Properties of Class of All Ordinals/Superinduction Principle
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Theorem
Let $\On$ denote the class of all ordinals.
Let $A$ be a class which satisfies the following $3$ conditions:
\((1)\) | $:$ | $A$ contains the zero ordinal $0$: | \(\ds 0 \in A \) | ||||||
\((2)\) | $:$ | $A$ is closed under successor mapping: | \(\ds \forall \alpha:\) | \(\ds \paren {\alpha \in A \implies \alpha^+ \in A} \) | |||||
\((3)\) | $:$ | $A$ is closed under chain unions: | \(\ds \forall C:\) | \(\ds \bigcup C \in A \) | where $C$ is a chain of elements of $A$ |
That is, let $A$ be a superinductive class under the successor mapping.
Then $A$ contains all ordinals:
- $\On \subseteq A$
Proof
We note that the zero ordinal, denoted $0$, is identified as the empty set:
- $0 \:= \O$
Hence by definition $A$ is indeed a superinductive class under the successor mapping.
By the definition of ordinal:
- $\alpha$ is an ordinal if and only if $\alpha$ is an element of every superinductive class.
Hence $\On$ is a subclass of every superinductive class.
The result follows.
$\blacksquare$
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: Theorem $1.2 \ (4)$