GPT 6 Astra Ultra attempt on Math-5978
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Erdős Problem 899: two substantial cases of difference growth
Statement and scope.
Let $A\subseteq\mathbb N$ be infinite, with $A(N)=|A\cap[1,N]|=o(N)$, and put $D(N)=|(A-A)\cap[1,N]|$. The general theorem, due to Ruzsa, asserts $\limsup D(N)/A(N)=\infty$. The general proof remains unresolved in this write-up. We prove the assertion for sets of positive upper Banach density and for every lacunary sequence. The first case includes very sparse unions of long, densely occupied intervals; the second includes sets with large multiplicative gaps.Positive upper Banach density forces many differences.
Suppose there are intervals $I_j\subset\mathbb Z$, with lengths tending to infinity, such that $|A\cap I_j|/|I_j|\to\delta>0$. If $A+t_1,\ldots,A+t_k$ are pairwise disjoint translates, then on the same intervals \[ |(A+t_i)\cap I_j|=|A\cap I_j|+O(|t_i|). \] Dividing their total by $|I_j|$ gives $k\delta\leq1$. Consequently a greedy family of pairwise disjoint translates terminates after finitely many steps, at shifts $t_1,\ldots,t_k$. Its maximality says that every translate $A+t$ meets some $A+t_i$, so \[ \mathbb Z=\bigcup_{i=1}^k((A-A)+t_i). \tag{1} \] The sign is harmless because $A-A$ is symmetric.Let $C=\max_i|t_i|$. Covering the integers $C+1,\ldots,N-C$ by $1$ uses differences in $[1,N]$, and therefore \[ N-2C\leq kD(N). \] Since $A(N)=o(N)$, this proves the stronger conclusion $D(N)/A(N)\to\infty$ in this case. In particular, if $A$ contains intervals of arbitrarily large length, then in fact $A-A=\mathbb Z$. Positive upper Banach density does not imply positive ordinary density, so the case is compatible with the hypothesis of the problem.
Lacunary sets have quadratically many distinct differences.
First suppose $b_1<b_2<\cdots$ and $b_{i+1}\geq3b_i$. For fixed $j\geq2$, the $j-1$ differences $b_j-b_i$, $i<j$, are distinct and lie in \[ [2b_j/3,b_j). \] The intervals for different $j$ are disjoint: the next one starts at least at $2b_j>b_j$. Thus the first $s$ terms have at least $\binom s2$ distinct positive differences, all at most $b_s$.Now let $A=\{a_1<a_2<\cdots\}$ satisfy $a_{i+1}\geq(1+\epsilon)a_i$, for fixed $\epsilon>0$. Choose $h$ with $(1+\epsilon)^h\geq3$. Among $a_1,\ldots,a_k$, one residue class of indices modulo $h$ has at least $\lceil k/h\rceil$ members. They form a 3-lacunary subsequence, so \[ D(a_k)\geq\binom{\lceil k/h\rceil}{2}, \qquad A(a_k)=k. \] Hence \[ \frac{D(a_k)}{A(a_k)} \geq\frac1k\binom{\lceil k/h\rceil}{2} \longrightarrow\infty. \] Every such sequence has $A(N)=O_\epsilon(\log N)$, so it satisfies the required zero-density hypothesis.