If $|A \times B|$ is finite, does it follow that $A$ and $B$ are finite?
This is a question from Munkres.
I was wondering if this is a reasonable argument. Without loss of generality let us first assume that $A$ and $B$ are nonempty. First note that $A=A \times \emptyset$. Then we see that $A \times \emptyset \subseteq A \times B$. Since $|A \times B|=M$ and since $A \times \emptyset$ is a subset of $A \times B$, then $|A \times \emptyset|\leq M$.
This is one of those problems which I believe to be true, but I'm not sure how to prove it rigorously.
Best Answer
The claim is false since $X\times \emptyset = \emptyset$ for all sets $X$. Thus, for instance, $A=\mathbb R$ and $B=\emptyset $ give a counter example.
It is true however, that if both $A$ and $B$ are non-empty and $A\times B$ is finite then both $A$ and $B$ are finite. Since, if without loss of generality, $A$ is infinite then take some $b\in B$ and then the set $A\times B$ contains the set $A\times \{b\}$ which has the same infinite cardinality as $A$. Thus the cardinality of $A\times B$ is not smaller than that of $A$.