A Universe of Sorts

Let $(J, <)$ be a well-ordered set.
Let $S(\alpha) \equiv J < \alpha$ , or $S(\alpha) \equiv \{ j \in J: j < \alpha \}$ . This is called as the section of $J$ by $\alpha$ .
Let a $J_0 \subseteq J$ be inductive iff for all $\alpha \in J$ , $S(\alpha) \subseteq J_0$ implies $\alpha \in J_0$ . That is:

\text{$J_0$ inductive} \equiv \forall \alpha \in J, S(\alpha) \subseteq J_0 \implies \alpha \in J_0

Then transfinite induction states that for any inductive set $J_0 \subseteq J$ , we have $J_0 = J$ .

Proof by contradiction. Suppose that $J_0$ is an inductive set such that $J_0 \neq J$ .
Let $W$ (for wrong) be the set $J_0 - J$ . That is, $W$ is elements that are not in $J_0$ .
$W$ is non-empty since $J_0 \neq J$ . Thus, consider $w \equiv \min(W)$ , which is possible since $J$ is well-ordered, thus the subset $W$ has a minimum element.
$w$ is the smallest element that is not in $J_0$ . So all elements smaller than $w$ are in $J_0$ . This, $S(w) \subseteq J_0$ . This implies $w \in J_0$ as $J_0$ is inductive.
This is contradiction, as we start with $w$ is the smallest element not in $J_0$ , and then concluded that $w$ is in $J_0$ .
Thus, the set $W \equiv J_0 - J$ must be empty, or $J_0 = J$ .

§ Transfinite Induction: Proof