Exercise 6.2

Proof. Suppose that $B$ is not finite and $B\subset A$ but that $A$ is finite. Since $B\subset A$, either $B=A$ or $B$ is a proper subset of $A$. In the former case, we clearly have a contradiction since $B$ would be finite since $A$ is and $B=A$. In the latter case, we have that there is a bijection from $A$ to $\left\{1,\dots ,n\right\}$ for some $n\in {\mathbb{Z}}_{\text{+}}$ by definition since $A$ is finite. Then, since $B$ is a proper subset of $A$, it follows from Theorem 6.2 that there is a bijection from $B$ to $\left\{1,\dots ,m\right\}$ for some $m<n$. However, then clearly $B$ is finite by definition, which is also a contradiction since we know $B$ is not finite. Hence in either case there is a contradiction so that $A$ must not be finite. □