# Is there a non-locally compact Hausdorff space in which all infinite compact sets (of which there is at least one) have nonempty interior?

Here is the background material from which I am working:

• The Cantor set is an uncountable compact Hausdorff space with empty interior.
• In a locally compact Hausdorff space, each countable set has empty interior.
• The rational numbers with the subspace topology is a non-locally compact Hausdorff space in which all compact sets have empty interior.

I am trying to find a non-locally compact Hausdorff space in which all infinite compact sets have nonempty interior, under the assumption that the space does have at least one infinite compact set. I am guessing the example will be an exotic function space.

I first posed this question without specifying that there should be at least one infinite compact set, and this was solved by Stefan H. on this site.

Let $X$ be the space from Stefan H.'s answer and let $Y=\{\frac1n|\;n\in\mathbb N\}\cup\{0\}$ with the subspace topology inherited from $\mathbb R$. Now, simply take their topological sum $Z=X+Y$. This space is Hausdorff, because $X$ and $Y$ are, and it is not locally compact, because $\infty\in X$ still doesn't have a compact neighborhood. Furthermore, $Y$ is an infinite compact subset. Every infinite compact subset is of the form $A+B$, with $A\subseteq X$ a finite set and $B\subseteq Y$ an infinite set containing $0$. But such a set necessarily has non-empty interior, since a singleton $\{y\}\subseteq Y$ is open iff $y\neq 0$.