# Minimal Infinite Successor Set is Ordinal/Proof 1

## Theorem

Let $\omega$ denote the minimal infinite successor set.

Then $\omega$ is an ordinal.

## Proof

The minimal infinite successor set is a set of ordinals by definition.

From the corollary of ordinals are well-ordered, it is seen that $\left({\omega, \Epsilon \! \restriction_\omega}\right)$ is a strictly well-ordered set.

It is to be shown by induction on minimal infinite successor set that $\forall n \in \omega: \omega_n = n$

### Basis for the Induction

 $\displaystyle \omega_\varnothing$ $=$ $\displaystyle \left\{ {x \in \omega: x \in \varnothing}\right\}$ Definition of initial segment $\displaystyle$ $=$ $\displaystyle \varnothing$ Definition of empty set

### Induction Hypothesis

Suppose that $\omega_n = n$ for some $n \in \omega$.

### Induction Step

 $\displaystyle \omega_{n^+}$ $=$ $\displaystyle \left\{ {x \in \omega: x \in n^+}\right\}$ $\displaystyle$ $=$ $\displaystyle \left\{ {x \in \omega: x \in n \lor x \in \left\{ {n}\right\} }\right\}$ Definition of successor set $\displaystyle$ $=$ $\displaystyle \left\{ {x \in \omega: x \in n \lor x = n}\right\}$ Definition of singleton $\displaystyle$ $=$ $\displaystyle \left\{ {x \in \omega: x \in n}\right\} \cup \left\{ {x \in \omega: x = n}\right\}$ Definition of set union $\displaystyle$ $=$ $\displaystyle \omega_n \cup \left\{ {n}\right\}$ $\displaystyle$ $=$ $\displaystyle n \cup \left\{ {n}\right\}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle n^+$ Definition of successor set

And so $\omega$ is an ordinal.

$\blacksquare$