We have the following definitions:
Definition 1. A set $P \subset \mathbb{N}$ is said to be a Poincaré sequence if for every finite measure-preserving system $(X, \mathcal{S}, \mu, T)$ and any set $A \in \mathcal{S}$ of positive measure there exists $n \in P, n \neq 0$, such that $\mu(T^{-n}(A) \cap A) > 0$.
Definition 2. A set $Q \subset \mathbb{N}$ is said to be a thick set if it contains intervals of integers of arbitrary length.
Problem. Thick sets are Poincaré.
I have a topological proof to the problem, my question is about whether or not we can find a purely combinatorial proof along these lines:
- From the measure-preserving property that for any infinite set $P \subset \mathbb{N}$, the set of differences $P - P$ is a Poincaré sequence.
- If every thick set contains a difference set of an infinite set, then we can conclude from (1.) that thick sets are Poincaré.
Specifically, my question is the following:
Let $Q \subset \mathbb{N}$ be thick. Does there always exist an infinite set $P \subset \mathbb{N}$ such that $P-P \subset Q$?
Any hints, discussion, and solutions are appreciated.
Note. For $A, B \subset \mathbb{N}$, we define $A - B = \{ a - b : a \in A, b \in B, a - b > 0 \}$.
No. There are thick sets with no $P \subset \mathbb{N}$ of size $3$ with $P-P \subset Q$. First note that if $P-P \subset Q$ and $|P| \ge 3$, then $Q$ is not sum-free. Indeed, if $P \ni a,a+b,a+b+c$ with $b,c \ge 1$ (obviously any 3 different numbers can be written in this form), then $P-P \subset Q$ implies $b,c,b+c \in Q$.
So we just have to exhibit a thick sum-free set. But this is easy. For example, take $Q = \cup_{k \ge 2} I_k$, where $I_k := \{2^{2^k}+1,2^{2^k}+2,\dots,2^{2^k}+k\}$.