Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-08 (Codex)
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  • 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.

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A Euclidean bump for a compact set inside an open set

Statement

If K⊆U⊆Rn with K compact and U open, then there exists a smooth function ρ:Rn→[0,1] such that ρ=1 on K and supp⁡(ρ)⊆U.

Facts & Assumptions

Proof

technique · direct
1.1givenF1construct

If K=∅, the zero function has empty support and satisfies the claim. Suppose henceforth that K≠∅. Consider all triples (p,r,R) with p∈K, 0<r<R, and BR(p)⊆U. Their inner balls Br(p) form an indexed open cover of K: openness gives a ball BR(p)⊆U at each p, and r=R/2 is admissible. This defines the whole family without making a simultaneous choice at every point.

2.1L1L2step 1.1

By [L1], finitely many admissible triples (pi,ri,Ri), 1≤i≤m, have inner balls covering K. For each of these finitely many triples use the explicit formula in [L2], with x replaced by x−pi, to define ρi. Translation preserves the coordinate-derivative formulas, so each ρi is smooth, lies in [0,1], equals 1 on B‾ri(pi), and has support contained in BRi(pi)⊆U.

3.1L3step 2.1construct

Define ρ=1−∏i=1m(1−ρi). Repeated use of [L3] shows that each iterated coordinate derivative is a finite sum of products of continuous derivatives of the ρi, so ρ is smooth. Since every factor lies in [0,1], so does ρ. On each inner ball one factor is zero; hence ρ=1 on their union, an open neighbourhood of K.

4.1F1step 2.1step 3.1construct

Let F=⋃i=1msupp⁡ρi. Each support is closed by [F1], so the finite union F is closed, and step 2.1 gives F⊆U. Off F, every ρi vanishes, so the formula in step 3.1 gives ρ=0. Therefore the closure of the nonzero locus of ρ lies in F: supp⁡ρ⊆F⊆U.

5.1step 1.1step 3.1step 4.1∎

Steps 3.1 and 4.1 give all three required properties; step 1.1 handled the empty case.

Depends on

Used by

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Sources