-%\begin{proof}
-%Let us prove it by contradiction, assuming that $\left(\X_{\tau_\infty},
-%f\right)$ is both transitive and regular.
-
-%Let $x \in \X$ and $\{x\}$ one of its neighborhood. This neighborhood must
-%contain a periodic point for $f$, if we want that $\left(\X_{\tau_\infty},
-%f\right)$ is regular. Then $x$ must be a periodic point of $f$.
-
-%Let $I_x = \left\{ f^{(n)}(x), n \in \mathds{N}\right\}$. This set is finite
-%because $x$ is periodic, and $\mathcal{X}$ is infinite, then $\exists y \in
-%\mathcal{X}, y \notin I_x$.
-
-%As $\left(\X_{\tau_\infty}, f\right)$ must be transitive, for all open nonempty
-%sets $A$ and $B$, an integer $n$ must satisfy $f^{(n)}(A) \cap B \neq
-%\varnothing$. However $\{x\}$ and $\{y\}$ are open sets and $y \notin I_x
-%\Rightarrow \forall n, f^{(n)}\left( \{x\} \right) \cap \{y\} = \varnothing$.
-%\end{proof}
-
-
-
-
-
-
-%\section{Chaos on the order topology}
-%\label{sec: chaos order topology}
-%\subsection{The phase space is an interval of the real line}
-
-%\subsubsection{Toward a topological semiconjugacy}
-
-%In what follows, our intention is to establish, by using a topological
-%semiconjugacy, that chaotic iterations over $\mathcal{X}$ can be described as
-%iterations on a real interval. To do so, we must firstly introduce some
-%notations and terminologies.
-
-%Let $\mathcal{S}_\mathsf{N}$ be the set of sequences belonging into $\llbracket
-%1; \mathsf{N}\rrbracket$ and $\mathcal{X}_{\mathsf{N}} = \mathcal{S}_\mathsf{N}
-%\times \B^\mathsf{N}$.
-
-
-%\begin{definition}
-%The function $\varphi: \mathcal{S}_{10} \times\mathds{B}^{10} \rightarrow \big[
-%0, 2^{10} \big[$ is defined by:
-%\begin{equation}
-% \begin{array}{cccl}
-%\varphi: & \mathcal{X}_{10} = \mathcal{S}_{10} \times\mathds{B}^{10}&
-%\longrightarrow & \big[ 0, 2^{10} \big[ \\
-% & (S,E) = \left((S^0, S^1, \hdots ); (E_0, \hdots, E_9)\right) & \longmapsto &
-%\varphi \left((S,E)\right)
-%\end{array}
-%\end{equation}
-%where $\varphi\left((S,E)\right)$ is the real number:
-%\begin{itemize}
-%\item whose integral part $e$ is $\displaystyle{\sum_{k=0}^9 2^{9-k} E_k}$, that
-%is, the binary digits of $e$ are $E_0 ~ E_1 ~ \hdots ~ E_9$.
-%\item whose decimal part $s$ is equal to $s = 0,S^0~ S^1~ S^2~ \hdots =
-%\sum_{k=1}^{+\infty} 10^{-k} S^{k-1}.$
-%\end{itemize}
-%\end{definition}
-
-
-
-%$\varphi$ realizes the association between a point of $\mathcal{X}_{10}$ and a
-%real number into $\big[ 0, 2^{10} \big[$. We must now translate the chaotic
-%iterations $\Go$ on this real interval. To do so, two intermediate functions
-%over $\big[ 0, 2^{10} \big[$ must be introduced:
-
-
-%\begin{definition}
-%\label{def:e et s}
-%Let $x \in \big[ 0, 2^{10} \big[$ and:
-%\begin{itemize}
-%\item $e_0, \hdots, e_9$ the binary digits of the integral part of $x$:
-%$\displaystyle{\lfloor x \rfloor = \sum_{k=0}^{9} 2^{9-k} e_k}$.
