In a book I am reading, the following fact is used without proof.
Let $M$ be a smooth manifold of dimension $m$ with an open covering $\Sigma$. Then there is a topological basis $\Sigma_0=\{U_\alpha\}$ such that each element $U_\alpha$ is a coordinate neighborhood, $\bar U_\alpha$ is compact, and there also exists $W_i\in\Sigma$ such that $\bar U_\alpha\subset W_i$.
I am having trouble proving the "topological basis" part. My attempts so far:
Let $\Sigma_1$ be the family of all coordinate neighborhoods. Intersect the members of $\Sigma_1$ with those in $\Sigma$ to obtain a new family of coordinate neighborhoods $\Sigma_1'$. This procedure makes true the following assertion:
$$\forall U\in\Sigma_1',\ \exists W\in\Sigma,\text{ such that }U\subset W$$
Hence for each point $p\in M$, we can find $U\in\Sigma_1'$ and $W\in\Sigma$ such that
$$p\in U\subset W$$
Since $U$ is a coordinate neighborhood, we can also find an open set $O\subset\mathbb R^m$ and a homeomorphism
$$\varphi_{_U}:U\to O$$
Choose a ball $B$ such that $\varphi_{_U}(p)\in B\subset\bar B\subset O$. Let $U_p=\varphi_{_U}^{-1}(B)$. Then
$$p\in U_p\subset \bar U_p\subset U\subset W$$
and $\bar U_p$ is compact. Finally let $\Sigma_0=\{U_p:p\in M\}$, where each $U_p$ is obtained as above. Then $\Sigma_0$ satisies everything but being a topological basis in the statement made at the beginning. How should I proceed from here?
p.s. A manifold in this book is by definition Hausdorff and second countable.
Best Answer
As you define it, the family $\Sigma_0$ is not necessarily a base because for each $p \in M$ you construct only one open neighborhood $U_p$ of $p$ having suitable properties. This will not give you a neigborhood base at $p$.
You correctedly argue that for each point $p\in M$ one can find $U(p)\in\Sigma_1'$ and $W(p)\in\Sigma$ such that $$p\in U(p)\subset W(p) .$$ Since $U(p)$ is a coordinate neighborhood, one can also find an open set $O(p)\subset\mathbb R^m$ and a homeomorphism $$\varphi_{p}:U(p)\to O(p) .$$
For $s >0$ let $B(p,s) \subset \mathbb R^m$ denote the open ball with center $\varphi_{p}(p)$ and radius $s$. Choose $R(p) >0$ such that $\overline{ B(p,R(p))} \subset O(p)$. For $0< r \le R(p)$ let $$U(p,r) =\varphi_{p}^{-1}(B(r,p)) .$$ Then $$p\in U(p,r)\subset \overline{U(p,r)} \subset U(p)\subset W(p)$$
and $\overline{U(p,r)}$ is compact. Finally let $\Sigma_0=\{U(p,r) : p\in M, 0 <r\le R(p)\}$. Then $\Sigma_0$ is the desired base.
Edited:
Here is a simpler proof. Start with any base $\mathcal B$. Define $$\Sigma_0 = \{ B \in \mathcal B \mid B \text{ is a coordinate neighborhood}, \bar B \text{ is compact and there exists } W\in\Sigma \text{ such that } \bar B\subset W \} .$$ We claim that $\Sigma_0$ is a base.
To see this, let $p \in M$ and $V$ be an open neigborhood of $p$. Choose $W \in \Sigma$ such that $p \in W$ and a homeomorphism $\varphi_p : U(p) \to O(p)$ such that $U(p)$ is an open neigborhood of $p$ and $O$ is open in $\mathbb R^m$. Then each open neighborhood $U'$ of $p$ which is contained in $U(p)$ is also a coordinate neigborhood of $p$.
For $r >0$ let $B(p,r) \subset \mathbb R^m$ denote the open ball with center $\varphi_{p}(p)$ and radius $r$. Choose $r(p) >0$ such that $\overline{B(p,r(p))} \subset \varphi_p(V \cap W \cap U(p))$. Then $V' = \varphi^{-1}(B(p,r(p)))$ is an open neigboorhood of $p$ which has compact closure. Choose $B \in \mathcal B$ such that $p \in B \subset V'$. By construction $B \in \Sigma_0$ and $B \subset V$.
By the way, using the balls $B(p,r(p))$ we have shown that each manifold $M$ is locally compact. If you already know this, then the argument can be simplified once more. It suffices to choose a coordinate neighborhood $U(p)$ of $p$ and then to choose an open neigborhood $V'$ of $p$ such that $\overline{V'}$ is compact and $\overline{V'} \subset V \cap W \cap U(p)$.