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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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    You can find counterexamples here" [A compact Hausdorff space that is not metrizable](http://math.stackexchange.com/questions/74923/a-compact-hausdorff-space-that-is-not-metrizable) and [Is a compact Hausdorff space metrizable? Maybe even complete?](http://math.stackexchange.com/questions/83652/is-a-compact-hausdorff-space-metrizable-maybe-even-complete/)2012-05-30

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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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    @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.2012-05-30