NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ

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1 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ Abstract. In this paper we study su-free subsets of the set {1,..., n}, that is, subsets of the first n positive integers which contain no solution to the equation x + y = z. Caeron and Erdős conjectured in 1990 that the nuber of such sets is O(2. This conjecture was confired by Green and, independently, by Sapozhenko. Here we prove a refined version of their theore, by showing that the nuber of su-free subsets of [n] of size is 2 O(n/(, for every 1. For n, this result is sharp up to the constant iplicit in the O(. Our proof uses a general bound on the nuber of independent sets of size in 3-unifor hypergraphs, proved recently by the authors, and new bounds on the nuber of integer partitions with sall suset. 1. Introduction What is the structure of a typical set of integers, of a given density, which avoids a certain arithetic sub-structure? This fundaental question underlies uch of Additive Cobinatorics, and has been ost extensively studied when the forbidden structure is a k-ter arithetic progression, see for exaple [22, 23, 29, 40, 48]. General systes of linear equations have also been studied, beginning with Rado [37] in 1933, and culinating in the recent advances of Green, Tao and Ziegler [30, 31]. The subject is extreely rich, and questions of this type have been attacked with tools fro a wide variety of areas of atheatics, fro Graph Theory to Nuber Theory, and fro Ergodic Theory to Haronic Analysis. See [49] for an excellent introduction to the area. In this paper we shall consider su-free sets of integers, that is, sets of integers which contain no solution of the equation x + y = z. It is easy to see that the odd nubers and the set { + 1,..., n} are the largest such subsets of [n] = {1,..., n}. Both of these sets have eleents, and therefore there are at least 2 su-free sets in [n]. In 1990, Caeron and Erdős [12] conjectured that this trivial lower bound is within a constant factor of the truth, that is, that the set [n] contains only O(2 su-free sets. Despite various attepts [2, 10, 19], their conjecture reained open for over ten years, until it was confired by Green [25] and, independently, by Sapozhenko [43]. We shall prove a natural generalization of the Caeron-Erdős Conjecture, by bounding the nuber of sufree subsets of [n] of size, for all 1. Moreover, we shall also give a quite Date: March 4, Research supported in part by: (NA an ERC advanced grant, a USA-Israeli BSF grant, and the Israeli I-Core progra; (JB NSF CAREER Grant DMS , UIUC Capus Research Board Grant 11067, and OTKA Grant K76099; (RM a CNPq bolsa de Produtividade e Pesquisa; (WS ERC Advanced Grant DMMCA, and a Trinity College JRF. 1

2 2 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ precise structural description of alost all su-free subsets of [n] of size C n log n. Our proof uses a general bound on the nuber of independent sets of size in 3-unifor hypergraphs, proved in [3], which allows one to deduce asyptotic structural results in the sparse setting (in fact, for all n fro stability results in the dense setting (see Theore 2.1. The dense stability result we shall use (see Proposition 2.2 was proved by Green [25]. The second ain ingredient in the proofs of our ain theores will be soe new bounds on the nuber of sets of integers with sall suset (see Theores 1.3 and 1.4. Finally, we shall use Freian s 3k 4 Theore (see below to count sets with an extreely sall suset. The study of su-free sets of integers dates back to 1916, when Schur [47] proved that if n is sufficiently large, then every r-colouring of [n] contains a onochroatic triple (x, y, z with x + y = z. (Such triples are thus often referred to as Schur triples. Su-free subsets of general Abelian groups have also been studied for any years, see e.g. [1, 4, 11, 50]. Diananda and Yap [16] and Green and Ruzsa [28] deterined the axiu density µ(g of a su-free set in any finite Abelian group G, and in [28] it was oreover shown that any such group G has 2 (1+o(1µ(G G su-free subsets. For the group Z p, Sapozhenko [44] deterined the nuber of su-free subsets up to a constant factor, and for finite Abelian groups of Type I (those for which G has a prie divisor q 2 (od 3 Green and Ruzsa [28] were able to deterine the asyptotic nuber of su-free subsets of G. One of the ost significant recent developents in Cobinatorics has been the forulation and proof of various sparse analogues of classical extreal, structural and Rasey-type results. Beginning over 20 years ago (see, e.g., [6, 34, 38, 39], and culinating in the recent breakthroughs of Conlon and Gowers [14] and Schacht [46] (see also [8, 21, 42, 45], enorous progress has been ade in understanding extreal structures in sparse rando objects. For exaple, it is now known (see [14, 46] that the theore of Szeerédi [48] on k-ter arithetic progressions extends to sparse rando sets of density p n 1/(k 1, but not to those of density p n 1/(k 1. A sparse analogue of Schur s Theore was proved by Graha, Rödl and Ruciński [24], who showed that if p 1/ n and B is a p-rando subset 1 of Z n, then with high probability every 2-colouring of B contains a onochroatic solution of x + y = z. A sharp version of this theore was proved by Friedgut, Rödl, Ruciński and Tetali [20], but the extreal version, that is, the proble of deterining the (asyptotic size of the largest su-free subset of a p-rando subset of Z n, was open for 15 years before being resolved by Conlon and Gowers [14] and Schacht [46]. Even ore recently, Balogh, Morris and Saotij [7] sharpened this result by proving that, for any finite Abelian group G of Type I(q 2, if G = n and pn C(q n log n, then with high probability every axiusize su-free subset of a p-rando subset of G is contained in soe su-free subset of G of axiu size. In the case G = Z 2n, they deterined the sharp threshold. For structural and enuerative probles, such as that addressed by our ain theore, results are known in only a few special cases. For exaple, Osthus, Pröel and Taraz [36] 1 A p-rando subset of a set X is a rando subset of X, where each eleent is included with probability p, independently of all other eleents. 2 An Abelian group is of Type I(q if q 2 (od 3 is the sallest such prie divisor of G.

