Initial Segment of Ordinals under Lexicographic Order

Theorem
Let $\operatorname{Le}$ denote the lexicographic order for the set $(\operatorname{On} \times \operatorname{On})$.

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

Then, the initial segment of $(1,\varnothing)$ with respect to the lexicographic order $\operatorname{Le}$ is a proper class.

This initial segment shall be denoted $( \operatorname{On} \times \operatorname{On} )_{(1,\varnothing)}$.

Proof
Define the mapping $F$ such that $F(x) = (\varnothing,x)$.

Then, $F:\operatorname{On} \to (\operatorname{On} \times \operatorname{On})_{(1,\varnothing)}$, since $\varnothing < 1$. That is, $F$ is a mapping from the class of all ordinals to the initial segment of $(1,\varnothing)$ 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 $(1,\varnothing)$ is a proper class as well.

Source

 * : $\S 7.54 \ (2)$