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

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A finite compactly supported partition of unity near a compact set

Statement

Assume the Axiom of Dependent Choice. Let $X$ be locally compact Hausdorff, let $K \subseteq X$ be compact, and let $U_{1}, \dots, U_{n}$ be open sets covering $K$ . Then there are nonnegative $φ_{i} \in C_{c} (X)$ with $supp φ_{i} \subseteq U_{i}$ such that $\sum_{i} φ_{i} = 1$ on an open neighbourhood of $K$ .

Facts & Assumptions

Given: Dependent Choice, and $K \subseteq ⋃_{i = 1}^{n} U_{i}$ , with $K$ compact and each $U_{i}$ open.

[L1]

Under Dependent Choice, LCH cutoffs exist between a compact set and an open neighbourhood. (LCH Urysohn cutoff)

Proof

technique · direct

1.1

Consider all triples $(V, W, i)$ with $V, W$ open, $1 \leq i \leq n$ , [given, L1] and $\overline{V} \subseteq W \subseteq \overline{W} \subseteq U_{i},$ where the displayed closures are compact. Local compactness and the Hausdorff property show that the sets $V$ occurring in these triples cover $K$ . Compactness therefore gives finitely many triples $(V_{j}, W_{j}, i_{j})$ whose $V_{j}$ cover $K$ . Apply [L1] to $\overline{V_{j}} \subseteq W_{j}$ to obtain $h_{j} \in C_{c} (X)$ with $0 \leq h_{j} \leq 1$ , $h_{j} = 1$ on $\overline{V_{j}}$ , and $h_{j} = 0$ off $W_{j}$ . Consequently $supp h_{j} \subseteq \overline{W_{j}} \subseteq U_{i_{j}}$ .

2.1

Put $h = \sum_{j} h_{j}$ . Then $h \geq 1$ on $K$ , so ${h > 1 / 2}$ is an [step 1.1, L1, choose] open neighbourhood of $K$ . Choose open sets $O, W$ with $K \subseteq O \subseteq \overline{O} \subseteq W \subseteq \overline{W} \subseteq {h > 1 / 2},$ where the displayed closures are compact. Apply [L1] to $\overline{O} \subseteq W$ to obtain $g \in C_{c} (X)$ with $0 \leq g \leq 1$ , $g = 1$ on $\overline{O}$ , and $g = 0$ off $W$ . Thus $supp g \subseteq \overline{W} \subseteq {h > 1 / 2}$ , and $g = 1$ on the open neighbourhood $O$ of $K$ .

step 1.1L1

3.1

For $1 \leq i \leq n$ , set $φ_{i} (x) = {\begin{cases} g (x) \frac{\sum_{j : i_{j} = i} h_{j} (x)}{h (x)}, & h (x) > 0, \\ 0, & h (x) = 0. \end{cases}$ The quotient is only used on $supp g \subseteq {h > 1 / 2}$ , so extension by zero is continuous. Each $φ_{i}$ is nonnegative, compactly supported in $U_{i}$ , and $\sum_{i} φ_{i} = g$ , hence the sum is $1$ on $O$ .

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

13 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

Donald L. Cohn, Measure Theory, 2nd ed., Chapter 7 (standard reference, not scraped)