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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Did you also see this?: http://math.stackexchange.com/questions/241530/if-x-is-a-topological-separable-space-then-every-open-set-is-a-union-of-a-coun – mathematics2x2life Nov 29 '13 at 04:22
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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 $2^{2^\omega}=2^{\mathfrak{c}}$.
Brian M. Scott
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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? – ghb Nov 29 '13 at 04:26
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@ghb: A countable set has $2^\omega$ different subsets. Moreover, two different open sets can have the same intersection with a dense subset. – Brian M. Scott Nov 29 '13 at 04:28
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If you don't demand an irreducible open cover, even the particular point topology on an infinite set will do the trick. – dfeuer Nov 29 '13 at 04:31
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@dfeuer: My default assumption is that all spaces are $T_1$. (And here I did better: $\Psi$ is Tikhonov, and $\kappa\Bbb N$ is Hausdorff.) – Brian M. Scott Nov 29 '13 at 04:34