TITLE: A lemma about $\sigma$-compact and locally compact
QUESTION [1 upvotes]: From a topology note I read this theorem: Let $X$ be a topological space which is $\sigma$-compact and locally compact, then $X$ can be expressed as $X=\bigcup_{i=1}^{\infty}K_{i}$ where each $K_{i}$ is compact and $K_{i}\subset\text{int}K_{i+1}$.
The author of the note did not give a proof of this theorem and I have no idea how to start. May anyone help?

REPLY [2 votes]: Since $X$ is $\sigma$-compact, we can write $X = \bigcup_{i=1}^\infty C_i$, with each $C_i$ compact. Now let's define compact sets $K_i$ by induction, with $C_i\subseteq K_i\subseteq \mathrm{int}( K_{i+1})$ for all $i$.
In the base case, define $K_1 = C_1$.
Now given $K_i$, we define $K_{i+1}$. Since $X$ is locally compact, for each point $x\in K_i$, there is an open set $U_x$ and a compact set $V_x$ with $x\in U_x\subseteq V_x$. The $U_x$ cover $K_i$, so since $K_i$ is compact, it is covered by $U_{x_1},\dots,U_{x_n}$ for finitely many $x_1,\dots,x_n\in K_i$. Let $U = \bigcup_{j=1}^n  U_{x_j}$ and $K_{i+1} = C_{i+1}\cup \bigcup_{j=1}^n V_{x_j}$. Then $U$ is a union of open sets, so it is open, and $K_{i+1}$ is a finite union of compact sets, so it is compact. And we have $C_{i+1}\subseteq K_{i+1}$ and $K_i\subseteq U \subseteq \mathrm{int}(K_{i+1})$, as desired.
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