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I was confused by Boothby's proof that a Manifold had a countable cover of compact sets. In his proof he says that as $M$ has a countable basis of open sets, we may assume that $M$ has a countable basis of relatively compact open sets $\{V_i\}$. I don't see why this is true. I don't know how to go about proving it. I've tried to come up with counter examples, but obviously I've failed (as the claim is true). I know I could prove it, if I could show that for each $(U,\phi)$ a coordinate chart there existed a basis element $B\subseteq U$, but can we show that? Any hints or help proving this?

I did look at some of the similar questions to my title, but I didn't understand them. For example, here

Does every manifold can be "covered" by compact sets?

The answer says we can cover the manifold by a countable collection of subseteq homeomorphic to $\mathbb{R}^n$. I know we can cover it by such sets, as we can take a coordinate chart and shrink so that it is homeomorphic to an open ball, which is in turn homeomorphic to $\mathbb{R}^n$, but I don't know why countable.

I would be able to prove the claim if I could verify the statement given here

https://mathworld.wolfram.com/TopologicalBasis.html

that:

"A topological basis is a subset B of a set T in which all other open sets can be written as unions or finite intersections of B. For the real numbers, the set of all open intervals is a basis."

But I don't see why that follows from the axioms they listen.

If anyone could provide any help it would be much appreciate, thank you.

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Each point of $M$ has an open nbhd homeomorphic to $\Bbb R^n$ for some $n\in\Bbb N$. $\Bbb R^n$ is locally compact, so each point of $M$ has a local base of relatively compact open sets, and therefore $M$ has a base of relatively compact open sets. $M$ has a countable base, so by the result that I proved here it has a countable base of relatively compact open sets.

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  • $\begingroup$ Forgive me for being ignorant, but why is $\bigcup \mathcal{B}(W)=W$? How do we know that we can choose $B\in\mathcal{B}$ such that $B\subseteq W$? Let along how do we know such $B$ cover $W$? Is there some condition on Base I'm not understanding, because I don't see how it follows from the two base axioms. $\endgroup$
    – Melody
    Commented Sep 28, 2020 at 2:25
  • $\begingroup$ @Melody: Since $\mathscr{B}$ is a base for the topology, by definition every open set is a union of elements of $\mathscr{B}$. In particular, then, it must be the union of all of the members of $\mathscr{B}$ contained in it. $\endgroup$
    – Brian M. Scott
    Commented Sep 28, 2020 at 2:29
  • $\begingroup$ Thank you, I guess I was misreading the page here mathworld.wolfram.com/TopologicalBasis.html It doesn't list that every open set is a union of of basis elements as one of the two axioms it gives, but I'll take your word for it. Thank you very much. $\endgroup$
    – Melody
    Commented Sep 28, 2020 at 2:33
  • $\begingroup$ @Melody: You’re welcome. It doesn’t list that as an axiom, but it mentions it in the first sentence. Those axioms are not the best way to think about a base for a topology: it’s much better to think of it as a subset of the topology whose unions generate the whole topology. $\endgroup$
    – Brian M. Scott
    Commented Sep 28, 2020 at 2:34

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