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So far, I have proved following two for a polish space $X$;

1.If $\{F_n\}$ is a family of closed subset of $X$, where $X=\bigcup_{n\in \omega} F_n$, then at least one $F_n$ has a nonempty inteior.

2.If $G_n$ is a dense open subset of $X$, then $\bigcap_{n\in \omega}G_n ≠ \emptyset$.

I have proved these two respectively, but can't prove the equivalence in ZF. ( I can prove the equivalence in ZF+AC$_\omega$ though)

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    @AsafKaragila: well, it looks like we agree on both things, in this case. :)2012-08-15

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First we observe that $G_n$ is open dense if and only if $F_n=X\setminus G_n$ is closed and has an empty interior. If $G_n$ is dense it intersects every open sets; so its complement does not contain any open set; and vice versa.

By De-Morgan laws we have that $\bigcap G_n=X\setminus\bigcup F_n$.

  • If the intersection of open dense is non-empty then the union of closed with empty interior is not everything.

  • If the union of closed sets with empty interior is not everything, the intersection of open dense is non-empty.