sub-family of cover with compact closure is still a cover?

Question

Here is a problem about existence of exhaustion of a manifold:

Denote manifold as $M$, with Hausdorff, second-countability and locally Euclidean property. Show there exists a exhaustion of $M$.

And first part of proof is here:

Since $M$ is second countable, there is a countable basis of the topology of $M$. Out of this countable sequence of open sets, we pick those that have compact closure, and denote them by $Y_1,Y_2,....$ Since $M$ is locally Euclidean it is easy to see that $\mathcal{Y}={Y_i}$ is a open cover of $M$.

And I’m confuse about the last part. How can locally Euclidean property grant $\mathcal{Y}$ be a sub-cover? I try to PBR but fail. Can anyone get me some hints?

Answer 1

All we have to show is that every point of $M$ is contained in some $Y_i\in \mathcal Y$. So let $x\in M$ be arbitrary. Because $M$ is locally Euclidean, there is some open subset $U\subseteq M$ containing $x$ and a homeomorphism $\phi\colon U\to \widehat U$, where $\widehat U$ is an open subset of $\mathbb R^n$. Let $\widehat x = \phi(x)$, and choose $r$ small enough that $B_{2r}(\widehat x)\subseteq \widehat U$. Then $\overline B_r(\widehat x)$ is compact and contained in $\widehat U$. Define \begin{align*} K & = \phi^{-1}\big(\overline B_r(\widehat x)\big),\\ W & = \phi^{-1}\big(B_r(\widehat x)\big). \end{align*} Since $\phi$ is a homeomorphism, $W$ is an open neighborhood of $x$ contained in the compact set $K$.

Now there is a set $Y$ in our original basis such that $x\in Y \subseteq W$. Because $\overline Y$ is a closed subset of the compact set $K$, it is compact. Thus $Y$ is one of the sets in $\mathcal Y$.