topology question about cartesian product I have a question about a proof I'm reading. It says: 
Suppose $A$ is an open set in the topology of the Cartesian product $X\times Y$, then you can write $A$ as the $\bigcup (U_\alpha\times V_\alpha)$ for $U_\alpha$ open in $X$ and for $V_\alpha$ open in $Y$. Why is this?
(I get that the basis of the product topology of $X \times Y$ is the collection of all sets of the form $U \times V$, where $U$ is an open subset of $X$ and $V$ is an open subset of $Y$, I just don't know why we can take this random union and say that it equals $A$.) 

It isn't a random union, it is the union for all open $U,V$ such that $U\times V$ is contained in $A$.  As a result we immediately have half of the containment, that the union is a subset of $A$.
To see why $A$ is contained in the union, consider a point $(x,y)$ in $A$.  Since $A$ is open, there must be a basic open set of the product topology that contains the point and is a subset of $A$.  But this is precisely a product $U\times V$ of open sets $U,V$ from $X,Y$ respectively.
QED
