# Successor Set of Ordinal is Ordinal/Proof 1

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## Theorem

Let $\On$ denote the class of all ordinals.

Let $\alpha \in \On$ be an ordinal.

Then its successor set $\alpha^+ = \alpha \cup \set \alpha$ is also an ordinal.

## Proof

We have the result that Class of All Ordinals is Minimally Superinductive over Successor Mapping.

Hence $\On$ is *a fortiori* a superinductive class with respect to the successor mapping.

Hence, by definition of superinductive class:

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

That is:

- $\forall \alpha \in \On: \alpha^+ \in \On$

$\blacksquare$

## 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 $1.2 \ (2)$