$(A _1 \times B _1 ) \cup (A _2 \times B _2 )= \textbf{?} $ $(A _1 \times B _1 ) \cup (A _2 \times B _2 )= \textbf{?} $
I tried as follows: 
$$
(x,y) \in [(A _1 \times B _1 ) \cup (A _2 \times B _2 )] \iff (x,y) \in (A _1 \times B _1 ) \vee (x,y) \in (A _2 \times B _2 ) \iff (x \in A _1 \wedge y \in B _1 ) \vee ( x \in A _2 \wedge y \in B _2) \iff ((x \in A _1 \wedge y \in B _1)\vee x \in A _2)\wedge ((x \in A _1 \wedge y \in B _1) \vee y \in B _2 )\iff (( x \in H _1 \vee x \in A _2)\wedge (x \in A _2 \vee y \in B _1)) ...
$$
And I don't know where this is going. Help would be much appreciated!
Thanks in advance!
 A: I don't think there's any better way of writing the set you describe. Take, for example, $A_1=B_1=[0,1]$ and $A_2=B_2=[2,3]$. The set $(A _1 \times B _1 ) \cup (A _2 \times B _2 )$ is a disjoint union of two squares, there's no better way of putting it.
A: The only intuitive possibility of the equation would be $ (A _1 \times B _1 ) \cup (A _2 \times B _2 )=(A _1 \cup A_2 ) \times (B _1 \cup B _2 ) $, but this don't hold in general: example, if $A_1=[0,2]; A_2=[1,3]; B_1=[3,5]; B_2=[4,6]$, than $(3,3) \in (A _1 \cup A_2 ) \times (B _1 \cup B _2 ) $, but (3,3) is not in $ (A _1 \times B _1 ) \cup (A _2 \times B _2 )$. So there is no such equality.
You can prove just the inclusion: $ (A _1 \times B _1 ) \cup (A _2 \times B _2 )$ is subset of  $(A _1 \cup A_2 ) \times (B _1 \cup B _2 ) $.
A: I would suggest that you simply stop after your initial translations, and perhaps use set builder notation to define $$(A_1\times B_1)\cup (A_2\times B_2) = \{(x, y)\mid (x, y) \in A_1 \times B_1 \lor (x, y) \in A_2\times B_2\} = \{(x, y)\mid (x \in A_1 \land y \in B_1) \lor (x \in A_2 \land y \in B_2)\}.$$
As Emin notes, we can say that $(A_1 \times B_1)\cup (A_2\times B_2) \subseteq (A_1 \cup A_2) \times (B_1 \cup B_2)$. But equality here will rarely hold.
