LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Every ultrafilter on a totally bounded uniform space is Cauchy

Statement

Every ultrafilter on a totally bounded uniform space is Cauchy.

Facts & Assumptions

Given: A totally bounded uniform space $X$ and an ultrafilter $\mathcal V$ on it.

[L1]

Total boundedness gives a finite cover by entourage balls (Totally bounded uniform space).

[L2]

An ultrafilter containing a finite union contains one member of the union (Ultrafilters are prime: a union in $\mathcal{U}$ has a member in $\mathcal{U}$ ).

[L3]

Cauchyness asks for an $E$ -small filter member for each entourage (Cauchy filter in a uniform space).

[L4]

Every entourage contains a symmetric entourage whose square lies in it (Every uniformity has a base of symmetric entourages).

Proof

technique · direct

1.1

Let $E$ be an entourage and choose a symmetric $D$ with $D^{-1}\circ D=D^{\circ2}\subseteq E$ .

L4choose

1.2

Total boundedness gives finite $F$ with $X=\bigcup_{x\in F}D[x]$ ; since $X\in\mathcal V$ , [L2] gives $D[x]\in\mathcal V$ for some $x\in F$ .

L1L2

2.1

Any two points of $D[x]$ are $E$ -related, so $D[x]\times D[x]\subseteq E$ .

step 1.1step 1.2

3.1

This supplies an $E$ -small member for every $E$ , so $\mathcal V$ is Cauchy by [L3].

step 2.1L3∎

Depends on

Used by

Assuming the ultrafilter lemma, every complete and totally bounded uniform space is compact Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 10 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

J. Wodzicki, Uniform Structure (standard reference, not scraped)
Encyclopedia of Mathematics, Uniform space (standard reference, not scraped)