Prove equivalent form of Baire's Category Theorem

Question

I'm trying to prove these two statements of Baire's Category Theorem are equivalent:

Let X complete metric space. A subset of X is meagre if it can be written as the countable union of nowhere dense sets.

1. The countable intersection of open dense sets in X is dense in X
2. The complement of a meagre set in X is dense in X

My approach is to apply complements to a meagre set and get a countable intersection of open dense sets. I understand that the complement of a nowhere dense set is dense and the complement of a closed set is open. However, for a set to be meagre it only needs to be the union of nowhere dense sets (no requirement to be closed).

I am struggling the meet the open requirement in 1.

Answer 1

$\newcommand{\IN}{\mathbb{N}}$Let us assume 1. and prove 2.: Let $M$ be a meagre set in $X$, i.e. $M = \bigcup_{n \in \IN} A_n$ where $A_n$ is nowhere dense. This means $(\overline{A_n})^\circ = \emptyset$. Taking complements as you suggest, we obtain $X \setminus M = \bigcap_{n \in \IN} X \setminus A_n$ and we want to show that this set is dense. Note that we have $A_n \subset \overline{A_n}$ and hence $X \setminus \overline{A_n} \subset X \setminus A_n$, whence $\bigcap_{n \in \IN} X \setminus \overline{A_n} \subset \bigcap_{n \in \IN} X \setminus A_n = X \setminus M$. If we can show that $\bigcap_{n \in \IN} X \setminus \overline{A_n}$ is dense, the bigger set $X \setminus M$ will clearly also be dense. (This is the trick you were missing. Now continue as follows:)

The sets $X \setminus \overline{A_n}$ as complements of closed sets are open. By assumption 1. it is enough to check that they are dense in order to conclude that there intersection is dense as well. In general a set is dense iff its complement has empty interior. But the complement of $X \setminus \overline{A_n}$ is $\overline{A_n}$ and this has empty interior because we assumed $A_n$ to be nowhere dense.