Equivalence of different forms of Baire Category Theorem

Question

The first form of Baire Category Theorem I learned is that in a complete metric space $X$, if $\{G_i\}$ is a set of open dense set, then $\cap_{i=1}^\infty G_i$ is dense.

The second form states that every complete metric space is a Baire space where Baire space means that every non-empty open set is second category, i.e. not first category.

(The definition of first category is that if A is first category, then A can be written as a countable union of nowhere dense sets.)

I wonder how the first form of Baire Category Theorem implies the second form.

Answer 1

Here is a proof that the first form implies the second, we want to show the contrapositive. Assume that there is some non-empty open set $U$ of the first category then I claim that we can find a countable collection of open dense sets whose intersection is not dense. Since $U=\bigcup_{i=1}^{\infty} A_i$ with $int(\overline{A_i})=\emptyset$ then we find that $\bigcap_{i=1}^{\infty} \left(\overline{A_i}\right)^c \subseteq \bigcap_{i=1}^{\infty} A_i^c \subseteq U^c$ furthermore $\overline{\left(\overline{A_i}\right)^c}=(int\left(\overline{A_i}\right))^c=X$ so they are all dense. However $\bigcap_{i=1}^{\infty} \overline{A_i}^c$ is contained in $U^c$ so it cannot be dense because $U$ was non-empty. Q.E.D.