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

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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.

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

1.1L2L3construct

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 $f [O_{x}] \subseteq W [f (x)]$ .

2.1step 1.1L1L3

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 $E [a]$ .

3.1step 1.1step 2.1L2∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources