Lexicographic Order Initial Segments
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Theorem
Let $\operatorname{Le}$ denote the lexicographic order for the set $\left({\operatorname{On} \times \operatorname{On} }\right)$.
Let the ordinal number $1$ denote the successor of $\varnothing$.
Then the initial segment of $ \left({1, \varnothing}\right)$ with respect to the lexicographic order $\operatorname{Le}$ is a proper class.
This initial segment shall be denoted $\left({\operatorname{On} \times \operatorname{On} }\right)_{\left({1, \varnothing}\right)}$.
The connection between these two statements is not clear.
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Proof
Define the mapping $F: \operatorname{On} \to \operatorname{On} \times \operatorname{On}$ as:
- $\forall x \in \operatorname{On}: F \left({x}\right) = \left({\varnothing, x}\right)$
Then, $F: \operatorname{On} \to \left({\operatorname{On} \times \operatorname{On}}\right)_{\left({1, \varnothing}\right)}$, since $\varnothing < 1$.
The above statement does not make linguistic sense. The entity $F: \operatorname{On} \to \left({\operatorname{On} \times \operatorname{On} }\right)_{\left({1, \varnothing}\right)}$ is a noun, and there is no verb in the "then" clause.
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That is, $F$ is a mapping from the class of all ordinals to the initial segment of $\left({1, \varnothing}\right)$ with respect to lexicographic order.
By Equality of Ordered Pairs, $F$ is injective.
But since $\operatorname{On}$ is a proper class by the Burali-Forti Paradox, the initial segment of $\left({1, \varnothing}\right)$ is a proper class as well.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 7.54 \ (2)$