Answer
Either or both of $A$ and $B$ is empty.
Work Step by Step
To answer this question, we will prove if $A$ and $B$ are sets and $A\times B=\emptyset $, then either $A=\emptyset$ or $B=\emptyset,$ where we are using the inclusive 'or'.
$Proof.$
Let $A$ and $B$ be sets. We will show $(A\times B=\emptyset) \rightarrow (A=\emptyset \text{ or } B=\emptyset)$ by contraposition. So suppose $A$ is nonempty and $B$ is nonempty. Then there exists an element $a$ in $A$ and an element $b$ in $B$. Thus $(a,b) \in A\times B$, which means $A\times B \neq \emptyset.$ Hence, if $A\times B=\emptyset $, then either $A=\emptyset$ or $B=\emptyset._\Box$
