# Are there Hausdorff spaces which are not locally compact and in which all infinite compact sets have nonempty interior?

Here is the background material from which I am working:

1. The Cantor set is an uncountable compact Hausdorff space with empty interior.
2. In a locally compact Hausdorff space, each countable set has empty interior.
3. 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. I am guessing the example will be an exotic function space.

• Your point 2 doesn't hold, $\mathbb{Z}$ is a locally compact Hausdorff space with the discrete topology. – Daniel Fischer Jul 11 '13 at 22:57
• But in a locally compact Hausdorff space each countable union of nowhere dense sets has empty interior, i.e. it is a Baire space. So if a point is nowhere dense, which means it is not open, then 2) is right. – Stefan Hamcke Jul 11 '13 at 23:02

Let $X=\Bbb R\cup\{\infty\}$ with the discrete topology on $\Bbb R$, and a neighborhood of $\infty$ is a set with a countable complement. This $X$ is clearly Hausdorff.
Now, the funny thing about this space is that each compact set is finite. Such a space is called anti-compact. To see this, let $S$ be an infinite subset. If it avoids $\infty$, then it has the discrete topology and is not compact. If $\infty\in S$, then choose a countable subset $Q$ of $S$ and note that $(S-Q)\cup\bigcup_{q\in Q}\{q\}$ is an infinite open cover of $S$ without a finite subcover. This means that no infinite set is compact. In particular, $\infty$ has no compact neighborhood, so $X$ is not locally compact.