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 that $\alpha^+$ is greater than every element of $S$.

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.