Equality of Successors implies Equality of Ordinals

Theorem

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

Then:

$\forall \alpha, \beta \in \On: \alpha^+ = \beta^+ \implies \alpha = \beta$

Proof

From Class of All Ordinals is Well-Ordered by Subset Relation:

$\alpha^+$ is the immediate successor of $\alpha$

$\beta^+$ is the immediate successor of $\beta$

and no two distinct elements of $\On$ can have the same immediate successor.

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 $1.17$