International Journal of Combinatorics
Volume 2014 (2014), Article ID 826141, 11 pages
http://dx.doi.org/10.1155/2014/826141

Research Article

On the General Erdős-Turán Conjecture

Georges Grekos,¹ Labib Haddad,² Charles Helou,³ and Jukka Pihko⁴

¹Université Jean Monnet, 23 rue Michelon, 42023 Saint-Étienne, France
²120 rue de Charonne, 75011 Paris, France
³Penn State University, 25 Yearsley Mill Road, Media, PA 19063, USA
⁴Kyläkunnantie 53, 00660 Helsinki, Finland

Received 28 July 2014; Accepted 14 October 2014; Published 17 November 2014

Academic Editor: Chris A. Rodger

Copyright © 2014 Georges Grekos et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

The general Erdős-Turán conjecture states that if is an infinite, strictly increasing sequence of natural numbers whose general term satisfies , for some constant and for all , then the number of representations functions of is unbounded. Here, we introduce the function , giving the minimum of the maximal number of representations of a finite sequence of natural numbers satisfying for all . We show that is an increasing function of and that the general Erdős-Turán conjecture is equivalent to . We also compute some values of . We further introduce and study the notion of capacity, which is related to the function by the fact that is the capacity of the set of squares of positive integers, but which is also of intrinsic interest.

1. Introduction

In 1941, Erdős and Turán conjectured [1] that if is an additive basis of the set of natural numbers, that is, if every is the sum of two elements of , then the number of representations functions of , defined by

is unbounded. As is well known [2], if is a basis of , then its general term satisfies , for a constant and for all positive integers . So, a few years later, Erdős stated a more general conjecture, known as the general Erdős-Turán conjecture, to the effect that if is an infinite (strictly increasing) sequence in whose general term satisfies , then the number of representations functions of is unbounded (e.g., see [3]).

In a previous paper [4], we defined an order relation between infinite (strictly increasing) sequences and of natural numbers, by setting if for all . We called an Erdős-Turán set if the relation implies that the number of representations functions of is unbounded. Then, denoting by the set of squares of positive integers, the general Erdős-Turán conjecture amounts to the following.

Conjecture 1 (GET). For any positive integer , the set is an Erdős-Turán set.

Moreover, we showed that (GET) is in fact equivalent to the statement.

Conjecture 2 (GET). is an Erdős-Turán set.

The present paper is a sequel to [4], to which we refer for broader background and further motivation, but the necessary prerequisite material is here recalled as needed, to make the present text self-contained and clear.

In Section 3, we define the capacity of an infinite set (identified to a sequence) of natural numbers as the infimum of the supremum of the number of representations functions of , for all infinite sequences such that ; that is,

It provides a measure of the deviation of from being an Erdős-Turán set, and it allows to characterize the Erdős-Turán sets as those whose capacity is infinite. The notion of capacity is also of intrinsic interest, and we give several of its properties. We thus prove that if the capacity of is finite, then there exists a positive integer such that, for any infinite sequence satisfying , we have . And if the capacity of is infinite, then, for any natural number , there exists a positive integer such that, for any infinite sequence satisfying , we have . Moreover, if two infinite sequences and are “close”, in the sense that there exists a positive integer such that for all , then we have

We also establish that the range of the capacity function is the interval of . There are therefore sets of any finite capacity ≥2, and those are the ones for which the established properties are most significant.

In another paper [5], we introduced a function , giving the minimum of the maximal number of representations of a set of natural numbers forming a basis of the interval of . We showed that is an increasing function of and that the original Erdős-Turán conjecture is equivalent to . Here, in Section 4, we introduce the analogue for the general Erdős-Turán conjecture of , namely, the function , giving the minimum of the maximal number of representations of a set of natural numbers satisfying for all . We show that is an increasing function of and that

so that the conjecture (GET) is equivalent to the statement.

Conjecture 3 (GET). Consider .

We also give the results of some computations of values of .

We note that there is an extensive literature related to the Erdős-Turán conjectures, for example, [6–16].

