Answer
See the solution.
Work Step by Step
To show this, we will show for any set $A$, $A\times \emptyset $ and $\emptyset \times A$ are both empty.
$Proof.$
Let $A$ be a set. Then $A\times \emptyset =\{(a,b): a \in A \text{ and } b\in \emptyset\}$. But there are no elements in the empty set. Hence there are no ordered pairs $(a,b)$ for which $a \in A$ and $b\in \emptyset $. Thus $A\times \emptyset $ must be empty. Similarly, there can be no elements in $\emptyset \times A=\{(b,a): b \in \emptyset \text{ and } a\in A\}.$ Thus, $\emptyset \times A$ must also be empty. Therefore, $A \times \emptyset=\emptyset = \emptyset \times A._\Box$
