Simple Order Product of Totally Ordered Sets may not be Totally Ordered

Theorem
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be totally ordered sets.

Let $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ denote the simple (order) product of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.

Then it is not necessarily the case that $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is also a totally ordered set.

Proof
From Simple Order Product of Ordered Sets is Ordered Set, we do have that $\struct {S_1, \preccurlyeq_1} \otimes^s \struct {S_2, \preccurlyeq_2}$ is an ordered set.

Let us take the simple product of the ordered set that is the natural numbers under the usual ordering with itself:


 * $\struct {\N, \le} \otimes^s \struct {\N, \le}$

We have from the Well-Ordering Principle that $\struct {\N, \le}$ is a well-ordered set, and so a fortiori a totally ordered set.

Consider the ordered pairs of natural numbers:
 * $s = \tuple {1, 4}$
 * $t = \tuple {2, 3}$

Both $s$ and $t$ are elements of $\N \times \N$.

We have that:
 * $1 \le 2$

but:
 * $4 \not \le 3$

So it is not the case either that:
 * $1 \le 2$ and $4 \le 3$

or:
 * $2 \le 1$ and $3 \le 4$

Hence $s$ and $t$ are non-comparable elements.

The result follows by definition of totally ordered set.