Ordinal equals its Initial Segment

Theorem
Let $\On$ denote the class of all ordinals.

Let $<$ denote the (strict) usual ordering of $\On$.

Let $\alpha$ be an ordinal.

Then $\alpha$ is equal to its own initial segment:
 * $\alpha = \set {\beta \in \On: \beta < \alpha}$

Proof
From Strict Ordering of Ordinals is Equivalent to Membership Relation:
 * $\forall \alpha, \beta \in \On: \beta < \alpha \iff \alpha \in \beta$

Hence the statement of the result is equivalent to:
 * $\alpha = \set {\beta \in \On: \beta \in \alpha}$

which is trivially true by definition of a set.