Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
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Every manifold has a compact exhaustion

Statement

Every smooth manifold admits a compact exhaustion.

Facts & Assumptions

Given: A smooth manifold M.

[L1]

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).

[A1]

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

Proof

technique · direct
1.1

Start from the countable cover U1,U2, of [L1]. Recursively choose integers 1m1<m2< such that U1im1Ui and, for each k1, the compact set imkUi is contained in imk+1Ui; this is possible by [A1].

L1A1choose
2.1

Put Kk:=imkUi. Each Kk is compact, step 1.1 gives Kkint(Kk+1), and every point of M lies in some Kk because the Ui cover M.

step 1.1
3.1

Therefore (Kk) is a compact exhaustion in the sense of Compact exhaustions of a manifold.

step 2.1

Depends on

Used by

Dependency tree · two levels

6 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