From 39211acd090848fea982c58fd56a95396913db0b Mon Sep 17 00:00:00 2001 From: Christophe Guyeux Date: Thu, 8 Sep 2011 16:19:10 +0200 Subject: [PATCH 1/1] =?utf8?q?Premi=C3=A8re=20sous-section=20ok?= MIME-Version: 1.0 Content-Type: text/plain; charset=utf8 Content-Transfer-Encoding: 8bit --- prng_gpu.tex | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/prng_gpu.tex b/prng_gpu.tex index 792d8ce..cd8bad3 100644 --- a/prng_gpu.tex +++ b/prng_gpu.tex @@ -66,6 +66,28 @@ $\mathcal{X}_\tau$ 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} + + + + \section{Chaos on the order topology} \subsection{The phase space is an interval of the real line} -- 2.39.5