A counter example which shows that $\Omega(f|_{\Omega}) \neq \Omega(f)$

Question

I'm looking for an example :

A counter example which shows that $\Omega(f|_{\Omega}) \neq \Omega(f)$

$(X,f)$ is a Dynamical System if $f:X \to X$ is a homeomorphism and $X$ is a compact space. $\Omega(f)$ 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 $f^n(U) \cap U \ne \emptyset$.

I think we have to take $f$ on a circle but I don't know how to compute the non wandering set.

Answer 1

My previous answer have an error. I will try differrently.

We will work on polar coordinate.

Our space will be the union of the sets $C_{n} = \{(\frac{n}{n+1},\frac{\pi}{k}) , k \in \{-n,...,-1,1,...,n \}\}$ for $n \in \mathbb{N}^* $ with $C_{\infty} = \{(1,\frac{\pi}{k}) k \in \mathbb{Z}^* \} \cup \{(1,0)\}$ and with $D=\{(n,0) ,n \geq 2 \}$ with the topology induced by $\mathbb{C}$.

Here a picture of the situation, the blue points are in the sets $C_k$ and the red one in $C_\infty$

We will consider the function $f$ defined by :

On $D$

• $f(n,0)=(n-1,0)$ if $n \ne 2$
• $f(2,0)=(1/2,\pi) \in C_{1}$

On $C_k$

• $f(\frac{n}{n+1},\frac{\pi}{k})=(\frac{n}{n+1},\frac{\pi}{k+1})$ if $k \ne n$ and $k \ne -1$
• $f(\frac{n}{n+1},\pi)=(\frac{n}{n+1},\frac{\pi}{2})$ if $k=-1$
• $f(\frac{n}{n+1},\frac{\pi}{n})=(\frac{n+1}{n+2},\frac{-\pi}{n}) \in C_{k+1}$ if $k =n $

And on $C_{\infty}$

• $f(1,\frac{\pi}{k})=(1,\frac{\pi}{k+1})$ if $k \ne -1$
• $f(1,\pi)=(1,\frac{\pi}{2})$ if $k =-1$
• $f(1,0) = (1,0)$

We have that $f$ is bijective. $\underset{k}{\cup} C_k \cup D$ and $C_{\infty}$ are dynamically speaking two bi-infinite shift and $(1,0)$ is a fixed point.

Let's show that $f$ is a homeomorphism.

$f$ is continuous and with continuous inverse at every point of $\underset{k}{\cup} C_k \cup D$ because the topology there is discrete.

Now on $C_{\infty}$, take a sequence $(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) \underset{i \to \infty}{\to} (1,\frac{\pi}{k})$. We should have that $n_i \underset{i \to \infty}{\to} \infty $ and $k_i \underset{i \to \infty}{\to} k $ so for every $i>I$ for $I$ large enough, $k_i \ne n_i $ and $f(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) = (\frac{n_i}{n_i +1},\frac{\pi}{k_i +1}) \underset{i \to \infty}{\to} (1,\frac{1}{k +1})= f(1,\frac{1}{k})$. We avoid the difficulties $k =-1 $ because $-\pi = \pi$ on polar coordinate.

If $(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) \underset{i \to \infty}{\to} (1,0)$ then we could have for infinitely many i, $k_i = n_i$ but then $f(\frac{n_i}{n_i+1},\frac{\pi}{k_i}) = (\frac{n_i+1}{n_i + 2},\frac{-\pi}{n_i +1 }) \underset{i \to \infty}{\to} (1,0)= f(1,0)$

So $f$ is continuous.

The same can be done for $f^{-1}$ it "just" rotate conterclockwise and sometimes can "gain" a level in $C_k$.

Now we have that $\underset{k}{\cup} C_k \cup D$ is not in $\Omega(f)$ since the topology there is discret, you can just take a singleton wich won't intersept itself after iteration of $f$.

$C_{\infty}$ is in $\Omega$, indeed for every $(1,\frac{\pi}{k})$, every open set which contain it, should contain $ \{ (\frac{n}{n+1},\frac{\pi}{k}) , n \geq N\}$ for a $N$ large enough. This subset intersect itself many times.

So $\Omega(f)=C_{\infty}$, but on $C_{\infty}$ $f$ is just a shift and a fixed point so $\Omega(f_{| \Omega}) = \{ (0,1) \}$

Answer 2

[DISCLAIMER] This answer is wrong, I keep it to show an exemple for a non continuous function. See the second answer below for a good one

---

Take $X= \ [0;1] \cup \{i ,-i\} \cup \{ i(1-\frac{1}{n}),n \in \mathbb{N}\} \cup \{ -i(1-\frac{1}{n}),n \in \mathbb{N}\}$ with the induce topology of $\mathbb{C}$. This is a compact set made of the segment $[0,1]$ and two sequences of point accumulating at $i$ and $-i$

Now take $f:x \mapsto x^2$ if $x \in ]0;1]$

$f(i)=i$

$f(-i)=-i$

$f(0)=\frac{i}{2}$

$f(i(1-\frac{1}{n}))=i(1-\frac{1}{n+1})$

$f(-i(1-\frac{1}{n}))=-i(1-\frac{1}{n-1})$, if $n \geq 2$

This function is indeed continuous and $\Omega(f)=\{1,0,i,-i\}$ but $\Omega(f_{| \Omega})= \{1,i,-i \}$

---

You can see The nonwandering set for an example of a counter-example on a connected space but with only a continuous map, not a homeomorphism.

The question is there a counter-example on connected space with an homeo is a good one. I will think about it and keep you informed.

---

I add an edit to discuss the discret dynamic case as ask in commentary.

In a discret space every singleton in open. So if $x \in \Omega(f)$, taking $U = \{ x \}$ we have that there exist $n$ such that $f^n(x)=x$. $x$ is a periodic point !

Of course periodic point for $f$ are also periodic for $f_{| \Omega}$ and therefor $\Omega(f) = \Omega(f_{| \Omega})$ in the discret case.