Antilexicographic Product of Totally Ordered Sets is Totally Ordered/General Result

Theorem

Let $S_1, S_2, \ldots, S_n$ all be totally ordered sets.

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

$\forall n \in \N_{>0}: T_n = \begin {cases} S_1 & : n = 1 \\ T_{n - 1} \otimes^a S_n & : n > 1 \end {cases}$

Then $T_n$ is a totally ordered set.

Proof

From Antilexicographic Product of Totally Ordered Sets is Totally Ordered, $S_1 \otimes^a S_2$ is a totally ordered set.

Suppose $T_{n - 1}$ is a totally ordered set.

Given that $S_n$ is a totally ordered set, it follows from Antilexicographic Product of Totally Ordered Sets is Totally Ordered that $T_{n - 1} \otimes^a S_n$ is also a totally ordered set.

The result follows by induction.

$\blacksquare$

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