2. Notation and Preliminaries

Notation 1. (1) The set of natural numbers is , while the set of positive integers is .
(2) A subset of , finite or infinite, is identified with the strictly increasing sequence of its elements (denoted by the corresponding lowercase letter).
(3) For any , let , where denotes the cardinality of a set . Further, let , considered as an element of .
(4) For any , let denote an initial segment of .
(5) For any , let denote a terminal segment of .
(6) For any , let denote a translate of .
(7) For any , let denote a homothetic of .
(8) The set of all infinite subsets of is denoted by .
(9) A partial order relation is defined in by if for all .
(10) An element of is called an Erdős-Turán set if, for any , the relation implies that .
(11) The class of all Erdős-Turán sets is denoted by .
(12) The set of squares of positive integers is denoted by ; that is, .

The following results, some of which are in [4], are used several times in the sequel.

Lemma 4. Let and .

(1)If , then .(2)In particular, , and .(3)Also, , and when .

For , with , so and for all .

Lemma 5. For any subset of   and any , one has .

We have , for all , and the result follows.

Lemma 6. Assume that has at least two elements.

(1)If   for all , then .(2)If and   for all , then .

Proof. If for some integer , there would exist indices such that . Therefore

Moreover, if , then the inequality becomes

In either case, this yields a contradiction.
Thus , and since , .

Example 7. Let . Then, by Lemma 6, we have .

Notation 2. For any subsets and of , and any , let

(1), ,(2), in ,(3); so , ,(4)An element of will be called an -representation of .

The following properties can be directly verified.

Lemma 8. For all subsets and of  , and all , one has

(1),(2),(3),(4),(5),(6)   both sums being equal to .

Lemma 9. Let and be nonempty subsets of such that is finite with largest element and is either finite or infinite, satisfying

(1),(2), for all .

Then .

Proof. Set . Since is the largest element in , we have

And, since , we also have

For , we therefore have

Given , set . We have . Thus , due to assumption and Lemma 6.
If , with in , and in , then and ; otherwise, we would have and

which is impossible. This means that any integer has at most one -representation. Moreover, if it has an -representation , then it cannot have a -representation; otherwise we would have .
Hence, in view of Lemma 8, and since is the union of four pairwise disjoint product sets, with replaced by or , we have

The result then follows.

3. The Capacity Function

Definition 10. For any , the capacity of is the element of defined by

Remark 11. For any , the capacity of is an attained minimum. More precisely, there exists such that

Indeed, if , then, for every in , we have

If is finite, then at least one of the ’s, for in , is finite, and the set

has smallest element . Taking one of those such that , we have

in view of Lemma 12, .

Lemma 12. For any and any , one has

(1),(2) if and only if ,(3)if then ,(4)if then ,(5),(6),(7)if , then .

Proof. These properties follow from the definitions and Lemma 4.

The following notion of caliber was introduced and studied in [4, item 5.5].

Definition 13. For any , the caliber of is the element of

defined by

We recall the following result from [4, item 5.9] that will be needed in the proof of the lemma following it.

Lemma 14. For any , one has

Lemma 15. For any , one has

Proof. For any such that , we have, from the definition,

Moreover, by the previous lemma, we have

Therefore , which is the infimum of all such , satisfies

Remark 16. It follows from Lemma 15 that if has zero caliber, that is, if , then ; that is, is an Erdős-Turán set. Thus, for example, if

where is the integer part of a real number , then and therefore is an Erdős-Turán set. In particular, any arithmetic progression is an Erdős-Turán set.
The infinite sets of natural numbers for which are the only known Erdős-Turán sets, because, in [4], we proved that the general Erdős-Turán conjecture (GET) is equivalent to the statement that there exists a subset of such that is an Erdős-Turán set and [4, item 5.16].

Remark 17. For any such that , there exists such that and .
In particular, there exists such that and , which is a nontrivially finite capacity set. We provide, below, an explicit example of such a set.

Indeed, since , there exists such that (e.g., for which , as in Example 7). Then, by [4, item 2.3], there exists such that and . Hence , while .

