LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

LCH Urysohn cutoff

Statement

Assuming Dependent Choice as in Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0, 1]$ , and conversely such a space is normal, if $K \subseteq U$ with $K$ compact and $U$ open in an LCH space $X$ , then some $f \in C_{c} (X)$ satisfies $1_{K} \leq f \leq 1_{U}$ .

Facts & Assumptions

Given: $K \subseteq U$ , with $K$ compact and $U$ open.

[L1]

Every compact set in an LCH space has an open neighbourhood $V$ with $K \subseteq V \subseteq \overline{V} \subseteq U$ and compact $\overline{V}$ . (In a locally compact Hausdorff space every open set containing a point contains an open set containing it whose closure is compact and still inside; such a space is regular)

[L2]

Under Dependent Choice, disjoint closed subsets of a normal space are separated by a continuous $[0, 1]$ -valued function. (Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0, 1]$ , and conversely such a space is normal)

Proof

technique · direct

1.1L1L2choose

Choose $V$ as in [L1]. The compact Hausdorff space $\overline{V}$ is normal; apply [L2] there to $K$ and $\overline{V} ∖ V$ , obtaining $h = 1$ on $K$ and $h = 0$ on $\overline{V} ∖ V$ .

2.1step 1.1construct∎

Extend $h$ by $0$ off $\overline{V}$ . Continuity of $h$ on $\overline{V}$ and its vanishing on the boundary $\overline{V} ∖ V$ make the extension continuous; it is supported in $\overline{V} \subseteq U$ , is $1$ on $K$ , and belongs to $C_{c} (X)$ .

Depends on

Dependency tree · 2 levels

Sources

Donald L. Cohn, Measure Theory, 2nd ed., Proposition 7.1.9