Ordinal equals its Initial Segment/Proof 4
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Theorem
Let $\On$ denote the class of all ordinals.
Let $<$ denote the (strict) usual ordering of $\On$.
Let $\alpha$ be an ordinal.
Then $\alpha$ is equal to its own initial segment:
- $\alpha = \set {\beta \in \On: \beta < \alpha}$
Proof
Let $\alpha$ be an ordinal by Definition 4.
$\alpha$ is an ordinal if and only if:
- $\alpha$ is an element of every superinductive class.
From Strict Ordering of Ordinals is Equivalent to Membership Relation:
- $\forall \alpha, \beta \in \On: \beta < \alpha \iff \alpha \in \beta$
Hence the statement of the result is equivalent to:
- $\alpha = \set {\beta \in \On: \beta \in \alpha}$
which is trivially true by definition of a set.
$\blacksquare$
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 $1.13$