Ordinal Membership is Trichotomy/Proof 1

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.

Proof

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

Hence:

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

From Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, 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.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.14$