Transfinite Induction/Principle 1

Theorem

 * Principle of Transfinite Induction

Let $\operatorname{On}$ denote the class of all Ordinals. Let $A$ denote a class.

For all elements $x$ of $\operatorname{On}$, if $x$ is a subset of class $A$, then $x$ is a member of $A$. Then, $\operatorname{On} \subseteq A$.


 * Transfinite Induction Schema

Suppose there is a property $\phi (x)$. Suppose that if $\phi (x)$ holds for all ordinals $x$ less than $y$, then $\phi (y)$ must hold as well. Then, $\phi (x)$ must hold for all ordinals $x$.

Proof

 * Principle of Transfinite Induction

Suppose that $\neg \operatorname{On} \subseteq A$. Then $( \operatorname{On} \setminus A ) \not = \varnothing$.

$( \operatorname{On} \setminus A )$ is also a subset of the ordinals, so it must also contain a minimal element, since Ordinal is Well-Ordered by Epsilon. Furthermore, this minimal element $y$ must be a subset of the ordinals by Ordinal Proper Subset Membership.

However, from the fact that it is a minimal element, $( \operatorname{On} \setminus A ) \cap y = \varnothing$. So by its subsethood of $\operatorname{On}$, $( \operatorname{On} \cap y ) \setminus A = ( y \setminus A ) = \varnothing$. Therefore $y \subseteq A$.

However, by the hypothesis, $y$ must also be an element of $A$, which contradicts the fact that it is an element of $( \operatorname{On} \setminus A )$. Therefore, $\operatorname{On} \subseteq A$.


 * Transfinite Induction Schema

$\forall x \in \operatorname{On} ( x \in y \to \phi (x) )$ is equivalent to $y \subseteq \{ x \in \operatorname{On} : \phi (x) \}$

Since $\forall x \in \operatorname{On} ( x \in y \to \phi (x) ) \to \phi (y)$, $y \subseteq \{ x \in \operatorname{On} : \phi (x) \} \to y \in \{ x \in \operatorname{On} : \phi (x) \}$. By the Principle of Transfinite Induction (above), $\operatorname{On} \subseteq \{ x \in \operatorname{On} : \phi (x) \}$, so $x \in \operatorname{On} \to \phi (x)$, for all $x$.

Source

 * :$7.12$