Ordinal Multiplication is Associative

Theorem

Let $x, y, z$ be ordinals.

Let $\times$ denote ordinal multiplication.

Then:

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

Proof

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

Basis for the Induction

Let $0$ denote the zero ordinal.

 $\displaystyle x \times \left({y \times 0}\right)$ $=$ $\displaystyle x \times 0$ Ordinal Multiplication by Zero $\displaystyle$ $=$ $\displaystyle 0$ Ordinal Multiplication by Zero $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) \times 0$ Ordinal Multiplication by Zero

This proves the basis for the induction.

Induction Step

 $\displaystyle x \times \left({y \times z}\right)$ $=$ $\displaystyle \left({x \times y}\right) \times z$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle x \times \left({y \times z^+}\right)$ $=$ $\displaystyle x \times \left({\left({y \times z}\right) + y}\right)$ Definition of Ordinal Multiplication $\displaystyle$ $=$ $\displaystyle x \times \left({y \times z}\right) + \left({x \times y}\right)$ Ordinal Multiplication is Left Distributive $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) \times z + \left({x \times y}\right)$ Inductive Hypothesis $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) \times z^+$ Definition of Ordinal Multiplication

This proves the induction step.

Limit Case

The inductive hypothesis for the limit case states that:

$\forall w \in Z: x \times \left({y \times w}\right) = \left({x \times y}\right) \times w$

where $z$ is a limit ordinal.

The proof shall proceed by cases:

Case 1

If $y = 0$, then:

 $\displaystyle x \times \left({y \times z}\right)$ $=$ $\displaystyle x \times 0$ Ordinal Multiplication by Zero $\displaystyle$ $=$ $\displaystyle 0$ Ordinal Multiplication by Zero $\displaystyle$ $=$ $\displaystyle 0 \times z$ Ordinal Multiplication by Zero $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) \times z$ Ordinal Multiplication by Zero

Case 2

If $y \ne 0$, then $y \times z$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication.

It follows that:

 $\displaystyle x \times \left({y \times z}\right)$ $=$ $\displaystyle \bigcup_{u \mathop < \left({y \times z}\right)} x \times u$ Definition of Ordinal Multiplication $\displaystyle \left({x \times y}\right) \times z$ $=$ $\displaystyle \bigcup_{w \mathop < z} \left({x \times y}\right) \times w$ Definition of Ordinal Multiplication

If $u < \left({y \times z}\right)$, then $u < \left({y \times w}\right)$ for some $w \in z$ by Ordinal is Less than Ordinal times Limit.

 $\displaystyle x \times u$ $\le$ $\displaystyle x \times \left({y \times w}\right)$ Membership is Left Compatible with Ordinal Multiplication $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) \times w$ Inductive Hypothesis $\displaystyle$ $\le$ $\displaystyle \left({x \times y}\right) \times z$ Subset is Right Compatible with Ordinal Multiplication

Generalizing, the result follows for all $u \in \left({y \times z}\right)$.

Therefore by Supremum Inequality for Ordinals:

$x \times \left({y \times z}\right) \le \left({x \times y}\right) \times z$

Conversely, take any $w < z$.

 $\displaystyle \left({x \times y}\right) \times w$ $=$ $\displaystyle x \times \left({y \times w}\right)$ Inductive Hypothesis $\displaystyle$ $\le$ $\displaystyle x \times \left({y \times z}\right)$ Membership is Left Compatible with Ordinal Multiplication

It follows by Supremum Inequality for Ordinals that:

$\left({x \times y}\right) \times z \le x \times \left({y \times z}\right)$

By definition of set equality:

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

This proves the limit case.

$\blacksquare$