## 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 .

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