# Ordinal Power of Power

## Contents

## Theorem

Let $x$, $y$, and $z$ be ordinals.

Then:

- $\left({x^y}\right)^z = x^{y \mathop \times z}$

## Proof

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

### Basis for the Induction

\(\displaystyle \left({x^y}\right)^0\) | \(=\) | \(\displaystyle 1\) | definition of ordinal exponentiation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^0\) | definition of ordinal exponentiation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{y \mathop \times 0}\) | definition of ordinal multiplication |

This proves the basis for the induction.

$\Box$

### Induction Step

Suppose that $\left({x^y}\right)^z = x^{y \mathop \times z}$.

Then:

\(\displaystyle \left({x^y}\right)^{z^+}\) | \(=\) | \(\displaystyle \left({x^y}\right)^z \times x^y\) | definition of ordinal exponentiation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{y \mathop \times z} \times x^y\) | by Inductive Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{\left({y \mathop \times z}\right) + y}\) | by Ordinal Sum of Powers | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{y \mathop \times z^+}\) | definition of ordinal multiplication |

This proves the induction step.

$\Box$

### Limit Case

Suppose $\left({x^y}\right)^w = x^{y \mathop \times w}$ for all $w \in z$ where $z$ is a limit ordinal.

Then:

\(\, \displaystyle \forall w \in z: \, \) | \(\displaystyle \left({x^y}\right)^w\) | \(=\) | \(\displaystyle x^{y \mathop \times w}\) | by Inductive Hypothesis | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \bigcup_{w \mathop \in z} \left({x^y}\right)^w\) | \(=\) | \(\displaystyle \bigcup_{w \mathop \in z} x^{y \mathop \times w}\) | by Indexed Union Equality | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({x^y}\right)^z\) | \(=\) | \(\displaystyle \bigcup_{w \mathop \in z} x^{y \mathop \times w}\) | definition of ordinal exponentiation |

To prove the statement, it is necessary and sufficient to prove that:

- $\displaystyle \bigcup_{w \mathop \in z} x^{y \mathop \times w} = x^{y \mathop \times z}$

The proof of this shall proceed by cases:

#### Case 1

If $y = 0$, it follows that:

\(\displaystyle \left({x^y}\right)^z\) | \(=\) | \(\displaystyle 1^z\) | definition of ordinal exponentiation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | by Exponent Base of One | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^0\) | definition of ordinal exponentiation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x^{y \mathop \times z}\) | by Ordinal Multiplication by Zero |

#### Case 2

If $y \ne 0$:

\(\displaystyle w\) | \(\in\) | \(\displaystyle z\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left({y \mathop \times w}\right)\) | \(\le\) | \(\displaystyle \left({y \mathop \times z}\right)\) | by Membership is Left Compatible with Ordinal Multiplication and Ordinal Multiplication by Zero | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x^{y \mathop \times w}\) | \(\le\) | \(\displaystyle x^{y \mathop \times z}\) | by Membership is Left Compatible with Ordinal Exponentiation and definition of ordinal exponentiation |

Therefore, by Supremum Inequality for Ordinals, it follows that:

- $\bigcup_{w \mathop \in z} x^{y \mathop \times w} \le x^{y \mathop \times z}$

Conversely:

\(\displaystyle v\) | \(\in\) | \(\displaystyle y \times z\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\, \displaystyle \exists w \in z: \, \) | \(\displaystyle v\) | \(\in\) | \(\displaystyle y \times w\) | by Ordinal is Less than Ordinal times Limit | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x^v\) | \(\le\) | \(\displaystyle x^{y \mathop \times w}\) | by Membership is Left Compatible with Ordinal Exponentiation |

By Supremum Inequality for Ordinals, it follows that:

- $\displaystyle \bigcup_{v \mathop \in y \mathop \times z} x^v \le \bigcup_{w \mathop \in z} x^{y \mathop \times w}$

Therefore, by the definition of ordinal exponentiation:

\(\displaystyle x^{y \mathop \times z}\) | \(=\) | \(\displaystyle \bigcup_{w \mathop \in z} x^{y \mathop \times w}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({x^y}\right)^z\) |

This proves the limit case.

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 8.42$