Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Every manifold has a compact exhaustion

Statement

Assume ACω (The Axiom of Countable Choice (ACω)). Every smooth manifold admits a compact exhaustion.

Facts & Assumptions

Given: ACω and a smooth manifold M.

[L1]

Under ACω, the manifold has a countable cover by relatively compact coordinate balls U1,U2,… (Every open cover of a manifold has a countable relatively compact coordinate-ball subcover). This is the only use of countable choice; the finite indices below are chosen canonically as least eligible integers.

[A1]

A compact subset of a space covered by open sets has a finite subcover.

Proof

technique · direct
1.1L1A1givenconstruct

If M=∅, the constant sequence Kk=∅ is already a compact exhaustion. Otherwise take the countable cover U1,U2,… of [L1], repeating a ball if the supplied cover is finite. Let m1 be the least integer with U1‾⊆⋃i≤m1Ui. Given mk, let mk+1>mk be the least integer such that the compact set ⋃i≤mkUi‾ lies in ⋃i≤mk+1Ui. Such integers exist by [A1], and selecting their least values uses no further choice.

2.1step 1.1

In the nonempty case put Kk:=⋃i≤mkUi‾. Each Kk is compact, step 1.1 gives Kk⊆int⁡(Kk+1), and every point of M lies in some Kk because the Ui cover M.

3.1step 1.1step 2.1∎

The empty case of step 1.1 and the nonempty case of step 2.1 both give a compact exhaustion in the sense of Compact exhaustions of a manifold.

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