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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Assuming dependent choice, every uniformizable space is completely regular

Statement

Assuming dependent choice, every uniformizable topological space is completely regular.

Facts & Assumptions

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

[L1]

A normal entourage sequence yields a uniformly continuous pseudometric with controlled balls (A normal sequence of entourages yields a uniformly continuous pseudometric with controlled dyadic balls).

[L2]

Complete regularity requires a continuous $[0, 1]$ -valued function equal to $1$ at $x$ and $0$ on $C$ (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces).

[L3]

Uniformly continuous maps are continuous for their induced topologies (Every uniformly continuous map is continuous for the induced topologies), and the usual metric uniformity on $R$ induces its usual topology (A metric on a nonempty set generates an entourage uniformity whose induced topology and uniformly continuous maps are the usual metric notions, and this uniformity is separated).

[L4]

Dependent choice produces the normal sequences used in the pseudometric construction (Assuming dependent choice, every entourage admits a normal symmetric sequence subordinate to it).

[L5]

Entourage balls form neighbourhood bases for the induced topology (The sets containing an entourage ball about each of their points form a topology).

Proof

technique · constructive

1.1

Choose an entourage $U$ with $U [x] \cap C = \emptyset$ by [L5]. Using dependent choice, take a normal sequence with $E_{0} = X \times X$ and $E_{1} \subseteq U$ .

L4L5chooseconstruct

2.1

Let $p$ be the controlled pseudometric from [L1]. Since ${p \leq 1 / 4} \subseteq E_{1} \subseteq U$ , every $y \in C$ satisfies $p (x, y) > 1 / 4$ .

step 1.1L1

3.1

Put $g (y) = \min {1, 4 p (x, y)}$ . The reverse triangle inequality for a pseudometric gives $∣ p (x, y) - p (x, z) ∣ \leq p (y, z),$ and truncation at $1$ does not increase absolute differences. Hence, for every $ε > 0$ , the entourage ${(y, z) : p (y, z) < ε / 4}$ forces $∣ g (y) - g (z) ∣ < ε$ ; $g$ is uniformly continuous. Also $g (x) = 0$ and $g [C] = {1}$ by step 2.1, so $1 - g$ has the orientation required in [L2].

step 2.1L1construct

4.1

By [L3], $1 - g$ is continuous, so [L2] proves complete regularity.

step 3.1L2L3discharge-construct∎

Depends on

Used by

Assuming dependent choice, a nonempty topological space is uniformizable if and only if it is completely regular Theorem

Dependency tree · two levels

46 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)
M. Megrelishvili, Lecture Notes in Topological Groups (standard reference, not scraped)