# Minimally Inductive Set is Ordinal

## Theorem

Let $\omega$ denote the minimally inductive set.

Then $\omega$ is an ordinal.

## Proof 1

The minimally inductive 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 minimally inductive set that $\forall n \in \omega: \omega_n = n$

### Basis for the Induction

 $\ds \omega_\O$ $=$ $\ds \set {x \in \omega: x \in \O}$ Definition of Initial Segment $\ds$ $=$ $\ds \O$ Definition of Empty Set

### Induction Hypothesis

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

### Induction Step

 $\ds \omega_{n^+}$ $=$ $\ds \set {x \in \omega: x \in n^+}$ $\ds$ $=$ $\ds \set {x \in \omega: x \in n \lor x \in \set n}$ Definition of Successor Set $\ds$ $=$ $\ds \set {x \in \omega: x \in n \lor x = n}$ Definition of Singleton $\ds$ $=$ $\ds \set {x \in \omega: x \in n} \cup \set {x \in \omega: x = n}$ Definition of Set Union $\ds$ $=$ $\ds \omega_n \cup \set n$ $\ds$ $=$ $\ds n \cup \set n$ Induction Hypothesis $\ds$ $=$ $\ds n^+$ Definition of Successor Set

And so $\omega$ is an ordinal.

$\blacksquare$

## Proof 2

Let $K_I$ denote the set of all nonlimit ordinals.

Let $\On$ denote the set of all ordinals.

Let $a \in \omega$.

It follows that $a^+ \subseteq K_I$, so $a \in K_I$.

Thus:

$\omega \subseteq K_I \subseteq \On$

We now must prove that $\omega$ is a transitive set, at which point it will satisfy the Alternative Definition of Ordinal.

Let $x \in y$ and $y \in \omega$.

Then:

$y \in \On \land y^+ \subseteq K_I$

Because $y$ is an ordinal, it is transitive.

Therefore:

$x \subseteq y$

and:

$x^+ \subseteq y^+ \subseteq K_I$

Therefore, $x^+ \subseteq K_I$.

Applying the definition of minimally inductive set:

$x \in \omega$

so $\omega$ is transitive.

$\blacksquare$