+
+
+\section{The relativity of disorder}
+\label{sec:de la relativité du désordre}
+
+\subsection{Impact of the topology's finenesse}
+
+Let us firstly introduce the following notations.
+
+\begin{notation}
+$\mathcal{X}_\tau$ will denote the topological space
+$\left(\mathcal{X},\tau\right)$, whereas $\mathcal{V}_\tau (x)$ will be the set
+of all the neighborhoods of $x$ when considering the topology $\tau$ (or simply
+$\mathcal{V} (x)$, if there is no ambiguity).
+\end{notation}
+
+
+
+\begin{theorem}
+\label{Th:chaos et finesse}
+Let $\mathcal{X}$ a set and $\tau, \tau'$ two topologies on $\mathcal{X}$ s.t.
+$\tau'$ is finer than $\tau$. Let $f:\mathcal{X} \to \mathcal{X}$, continuous
+both for $\tau$ and $\tau'$.
+
+If $(\mathcal{X}_{\tau'},f)$ is chaotic according to Devaney, then
+$(\mathcal{X}_\tau,f)$ is chaotic too.
+\end{theorem}
+
+\begin{proof}
+Let us firstly establish the transitivity of $(\mathcal{X}_\tau,f)$.
+
+Let $\omega_1, \omega_2$ two open sets of $\tau$. Then $\omega_1, \omega_2 \in
+\tau'$, becaus $\tau'$ is finer than $\tau$. As $f$ is $\tau'-$transitive, we
+can deduce that $\exists n \in \mathds{N}, \omega_1 \cap f^{(n)}(\omega_2) =
+\varnothing$. Consequently, $f$ is $\tau-$transitive.
+
+Let us now consider the regularity of $(\mathcal{X}_\tau,f)$, \emph{i.e.}, for
+all $x \in \mathcal{X}$, and for all $\tau-$neighborhood $V$ of $x$, there is a
+periodic point for $f$ into $V$.
+
+Let $x \in \mathcal{X}$ and $V \in \mathcal{V}_\tau (x)$ a $\tau-$neighborhood
+of $x$. By definition, $\exists \omega \in \tau, x \in \omega \subset V$.
+
+But $\tau \subset \tau'$, so $\omega \in \tau'$, and then $V \in
+\mathcal{V}_{\tau'} (x)$. As $(\mathcal{X}_{\tau'},f)$ is regular, there is a
+periodic point for $f$ into $V$, and the regularity of $(\mathcal{X}_\tau,f)$ is
+proven.
+\end{proof}
+
+\subsection{A given system can always be claimed as chaotic}
+
+Let $f$ an iteration function on $\mathcal{X}$ having at least a fixed point.
+Then this function is chaotic (in a certain way):
+
+\begin{theorem}
+Let $\mathcal{X}$ a nonempty set and $f: \mathcal{X} \to \X$ a function having
+at least a fixed point.
+Then $f$ is $\tau_0-$chaotic, where $\tau_0$ is the trivial (indiscrete)
+topology on $\X$.
+\end{theorem}
+
+
+\begin{proof}
+$f$ is transitive when $\forall \omega, \omega' \in \tau_0 \setminus
+\{\varnothing\}, \exists n \in \mathds{N}, f^{(n)}(\omega) \cap \omega' \neq
+\varnothing$.
+As $\tau_0 = \left\{ \varnothing, \X \right\}$, this is equivalent to look for
+an integer $n$ s.t. $f^{(n)}\left( \X \right) \cap \X \neq \varnothing$. For
+instance, $n=0$ is appropriate.
+
+Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V =
+\mathcal{X}$, so $V$ has at least a fixed point for $f$. Consequently $f$ is
+regular, and the result is established.
+\end{proof}
+
+
+
+
+\subsection{A given system can always be claimed as non-chaotic}
+
+\begin{theorem}
+Let $\mathcal{X}$ be a set and $f: \mathcal{X} \to \X$.
+If $\X$ is infinite, then $\left( \X_{\tau_\infty}, f\right)$ is not chaotic
+(for the Devaney's formulation), where $\tau_\infty$ is the discrete topology.
+\end{theorem}
+
+\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}
+
+\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.
+
+\begin{theorem}
+\label{th:IC et topologie de l'ordre}
+The chaotic iterations $g$ on $\mathds{R}$ are chaotic according to the
+Devaney's formulation, when $\mathds{R}$ has his usual topology, which is the
+order topology.
+\end{theorem}
+
+Indeed this result is weaker than the theorem establishing the chaos for the
+finer topology $d$. However the Theorem \ref{th:IC et topologie de l'ordre}
+still remains important. Indeed, we have studied in our previous works a set
+different from the usual set of study ($\mathcal{X}$ instead of $\mathds{R}$),
+in order to be as close as possible from the computer: the properties of
+disorder proved theoretically will then be preserved when computing. However, we
+could wonder whether this change does not lead to a disorder of a lower quality.
+In other words, have we replaced a situation of a good disorder lost when
+computing, to another situation of a disorder preserved but of bad quality.
+Theorem \ref{th:IC et topologie de l'ordre} prove exactly the contrary.
+
+
+
+
+
+
+
+
+\section{Conclusion}
+\bibliographystyle{plain}
+\bibliography{mabase}
