Exists Ordinal Greater than Set of Ordinals

Theorem

Let $S$ be a set of ordinals.

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 $1.18$