most recent 30 from http://mathoverflow.net 2013-06-20T00:08:14Z http://mathoverflow.net/feeds/question/108420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108420/density-of-a-difference-set-of-a-set-wih-zero-upper-banach-density maliheh 2012-09-29T19:20:06Z 2013-06-12T14:22:00Z

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

http://mathoverflow.net/questions/108420/density-of-a-difference-set-of-a-set-wih-zero-upper-banach-density/114664#114664 Joel Moreira 2012-11-27T15:41:54Z 2012-11-27T15:41:54Z

<p>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 <code>$\Delta(a_1,...,a_r)=\{a_i-a_j:1\leq i&lt;j\leq r\}$</code>.</p> <p>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 <code>$nX:=\left\{nx:x\in X\right\}$</code>). Thus $P$ has upper density $1$ (and actually Banach <em>lower</em> density $1$), but for any $A$ with positive Banach upper density $A-A\not\subset P$.</p>