Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A locally finite open cover by subspaces with σ\sigma-locally-finite bases yields a σ\sigma-locally-finite basis of the whole space

Statement

Let U\mathcal U be a locally finite open cover of XX. If every UUU\in\mathcal U, with its subspace topology, has a σ\sigma-locally-finite open basis nBU,n\bigcup_n\mathcal B_{U,n}, then XX has a σ\sigma-locally-finite open basis.

Facts & Assumptions

Given: A locally finite open cover U\mathcal 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).

Proof

technique · direct
1.1

Since every UUU\in\mathcal U is open in XX, every member of a relative open basis BU,n\mathcal B_{U,n} is open in XX by [L2]. Put Bn=UUBU,n\mathcal B_n=\bigcup_{U\in\mathcal U}\mathcal B_{U,n}.

L2construct
2.1

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

L1step 1.1
2.2

If OO is open and xOx\in O, choose UUU\in\mathcal U containing xx and then a member of the basis of UU containing xx and contained in OUO\cap U. Thus nBn\bigcup_n\mathcal B_n is a basis of XX.

step 1.1
3.1

Steps 2.1 and 2.2 prove the result.

step 2.1step 2.2

Depends on

Used by

Dependency tree · next 3 levels

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Sources