Ordinal Exponentiation is Closed

Theorem
Let $x$ and $y$ be ordinals.

Then:


 * $x^y \in \operatorname{On}$

Proof
Suppose that $x = 0$.

If $x = 0$ and $y = 0$, then $x^0 = 1$ by the definition of ordinal exponentiation.

If $x = 0$ and $y \ne 0$, then $x^0 = 0$ by the definition of ordinal exponentiation.

In either case, $x^y$ is an ordinal.

Now suppose that $x \ne 0$.

The proof shall proceed by Transfinite Induction on $y$.

Basis for the Induction

 * $x^0 = 1$ by the definition of ordinal exponentiation.


 * $x^0 = 1$, which is an ordinal.

This proves the basis for the induction.

Induction Step
The inductive hypothesis states that $x^y \in \operatorname{On}$.

If the inductive hypothesis holds, then:

This proves the induction step.

Limit Case
The inductive hypothesis states that $x^z \in \operatorname{On}$ for all $z \in y$.

This proves the limit case.