The product topology on $X \times Y$ is defined to be that topology whose basis consists of sets of the form $U \times V$, where $U$ is open in $X$ and $V$ is open in $Y$. According to a lemma in my book, the topology generated by a basis is in fact just the collection of all arbitrary unions of basis elements.
I realize that what is to follow is simple, but I just want my reasoning to corroborated, as I am studying topology on my own. If $\mathcal{O}$ is open in the product topology, then $\mathcal{O} = \bigcup_{i \in I} U_i \times V_i = \left(\bigcup_{i \in I} U_i \right) \times \left(\bigcup_{i \in I} V_i \right)$. Since $\bigcup_{i \in I} U_i$ is a union of open sets in $X$, it, too, must be some open set in $X$, call it $U$; the same goes for $\bigcup_{i \in I} V_i$. Hence, $\mathcal{O} = U \times V$. So, open sets in the product topology really amount to cartesian products of open sets in the factor spaces, and so we don't have to concern ourselves with arbitrary unions.
Does this seem right?