Transfinite Induction/Principle 2

Theorem
Let $A$ be a class satisfying the following conditions:


 * $\varnothing \in A$.
 * $\forall x \in A: x^+ \in A$
 * If $y$ is a limit ordinal, then $ ( \forall x \lt y: x \in A ) \implies y \in A $

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

Proof
We shall prove this using the first principle of transfinite induction.

Assume that $\forall x < y: x \in A$.

If $y$ is a limit ordinal, then $y \in A$.

If $y$ is not a limit ordinal, then by definition it is either $\varnothing$ or $x^+$ for some ordinal $x$.

If $y = \varnothing$, then $y \in A$

If $y = x^+$, then since $x < y$ and $x \in A$, we have that $x^+ \in A$. Therefore, $y \in A$.

In any case, $y \in A$. By applying induction, we can infer that $y \in A$ for all ordinals $y$. Therefore, $\operatorname{On} \subseteq A$.