Basis of the product topology Let $A,B$ be topological spaces with $T_A,T_B$ their respective topologies and $\mathcal{B}_A,\mathcal{B}_B$ their respective bases.
How do I show that

$\{F\times G|F\in \mathcal{B}_A, G\in \mathcal{B}_B\}$ is a basis of $A\times B$?

What I thought:
Let $U\times V$ be open in $A\times B$, then $U=\cup_{i\in I}F_i$ and $V=\cup_{j\in J}G_j$.
I suppose that $U\times V=\cup_{i\in I}F_i\times\cup_{j\in J}G_j$. Is this correct? And how can I write this as a union of $F\times G$?

Start by proving that a set $U\subseteq A\times B$ that is open in the product topology can be written as a union of elements of the collection $\mathcal V:=\{F\times G\mid F\in\mathcal B_A,G\in\mathcal B_B\}$.
Let $\langle x,y\rangle\in U$. 
On base of the definition of "product topology" we conclude that some $V_x\in T_A$ and some $W_y\in T_B$ exist with:$$\langle x,y\rangle\in V_x\times W_y\subseteq U$$
$\mathcal B_A$ is a base of $T_A$ so $x\in F_x\subseteq V_x$ for some $F_x\in\mathcal B_A$.
$\mathcal B_B$ is a base of $T_B$ so $y\in G_y\subseteq W_y$ for some $G_y\in\mathcal B_B$.
Then we have:$$\langle x,y\rangle\in F_x\times G_y\subseteq V_x\times W_y\subseteq U$$
We can do this for every $\langle x,y\rangle\in U$ enabling us to write $U$ as a union of elements of the collection:$$U=\bigcup_{\langle x,y\rangle\in U}F_x\times G_y$$
A second thing that has to be shown is that the intersection of two elements $F_1\times G_1$ and $F_2\times G_2$ of $\mathcal V$ can be written as a union of elements of $\mathcal V$. But it is evident that this is the intersection of two sets that are open in the product topology, hence the intersection is open in the product topology also. Above we allready proved that such a set can be written as union of elements of $\mathcal V$, so we are ready.
