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I understand that a space $X$ is said to be paracompact if any open cover $\{O_\alpha\}$ of $X$ has a locally finite refinement. In the book I'm reading (General relativity, Robert M.Wald, apendix A), the author says that he wants his definition of manifolds to satisfy paracompactness because it "prevents them from being too large", but doesn't explain further. Why does that mean and why would that happen? My understanding of these topics is rather elementary, so if you could also provide a bit of intuition, I would really be grateful.

Ash
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    Take a look at the long line, possibly the simplest non-paracompact manifold. That may convince you that the long line at least is too big. Another reason might be that the connected paracompact manifolds are the ones embeddable in Euclidean spaces; the long line is too "long" to be embedded in some $\Bbb R^n$. – Angina Seng Aug 20 '17 at 15:47
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    I think it has in mind the example of the long line, a non-paracompact space a space somewhat similar to the real line, but in a certain way "longer".

    https://en.wikipedia.org/wiki/Long_line_(topology)

     – Francesco Polizzi Aug 20 '17 at 15:48
  • Thank you guys. This... Kinda blew my mind. @LordSharktheUnknown, about the second reason you give (sorry if the question is trivial, this is really not my domain), the Nash embedding theorem says you can embedd any riemannian manifold in an Euclidian space. But aren't some reimannian manifolds disconnected? – Ash Aug 20 '17 at 17:06

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