Ordinal equals Successor of its Union
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Theorem
Let $\alpha$ be an ordinal.
Then:
- $\bigcup \alpha^+ = \alpha$
where:
- $\alpha^+$ denotes the successor set of $\alpha$
- $\bigcup \alpha^+$ denotes the union of $\alpha^+$.
Proof
Let $x \in \bigcup \alpha^+$.
Then there exists an ordinal $\beta$ such that:
- $x \in \beta$
and:
- $\beta \in \alpha^+$
By definition of the usual ordering of ordinals:
- $\beta < \alpha^+$
Thus:
- $\beta \le \alpha$
Hence because $x < \beta$:
- $x < \alpha$
and thus:
- $x \in \alpha$
That is:
- $\bigcup \alpha^+ \subseteq \alpha$
Now suppose $x \in \alpha$.
Since we have:
- $\alpha \in \alpha^+$
we have:
- $x \in \bigcup \alpha^+$
Thus:
- $a \subseteq \bigcup \alpha^+$
Hence by set equality:
- $a = \bigcup \alpha^+$
$\blacksquare$
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 $1.20 \ (2)$