Is _the_ wandering set _a_ wandering set?

Question

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?

Answer 1

The question was somehow ill-posed. Clearly the continuous map $T \colon x \in [0,1] \to x/2$ has for wandering set $(0,1]$, however for any integer $n$, $$ T^n((0,1]) \cap (0,1] = (0, 2^{-n}] \neq \emptyset. $$

I had aimed something else, which is resolved in this post.

Answer 2

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.