(May 2016) Given any topological space \(Z\) and subset \(D\subseteq Z\), let \(\operatorname{Cl}_Z(D)\) denote the closure of \(D\) in \(Z\). Show that if \(X\) and \(Y\) are topological spaces and \(A\subseteq X\), \(B\subseteq Y\), then \[\operatorname{Cl}_{X\times Y}(A\times B)=\operatorname{Cl}_X(A)\times\operatorname{Cl}_Y(B).\]
Solution
We prove both inclusions by the neighborhood characterization of closure.
Suppose \((x,y)\in\operatorname{Cl}_{X\times Y}(A\times B)\). If \(U\subset X\) is any open neighborhood of \(x\), then \(U\times Y\) is an open neighborhood of \((x,y)\), so it meets \(A\times B\). Hence \(U\cap A\ne\varnothing\). Thus \(x\in\operatorname{Cl}_X(A)\). Similarly, using \(X\times V\), we get \(y\in\operatorname{Cl}_Y(B)\). Therefore \[ \operatorname{Cl}_{X\times Y}(A\times B) \subseteq \operatorname{Cl}_X(A)\times\operatorname{Cl}_Y(B). \]
Conversely, suppose \(x\in\operatorname{Cl}_X(A)\) and \(y\in\operatorname{Cl}_Y(B)\). Let \(W\) be any open neighborhood of \((x,y)\) in \(X\times Y\). There are open sets \(U\ni x\) and \(V\ni y\) with \[ (x,y)\in U\times V\subseteq W. \] Since \(x\in\operatorname{Cl}_X(A)\) and \(y\in\operatorname{Cl}_Y(B)\), choose \(a\in U\cap A\) and \(b\in V\cap B\). Then \[ (a,b)\in W\cap(A\times B). \] Thus every neighborhood of \((x,y)\) meets \(A\times B\), so \((x,y)\in\operatorname{Cl}_{X\times Y}(A\times B)\). Hence equality holds.