Compact subset of locally compact $\sigma$-compact

Question

Let $X$ be a non-compact, locally compact space. Also suppose that there is a sequence of compact non-empty sets $\{K_n\}_{n\in N}$ such that $$X=\bigcup_{n\in N} K_n,\quad K_n\subset K_{n+1}.$$ Now could we say that for any compact subset $K\subset X$ there is a number $n\in N$ such that $K\subset K_n$? If not, which conditions can help? or, is it possible if we choose sequence of compact sets? How about topological groups?

Answer 1

No, not as stated.

Let $X = [0,1)$ be the half-open unit interval, which is locally compact but not compact. Let $K_n = \{0\} \cup [\frac{1}{n}, 1-\frac{1}{n}]$ which is compact. Then we have $K_n \subset K_{n+1}$ and $X = \bigcup_n K_n$. But the compact set $K = [0, \frac{1}{2}]$ is not contained in any of the $K_n$.

However, under these assumptions here is something we can prove.

There exists a sequence of compact sets $K_n'$ such that $\bigcup_n K_n' = X$ and $K_{n-1}' \subset (K_{n}')^\circ$ for each $n$. In particular, for any compact set $K$, there is some $n$ with $K \subset K_n'$.

Proof. By local compactness, for each $x \in X$ there is an open set $U_x$ such that $x \in U_x$ and $\overline{U_x}$ is compact. We construct $K_n'$ recursively. To get started, let $K_0'=\emptyset$. Now to construct $K_n'$, suppose $K_{n-1}'$ is already constructed. Since $K_n \cup K_{n-1}'$ is compact, there exist $x_1, \dots, x_r$ such that $K_n \cup K_{n-1}' \subset U_{x_1} \cup \dots \cup U_{x_r}$. Set $K_n' = \overline{U_{x_1}} \cup \dots \cup \overline{U_{x_r}}$ which is compact. Then $K_{n-1}' \cup K_n \subset (K_n')^\circ$.

In particular, $\bigcup_n K_n' \supset \bigcup (K_n')^\circ \supset \bigcup_n K_n = X$. So the $K_n'$ cover $X$.

Moreover, the sets $(K_n')^\circ$ are an increasing open covering of $X$. So if $K$ is any compact set, we must have $K \subset (K_n')^\circ \subset K_n'$ for some $n$.