# Minimal Hausdorff

A Hausdorff space $(X,\tau)$ is said to be minimal Hausdorff if for each topology $\tau' \subseteq \tau$ with $\tau' \neq \tau$ the space $(X,\tau')$ is not Hausdorff.

Every compact Hausdorff space is minimal Hausdorff.

I would like to know:

1) Is every minimal Hausdorff space compact?

2) Does every Hausdorff topology contain a minimal Hausdorff topology?

Many thanks!

• Related question: mathoverflow.net/questions/15841/… – Joel David Hamkins Aug 19 '10 at 15:17
• The counterpart question deals with "maximal compact" topologies. Must they be Hausdorff? (no) – Gerald Edgar Aug 19 '10 at 20:58
• Katetov, 1940.. – Gerald Edgar Aug 19 '10 at 21:02
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My instinct (need to sit down with a piece of paper to confirm it) is No and No. For the first, I'm pretty sure that any compactly generated non-compact Hausdorff space will provide a counter-example: thus, in particular, $\mathbb{R}$ with its usual topology. For the second, I'd start looking at spaces where there are two places where there's redundancy but where you can't remove both lots of redundancy at the same time.
• I don't think $\mathbb R$ works. There's a coarser Hausdorff topology obtained by making everything that used to diverge to $+\infty$ converge to 0 instead. (Pictorially, take the $+\infty$ end of $\mathbb R$ and bend it back to approach 0.) – Andreas Blass Nov 21 '10 at 23:32