Initial Segment of Ordinal is Ordinal

Theorem

Let $S$ be an ordinal.

Let $a \in S$.

Then the initial segment $S_a = a$ of $S$ determined by $a$ is also an ordinal.

In other words, every element of a (non-empty) ordinal is also an ordinal.

Suppose that $b \in S_a$.

From Ordering on Ordinal is Subset Relation, and the definition of an initial segment, it follows that $b \subset a$.

Then:

$\ds \paren {S_a}_b$

$=$

$\ds \set {x \in S_a: x \subset b}$

$\ds $

$=$

$\ds \set {x \in S: x \subset a \land x \subset b}$

$\ds $

$=$

$\ds \set {x \in S: x \subset b}$

as $b \subset a$

$\ds $

$=$

$\ds S_b$

$\ds $

$=$

$\ds b$

as $S$ is an ordinal

The result follows from the definition of an ordinal.

$\blacksquare$