-%\item $(s^k)_{k\in \mathds{N}}$ the digits of $x$, where the chosen decimal
-%decomposition of $x$ is the one that does not have an infinite number of 9:
-%$\displaystyle{x = \lfloor x \rfloor + \sum_{k=0}^{+\infty} s^k 10^{-k-1}}$.
-%\end{itemize}
-%$e$ and $s$ are thus defined as follows:
-%\begin{equation}
-%\begin{array}{cccl}
-%e: & \big[ 0, 2^{10} \big[ & \longrightarrow & \mathds{B}^{10} \\
-% & x & \longmapsto & (e_0, \hdots, e_9)
-%\end{array}
-%\end{equation}
-%and
-%\begin{equation}
-% \begin{array}{cccc}
-%s: & \big[ 0, 2^{10} \big[ & \longrightarrow & \llbracket 0, 9
-%\rrbracket^{\mathds{N}} \\
-% & x & \longmapsto & (s^k)_{k \in \mathds{N}}
-%\end{array}
-%\end{equation}
-%\end{definition}
-
-%We are now able to define the function $g$, whose goal is to translate the
-%chaotic iterations $\Go$ on an interval of $\mathds{R}$.
-
-%\begin{definition}
-%$g:\big[ 0, 2^{10} \big[ \longrightarrow \big[ 0, 2^{10} \big[$ is defined by:
-%\begin{equation}
-%\begin{array}{cccc}
-%g: & \big[ 0, 2^{10} \big[ & \longrightarrow & \big[ 0, 2^{10} \big[ \\
-% & x & \longmapsto & g(x)
-%\end{array}
-%\end{equation}
-%where g(x) is the real number of $\big[ 0, 2^{10} \big[$ defined bellow:
-%\begin{itemize}
-%\item its integral part has a binary decomposition equal to $e_0', \hdots,
-%e_9'$, with:
-% \begin{equation}
-%e_i' = \left\{
-%\begin{array}{ll}
-%e(x)_i & \textrm{ if } i \neq s^0\\
-%e(x)_i + 1 \textrm{ (mod 2)} & \textrm{ if } i = s^0\\
-%\end{array}
-%\right.
-%\end{equation}
-%\item whose decimal part is $s(x)^1, s(x)^2, \hdots$
-%\end{itemize}
-%\end{definition}
-
-%\bigskip
-
-
-%In other words, if $x = \displaystyle{\sum_{k=0}^{9} 2^{9-k} e_k +
-%\sum_{k=0}^{+\infty} s^{k} ~10^{-k-1}}$, then:
-%\begin{equation}
-%g(x) =
-%\displaystyle{\sum_{k=0}^{9} 2^{9-k} (e_k + \delta(k,s^0) \textrm{ (mod 2)}) +
-%\sum_{k=0}^{+\infty} s^{k+1} 10^{-k-1}}.
-%\end{equation}
-
-
-%\subsubsection{Defining a metric on $\big[ 0, 2^{10} \big[$}
-
-%Numerous metrics can be defined on the set $\big[ 0, 2^{10} \big[$, the most
-%usual one being the Euclidian distance recalled bellow:
-
-%\begin{notation}
-%\index{distance!euclidienne}
-%$\Delta$ is the Euclidian distance on $\big[ 0, 2^{10} \big[$, that is,
-%$\Delta(x,y) = |y-x|^2$.
-%\end{notation}
-
-%\medskip
-
-%This Euclidian distance does not reproduce exactly the notion of proximity
-%induced by our first distance $d$ on $\X$. Indeed $d$ is finer than $\Delta$.
-%This is the reason why we have to introduce the following metric:
-
-
-
-%\begin{definition}
-%Let $x,y \in \big[ 0, 2^{10} \big[$.
-%$D$ denotes the function from $\big[ 0, 2^{10} \big[^2$ to $\mathds{R}^+$
-%defined by: $D(x,y) = D_e\left(e(x),e(y)\right) + D_s\left(s(x),s(y)\right)$,
-%where:
-%\begin{center}
-%$\displaystyle{D_e(E,\check{E}) = \sum_{k=0}^\mathsf{9} \delta (E_k,
-%\check{E}_k)}$, ~~and~ $\displaystyle{D_s(S,\check{S}) = \sum_{k = 1}^\infty
-%\dfrac{|S^k-\check{S}^k|}{10^k}}$.
