Unique attracting set is the whole space?

Question

I'm looking for a simple example in which $\Omega(f) \neq X$ but the unique attracting set is $X$.

$(X,f)$ 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 $f^n(U) \cap U \ne \emptyset$.

An open set $U \subseteq X$ is called a trapping region of $f$ if $f(\overline{U}) \subseteq U$. A compact invariant set $A \subseteq X$ is said to be attracting of $f$ if it has a neighborhood $U$ which is a trapping region such that $A = \bigcap_{n \geq 0}f^{n}(\overline{U}) $.

Answer 1

I think you can take $x \mapsto x^2$ on the torus $\mathbb{R}/\mathbb{Z}$.

The only point in $\Omega(f)$ is $0$. Indeed, every fixed point is in the non-wandering set, and $0$ is a fixed point. For every over point, $x \in ]0;1[$ you can take a small intervall $I=]x- \epsilon; x+ \epsilon[$ aroud it, such that $0 \not\in I$. Then for $N$ big enough, $(x+\epsilon)^N \leq x-\epsilon$ so $N \geq \frac{\ln(x-\epsilon)}{\ln(x+\epsilon)}$ and for all $n \geq N$, $f^n(I) \subset[0;x-\epsilon[$ which has empty intersection with $I$. In conclusion $\Omega(f)=x$

Now if $U$ is atrapping region, then we have two case

• $0 \in U$, then there is a $\epsilon$ such that $]1- \epsilon;1] \subset U$, taking image with $f$ you find that $U = X$
• $0 \not\in U$ in this case you can do the same argument as above and this case is not

So the only trapping region is $X$ and then the only attractor is $X$.