Let \(X\) be a topological space. A set \(A \subseteq X\) is called nowhere dense if the closure \(\overline{A}\) of \(A\) has empty interior, i.e., \(\operatorname{int}(\overline{A}) = \varnothing\). Show that if \(U \subseteq X\) is open, then \(A = \overline{U} \setminus U\) is nowhere dense.
Solution
1. The set \(A=\overline U\setminus U\) is closed in \(X\), and therefore \(\overline A=A\).
Since \(U\) is open, \(X\setminus U\) is closed. Also \(\overline U\) is closed by definition of closure. Hence \(A=\overline U\cap(X\setminus U)\) is an intersection of closed sets, so it is closed. Thus \(\overline A=A\).
2. The interior of \(A\) is empty.
Suppose instead that \(\operatorname{int}(A)\neq\varnothing\). Choose \(x\in\operatorname{int}(A)\) and put \(V=\operatorname{int}(A)\). Then \(V\) is an open neighborhood of \(x\) and \(V\subseteq A\subseteq\overline U\). In particular \(x\in\overline U\). By the neighborhood characterization of closure, every open neighborhood of \(x\) meets \(U\); since \(V\) is such a neighborhood, \(V\cap U\neq\varnothing\). But also \(V\subseteq A=\overline U\setminus U\subseteq X\setminus U\), so \(V\cap U=\varnothing\), a contradiction. Hence \(\operatorname{int}(A)=\varnothing\).
3. Q.E.D.
By steps 1 and 2, \(\operatorname{int}(\overline A)=\operatorname{int}(A)=\varnothing\), so \(A\) is nowhere dense.