Let $X$ be a polish space and $A \subset X$ be an open subset. Can we find a sequence of compact subsets $(K_n)_{n \in \mathbb{N}} \subset X$ such that $A= \cup_{n \in \mathbb{N}} K_n$?
I read a post where the following conditions were mentioned: "In fact the statement holds in all hereditarily Lindelöf locally compact Hausdorff spaces...".
But there are polish space which are not locally compact. What is the reason for this disparity.
$C[0,1]$ with the usual sup norm is counter-example. This is Polish and the whole space is an open subset. If it is written as a countable union of compact sets, then Baire Category Theorem would show that one of those compact sets must have an interior. But this makes the space finite dimensional, a contradiction.
[ If a compact set $K$ contains an interior point then some closed ball of (some postive radius) is contained in $K$. By transaltion, this makes the closed ball around $0$ compact. But this cannot happen in an infinite dimensional normed linear space].
$C[0,1]$ can be replaced with any separable infinite dimensional Banach space.