Note that $\displaystyle S_\alpha = \{j |j\in J\quad and\quad j<\alpha\}.$

Let $\displaystyle J$ be a well-ordered set. A subset $\displaystyle J_0$ of $\displaystyle J$ is said to be inductive if for every $\displaystyle \alpha\inJ,$

$\displaystyle

(S_{\alpha} \subset J_0) ==> \alpha \in J_0.

$

Theorem (The principal of transitive induction). If J is a well-ordered set and $\displaystyle J_0$ is an inductive subset of $\displaystyle J$, then $\displaystyle J_0=J$.

Can someone help me get started on proving this. I am trying to show that [tex] J \subseteq J_0 [/Math] by choosing $\displaystyle \alpha \in J, $ assuming that $\displaystyle \alpha \notin J_0$ and showing that there is a contradiction. Is this a good approach and do you have any hints that I can use to get anywhere?

Ok so I made some progress, but I am not sure if I am moving in the right direction.

Proof: Choose $\displaystyle \alpha \in J $ and assume that $\displaystyle \alpha \notin J_0.$ Then $\displaystyle J_0\subseteq S_\alpha. $

If $\displaystyle J_0\subset S_\alpha $ then consider $\displaystyle \{\beta\in S_\alpha }| \beta\notin J_0\}.$

Then this set has a least-element, call it $\displaystyle \beta_0.$ It follows that $\displaystyle S_{\beta_0}\subset J_0 ==> \beta_0\in J_0.$

This is a contradiction.

If $\displaystyle J_0 = S_\alpha,$ for this case, I can not find any kind of contradiction. Any suggestions?