# Ordinal Subset of Ordinal is Initial Segment

## Theorem

Let $S$ be an ordinal.

Let $T \subset S$ also be an ordinal.

Then $\exists a \in S: T = S_a$, where $S_a$ is the initial segment of $S$ determined by $a$.

That is, $T = S_a = a \in S$.

## Proof

Recall that the Ordering on Ordinal is Subset Relation.

Let $a$ be the minimal element of $S \setminus T$.

By definition of minimal element:

$\forall x \in S: x \subset a$

Hence $x \notin S \setminus T$.

By definition of set difference:

$x \in T$

Thus by definition of initial segment:

$S_a \subseteq T$

Now let $b \in T$.

Then by definition of ordinal:

$T_b = b = S_b$

Now, if $a \subset b$ then $a \in S_b$.

So $a \in T_b$ and hence $a \in T$.

But $a \in S \setminus T$, so by definition of set difference, this is not the case.

So $a \not \subset b$.

We have that an ordinal is well-ordered by $\subseteq$.

So by the definition of well-ordering, $\subseteq$ is a total ordering on $S$

So it follows that $b \subseteq a$.

But $b \ne a$ since $b \in T$.

Hence $b \subset a$, and so $b \in S_a$.

This proves that $T \subseteq S_a$.

Thus $T = S_a$.

$\blacksquare$