Is A set with upper density 1 a set of recurrence?

Question

I'm trying to show that If $E\subset \mathbb{N}$, satisfies $\bar{d}(E)$ then E is a set of recurrence.

Here is some definitions :

$\bar{d}(E)=\limsup \frac{|E \cap {1,2,...,n}|}{n}$

E is a set of recurrence if for any measure preserving system $(X,B,\mu,T)$ and $A\in B $ with $\mu(A)>0$, $\exists n \in E $ such that $\mu(A\cap T^{-n}A)>0$

My Attempt : I already proved that $E-E=({n: x,x+n \in E})$ is a set of recurrence if $E$ is an infinite set. So I'm trying to connect this fact to the problem. It suffices to show that any set of density 1 contains a difference set( $i.e. S-S$ ) for some infinite subset S of $\mathbb{N}$.

How can I show this? Any hint would be appreciated.

Answer 1

Note that E contains a set of consecutive integers with arbitrarility long length.

Let $a_1 \in E$, then $\exists a_2$ such that $\{a_2,a_2+1,\cdots,a_2+a_1\}\subset E$. Continue this, we have $\{a_k,\cdots,a_k+a_{k-1}+\dots+a_1\}\subset E$. Pick $S=\{\sum_{i=1}^la_i:l\in \mathbb{N}\}$, then $S-S\subset E$ for an infinite set $S$.