Showing the existence of a special basis for a second countable, locally compact Hausdorff space I am studying Abraham and Marsden's Foundations of Mechanics.
In proving that second countable, locally compact Hausdorff spaces are paracompact, they begin by stating that

$S$ [the topological space] is the countable union of open sets $U_n$ such that $cl(U_n)$ is compact and $cl(U_n)\subset U_{n+1}$.

However, I can't see why this statement holds.
I do know that locally compact Hausdorff spaces are regular and in such spaces, for any open neighborhood $U$ of a point $u$, there is an open set $V$ containing $u$ such that $cl(V)$ is compact and $cl(V)\subset U$.
 A: HINT: First show that if $\mathscr{B}$ is a countable base for $S$, then
$$\{B\in\mathscr{B}:\operatorname{cl}B\text{ is compact}\}$$
is also a base for $S$. Thus, we may as well assume that $\mathscr{B}=\{B_n:n\in\Bbb N\}$, where each $B_n$ has compact closure. The sets $U_n$ are then constructed recursively.
Let $U_0=B_0$; then $\operatorname{cl}U_0=\operatorname{cl}B_0$ is compact, so there is a finite $\mathscr{B}_0\subseteq\mathscr{B}$ such that $\operatorname{cl}U_0\subseteq\bigcup\mathscr{B}_0$. Let $U_1=B_1\cup\bigcup\mathscr{B}_0$; clearly $U_1$ is open, and $\operatorname{cl}U_0\subseteq U_1$. Moreover,
$$\operatorname{cl}U_1=\operatorname{cl}B_1\cup\bigcup_{B\in\mathscr{B}_0}\operatorname{cl}B$$
is the union of finitely many compact sets, so it is compact, and there is therefore a finite $\mathscr{B}_1\subseteq\mathscr{B}$ such that $\operatorname{cl}U_1\subseteq\bigcup\mathscr{B}_1$. Let $U_2=B_2\cup\bigcup\mathscr{B}_1$, and continue.
At this point you might want to say and so on or the like, but that’s not really good enough. To carry out the recursive construction, you need to do say exactly how to carry out the general step of the construction. You might begin like this:

Suppose that we’ve constructed open sets $U_0,\ldots,U_n$ for some $n\in\Bbb N$ so that $\operatorname{cl}U_k$ is compact and $B_k\subseteq U_k$ for $k=0,\ldots,n$, and $\operatorname{cl}U_k\subseteq U_{k+1}$ for $k=0\le k<n$.

Then you explain how to construct an open set $U_{n+1}$ so that $\operatorname{cl}U_n\subseteq U_{n+1}$, $\operatorname{cl}U_{n+1}$ is compact, and $B_{n+1}\subseteq U_{n+1}$, so that all of the necessary conditions are met for the next step in the construction. Once you’ve done that, you’re entitled to assert that there are open sets $U_n$ for $n\in\Bbb N$ such that

*

*$\operatorname{cl}U_n$ is compact for each $n\in\Bbb N$;

*$\operatorname{cl}U_n\subseteq U_{n+1}$ for each $n\in\Bbb N$; and

*$B_n\subseteq U_n$ for each $n\in\Bbb N$.

At this point all that remains is to show that $S=\bigcup_{n\in\Bbb N}U_n$, and I did something to ensure that this would be the case; what is it?
