Let \(A \subset X\) and \(B \subset Y\). Show that in the space \(X \times Y\),
\[ \overline{A \times B} = \overline{A} \times \overline{B}. \]
Solution
First, \(A\times B\subseteq\overline A\times\overline B\), and the latter set is closed by E-PBT3I. Hence \[ \overline{A\times B}\subseteq\overline A\times\overline B. \] For the reverse inclusion, let \((x,y)\in\overline A\times\overline B\). A basic neighborhood of \((x,y)\) has the form \(U\times V\) with \(x\in U\), \(y\in V\). Since \(x\in\overline A\) and \(y\in\overline B\), \[ U\cap A\neq\varnothing,\qquad V\cap B\neq\varnothing. \] Choose \(a\in U\cap A\) and \(b\in V\cap B\). Then \((a,b)\in(U\times V)\cap(A\times B)\). Thus every basic neighborhood of \((x,y)\) meets \(A\times B\), so \[ (x,y)\in\overline{A\times B}. \] Therefore \[ \boxed{\overline{A\times B}=\overline A\times\overline B}. \]