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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Smooth partitions of unity exist on manifolds with boundary

Statement

Assume ACω. Every open cover of a smooth manifold with boundary admits a smooth partition of unity subordinate to it.

Facts & Assumptions

Given: ACω, a smooth manifold with boundary, and an open cover.

[L1]

Relatively compact coordinate balls and half-balls form a basis at interior and boundary points (Relatively compact coordinate balls and half-balls form a boundary-manifold basis).

[L2]

Under ACω (The Axiom of Countable Choice (ACω)), second countability gives a countable subcover of every open cover (Assuming countable choice, every second countable space is Lindelöf).

[L3]

Under ACω, a countable relatively compact boundary coordinate cover has an at-most-countable locally finite shrinking Wk‾⊆Vk⊆Vk‾⊆Bn(k) (A countable boundary coordinate cover has a locally finite shrinking).

[L4]

A compact set inside an open Euclidean set has a smooth [0,1]-valued bump equal to one near it and supported inside the open set (A Euclidean bump for a compact set inside an open set). Its proof constructs the support inside a finite union of bounded closed balls, so that support is compact.

[L5]

Smoothness in half-space charts uses local smooth extension and is preserved by smooth composition (Smooth functions on relatively open half-space sets, Chain rule for smooth half-space maps).

[L6]

The boundary partition contract requires smoothness in half-space charts, locally finite supports subordinate to the indexed cover, and sum one (Smooth partition of unity on a manifold with boundary).

Proof

technique · direct
1.1L1L2L6givenconstruct

Form the set of all pairs (B,j) where j indexes a member Uj of the given cover, B is a coordinate ball or half-ball with compact closure, and B‾⊆Uj. By [L1] the balls in these pairs cover M, without choosing one pair for every point. By [L2], their projection to balls has an at-most-countable subcover (Bn). Use ACω to choose for each listed Bn an index j(n) witnessing Bn‾⊆Uj(n). If M=∅, the empty partition already satisfies [L6]; henceforth assume it is nonempty, repeating one selected pair if the subcover is finite.

2.1L3step 1.1

Apply [L3] to (Bn), obtaining an at-most-countable family (Wk,Vk)k∈I with ⋃kWk=M, Wk‾⊆Vk⊆Vk‾⊆Bn(k), and (Vk) locally finite. Assign each k to the original cover index j(n(k)) from step 1.1.

3.1L4L5step 2.1construct

Fix k. A chart for Vk identifies the compact set Wk‾⊆Vk with a compact subset Kk of a relatively open set Qk⊆Hn. Choose an open Ok⊆Rn with Ok∩Hn=Qk. For n>0, [L4] gives a smooth Euclidean bump equal to one near Kk and supported inside Ok. Its support is compact by the finite-ball construction in [L4], so its intersection with the closed half-space is compact and lies inside Qk. Restrict the bump to Hn, pull it back through the chart, and extend by zero outside Vk. Its support is a compact subset of Vk, so this extension is smooth both at the manifold boundary and across the edge of Vk by [L5]. It is a [0,1]-valued smooth function bk, equal to one on Wk‾ and supported inside Vk. For n=0, use the constant-one function on the singleton chart Vk, extended by zero outside it.

4.1L5step 2.1step 3.1choose

The nonempty set of valid bumps in step 3.1 is specified independently for each k in an at-most-countable family. Use ACω once to select all bk (finite choice if I is finite). Because supp⁡bk⊆Vk and (Vk) is locally finite, a:=∑k∈Ibk is locally a finite smooth sum. It is positive everywhere because the Wk cover M and bk=1 on Wk. By [L5], 1/a is smooth in boundary charts: near each point a smooth local extension of a stays positive after shrinking the chart, so the ordinary reciprocal composes smoothly. Set ck=bk/a.

5.1L6step 1.1step 2.1step 4.1∎

For each original cover index j, put ϕj=∑k:j(n(k))=jck, taking the zero function when there are no such k. The family is locally finite because the Vk are, and every ϕj is smooth and nonnegative. The union of the locally finite closed supports assigned to j is closed and lies in Uj, so supp⁡ϕj⊆Uj. Also ∑jϕj=∑kck=1, and each ϕj≤1. By [L6], this is a smooth partition subordinate to the original cover. The empty case was handled in step 1.1.

Depends on

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