Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
How 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.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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 XX.

[L1]

Under countable choice, first countability gives seqcl(A)=A\operatorname{seqcl}(A)=\overline A for every AXA\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 (ACω\mathrm{AC}_\omega)).

[L2]

Aseqcl(A)AA\subseteq\operatorname{seqcl}(A)\subseteq\overline A for every AA (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)=A\operatorname{seqcl}(A)=\overline A for every subset AA, 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 XX is Fréchet–Urysohn and CC is sequentially closed. Then seqcl(C)=C\operatorname{seqcl}(C)=C, because the constant sequence gives Cseqcl(C)C\subseteq\operatorname{seqcl}(C) and sequential closedness gives the reverse inclusion.

L2
2.1

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

step 1.2A1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 45 results over 18 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