This entry provides another example of a nowhere dense set.

Proof. Let $B=\partial A$ . Since $B = \overline{A}\cap \overline{A^\complement}$ , it is closed, so all we need to show is that $B$ has empty interior $\int(B)=\varnothing$ . First notice that $B= \overline{A}\cap A^\complement$ , since $A$ is open. Now, we invoke one of the interior axioms, namely $\int(U\cap V)=\int(U)\cap \int(V)$ . So, by direct computation, we have $$\int(B)=\int(\overline{A})\cap \int(A^\complement) = \int(\overline{A})\cap \overline{A}^\complement \subseteq \overline{A}\cap \overline{A}^\complement =\varnothing.$$ The second equality and the inclusion follow from the general properties of the interior operation, the proofs of which can be found here.

Remark. The fact that $A$ is open is essential. Otherwise, the proposition fails in general. For example, the rationals $\mathbb{Q}$ , as a subset of the reals $\mathbb{R}$ under the usual order topology, is not open, and its boundary is not nowhere dense, as $\overline{\mathbb{Q}}\cap \overline{\mathbb{Q}^\complement} = \mathbb{R}\cap \mathbb{R}=\mathbb{R}$ , whose interior is $\mathbb{R}$ itself, and thus not empty.