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 $1.18$