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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 read topics about closed or open sets, But I can't find about locally closed.

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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 $X$).

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