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.

[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.