-%\end{center}
-%\end{definition}
-
-%\begin{proposition}
-%$D$ is a distance on $\big[ 0, 2^{10} \big[$.
-%\end{proposition}
-
-%\begin{proof}
-%The three axioms defining a distance must be checked.
-%\begin{itemize}
-%\item $D \geqslant 0$, because everything is positive in its definition. If
-%$D(x,y)=0$, then $D_e(x,y)=0$, so the integral parts of $x$ and $y$ are equal
-%(they have the same binary decomposition). Additionally, $D_s(x,y) = 0$, then
-%$\forall k \in \mathds{N}^*, s(x)^k = s(y)^k$. In other words, $x$ and $y$ have
-%the same $k-$th decimal digit, $\forall k \in \mathds{N}^*$. And so $x=y$.
-%\item $D(x,y)=D(y,x)$.
-%\item Finally, the triangular inequality is obtained due to the fact that both
-%$\delta$ and $\Delta(x,y)=|x-y|$ satisfy it.
-%\end{itemize}
-%\end{proof}
-
-
-%The convergence of sequences according to $D$ is not the same than the usual
-%convergence related to the Euclidian metric. For instance, if $x^n \to x$
-%according to $D$, then necessarily the integral part of each $x^n$ is equal to
-%the integral part of $x$ (at least after a given threshold), and the decimal
-%part of $x^n$ corresponds to the one of $x$ ``as far as required''.
-%To illustrate this fact, a comparison between $D$ and the Euclidian distance is
-%given Figure \ref{fig:comparaison de distances}. These illustrations show that
-%$D$ is richer and more refined than the Euclidian distance, and thus is more
-%precise.
-
-
-%\begin{figure}[t]
-%\begin{center}
-% \subfigure[Function $x \to dist(x;1,234) $ on the interval
-%$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien.pdf}}\quad
-% \subfigure[Function $x \to dist(x;3) $ on the interval
-%$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien2.pdf}}
-%\end{center}
-%\caption{Comparison between $D$ (in blue) and the Euclidian distane (in green).}
-%\label{fig:comparaison de distances}
-%\end{figure}
-
-
-
-
-%\subsubsection{The semiconjugacy}
-
-%It is now possible to define a topological semiconjugacy between $\mathcal{X}$
-%and an interval of $\mathds{R}$:
-
-%\begin{theorem}
-%Chaotic iterations on the phase space $\mathcal{X}$ are simple iterations on
-%$\mathds{R}$, which is illustrated by the semiconjugacy of the diagram bellow:
-%\begin{equation*}
-%\begin{CD}
-%\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right) @>G_{f_0}>>
-%\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right)\\
-% @V{\varphi}VV @VV{\varphi}V\\
-%\left( ~\big[ 0, 2^{10} \big[, D~\right) @>>g> \left(~\big[ 0, 2^{10} \big[,
-%D~\right)
-%\end{CD}
-%\end{equation*}
-%\end{theorem}
-
-%\begin{proof}
-%$\varphi$ has been constructed in order to be continuous and onto.
-%\end{proof}
-
-%In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N}
-%\big[$.
