Basic Set Theory (Cartesian Products): a simple exercise.

**MadMikey** · November 25th 2012, 04:59 PM

Hey guys! I'm brand new to set theory; I just started learning it as a hobby. I just finished the section on cartesian products, and I'm a bit hung up on one of the exercises.

Let A,B,X, and Y be sets. A contains X, and B contains Y.
Prove that C(XxY) = AxC(Y) U C(X)xB

Mendelson's Intro to Topology was a bit vague on what the complement of a cartesian product would be (and by a bit vague, I mean completely silent). I'm guessing that C(XxY) would = {(q₁,w₁₎,(q₂,w₂),...(q_n,w_n)} where q is not in X and w is not in Y, yes?

**Plato** · November 25th 2012, 05:48 PM

Originally Posted by MadMikey

Let A,B,X, and Y be sets. A contains X, and B contains Y. Prove that C(XxY) = AxC(Y) U C(X)xB

I find this post almost unreadable. I know Bert's textbook very well having taught from it several times. You description is hard to follow. But this must be what it means: $X\subset A~\&~Y\subset B$ all of which must be in a larger space.

By definition $(p,q)\in X\times Y$ means that $p\in X\text{ AND }q\in Y$ .

So the negation is $(p,q)\notin X\times Y$ means that $p\notin X\text{ OR }q\notin Y$ .

Now $\text{ OR }$ means $\cup$ , union.

So it is very clear $(p,q)\notin X\times Y$ means that $(p,q)\in (X\times Y^c)\cup (X^c\times Y).$

**jakncoke** · November 25th 2012, 05:56 PM

Lets see, if X and Y are sets, then $X \times Y$ is the set of all 2 tuples (x, y) where $x \in X$ AND $y \in Y$ . Now the logical negation of the statement $x \in X & y \in Y$ is $x \not \in X or y \not \in Y$ . Which means that atleast one of the statements have to be true. So either $x \not \in X$ or $y \not \in Y$ or $x \not \in X$ and $y \not \in Y$ have to be true. So we have the union of 3 sets. The first case, if $x \not \in X$ is true. Now it dosent matter where y is in, it can be anywhere in its universe, since the we only care about the statement $x \not \in X$ . So the first cartesian product is $X^\complement \times B$ . The next case is if $y \not \in Y$ is true. Then it does not matter where x is in its universe, which is A, so the second cartesian product is $A \times Y^\complement$ . Now the third case is if both statements are true. $a \not \in A$ and $b \not \in B$ . which means $X^\complement \times Y^\complement$ . So ${X \times Y}^\complement = X^\complement \times B \cup A \times Y^\complement \cup X^\complement \times Y^\complement$ . Notice that $X^\complement \times Y^\complement \subset X^\complement \times B \cup A \times Y^\complement$ So, we have only $(X \times Y)^\complement = X^\complement \times B \cup A \times Y^\complement$

**MadMikey** · November 25th 2012, 06:05 PM

Originally Posted by Plato

I find this post almost unreadable.

I apologize, I don't yet know how to use the LaTex programming. I'll go check out the help forum.

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