Class of All Ordinals is Minimally Superinductive over Successor Mapping

Theorem
The class of all ordinals $\On$ is the unique class which is minimally superinductive under the successor mapping.

Proof
We need to show that:
 * $\On$ is a superinductive class under the successor mapping

and:
 * no proper subclass of $\On$ is superinductive class under the successor mapping.

We recall immediately that Successor Mapping is Progressing.

This validates the definition of superinductive class under the successor mapping.

By definition of ordinal:


 * $S \in \On$


 * $S$ is an element of every superinductive class under the successor mapping.
 * $S$ is an element of every superinductive class under the successor mapping.

Hence $\On$ is a subclass of every superinductive class under the successor mapping.

Let $\cap S$ denote the intersection of every superinductive class under the successor mapping.

Thus by definition of class intersection:
 * $\On$ is a subclass of $\cap S$.

Let $M$ be an arbitrary superinductive class under the successor mapping.

We have by definition of superinductive class:
 * $\O \in M$

Thus as $M$ is arbitrary:
 * $\O \in \cap S$

Thus $\O$ is an element of every superinductive class.

Hence:
 * $\O \in \On$

Now let $\alpha \in \On$.

Then by definition of ordinal:
 * $\alpha$ is an element of every superinductive class under the successor mapping.

Hence by definition of superinductive class:
 * $\alpha^+$ is an element of every superinductive class under the successor mapping

where $\alpha^+$ denotes the successor set of $\alpha$.

Hence by definition:
 * $\alpha^+ \in \On$

Thus $\On$ is closed under the successor mapping.

Let $C$ be a chain in $\On$.

Then $C$ is a chain in every superinductive class under the successor mapping.

By definition of superinductive class:


 * $\cup C$ is an element of every superinductive class under the successor mapping

where $\cup C$ denotes the union of $C$.

That is:
 * $\cup C \in \On$

demonstrating that $\On$ is closed under chain unions.

Thus $\On$ is a superinductive class.

$A$ is a superinductive class which is a proper subclass of $\On$.

Then:
 * $\exists x \in \On: x \notin A$

Hence $\On \not \subseteq A$

But this contradicts the statement that $\On$ is a subclass of every superinductive class.

Hence by Proof by Contradiction no proper subclass of $\On$ is superinductive class under the successor mapping.

Hence by definition $\On$ is a minimally superinductive under the successor mapping.

It remains to prove uniqueness.

Suppose $\On'$ is another minimally superinductive class under the successor mapping.

Then we have:
 * $\On' \subseteq \On$

but also we have:
 * $\On \subseteq \On'$

and so:
 * $\On' = \On$

Thus $\On$ is the unique minimally superinductive class under the successor mapping.