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

1. 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?

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

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: $\displaystyle X\subset A~\&~Y\subset B$ all of which must be in a larger space.

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

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

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

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

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

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

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

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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complement of cartesian product

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