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 $x \in W$, $d(T^n(x),M) \to_n 0$.
Best Answer
There is a nice proof in Birkhoff's Dynamical Systems (1927), page 192. With more modern notation it gives the following.
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 $M$.