# Ordinal Multiplication by One

## Theorem

Let $x$ be an ordinal.

Let $1$ denote the successor of $\varnothing$.

 $\displaystyle \left({x \cdot 1}\right)$ $=$ $\displaystyle x$ $\displaystyle \left({1 \cdot x}\right)$ $=$ $\displaystyle x$

## Proof

 $\displaystyle \left({x \cdot 1}\right)$ $=$ $\displaystyle \left({x \cdot \varnothing^+}\right)$ Definition of One ($1$) $\displaystyle$ $=$ $\displaystyle \left({\left({x \cdot \varnothing}\right) + x}\right)$ Definition of Ordinal Multiplication $\displaystyle$ $=$ $\displaystyle \left({\varnothing + x}\right)$ Definition of Ordinal Multiplication $\displaystyle$ $=$ $\displaystyle x$ Ordinal Addition by Zero

$\Box$

The proof of the other equality shall proceed by Transfinite Induction.

### Basis for the Induction

 $\displaystyle \left({1 \cdot \varnothing}\right)$ $=$ $\displaystyle \varnothing$ Definition of Ordinal Multiplication

This proves the basis for the induction.

### Induction Step

 $\displaystyle \left({1 \cdot x}\right)$ $=$ $\displaystyle x$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle \left({\left({1 \cdot x}\right) + 1}\right)$ $=$ $\displaystyle x^+$ Ordinal Addition by One $\displaystyle \left({1 \cdot x^+}\right)$ $=$ $\displaystyle \left({\left({1 \cdot x}\right) + 1}\right)$ Definition of Ordinal Multiplication $\displaystyle \implies \ \$ $\displaystyle \left({1 \cdot x^+}\right)$ $=$ $\displaystyle x^+$ Equality is Transitive

This proves the induction step.

### Limit Case

 $\displaystyle \forall y \in x: \ \$ $\displaystyle \left({1 \cdot y}\right)$ $=$ $\displaystyle y$ Hypothesis $\displaystyle \implies \ \$ $\displaystyle \bigcup_{y \mathop \in x} \left({1 \cdot y}\right)$ $=$ $\displaystyle \bigcup_{y \mathop \in x} y$ Indexed Union Equality $\displaystyle \implies \ \$ $\displaystyle \left({1 \cdot x}\right)$ $=$ $\displaystyle \bigcup_{y \mathop \in x} y$ Definition of Ordinal Multiplication $\displaystyle \implies \ \$ $\displaystyle \left({1 \cdot x}\right)$ $=$ $\displaystyle x$ Union of Limit Ordinal

This proves the limit case.

$\blacksquare$