Second Principle of Transfinite Induction

Theorem

Let $A$ be a class.

Let $\preccurlyeq$ be a well-ordering on $A$.

Let $P$ be a property that satisfies the following $3$ conditions:

$(1): \quad P$ holds for the smallest element of $A$.

$(2): \quad$ For all $x \in A$ which have an immediate successor $\map S x$, if $P$ holds for $x$, then $P$ holds for $\map S x$.

$(3): \quad$ For all limit element $y$ of $A$, if $P$ holds for all $x \prec y$, then $P$ holds for $y$.

Then $P$ holds for all $x \in A$.

Aiming for a contradiction, suppose $P$ fails to hold for some $z \in A$.

The class of elements of $A$ for which $P$ does not hold is therefore non-empty.

We have that $A$ is a well-ordered class.

Hence there must be some smallest element $x \in A$ for which $P$ fails to hold.

By $(1)$, this cannot be the smallest element of $A$.

By $(2)$, this cannot be an immediate successor element.

By $(3)$, this cannot be a limit element.

Hence for all $x \in A$, by hypothesis, $P$ holds for $x$.

This contradicts our definition of $x$.

Hence there can be no such $z \in A$ for which $P$ fails to hold.

$\blacksquare$

2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.9$ (Transfinite Induction Principle $2$)