Ordinal Exponentiation is Closed
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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.
$\blacksquare$