Initial Segment of Ordinals under Lexicographic Order

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Theorem

Let $\preccurlyeq_l$ denote the lexicographic order for the set $\paren {\On \times \On}$.

Let the ordinal number $1$ denote the successor of $\O$.

Then the initial segment of $\tuple {1, \O}$ with respect to the lexicographic order $\preccurlyeq_l$ is a proper class.

This initial segment shall be denoted $\paren {\On \times \On}_{\tuple {1, \O} }$.


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Proof

Define the mapping $F: \On \to \On \times \On$ as:

$\forall x \in \On: \map F x = \tuple {\O, x}$

Then, $F: \On \to \paren {\On \times \On}_{\tuple {1, \O} }$, since $\O < 1$.


This article, or a section of it, needs explaining.
In particular: The above statement does not make linguistic sense. The entity $F: \On \to \paren {\On \times \On}_{\tuple {1, \O} }$ is a noun, and there is no verb in the "then" clause.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.
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That is, $F$ is a mapping from the class of all ordinals to the initial segment of $\tuple {1, \O}$ with respect to lexicographic order.

By Equality of Ordered Pairs, $F$ is injective.

But since $\On$ is a proper class by the Burali-Forti Paradox, the initial segment of $\tuple {1, \O}$ is a proper class as well.

$\blacksquare$


Sources

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