# Proof of Sets involving the Cartesian Product

The question goes:
Let $A,B,C,D$ be sets. Prove that $\big(A\times B\big)\bigcup \big(C\times D\big)=\big(A\bigcap C\big)\times \big(B\bigcap D\big)$
I started with the definition of the cartesian product $A \times B=\left\{\big(a,b\big): a\in A \wedge b \in B\right\}$
The union of $A\times B$ and $C \times D$ would be a set of coordinates where $\big(x,y\big): x\in A\bigcup C \wedge y\in B\bigcup D$
I am stuck here because the union of the two cartesian products on the left doesn't seem to imply that $x\in A\bigcap C$ and $y \in B\bigcap D$. Could anyone give me a hint or a possible step I might not be seeing? Please and thank you.

## 2 Answers

You are going to have a bad time proving that.

Consider A={0}, B={0}, C={1}, D={1}.

Then $A\times B=\{(0,0)\}$, $C\times D = \{(1,1)\}$,

$\big(A\times B\big)\bigcup \big(C\times D\big)=\{(0,0),(1,1)\}\ne \big(A\bigcap C\big)\times \big(B\bigcap D\big)=\emptyset$

• Thank you, I did not even consider finding counter examples before starting to prove it. Feb 6 '16 at 20:23

Yet you can actually show that

$$(A\times B)\cup(C\times D)\subset(A\cup C)\times(B\cup D)$$

or in other words:

$$(x,y)\in(A\times B)\cup(C\times D)\implies(x,y)\in(A\cup C)\times(B\cup D).$$

Perhaps you confused $\cap$ with $\cup$ ?