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.



Locally compact separable metric spaces
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 $\bigcup B\_n=E$. 


