Ordinal Exponentiation is Closed

Theorem

Let $x$ and $y$ be ordinals.

Then:

$x^y \in \On$

That is, ordinal exponentiation is closed.

Proof

Let $x = 0$ and $y = 0$.

Then by the definition of ordinal exponentiation:

$x^y = 1$

Let $x = 0$ and $y \ne 0$.

Then by the definition of ordinal exponentiation:

$x^y = 0$

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

By 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 $x^y \in \On$.

Suppose the inductive hypothesis holds.

Then:

\(\ds x^{y^+}\)

\(=\)

\(\ds x^y \times x\)

Definition of Ordinal Exponentiation

\(\ds \)

\(\in\)

\(\ds \On\)

Ordinal Multiplication is Closed

This proves the induction step.

Limit Case

The inductive hypothesis states that $x^z \in \On$ for all $z \in y$.

\(\ds x^y\)

\(=\)

\(\ds \bigcup_{z \mathop \in y} x^z\)

Definition of Ordinal Exponentiation

\(\ds \)

\(\in\)

\(\ds \On\)

Union of Set of Ordinals is Ordinal: Corollary

This proves the limit case.

$\blacksquare$