Lemma $X$ is a Baire space if and only if given any countable collection $\{U_n\}$ of open sets in $X$, each of which is dense in $X$, their intersection $\cap U_n$ is also dense in X.
A space X is said to be Baire space when for given any countable collection of closed sets of X, where each closed set has empty interior in X, their union also has empty interior in X.
I can't prove why the intersection definition become equivalent to the formal Baire space definition. Any hint to prove it?
Suppose we have such a countable collection $U_n$ of dense open subsets of $X$. Consider $C_n = X \setminus U_n$, each of which are closed. Note that $\operatorname{int} C_n = \emptyset$, otherwise $\operatorname{int} C_n$ would be an open neighbourhood that $U_n$ failed to intersect, contradicting density.
By definition of a Baire space, this means that $\operatorname{int}(\bigcap_n C_n) = \emptyset$. Taking complements of both sides, $$X = X \setminus \emptyset = X \setminus \operatorname{int}\left(\bigcap_n C_n\right) = \operatorname{cl} \bigcup_n \left(X \setminus C_n\right) = \operatorname{cl} \bigcup_n U_n,$$ i.e. $\bigcup_n U_n$ is dense in $X$.
The converse follows in much the same fashion.
