How do the open sets look in the product topology? 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?
 A: 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?
A: Note that the following are true (in all spaces): (fix a base $\mathcal{B}$ for a space $X$)


*

*$f: X \to Y$ is open iff $f[B]$ is open in $Y$ for every $B \in \mathcal{B}$.

*$f: X \to Y$ is continuous at $x$ iff for every open set $O$ that contains $f(x)$, there exists some $B \in \mathcal{B}$ such that $x \in B$ and $f[B] \subset O$.

*$X$ is compact iff every cover of $X$ with elements from $\mathcal{B}$ has a finite subcover. 

*$D \subset X$ is dense iff every non-empty $B \in \mathcal{B}$ intersects $D$.

*$f : Y \to X$ is continuous iff $f^{-1}[B]$ is open in $Y$ for all $B \in \mathcal{B}$.
Note that we can reason about continuity, openness, compactness, knowing only a base for the topology. So in most cases, all we really need is a good description for a base.
This is analogous to the situation of metric spaces $(X,d)$, where a base is specified (all sets of the form $B(x,r) = \{ y \in X: d(x,y) < r \}$, where $r>0$) and continuity between metric spaces is often expressed using the $\epsilon$-$\delta$ definition, which is just like 2., except using this base in both spaces. For product spaces as well, all proofs involving them essentially uses this base (or the subbase of all $\pi_s^{-1}[O]$ for open sets $O$ in $X_s$). We really do not need a description beyond the fact that they are unions of basic sets.
