Antilexicographic Product of Totally Ordered Sets is Totally Ordered

Theorem
Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be tosets.

Let $S_1 \cdot S_2 = \struct {S_1 \times S_2, \preceq}$ be the ordered product of $S_1$ and $S_2$.

Then $\struct {S_1 \times S_2, \preceq}$ is itself a toset.

Proof
By definition of ordered product, we have that:


 * $b_1 \prec_2 b_2 \implies \tuple {a_1, b_1} \prec \tuple {a_2, b_2}$
 * $b_1 = b_2, a_1 \prec_1 a_2 \implies \tuple {a_1, b_1} \prec \tuple {a_2, b_2}$
 * $b_1 = b_2, a_1 = a_2 \implies \tuple {a_1, b_1} = \tuple {a_2, b_2}$

We note that as $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are both tosets, then $\preceq_1$ and $\preceq_2$ are both connected.

Thus it is clear that $\preceq$ is connected.

Now we check in turn each of the criteria for an ordering:

Reflexivity
We have by definition of set union:
 * $\forall \tuple {a, b} \in S_1 \times S_2: a = a \land b = b$

and so $\tuple {a, b} = \tuple {a, b}$.

Thus $\tuple {a, b} \preceq \tuple {a, b}$ and so $\preceq$ is reflexive.

Transitivity
Suppose $\tuple {a_1, b_1} \preceq \tuple {a_2, b_2} \preceq \tuple {a_3, b_3}$.

Suppose $\tuple {a_3, b_3} \prec \tuple {a_1, b_1}$.

This would happen because:


 * $b_3 \prec_2 b_1$, which can't happen because $\preceq_2$ is transitive;
 * $b_3 = b_1$ and $a_3 \prec_1 a_1$, which can't happen because $\preceq_1$ is transitive.

So we have shown that $\tuple {a_1, b_1} \preceq \tuple {a_3, b_3}$ and so $\preceq$ is transitive.

Antisymmetry
Suppose $\tuple {a_1, b_1} \preceq \tuple {a_2, b_2} \preceq \tuple {a_1, b_1}$.

Suppose $\tuple {a_1, b_1} \ne \tuple {a_2, b_2}$.

Then one of two cases holds:
 * $b_1 \ne b_2$, which can't happen because then either $\tuple {a_1, b_1} \prec \tuple {a_2, b_2}$ or $\tuple {a_2, b_2} \prec \tuple {a_1, b_1}$;
 * $b_1 = b_2$, and $a_1 \ne a_2$, which can't happen because then either $\tuple {a_1, b_1} \prec \tuple {a_2, b_2}$ or $\tuple {a_2, b_2} \prec \tuple {a_1, b_1}$.

Thus in all cases it can be seen that $\tuple {a_1, b_1} \preceq \tuple {a_2, b_2} \preceq \tuple {a_1, b_1} \implies \tuple {a_1, b_1} = \tuple {a_2, b_2}$.

So $\preceq$ is antisymmetric.

So we have shown that:
 * $\preceq$ is connected
 * $\preceq$ is reflexive, transitive and antisymmetric.

Thus by definition, $\preceq$ is a total ordering and so $\struct {S_1 \times S_2, \preceq}$ is a toset.