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.

Take $a \in \omega$. 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. Take $x \in y \land y \in \omega$. Then, $y \in \operatorname{On} \land y^+ \subseteq K_I$. Because $y$ is an ordinal, it is transitive, so $x \subseteq y$ and $x^+ \subseteq y^+ \subseteq K_I$. Therefore, $x^+ \subseteq K_I$. Applying the definition of Minimal Infinite Successor Set, it follows that $x \in \omega$, 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.