Ordinal Subset of Ordinal is Initial Segment
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, $T = S_a = a \in S$.
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 $a \in S - T$, 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 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 $T = S_a$.
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.7$