LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). 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: ${A C}_{ω}$ , 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).

[L2]

Under ${A C}_{ω}$ , second-countable spaces are Lindelof (Assuming countable choice, every second countable space is Lindelöf).

[A1]

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

Proof

technique · direct

1.1L1givenconstruct

Form the set $B$ of all coordinate balls $B$ for which $\overline{B}$ is compact and $\overline{B} \subseteq U$ for at least one $U \in U$ . No member is chosen for each point. This family covers $M$ : for a given $p$ , one cover member $U$ contains it, and [L1] supplies one such ball within $U$ .

2.1A1L2step 1.1given

The family $B$ is an open cover, so [A1] and [L2] give a finite or countable subcover $(B_{n})$ . By its definition, each $B_{n}$ has a nonempty set of witness members $U \in U$ with $\overline{B_{n}} \subseteq U$ . Use ${A C}_{ω}$ once to choose a witness $U_{n}$ for each listed ball (finite choice suffices in the finite case). Thus $\overline{B_{n}}$ is compact and contained in $U_{n}$ .

3.1step 2.1∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources