Minimally Inductive Set is Ordinal/Proof 2
Theorem
Let $\omega$ denote the minimally inductive set.
Then $\omega$ is an ordinal.
Proof
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.
$\blacksquare$
Motivation
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.