Transfinite Induction/Principle 2

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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 $\left({ \forall x < y: x \in A }\right) \implies y \in A$

where $x^+$ denotes the successor of $x$.


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


Proof

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

Assume $y$ is an ordinal such that $y \subseteq A$.

This implies $\forall x < y: x \in A$.

By definition, $y$ is a limit ordinal, $y = \varnothing$, or $\exists x: y = x^+$.


If $y$ is a limit ordinal, then, since:

$\forall x < y: x \in A$

it follows that $y \in A$.


If $y = \varnothing$, then $y \in A$ by hypothesis.


Finally, assume $y = z^+$ for some ordinal $z$.

Since $z < z^+ = y$ and $\forall x < y: x \in A$, it follows that $z \in A$.

By hypothesis, we conclude $y = z^+ \in A$.


In any case, $y \in A$.

Therefore, for all ordinals $y$, if $y \subseteq A$, then $y \in A$.

By the first principle of transfinite induction, $\operatorname{On} \subseteq A$.

$\blacksquare$