Alternative Definition of Ordinal in Well-Founded Theory

Theorem
A set $S$ is an ordinal iff $S$ is transitive and $\forall x, y \in S: \left({x \in y \lor x = y \lor y \in x}\right)$.

Forward Implication
Let $S$ be an ordinal.

By Alternative Definition of an Ordinal, $S$ is transitive and strictly well-ordered by the epsilon relation.

By Strict Well-Ordering is Strict Total Ordering, $S$ is strictly totally ordered by $\in$.

Thus:


 * $\forall x, y \in S: \left({ x \in y \lor x = y \lor y \in x }\right)$

Reverse Implication
Let $S$ be a transitive set such that for any $x, y \in S$, $x \in y \lor y \in x \lor x = y$.

We first show that $\in$ is a strict ordering of $S$.

Asymmetric: Let $x, y \in S$.

By Epsilon is Foundational, $\{ x,y \}$ has an $\Epsilon$-minimal element.

Thus $x \notin t$ or $y \notin x$.

Transitive: Let $x, y, z \in S$ with $x \in y$ and $y \in z$.

By assumption, $x = z$, $x \in z$, or $z \in x$.

Suppose for the sake of contradiction that $x = z$.

Then $x \in y$ and $y \in x$, contradicting the fact that $\Epsilon$ is asymmetric.

Suppose that $z \in x$.

Then $x \in y$, $y \in z$, and $z \in x$.

Thus the set $\{x, y, z\}$ has no $\Epsilon$-minimal element, contradicting Epsilon is Foundational. Thus $x \in z$.

Thus $\in$ is a strict ordering of $S$.

Let $T$ be a non-empty subset of $S$.

By Epsilon is Foundational, $S$ has an $\Epsilon$-minimal element, $m$.

Since a minimal element of a strictly totally ordered set is the smallest element, $\Epsilon$ strictly well-orders $S$.

Source

 * : $\S 7.3$, $\S 7.4$