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:

$a = \bigcup \alpha^+$

$\blacksquare$


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