# An example which shows that $F(f) \subset \overline{Per(f)} \subset L(f) \subset \Omega(f)$ is restrict?

I have some questions on Dynamical Systems.

$$(X,f)$$ is a Dynamical System if $$f:X \to X$$ is a homeomorphism and $$X$$ is a compact space. Let's define \begin{align} &Per(f):=\{ x \in X ; f^n(x)=x ,\text{ for some } n \in \mathbb{Z} \}\\ &L(f):=L_{+}(f) \bigcup L_{-}(f) \end{align} where $$L_{+}(f)=\overline{\bigcup_{x\in X}\omega(x)}\quad\text{ and }\quad L_{-}(f)=\overline{\bigcup_{x\in X}\alpha(x)},$$ where

• $$\omega(x):=\{y : \exists (n_i), f^{n_i}(x) \to y \}$$ and
• $$\alpha(x):=\{y : \exists (n_i), f^{ -n_i}(x) \to y \}$$

and finally where $$\Omega(f)$$ is the set of non wandering points, i.e. all $$x$$ such that $$\forall$$ U open containing $$x$$ and $$\forall$$ $$N>0$$ there exists some $$n>N$$ such that $$f^n(U) \cap U \ne \emptyset$$ and $$F(f) := \{x \in X : f(x)=x\}$$

1 - I was looking for an example which shows that $$Per(f)$$ could be closed. I think I found a simple example which shows that : \begin{align} &f : [0,1] \to [0,1]\\ &f(x)=x \quad , \forall x\in [0,1] \end{align} In this case the set $$Per(f)$$ is closed because we know the set of all fixed points is closed. am I right? Now I'm looking for an example which shows that $$Per(f)$$ is not closed in general, I think If we find an example whose periodic points set is empty the question will be solved.

2- I already proved this; \begin{align} F(f) \subseteq \overline{Per(f)} \subseteq L(f) \subseteq \Omega(f) \end{align} Now I'm looking for an example which shows that the relation may be restrict what I mean is that; \begin{align} F(f) \subset \overline{Per(f)} \subset L(f) \subset \Omega(f) \end{align}

3- The third question is also an example just like question $$1,2$$ I want an example which shows that; \begin{align} \overline{Per(f)} \neq \Omega(f) \end{align} For this we have to find an example which shows that; \begin{align} \Omega(f) \nsubseteq \overline{Per(f)} \end{align}

I consider discrete dynamical systems. Can any one help?

• The empty set is closed, so your strategy for #1 isn't the right approach
Commented Nov 16, 2020 at 22:52
• Do you have an example ? Commented Nov 17, 2020 at 4:29
• Sorry I forgot to write it I mean of $F(f)$ is the set of fixed point of the system $f$. Commented Nov 17, 2020 at 5:07

In this case the set $$Per(f)$$ is closed because we know the set of all fixed points is closed.

Yes (when $$X$$ is Hausdorff).

Examples such that:

1)) $$F(f) \subset \overline{Per(f)}$$. Let $$X$$ be a set $$\{1,2,3\}$$ endowed with the discrete topology and $$f(1)=2$$, $$f(2)=3$$, and $$f(3)=1$$. Then $$F(f)=\varnothing$$ but $$Per(f)=X$$.

2)) $$\Omega(f) \nsubseteq\overline{Per(f)} \subset L(f)$$. Let $$X$$ be the unit circle $$\{z\in\Bbb C: |z|=1\}$$ endowed with the topology inherited from $$\Bbb C$$, $$\varphi$$ be a real number such that $$\varphi/\pi$$ is irrational and $$f:X\to X$$, $$x\mapsto xe^{\varphi i}$$ be a rotation at the angle $$\varphi$$. It is easy to check that $$Per(f)=\varnothing$$ and $$L(f)=\Omega(f)=X$$.

3)) $$L(f) \subset \Omega(f)$$. I assume that $$(n_i)$$ in the definitions of $$\alpha(x)$$ and $$\omega(x)$$ is a strictly increasing sequence of natural numbers.

