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If a topological space $X$ is separable, then every open cover of $X$ must be countable? since $X$ is separable , then there exists a countable dense subset $S$. This implies, in every open cover any set must intersect with $S$.

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No, definitely not. The Mrówka space $\Psi$ is a separable space with an irreducible open cover of cardinality $2^\omega=\mathfrak{c}$. (Irreducible means that it has no proper subcover.) The Katětov extension of $\Bbb N$ is a separable space that has an irreducible open cover of cardinality $2^{2^\omega}=2^{\mathfrak{c}}$.

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  • $\begingroup$ But each member of the cover must be intersect with dense subset. so, where is my mistake? ı could not see. I mean that which set must be countable? $\endgroup$ – ghb Nov 29 '13 at 4:26
  • $\begingroup$ @ghb: A countable set has $2^\omega$ different subsets. Moreover, two different open sets can have the same intersection with a dense subset. $\endgroup$ – Brian M. Scott Nov 29 '13 at 4:28
  • $\begingroup$ I see now, thanks. $\endgroup$ – ghb Nov 29 '13 at 4:29
  • $\begingroup$ @ghb: You’re welcome. $\endgroup$ – Brian M. Scott Nov 29 '13 at 4:30
  • $\begingroup$ If you don't demand an irreducible open cover, even the particular point topology on an infinite set will do the trick. $\endgroup$ – dfeuer Nov 29 '13 at 4:31

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