2
$\begingroup$

I was looking for something about nonwandering sets and i saw the following definition here: A question about non-wandering points :

Let $f:X\to X$ be a homeomorphism of a compact metric space $(X, d)$. A point $x\in X$ is called a nonwandering point, $x\in\Omega(f)$, if for every open set $x\in U$ there is $n\in\mathbb{N}$ such that $f^n(U)\cap U\neq \emptyset$. A point $x\in X$ is strong non-wandering, $x\in\Omega_s(f)$, if for every open set $x\in U$, there is $n\in\mathbb{N}$ such that $f^{nk}(U)\cap U\neq\emptyset$ for all $k\in\mathbb{N}$.

Also:

$\Omega_s(f)$ is a closed set. Let $x_n\to x$ and $x_n\in\Omega_s(f)$ for all $n$. Let $U$ be an pen set of $x$, hence there is $x_n\in U$, this implies that there is $m\in\mathbb{N}$ such that $f^{mk}(U)\cap U\neq \emptyset$ for all $k\in\mathbb{N}$, i.e. $x\in\Omega_s(f)$.

It is easy to see that $cl\big( Per(f) \big)$ the closer of priodic points is contained in $\Omega_s(f)$. Since if $\{x_n\}_{n\in \mathbb{N}}$ is a sequence of priodic points with limit $x$, then for every open set $x\in U$ there must be $m\ge 1 $ for which for every $n \ge m$, we can conclude $x_n \in U$. For some $ p \ge m$ let $x_p \in U$ be arbitrary. Then if for some $q$ it is happen that $f^q(x_p)=x_p$, then for every $k\in \mathbb{N}$ we have $f^{kq}(U)\cap U\neq \emptyset$. Now my question is :

Is it true that $cl\big( Per(f) \big) = \Omega_s(f)$ ? Is there a dynamical system on some compact space with $cl\big( Per(f) \big) \neq \Omega_s(f)$?

$\endgroup$
1
$\begingroup$

Is false see this paper, maybe help, in general diffeomorphisms that satisfy Axiom A have the property of equality between these sets, for example, Smale's horseshoe.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.