From: Christophe Guyeux <christophe.guyeux@univ-fcomte.fr>
Date: Thu, 8 Sep 2011 14:37:41 +0000 (+0200)
Subject: Fin de la section de chaos relatif
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/prng_gpu.git/commitdiff_plain/12410996f49b94978103d0ba32d17f85c89fa51a

Fin de la section de chaos relatif
---

diff --git a/prng_gpu.tex b/prng_gpu.tex
index 6ebbcd9..082c379 100644
--- a/prng_gpu.tex
+++ b/prng_gpu.tex
@@ -105,6 +105,25 @@ Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V = \
 
 
 
+\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}
+
+
+