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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Under choice, every metric space has a $σ$ -locally-finite basis

Statement

Assume the Axiom of Choice. Every metric space has a $σ$ -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 $n \in N$ , let $C_{n}$ be the cover by balls of radius $2^{- n - 3}$ . By [L1], choose a locally finite open refining cover $V_{n}$ of $C_{n}$ .

L1choose

2.1

The family $B = ⋃_{n} V_{n}$ is $σ$ -locally finite. It is a basis: if $x \in O$ with $O$ open, [L2] gives $ε > 0$ with $B_{d} (x, ε) \subseteq O$ ; choose $n$ with $2^{- n - 2} < ε$ , and a member $V \in V_{n}$ 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 $B$ is the asserted $σ$ -locally-finite basis.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

17 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

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

Under choice, every metric space has a σσ-locally-finite basis