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It is a known fact that a 2nd countable compact Hausdorff space is metrizable. What if we weaken the 2nd countable to separable only - is the space still metrizable?

The core of the question, or a better formulation, is this: Is separability equivalent to 2nd countability in the class of compact Hausdorff spaces? (Similarly as it is in some classes of topological spaces and this certainly is not answered by Wikipedia as suggested below.)

The answer is NO. You can find examples of non-metrizable separable compact Hausdorff spaces below.

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    $\begingroup$ I apologize for the question and wasting your time - I should have easily found the answer myself. $\endgroup$
    – Roman Hric
    Commented Aug 17, 2018 at 17:29
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    $\begingroup$ It was fun to several of us. $\endgroup$
    – Wlod AA
    Commented Aug 17, 2018 at 17:38
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    $\begingroup$ Standard place to look is the book Counterexamples in Topology. π-Base is a more high-tech resource. $\endgroup$
    – Nate Eldredge
    Commented Aug 17, 2018 at 18:16
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    $\begingroup$ Next ask whether a first-countable compact Hausdorff space is metrizable. $\endgroup$
    – Gerald Edgar
    Commented Aug 17, 2018 at 21:32

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The Cech-Stone compactification of the integers is by the very definition separable (as it contains the countable dense set of integers) yet it is hardly metrisable as it contains a non-separable subspace (remove the integers).

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From Wikipedia: a compact Hausdorff space is metrizable if and only if it is second-countable. See the proof on PlanetMath.

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    $\begingroup$ See also math.stackexchange.com/q/83652, math.stackexchange.com/q/1642751, math.stackexchange.com/q/573805, math.stackexchange.com/q/234018. Even more counterexamples here: math.stackexchange.com/q/74923. A web search (like I did) would have made posting your question needless. $\endgroup$
    – Alex M.
    Commented Aug 17, 2018 at 17:19
  • $\begingroup$ I should have formulated my question as follows: Is separability equivalent to 2nd countability in compact Hausdorff spaces? Sure I did the web search but didn't succeed to find the counterexamples. $\endgroup$
    – Roman Hric
    Commented Aug 17, 2018 at 18:37
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    $\begingroup$ Since second-countability is equivalent to metrizability, and since there are separable compact Hausdorff spaces that are not metrizable (see my couterexamples link), it follows that separability is weaker than second-countability. $\endgroup$
    – Alex M.
    Commented Aug 17, 2018 at 18:42
  • $\begingroup$ Of course, it was clear after opening your link to counterexamples. Anyway, even the question might be dumb, the answers might be useful also to someone else. $\endgroup$
    – Roman Hric
    Commented Aug 17, 2018 at 18:49
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Alexandrov double arrow space is a standard example of a compact Hausdorff separable space that is not metrizable.

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I can't resist.

Urysohn (followed by Bing and others) has constructed a connected Hausdorff space which has only countably many points but more than one. Of course, such spaces are separable.

On the other hand, every connected component of an arbitrary countable metric space has only one point.

Also, countable connected spaces cannot be Hausdorff compact since every connected component of an arbitrary Hausdorff compact countable space has only one point.

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