The lexicographic ordering on On$\times$ On, the class of all pairs of ordinals, is a well-order in the broad sense, in that every subclass of On$\times$ 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$\times$ On such that every initial segment of On$\times$ On is a set? The answer is yes. The construction of such a well-ordering in the following discussion is what is known as the canonical well-ordering of On$\times$ On.

To begin, let us consider a strictly linearly ordered set $(A,<)$ . We construct a binary relation $\prec$ on $A\times A$ as follows:

   iff

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 $\lbrace (m,n)\mid \max\lbrace m,n\rbrace < q\rbrace \cup \lbrace (m,q)\mid m < p \rbrace \cup \lbrace (p,n)\mid n < q\rbrace$ .

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. $\max \lbrace a_1,a_2\rbrace < \max \lbrace b_1, b_2\rbrace$ and $\max \lbrace b_1,b_2\rbrace \le \max \lbrace c_1, c_2\rbrace$ ,
2. $\max \lbrace a_1,a_2\rbrace \le \max \lbrace b_1, b_2\rbrace$ and $\max \lbrace b_1,b_2\rbrace < \max \lbrace c_1, c_2\rbrace$ ,

produce $\max \lbrace a_1,a_2\rbrace < \max \lbrace c_1, c_2\rbrace$ . Now, assume $\max \lbrace a_1,a_2\rbrace = \max \lbrace b_1, b_2\rbrace = \max \lbrace c_1, c_2\rbrace$ , which result in three more cases

1. $a_1 < b_1$ and $b_1 \le c_1$ ,
2. $a_1 \le b_1$ and $b_1 < c_1$ ,
3. $a_1=b_1=c_1$ , and $a_2 < b_2$ and $b_2 < c_2$ ,

the first two produce $a_1 < c_1$ , and the last $a_1=c_1$ and $a_2<c_2$ . In all cases, we get $(a_1,a_2)\prec (c_1,c_2)$ .

The ordering relation above can be generalized to arbitrary classes. Since On is well-ordered by $\in$ , the canonical ordering on On$\times$ On is a well-ordering by proposition 2. Moreover,

Proof. Given ordinals $\alpha, \beta\in $ On, suppose $\lambda = \max\lbrace \alpha,\beta\rbrace$ . The initial segment of $(\alpha,\beta)$ is the union of the following collections

1. $\lbrace (\gamma,\delta)\mid \max\lbrace \gamma,\delta\rbrace < \lambda \rbrace$ , which is a subcollection of $\lambda \times \lambda$ ,
2. $\lbrace (\gamma,\delta) \mid \max\lbrace \gamma,\delta\rbrace = \lambda, \mbox{ and }\gamma < \alpha \rbrace$ , which again is a subcollection $\lambda\times \lambda$ , and
3. $\lbrace (\alpha,\delta) \mid \max\lbrace \alpha,\delta \rbrace = \lambda, \mbox{ and }\delta < \beta \rbrace$ , which is a subcollection of $\lbrace \alpha \rbrace \times \beta $ .

Since $\lambda\times \lambda$ and $\lbrace \alpha \rbrace \times \beta $ are both sets, so is the initial segment of $(\alpha,\beta)$ .

Remark. The canonical well-ordering on On$\times$ On can be used to prove a well-known property of alephs: $\aleph_{\alpha} \cdot \aleph_{\alpha} = \aleph_{\alpha}$ , for any ordinal $\alpha$ .