From: Christophe Guyeux <christophe.guyeux@univ-fcomte.fr>
Date: Thu, 8 Sep 2011 14:19:10 +0000 (+0200)
Subject: Première sous-section ok
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/prng_gpu.git/commitdiff_plain/39211acd090848fea982c58fd56a95396913db0b?hp=5ed99022e098863388088d7f23852378a56cdb5b

Première sous-section ok
---

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}