Minimally Inductive Set is Ordinal/Proof 2

Theorem

Let $\omega$ denote the minimally inductive set.

Then $\omega$ is an ordinal.

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.

$\blacksquare$

This demonstrates that $\omega$ can be shown to be an ordinal without use of the Axiom of Infinity.

By Ordinal is Member of Class of All Ordinals, it follows that $\omega \in \On$ or $\omega = \On$.

The Axiom of Infinity rejects the latter option in favor of the former.