How to prove that $(A×B) ∩ (B×A)=(A ∩ B) × (B ∩ A)$? It is known that, to prove that two sets are equal, we need to show that they are subset of each other. Here we have, intersection of cartesian products on the left side and cartesian product of intersections of sets on the right side, I know the definitions here that is:

Cartesian Product: If $A$ and $B$ are any two sets, then $A\times B=$ {$ (a,b), a \in A, b \in B$ }.
Intersection of sets: $A \cap B=${$ x\colon x \in A, x \in B $}.

but don't know how to prove that.
 A: Let $(a,b) \in (A \times B) \cap(B \times A),$ then we have that $a \in B$ and $b \in A.$ This gives $a,b \in A \cap B.$ Since $A \cap B=B \cap A, $ we derive $(a,b) \in (A \cap B) \times (B \cap A).$
Thus we have shown that
$$(A \times B) \cap(B \times A) \subseteq (A \cap B) \times (B \cap A).$$
It is now your turn to show that
$$(A \times B) \cap(B \times A) \supseteq (A \cap B) \times (B \cap A).$$
A: $\Rightarrow: $ If $(x, y) \in A \times  B \cap B \times A$, then $x$ must be in both $A$ and $B$, and likewise for $y$. So $(x,y) $ must be in the right side as well.
$\Leftarrow$: The reasoning is nearly identical to the previous reasoning, just reversed.
A: Let $(a,b)\in (A\times B)\cap (B\cap A).$
Then $$(a,b)\in (A\times B) \text{ and } (a,b)\in (B\times A)$$
iff $a\in A, b\in B \text{ and } a\in B, b\in A$
iff $a\in (A\cap B), b\in (B\cap A)$
iff $(a,b)\in (A\cap B)\times (B\cap A).$
A: You need to use a double inclusion demonstration.
I'll help you for the first one :
I note $C=(A×B) ∩ (B×A)$ and $D=(A ∩ B) × (B ∩ A)$.
Let's show that $C \subset D$.
To prove that, it is enough to prove that all elements $x$ in $C$ belongs to $D$.
Let $x$ be an element in $C$. By definition of $C$, it means that
$$x \in A \times B \text{ and } x \in B \times A.$$
The first inclusion means that there exists $a_1 \in A$ and $b_1 \in B$ such that $x=(a_1,b_1)$.
The second one means that there exists $a_2 \in A$ and $b_2 \in B$ such that $x=(b_2,a_2)$.
Therefore we have $x=(a_1,b_1)=(b_2,a_2)$, which gives that $\underbrace{a_1}_{\in A}=\underbrace{b_2}_{\in B}$ and $\underbrace{b_1}_{\in B}=\underbrace{a_2}_{\in A}$.
This gives you that $a_1 \in A$ and $a_1 \in B$, which means $a_1 \in A \cap B$. For the same reason, $b_1 \in A \cap B$.
Conclusion : we have proved that $x=(a_1,b_1)$ with $a_1$ and $b_1$ in $A \cap B$, which means exactly that $x \in (A \cap B) \times (A \cap B)=D$. Since we did this for any $x$ in $C$, it proves that $C \subset D$.
I hope this help, now try to do the other way around : $D \subset C$, which will give you $D=C$.
A: Apply those definitions like so:
$$\begin{align}&(A\times B)\cap(B\times A)\\[1ex]=~&\{z: z\in (A\times B)\land z\in(B\times A) \}&&\textsf{definition of }\cap\\[1ex]=~&\{\langle x,y\rangle: (x\in A\land y\in B)\land(x\in B\land y\in A) \}&&\textsf{definition of }\times\\[-1ex]\vdots~~~&&&\textsf{comutation and association}\\[-1ex]\vdots~~~&&&\textsf{definition of }\cap\\[0ex]=~&(A\cap B)\times (B\cap A)&&\textsf{definition of }\times\end{align}$$
