Let $\{ X_s \}_{s \in S}$ be a family of topological spaces, then the product topology defined on the cartesian product $X := \prod_{s\in S} X_s$ is the coarsest (i.e. smallest) topology such that every projection map $\tau_s: X \to X_s$ is continuous (see PlanetMath).
Now I am interested how the open sets look like in this product topology. In my notes and textbook's I find, that the sets of the form
$$
\prod_{s \in S} W_s
$$
with $W_s$ open in $X_s$ and $W_s \ne X_s$ only for finitely many $s \in S$ form a base of this topology. Now I know how does the base sets look, but how does the open sets look? I know every open set could be written as an union of base sets, but because in general
$$
(A \times B) \cup (C \times D) \ne (A \cup B) \times (C \cup D)
$$
(just $\subset$ holds) I can not say for example that the open sets are the sets $\prod_{s \in S} W_s$ with $W_s$ open and $W_s \ne X_s$ only for finitely many $s \in S$. So, could something be said about the form of the open sets?
Best Answer
As Hui Yu says, you are not going too far with just the definition and set-theoretic operations: think about specific examples.
For instance, before fighting against scary monsters like infinite arbitrary products, how about looking for examples in the humble $\mathbb{R}^2$? Are you sure you could find out a simple characterization of the open sets there (which happen to be the same for the product topology and for the Euclidian, usual one)?
E.g., what about a set like this one:
$$ U = \left\{ (x,y) \in \mathbb{R}^2 \ \vert \ xy > 1 \ , \ x > 0 \right\} \quad \text{?} $$
It's an open set. Do you think you could describe it easily (I mean, without just repeating the definition of open sets in $\mathbb{R}^2$) in terms of the open sets of the basis of the product topology?