Successor Set of Ordinal is Ordinal/Proof 1

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 $1.2 \ (2)$