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 well-known construction using Partition of unity). Does anybody know a reference or Proof for paracompactness?
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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 BudneyCommented May 12, 2012 at 16:09
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$\begingroup$ More generally, a regular Lindelöf space is paracompact. This should be proved in general topology texts. $\endgroup$– Mariano Suárez-ÁlvarezCommented May 12, 2012 at 16:39
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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 self-contained answer. Yes, the answer in Conlon is similar, but even there, more spread out. $\endgroup$– Greg KuperbergCommented May 12, 2012 at 16:53
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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 BudneyCommented May 13, 2012 at 17:31
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3$\begingroup$ @Ryan Fair enough. Still, if MathOverflow builds up a Wikipedia-like library of answers, even those that appear in textbooks, that's not such a bad thing. $\endgroup$– Greg KuperbergCommented May 20, 2012 at 2:24
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1 Answer
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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 $K_{1,i} \cup \cdots \cup K_{i-1,i}$ from $U_i$ to make a new chart $V_i$. Some of the $V_i$ might be empty, but this is no problem.
answered May 12, 2012 at 16:35
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1$\begingroup$ This does not seem to work. Take $M = \mathbb{R}$; let $U_i = \mathbb{R}$ for all $i$, mapped identically into $M$; let $K_{i, j} = K_j = [-j, -\frac{1}{j}] \cup [\frac{1}{j}, j]$, independently of $i$. Then $V_i$ is simply the complement of $K_i$, but the family formed by the $V_i$ is not locally finite, as they all contain $0$. Or did I miss something? $\endgroup$ Commented Oct 29, 2019 at 10:48
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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 K--and so, a fortiori, the interior of any given K--meets 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. $\endgroup$ Commented Oct 31, 2019 at 17:56
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1$\begingroup$ @IliaSmilga You need to take compact sets whose union is $U_i$, not just whose union is dense in $U_i$. $\endgroup$ Commented Oct 31, 2019 at 18:11
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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. $\endgroup$ Commented Oct 31, 2019 at 20:12
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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. $\endgroup$ Commented Nov 1, 2019 at 13:32