ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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Assuming the Ultrafilter Lemma and Countable Choice, an uncountable Cantor cube is compact Hausdorff and uniformizable but not first countable, hence not metrizable

Example

Assume the ultrafilter lemma and countable choice. For an uncountable index set $I$ , the Cantor cube $2^{I}$ is compact Hausdorff and uniformizable, but it is not first countable and therefore not metrizable.

Facts & Assumptions

Given: An uncountable set $I$ and the product $2^{I}$ of discrete two-point spaces.

[L1]

Under the ultrafilter lemma, arbitrary products of compact Hausdorff spaces are compact (Assuming the ultrafilter lemma, an arbitrary product of compact Hausdorff spaces is compact).

[L2]

Basic product-open sets restrict only finitely many coordinates (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).

[L3]

A first-countable space has a countable local base, while every metric space is first countable (First countable space: a countable neighbourhood base at every point, The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable).

[L4]

Cantor's theorem supplies uncountable power sets (Cantor's theorem: $A ≺ P (A)$ ).

[L5]

A compact Hausdorff space has a unique compatible uniformity (A nonempty compact Hausdorff space carries exactly one compatible uniformity).

[L6]

Under countable choice, a countable union of finite sets is at most countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L7]

An arbitrary product of Hausdorff spaces is Hausdorff (Arbitrary products preserve $T_{0}$ , $T_{1}$ , and Hausdorffness).

Verification

technique · contradiction

1.1

Each two-point factor is compact Hausdorff, so [L1] makes $2^{I}$ compact and [L7] makes it Hausdorff; [L5] then makes it uniformizable.

L1L5L7

1.2

At the constant-zero point, suppose $(B_{n})$ were a countable local base. For each $n$ , choose a finite coordinate set $F_{n}$ such that the basic zero-cylinder restricting $F_{n}$ is contained in $B_{n}$ ; countable choice licenses these selections. The union $⋃_{n} F_{n}$ is at most countable by [L6], so choose $i \in I ∖ ⋃_{n} F_{n}$ (for instance take $I = P (N)$ , uncountable by [L4]). The one-coordinate neighbourhood requiring coordinate $i$ to be zero contains no $B_{n}$ , because the cylinder inside $B_{n}$ permits coordinate $i$ to be $1$ . This contradicts the local-base property.

L2L4L6choose

2.1

Suppose $2^{I}$ were metrizable. Then [L3] would make it first countable, contradicting step 1.2.

assume-contrastep 1.2L3

3.1

Hence it is not metrizable, and step 1.2 gives failure of first countability.

step 2.1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

J. Wodzicki, Uniform Structure (standard reference, not scraped)