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Let $(X,d)$ be a metric space. When can we guarantee that there exists a sequence of compact subsets $(K_n)\subset X$ such that $K_{n}\subset\text{int}(K_{n+1} )$ and $X =\bigcup_n K_n$?

John Paul Littlefield
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2 Answers2

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A necessary and sufficient condition (for topological spaces, not just metric spaces) is that $X$ is $\sigma$-compact and locally compact.

This is obviously necessary. On the other hand, suppose $X$ is $\sigma$-compact and locally compact. Since it is $\sigma$-compact, there is a sequence $J_n$ of compact sets with $\cup_n J_n = X$. Construct the sequence $K_n$ as follows, starting with $K_1 = J_1$. Given $K_n$, for each $x \in K_n \cup J_{n+1}$ take a compact neighbourhood $U(x)$. The open cover $\{U(x): x \in K_n \cup J_{n+1}\}$ of the compact set $K_n \cup J_{n+1}$ has a finite subcover: let $K_{n+1}$ be its union. Thus $K_{n+1}$ is a compact set whose interior contains $K_n \cup J_{n+1}$.

Robert Israel
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Just a partial answer for Banach spaces: If $X$ is a finite dimensional Banach Space, this is true, just take $K_n:=\overline{B_n(0)}$.

On the other hand if $X$ is an infinite dimensional Banach space this is wrong, because for $K_N\subset \operatorname{int}(K_{n+1})$ to be correct, the interior of $K_{n+1}$ would have to be nonempty. Hence there would be a closed Ball $\overline{B_r(x)}\subset K_{n+1}$ which would have to be compact as well. This is a contradiction to $X$ being infinite dimensional (see e.g. this question Is any closed ball non-compact in an infinite dimensional space?)

humanStampedist
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