Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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Under the ultrafilter lemma and the Axiom of Choice, a metrizable space is Čech-complete exactly when it is completely metrizable

Statement

Assume the ultrafilter lemma and the Axiom of Choice. A metrizable space is Čech-complete if and only if it is completely metrizable.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

Assume the ultrafilter lemma and the Axiom of Choice. Every completely metrizable space is Čech-complete. (Under the ultrafilter lemma and the Axiom of Choice, every completely metrizable space is Čech-complete).

[F2]

Assume the ultrafilter lemma. Every metrizable Čech-complete space is completely metrizable. (Under the ultrafilter lemma, every metrizable Čech-complete space is completely metrizable).

Proof

technique · direct
1.1

The empty space lies in both classes.

givenF1F2
2.1

For a nonempty metrizable space combine the two preceding implications, retaining metrizability only for the direction from Čech-completeness to complete metrizability.

step 1.1F2F1
3.1

The preceding construction and implications establish the assertion.

step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 17 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