# Exists Ordinal Greater than Set of Ordinals

## Theorem

Then there exists an ordinal greater than every element of $S$:

- If $S$ contains a greatest ordinal $\alpha$, then $\alpha^+$ is greater than every element of $S$

- If $S$ does not contain a greatest ordinal, then $\bigcup S$ is greater than every element of $S$.

## Proof

Recall that Class of All Ordinals is Well-Ordered by Subset Relation.

Suppose $S$ contains a greatest ordinal $\alpha$.

Because $\alpha^+$ is greater than $\alpha$ by definition, it follows *a priori* that $\alpha^+$ is greater than every element of $S$.

$\Box$

Suppose $S$ does not contain a greatest ordinal.

Consider the union $\bigcup S$ of $S$.

If $\bigcup S \in S$ it would be the greatest element of $S$.

Hence $\bigcup S \notin S$.

Let $\alpha \in S$.

Then:

- $\alpha \ne \bigcup S$

But:

- $\alpha \subseteq \bigcup S$

and so:

- $\alpha \subsetneqq S$

From Union of Set of Ordinals is Ordinal, $\bigcup S$ is an ordinal.

Hence by definition of the usual ordering on ordinals:

- $a < \bigcup S$

Thus $\bigcup S$ is greater than every element of $S$.

Hence the result.

$\blacksquare$

## 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 $1.18$