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\begin{document}

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\section*{lexicographic order}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{category_theory}{}\paragraph*{{$(0,1)$-Category theory}}\label{category_theory}

[[!include (0,1)-category theory - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{corecursive_definition}{Corecursive definition}\dotfill \pageref*{corecursive_definition} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

The \emph{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.

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\begin{defn}
\label{}\hypertarget{}{}
Let $\{L_i\}_{i \in I}$ be a [[well-ordered set|well-ordered]] family of [[linear order|linearly ordered sets]]. The \textbf{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 $\{j \in I: x_j \neq y_j\}$.

\end{defn}
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 [[linear order|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

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

\begin{displaymath}
i(s_1, \ldots, s_k) = (s_1, s_2, \ldots, s_k, e, e, e, \ldots).
\end{displaymath}
Then the lexicographic order on $S^\ast$ is the one inherited from its embedding into the lexicographically ordered set $(1 + S)^\mathbb{N}$.

\begin{remark}
\label{antidictionary}\hypertarget{antidictionary}{}
The decision to freely adjoin a \emph{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.

\end{remark}
\hypertarget{corecursive_definition}{}\subsubsection*{{Corecursive definition}}\label{corecursive_definition}

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 [[corecursion|corecursively]]:

\begin{itemize}%
\item $c \lt c'$ if $h(c) \lt h(c')$ or ($h(c) = h(c')$ and $t(c) \lt t(c')$).

\end{itemize}
For the special case $L = \mathbb{N}$, the terminal coalgebra $\mathbb{N}^\mathbb{N}$ with this lexicographic order is order-isomorphic to the real interval $[0, \infty)$. This isomorphism is described more precisely \href{/nlab/show/continued+fraction#PavPratt}{here}.

\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\begin{prop}
\label{}\hypertarget{}{}
Given a well-ordered family of well-ordered sets $\{L_i\}_{i: \alpha}$, the product $\prod_{i: \alpha} L_i$ with the lexicographic order is also well-ordered.

\end{prop}
\begin{proof}
Represent elements of the set $\prod_{i: \alpha} L_i$ as sections $w: \alpha \to \sum_{i: \alpha} L_i$ of the canonical projection $\pi: \sum_{i: \alpha} L_i \to \sum_{i: \alpha} 1 \cong \alpha$. If $S \subseteq \prod_{i: \alpha} L_i$ is an inhabited subset, then its least element $s$ is defined recursively as follows: given $j \in \alpha$ and elements $s(i)$ for $i \lt j$, the element $s(j) \in L_j$ is the least element of the (nonempty!) set

\begin{displaymath}
\{x \in L_j: \exists_{w \in S} \left(\forall_{i \lt j} w(i) = s(i)\right) \wedge w(j) = x\}.
\end{displaymath}
With the section $s: \alpha \to \sum_{i: \alpha} L_i$ thus defined, if $w \in S$ and $w \lt s$ in lexicographic order, with $j \in \alpha$ the least element such that $w(j) \lt s(j)$, then $w(i) = s(i)$ for all $i \lt j$, and we reach a contradiction in view of how $s(j)$ was defined. Hence $s$ is the minimal element of $S$.

\end{proof}
[[!redirects lexicographic ordering]] [[!redirects dictionary order]] [[!redirects dictionary ordering]]



\end{document}