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!

  • $\begingroup$ Related question: mathoverflow.net/questions/15841/… $\endgroup$ – Joel David Hamkins Aug 19 '10 at 15:17
  • $\begingroup$ The counterpart question deals with "maximal compact" topologies. Must they be Hausdorff? (no) $\endgroup$ – Gerald Edgar Aug 19 '10 at 20:58
  • $\begingroup$ Katetov, 1940.. $\endgroup$ – Gerald Edgar Aug 19 '10 at 21:02
  • $\begingroup$ Can I ask that you try to find a way to ask (a shortened version of) your question in the title? It will help other users know what question you want to ask, making it more likely they follow the link, and it will also help Google index your question and raise it to the top of the search page, making it more likely to help future mathematicias. Remember that you get 240 characters for the title --- more than a tweet and a half. Even "Is every minimal Hausdorff space compact, and does every Hausdorff topology contain a minimal one?" fits with room to spare. $\endgroup$ – Theo Johnson-Freyd Aug 19 '10 at 21:36

The answer to both questions is no - see 7.5 in Porter and Woods book, "Extensions and Absolutes of Hausdorff Spaces", Springer-Verlag, 1988. The space of rational numbers with the usual topology has no coarser minimal Hausdorff topology. Every Hausdorff space can be embedded in a minimal Hausdorff space; in particular, if you start with a Hausdorff space X that is not Tychonoff and embed it in a minimal Hausdorff space Y, then Y can not be compact Hausdorff.


The answer to this MO question, provided by François G. Dorais, shows that there is a minimal Hausdorff topology on a countable space that is not compact.
See Steen & Seebach 100.


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.

  • 2
    $\begingroup$ 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.) $\endgroup$ – Andreas Blass Nov 21 '10 at 23:32

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