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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?

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    $\begingroup$ What is your universe? $(A \times B)^c$ doesn't make sense unless you know what your universal set is. $\endgroup$
    – Xander Henderson
    Sep 12 '17 at 3:15
  • $\begingroup$ just edited the question. A and B are arbitrary non empty sets $\endgroup$
    – user510
    Sep 12 '17 at 3:18
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    $\begingroup$ $A$ and $B$ are arbitrary non-empty sets where? What is the universe that $A$ and $B$ live in? Do they live in the same universe? What is $A^c$? What is $B^c$? $\endgroup$
    – Xander Henderson
    Sep 12 '17 at 3:18
  • $\begingroup$ what do you mean? A and B are arbitrary sets $\endgroup$
    – user510
    Sep 12 '17 at 3:23
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    $\begingroup$ If $A$ is an "arbitrary set", then what is $A^c$? $\endgroup$
    – Xander Henderson
    Sep 12 '17 at 3:25
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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 $$(A{\times}B)^{\complement_{X{\times}Y}} = (A^{\complement_X}{\times}Y) \cup (X{\times}B^{\complement_Y}) \cup (A^{\complement_X}{\times}Y^{\complement_Y})$$

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  • $\begingroup$ why you don't conclude that: $A x B^c=(A^c x B^c)\cup(A^c x B)\cup(A x B^c)$ $\endgroup$
    – Mary Maths
    Jan 26 '20 at 12:22
  • $\begingroup$ I think in the last line, you did a mistake,, in the last it would be $A^{c} \times B^{c}$ $\endgroup$
    – A learner
    May 30 at 4:55
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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 $A^c \times B^c$.

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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:

$$S{\times}T ~\smallsetminus~ A{\times}B ~=~ (S \smallsetminus A){\times}T ~\cup~ S{\times}(T \smallsetminus B)$$

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