CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every continuous map from a nonempty compact Hausdorff space to a uniform space is uniformly continuous

Statement

Every continuous map from a nonempty compact Hausdorff space to a uniform space is uniformly continuous.

Facts & Assumptions

Given: A continuous map $f : X \to Y$ with $X$ nonempty compact Hausdorff and $Y$ uniform.

[L1]

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

[L2]

Continuity means that every neighbourhood of $f (x)$ contains the image of some neighbourhood of $x$ , while uniform continuity is the entourage condition (Continuity of a map of topological spaces at a point and globally, Uniformly continuous map between uniform spaces).

[L3]

Every entourage ball is a neighbourhood in the induced topology (The sets containing an entourage ball about each of their points form a topology). Every open cover of a nonempty compact Hausdorff space is uniform (The covers admitting an open refinement form a compatible uniform-cover structure on a nonempty compact Hausdorff space; in particular every open cover is uniform), and every uniform cover has an entourage-ball cover refining it (On a nonempty set, entourage uniformities and uniform-cover structures determine one another); every target entourage has a symmetric square root (Every uniformity has a base of symmetric entourages).

Proof

technique · direct

1.1

Let $V$ be a target entourage and choose a symmetric $W$ with $W^{- 1} \circ W = W^{\circ 2} \subseteq V$ . For each $x \in X$ , let $O_{x}$ be the union of all open sets $O$ such that $x \in O$ and $f [O] \subseteq W [f (x)]$ . Continuity makes this family nonempty, and its union is an open neighbourhood of $x$ satisfying $f [O_{x}] \subseteq W [f (x)]$ .

L2L3construct

2.1

The open cover $(O_{x})_{x \in X}$ is uniform by [L3]. Hence there is a source entourage $E$ whose ball cover refines it: for each $a \in X$ , some $O_{x}$ contains $E [a]$ .

step 1.1L1L3

3.1

If $(a, b) \in E$ , then $a, b \in E [a] \subseteq O_{x}$ for some $x$ . Thus $f (a), f (b) \in W [f (x)]$ , so $(f (a), f (b)) \in W^{- 1} \circ W \subseteq V$ . This is uniform continuity.

step 1.1step 2.1L2∎

Depends on

Used by

Dependency tree · two levels

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