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 $y\neq 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.