# Strong non-wandering points and the periodic points

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