Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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Assuming countable choice, every first countable space is Fréchet–Urysohn; in ZF every Fréchet–Urysohn space is sequential

Statement

Assume countable choice. Every first countable space is Fréchet–Urysohn. In ZF, every Fréchet–Urysohn space is sequential.

Facts & Assumptions

Given: A topological space X.

[A1]

A space is Fréchet--Urysohn when seqcl⁡(A)=A‾ for every subset A, and it is sequential when every sequentially closed subset is closed (Fréchet–Urysohn spaces and sequential spaces).

Proof

technique · direct
1.1

Under countable choice, [L1] is exactly the defining equality for a first countable space to be Fréchet–Urysohn.

L1A1
1.2

Now suppose X is Fréchet–Urysohn and C is sequentially closed. Then seqcl⁡(C)=C, because the constant sequence gives C⊆seqcl⁡(C) and sequential closedness gives the reverse inclusion.

L2
2.1

Fréchet–Urysohnness gives C‾=seqcl⁡(C)=C, so C is closed. Therefore X is sequential.

step 1.2A1∎

Depends on

Used by

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Dependency tree · two levels

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Sources