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

  1. #1
    Newbie MadMikey's Avatar
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    Basic Set Theory (Cartesian Products): a simple exercise.

    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),...(qn,wn​)} where q is not in X and w is not in Y, yes?
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    Re: Basic Set Theory (Cartesian Products): a simple exercise.

    Quote Originally Posted by MadMikey View Post
    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).
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    Senior Member jakncoke's Avatar
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    Re: Basic Set Theory (Cartesian Products): a simple exercise.

    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
    Last edited by jakncoke; November 25th 2012 at 06:03 PM.
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    Newbie MadMikey's Avatar
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    Re: Basic Set Theory (Cartesian Products): a simple exercise.

    Quote Originally Posted by Plato View Post
    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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