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The upper Banach density of a subset $A\subset\mathbb N$ is defined $$d^*(A)=\limsup_{M-N\to\infty}\frac{|A\cap[M,M+1,\cdots,N]|}{M-N}$$ also the the upper density and lower density of a set $A$ is defined as following respectively $$\bar{d}(A)=\limsup_{n\to\infty}\frac{|A\cap[1,\cdots,n]|}{n}$$ $$\underline{d}(A)=\liminf_{n\to\infty}\frac{|A\cap[1,\cdots,n]|}{n}.$$ It is known that if $d^\ast(A)\gt 0$, then $A-A$ is $\Delta^*$. It means that $A-A$ has non-empty intersection with difference set of any sequence.

Now I want to know that Is there any subset $A\subset\mathbb N$ such that $\bar{d}(A-A)=1$?

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  • $\begingroup$ Sorry, but this question is not appropriate to MO. What about $A = \mathbb N \setminus S$ with $S$ being any finite subset of $\mathbb N$? $\endgroup$
    – Salvo Tringali
    Commented Sep 29, 2012 at 20:11
  • $\begingroup$ What is the assumption on $A$? As stated the question is trivial. Do you want to know if $\overline{d}(A-A)=1$ is possible even if $d^*(A)=0$? (the answer is yes, take $A$ as the set of natural numbers with coefficients $0$ and $1$ in their base $3$ expansion. Then $d^*(A)=0$ but $A-A=\mathbb{Z}$). There's also a typo in the definition of $\underline{d}(A)$. $\endgroup$
    – Pablo Shmerkin
    Commented Sep 29, 2012 at 20:12
  • $\begingroup$ Thanks alot for your answer and I'm really sorry for my typos. But I want both $A$ and $A-A$ as a subset of natural numbers.I need a subset of $\mathbb N$ with zero Banach density such that $A-A=\{a_i-a_j: a_i>a_j\}\subset\mathbb N$ has full upper density. $\endgroup$
    – maliheh
    Commented Oct 10, 2012 at 7:02
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    $\begingroup$ I think it is better that I ask in other way. I need a subset of natural numbers $P$ with $\bar{d}(P)=1$, but there is no $A\subset\mathbb N$ with positive Banach density such that $A-A\subset P$? In $\mathbb N$ or $\mathbb Z$ we can find other sequence by positive density where their difference set is in $\mathbb N$. $\endgroup$
    – maliheh
    Commented Oct 10, 2012 at 7:07

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Answering to the question formulated in the comments ("Is there a set $P$ with upper density $1$ such that $A-A$ is not contained in $P$ for any set $A$ with positive upper Banach density?"), note that for any $A$ with $\bar d(A)>0$, if $r>1/\bar d(A)$ then for any integers $a_1,...,a_r$ the total density of the sets $A-a_1,A-a_2,...,A-a_r$ is greater than $1$. Therefore $(A-a_i)\cap(A-a_j)$ is non-empty for some pair $(i,j)$. In other words, $A-A$ intersects the set of differences $\Delta(a_1,...,a_r)=\{a_i-a_j:1\leq i<j\leq r\}$.

Now construct $P$ as to avoid difference sets of arbitrarily large sets, but still have upper density $1$, for instance: $$P:=\mathbb{N}\setminus\left[\bigcup_{r=1}^\infty10^r\Delta(1,2,...,r)\right]$$ (here I used the usual notation, for $n\in\mathbb{N}$ and $X\subset\mathbb{N}$ denote by $nX:=\left\{nx:x\in X\right\}$). Thus $P$ has upper density $1$ (and actually Banach lower density $1$), but for any $A$ with positive Banach upper density $A-A\not\subset P$.

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