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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?

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

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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.

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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

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