$A \subseteq X$ and $B \subseteq Y$, show that $(X \times Y) \setminus (A \times B) = (X \setminus A) \times (Y \setminus B)$ is not always true. I want to prove that $(X \times Y) \setminus (A \times B) = (X \setminus A) \times (Y \setminus B)$ with $A \subseteq X$ and $B \subseteq Y$ is not always true. However, I have difficulties to understand how this can be shown.
The solution of the right side of the equation is a group of Cartesian products of the form: $\{(x,y) | (x \in X \wedge y \in Y) \wedge (x \notin A \wedge y \notin Y)\}$, isn't it? 
I would write the solution of the left side of the equation the same way, so I do not know how they are different and so I can not find an example that shows that the equation does not always have to be true. Can somebody explain the difference?
 A: A point is in the left hand side if it is in $X \times Y$, so of the form $(x,y)$ with $x \in X, y \in Y$, but not in $A \times B$, so it is not of the form $(a,b)$ with $a \in A, b \in B$. But this can happen in two diferent ways: $x \notin A$ or $y \notin B$. In the right hand side both are the case, while either one suffices not to be in $A \times B$.
A: $X=Y=\mathbb{R},A=B=[0,1]$,
$(X \times Y) \setminus (A \times B) =\Big((-\infty,0)\cup (1,\infty)\times\mathbb{R}\Big)\cup \Big(\mathbb{R}\times(-\infty,0)\cup (1,\infty)\Big)$
$(X \setminus A) \times (Y \setminus B)=(-\infty,0)\cup (1,\infty)\times (-\infty,0)\cup (1,\infty)$ 
In particular, $(-\frac{1}{2},\frac{1}{2})$ is in the first not the second.
The correct case is that you delete a rectangle, but the incorrect one you delete a cross.
A: The set :

$(X \times Y) \backslash (A \times B)$ 

is obtained from the set $\{ (x,y) | (x∈X ∧ y∈Y) \}$, "deleting" all couples from the set $\{ (x,y) | (x∈A ∧ y∈B) \}$.
Thus, a couple $(x,b)$, with $x \in X$ and $b \in B$ will be in it, while the same couple will not belong to $(X \backslash A) \times (Y \backslash B)$.

Let :

$A = \{ 1, 2 \} \subseteq X = \{ 1, 2, 3 \}$

and :

$B = \{ v, w \} \subseteq Y = \{ v, w, z \}$.

$\langle 3, c \rangle$ is the unique element in $(X \backslash A) \times (Y \backslash B)$, while $\langle 3, a \rangle , \langle 3, b \rangle , \langle 3, b \rangle \in (X \times Y) \backslash (A \times B)$.
