# Convergence to the non-wandering set (for a compact dynamical system)

Let $$X$$ be a compact metric space and let $$T\colon X \to X$$ be continuous and injective. A point $$x$$ is said to be wandering if there exists an open neighborhood $$V \ni x$$ and a time $$N \in \mathbb{N}^*$$ such that, for all $$n \geq N$$, $$T^n(V) \cap V = \emptyset.$$ A point is said to be non-wandering, well, if it is not wandering. Denote by $$W$$ the set of wandering points and $$M$$ its complement. As a matter of fact, $$W$$ is open and positively invariant ($$T(W) \subset W$$), while $$M$$ is closed (thus compact) and invariant ($$T(M) = M$$).

The question is whether or not $$\bigcap_{n \in \mathbb{N}} T^n(W) = \emptyset$$, or in other words is it true that for any $$x \in W$$, $$d(T^n(x),M) \to_n 0$$.

• With your reformulation, it's looks like you'll want to investigate the $\omega$-limit set and the set of recurrent points. Jul 25 '19 at 12:47

Let $$\epsilon > 0$$ and define the following compact set, $$K = \{ x \in X, ~d(x,M) \geq \epsilon\} \subset W.$$ For all $$x \in K$$, there exists an open neighbourhood $$V_x$$ of $$x$$ and $$n_x > 0$$ such that for all $$n \geq n_x$$, $$\varphi^n(V_x) \cap V_x = \emptyset.$$ By compactness of $$K$$, we can extract a finite cover of $$K$$ from $$(V_x)_{x \in K}$$, say given by $$E = \{x_1, \dots, x_m \}$$. We note $$N = \max_{x \in E} n_x$$. If $$x \in K$$, then $$x$$ belongs to some $$V_y$$ with $$y \in E$$. However $$x$$ can only stay for at most $$N$$ times, and never comes back. It might reach another $$V_{y'}$$ with $$y' \in E$$ different than $$y$$, and again can only stay at most $$N$$ times and never visit again $$V_{y'}$$. Eventually, all $$V_z$$ with $$z \in E$$ are exhausted, so that $$x$$ reaches and remains in $$X \setminus K$$.
If $$x \in X \setminus K$$, then either it remains in $$X \setminus K$$, or it leaves and the previous reasoning brings $$x$$ in $$X \setminus K$$ forever. As a result, any initial condition tends toward $$M$$.