I have the following doubt. Consider a manifold $(M,\tau)$ and its product topology $\tau^2$.
How is then an open set $U\in\tau^2$ defined?
Is it $U=\bigcup_{i\in I} U_i\times V_i$, where $U_i,V_i\in\tau$ or $U=\bigcup_{i,j\in I}U_i\times V_j$?
It seems to me that it is the first one because in the second one
\begin{align} U=\bigcup_{i,j\in I}U_i\times V_j=\bigcup_{i\in I}(\bigcup_{j\in I}U_i\times V_j)=\bigcup_{i\in I}U_i\times(\bigcup_{j\in I}V_j)=(\bigcup_{i\in I}U_i)\times\text{Im }U=\text{Dom }U\times\text{Im }U \end{align}
And, so, for example, $U=A\times B\cup C\times D$ could not be an open set if $A$ and $C$, and $B$ and $D$ are disjoint open sets in $\tau$.
Am I right? Thanks.
It should be the first. An open set is a union of products of open sets, so it would take the form $\bigcup_{i \in I} U_i \times V_i$. It is a union of single terms (the terms just happen to have two factors) so only one index is needed. The second option you present is ``too big."
You should think of an open set as $\bigcup_i \mathcal{O_i}$, where each $\mathcal{O}_i$ has the form $\mathcal{O}_i = U_i \times V_i$. This makes it clear that one index suffices.