Baire spaces (Definition) Topology In my textbook the definition of Baire space is the following: Given any countable collection $\left\lbrace A_n \right\rbrace$ of closed sets of X each of which has empty interior in X, their union $\cup A_n$ also has empty interior in X.
I have to find the equivalence with given any countable collection $\left\lbrace U_n \right\rbrace$ of open sets in X, each of which is dense in X, their intersection $\cap U_n$ is also dense in X.
Well, I did this.
⇒)Let X an Baire space, by definition let a countable collection $\left\lbrace A_n \right\rbrace$ of closed sets of X each of which has empty interior in X, their union $\cup A_n$ also has empty interior in X, that's why $X−\left\lbrace A_n \right\rbrace$ is open, $\left\lbrace A_n \right\rbrace$  has empty interior  then $X−\left\lbrace A_n \right\rbrace$ is dense in X ... So I have to see its intersection, right? 
I don't have any idea to the another.
Thanks for your help.
 A: I think this is more or less correct. The complement of $\{U_i\}$ has empty interior, so they are dense.
A slightly re-organized argument: let $\{U_i\}$ be some collection of open sets. Let $A_i:=(U_i)^{c}$. By assumption, $A_i$ are closed and have empty interior. Then $(\bigcap_{i \in \mathbb N} U_i)^{c}=\bigcup_{i \in \mathbb N} A_i$, which also has empty interior by baire category.
A: Essentially the argument is as follows:
$A$ is closed with empty interior iff $X\setminus A$ is open and dense.
Taking complements we also see that $U$ is open and dense iff $X\setminus U$ is closed with empty interior.
$\bigcup_n A_n$ has empty interior iff $X\setminus (\bigcup A_n) = \bigcap_n (X\setminus A_n)$ is dense. And dually $\bigcap U_n$ is dense iff $X\setminus (\bigcap_n U_n) = \bigcup_n (X\setminus U_n)$ has empty interior.
So if Baireness fails in one way (the union of closed empty interior way), then the complements of those sets show it fails in the other (the intersection of open dense sets) way, and vice versa. So, non-Baireness "type 1" is equivalent to non-Baireness "type 2", so the same is true for Baireness type 1 and 2 as well.
