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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.

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Do we need at all that the space is metrizable? – Martin Brandenburg Nov 11 '10 at 13:10
up vote 3 down vote accepted

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 $\bigcup B_n=E$.

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You beat me to it, but I posted mine anyhow since it seems to use less machinery. Let the voters, uh, vote. – Harald Hanche-Olsen Nov 11 '10 at 12:52
I like your robust, no-nonsense use of the definition of voter, Harald :) – Georges Elencwajg Nov 11 '10 at 13:19
How do you construct such a basis of the topology? It seems that you need to do something like in the answer of Harald. – Guillaume Brunerie Nov 11 '10 at 13:41
The collection of $B_{n,m}= B(x_n, 1/m)$ is a basis of the topology (being $x_n$ a dense countable set). Let $m_n$ be the first $m$ such that $B(x_n,1/m)$ becomes precompact. Then, $m_n$ varies semicontinuously so, the balls which are precompact also form a basis. (Yes, it is something like Harald's argument). – rpotrie Nov 11 '10 at 13:56

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 $y\in K_n$.

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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.

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