In particular, taking and yields the latter special case. We next proceed to explicitly construct such a set having and , following the procedure described in the proof of [4, item 2.3].

We start with the set

For any , let be the unique natural numbers such that

and let

Then

while if , then

Thus

Moreover, for , with ,

So , and therefore

Finally, if

then

so that . In particular, taking , we get

Hence .

Proposition 18. If and are such that , then there exists an integer such that, for any satisfying , one has for all .

Proof. We argue by contradiction. Assume that, for every , the set

is not empty. Clearly, . For , let

We have and for all , in view of the corresponding properties of the ’s. Moreover, each is finite since it is contained in . We thus have, for every , a decreasing sequence of finite nonempty sets , which is therefore stationary; that is, there exists some such that, for all , we have , and therefore the set

Now, consider an and fix an . Note that if

then

Conversely, given any , there is some whose first terms are the given ones in , and then, since , the first terms of lie in .
Therefore consists exactly of the first terms of the elements of . Hence the existence, by induction on , of a set such that and for all , so that for all . It follows that

while , contradicting the definition of .
So the original assumption does not hold; that is, there exists some such that is empty, which means that for all satisfying we have , and thus for all .

Corollary 19. Let .

(1)If , then for every , there exists an integer such that, for any satisfying , we have for all .(2)If , then there exists an integer such that, for any satisfying , we have for all .

Proof. If , then and Proposition 18 applies with any to give .
If , then Proposition 18 applies with to give .

Lemma 20. For every integer , as well as for , there exists some set such that .
Thus the range of the capacity function is the closed interval of  .

Proof. For a given integer , let

Thus is the union of the finite interval

for which , and of the infinite set with elements

Since

where is the largest element of , and since, for all , we have

Lemma 9 applies and gives

Moreover, if is such that , then, by definition and in view of Lemma 4, we have

In particular, for , we get

so that

and therefore

It follows that

For , we may take . Indeed, being the smallest element of for the order , the definition immediately yields

Lemma 21. For any , one has

Proof. As the sequence is strictly increasing, the inequalities

hold for all , and they amount to

So, by Lemma 12,

Now, let be such that

There are two possible cases.

(i)Either for all , that is, , and thus .(ii)Or there exists such that , and we let be the smallest such , so that

In this case, we define a sequence by

Since the sequence is strictly increasing and since

(assuming for the first two inequalities, otherwise they are unnecessary), the sequence is also strictly increasing; that is, lies in . Also, from the definition of , we have for all ; that is, , so that . Moreover, and therefore, by [4, item 2.9], . Hence

Thus, in any case, and for all such that , we have

It follows that

Corollary 22. For any and for any , one has

Proof. For , the stated inequalities are given by Lemma 21. For general , the result follows by induction on .

Remark 23. For any , there exists , such that, for all , we have

Indeed, first, if , then, in view of Lemma 12, , and therefore, by [4, item 2.14] and [4, item 2.16], both and , so that

Second, if , then and are decreasing sequences in , since if , then

and therefore, in view of Lemma 12,

Thus, the two sequences are stationary.

The following definition is taken from [4, 3.1].

Definition 24. For , one sets

If for some , we say that and are -close.
More generally, if , we say that and are close.

Lemma 25. Let and . If and are -close, then

Proof. Let be such that . Define a sequence in by

Since the sequences and are strictly increasing, so is ; that is,

Since and are -close, and , we have

and since (Lemma 4), , for all . So , and . Moreover, from the definition of (and using Lemma 4 again),

so that and are -close. It then follows from [4, item 3.3] that

Thus

and therefore

Furthermore, by exchange of and , we similarly get

Corollary 26. For any , one has

(1) and ,(2)if and are finite, then

Proof. If , then all the inequalities hold trivially. We may thus assume that , with , so that and are -close.
The inequalities in immediately follow from Lemma 25.
Assuming and are finite, we deduce from the two inequalities

of which one is precisely the inequality in .

