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.

  • 5
    $\begingroup$ What’s your definition of a manifold? The most general definition allows some non-paracompact manifolds. $\endgroup$ Commented Jan 9, 2013 at 17:49
  • $\begingroup$ @BrianM.Scott The definition I'm using is: A manifold is a locally Euclidean second countable Hausdorff space. $\endgroup$
    – user54938
    Commented Jan 9, 2013 at 17:50
  • $\begingroup$ I suspected as much; then @Mariano’s answer is what you want. $\endgroup$ Commented Jan 9, 2013 at 17:54
  • $\begingroup$ 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". $\endgroup$ Commented Jan 9, 2013 at 17:56
  • $\begingroup$ Nils and Brian: thank you, it is becoming clearer now. $\endgroup$
    – user54938
    Commented Jan 9, 2013 at 17:57

1 Answer 1


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)

  • 6
    $\begingroup$ A minor note to the regularity: probably the easiest way to observe this is to note that locally compact Hausdorff spaces are regular. $\endgroup$
    – T. Eskin
    Commented Jan 9, 2013 at 20:49
  • $\begingroup$ @ThomasE. Thank you, your comment is very helpful to me. $\endgroup$
    – user54938
    Commented Jan 13, 2013 at 15:52

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