By Definition, smooth manifolds are assumed to be Hausdorff and to satisfy the second countability axiom. I have heard (but never seen written) that these assumptions imply paracompactness (and thus the existence of a Riemannian metric by the wellknown construction using Partition of unity). Does anybody know a reference or Proof for paracompactness?

4$\begingroup$ The proof is in most introductory manifold theory textbooks, usually immediately before the construction of partitions of unity. Try Conlon's Differentiable Manifolds for example. $\endgroup$ – Ryan Budney May 12 '12 at 16:09

$\begingroup$ More generally, a regular Lindelöf space is paracompact. This should be proved in general topology texts. $\endgroup$ – Mariano SuárezÁlvarez May 12 '12 at 16:39

7$\begingroup$ Oh come on guys, it's all too easy to click on the close button instead of answering the question. I thought it was useful to give a selfcontained answer. Yes, the answer in Conlon is similar, but even there, more spread out. $\endgroup$ – Greg Kuperberg May 12 '12 at 16:53

3$\begingroup$ @Greg, that seems like a rather narrow reading of what happened here. I'm having a hard time thinking of an introductory manifold theory textbook that does not cover this topic, one way or another. $\endgroup$ – Ryan Budney May 13 '12 at 17:31

3$\begingroup$ @Ryan Fair enough. Still, if MathOverflow builds up a Wikipedialike library of answers, even those that appear in textbooks, that's not such a bad thing. $\endgroup$ – Greg Kuperberg May 20 '12 at 2:24
Theorem: A countable atlas of charts for a Hausdorff $n$manifold $M$ can be refined to a locally finite atlas. In fact, each chart only needs to be trimmed.
Proof: Let $U_1,U_2,\ldots$ be the charts. Each $U_i$, as a subset of $\mathbb{R}^n$, is the limit of a nested sequence of compact subsets $K_{i,1} \subseteq K_{i,2} \subseteq \ldots$. Since $M$ is Hausdorff, each $K_{i,j}$ is closed in $M$. So it suffices to delete $K_{1,i} \cup \cdots \cup K_{i1,i}$ from $U_i$ to make a new chart $V_i$. Some of the $V_i$ might be empty, but this is no problem.

8$\begingroup$ The condition on the K's means that each U is the union of the corresponding K's; yours don't satisfy this requirement. Your revised example does clearly show that an emendation to the original argument is needed; the following seems to suffice: Require of the sets K that the interiors of those associated with any U exhaust the corresponding U. One easily sees that any given Kand so, a fortiori, the interior of any given Kmeets only finitely many V's. Since any point belongs to the interior of some K, local finiteness of the covering by the V's is thus established. $\endgroup$ – Howard Stein Oct 31 '19 at 17:56

1$\begingroup$ @IliaSmilga You need to take compact sets whose union is $U_i$, not just whose union is dense in $U_i$. $\endgroup$ – Will Sawin Oct 31 '19 at 18:11

1$\begingroup$ Oh, sorry, my mistake. So let's take $K_j = [j,0] \cup [\frac{1}{j},j]$. Then they do indeed exhaust $\mathbb{R}$; but the family of their complements $V_i$ is still not locally finite, since any neighborhood of $0$ intersects infinitely many of them. $\endgroup$ – Ilia Smilga Oct 31 '19 at 20:12

2$\begingroup$ Equivalently, since we are working on $\mathbb{R}^n$, we can just change one symbol in the proof: instead of $\subset$ use $\Subset$ for the relation between the $K$s. $\endgroup$ – Willie Wong Nov 1 '19 at 13:32

1$\begingroup$ What does the notation $X \Subset Y$ mean? (Guessing from context: it means that $X$ is contained in the interior of $Y$, right?) $\endgroup$ – Ilia Smilga Nov 4 '19 at 18:30