LemmaStatement: AI-adaptedProof: AI-generatedverified 2026-09-24 (gpt-6-sol)

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.

A closure-controlled locally finite open cover transfers relative $σ$ -locally-finite bases to the whole space

Statement

Let $(W_{s})_{s \in S}$ be a locally finite open cover of $X$ . For each $s$ , suppose an open set $U_{s}$ satisfies $\overline{W_{s}} \subseteq U_{s}$ and has a supplied relative $σ$ -locally-finite open basis $⋃_{n < ω} B_{s, n}$ . Then $X$ has a $σ$ -locally-finite open basis. No choice principle is needed beyond the supplied indexed bases and assignments.

Facts & Assumptions

Given: The locally finite cover, closure-controlled assignments, and indexed relative bases in the statement.

[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

1.1L2

Put $C_{n} = {B \cap W_{s} : s \in S, B \in B_{s, n}}$ . Each member is open in $X$ , since $W_{s}$ is open and a set open in the open subspace $U_{s}$ is ambient open.

2.1L1step 1.1

Fix $x \in X$ . By [L1], choose a neighborhood $N$ meeting only finitely many $W_{s}$ . For each of these indices, if $x \in U_{s}$ , relative local finiteness supplies an ambient neighborhood $N_{s}$ whose intersection with $U_{s}$ meets only finitely many $B \in B_{s, n}$ . If $x \notin U_{s}$ , then $x \notin \overline{W_{s}}$ , so choose $N_{s}$ disjoint from $W_{s}$ . Intersect $N$ with this finite collection of $N_{s}$ . It meets only finitely many members of $C_{n}$ ; hence $C_{n}$ is locally finite.

2.2step 1.1

If $O$ is open and $x \in O$ , choose $s$ with $x \in W_{s}$ . A relative basis member $B \in B_{s, n}$ contains $x$ and lies in $O \cap U_{s}$ . Then $x \in B \cap W_{s} \subseteq O$ . Thus $⋃_{n} C_{n}$ is a basis.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the result.

Depends on

Used by

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Sources

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

A closure-controlled locally finite open cover transfers relative σσ-locally-finite bases to the whole space