# 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.

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:

- $\O \in \On$

$\Box$

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.

$\Box$

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.

$\Box$

Thus $\On$ is a superinductive class.

$\Box$

Aiming for a contradiction, suppose $A$ 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.

$\Box$

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.

$\blacksquare$

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers