Antilexicographic Product of Totally Ordered Sets is Totally Ordered

Theorem
Let $$\left({S_1, \preceq_1}\right)$$ and $$\left({S_2, \preceq_2}\right)$$ be tosets.

Let $$S_1 \cdot S_2 = \left({S_1 \times S_2, \preceq}\right)$$ be the ordered product of $$S_1$$ and $$S_2$$.

Then $$\left({S_1 \times S_2, \preceq}\right)$$ is itself a toset.

General Definition
Let $$S_1, S_2, \ldots, S_n$$ all be tosets.

Let $$T_n$$ be the ordered product of $$S_1, S_2, \ldots, S_n$$:


 * $$\forall n \in \N^*: T_n = \begin{cases}

S_1 & : n = 1 \\ T_{n-1} \cdot S_n & : n > 1 \end{cases}$$

Then $$T_n$$ is a toset.

Proof
By definition of ordered product, we have that:


 * $$b_1 \prec_2 b_2 \implies \left({a_1, b_1}\right) \prec \left({a_2, b_2}\right)$$
 * $$b_1 = b_2, a_1 \prec_1 a_2 \implies \left({a_1, b_1}\right) \prec \left({a_2, b_2}\right)$$
 * $$b_1 = b_2, a_1 = a_2 \implies \left({a_1, b_1}\right) = \left({a_2, b_2}\right)$$

We note that as $$\left({S_1, \preceq_1}\right)$$ and $$\left({S_2, \preceq_2}\right)$$ 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:
 * $$\forall \left({a, b}\right) \in S_1 \times S_2: a = a \and b = b$$

and so $$\left({a, b}\right) = \left({a, b}\right)$$.

Thus $$\left({a, b}\right) \preceq \left({a, b}\right)$$ and so $$\preceq$$ is reflexive.

Transitivity
Suppose $$\left({a_1, b_1}\right) \preceq \left({a_2, b_2}\right) \preceq \left({a_3, b_3}\right)$$.

Suppose $$\left({a_3, b_3}\right) \prec \left({a_1, b_1}\right)$$.

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 $$\left({a_1, b_1}\right) \preceq \left({a_3, b_3}\right)$$ and so $$\preceq$$ is transitive.

Antisymmetry
Suppose $$\left({a_1, b_1}\right) \preceq \left({a_2, b_2}\right) \preceq \left({a_1, b_1}\right)$$.

Suppose $$\left({a_1, b_1}\right) \ne \left({a_2, b_2}\right)$$.

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

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

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 $$\left({S_1 \times S_2, \preceq}\right)$$ is a toset.

Proof of General Result
We have that $$S_1 \cdot S_2$$ is a toset from the main result.

Suppose $$T_{n-1}$$ is a toset.

Given that $$S_n$$ is a toset, it follows from the main result that $$T_{n-1} \cdot S_n$$ is also a toset.

The result follows by induction.