-\section{The relativity of disorder}
-\label{sec:de la relativité du désordre}
-
-In the next two sections, we investigate the impact of the choices that have
-lead to the definitions of measures in Sections \ref{sec:chaotic iterations} and \ref{deuxième def}.
-
-\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}
