Locally compact separable metric spaces Hi,
Is it true that for every locally compact separable metric space $E$ there exists a sequence $(K_n)_{n\in\mathbb{N}}$ of compact subsets of $E$ such that $K_n\subset\stackrel{\circ}{K_{n+1}}$ and $\cup K_n = E$ ?
I’m almost sure this is false but I can’t find a counterexample.
Thank you.
 A: I think the following argument works under your hypothesis:
Consider $\mathcal{B}=\{B_n\}$ a countable basis of the topology of $E$ such that $\overline{B_n}$ is compact for any $n$ (this exists since $E$ is a separable metric space, thus, it has a countable basis and then a basis like this is constructed using local compactness). 
Now, start with $K_0= \overline{B_0}$. Now, given $K_n$, define $K_{n+1}$ by the union of $\overline{B_{n+1}}$ with the closure of a finite subcovering of $K_n$ so, it is compact (being finite union of compact sets) and $K_n \subset int(K_{n+1})$. 
We get that $\bigcup K_n =E$ since it contains $\bigcup B_n=E$. 
A: Sure. Let $\{x_1,x_2,\ldots\}$ be dense in $E$, let $K_0=\emptyset$ and let $K_n$ be a compact neighbourhood of $K_{n-1}\cup\{x_n\}$ for $n=1,2,\dots$. Just make sure that $K_n$ contains a sufficiently large ball around $x_n$: Say if $\epsilon<1$ and $B_{2\epsilon}(x_n)$ is precompact, then insist on $B_\epsilon(x_n)\subseteq K_n$. Now if $y$ is not in the union and $B_{3\epsilon}(y)$ is precompact, then $x_n\in B_\epsilon(y)$ will imply $y\in K_n$.
A: If I remind correctly Bourbaki's General Topology, a locally compact space is paracompact if and only if any connected component is countable at infinity. Since your $E$ is metrizable, it is paracompact, and since it is separable, it has countably many components. Conclusion : $(K_n)$ exists.    
