If $W$ is a Poincaré sequence then $W \cap m\mathbb{Z}$ is also a Poincaré sequence.

Question

I'm studying ergodic theory and dynamical systems, and I've encounterd the following problem:

Definition: A set $W \subset \mathbb{Z}$ is called a Poincaré sequence if for any m.p.s $(X,\mathcal{B},\mu,T)$ and $A \in \mathcal{B}$ with $\mu(A)>0$, with $\mu(T^{-n}A \cap A)>0$ for some $n \in W, n\neq 0$.

Problem: Show that $W \cap m\mathbb{Z}$ is also a Poincaré sequence for any $m \in \mathbb{Z}, m \neq 0$.

The question has a solution in the following book "Recurrence in Ergodic Theory and Combinatorial Number Theory" by H. Furstenberg and I'm having problems following the proof. It says that:

We merely observe $(X,\mathcal{B},\mu,T^{m})$ is a m.p.s whenever $(X,\mathcal{B},\mu,T)$ is a m.p.s. In particular $W \cap m \mathbb{Z}$ must contain nonzero elements if $W$ is a Poincaré sequence.

My doubt is in the assertion "$W \cap m \mathbb{Z}$ must contain nonzero elements" I don't see how one can conclude that.

As far as I understand since $(X,\mathcal{B},\mu,T^{m})$ is a m.p.s and $W$ is a Poincaré sequence we have that, for a given set $A \in \mathcal{B}$ with $\mu(A) >0$ there is a $n \in W$ such that $\mu(T^{-mn}A \cap A) >0$ and so the sequence $mW := \{m n \in \mathbb{Z} \mid n \in W\}$ is the set that we can conclude is a Poincaré sequence, and I don't see how we can conclude that $nm \in W \cap m\mathbb{Z}$ or that $W \cap m \mathbb{Z}$ has nonzero elements.

So I don't know if I'm not understanding the meaning of $W \cap m \mathbb{Z}$, that I think of $W \cap m \mathbb{Z}:=\{p \in \mathbb{Z} \mid p \in W \land p \in m\mathbb{Z}\}$ or I'm not seeing how to end the argument.

Any help would be gladly appreciated.

Edit

I've just manage to show that $W \cap m \mathbb{Z}$ has a nonzero element. But I'm still not seeing clearly how to conclude that the set is a Poincaré sequence.

Answer 1

Here is a proof that doesn't exactly follow the outline indicated in the question.

Let $W$ be a Poincaré sequence, $(X,\mathcal B,\mu,T)$ an m.p.s., and let $A\subset X$ have $\mu(A)>0$. We will find $n\in W\cap m\mathbb Z$ such that $\mu(T^{-n}A\cap A)>0$.

Let $(Y,\mathcal D,\nu,S)$ be the rotation on $m$ points: $Y= \{0,\dots,m-1\}$, $Sy = y+1$ mod $m$, with $\nu=$ normalized counting measure. Let $B = \{0\} \subset Y$, so that $S^{-n}B\cap B\neq \varnothing$ iff $n\in m\mathbb Z$.

The product $(X\times Y, \mathcal B\otimes \mathcal D, \mu\times \nu, T\times S)$ is an m.p.s., and $\mu\times \nu (A\times B) = \mu(A)\nu(B)>0$. Since $W$ is a Poincaré sequence, there is an $n\in W$ such that $\mu((T\times S)^{-n}(A\times B) \cap (A\times B))> 0$. The measure on the left can be rewritten as $\mu(T^{-n}A\cap A)\nu(S^{-n}B\cap B)$, so we have $\mu(T^{-n}A\cap A)>0$, and $n\in m\mathbb Z$, since $\nu(S^{-n}B\cap B)>0$.