Transfinite Induction/Principle 1

Theorem
Let $\operatorname{On}$ denote the class of all ordinals.

Let $A$ denote a class.

Suppose that:
 * For all elements $x$ of $\operatorname{On}$, if $x$ is a subset of $A$, then $x$ is an element of $A$.

Then $\operatorname{On} \subseteq A$.

Transfinite Induction Schema
Let $\phi \left({x}\right)$ be a property

Suppose that:
 * If $\phi \left({x}\right)$ holds for all ordinals $x$ less than $y$, then $\phi \left({y}\right)$ also holds.

Then $\phi \left({x}\right)$ holds for all ordinals $x$.

Proof
Suppose that $\neg \operatorname{On} \subseteq A$.

Then:
 * $\left({\operatorname{On} \setminus A}\right) \ne \varnothing$

From Set Difference Subset, $\operatorname{On} \setminus A$ is also a subset of the ordinals.

As Ordinal is Well-Ordered by Epsilon, $\operatorname{On} \setminus A$ it must contain a minimal element $y$.

By Ordinal Proper Subset Membership, this minimal element $y$ must be a subset of the ordinals.

However, from the fact that $y$ is a minimal element, $\left({\operatorname{On} \setminus A}\right) \cap y = \varnothing$.

So by its subsethood of $\operatorname{On}$:
 * $\left({\operatorname{On} \cap y}\right) \setminus A = \left({y \setminus A}\right) = \varnothing$

Therefore $y \subseteq A$.

However, by the hypothesis, $y$ must also be an element of $A$.

This contradicts the fact that $y$ is an element of $\operatorname{On} \setminus A$.

Therefore $\operatorname{On} \subseteq A$.

Proof of Transfinite Induction Schema
The statement:
 * $\forall x \in \operatorname{On}: x \in y \implies \phi \left({x}\right)$

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

Since $\forall x \in \operatorname{On}: \left({x \in y \implies \phi \left({x}\right)}\right) \implies \phi \left({y}\right)$:
 * $y \subseteq \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\} \implies y \in \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\}$

By the Principle of Transfinite Induction (above):
 * $\operatorname{On} \subseteq \left\{{x \in \operatorname{On} : \phi \left({x}\right)}\right\}$

Therefore:
 * $x \in \operatorname{On} \implies \phi \left({x}\right)$

for all $x$.

Source

 * : $7.12$