Transfinite Induction/Principle 1

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

Let $A$ denote a class.

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

Then $\On \subseteq A$.

Proof
that $\neg \On \subseteq A$.

Then:
 * $\paren {\On \setminus A} \ne \O$

From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.

By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\On \setminus A$ must have a strictly minimal element $y$ under $\in$.

By Element of Ordinal is Ordinal, $y$ must be a subset of $\On$, the class of all ordinals.

However, from the fact that $y$ is a strictly minimal element under $\in$ of $\On \setminus A$:


 * $\paren {\On \setminus A} \cap y = \O$

So by its subsethood of $\On$:
 * $\paren {\On \cap y} \setminus A = \paren {y \setminus A} = \O$

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 $\On \setminus A$.

Therefore $\On \subseteq A$.