# 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?

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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-Alvarez 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.