Is it true that the interior of a product is the product of the interiors? Specifically, for topological spaces \(X\) and \(Y\) and subsets \(A \subseteq X\) and \(B \subseteq Y\), is \(\operatorname{int}(A \times B) = \operatorname{int}(A) \times \operatorname{int}(B)\)? What about infinite products?
Solution
1. For two factors,\[ \operatorname{int}_{X\times Y}(A\times B) =\operatorname{int}_X(A)\times\operatorname{int}_Y(B). \]
1.1. The right-hand side is open in \(X\times Y\) and contained in \(A\times B\), so it is contained in the left-hand side.
1.2. Conversely, if \((x,y)\in\operatorname{int}(A\times B)\), then some basic open set \(U\times V\) satisfies\[ (x,y)\in U\times V\subseteq A\times B. \]Hence \(U\subseteq A\) and \(V\subseteq B\), so \(x\in\operatorname{int}(A)\) and \(y\in\operatorname{int}(B)\).
2. The same argument gives the finite-product identity\[ \operatorname{int}\!\left(\prod_{i=1}^n A_i\right) =\prod_{i=1}^n\operatorname{int}(A_i). \]
3. Let now \(\prod_{i\in I}X_i\) have the product topology and put \(A=\prod_{i\in I}A_i\).
3.1. If \(A\) has nonempty interior, choose a basic open set\[ \prod_{i\in I}U_i\subseteq A, \]where \(U_i=X_i\) for all but finitely many \(i\).
3.2. For every index with \(U_i=X_i\), the inclusion \(U_i\subseteq A_i\) forces \(A_i=X_i\). Thus \(A_i=X_i\) for all but finitely many \(i\).
3.3. Conversely, if \(A_i=X_i\) for all but finitely many \(i\), then\[ \operatorname{int}(A)=\prod_{i\in I}\operatorname{int}(A_i), \]because the right-hand side is then a basic open product (possibly empty), and the finite-coordinate argument from 1 proves maximality.
3.4. Therefore\[ \operatorname{int}\!\left(\prod_{i\in I}A_i\right)= \begin{cases} \prod_{i\in I}\operatorname{int}(A_i),&A_i=X_i\text{ for all but finitely many }i,\\ \varnothing,&\text{otherwise}. \end{cases} \]
4. Hence the finite-product identity always holds, whereas for infinite products it can fail. For example,\[ \operatorname{int}_{\mathbb R^{\mathbb N}}([0,1]^{\mathbb N})=\varnothing \quad\text{but}\quad \prod_{n\ge1}\operatorname{int}[0,1]=(0,1)^{\mathbb N}\ne\varnothing. \]