# Alternative Definition of Ordinal in Well-Founded Theory It has been suggested that this article or section be renamed: Definitions are treated as "equal" in that they are entered as definitions. Then an equivalence proof page is built to demonstrate that those two definitions are indeed equivalent. One may discuss this suggestion on the talk page. It has been suggested that this page or section be merged into Equivalence of Definitions of Ordinal. (Discuss)

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## Theorem

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

## Proof

### Forward Implication

Let $S$ be an ordinal.

By Alternative Definition of 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:

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

$\Box$

### 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 $\left\{ {x, y, z}\right\}$ 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 $S$.

Thus by Alternative Definition of Ordinal, $S$ is an ordinal.

$\blacksquare$