Let $H$ be a closed set then, $Cl(H) =H$ and hence the $\partial H \subset H$.

Now to show that the boundary is nowhere dense, it would suffice to show that $Int(Cl(\partial H)) =\emptyset$, i.e., $Int(\partial H) = \emptyset$, but how do I proceed further in order to show this?


1 Answer 1


Let $U$ be an open set such that $U\subset\partial H$. We'll show that $U=\emptyset$:

Since $\partial H\subset H$ (since $H$ is closed), we must have $U\subset H$. Since $U$ is open, this implies that $U\subset\operatorname{Int}(H)$.

Hence $U\subset\partial H\cap\operatorname{Int}(H)=\emptyset$.


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