TITLE: Why are smooth manifolds defined to be paracompact?
QUESTION [49 upvotes]: The way I understand things, roughly speaking, the importance of smooth manifolds is that they form the category of topological spaces on which we can do calculus. The definition of smooth manifolds requires that they be paracompact. I've looked all over, but I haven't found a clean statement for how paracompactness is a necessary condition to do calculus.
I understand that, by a theorem of Stone, every metric space is paracompact, but I'm not sure why we need global metrizability either.

Question: In what sense is paracompactness exactly the right condition to impose on a topological manifold to allow us to do calculus on it? Is there some theorem of the form "X has [some structure we strictly need in calculus] if and only if it is paracompact"?

REPLY [2 votes]: I've thought about this a bit myself, and I think that, calculus-wise, a key point might be that a locally Euclidean space has to be paracompact in order to be completely metrizable. I think that a complete metric is indispensible to any attempt to do calculus, because without it, the analogues of the IVT (which is a direct consequence of completeness of the reals) and MVT, and Stokes Theorem (if you could even formulate it!), would fail.