Sets of large doubling and a question of Rudin
Update (May 2, 2010): After posting this preprint, Stefan Neuwirth informed us that Rudin’s question had been previously answered by Y. Meyers in 1968. It appears that Meyers’ construction doesn’t, however, say anything about the anti-Freiman problem. Indeed Meyers’ set (and all of its subsets) contains a set of density
. Hence, the construction of a
set that doesn’t contain a large
set still appears to be new. A revised version of the paper has been posted reflecting this information. Most notably, we have changed the title to “On the Structure of Sets of Large Doubling”.
Allison Lewko and I recently arXiv’ed our paper “Sets of Large Doubling and a Question of Rudin“. The paper (1) answers a question of Rudin regarding the structure of sets (2) negatively answers a question of O’Bryant about the existence of a certain “anti-Freiman” theorem (3) establishes a variant of the (solved) Erdös-Newman conjecture. I’ll briefly describe each of these results below.
— Structure of sets —
Before describing the problem we will need some notation. Let and define
to be the number of unordered solutions to the equation
with
. We say that
is a
set if
for all
. There is a similar concept with sums replaced by differences. Since this concept is harder to describe we will only introduce it in the case
. For
we define
to be the number of solutions to the equation
with
. If
for all nonzero
we say that
is a
set.
Let be a subset of the integers
, and call
an
-polynomial if it is a trigonometric polynomial whose Fourier coefficients are supported on
(i.e.
if
). We say that
is a
set (for
) if
holds for all -polynomials where the constant
only depends on
and
. If
is an even integer, we can expand out the
norm in 1. This quickly leads to the following observation: If
is a
set then
is also an
set (
,
). One can also easily show using the triangle inequality that the union of two
sets is also a
set. It follows that the finite union of
sets is a
set. In 1960 Rudin asked the following natural question: Is every
set is a finite union of
sets?
In this paper we show that the answer is no in the case of sets. In fact, we show a bit more than this. One can easily show that a
set is also a
set. Our first counterexample to Rudin’s question proceeded (essentially) by constructing a
set which wasn’t the finite union of
sets. This however raised the following variant of Rudin’s question: Is every
set the mixed finite union of
and
sets? We show that the answer to this question is no as well. To do this we construct a
set, A, which isn’t a finite union of
sets, and a
set,
, which isn’t the finite union of
sets. We then consider the product set
which one can prove is a
subset of
. It isn’t hard to deduce from this that
is a
subset of
that isn’t a mixed finite union of
and
sets. Moreover, one can (essentially) map this example back to
while preserving all of the properties stated above. Generalizing this further, we show that there exists a
set that doesn’t contain (in a sense that can be made precise) a large
or
. This should be compared with a related theorem of Pisier which states that every Sidon set contains a large independent set (it is conjectured that a Sidon set is a finite union of independent sets, however this is open).
We have been unable to extend these results to sets for
. Very generally, part of the issue arises from the fact that the current constructions hinges on the existence of arbitrary large binary codes which can correct strictly more than a
fraction of errors. To modify this construction (at least in a direct manner) to address the problem for, say,
sets it appears one would need arbitrary large binary codes that can correct strictly more than a
fraction of errors. However, one can show that such objects do not exist.
— Is there an anti-Freiman theorem? —
Let be a finite set of integers and denote the sumset of
as
. A trivial inequality is the following
In fact, it isn’t hard to show that equality only occurs on the left if is an arithmetic progression and only occurs on the right if
is a
set. A celebrated theorem of Freiman states that if
then
is approximately an arithmetic progression. More precisely, if
is a finite set
satisfying
for some constant
, then
is contained in a generalized arithmetic progression of dimension
and size
where
and
depend only on
and not on
.
It is natural to ask about the opposite extreme: if , what can one say about the structure of
as a function only of
? A first attempt might be to guess that if
for some positive constant
, then
can be decomposed into a union of
sets where
and
depend only on
. This is easily shown to be false. For example, one can start with a
of
elements contained in the interval
and take its union with the arithmetic progression
. It is easy to see that
regardless of
. However, the interval
cannot be decomposed as the union of
sets with
and
independent of
.
There are two ways one might try to fix this problem: first, we might ask only that contains a
set of size
, where
and
depend only on
. (This formulation was posed as an open problem by O’Bryant here). Second, we might ask that
hold for all subsets
for the same value of
. Either of these changes would rule out the trivial counterexample given above. In this paper we show that even applying both of these modifications simultaneously is not enough to make the statement true. We provide a sequence of sets
where
holds for all of their subsets for the same value of
, but if we try to locate a
set,
, of density
in
then
must tend to infinity with the size of
. As above, our initial construction of such a sequence of
‘s turned out to be
sets. This leads us to the even weaker anti-Freiman conjecture:
(Weak Anti-Freiman) Suppose that satisfies
and
for all subsets
. Then
contains either a
set or a
set of size
, where
and
depend only on
.
We conclude by showing that even this weaker conjecture fails. The constructions are the same as those used in the results above. The two problems are connected by the elementary observation that if
is a subset of a
set
then
holds where
only depends on the
constant
of the set
.
— A variant of the Erdös-Newman conjecture —
In the early 1980′s Erdös and Newman independently made the following conjecture: For every there exists a
that isn’t a finite union of
sets for any
. This conjecture was later confirmed by Erdös for certain values of
using Ramsey theory, and finally resolved completely by Nešetřil and Rödl using Ramsey graphs. One further application of our technique is the following theorem which can be viewed as an analog of the Erdös-Newman problem with the roles of the union size and
reversed.
Theorem 1 For every there exists a union of
sets that isn’t a finite union of
sets for any
.
leave a comment