X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/prng_gpu.git/blobdiff_plain/5ed99022e098863388088d7f23852378a56cdb5b..716f5cce1f0b4cef3c6dfe87f218f6ad04d0ab97:/prng_gpu.tex

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@@ -66,6 +66,48 @@ $\mathcal{X}_\tau$ will denote the topological space $\left(\mathcal{X},\tau\rig
 
 
 
+\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}
+
+
+
+
+
+
+
 \section{Chaos on the order topology}
 
 \subsection{The phase space is an interval of the real line}