How to do this simple proof on intersection and union of sets? 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
 A: $$
(x,y)\in(A\times B)\cap (C\times D)
$$
$$
\iff (x,y)\in A\times B\;\mbox{and}\; (x,y)\in C\times D
$$
$$
\iff (x\in A \;\mbox{and}\; y\in B) \;\mbox{and}\;(x\in C\;\mbox{and}\;y\in D)
$$
$$
\iff (x\in A\;\mbox{and}\;x\in C) \;\mbox{and}\;(y\in B\;\mbox{and}\;y\in D)
$$
$$
\iff x\in A\cap C \;\mbox{and}\; y\in B\cap D
$$
$$
\iff (x,y)\in (A\cap C) \times (B\cap D)
$$
A: I'll point you to the tools you can use in this case, and in other problems:
You'll want to use the definitions of the set operations: the cross-product $P\times Q$, e.g., and intersection of sets: $(P \cap Q)$, e.g.


*

*$P\times Q = \{(p, q)\mid p \in P \text{ and}\; q \in Q\}$

*$P\cap Q = \{x\mid x \in P \text{ and}\;x \in Q\}$


To show set equality, say $\;P = Q,\;$ the "tried-and-true" approach to use is to show that each side is a subset of the other. (There are more direct approaches, when appropriate, but not all set equalities are obvious and/or straightforward to prove bi-directionally): 
So, use the definitions of the necessary set operations, to show that $$(A\times B) \cap (C\times D) \subseteq (A\cap C) \times (B \cap D)\tag{1}$$
and that $$(A\cap C)\times (B\cap D) \subseteq (A\times B)\cap (C\times D)\tag{2}$$
Once you've shown $(1)$ and $(2)$, you can conclude that the sets are equal.

The general strategy for showing that one set, $P$ is a subset of another set $Q$, i.e., to show that $P \subseteq Q,\;$ is to start by picking an arbitrary element of the set $P$, with the assumption: $x \in P,\,$ and then argue your way to conclude that this assumption implies $x \in Q$.  In your case, an arbitrary element in $A\times B$, e.g., would be an ordered pair: Let $(a, b) \in  (A\times B) \cap (C\times D)$. Then ...
A: Show that $(A\times B)\cap (C\times D)\subseteq (A\cap C)\times (B\cap D)$ first, and then show the other inclusion $\supseteq$.
For $\subseteq$:
Let $z=(x,y)\in (A\times B)\cap (C\times D)$. That means $(x,y)\in A\times B$ and $(x,y)\in C\times D$. Now finish it and conclude that $z=(x,y)\in (A\cap C)\times (B\cap D)$.
For $\supseteq$: Let $z=(x,y)\in  (A\cap C)\times (B\cap D)$. That means $x\in A\cap C$ and $y\in B\cap D$. Now conclude.
A: 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 $(2)$.
A: Let $x\in (A\times B)\cap (C\times D)$. Then $x\in A\times B$ and $x\in C\times D$, so $x=(a,b)$ for some $a\in A$ and $b\in B$, and $x=(c,d)$ for some $c\in C$ and $d\in D$. This means that $a=c\in A\cap C$ and $b=d\in B\cap D$ so $x\in (A\cap C)\times (B\cap D)$.
Let $y\in (A\cap C)\times (B\cap D)$. Then $y=(p,q)$ for some $p\in A\cap C$ and $q\in B\cap D$. But $p\in A$, $p\in C$, $q\in B$ and $q\in D$. Hence $y\in A\times B$ and $y\in C\times D$ so $y\in (A\times B)\cap (C\times D)$.
