Operations on sets: how to transform the left side to the right side? Could you please give me a detailed description about what is happening here:
$(X_1 \times X_2)\backslash (A_1 \times A_2)=((X_1 \backslash A_1) \times X_2) \cup (X_1 \times(X_2 \backslash A_2))$
?
I know that I can rewrite $(X_1 \times X_2)\backslash (A_1 \times A_2)$ as $(X_1 \times X_2)\cap \overline{(A_1 \times A_2)}$, but I do not understand why the sets on the right side are grouped this way, as much as where the $\cup$ comes from. Moreover, I have never used set operations with the product $\times$. In other words - I do not see how the left side is transformed to the right side. 
Will be glad if you help.
p.s. btw. $A_i \subset X_i$.
 A: An arbitrary element of $(X_1 \times X_2) \setminus (A_1 \times A_2)$ is a pair $(x_1,x_2) \in X_1 \times X_2$ such that $(x_1,x_2) \not \in A_1 \times A_2$. To say that $(x_1,x_2) \in A_1 \times A_2$ is precisely to say $x_1 \in A_1 \wedge x_2 \in A_2$, so negating this gives $x_1 \not \in A_1 \vee x_2 \not \in A_2$. This is ultimately where the $\cup$ comes from, as you shall soon see, since we have to split into cases:


*

*If $x_1 \not \in A_1$, then $x_1 \in X_1 \setminus A_1$ and $x_2 \in X_2$ (regardless of whether it's in $A_2$ or not), so that $(x_1,x_2) \in (X_1 \setminus A_1) \times X_2$;

*If $x_2 \not \in A_2$, then $x_2 \in X_2 \setminus A_2$ and $x_1 \in X_1$ (regardless of whether it's in $A_1$ or not), so that $(x_1,x_2) \in X_1 \times (X_2 \setminus A_2)$.


Putting this together yields $(x_1,x_2) \in ((X_1 \setminus A_1) \times X_2) \cup (X_1 \times (X_2 \setminus A_2))$, and hence
$$(X_1 \times X_2) \setminus (A_1 \times A_2) \subseteq ((X_1 \setminus A_1) \times X_2) \cup (X_1 \times (X_2 \setminus A_2))$$
Similar reasoning proves the other direction of containment, so the two sets are equal.
As a general tip, proving set equalities by double containment (as is (half-)done above) is a good thing to do if it's not immediately clear how a string of algebraic manipulations would help you to arrive at the answer.