3 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 3 proved that if ( 3 4 +ε n 3/2 log n, then alost all triangle-free graphs with edges are bipartite, and that the constant 3/4 is best possible. This can be seen as a sparse version of the classical theore of Erdős, Kleitan and Rothschild [17], which states that alost all triangle-free graphs are bipartite. In [3], the authors proved a sparse analogue of the theore of Green and Ruzsa [28] entioned above, by showing that if C(q n log n, then alost every su-free -subset 3 of G is contained in soe axiu-size su-free set. We reark that there are only at ost G axiu-size su-free subsets of such a group G, and that oreover they adit an elegant description. In this paper we shall be interested in the corresponding question for the set [n]. As noted above, Caeron and Erdős [12] conjectured, and Green [25] and Sapozhenko [43] proved, that there are only O(2 su-free subsets of [n]. Our ain result is the following sparse analogue of this theore. Theore 1.1. There exists a constant C > 0 such that, for every n N and every 1, the set [n] contains at ost 2 Cn/( su-free sets of size. If n, then Theore 1.1 is sharp up to the value of C, since in this case there is a constant c > 0 such that there are at least 2 cn/( su-free -subsets of [n] (see Proposition 3.1. Note that if n then the result is trivial, since in this case our upper bound is greater than ( n ; however, if n then the proble is less interesting, since in this case it is relatively straightforward 4 to show that there are roughly e µ( n su-free -subsets of [n], where µ denotes the expected nuber of Schur triples in a rando -set. Since there are fewer than 2 n/3 subsets of [n] with at ost n/100 eleents, Theore 1.1 easily iplies the Caeron-Erdős Conjecture. However, Theore 1.1 only iplies that there are O(2 su-free subsets of [n], whereas Green [25] and Sapozhenko [43] proved that there are asyptotically c(n2 such sets, where c(n takes two different constant values according to whether n is even or odd. Since for us the parity of n will not atter, we shall assue for siplicity throughout the paper that n is even; the proof in the case n is odd is identical. We shall also prove the following structural description of a typical su-free -subset of [n]. Let O n denote the set of odd nubers in [n]. Theore 1.2. There exists C > 0 such that if n N and C n log n, then alost every su-free subset I [n] of size satisfies either I O n, or S(I Cn + ω(n and ( n 2 a Cn3 + ω(n, 3 a S(I where S(I = {x I : x }, and ω(n arbitrarily slowly as n. We reark that the upper bounds on S(I and k(i := a S(I ( a in Theore 1.2 are sharp up to a constant factor (see Section 6. Indeed, we shall show that if = o(n, then alost all su-free -sets I [n] have S(I = Ω(n/ and k(i = Ω(n 3 / 3. 3 An -subset of a set X is siply a subset of X of size. 4 The proof is a standard application of the FKG and Janson inequalities, see Section 4, or, e.g., [5, 33].

