# Ordinal Multiplication by Zero

## Theorem

Let $x$ be an ordinal.

 $\displaystyle \left({x \cdot \varnothing}\right)$ $=$ $\displaystyle \varnothing$ $\displaystyle \left({\varnothing \cdot x}\right)$ $=$ $\displaystyle \varnothing$

## Proof

 $\displaystyle \left({x \cdot \varnothing}\right)$ $=$ $\displaystyle \varnothing$ Definition of Ordinal Multiplication

For $\left({\varnothing \cdot x}\right) = \varnothing$, the proof shall proceed by Transfinite Induction on $x$.

### Basis for the Induction

 $\displaystyle \left({\varnothing \cdot \varnothing}\right)$ $=$ $\displaystyle \varnothing$ Definition of Ordinal Multiplication

This proves the basis for the induction.

### Induction Step

 $\displaystyle \left({\varnothing \cdot x}\right)$ $=$ $\displaystyle \varnothing$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle \left({\left({\varnothing \cdot x}\right) + \varnothing}\right)$ $=$ $\displaystyle \varnothing$ Definition of Ordinal Addition $\displaystyle \left({\varnothing \cdot x^+}\right)$ $=$ $\displaystyle \left({\left({\varnothing \cdot x}\right) + \varnothing}\right)$ Definition of Ordinal Multiplication $\displaystyle \implies \ \$ $\displaystyle \left({\varnothing \cdot x^+}\right)$ $=$ $\displaystyle \varnothing$ Equality is Transitive

This proves the induction step.

### Limit Case

 $\, \displaystyle \forall y \in x: \,$ $\displaystyle \left({\varnothing \cdot y}\right)$ $=$ $\displaystyle \varnothing$ Hypothesis $\displaystyle \implies \ \$ $\displaystyle \bigcup_{y \mathop \in x} \left({\varnothing \cdot y}\right)$ $=$ $\displaystyle \varnothing$ Indexed Union Equality $\displaystyle \left({\varnothing \cdot x}\right)$ $=$ $\displaystyle \bigcup_{y \mathop \in x} \left({\varnothing \cdot y}\right)$ Definition of Ordinal Multiplication $\displaystyle \implies \ \$ $\displaystyle \left({\varnothing \cdot x}\right)$ $=$ $\displaystyle \varnothing$ Equality is Transitive

This proves the limit case.

$\blacksquare$