# Manifolds are paracompact

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?

• 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. – Ryan Budney May 12 '12 at 16:09
• More generally, a regular Lindelöf space is paracompact. This should be proved in general topology texts. – Mariano Suárez-Álvarez May 12 '12 at 16:39
• 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. – Greg Kuperberg May 12 '12 at 16:53
• @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. – Ryan Budney May 13 '12 at 17:31
• @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. – 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 $K_{1,i} \cup \cdots \cup K_{i-1,i}$ from $U_i$ to make a new chart $V_i$. Some of the $V_i$ might be empty, but this is no problem.
• @IliaSmilga You need to take compact sets whose union is $U_i$, not just whose union is dense in $U_i$. – Will Sawin Oct 31 '19 at 18:11
• 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. – Ilia Smilga Oct 31 '19 at 20:12
• 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. – Willie Wong Nov 1 '19 at 13:32
• What does the notation $X \Subset Y$ mean? (Guessing from context: it means that $X$ is contained in the interior of $Y$, right?) – Ilia Smilga Nov 4 '19 at 18:30