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suppose $M$ is a locally Euclidean Hausdorff space, show that $M$ is second countable if and only if it is paracompact and has countably many components. This is Problem 2-15 p.59 (or 1-5 p.30 in the new version) in: Introduction to smooth manifolds by John M. Lee.

In the hint he said if $M$ is paracompact, show that each component of $M$ has a locally finite cover by precompact open balls, and extract a subcover. But since we don't know about the metrizability of the space, every open cover has a subcover doesn't necessarily mean the space is second countable? Am I missing something here?

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  • $\begingroup$ Have you got the answer? I am also curious about this problem. $\endgroup$ – Danielsen Jul 30 '14 at 11:44
  • $\begingroup$ you probably do not mean that each component is countable, but that the space has countably many components. If so, please edit to say "has countably many components". Show that the subcover of each component that you extract, being locally finite, is in fact countable (use connectedness of each component). You may just assume at first that the space itself is connected, since once you do this case, the the case with countably many components will follow easily. $\endgroup$ – Mirko Nov 28 '15 at 21:24
  • $\begingroup$ As a hint for the ones who want to try the problem: One can in general prove that a path connected, locally compact and paracompact space is second countable. $\endgroup$ – yamete kudasai Aug 18 at 13:01
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First of all, a topological space that is locally homeomorphic to a Euclidean space is locally compact, cf. this, and locally connected as those properties are local and preserved by homeomorphisms (and they hold for Euclidian spaces). Then

$M$ second-countable $\Longrightarrow\ M$ paracompact with countably many connected components:

This is Lemma 1.9 p.9 Foundations of Differentiable Manifolds and Lie Groups, Frank W. Warner or Thm 4.77 p.110 Introduction to Topological Manifolds, John Lee:

The authors show in the first part that $M$ is countable at infinity/$\sigma$-compact, i.e. $M = \bigcup_{n\in \mathbb{N}} K_n$ with $K_n \subset \overset{\circ}{K}_{n+1}$ compact. Let now $ \mathcal{U}:=(U_i)_{i\in I}$ be any open cover of $M$:

$K_0$ is compact and covered by the family open subsets $(U_i\cap \overset{\circ}{K}_1)_{i\in I}$ so one can extract a finite subsequence. (Similarly for $K_1$)

Then by induction: for any $n \geq 2,\enspace \left(U_i\cap (\overset{\circ}{K}_{n+1}\backslash \overline{K}_{n-2})\right)_{i\in I}$ is an open cover of $\overline{K}_n\backslash K_{n-1}$, from which one extracts a finite cover.

By such a construction, one will cover every $K_n$ and hence all of $M$; the open subset of the cover are of the form $U_i\cap (\overset{\circ}{K}_{n+1}\backslash \overline{K}_{n-2})$ and thus included in $U_i$ (refinement of a cover) and any point is in one of the $\overset{\circ}{K}_n$ which is covered by finitely many $U_i$ (locally finite cover), i.e. $M$ is paracompact.

Edit: this last implication is not well justified here because a cover of $K_{n+1}$ or $K_{n-1}$ may also intersect with points of $K_n$ ... although the intuition is correct...

In this post, one uses the fact that a second countable space is separable, together with local connectedness to show that $M$ has countably many connected components.

$M$ paracompact with countably many connected components $\Longrightarrow\ M$ second-countable :

Since $M$ has countably many connected components $(C_k)_{k\in \mathbb{N}}$, it suffices to show that each of these components has a countable basis ($\mathbb{N}\times \mathbb{N}$ is still countable).

The idea is to use the fact that for any chart $(U,\varphi)$ (from definition of $M$ locally Euclidian), $ U$ is second-countable as it is homeomorphic to a subset of $\mathbb{R}^n$ which is second-countable.

One shows again that $M$ is countable at infinity / $\sigma$-compact but starting with the hypothesis that $M$ paracompact. It is similar as above, cf. Theorem 12.11 p.38 of Topology and Geometry, Glen E. Bredon.

Finally with $\sigma$-compactness, $M$ is covered by countably many open subsets, each of which are second-countable.


I found these notes which also answers precisely this question link

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