# Ordinal Multiplication by Zero

## Theorem

Let $x$ be an ordinal.

 $\ds x \cdot \O$ $=$ $\ds \O$ $\ds \O \cdot x$ $=$ $\ds \O$

## Proof

 $\ds x \cdot \O$ $=$ $\ds \O$ Definition of Ordinal Multiplication

For $\O \cdot x = \O$, the proof shall proceed by Transfinite Induction on $x$.

### Basis for the Induction

 $\ds \O \cdot \O$ $=$ $\ds \O$ Definition of Ordinal Multiplication

This proves the basis for the induction.

### Induction Step

 $\ds \O \cdot x$ $=$ $\ds \O$ Inductive Hypothesis $\ds \leadsto \ \$ $\ds \paren {\O \cdot x} + \O$ $=$ $\ds \O$ Definition of Ordinal Addition $\ds \O \cdot x^+$ $=$ $\ds \paren {\O \cdot x} + \O$ Definition of Ordinal Multiplication $\ds \leadsto \ \$ $\ds \O \cdot x^+$ $=$ $\ds \O$ Equality is Transitive

This proves the induction step.

### Limit Case

 $\ds \forall y \in x: \,$ $\ds \O \cdot y$ $=$ $\ds \O$ Hypothesis $\ds \leadsto \ \$ $\ds \bigcup_{y \mathop \in x} \paren {\O \cdot y}$ $=$ $\ds \O$ Indexed Union Equality $\ds \O \cdot x$ $=$ $\ds \bigcup_{y \mathop \in x} \paren {\O \cdot y}$ Definition of Ordinal Multiplication $\ds \leadsto \ \$ $\ds \O \cdot x$ $=$ $\ds \O$ Equality is Transitive

This proves the limit case.

$\blacksquare$