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In couple of places on that site we can find the following statement:

A smooth manifold $M$ admits a finite atlas.

Look for example here Definition of manifolds with finite atlas or here Topology of a manifold, yet with no reference for a proof. I would appreciate if someone give any since I could'n found.

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J.E.M.S
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  • A manifold with a finite atlas must be second countable. So it depends on the definition of manifold. If you don't require a manifold to be paracompact, the long line would be a counterexample. If you require manifolds paracompact, you still need to require that they have at most countably many connected components. – Daniel Fischer Oct 21 '16 at 12:35
  • I require the manifold to be paracompact. I don't know the proof but I have read that if dim M = n than it needs at most n+1 charts so we take such on each component say $U_\alpha$. Let that component is overed by $U_\alpha^i$ and now we have a cove by $V^i = \cup U_\alpha^i$. So I dont see we we need countabely many components? – J.E.M.S Oct 21 '16 at 13:24

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