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I am a physics student trying to study differential geometry. I am trying to work out the following exercise. Please give me some help.

Let $M$ be be a connected Hausdorff space which is locally Euclidean. Show that $M$ is paracompact iff $M$ has a countable atlas.

Currently, I can show that $M$ has has a countable atlas iff $M$ is $2^{nd}$ countable. So it seems that I need to prove that $M$ is $2^{nd}$ countable iff $M$ is paracompact. But this still seems very hard.

Xiaoyi Jing
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  • See https://math.stackexchange.com/questions/527642/the-equivalence-between-paracompactness-and-second-countablity-in-a-locally-eucl – H1ghfiv3 Sep 22 '17 at 10:19

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In Munkres, lemma 41.3 (I proved part of it here or completely here) it is shown that for a regular space the following two facts are equivalent:

  1. Every open cover of $X$ has a locally finite open refinement. (i.e. $X$ is paracompact)

  2. Ever open cover of $X$ has a $\sigma$-locally finite open refinement.

It's easy to show 2. for a second countable space (every open cover has a refinement by base elements so a countable refinement, which is trivially $\sigma$-locally finite.).

In fact, this argument shows that a regular Lindelöf space is paracompact.

Henno Brandsma
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  • I don't know what $\sigma$-finite is, but I will read the book. Thank you very much. – Xiaoyi Jing Sep 23 '17 at 07:13
  • A collection $\mathcal{B}$ is $\sigma$-locally finite if we can write $\mathcal{B} = \cup_{n \in \mathbb{N}} \mathcal{B}_n$ where each $\mathcal{B}_n$ is locally finite. Munkres calls it "countably locally finite", which is non-standard. – Henno Brandsma Sep 23 '17 at 08:03