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 $\left({\operatorname{On} \times \operatorname{On} }\right)_{\left({1, \varnothing}\right)}$.


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 $\varnothing < 1$.



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.

$\blacksquare$


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