Sum-avoiding sets in groups

• Main Authors: Terence Tao, Van Vu
• Format: Article
• Language: English
• Published: Diamond Open Access Journals
• Series: Discrete Analysis
• Online Access: http://discrete-analysis.scholasticahq.com/article/887-sum-avoiding-sets-in-groups.pdf

Sum-avoiding sets in groups, Discrete Analysis 2016:15, 27 pp. Let $A$ be a subset of an Abelian group $G$. A subset $B\subset A$ is called _sum-avoiding in $A$_ if no two elements of $B$ add up to an element of $A$. Write $\phi(A)$ for the size of the largest sum-avoiding subset of $A$. If $G=\mathbb Z$ and $|A|=n$, then it is known that $\phi(A)$ must be at least $\log n(\log\log n)^{1/2-o(1)}$, and examples are known of sets for which $\phi(A)$ is at most $\exp(O(\sqrt{\log n}))$. These results are due to Xuancheng Shao and Imre Ruzsa, respectively. Reducing this gap to a reasonable size appears to be a very hard problem. If on the other hand, $G$ has torsion, then it is possible for $A$ to be large and finite while $\phi(A)$ is bounded. Indeed, if $A$ is any subgroup, then $\phi(A)=1$. However, this is not the end of the story, as the authors show. Suppose we know that $A$ is a subset of an Abelian group and that $\phi(A)\leq k$ for some fixed $k$. What can we say about $A$? Note that this condition expresses a kind of weak closure property: instead of saying that any two elements of $A$ add up to an element of $A$, it says that from any $k+1$ elements of $A$, two must add up to an element of $A$. A simple example of such a set that isn't itself a subgroup is a union of at most $k$ subgroups. Then given more than $k$ elements of $A$, two must belong to the same subgroup and hence add up to an element of $A$. In this paper, the authors show a converse to this easy observation: if $\phi(A)\leq k$, then there exist subgroups $H_1,\dots,H_m$ with $m\leq k$ such that $|A\cap H_i|\geq c|H_i|$ for each $i$, and all but at most $C$ elements of $A$ belong to $H_1\cup\dots\cup H_m$. Here, $c>0$ and $C$ are constants that depend on $k$ only. If you are familiar with the Balog-Szemerédi theorem and Freiman's theorem, then you might expect the proof of this result to be a fairly straightforward use of those tools. However, when one attempts to turn this thought into a proof, a significant difficulty arises, which the authors explain in their introduction. They resolve this difficulty by means of a complicated iterative argument -- in fact, it is sufficiently complicated that instead of desperately trying to keep control of all the parameters that would arise, they resort to the language of non-standard analysis. This tidies up the argument considerably, but at the price of yielding no bound at all for how the constants $c$ and $C$ depend on $k$. However, this is not a huge price, as they also say that if they had avoided non-standard analysis, then the bounds they would have obtained would have been extremely weak. The paper also contains a construction of arbitrarily large sets $A$ with $\phi(A)\leq k$ that contain no inverse pairs. This gives a negative answer to a question of Erdős. The construction, which is surprisingly simple, makes heavy use of the fact that their set lives in a group with order divisible by a small prime. They go on to show that this is necessary: if the order of $G$ has no small prime divisors, then Erdős's question has a positive answer.