- What is a nowhere dense subspace?
Solution
1. A subset \(A\subseteq X\) is nowhere dense in \(X\) if\[ \operatorname{int}_X(\overline A)=\varnothing. \]
2. Equivalently,\[ X\setminus\overline A \]is dense in \(X\).Indeed,\[ \overline{X\setminus\overline A} =X\setminus\operatorname{int}(\overline A). \]
3. Equivalently, every nonempty open set \(U\subseteq X\) contains a nonempty open set \(V\subseteq U\) with \(V\cap A=\varnothing\).
3.1. If \(A\) is nowhere dense, take\[ V=U\cap(X\setminus\overline A), \]which is nonempty because \(X\setminus\overline A\) is dense.
3.2. Conversely, if every nonempty open \(U\) contains such a \(V\), then no nonempty open set can be contained in \(\overline A\); hence \(\operatorname{int}(\overline A)=\varnothing\).
4. For example, finite subsets of \(\mathbb R\), the set \(\mathbb Z\), and the Cantor set are nowhere dense in \(\mathbb R\). The rationals are not: \(\overline{\mathbb Q}=\mathbb R\).