TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-29

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Topological manifolds are sigma-compact

Statement

Every topological manifold is $σ$ -compact.

Facts & Assumptions

Given: A topological manifold $M$ .

[F1]

A topological manifold is second countable (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces).

[F2]

Coordinate balls form a basis of the topology, and each coordinate ball has compact closure in $M$ (Coordinate balls form a basis of a topological manifold).

Proof

technique · direct

1.1F1F2choose

By [F1] choose a countable basis $B = (B_{n})_{n \in N}$ for the topology of $M$ . By [F2], for each $n$ and each point of $B_{n}$ there is a coordinate ball contained in $B_{n}$ whose closure is compact. Replacing each $B_{n}$ by all coordinate balls it contains, we obtain a countable basis $(U_{k})_{k \in N}$ of coordinate balls with compact closures.

2.1F2step 1.1

Every $\overline{U_{k}}$ is compact by [F2], and the family $(\overline{U_{k}})_{k \in N}$ covers $M$ because the basis $(U_{k})$ does.

$M = ⋃_{k \in N} \overline{U_{k}}$

is a countable union of compact subsets.

3.1step 2.1∎

Therefore $M$ is $σ$ -compact.

Topological manifolds are sigma-compact

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources