# Is dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?

The answer to Cardinality of a locally compact space without isolated point shows that AC is required to show that if $X$ is a compact Hausdorff space with no isolated points then $|X| \ge 2^{\aleph_0}$.

But I haven't been able to find anything sensible about whether dependent choice is needed to prove the weaker statement that $|X| \not\le \aleph_0$.

http://mathoverflow.net/questions/38450/compact-hausdorff-spaces-without-isolated-points-in-zf seems to try to answer the question, but it falls short. Choosing points is not a problem, but choosing open sets is.

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Oh good! This is better than sleep! gets out his pen and paper... –  Asaf Karagila May 7 '13 at 0:59
(In my deleted answer, to those who can read it, there are a couple of mistakes that I have yet to correct. When those are cleared up, I'll fix and undelete it.) –  Asaf Karagila May 7 '13 at 20:33
I avoid using the built in chat system. Feel free to send me an email, my address is not very difficult to find... –  Asaf Karagila May 19 '13 at 17:14