Definition:Ordered Product

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Definition

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

Let:

Let $T = S_1 \times S_2$ be the cartesian product of $S_1$ and $S_2$.

Consider the relation $\preceq$ defined on $T$ as follows.

Let $a_1$ and $a_2$ be arbitrary elements of $S_1$, and $b_1$ and $b_2$ be arbitrary elements of $S_2$.

Then:

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

The ordered set $\left({S_1 \times S_2, \preceq}\right)$ is called the ordered product of $S_1$ and $S_2$, and is denoted $S_1 \cdot S_2$.

The order type of $S_1 \cdot S_2$ is denoted $\theta_1 \cdot \theta_2$.

We can define the ordered product of any finite number of tosets as follows.

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

Then we define $T_n$ as the ordered product of $S_1, S_2, \ldots, S_n$ as:

$\forall n \in \N^*: T_n = \begin{cases} S_1 & : n = 1 \\ T_{n-1} \cdot S_n & : n > 1 \end{cases}$

The ordered product is defined only for totally ordered sets.