Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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Assuming countable choice, a countable product of second countable spaces is second countable

Statement

Assuming ACω\mathrm{AC}_\omega, a countable product of second countable spaces is second countable.

Facts & Assumptions

Given: Second countable factors (Xn)(X_n) indexed by a countable set.

[A1]

Countable choice selects a countable basis in each factor (The Axiom of Countable Choice (ACω\mathrm{AC}_\omega)).

[L1]

A product of two at most countable sets is at most countable, and under countable choice a countable union of at most countable sets is at most countable (A product of two at most countable sets is at most countable, Countable unions of at most countable sets, assuming ACω\mathrm{AC}_\omega).

Proof

technique · constructive
1.1

Use [A1] to choose countable factor bases.

A1construct
2.1

Finite-support boxes with selected basic coordinates form a basis for the product by [F1].

step 1.1F1
3.1

The finite subsets of a countable index set form an at most countable family: after enumerating the index set, subsets of size at most nn are coded by nn-tuples of natural numbers, which are countable by finite induction using the product theorem in [L1], and their union over nn is countable by the union theorem in [L1]. For each fixed finite support FF, the choices of one member of the selected basis in every coordinate of FF form a finite product of countable sets and are countable by the same induction. A final application of the countable-union theorem shows that all finite-support boxes form an at most countable family.

step 2.1L1
4.1

Thus the product is second countable.

step 3.1discharge-construct

Depends on

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Sources