)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 1 statement not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Under choice, a regular T₁ space with a σ-locally-finite basis has a compatible normal sequence. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Under choice, a space is metrizable if and only if it is paracompact, Hausdorff, and locally metrizable

Statement

Assume the Axiom of Choice. A space is metrizable if and only if it is paracompact, Hausdorff, and locally metrizable.

Facts & Assumptions

Given: The Axiom of Choice and a topological space $X$ .

[L1]

Under choice, every metric space is paracompact and has a $σ$ -locally-finite basis (Stone's theorem, under choice: every metric space is paracompact, Under choice, every metric space has a $σ$ -locally-finite basis).

[L2]

A paracompact Hausdorff space is regular, and Nagata–Smirnov applies to a regular $T_{1}$ space with a $σ$ -locally-finite basis (Every paracompact Hausdorff space is regular, Under choice, a space is metrizable if and only if it is regular, $T_{1}$ , and has a $σ$ -locally-finite basis).

Proof

technique · cases

1.1

If $X$ is metrizable, it is Hausdorff and locally metrizable by taking $X$ itself as the open neighbourhood, and it is paracompact by [L1].

assume-case forwardL1

1.2

Conversely, local metrizability gives an open cover by metrizable subspaces. Paracompactness refines it by a locally finite open cover; every refining member is a metrizable subspace and has a $σ$ -locally-finite basis by [L1]. The merger lemma A locally finite open cover by subspaces with $σ$ -locally-finite bases yields a $σ$ -locally-finite basis of the whole space gives such a basis for $X$ .

assume-case reverseL1

2.1

Hausdorffness implies $T_{1}$ , and [L2] makes $X$ regular; applying Nagata–Smirnov in [L2] to the basis from step 1.2 yields a metric.

L2step 1.2

3.1

The two cases prove the equivalence.

step 1.1step 2.1cases-exhaustive∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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

UCR, Partitions of Unity and a Metrization Theorem of Smirnov (standard reference, not scraped)