11
$\begingroup$

For $S\subset \mathbb{N}$ define the upper density as $D^{\ast }(S)=\limsup_{n\rightarrow \infty }\frac{\left\vert S\cap \{1,2,\ldots,n\} \right\vert }{% \left\vert n\right\vert }.$

Question: Suppose $D^{\ast }(S)>0$. Is there $n\in\mathbb{N}$ such that $(S-n )\cap S$ has positive upper density?

$S-n=\{s-n| s\in S\}$

$\endgroup$
12
$\begingroup$

Suppose that $(S-n)\cap S$ has upper density 0 for all $n\leq N$. Then the density of the set of integers $x$, such that $|[xN, (x+1)N]\cap S|\geq 2$ is also 0, hence the upper density of $S$ is $\leq \frac{1}{N}$. Taking $N\rightarrow\infty$ gives a positive answer to your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.