# 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.

$A$ 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.
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.