# Ordinal Membership is Trichotomy

## Theorem

Let $\alpha$ and $\beta$ be ordinals.

Then:

$\paren {\alpha = \beta} \lor \paren {\alpha \in \beta} \lor \paren {\beta \in \alpha}$

where $\lor$ denotes logical or.

### Corollary

Let $\alpha$ be an ordinal.

Let $x, y \in \alpha$ such that $x \ne y$.

Then either:

$x \in y$

or:

$y \in x$

## Proof 1

From Class of All Ordinals is Well-Ordered by Subset Relation, $\On$ is a nest.

Hence:

$\forall \alpha, \beta \in \On: \paren {\alpha \subsetneqq \beta} \lor \paren {\beta \subsetneqq \alpha} \lor \paren {\alpha = \beta}$

As Ordinal is Transitive, this is equivalent to:

$\forall \alpha, \beta \in \On: \paren {\alpha \in \beta} \lor \paren {\beta \in \alpha} \lor \paren {\alpha = \beta}$

Hence the result.

$\blacksquare$

## Proof 2

By Relation between Two Ordinals, it follows that:

$\paren {\alpha = \beta} \lor \paren {\alpha \subset \beta} \lor \paren {\beta \subset \alpha}$

By Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, the result follows.

$\blacksquare$