# Equivalent statements of Baire Category theorem

Prove that the following two versions of the Baire Category theorem are equivalent

1. Countable union of nowhere dense sets can not contain a ball.

2. Countable intersections of dense open sets in a complete metric space are dense.

• The second assertion is in fact a theorem, by the completeness assumption. It dropped both are equivalent, just by complementation, as does not contain a ball means has empty interior. – Jesus RS Feb 23 '15 at 19:17

Let $\{ \overline{E}_i \}_{i=1}^{\infty}$ be a countable collection of nowhere dense sets. Then $\{ \overline{E_i}^c \}_{i=1}^{\infty}$ is a countable collection of open dense sets, thus $\bigcap_{i=1}^{\infty} \overline{E_i}^c$ is dense. Let $D = \bigcap_{i=1}^{\infty} \overline{E_i}^c$. We show that $D^c = \bigcup_{i=1}^{\infty} \overline{E}_i$ contains no balls. Assume $D^c$ does contain a ball, $B$, since every ball intersects $D$ we must have that $B \cap D \neq \emptyset$ However this contradiction since $B$ is contained inside of the complement and thus $D^c$ contains no balls, i.e the countable union of nowhere dense sets contains no balls. \n
Conversely, suppose, $\{\overline{E}_n \}_{n=1}^{\infty}$ is countable collection of nowhere dense sets such that the countable union, $$\bigcup_{n=1}^{\infty}\overline{E}_n$$ contains no balls. Then $$\bigcap_{n=1}^{\infty}\overline{E_n}^c$$ is the countable intersection of open dense sets. Now, since, $$\left (\bigcup_{n=1}^{\infty}\overline{E}_n \right)^c = \bigcap_{n=1}^{\infty}\overline{E_n}^c$$ We have that, $$\bigcup_{n=1}^{\infty}\overline{E}_n \cup \bigcap_{n=1}^{\infty}\overline{E_n}^c = X$$
Let $B \subset X$ be an arbitrary ball in $X$. Then since $B \not \subset \bigcup_{n=1}^{\infty}\overline{E}_n$ we have that there exists at least one $x \in B$ such that $x \in \bigcap_{n=1}^{\infty}\overline{E_n}^c$. Since this is true for all balls we have show that $$B \cap \bigcap_{n=1}^{\infty}\overline{E_n}^c \neq \emptyset$$ Thus we have that every ball in $X$ intersects $\bigcap_{n=1}^{\infty}\overline{E_n}^c$ and thus is dense.