Let $$X={\Bbb T}^{\Bbb T}$$. By Tychonov Theorem, $$X$$ is a compact topological group with coordinate-wise multiplication. Let $$g=(g_z)_{z\in\Bbb T}\in X$$ be an element such that $$g_z=z$$ for each $$z\in\Bbb T$$. For each $$x\in X$$ put $$f(x)=gx$$.

We claim that $$\Omega(f)=X$$. Indeed, let $$x$$ be any element of $$X$$, $$U$$ be any neighborhood of $$x$$, and $$N$$ be any natural number. Since $$X$$ is a topological group, there exists a neighborhood $$V$$ of the identity of $$G$$ such that $$xVV{-1}\subset U$$. A family $$\{yV:y\in X\}$$ is an open cover of a compact space $$X$$, so there exists a finite subset $$F$$ of $$X$$ such that $$FV=X$$. Therefore there exist an element $$y\in F$$ and natural numbers $$n,m$$ such that $$g^n,g^m\in yV$$ and $$n>m+N$$. Then $$g^m\in g^nVV^{-1}$$ and so $$x\in x g^{n-m}VV^{-1}\subseteq g^{n-m}U$$.

We claim that sets $$\alpha(x)$$ and $$\omega(x)$$ are empty for each $$x=(x_z)_{z\in\Bbb T}\in X$$. Since the group $$G$$ is Abelian, $$\omega(x)=\alpha(x^{-1})^{-1}$$, it suffices to show that the set $$\alpha(x)$$ is empty.

Suppose to the contrary that there exists an increasing sequence $$\{n_i\}$$ of natural numbers such that a sequence $$\{g^{n_i}x\}$$ converges to a point $$y=(y_z)_{z\in\Bbb T}\in X$$. Let $$U_0=\{z\in\Bbb T: |z-1|\le 1/\sqrt{2} \}$$ be a neighborhood of the identity of the group $$\Bbb T$$. For each natural number $$i$$ put $$T_i=\{z\in\Bbb T: (n_j-n_k)z\in U_0^2\mbox{ for each }j,k>i\}.$$ Since the set $$U_0^2$$ is closed in $$\Bbb T$$, the continuity of power on the group $$\Bbb T$$ implies that the set $$T_i$$ is closed for each natural number $$i$$. We claim that $$\Bbb T=\bigcup_{i\in\Bbb N} T_i$$. Indeed, let $$z\in \Bbb T$$ be any element. There exists a natural number $$i$$ such that $$n_jzx_z\in y_zU_0$$ for each $$j>i_z$$. Then $$(n_i-n_k)z\in U_0^2$$ for each $$j,k>i$$, so $$z\in T_i$$.

Since $$\Bbb T=\bigcup_{i\in\Bbb N} T_i$$ and $$\Bbb T$$ is a compact metrizable space, Baire theorem implies that there exists a natural number $$i$$ such that a set $$T_i$$ has non-empty interior. Therefore there exists an open arc $$U\subseteq T_i$$ of the circle $$\Bbb T$$. Let $$\ell$$ be the length of $$U$$. Since the sequence $$\{n_i\}$$ is increasing, there exists a number $$j>i$$ such that $$n_j-n_{i+1}>2\pi/\ell$$. But then $$U_0^2\supseteq (n_j-n_{i+1})T_i\supseteq (n_j-n_{i+1}) U=\Bbb T$$, a contradiction.

4)) $$Per(f)$$ is not closed. Let $$X$$ be the unit disk $$\{z\in\Bbb C: |z|\le 1\}$$ and $$f:X\to X$$, $$x\mapsto xe^{|x|i}$$. Then $$Per(f)=\{x\in X: |x|/\pi\in\Bbb Q\}.$$

• Thanks, I think the last inclusion is remained to answer Commented Nov 17, 2020 at 10:01
• You first example is also saying that $per(f)$ may be open. isn't it ? Commented Nov 17, 2020 at 11:22
• @Reza Yes, if $Per(f)=X$ then it is open. Commented Nov 17, 2020 at 11:28
• in your last example why $Per(f)=\{x\in X : \dfrac{|x|}{\pi} \in \mathbb{Q}\}$ and why is it open ? Commented Nov 17, 2020 at 12:12
• @Reza In the last example $Per(f)\ne X$ and it is non-open. Commented Nov 17, 2020 at 12:19