# Class of All Ordinals is Proper Class/Proof 1

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## Theorem

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

Then $\On$ is a proper class.

That is, $\On$ is not a set.

## Proof

We have that Successor Mapping on Ordinals is Strictly Progressing.

The result follows from Superinductive Class under Strictly Progressing Mapping is Proper Class.

$\blacksquare$

## 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 $1.10$