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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Every open cover of a manifold has a countable relatively compact coordinate-ball subcover

Statement

Every open cover of a smooth manifold has a countable cover by coordinate balls with compact closures, each closure contained in one member of the original cover.

Facts & Assumptions

Given: A smooth manifold M and an open cover U of M.

[L1]

Coordinate balls form a basis of the underlying topological manifold (Coordinate balls form a basis of a topological manifold).

[A1]

By the library convention in Smooth manifolds and their smooth charts, every smooth manifold is second countable.

Proof

technique · direct
1.1

For each pM, choose UpU containing p, then choose a coordinate ball Bp with pBpBpUp by [L1].

L1givenchoose
2.1

The family (Bp)pM is an open cover of M, so [A1] and [L2] give a countable subcover B1,B2,. Each Bn is compact and lies in some member of U by step 1.1.

A1L2step 1.1
3.1

Thus the original cover has a countable subordinate cover by relatively compact coordinate balls.

step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources