 
Summary: Nonaveraging subsets and nonvanishing transversals
Noga Alon
Imre Z. Ruzsa
Abstract
It is shown that every set of n integers contains a subset of size (n1/6
) in which no element is
the average of two or more others. This improves a result of Abbott. It is also proved that for every
> 0 and every m > m( ) the following holds. If A1, . . . , Am are m subsets of cardinality at least
m1+
each, then there are a1 A1, . . . , am Am so that the sum of every nonempty subset of the set
{a1, . . . , am} is nonzero. This is nearly tight. The proofs of both theorems are similar and combine
simple probabilistic methods with combinatorial and number theoretic tools.
1 Introduction
In this paper we consider two problems in additive number theory. The problems are not directly related,
but the methods we use in tackling them are similar. The first problem deals with the existence of large
nonaveraging subsets in sets of integers. A set of integers is called nonaveraging if no member of the set
is the average of two or more others. Answering a problem of Erdos, Abbott proved in [4] that every set
of n integers contains a nonaveraging subset of cardinality (n1/13/(log n)1/13). His method, together
with the result of Bosznay [8] mentioned in the next section, can be used to get an (n1/7 ) bound, for
any > 0. Here we improve this estimate and prove the following.
