# On the non-wandering set of a map

Let $$f: X \to X$$ be a homeomorphism from a compact metric space $$X$$ to itself.

$$x$$ is said to be a non-wandering point of $$f$$ if for all open neighborhood of $$x$$ say $$U$$ there exists $$n \in \mathbb{Z}$$ such that $$f^n(U) \cap U \neq \emptyset$$. The set of all non-wandering points of $$f$$ is denoted by $$\Omega(f)$$. I'm trying to prove the following exercise :

Show that if $$x \in \Omega(f)$$ and $$U$$ is a neighborhood of $$x$$, there is a sequence of integers $$n_i$$ tending to infinity such that the intersection $$f^{n_i}(U) \cap U$$ is non-empty.

There is a proposition in the book which says if $$U$$ is a neighborhood of $$\Omega(f)$$ then for every $$x \in X$$ there is $$N>0$$ such that for every $$n \geq N$$ , $$f^n(x) \in U$$.

May I use the proposition above to prove the main problem?