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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⊆X be compact, and let U1,…,Un be open sets covering K. Then there are nonnegative φi∈Cc(X) with supp⁡φi⊆Ui such that ∑iφi=1 on an open neighbourhood of K.

Facts & Assumptions

Given: Dependent Choice, and K⊆⋃i=1nUi, with K compact and each Ui open.

[L1]

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

Proof

technique · direct
1.1L1

Consider all triples (V,W,i) with V,W open, 1≤i≤n, [given, L1] and V‾⊆W⊆W‾⊆Ui, 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 (Vj,Wj,ij) whose Vj cover K. Apply [L1] to Vj‾⊆Wj to obtain hj∈Cc(X) with 0≤hj≤1, hj=1 on Vj‾, and hj=0 off Wj. Consequently supp⁡hj⊆Wj‾⊆Uij.

2.1step 1.1L1

Put h=∑jhj. Then h≥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⊆O⊆O‾⊆W⊆W‾⊆{h>1/2}, where the displayed closures are compact. Apply [L1] to O‾⊆W to obtain g∈Cc(X) with 0≤g≤1, g=1 on O‾, and g=0 off W. Thus supp⁡g⊆W‾⊆{h>1/2}, and g=1 on the open neighbourhood O of K.

3.1step 1.1step 2.1∎

For 1≤i≤n, set φi(x)={g(x)∑j:ij=ihj(x)h(x),h(x)>0,0,h(x)=0. The quotient is only used on supp⁡g⊆{h>1/2}, so extension by zero is continuous. Each φi is nonnegative, compactly supported in Ui, and ∑iφi=g, hence the sum is 1 on O.

Depends on

Used by

Dependency tree · two levels

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Sources