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 $cl\big( Per(f) \big) \neq \Omega_s(f)$?
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.