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 $A$.) 

This is because the sets of type $U\times V$ ($U$ open in $X$, $V$ open in $Y$) form a basis of the product topology on $X\times Y$.
In general, if you have any arbitrary product of topological spaces $\times_{\alpha\in I}X_\alpha$, let the projection on $X_\alpha$ be denoted by $\pi_\alpha$. Then a subbasis for the product topology is given by $\{\pi^{-1}_\alpha(U_\alpha)|\alpha\in I,\ U_\alpha\ \mathrm{is\ open\ in\ }X_\alpha\}$ (this is actually the definition of the product topology; notice that the projection maps are continuous wrt this topology). Now, take $U$ open in $X$, $V$ open in $Y$. Then $U\times V = \pi_X^{-1}(U)\cap\pi_Y^{-1}(V)$.
