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A compact Hausdorff space that is not metrizable

Is it true that every topological space $X$ that is Hausdorff and compact is also metrizable? Maybe even complete? What is the relationship between the completeness and the compactness in a metric space ?

I'm not asking this question out of nowhere. The reason is that almost every theorem I have encountered lately about compact metric spaces could easily be generalized to compact Hausdorff space but also some theorems I encountered about complete metrics have a counterpart for compact Hausdorff spaces (e.g. Baire's theorem).

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    By Urysohn's metrization theorem, a second countable compact Hausdorff space is metrizable. Since every compact metric space is second countable and Hausdorff, this means that a compact Hausdorff space is metrizable iff it is second-countable. There is also a result that a metrizable space is compact if and only if every metric metrizing it is complete. I don't know a reference for this result.2012-01-04

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No, take $\omega_1+1=[0,\omega_1]$ with the order topology ($\omega_1$ is the smallest uncountable ordinal). It is not metrizable, since the character of $\omega_1$ is uncountable. Another example is the Stone-Čech compactification $\beta \mathbb{N}$ of natural numbers.

Each compact space is Čech-complete. A metrizable space space is Čech-complete if and only if it is homeomorphic to a complete metric space. Any Čech-complete space is a Baire space.

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    ok, thank you very much.2011-11-19
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To address your second question in perhaps a simpler way than Daria's answer: every compact metric space is complete. This is easy to prove using the Bolzano-Weierstrass theorem: any Cauchy sequence $\{x_n\}$ has a subsequence that converges to some point $x$. Then use the triangle inequality to show that the original sequence $\{x_n\}$ must converge to $x$.

Of course the converse is false: there are complete metric spaces which are not compact. $\mathbb{R}$, for example.