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