4 4 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ Our proof of Theores 1.1 and 1.2 has two ain coponents. The first is a bound on the nuber of independent -sets in 3-unifor hypergraphs (see Theore 2.1, which was proved in [3], and used there to deterine the asyptotic nuber of su-free -subsets of a finite Abelian group G such that G has a prie factor q 2 (od 3, for every C(q n log n. Using this theore, together with a stability result fro [25] (which follows fro a result of Lev, Luczak and Schoen [35] it will be straightforward to bound the nuber of su-free -sets which contain at least δ even nubers, and at least δ eleents less than. The second coponent involves counting restricted integer partitions with sall suset. Let p(k denote the nuber of integer partitions of k, so, for exaple, p(3 = 3 since 3 = = In 1918, Hardy and Raanujan [32] obtained an asyptotic forula for p(k, proving that p(k = 1 + o(1 4k 3 eπ 2k/3. We shall study the following type of restricted partition. Let p l (k denote the nuber of integer partitions of k into l distinct parts, i.e., the nuber of sets S N such that S = l and a S a = k. Thus, for exaple, p 3(8 = 2, since 8 = = It is straightforward to show that p l (k ( e 2 k l, l see Lea We shall bound the nuber of such partitions under the following ore restrictive condition. Recall that, given sets A, B N, the suset A + B is defined to be the set {a + b : a A, b B}. The following theore bounds the nuber of partitions of k into l distinct parts, such that the resulting set S has sall suset S + S. Theore 1.3. For every c 0 > 0 and δ > 0, there exists a C = C(δ, c 0 > 0 such that the following holds. If l 3 Ck and c c 0, then there are at ost 2 δl ( 2cek 3l 2 sets S N with S = l, a S a = k and S + S ck/l. Sets with sall suset are a central object of interest in Cobinatorial Nuber Theory, and have been extensively studied in recent years (see, e.g., [49]. It is easy to see that if A, B Z, then A + B A + B 1, with equality if and only if A and B are arithetic progressions with the sae coon difference. The Cauchy-Davenport Theore, proved by Cauchy [13] in 1813 and rediscovered by Davenport [15] in 1935, says that this result extends to the group Z p ; ore precisely, that l A + B in { A + B 1, p }. Many extensions of these results are now known; for exaple, the Freian-Ruzsa Theore (see [18, 41] states that if A Z and A + A C A, then A is contained in a O(1- diensional generalized arithetic progression of size O( A. This result itself has any generalizations, culinating in the very recent theore of Breuillard, Green and Tao [9], which is stated in the language of approxiate groups.

5 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 5 Despite the enorous interest in such probles, very little sees to be known about the nuber of different sets with sall suset (see [26], for exaple. The following classical result, proved by Freian [18] in 1959, iplies a bound for sets with so-called doubling constant less than 3. Freian s 3k 4 Theore. If A Z satisfies A + A 3 A 4, then A is contained in an arithetic progression of size at ost A + A A + 1. Observe that this iplies that, for all λ < 3, there are at ost 2 o(l( (λ 1l l sets S Z such that S = l and S + S λl, up to equivalence under translation and dilation. (That is, if we assue that in(s = 0 and S has no coon divisor greater than one. Our final theore, which also follows fro the proof of Theore 1.3, provides an upper bound of this type whenever S + S = O( S. 5 The following result will be crucial in the proof of Theores 1.1 and 1.2 in the case = Θ(n. Theore 1.4. Let δ > 0, and suppose that l N is sufficiently large and that k l 2 /δ. Then for each λ 2, there are at ost 2 δl ( (4λ 3e 6 sets S N with S = l, a S a = k, and S + S λl. Theores 1.3 and 1.4 are sufficient for our purposes; however, we believe the following stronger bound to be true. Conjecture 1.5. For every δ > 0, there exists C > 0 such that the following holds. C N and C log n, then there are at ost ( N/2 2 δ sets S [n] with S = and S + S N. Since S + S N for every -subset S [N/2], the conjecture (if true is close to optial. Note that the condition C log n iplies that n 2 /C, and thus guarantees that the nuber of translates of a given set S is negligible. The rest of the paper is organised as follows. In Section 2, we shall recall the general structural theore fro [3] and deduce fro it a bound on the nuber of su-free -sets which contain at least δ even nubers, and at least δ eleents less than. In Section 3 we shall prove a lower bound on the nuber of su-free -subsets of [n], and in Section 4 we shall use Janson s inequality to bound the nuber of su-free sets which contain at ost δ even nubers. In Section 5 we shall prove Theores 1.3 and 1.4. Finally, in Sections 6 and 7, we shall prove Theores 1.1 and Throughout the paper, f(n = O(g(n eans there exists an absolute constant, independent of all other variables, such that f(n Cg(n. l If

