# (ZF) Equivalent statements to Baire Category Theorem

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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If two statements are true, then they are equivalent. However, those two are equivalent for all topological spaces $X$ (as per Asaf's answer), which is something quite more substantial. – tomasz Aug 15 '12 at 15:40
@tomasz: While you are very correct about the equivalence of two true statements (or even two provable statements); it is a good idea to know that there is an actual proof of the equivalence, so if in the future you want to prove either of them, one would be enough. – Asaf Karagila Aug 15 '12 at 16:16
@AsafKaragila: But proving equivalence of two statements using assumptions strong enough to actually prove both of them is not really substantial, is it (except as a stepping stone in proof of one of them, perhaps)? That's why I said that what is somewhat substantial is the fact that they are equivalent even with weaker assumptions, i.e. without assuming that $X$ is Polish. – tomasz Aug 15 '12 at 16:50
@tomasz: Don't get me wrong. I agree with you in general. However I also see the value of understanding the equivalence. – Asaf Karagila Aug 15 '12 at 17:05
@AsafKaragila: well, it looks like we agree on both things, in this case. :) – tomasz Aug 15 '12 at 17:19

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 $\bigcap G_n=X\setminus\bigcup F_n$.