I tried to prove the following theorem whilst having $\mathbb{Z}$ in mind. However, it is by no means a rigorous proof, and because of that, I have the feeling that I may be missing something that cannot be shown by intuitive arguments:
Let $G\subseteq\mathbb{R}$, let $\bar{G}$ stand for the closure of $G$, and let $G^c$ stand for the complement of $G$.
Definitions.
- $G$ is dense in $\mathbb{R}$ if and only if $\bar{G}=\mathbb{R}$.
- $G$ is nowhere-dense in $\mathbb{R}$ if $\bar{G}$ contains no nonempty open intervals.
Theorem. A set $G$ is nowhere-dense in $\mathbb{R}$ if and only if the complement of $\bar{G}$ is dense in $\mathbb{R}$.
Proof. Let $G$ be nowhere-dense in $\mathbb{R}$. Then $\bar{G}$ contains no nonempty open intervals. This implies that $G$ must be composed of isolated points, and that $\bar{G}^c$ must contain infinitely many nonempty open intervals separated by these isolated points. Therefore, $\bar{\bar{G}^c}=\mathbb{R}$, and $\bar{G}^c$ is dense in $\mathbb{R}$. Conversely, let $\bar{\bar{G}^c}=\mathbb{R}$. It follows that either $\bar{G}^c=\mathbb{R}$ or $\bar{G}^c$ is composed of infinitely many open intervals that are separated by isolated points. If $\bar{G}^c=\mathbb{R}$, then $\bar{G}=G=\emptyset$, which is nowhere-dense in $\mathbb{R}$. If $\bar{G}^c$ is composed of infinitely many open intervals separated by isolated points, then $\bar{G}=G$ is a set composed of isolated points, which is nowhere-dense in $\mathbb{R}$. $\square$
What do you guys think?