Exercise 6.5

Proof. As a counterexample, let $A={\mathbb{Z}}_{\text{+}}$ and $B=\varnothing$. Clearly, $A$ is infinite by Corollary 6.4 so that not both $A$ and $B$ are finite. It also follows from Exercise 5.3 part (c) that $A\times B=\varnothing$ since $B$ is empty. Hence clearly $A\times B$ is finite. □

Proof. Since $A\times B$ is finite there is a bijective function $f:A\times B\to \left\{1,\dots ,n\right\}$ for some $n\in {\mathbb{Z}}_{\text{+}}$. We then show that $A$ is finite by first constructing an injective function $g$ from $A$ to $A\times B$. Since $B\ne \varnothing$, there is a $b\in B$. So, for any $x\in A$, set $g(x)=(x,b)$, which is clearly in $A\times B$ so that $g$ is a function from $A$ to $A\times B$. Now consider $x$ and $x^{\prime }$ in $A$ where $x\ne x^{\prime }$. Then clearly $g(x)=(x,b)\ne (x^{\prime },b)=g(x^{\prime })$. This shows that $g$ is injective since $x$ and $x^{\prime }$ were arbitrary.

We then have that the composition $f\circ g$ is an injective function from $A$ to $\left\{1,\dots ,n\right\}$ by Exercise 2.4 part (b) since $f$ is injective as well (since it is a bijection). Therefore $A$ is finite by Corollary 6.7. An analogous argument uses the fact that $A\ne \varnothing$ to show that $B$ is also finite. □