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.

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