Baire: Equivalent Statements How to proof that these statements are equivalent:


*

*Every intersection of countably many dense open sets is dense.

*The interior of every union of countably many closed nowhere dense sets is empty.


Moreover, why does it follow for Baire Spaces and why is it strictly weaker that:


*

*$X=\bigcup_{k=1}^\infty A_k\quad\Rightarrow\quad\exists k_0\in\mathbb{N},x_0\in X: \operatorname{cl}(A_{k_0})\in\mathcal{N}_{x_0}$


...in words, the space is not meagre iff whenever it is a given by a countable union of subsets then at least one of them is large.
 A: Hint: 
Take the complement. The complement of a dense open set is a closed nowhere dense set.
Solution:
Assume that $\bigcap_n U_n$ is dense where $(U_n)_{n\in I}$ are open sets. Then
the interior of $$\left(\bigcap_n U_n\right)^c = \bigcup_n U_n^c$$ is empty (since its complement is a dense set). But $U_n^c$ are closed nowhere dense sets (closed because they are complements of open sets and nowhere dense, since they are closed and have empty interior).
The other direction goes the same.
A: $A_n, n\in\Bbb N$, are countably many closed nowhere dense sets if and only if $D_n:=X-A_n$ are open dense sets (More generally $A_n$ is nowhere dense iff $D_n$ has a dense interior). Now 
$$\overline{\bigcap_n D_n}=\overline{X-\bigcup_n A_n}
=X-\text{int}\left(\bigcup_nA_n\right)$$ So $\bigcap D_n$ is dense iff $\bigcup A_n$ has empty interior.
Note that in the second statement is turned into an equivalent one by omitting the word "closed". In one direction this is trivial. For the other direction, note that the closure of a nowhere dense  set is a closed nowhere dense set. And if the union of their closures has empty interior, then so does the union of the original sets.
