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Every manifold is paracompact. I tried:

$M$ is an $n$--manifold with open covering $U_\alpha$ and $\varphi_\alpha$ local homeomorphisms; $\varphi_\alpha (U_\alpha)$ are open in $\mathbb R^n$. Adding $B(x, \varepsilon)$ for $x \in (\bigcup_\alpha \varphi_\alpha (U_\alpha))^c$ yields an open covering of $\mathbb R^n$. $\mathbb R^n$ is paracompact hence there is a refinement $V_\alpha$. We discard $V_\alpha \subseteq B(x,\varepsilon)$ and observe that $\varphi_\alpha^{-1}(V_\alpha)$ are a refinement of $U_\alpha$. Fix $p \in M$ and $\alpha_0$ with $p \in U_{\alpha_0}$. Then there is an open nbhd $N$ of $\varphi_{\alpha_0} (p)$ such that $N$ intersects only finitely many $V_\alpha$. Let $N' = \varphi_{\alpha_0}^{-1}(N \cap \varphi_{\alpha_0} (U_{\alpha_0}))$. Then $N'$ is an open nbhd of $p$.

My intended finish was "$N'$ only intersects finitely many $\varphi_\alpha^{-1}(V_\alpha)$". Alas, it appears that one cannot argue like this since $\varphi_\alpha$ and $\varphi_{\alpha_0}$ map $\varphi_\alpha^{-1}(V_\alpha)$ to different sets. How to salvage the proof? Thank you.

Asaf Karagila
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user54938
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    What’s your definition of a manifold? The most general definition allows some non-paracompact manifolds. – Brian M. Scott Jan 09 '13 at 17:49
  • @BrianM.Scott The definition I'm using is: A manifold is a locally Euclidean second countable Hausdorff space. – user54938 Jan 09 '13 at 17:50
  • I suspected as much; then @Mariano’s answer is what you want. – Brian M. Scott Jan 09 '13 at 17:54
  • Somewhere in your proof you have to use second-countability of $M$, because if you drop the second-countability condition on $M$, then, as Brian noted, there are "manifolds", which are not paracompact, e.g. the "long line". – Nils Matthes Jan 09 '13 at 17:56
  • Nils and Brian: thank you, it is becoming clearer now. – user54938 Jan 09 '13 at 17:57
  • Second-countable regular spaces and also regular Lindelöf spaces are paracompact. You can use Hausdorff and locally Euclidean to show the regularity of a manifold. – Stefan Hamcke Jan 09 '13 at 18:02
  • See http://mathoverflow.net/questions/96780/manifolds-are-paracompact – Mariano Suárez-Álvarez Jan 09 '13 at 18:02
  • Stefan and Mariano: thank you, you are very helpful. – user54938 Jan 09 '13 at 18:03
  • See also http://math.stackexchange.com/q/98105/. A curiosity (IMO): Andrew Stacey points out on mathoverflow http://mathoverflow.net/questions/229 that there are no less than 107 (!!) equivalent conditions for metrizability of a connected locally Euclidean Hausdorff space. – Martin Jan 09 '13 at 21:28
  • Martin: thank you, you are very helpful. – user54938 Jan 11 '13 at 08:51

1 Answers1

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Every Hausdorff second-countable regular space is metrizable —this is Urysohn's metrization theorem— and metrizable spaces are paracompact because metric spaces are.

(And manifolds are regular spaces, of course)

Mariano Suárez-Álvarez
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