Royden 10.2.17

$\textbf{Problem 10.2.17}$ In a complete metric space, is the union of a countable collection of nowhere dense sets also nowhere dense?

$\textit{Solution}$: No. Consider $(\mathbb{R}, |\cdot |)$ which is complete as a metric space. A singleton point $\{x\} \in{\mathbb{R}}$ is nowhere dense. $closure\{x\}=\{x\}$ and $interior\{x\}= \emptyset$ since $\not \exists B_r(x) \subseteq \{x\}$. (Every ball about $x$ contains $x$, not the other way around). $\mathbb{Q}$ is countable, so we can write $ \mathbb{Q}= \cup_{i=1}^\infty \{q_i\}$ while $closure(\mathbb{Q})=\mathbb{R}$ and $interior(\mathbb{R})=\mathbb{R} \neq \emptyset$ ($\mathbb{R}$ is both open and closed). So, without additional conditions, we cannot determine if the countable union of nowhere dense subsets will be nowhere dense, dense, or neither.