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 $X$ .

[L1]

Under countable choice, first countability gives $seqcl (A) = \overline{A}$ for every $A \subseteq X$ (Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

$A \subseteq seqcl (A) \subseteq \overline{A}$ for every $A$ (The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique).

[A1]

A space is Fréchet--Urysohn when $seqcl (A) = \overline{A}$ for every subset $A$ , and it is sequential when every sequentially closed subset is closed (Fréchet–Urysohn spaces and sequential spaces).

Proof

technique · direct

1.1L1A1

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

1.2L2

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

2.1step 1.2A1∎

Fréchet–Urysohnness gives $\overline{C} = seqcl (C) = C$ , so $C$ is closed. Therefore $X$ is sequential.

Assuming countable choice, every first countable space is Fréchet–Urysohn; in ZF every Fréchet–Urysohn space is sequential

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources