lexicographic order

The *lexicographic order* is a generalization of the order in which words are listed in a dictionary, according to the order of letters where the spelling of two words first differs.

Let $\{L_i\}_{i \in I}$ be a well-ordered family of linearly ordered sets. The **lexicographic order** on the product of sets $L = \prod_{i \in I} L_i$ is the linear order defined as follows: if $x, y \in L$ and $x \neq y$, then $x \lt y$ iff $x_i \lt y_i$ where $i$ is the least element in the set $\{j \in I: x_j \neq y_j\}$.

While this notion is most often seen for linear orders, it can be applied also toward more general relations. For example, one might apply the construction to sets equipped with a transitive relation $\lt$, dropping the trichotomy assumption.

Often this notion is extended to subsets of $\prod_{i \in I} L_i$ as well. For instance, the free monoid $S^\ast$ on a linearly ordered set $S$ can be embedded in a countable power

$i \colon S^\ast \hookrightarrow (1 + S)^\mathbb{N} = \prod_{n \in \mathbb{N}} (1 + S)$

where $1 + S$ is the result of freely adjoining a bottom element $e$ to $S$, and for each finite list $(s_1, \ldots, s_k)$ we have

$i(s_1, \ldots, s_k) = (s_1, s_2, \ldots, s_k, e, e, e, \ldots).$

Then the lexicographic order on $S^\ast$ is the one inherited from its embedding into the lexicographically ordered set $(1 + S)^\mathbb{N}$.

The decision to freely adjoin a *bottom* element $e$ is of course purely a convention, based on the ordinary dictionary convention that the Scrabble word AAH should come after AA. Alternatively, we could equally well deem that $e$ is a freely adjoined top element, so that AA comes after AAH; this might be called the “anti-dictionary” convention.

if $L$ is linearly ordered and the underlying set $C = L^\mathbb{N}$ is regarded as the terminal coalgebra for the functor $L \times - \colon Set \to Set$, with coalgebra structure $\langle h, t \rangle \colon C \to L \times C$, then the lexicographic order on $C$ may be defined corecursively:

- $c \lt c'$ if $h(c) \lt h(c')$ or ($h(c) = h(c')$ and $t(c) \lt t(c')$).

Revised on April 24, 2014 13:56:35
by Todd Trimble
(67.80.196.20)