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

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.

• $\Omega$ is an invariant set. Dec 2 '20 at 12:45
• I don't know what you mean! It's an exercise in differential dynamical systems book! $\omega$ in an invariant set and what I've written is compeletly correct! Dec 2 '20 at 16:56
• I already proved that $f(\Omega) \subset \Omega$. I'll add it Dec 3 '20 at 4:38
• @TSF $\Omega$ is an invariant set for trivial reasons, it even makes sense right now! Dec 3 '20 at 12:08

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 $$\Omega(f_{| \Omega}) = \{ (0,1) \}$$

[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.

• Thank you, could you please explain how to compute $\Omega(f)$ above ? I want to know how you computed this. Dec 3 '20 at 15:54
• Could you give me an example in discrete dynamical system?? Dec 3 '20 at 16:55
• $1$, $i$, $-i$ are fixed points so in $\Omega(f)$, for $0$ any open set $U$ containing $0$ containt an interval of the form $]0; \epsilon[$ which intercept itself after iterationf of \$ff. Dec 3 '20 at 22:10
• I edited my answer for discret case. Dec 9 '20 at 9:59
• I'm working on a new one. For now do you think I should delete this one or just edit it with a warning? Dec 9 '20 at 10:07