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While looking at proofs here and there on the existence of a Riemannian metric, I always miss something.

Given a manifold $M$ I would like to give myself an atlas $\{(U_\alpha,\phi_\alpha)\}$ with a countable number of maps. Then on each of the open sets of the atlas I can define my metric $g_i$. Then I know there exists a partition of unity subordinate to $U_\alpha$. Then I define my Riemannian metric as $g = \sum_ig_i\theta_i$.

But How do I get my locally finite countable cover?

user405156
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    "Riemannian metric" (not "riemannien metric"). – Dietrich Burde Nov 21 '17 at 09:42
  • The existence of the smooth partition of unity is guaranteed for a smooth manifold. The property of locally finite is in fact one of the requirement of partition of unity. So when we claimed that we have a partition of unity subordinate so a open cover, it automatically understood that the family of supports of our function $\psi_{\alpha}$ is locally finite. – Kelvin Lois Nov 21 '17 at 09:42
  • What I understand is every manifold is supposed paracompact so this gives me the locally finite cover. But how do I get the countability? because the partition of unity is a countable but nothing guarantees that I can say that I have a countable atlas. – user405156 Nov 21 '17 at 10:07
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    Look https://math.stackexchange.com/questions/2440003/paracompact-iff-countable-atlas – Kelvin Lois Nov 21 '17 at 11:53
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    What is your definition of manifold? Paracompactness alone is not enough to get a countable atlas - consider an uncountable disjoint union of copies of $\mathbb R$. However, I don't think this is a problem for partitions of unity - an uncountable sum can still make sense thanks to the local finiteness. – Anthony Carapetis Nov 21 '17 at 12:47
  • Hausdorff and second countable – user405156 Nov 21 '17 at 15:41

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Generally, we suppose that a manifold is paracompact t ensure the existence of partitions of unity.

https://en.wikipedia.org/wiki/Paracompact_space

Tsemo Aristide
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