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 Pair 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.
$\blacksquare$
Sources
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $29 \ \text {(b)}$