-
-
-
-
-
-
-%\subsection{Study of the chaotic iterations described as a real function}
-
-
-%\begin{figure}[t]
-%\begin{center}
-% \subfigure[ICs on the interval
-%$(0,9;1)$.]{\includegraphics[scale=.35]{ICs09a1.pdf}}\quad
-% \subfigure[ICs on the interval
-%$(0,7;1)$.]{\includegraphics[scale=.35]{ICs07a95.pdf}}\\
-% \subfigure[ICs on the interval
-%$(0,5;1)$.]{\includegraphics[scale=.35]{ICs05a1.pdf}}\quad
-% \subfigure[ICs on the interval
-%$(0;1)$]{\includegraphics[scale=.35]{ICs0a1.pdf}}
-%\end{center}
-%\caption{Representation of the chaotic iterations.}
-%\label{fig:ICs}
-%\end{figure}
-
-
-
-
-%\begin{figure}[t]
-%\begin{center}
-% \subfigure[ICs on the interval
-%$(510;514)$.]{\includegraphics[scale=.35]{ICs510a514.pdf}}\quad
-% \subfigure[ICs on the interval
-%$(1000;1008)$]{\includegraphics[scale=.35]{ICs1000a1008.pdf}}
-%\end{center}
-%\caption{ICs on small intervals.}
-%\label{fig:ICs2}
-%\end{figure}
-
-%\begin{figure}[t]
-%\begin{center}
-% \subfigure[ICs on the interval
-%$(0;16)$.]{\includegraphics[scale=.3]{ICs0a16.pdf}}\quad
-% \subfigure[ICs on the interval
-%$(40;70)$.]{\includegraphics[scale=.45]{ICs40a70.pdf}}\quad
-%\end{center}
-%\caption{General aspect of the chaotic iterations.}
-%\label{fig:ICs3}
-%\end{figure}
-
-
-%We have written a Python program to represent the chaotic iterations with the
-%vectorial negation on the real line $\mathds{R}$. Various representations of
-%these CIs are given in Figures \ref{fig:ICs}, \ref{fig:ICs2} and \ref{fig:ICs3}.
-%It can be remarked that the function $g$ is a piecewise linear function: it is
-%linear on each interval having the form $\left[ \dfrac{n}{10},
-%\dfrac{n+1}{10}\right[$, $n \in \llbracket 0;2^{10}\times 10 \rrbracket$ and its
-%slope is equal to 10. Let us justify these claims:
-
-%\begin{proposition}
-%\label{Prop:derivabilite des ICs}
-%Chaotic iterations $g$ defined on $\mathds{R}$ have derivatives of all orders on
-%$\big[ 0, 2^{10} \big[$, except on the 10241 points in $I$ defined by $\left\{
-%\dfrac{n}{10} ~\big/~ n \in \llbracket 0;2^{10}\times 10\rrbracket \right\}$.
-
-%Furthermore, on each interval of the form $\left[ \dfrac{n}{10},
-%\dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$,
-%$g$ is a linear function, having a slope equal to 10: $\forall x \notin I,
-%g'(x)=10$.
-%\end{proposition}
-
-
-%\begin{proof}
-%Let $I_n = \left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket
-%0;2^{10}\times 10 \rrbracket$. All the points of $I_n$ have the same integral
-%prat $e$ and the same decimal part $s^0$: on the set $I_n$, functions $e(x)$
-%and $x \mapsto s(x)^0$ of Definition \ref{def:e et s} only depend on $n$. So all
-%the images $g(x)$ of these points $x$:
-%\begin{itemize}
-%\item Have the same integral part, which is $e$, except probably the bit number
-%$s^0$. In other words, this integer has approximately the same binary
-%decomposition than $e$, the sole exception being the digit $s^0$ (this number is
-%then either $e+2^{10-s^0}$ or $e-2^{10-s^0}$, depending on the parity of $s^0$,
-%\emph{i.e.}, it is equal to $e+(-1)^{s^0}\times 2^{10-s^0}$).
-%\item A shift to the left has been applied to the decimal part $y$, losing by
-%doing so the common first digit $s^0$. In other words, $y$ has been mapped into
-%$10\times y - s^0$.
-%\end{itemize}
-%To sum up, the action of $g$ on the points of $I$ is as follows: first, make a
-%multiplication by 10, and second, add the same constant to each term, which is
-%$\dfrac{1}{10}\left(e+(-1)^{s^0}\times 2^{10-s^0}\right)-s^0$.
-%\end{proof}
-
-%\begin{remark}
-%Finally, chaotic iterations are elements of the large family of functions that
-%are both chaotic and piecewise linear (like the tent map).
