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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?

Cryme
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2 Answers2

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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.

Captain Lama
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  • I see, nice example. I indeed want to consider $X$ compact, but I was wondering if it was true in general as well. Do you have a hint for the compact case?

    I was thinking something like the union of two wandering sets is wandering (is that true?), then something in the flavor of an open cover of the wandering set (but it is open...) would propagate by compacity a global $n$, but I can't seem to figure it out.

     – Cryme Jul 25 '19 at 08:51
  • Well, I figure it's not true actually.

    Just consider the map $T : x \in [0,1] \to x/2$. Then the wandering set is $(0,1]$, while for any integer $n > 0$, $T^n((0,1]) \cap (0,1] = (0, 2^{-n}] \neq \emptyset$.

    I should like to add that I was looking for something else writing this question, basically I was wondering if the iteration of a wandering point would converge to the non-wandering set. I think I should rather create a new question, better framed.

     – Cryme Jul 25 '19 at 09:05
  • Yes, it's $W$ that should be compact for the proof to work, but I guess it's not a very useful condition. I just wanted to emphasize the local/global phenomenon. – Captain Lama Jul 25 '19 at 09:48
  • Well, $W$ is open in $X$ compact, so I wouldn't bet on $W$ compact. – Cryme Jul 25 '19 at 11:29
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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.

Cryme
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