Say we have a dynamical system $T : X \to X$ ($T$ continuous injective on $X$ topological). The wandering set is the set of wandering points. We say $x$ is wandering if there exits an open neighborhood $U$ of $x$ and a time $n \in \mathbb{N}^*$ such that,
$$
T^n(U) \cap U = \emptyset.
$$
Furthermore, a set $W$ is said to be wandering if there exists a time $n \in \mathbb{N}^*$ such that,
$$
T^n(W) \cap W = \emptyset.
$$
My question is, is the wandering set wandering?
Edit: If $X$ is not compact, it is not true in general as pointed Captain Lama, with the concise counter-example $T : x \in \mathbb{R} \to 2x$, where the wandering set is $\mathbb{R}^*$, clearly not wandering as $T^n(\mathbb{R}^*) \cap \mathbb{R}^* = \mathbb{R}^* \neq \emptyset$, for all integer $n$.
Adding the hypothesis $X$ is compact, does the result now hold?
Best Answer
It is not in general: take $T: x\mapsto 2x$ for $X=\mathbb{R}$. Then every point is wandering except for $0$, but $T^n(\mathbb{R}^*)=\mathbb{R}^*$ for all $n\in \mathbb{N}$.
The point is that you have a $n$ that works locally for each wandering point, but you can't always find a global $n$ that works for all such points. Unless your space is compact, or some other similar hypothesis.