canonical ordering on pairs of ordinals
The lexicographic ordering on On$\mathrm{\times}$On, the class of all pairs of ordinals^{}, is a wellorder in the broad sense, in that every subclass of On$\mathrm{\times}$On has a least element, as proposition^{} 2 of the parent entry readily shows. However, with this type of ordering^{}, we get initial segments which are not sets. For example, the initial segment of $(1,0)$ consists of all ordinal pairs of the form $(0,\alpha )$, where $\alpha \in $ On, and is easily seen to be a proper class. So the questions is: is there a way to order On$\mathrm{\times}$On such that every initial segment of On$\mathrm{\times}$On is a set? The answer is yes. The construction of such a wellordering in the following discussion is what is known as the canonical wellordering of On$\mathrm{\times}$On.
To begin, let us consider a strictly linearly ordered set $$. We construct a binary relation^{} $\prec $ on $A\times A$ as follows:
$$ 
For example, consider the usual ordering on $\mathbb{Z}$. Given $(p,q)\in \mathbb{Z}\times \mathbb{Z}$. Suppose $p\le q$. Then the set of all $(m,n)\in \mathbb{Z}\times \mathbb{Z}$ such that $(m,n)\prec (p,q)$ is the union of the three pairwise disjoint sets $$.
Proposition 1.
. $\mathrm{\prec}$ is a strict linear ordering on $A\mathrm{\times}A$.
Proof.
It is irreflexive^{} because $({a}_{1},{a}_{2})$ is never comparable^{} with itself. It is linear because, first of all, given $({a}_{1},{a}_{2})\ne ({b}_{1},{b}_{2})$, exactly one of the three conditions is true, and hence either $({a}_{1},{a}_{2})\prec ({b}_{1},{b}_{2})$, or $({b}_{1},{b}_{2})\prec ({a}_{1},{a}_{2})$. It remains to show that $\prec $ is transitive^{}, suppose $({a}_{1},{a}_{2})\prec ({b}_{1},{b}_{2})$ and $({b}_{1},{b}_{2})\prec ({c}_{1},{c}_{2})$.
The two cases

1.
$$ and $\mathrm{max}\{{b}_{1},{b}_{2}\}\le \mathrm{max}\{{c}_{1},{c}_{2}\}$,

2.
$\mathrm{max}\{{a}_{1},{a}_{2}\}\le \mathrm{max}\{{b}_{1},{b}_{2}\}$ and $$,
produce $$. Now, assume $\mathrm{max}\{{a}_{1},{a}_{2}\}=\mathrm{max}\{{b}_{1},{b}_{2}\}=\mathrm{max}\{{c}_{1},{c}_{2}\}$, which result in three more cases

1.
$$ and ${b}_{1}\le {c}_{1}$,

2.
${a}_{1}\le {b}_{1}$ and $$,

3.
${a}_{1}={b}_{1}={c}_{1}$, and $$ and $$,
the first two produce $$, and the last ${a}_{1}={c}_{1}$ and $$. In all cases, we get $({a}_{1},{a}_{2})\prec ({c}_{1},{c}_{2})$. ∎
Proposition 2.
If $$ is a wellorder on $A$, then so is $\mathrm{\prec}$ on $A\mathrm{\times}A$.
Proof.
Let $R\subseteq A\times A$ be nonempty. Let
$$B:=\{\mathrm{max}\{{b}_{1},{b}_{2}\}\mid ({b}_{1},{b}_{2})\in R\}.$$ 
Then $\mathrm{\varnothing}\ne B\subseteq A$, and therefore has a least element $b$, since $$ is a wellorder on $A$. Next, let
$$C:=\{{c}_{1}\mid \mathrm{max}\{{c}_{1},{c}_{2}\}=b,\text{where}({c}_{1},{c}_{2})\in R\}.$$ 
Then $C\ne \mathrm{\varnothing}$, and has a least element $c$. Finally, let
$$D:=\{{d}_{2}\mid \mathrm{max}\{c,{d}_{2}\}=b,\text{where}(c,{d}_{2})\in R\}.$$ 
Again, $D\ne \mathrm{\varnothing}$, so has a least element $d$. So $(c,d)\in R$. We want to show that $(c,d)$ is the least element of $R$.
Pick any $(x,y)\in R$ distinct from $(c,d)$. Then $\mathrm{max}\{x,y\}\in B$ is at least $b=\mathrm{max}\{c,d\}$. If $$, then $(c,d)\prec (x,y)$. Otherwise, $b=\mathrm{max}\{x,y\}$, so that $x\in C$ is at least $c$. If $$, then again we have $(c,d)\prec (x,y)$. But if $c=x$, then $y\in D$, so that $d\le y$. Since $(x,y)\ne (c,d)$, and $x=c$, $y\ne d$. Therefore $$, and $(c,d)\prec (x,y)$ as a result. ∎
The ordering relation above can be generalized to arbitrary classes. Since On is wellordered by $\in $, the canonical ordering on On$\mathrm{\times}$On is a wellordering by proposition 2. Moreover,
Proposition 3.
Given the canonical ordering $\mathrm{\prec}$ on On$\mathrm{\times}$On, every initial segment is a set.
Proof.
Given ordinals $\alpha ,\beta \in $On, suppose $\lambda =\mathrm{max}\{\alpha ,\beta \}$. The initial segment of $(\alpha ,\beta )$ is the union of the following collections^{}

1.
$$, which is a subcollection of $\lambda \times \lambda $,

2.
$$, which again is a subcollection $\lambda \times \lambda $, and

3.
$$, which is a subcollection of $\{\alpha \}\times \beta $.
Since $\lambda \times \lambda $ and $\{\alpha \}\times \beta $ are both sets, so is the initial segment of $(\alpha ,\beta )$. ∎
Remark. The canonical wellordering on On$\mathrm{\times}$On can be used to prove a wellknown property of alephs: ${\mathrm{\aleph}}_{\alpha}\cdot {\mathrm{\aleph}}_{\alpha}={\mathrm{\aleph}}_{\alpha}$, for any ordinal $\alpha $.
Title  canonical ordering on pairs of ordinals 

Canonical name  CanonicalOrderingOnPairsOfOrdinals 
Date of creation  20130322 18:50:02 
Last modified on  20130322 18:50:02 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03E10 
Classification  msc 06A05 
Synonym  canonical wellordering 
Related topic  IdempotencyOfInfiniteCardinals 
Defines  canonical ordering 