Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
What’s your definition of a manifold? The most general definition allows some non-paracompact manifolds. – Brian M. Scott Jan 9 '13 at 17:49
@BrianM.Scott The definition I'm using is: A manifold is a locally Euclidean second countable Hausdorff space. – user54938 Jan 9 '13 at 17:50
I suspected as much; then @Mariano’s answer is what you want. – Brian M. Scott Jan 9 '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 9 '13 at 17:56
Nils and Brian: thank you, it is becoming clearer now. – user54938 Jan 9 '13 at 17:57
up vote 9 down vote accepted

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)

share|cite|improve this answer
A minor note to the regularity: probably the easiest way to observe this is to note that locally compact Hausdorff spaces are regular. – T. Eskin Jan 9 '13 at 20:49
@ThomasE. Thank you, your comment is very helpful to me. – user54938 Jan 13 '13 at 15:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.