If \(A \times B\) is finite, does it follow that \(A\) and \(B\) are finite?
Solution
Goal: Determine whether finiteness of \(A \times B\) implies finiteness of \(A\) and \(B\).
1. The claim is FALSE in general if one of the sets is empty.
1.1. Suppose \(B = \emptyset\).
1.2. By definition of the Cartesian product, \(A \times \emptyset = \emptyset\).
1.3. The empty set \(\emptyset\) has cardinality \(0\), which is finite.
1.4. However, \(A\) can be chosen to be an arbitrary infinite set (for example, \(A = \mathbb{N}\)).
1.5. Thus \(A \times B\) is finite while \(A\) is infinite.
2. If both \(A\) and \(B\) are nonempty, then BOTH \(A\) and \(B\) must be finite.
2.1. Since \(B \neq \emptyset\), fix an element \(b_0 \in B\).
2.2. Define the inclusion map \(\iota: A \to A \times B\) by \(\iota(a) = (a, b_0)\).
2.3. The map \(\iota\) is injective: if \(\iota(a_1) = \iota(a_2)\), then \((a_1, b_0) = (a_2, b_0)\), so \(a_1 = a_2\).
2.4. Since \(A \times B\) is finite and \(\iota(A) \subseteq A \times B\), the image \(\iota(A)\) is finite.
2.5. Since \(\iota: A \to \iota(A)\) is a bijection, \(A\) is finite.
2.6. Symmetrically, fixing \(a_0 \in A\) and considering \(b \mapsto (a_0, b)\) shows that \(B\) is finite.
3. Conclusion:
It does not follow in general, because if one factor is empty the other may be infinite. It does follow whenever both \(A\) and \(B\) are nonempty.