Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
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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.
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).
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
The empty space lies in both classes.
For a nonempty metrizable space combine the two preceding implications, retaining metrizability only for the direction from Čech-completeness to complete metrizability.
The preceding construction and implications establish the assertion.
Depends on
Used by
- FALSE: every metrizable space is Čech-complete False statement
Dependency tree · two levels
10 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
- David Marker, Descriptive Set Theory, §§1–2 (standard reference, not scraped)
- Michael Kunzinger, General Topology, §§11.3–11.4 (standard reference, not scraped)
- MFF General Topology course summary, §4.3 (standard reference, not scraped)
- Jesse Peterson, Real Analysis, §§3.6–3.7 (standard reference, not scraped)