TITLE: How to do this simple proof on intersection and union of sets?
QUESTION [5 upvotes]: The proof example preceding the question seems a bit random to me, if someone could explain the proof I would understand all questions of the type

(a) Let $A, B, C, D$ be arbitrary sets. Prove that $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$

Regards

REPLY [6 votes]: $$
(x,y)\in(A\times B)\cap (C\times D)
$$
$$
\iff (x,y)\in A\times B\;\mbox{and}\; (x,y)\in C\times D
$$
$$
\iff (x\in A \;\mbox{and}\; y\in B) \;\mbox{and}\;(x\in C\;\mbox{and}\;y\in D)
$$
$$
\iff (x\in A\;\mbox{and}\;x\in C) \;\mbox{and}\;(y\in B\;\mbox{and}\;y\in D)
$$
$$
\iff x\in A\cap C \;\mbox{and}\; y\in B\cap D
$$
$$
\iff (x,y)\in (A\cap C) \times (B\cap D)
$$

REPLY [3 votes]: The standard approach to such problems works fine here: show that
$$(A\times B)\cap(C\times D)\subseteq(A\cap C)\times(B\cap D)\tag{1}$$
and that
$$(A\cap C)\times(B\cap D)\subseteq(A\times B)\cap(C\times D)\;.\tag{2}$$
Each of these can be done by what I call element-chasing: let $x$ be an arbitrary element of the lefthand side, and show that it must belong to the righthand side as well. To get you started, I’ll do $(1)$.

Let $x\in(A\times B)\cap(C\times D)$. Then $x\in A\times B$ and $x\in C\times D$. Since $x\in A\times B$, there must be an $a\in A$ and a $b\in B$ such that $x=\langle a,b\rangle$. On the other hand, this ordered pair $\langle a,b\rangle$ is also an element of $C\times D$, so it must be the case that $a\in C$ and $b\in D$. Now we know that $a\in A$ and $a\in C$, so $a\in A\cap C$. Similarly, $b\in B$ and $b\in D$, so $b\in B\cap D$. Thus, the ordered pair $\langle a,b\rangle$ belongs to the Cartesian product $(A\cap C)\times(B\cap D)$: $x=\langle a,b\rangle\in(A\cap C)\times(B\cap D)$. And since $x$ was an arbitrary element of $(A\times B)\cap(C\times D)$, it follows that $$(A\times B)\cap(C\times D)\subseteq(A\cap C)\times(B\cap D)\;.$$

Now see if you can make the same sort of argument to prove $(2)$.