# Is every manifold sigma-compact?

Thanks for reading my post. Here is my question. I want to know if every surface is hemicompact, i.e., there is a compact exhaustion. I think that question could be asked for every manifold. I know that every locally compact and sigma compact space is hemicompact. Any manifold is locally compact, and second countable, so I wonder if these imply that the countable basis could be choosen in such way that every open subset in the basis is precompact, and therefore sigma compact. Thanks in advance for your answer.

• There is a basis formed by standardly embedded balls. – Mariano Suárez-Álvarez Dec 20 '12 at 17:21
• Every metrizable manifold is $\sigma$-compact, and your definition of manifold implies metrizability. The long line is a non-metrizable manifold that is not $\sigma$-compact. – Brian M. Scott Dec 20 '12 at 17:26

## 2 Answers

Yes.

Every topological manifold has a countable basis of precompact coordinate balls. By taking the closure of these coordinate balls, the manifold can be shown to equal the union of countably many compact sets.

You may refer to the answer to another question. Locally Euclidean Hausdorff topological space is topological manifold iff $\sigma$-compact.

Really just a remark but an uncountable disjoint union of circles is locally-Euclidean, Hausdorff and not $\sigma$-compact. Any compact subset intersects at most finitely many circles so a countable union intersects at most countably many circles.

• Your example is not second countable, so it doesn't match the definition of a manifold used by math_failure. – gpr1 Aug 8 '20 at 19:07