Complement Set of Cartesian product I have a hard time knowing what it means for a thing to be a complement of a cartesian coordinate.
So let's say I have the arbitrary non empty sets $A$ and $B$:
$$ (A \times B)^c$$ 
does this equal the individual parts:
$$(A \times B)^c = A^c \times B^c?$$
If so then why? 
 A: When considering subsets of $X{\times}Y$, then for all not empty $S,A \subseteq X~$ and for all not empty $T,B \subseteq Y$, we have that the relative complement of the cartesian product is:
$$S{\times}T ~\smallsetminus~ A{\times}B ~=~ (S \smallsetminus A){\times}T ~\cup~ S{\times}(T \smallsetminus B)$$
A: Assume $A{\times}B\subseteq X{\times}Y$ and so using set builder notation: $$A{\times}B = \{(x,y)\in X{\times}Y\mid x\in A~\wedge~ y\in B\}$$
Due to de Morgan's Laws of Dual Negation, the complement (relative to $X{\times}Y$) would therefore be : $$(A{\times}B)^{\complement_{X{\times}Y}} = \{(x,y)\in X{\times}Y\mid x\notin A~\vee~ y\notin B\}$$
Thus we have $$(A{\times}B)^{\complement_{X{\times}Y}} = (A^{\complement_X}{\times}Y) \cup (X{\times}B^{\complement_Y}) \cup (A^{\complement_X}{\times}Y^{\complement_Y})$$
A: Typically,
$$ (A\times B)^{c} \ne A^c \times B^c. $$
For example, suppose that $A = B = [0,1] \subseteq \mathbb{R}$.  Then $A \times B$ is the unit square in $\mathbb{R}^2$.  The complement of the unit square in $\mathbb{R}^2$ is the Euclidean plane with a square shaped hole near the origin.  On the other hand,
$$A^c = B^c = \mathbb{R} \setminus [0,1] = (-\infty,0)\cup(1,\infty).$$
Then $A^c \times B^c$ ends up being the Euclidean plane, minus a "cross" that is centered near the origin.  Specifically, it is the set of points
$$ \{ (x,y) : (x < 0 \text{ or } x > 1) \text{ and } (y < 0 \text{ or } y > 1) \}.$$
This is not the same as the plane minus a square, and so $(A\times B)^c$ is not equal to $A^c \times B^c$.
