LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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Binary-sequence space is compact without Tychonoff

Statement

On $Ω = {0, 1}^{N_{0}}$ put $d (x, x) = 0$ and $d (x, y) = 2^{- \min {j : x_{j} \neq y_{j}}}$ for $x \neq y$ . This is a metric whose topology is generated by cylinders; it is compact in ZF.

Facts & Assumptions

[F1]

Finite prescriptions define nonempty cylinders and each prefix has two disjoint children. Binary-sequence cylinders and fair-coin content.

[F2]

Compactness means every open cover has a finite subcover. Open cover, subcover, compact metric space, and compact subset of a metric space.

Proof

Given: On $Ω = {0, 1}^{N_{0}}$ put $d (x, x) = 0$ and $d (x, y) = 2^{- \min {j : x_{j} \neq y_{j}}}$ for $x \neq y$ . This is a metric whose topology is generated by cylinders; it is compact in ZF.

1.1

Positive definiteness and symmetry follow immediately from the first disagreement. If x,y agree before index r and y,z agree before index s, then x,z agree before min(r,s). Thus $d (x, z) \leq \max (d (x, y), d (y, z)) \leq d (x, y) + d (y, z)$ , with equal pairs included. A prefix cylinder fixing indices 0 through m-1 is the open ball of radius $2^{- (m - 1)}$ about any of its points when m>=1. Arbitrary finite-coordinate cylinders are finite unions of sufficiently long prefixes. Conversely prefixes of arbitrarily small diameter fit in every ball. Hence cylinders generate exactly the metric topology and are clopen.

2.1

Fix an open cover with no finite subcover. The empty prefix cylinder has no finite subcover. If a prefix cylinder has no finite subcover, at least one of its two children also lacks one; otherwise the union of the two finite subcovers would cover it. Choose the zero child if it lacks a finite subcover, and the one child otherwise. This is a uniquely specified recursion, producing a binary sequence x whose every prefix cylinder lacks a finite subcover.

step 1.1F1F2

3.1

Some member U of the cover contains x. Since U is open, it contains a prefix cylinder about x. That cylinder has the one-element subcover {U}, contradicting its construction. Therefore no open cover without a finite subcover exists, proving compactness. The recursion makes no arbitrary infinite choices; neither Tychonoff nor dependent choice is used.

step 1.1step 2.1F2∎

Depends on

Used by

Fair-coin cylinder content is a premeasure Lemma

Dependency tree · two levels

9 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

E–W Examples 2.8–2.9 compact metric footnote; elementary binary-tree proof (standard reference, not scraped)