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
This proves the induction step.

Limit Case
This proves the limit case.

Also see

 * Subset is Right Compatible with Ordinal Addition