Let \(A\) be a subset of a topological space \(X\) and let \(B\) be a subset of a topological space \(Y\). Let \(X\times Y\) be the product space, and let \(\operatorname{int}_Z(C)\) denote the interior of the set \(C\) in the space \(Z\). Prove or give a counterexample to \[\operatorname{int}_{X\times Y}(A\times B)=\operatorname{int}_X(A)\times\operatorname{int}_Y(B).\]
Solution
1. We have\[ \operatorname{int}_X(A)\times\operatorname{int}_Y(B) \subseteq \operatorname{int}_{X\times Y}(A\times B). \]
Both \(\operatorname{int}_X(A)\) and \(\operatorname{int}_Y(B)\) are open, so their product is open in \(X\times Y\). Also \[ \operatorname{int}_X(A)\times\operatorname{int}_Y(B)\subseteq A\times B. \] The interior of \(A\times B\) is the largest open subset of \(A\times B\), which proves the inclusion.
2. We also have\[ \operatorname{int}_{X\times Y}(A\times B) \subseteq \operatorname{int}_X(A)\times\operatorname{int}_Y(B). \]
Take \[ (x,y)\in\operatorname{int}_{X\times Y}(A\times B). \] Since products of open sets form a basis for the product topology, there are open sets \(U\subseteq X\) and \(V\subseteq Y\) such that \[ (x,y)\in U\times V\subseteq A\times B. \] In particular \(U\) and \(V\) are nonempty. For any \(u\in U\), choose the fixed point \(y\in V\); then \[ (u,y)\in U\times V\subseteq A\times B, \] so \(u\in A\). Thus \(U\subseteq A\). Similarly, using the fixed point \(x\in U\), we obtain \(V\subseteq B\). Therefore \[ x\in\operatorname{int}_X(A), \qquad y\in\operatorname{int}_Y(B), \] and hence \[ (x,y)\in\operatorname{int}_X(A)\times\operatorname{int}_Y(B). \]
3. Consequently\[ \operatorname{int}_{X\times Y}(A\times B) = \operatorname{int}_X(A)\times\operatorname{int}_Y(B). \]
Combine 1 and 2.