Let $(X, \mathcal{U})$ be a compact Hausdorff space and it has no isolated point. Let $A\subseteq X$ is a closed and infinite set with no isolated point . Is it true that $A$ is uncountable set?

Thanks for your help.

  • $\begingroup$ What are your thoughts? $\endgroup$ – Math1000 Dec 13 '18 at 20:52
  • $\begingroup$ @Math1000,I know that in a complete metric space, every perfect set is uncountable. But The set of rationals with the usual subspace topology is a countable perfect set. In recently, I study uniform spaces and I need to know that when a closed and infinite set is uncountable. $\endgroup$ – user479859 Dec 13 '18 at 21:00

No. Consider $X=[0,1]$, $A=\{0\}\cup\{\frac1n:n\in\mathbb{N}\}$.

EDIT: However, any (non-empty) compact Hausdorff space with no isolated points is uncountable. Show the given space is uncountable. And if you have a closed subset, it is also a compact Hausdorff space, so if it (is non-empty and) has no isolated points, it must be uncountable.

  • $\begingroup$ Thanks. What can say about $A$, if $A$ has no isolated point? $\endgroup$ – user479859 Dec 13 '18 at 21:14
  • $\begingroup$ Made an edit to answer that question. $\endgroup$ – SmileyCraft Dec 13 '18 at 21:24

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