TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Under choice, every Cantor cube $2^{I}$ satisfies ccc

Statement

Assuming choice, every Cantor cube $2^{I}$ is ccc.

Facts & Assumptions

Given: The Axiom of Choice and the Cantor cube $2^{I}$ with its product topology.

[A1]

Choice selects from an arbitrary family of nonempty sets (The Axiom of Choice).

[F1]

A basic cylinder specifies values in only finitely many coordinates, and these cylinders form a basis for the product topology (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L1]

Every uncountable family of finite sets has an uncountable $Δ$ -subfamily (Under choice, the uncountable $Δ$ -system lemma for finite sets).

Proof

technique · contradiction

1.1

Suppose $U$ is an uncountable pairwise-disjoint family of nonempty open sets. By [A1] and [F1], choose for every $U \in U$ a nonempty basic cylinder $[p_{U}] \subseteq U$ , where $p_{U}$ is a function from a finite support $F_{U} \subseteq I$ to $2$ . Distinct $U$ give distinct cylinders.

A1F1assume-contraconstruct

2.1

By [L1], after passing to an uncountable subfamily the supports form a $Δ$ -system with finite root $R$ . There are only finitely many functions $R \to 2$ , so one further uncountable subfamily has the same restriction $p_{U} ↾ R$ .

L1step 1.1

3.1

Choose two members of that subfamily. Their supports meet exactly in $R$ and their partial functions agree there, so the union of the two partial functions extends—by assigning $0$ elsewhere—to a point of $2^{I}$ lying in both cylinders. The corresponding members of $U$ intersect, a contradiction. Thus every such family is at most countable and $2^{I}$ is ccc.

step 1.1step 2.1discharge-contradiction∎

Depends on

Used by

Under choice, a concrete ccc nonseparable Cantor cube indexed above 2^ℵ₀ Example
Assuming choice, refuted: every ccc space is separable False statement

Dependency tree · two levels

22 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

UCR General Topology Notes (standard reference, not scraped)
Cantor cube (Wikipedia) (standard reference, not scraped)
Countable chain condition (Wikipedia) (standard reference, not scraped)

Under choice, every Cantor cube 2I2I satisfies ccc