Exercise 10.7

Proof. Suppose that $J_0$ is an inductive subset of the well-ordered set $J$. Also, suppose that $J_0\ne J$. Since $J_0\subset J$, it follows that there must be an $x\in J$ such that $x\notin J_0$. Thus the set $J-J_0$ is nonempty. Since clearly, this is also a subset of $J$, it must have a smallest element $\alpha$ since $J$ is well-ordered. Consider any $y\in S_{\alpha }$ so that $y<\alpha$. Then it cannot be that $y\in J-J_0$ since otherwise $\alpha$ would not be the smallest element of $J-J_0$. Since clearly $y\in J$ (since $S_{\alpha }\subset J$) it has to be that $y\in J_0$. Since $y$ was arbitrary this shows that $S_{\alpha }\subset J_0$. It then follows that $\alpha \in J_0$ since $J_0$ is inductive. However, this contradicts the fact that $\alpha \in J-J_0$ so that our initial supposition that $J_0\ne J$ must be incorrect. Hence $J_0=J$ as desired. □