LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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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.

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A locally finite open cover by subspaces with $σ$ -locally-finite bases yields a $σ$ -locally-finite basis of the whole space

Statement

Let $U$ be a locally finite open cover of $X$ . If every $U \in U$ , with its subspace topology, has a $σ$ -locally-finite open basis $⋃_{n} B_{U, n}$ , then $X$ has a $σ$ -locally-finite open basis.

Facts & Assumptions

Given: A locally finite open cover $U$ and the stated relative bases.

[L1]

A locally finite family has a neighbourhood at each point meeting only finitely many members (Refinements, locally finite families, point-finite families, and star refinements).

[L2]

A subspace-open set is the intersection of the subspace with an ambient open set (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Proof

technique · direct

1.1

Since every $U \in U$ is open in $X$ , every member of a relative open basis $B_{U, n}$ is open in $X$ by [L2]. Put $B_{n} = ⋃_{U \in U} B_{U, n}$ .

L2construct

2.1

The family $B_{n}$ is locally finite. At $x$ , take from [L1] a neighbourhood meeting only finitely many $U$ ; within each of those finitely many $U$ , local finiteness of $B_{U, n}$ supplies a neighbourhood meeting finitely many members, and their finite intersection meets only finitely many members of $B_{n}$ .

L1step 1.1

2.2

If $O$ is open and $x \in O$ , choose $U \in U$ containing $x$ and then a member of the basis of $U$ containing $x$ and contained in $O \cap U$ . Thus $⋃_{n} B_{n}$ is a basis of $X$ .

step 1.1

3.1

Steps 2.1 and 2.2 prove the result.

step 2.1step 2.2∎

Depends on

Used by

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

Dependency tree · two levels

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Sources

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

A locally finite open cover by subspaces with σσ-locally-finite bases yields a σσ-locally-finite basis of the whole space