Minimally Inductive Set is Ordinal

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

Then, $\omega$ is an ordinal.

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

Let $\operatorname{On}$ denote the set of all ordinals.

Let $a \in \omega$.

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

Thus:
 * $\omega \subseteq K_I \subseteq \operatorname{On}$

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:
 * $y \in \operatorname{On} \land y^+ \subseteq K_I$

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 Minimal Infinite Successor Set:
 * $x \in \omega$

so $\omega$ is transitive.

Remark
This demonstrates that $\omega$ can be shown to be an ordinal without use of the axiom of infinity. By Ordinal is Member of Ordinal Class, it follows that $\omega \in \operatorname{On} \lor \omega = \operatorname{On}$. The Axiom of Infinity rejects the latter option in favor of the former.