Ordinals are Well-Ordered
Theorem
The ordinals are well-ordered.
Corollary
Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$.
Then $A$ is strictly well-ordered by $\Epsilon {\restriction_A}$.
Proof 1
Recall that the Ordinals are Totally Ordered.
Let $A$ be a non-empty set of ordinals.
Let $\alpha \in A$.
Let $B = \alpha^+ \cap A$, where $\alpha^+$ denotes the successor set of $\alpha$.
Recall that $\alpha^+$ is an ordinal.
Note that $\alpha \in B$, so $B$ is non-empty.
By Intersection is Subset, $B \subseteq \alpha^+$.
It follows that there exists a smallest element $\kappa$ of $B$.
We claim that $\kappa$ is the smallest element of $A$.
So let $\beta \in A$. We need to show that $\kappa = \beta$ or $\kappa \in \beta$.
By Ordinal Membership is Trichotomy, either $\beta \in \alpha^+$, $\alpha^+ = \beta$, or $\alpha^+ \in \beta$.
If $\beta \in \alpha^+$, then $\beta \in B$; it follows by the definition of $\kappa$ that $\kappa = \beta$ or $\kappa \in \beta$.
If $\alpha^+ = \beta$ or $\alpha^+ \in \beta$, then it follows by the transitivity of $\beta$ that $\alpha^+ \subseteq \beta$.
Since $\kappa \in \alpha^+$, it follows that $\kappa \in \beta$.
$\blacksquare$
Proof 2
By Ordinals are Totally Ordered, the ordinals are totally ordered by $\subseteq$:
\(\ds X\) | \(\subsetneqq\) | \(\ds Y\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists a \in Y: \, \) | \(\ds X\) | \(=\) | \(\ds Y_a\) | where $Y_a$ denotes the initial segment of $Y$ determined by $a$ | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds X\) | \(=\) | \(\ds a\) | since $Y_a = a$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds X\) | \(\in\) | \(\ds Y\) |
Thus:
- the strict ordering $\subsetneqq$ on ordinals
and
- the strict ordering $\in$ on ordinals
are the same.
Aiming for a contradiction, suppose the ordinals were not well-ordered by $\subsetneqq$.
Then we could find a sequence $\sequence {X_n}_{n \mathop = 0}^\infty$ of ordinals such that:
- $X_0 \supsetneqq X_1 \supsetneqq X_2 \cdots$
So for all $n > 0$:
- $X_n \subsetneqq X_0$
so:
- $X_n \in X_0$
Thus $\sequence {X_{n + 1} }_{n \mathop = 0}^\infty$ is a decreasing sequence under $\subsetneqq$ of elements of $\sequence {X_n}$.
But since $X_0$ is an ordinal it is well-ordered by $\subsetneqq$.
From Infinite Sequence Property of Well-Founded Relation, this demonstrates a contradiction.
$\blacksquare$