6 6 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ 2. Preliinaries In this section we shall recall soe of the ain tools we shall use in the proofs of Theores 1.1 and 1.2, and deduce that alost all su-free -sets I [n] either contain at ost δ even eleents, or satisfy I \ B δ for soe interval B of length. We reark that it will be fairly easy to bound fro above the nuber of such sets if B contains ore than δn eleents less than (see Sections 4 and 6; the ain difficulty will be counting su-free sets that are alost contained in the interval { + 1,..., n} A structural theore for 3-unifor hypergraphs. We begin by recalling fro [3] our ain tool: a theore which allows one to deduce asyptotic structural results for sparse su-free sets fro stability results for dense su-free sets. It is stated in the language of (sequences of general 3-unifor hypergraphs H = (H n n N, where V (H n = n. Throughout this section, the reader should think of the hypergraph H n as encoding the Schur triples (that is, triples (x, y, z with x + y = z in [n]. We next recall the type of dense stability property which we shall consider. Let α (0, 1 and let B = (B n n N, where B n is a faily of subsets of V (H n. Definition. A sequence of hypergraphs H = (H n n N is said to be (α, B-stable if for every γ > 0 there exists β > 0 such that the following holds for every n N. If A V (H n satisfies A (α βn, then either e(h n [A] βe(h n, or A \ B γn for soe B B n. Roughly speaking, a sequence of hypergraphs (H n is (α, B-stable if for every A V (H n such that A is alost as large as the independence nuber for H n, the set A is either very close to soe extreal set B B n, or it contains any (i.e., a positive fraction of all of the edges of H n. If H n is a hypergraph and N, then let SF(H n, denote the collection of independent sets of size in H n. Given a faily of sets B n and δ > 0, we define SF (δ (H n, B n, = {I SF(H n, : I \ B δ for every B B n }. Roughly speaking, the collection SF (δ contains all independent sets that are far fro all extreal sets. Finally, for each T V (H n, let d Hn (T = { e H n : T e } and define 2 (H n = ax { d Hn (T : T V (H n, T = 2 }. Note that if H n encodes Schur triples in [n], then 2 (H n 2. The following theore, which was proved in [3], shows that if H is (α, B-stable and n, then there are very few independent sets (i.e., su-free sets of size in H n which are far fro every set B B n. We write B n = ax { B : B B n }. Theore 2.1 (Theore 4.1 of [3]. Let α > 0 and let H = (H n n N be a sequence of 3-unifor hypergraphs which is (α, B-stable, has e(h n = Θ(n 2 and 2 (H n = O(1. If B n αn, then for every δ > 0, there exists a C > 0 such that the following holds. If

