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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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marked as duplicate by Chris Eagle, Martin Sleziak, t.b., J. M. is a poor mathematician, Rudy the Reindeer Jan 4 '12 at 15:03

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    $\begingroup$ A simple example is an uncountable product of a two-point discrete space with itself. This is Hausdorff and compact, because these are preserved by products, but it is not second-countable, and thus not metrizable, because a compact metric space is second-countable. $\endgroup$ – Carl Mummert Nov 19 '11 at 16:04
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    $\begingroup$ 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. $\endgroup$ – Michael Greinecker Jan 4 '12 at 14:02
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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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  • $\begingroup$ Thanks. But could you maybe also provide an answer to the questions, as to what is the connection between the completeness and compactness of a metric space ? $\endgroup$ – resu Nov 19 '11 at 14:58
  • $\begingroup$ Done (to the best of my knowledge). $\endgroup$ – Daria Morys Nov 19 '11 at 15:05
  • $\begingroup$ @resu: Čech-completeness is the best answer to what compact and completely metrizable spaces have in common, but it might also be pointed out that a metric (not just metrizable) space is compact iff it is both complete and totally bounded. $\endgroup$ – Brian M. Scott Nov 19 '11 at 15:19
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    $\begingroup$ "General topology" by R. Engelking. $\endgroup$ – Daria Morys Nov 19 '11 at 15:36
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    $\begingroup$ @resu: There’s a good discussion in Ryszard Engelking, General Topology, which is one of the most comprehensive reference texts around. Unfortunately, it may be hard to find. I don’t immediately find anything useful on-line. $\endgroup$ – Brian M. Scott Nov 19 '11 at 15:46
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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.

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