LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Every uniformizable space is regular

Statement

Every uniformizable topological space is regular, in ZF.

Facts & Assumptions

Given: A topology induced by a uniformity, a closed $C$ , and $x \notin C$ .

[L1]

Entourage balls form neighbourhood bases and entourages have iterated square roots (The sets containing an entourage ball about each of their points form a topology, Uniform space in the entourage formulation).

[L2]

Regularity separates a point from a closed set by disjoint open sets (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).

[L3]

Symmetric entourages have square roots, and a point is outside the closure of a set when it has a neighbourhood disjoint from that set (Every uniformity has a base of symmetric entourages, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

Proof

technique · direct

1.1

Since $X ∖ C$ is an open neighbourhood of $x$ , choose an entourage $E$ with $E [x] \subseteq X ∖ C$ , then choose a symmetric $D$ with $D^{\circ 2} \subseteq E$ .

L1L3choose

1.2

Let $O$ be the union of all open subsets of $D [x]$ containing $x$ . Then $O$ is open and $x \in O \subseteq D [x]$ because $D [x]$ is a neighbourhood.

L1construct

2.1

One has $\overline{O} \subseteq E [x]$ . Indeed, if $y \notin E [x]$ , then the neighbourhood $D [y]$ is disjoint from $O$ : a point $z \in D [y] \cap O \subseteq D [y] \cap D [x]$ would give $(x, y) \in D^{\circ 2} \subseteq E$ by symmetry. Hence $y \notin \overline{O}$ by [L3].

step 1.1step 1.2L1L3

3.1

Since $E [x] \cap C = \emptyset$ , step 2.1 gives $C \subseteq X ∖ \overline{O}$ . The two open sets $O$ and $X ∖ \overline{O}$ are disjoint neighbourhoods of $x$ and $C$ , so the space is regular by [L2].

step 1.1step 2.1L2∎

Depends on

Used by

The K-topology on ℝ is not uniformizable Counterexample

Dependency tree · two levels

15 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

J. Wodzicki, Uniform Structure (standard reference, not scraped)
M. Kunzinger, General Topology (standard reference, not scraped)