If we have a Cartesian product of topological spaces $X_i$, i.e. $\prod_{i\in I}X_i$, then for the product topology we have that products of open sets are open in case $I$ is finite.
Let A be a finite set equipped with discrete topology and let $X_1:=A^{\mathbb{Z}}$ the product space with associated product topology.
Moreover let $X_2$ be some other topological space.
Now, I am wondering if we have again that for $X_1\times X_2$ we have that products of open products are open? $I=\left\{1,2\right\}$ is finite but if we write $X_1$ as $X_1=\prod_{i\in\mathbb{Z}}A_i$ with $A_i=A$ for all $i\in\mathbb{Z}$, we have
$$
X_1\times X_2=\left(\prod_{i\in\mathbb{Z}}A_i\right)\times X_2
$$
which does not look as a finite product.
On the other Hand, we know rather good how the product topology on $X_1$ Looks like, namely that it has the set of cylindersets as a base.
So, in this case I do not know if the product of open sets are again open.
Best Answer
The product you wrote does not have to be finite because it equals the whole space $X_1\times X_2$, which is always open. An example of an open subset of $X_1 \times X_2$ would be \begin{equation} \left(\prod_{i\in \mathbb{Z}}A_i \right)\times X_2 \end{equation} where $A_i$ is an open subset of $X_i$, and $A_i = X_i$ for all but finitely many $i$.