Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Is a compact Hausdorff space metrizable? Maybe even complete?

We know that a second countable locally compact Hausdorff space is a Polish space. Does compact Hausdorff also imply Polish?

share|improve this question

marked as duplicate by t.b., Martin Sleziak, Asaf Karagila, Michael Greinecker, Leonid Kovalev Aug 14 '12 at 18:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer 1

up vote 6 down vote accepted

No: $\omega_1+1$ with the order topology is compact and Hausdorff but not even first countable, let alone metrizable.

Added: $\beta\Bbb N$, the Čech-Stone compactification of the natural numbers with the discrete topology, is another example, and it’s even separable.

share|improve this answer
Thanks, but what is the "weakast" assumption we have to add in order that a compact Hausdorff space is polish? As already mentioned, we know that second countability suffices, but is it also necessary? –  Andy Teich May 30 '12 at 10:19
@Andy: Yes, in the sense that every Polish space is automatically second countable. Any condition that (together with compactness and Hausdorffness) implies metrizability would do, and second countability is the simplest choice. –  Brian M. Scott May 30 '12 at 10:21

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