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.

  • $\begingroup$ 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. $\endgroup$ – 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.


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