Note that $\mathbb{Q}$ is meager in $\mathbb{R}$ under the usual topology. Let $\{r_n\}_{n \in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$ . Then $\displaystyle \mathbb{Q}=\bigcup_{n \in \mathbb{N}} \{r_n\}$ and, for every $n \in \mathbb{N}$ , $\operatorname{int} \overline{\{r_n\}}=\operatorname{int} \{r_n\}=\emptyset$ .