False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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False: BPI proves Stone's theorem for metric spaces

Statement

False: over $Z F$ , BPI implies that every metrizable space is paracompact.

More precisely, the universal implication from BPI to Stone's theorem for metric spaces is not provable over $Z F$ : relative to $Con (Z F)$ there is a model of $Z F + B P I$ containing a metrizable space that is not paracompact (The Boolean prime ideal principle, Paracompactness: every open cover has a locally finite open refinement, with no separation axiom built into the word, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Refutation

Facts & Assumptions

Given: The relative-consistency theorem for BPI with a metrizable nonmetacompact space, and the assumed consistency of $Z F$ .

[F1]

Relative to $Con (Z F)$ there is a model of $Z F + B P I$ containing a metrizable nonmetacompact space; by definition, that space has an open cover with no point-finite open refining cover (Relative consistency of BPI with failure of Stone's theorem, Metacompactness: every open cover has a point-finite open refinement, Refinements, locally finite families, point-finite families, and star refinements).

[F2]

A paracompact space is one in which every open cover has a locally finite open refinement, and a locally finite family is point-finite (Paracompactness: every open cover has a locally finite open refinement, with no separation axiom built into the word, Refinements, locally finite families, point-finite families, and star refinements).

Proof

technique · contradiction

1.1

Assume, for the sake of contradiction, that over $Z F$ BPI implies that every metrizable space is paracompact, and assume $Con (Z F)$ .

assume-contragiven

2.1

In the model of [F1] the theory $Z F + B P I$ holds, so by the assumed implication every metrizable space in that model is paracompact; in particular its metrizable nonmetacompact space $X$ would be paracompact.

step 1.1F1

3.1

By [F1], some open cover $U$ of $X$ has no point-finite open refining cover. Paracompactness would give a locally finite open refinement $V$ covering $X$ , and $V$ would be point-finite by [F2], a contradiction.

step 2.1F1F2

4.1

The contradiction shows that BPI does not imply Stone's theorem for metric spaces over $Z F$ , conditionally on $Con (Z F)$ ; the refutation is relative-consistency based and does not exhibit an outright counterexample in ZF.

step 3.1F1discharge-contradiction∎

Depends on

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Dependency tree · two levels

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Sources

Samuel Corson, The Independence of Stone's Theorem from the Boolean Prime Ideal Theorem (standard reference, not scraped)