-%\end{remark}
-
-
-
-%\subsection{Comparison of the two metrics on $\big[ 0, 2^\mathsf{N} \big[$}
-
-%The two propositions bellow allow to compare our two distances on $\big[ 0,
-%2^\mathsf{N} \big[$:
-
-%\begin{proposition}
-%Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,\Delta~\right) \to \left(~\big[ 0,
-%2^\mathsf{N} \big[, D~\right)$ is not continuous.
-%\end{proposition}
-
-%\begin{proof}
-%The sequence $x^n = 1,999\hdots 999$ constituted by $n$ 9 as decimal part, is
-%such that:
-%\begin{itemize}
-%\item $\Delta (x^n,2) \to 0.$
-%\item But $D(x^n,2) \geqslant 1$, then $D(x^n,2)$ does not converge to 0.
-%\end{itemize}
-
-%The sequential characterization of the continuity concludes the demonstration.
-%\end{proof}
-
-
-
-%A contrario:
-
-%\begin{proposition}
-%Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,D~\right) \to \left(~\big[ 0,
-%2^\mathsf{N} \big[, \Delta ~\right)$ is a continuous fonction.
-%\end{proposition}
-
-%\begin{proof}
-%If $D(x^n,x) \to 0$, then $D_e(x^n,x) = 0$ at least for $n$ larger than a given
-%threshold, because $D_e$ only returns integers. So, after this threshold, the
-%integral parts of all the $x^n$ are equal to the integral part of $x$.
-
-%Additionally, $D_s(x^n, x) \to 0$, then $\forall k \in \mathds{N}^*, \exists N_k
-%\in \mathds{N}, n \geqslant N_k \Rightarrow D_s(x^n,x) \leqslant 10^{-k}$. This
-%means that for all $k$, an index $N_k$ can be found such that, $\forall n
-%\geqslant N_k$, all the $x^n$ have the same $k$ firsts digits, which are the
-%digits of $x$. We can deduce the convergence $\Delta(x^n,x) \to 0$, and thus the
-%result.
-%\end{proof}
-
-%The conclusion of these propositions is that the proposed metric is more precise
-%than the Euclidian distance, that is:
-
-%\begin{corollary}
-%$D$ is finer than the Euclidian distance $\Delta$.
-%\end{corollary}
-
-%This corollary can be reformulated as follows:
-
-%\begin{itemize}
-%\item The topology produced by $\Delta$ is a subset of the topology produced by
-%$D$.
-%\item $D$ has more open sets than $\Delta$.
-%\item It is harder to converge for the topology $\tau_D$ inherited by $D$, than
-%to converge with the one inherited by $\Delta$, which is denoted here by
-%$\tau_\Delta$.
-%\end{itemize}
-
-
-%\subsection{Chaos of the chaotic iterations on $\mathds{R}$}
-%\label{chpt:Chaos des itérations chaotiques sur R}
-
-
-
-%\subsubsection{Chaos according to Devaney}
-
-%We have recalled previously that the chaotic iterations $\left(\Go,
-%\mathcal{X}_d\right)$ are chaotic according to the formulation of Devaney. We
-%can deduce that they are chaotic on $\mathds{R}$ too, when considering the order
-%topology, because:
-%\begin{itemize}
-%\item $\left(\Go, \mathcal{X}_d\right)$ and $\left(g, \big[ 0, 2^{10}
-%\big[_D\right)$ are semiconjugate by $\varphi$,
-%\item Then $\left(g, \big[ 0, 2^{10} \big[_D\right)$ is a system chaotic
-%according to Devaney, because the semiconjugacy preserve this character.
-%\item But the topology generated by $D$ is finer than the topology generated by
-%the Euclidian distance $\Delta$ -- which is the order topology.
-%\item According to Theorem \ref{Th:chaos et finesse}, we can deduce that the
-%chaotic iterations $g$ are indeed chaotic, as defined by Devaney, for the order
-%topology on $\mathds{R}$.
-%\end{itemize}
-
-%This result can be formulated as follows.
