# Why boundary of a locally closed set is nowhere dense?

Let $X$ is locally closed , i.e. exist open $U$ S.t. $X=\overline{X} \cap U$ , and $bd (X) = \overline{X} \setminus \mathring{X}$.

How can I show that $bd(X)$ is nowhere dense?

I will denote the topological space in which we work by $\Omega$. First we note that the boundary $bd(X)$ is a closed set, since it is the intersection of the two closed sets $$bd(X) = \overline{X} \cap \Omega \setminus Int(X)$$ Thus we get that $$bd(X) = \overline{bd(X)}$$ Now we recall that a definition of nowhere dense states: ''A set is nowhere dense if and only if its closure has empty interior''. Thus we have to prove that $$Int(bd(X)) = \varnothing$$ Suppose for contradictions sake that $$x \in Int(bd(X))$$ Then we know that there is an open neighbourhood $$W \in \mathcal{V}(x): W \subseteq bd(X)$$ Now if we let $U$ be the open set such that $$X = \overline{X} \cap U$$ we have that $$W \cap X = W \cap U \cap \overline{X} \subseteq X \setminus Int(X)$$ Now Since $W \subseteq \overline{X}$ we get a refinement of the last statement: $$W \cap U \subseteq X \setminus Int(X)$$ Now we can see that $W \cap U$ is non-empty, due to the fact that the $W \cap X$ is non-empty (and $U$ contains $X$). Now if $W\cap U$ is non-empty we have a contradiction (since every element in it has this set as an open neigbourhood and thus would be in the interior of $X$).