Strict Ordering of Ordinals is Equivalent to Membership Relation
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Theorem
Let $\On$ denote the class of all ordinals.
Let $<$ denote the (strict) usual ordering of $\On$.
Then:
- $\forall \alpha, \beta \in \On: \alpha < \beta \iff \alpha \in \beta$
Proof
Necessary Condition
Let $\alpha \in \beta$.
Then from Ordinal is Transitive:
- $\alpha \subseteq \beta$
But if $\alpha = \beta$ we would have $\alpha \in \alpha$.
This is contrary to Ordinal is not Element of Itself.
Hence we have:
- $\alpha \subseteq \beta$
and:
- $\alpha \ne \beta$
That is:
- $\alpha \subsetneqq \beta$
Hence by definition of the (strict) usual ordering of $\On$:
- $\alpha < \beta$
$\Box$
Sufficient Condition
Let $\alpha < \beta$.
Then by Sandwich Principle for $g$-Towers: Corollary 1:
- $\alpha^+ \subseteq \beta$
where $\alpha^+$ denotes the successor set of $\alpha$.
By definition of successor set:
- $\alpha^+ = \alpha \cup \set \alpha$
and so:
- $\alpha \in \alpha^+$
That is:
- $\alpha \in \beta$
$\blacksquare$
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: Theorem $1.12$