LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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Under choice, every metric space has a $\sigma$ -locally-finite basis

Statement

Assume the Axiom of Choice. Every metric space has a $\sigma$ -locally-finite open basis.

Facts & Assumptions

Given: A metric space $(X,d)$ and the Axiom of Choice.

[L1]

Under choice every metric space is paracompact, so every open cover has a locally finite open refining cover (Stone's theorem, under choice: every metric space is paracompact).

[L2]

Open balls form a basis for the metric topology (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Proof

technique · direct

1.1

For each $n\in\mathbb N$ , let $\mathcal C_n$ be the cover by balls of radius $2^{-n-3}$ . By [L1], choose a locally finite open refining cover $\mathcal V_n$ of $\mathcal C_n$ .

L1choose

2.1

The family $\mathcal B=\bigcup_n\mathcal V_n$ is $\sigma$ -locally finite. It is a basis: if $x\in O$ with $O$ open, [L2] gives $\varepsilon>0$ with $B_d(x,\varepsilon)\subseteq O$ ; choose $n$ with $2^{-n-2}<\varepsilon$ , and a member $V\in\mathcal V_n$ containing $x$ . As $V$ lies in some $B_d(c,2^{-n-3})$ containing $x$ , the triangle inequality gives $V\subseteq B_d(x,2^{-n-2})\subseteq O$ .

L2step 1.1

3.1

Thus $\mathcal B$ is the asserted $\sigma$ -locally-finite basis.

step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Umeå University, The Smirnov- and Bing–Nagata–Smirnov Metrization Theorems (standard reference, not scraped)

Under choice, every metric space has a σ\sigmaσ-locally-finite basis