Lemma 27. For any and any , one has

Proof. Let . The case where being trivial, we assume that . Let be the smallest positive integer such that , so that , where if , we set . Thus

Since , by Lemma 12, , which yields the first inequality.
On the other hand, by Remark 11, there exists such that and . Thus, to prove the second inequality, it is enough to prove the existence of a set such that and . To this end, we distinguish three cases.
(i) If and for , then , and we may take .
(ii) If , then

and we may take .
(iii) If and there exists such that , let be the smallest such , so that

In particular,

so that

and we may take .
In all three cases, we have

and therefore, by [4, item 2.9], . Since, in addition, , we conclude that

Corollary 28. For any and any finite subset of , one has

Proof. This follows from Lemma 27 by induction on the cardinality of .

4. The Function and an Equivalent Formulation of (GET)

Definition 29. For , let denote the set of all subsets of satisfying

Also, define a function by

Lemma 30. The following properties hold.

(1)The function is increasing(2)It has a limit in satisfying

Proof. For any and any , the set is a subset of lying in , so that . Hence

Thus is an increasing function.
The sequence is increasing in and is thus either stationary or unbounded, so that

exists in . Moreover, for any such that , the truncated set

consisting of the first elements of , lies in . Therefore

and thus . Hence

We recall the “diagonal lemma” and part of its corollary from [5, items 2.2 and 2.3] that will be used in the proof of the following proposition.

Lemma 31 (diagonal lemma and corollary). Let be a family of subsets of   indexed by an infinite set . There exists a subset of  , called a diagonal of the family , satisfying the following condition.

(DL)For any , there are infinitely many indices such that .

Moreover,

(CDL)If , for some and all , then .

Proposition 32. One has ; that is,

Proof. For every , let be such that .
By the diagonal lemma, since is a family of subsets of , indexed by the infinite set , there exists a subset of (called a diagonal of this family) satisfying the condition.
(DL) For any , there exists an infinite subset of such that for every .
Moreover, since for all , we have , by (CDL) in the previous lemma.
Furthermore, for every , taking and , we get

and since , we conclude that . Therefore .
Thus

and since, by Lemma 30, , it follows that .

Corollary 33. The conjecture () is equivalent to the following condition:

Proof. By [4, item 4.14], (GET) is equivalent to the condition , which, by Lemma 12, is equivalent to , which, by Proposition 32, amounts to .

Remark 34. The function is the analogue, for the conjecture (GET), of the function , for the conjecture (ET), introduced in [5, item 3.12].
Similarly, the statement (GET ), equivalent to (GET), is the analogue of the statement (ET ), equivalent to (ET).

Indeed, we just need to recall some notations and results.

(i) For any , denotes the set of all subsets of , called bases of , satisfying .

(ii) For any , denotes the set of all the subsets of , called finite bases of cardinality , satisfying and .

(iii) Furthermore, denotes the set of all subsets of , called bases of , satisfying .

(iv) The function is defined, in [5], by

It is an increasing function, like .

(v) In [5], we proved that

and that the conjecture (ET) is equivalent to

(vi) Here, we proved that

and that the conjecture (GET) is equivalent to

Hence the analogy.

Lemma 35. The following properties hold.

(1)If lies in , then

(2)For any , we have .(3)For any , we have .

Proof. Let , so that and . Then, for every , the truncated set

Therefore, by [5, item 3.15], we have

so that

In particular, . Hence , and therefore

Corollary 36. One has ; that is,

Corollary 37. If is an additive basis of , then

so that .

Remark 38. Either Corollary 36 or Corollary 37 confirms that (GET) implies (ET), as stated in Section 1.

Remark 39. It would be interesting to find a large lower bound for . However, the function seems to increase very slowly. We found by computer calculations that

and that

Indeed, we have

Moreover, the set

lies in and satisfies , so that

Also, the following set lies in and satisfies , so that

This set was constructed recursively by setting and then systematically taking to be the least integer satisfying the two conditions

Remark 40. In view of the slow increase of the function , it would be interesting to have a probabilistic model of the behavior of , defined on the space . A first step is to determine the cardinality of , which is, in itself, a nontrivial problem of intrinsic interest.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors thank an anonymous referee for careful reading of the paper and for several pertinent suggestions which helped improve the exposition.

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