Ordinal equals Successor of its Union

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:
 * $a = \bigcup \alpha^+$