# Subset is Right Compatible with Ordinal Multiplication

## Theorem

Let $x, y, z$ be ordinals.

Then:

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

## Proof

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

### Basis for the Induction

By definition of ordinal multiplication:

$\left({x \cdot 0}\right) = 0$

so the statement simply reduces to:

$0 \le 0$

This proves the basis for the induction.

### Induction Step

 $\displaystyle x$ $\le$ $\displaystyle y$ $\displaystyle \implies \ \$ $\displaystyle \left({x \cdot z}\right)$ $\le$ $\displaystyle \left({y \cdot z}\right)$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle \left({\left({x \cdot z}\right) + x}\right)$ $\le$ $\displaystyle \left({\left({y \cdot z}\right) + y}\right)$ Membership is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition $\displaystyle \implies \ \$ $\displaystyle \left({x \cdot z^+}\right)$ $\le$ $\displaystyle \left({y \cdot z^+}\right)$ Definition of Ordinal Multiplication

This proves the induction step.

### Limit Case

 $\displaystyle x$ $\le$ $\displaystyle y$ $\displaystyle \implies \ \$ $\, \displaystyle \forall w \in z: \,$ $\displaystyle \left({x \cdot w}\right)$ $\le$ $\displaystyle \left({y \cdot w}\right)$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle \bigcup_{w \mathop \in z} \left({x \cdot w}\right)$ $\le$ $\displaystyle \bigcup_{w \mathop \in z} \left({y \cdot w}\right)$ Indexed Union Subset $\displaystyle \implies \ \$ $\displaystyle \left({x \cdot z}\right)$ $\le$ $\displaystyle \left({y \cdot z}\right)$ Definition of Ordinal Multiplication

This proves the limit case.

$\blacksquare$