Ordinal Membership is Trichotomy/Proof 1
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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 $1.14$