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 $S$.
1
-
$\begingroup$ Did you also see this?: math.stackexchange.com/questions/241530/… $\endgroup$– mathematics2x2lifeCommented Nov 29, 2013 at 4:22
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
6
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 $2^{2^\omega}=2^{\mathfrak{c}}$.
answered Nov 29, 2013 at 4:22
-
$\begingroup$ 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? $\endgroup$– ghbCommented Nov 29, 2013 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. $\endgroup$ Commented Nov 29, 2013 at 4:28
-
-
-
$\begingroup$ If you don't demand an irreducible open cover, even the particular point topology on an infinite set will do the trick. $\endgroup$– dfeuerCommented Nov 29, 2013 at 4:31