# Initial Segment of Canonical Order is Set

## Theorem

Let $R_0$ denote the canonical ordering of $\left({\operatorname{On} \times \operatorname{On} }\right)$.

Then, for all $\left({x, y}\right) \in \left({\operatorname{On} \times \operatorname{On} }\right)$, the $R_0$-initial segment is a set.

## Proof

Let $z = \max \left({x, y}\right)^+$.

Let $\left({v, w}\right) R_0 \left({x, y}\right)$.

Then:

 $\displaystyle \max \left({v, w}\right)$ $\le$ $\displaystyle \max \left({x, y}\right)$ $\displaystyle$ $\lt$ $\displaystyle z$

Thus, the initial segment:

$\left({\operatorname{On} \times \operatorname{On} }\right)_{\left({x, y}\right)} \subseteq \left({z \times z}\right)$

By Axiom of Subsets Equivalents, the initial segment of $\left({x, y}\right)$ is a set.

$\blacksquare$