7 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 7 C n and n is sufficiently large, then (δ SF (H n, B n, ( ( Bn 1 ε for soe ε = ε(h, δ > 0. In the next subsection, we shall use this theore, together with a result of Green [25], to deduce an approxiate version of Theore Green s stability theore. Let H = (H n n N be the sequence of hypergraphs which encodes Schur triples in [n]; that is, V (H n = [n] and {x, y, z} E(H n whenever x + y = z. The following stability result, due to Green [25], iplies that H is (α, B-stable, where α = 1/2 and { {a } } B n = + 1,..., a + : 0 a { } O n, (1 where, as before, O n denotes the odd nubers in [n]. Proposition 2.2 ([25, Proposition 7]. For any γ > 0, if β = β(γ > 0 is sufficiently sall, then the following holds. If A [n] with A (1/2 βn, then either A contains at least βn 2 Schur triples, or A \ B γn for soe B B n. Using Theore 2.1 and Proposition 2.2, we easily obtain the following corollary. Proposition 2.3. For every δ > 0, there exist constants C > 0 and ε > 0 such that the following holds for every n N and C n. There are at ost ( 2 ε su-free subsets I [n] of size such that I \ B > δ for every B B n. In particular, for alost every su-free set I [n] of size, either I \ O n δ, or I \ B δ for soe interval B of length. Proof. Let H = (H n n N be the sequence of hypergraphs which encodes Schur triples in [n], as above, and let B = (B n n N be the collection of intervals of length, plus the odds, as in (1. Set α = 1/2, and observe that H is (α, B-stable, by Proposition 2.2. Now, by Theore 2.1, if C = C(δ > 0 is sufficiently large, and C n, then (δ SF (H n, B n, ( ( 2 ε Bn = 2 ε for soe ε = ε(δ > 0. Since there are at least ( su-free -subsets of [n], it follows that for alost every such set I we have I \ B δn for soe B B n, as required. As we rearked above, it will be relatively straightforward to count the sets I that contain fewer than δ even eleents, using Janson s inequality (see Section 4, and those that contain ore than δ eleents less than, using induction on n (see Section 6. Thus, Proposition 2.3 essentially reduces the proble of counting su-free -sets in [n] to counting the su-free sets that are alost contained in the interval { + 1,..., n}.

8 8 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ 2.3. Binoial coefficient inequalities. We shall ake frequent use of soe siple inequalities involving binoial coefficients; for convenience, we collect the here. Note first that ( ( a b ea b ( b and that a b is increasing in a. Next, observe that if a > b > c 0, then ( ( c ( ( ( b ( a b a a c a c a and, (2 b c a b b b a b and hence ( ( b d ( d ( a c a c b a. (3 b d a a b b We shall also use several ties the observation that, for every a 1 and b > 0, k a e bk Γ(a + 1 ( a a+1 c C e (a+1, (4 b a+1 b k=1 where Γ( is Euler s Gaa function, for soe absolute constants C > c > 0. For other standard probabilistic bounds, such as the FKG inequality and Chernoff s inequality, we refer the reader to [5]. 3. A lower bound on the nuber of su-free sets In this section we shall prove the following siple proposition, which shows that the bound in Theore 1.1 is tight. Proposition 3.1. If n, then there are 2 Ω(n/( su-free subsets of [n] of size. Proof. Let c > 0 be a sufficiently sall absolute constant and set a = cn 2 / 2. We clai that if S is a uniforly chosen rando -subset of U = { a,..., n}, then P ( S is su-free ( exp cn. (5 2 In order to prove (5, we shall in fact choose the eleents of S independently at rando with probability p = 4/n, and bound P p ( S is su-free S =, which is clearly equivalent. (Note that the proposition is trivial if = Ω(n, so we ay assue that p is sufficiently sall. First observe that there are at ost a 2 + a triples {x, y, z} in U with x + y = z, and at ost a + 1 pairs {x, y} in U with 2x = y. Thus, by the FKG inequality, ( ( P p S is su-free 1 p 3 a 2 +a( ( 1 p 2 a+1 exp cn 3 since c > 0 is sufficiently sall, by our choices of a and p. Next, note that, by Chernoff s inequality, P p ( S < e c e cn/, since ( n. Finally, observe that g(t = P p S is su-free S = t is decreasing in t. It follows iediately that ( ( P p S is su-free S = exp cn, 2

9 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 9 which proves (5. Hence the nuber of su-free -sets in { a,..., n} is at least ( ( + a exp cn ( ( a exp 2 n cn ( ( cn = exp, 2 2 where the inequality follows fro (2 and the fact that ( 1 + 2a n e a/n. 4. Janson arguent In this section, we shall count the su-free sets that have few even eleents. Recall that O n denotes the odd nubers in [n]. Proposition 4.1. If δ > 0 is sufficiently sall, then there are at ost 2 O(n/( su-free subsets I [n] with I = and I \ O n δ, for every, n N. We reark that an arguent siilar to the one presented in this section was used in [3] in a soewhat ore general context, see also [7]. Indeed, the following result was proved in [3]. Proposition 4.2 ([3, Proposition 5.1]. There exists constants δ > 0 and C > 0 such that the following holds for every C n log n. There are at ost ( ( n 3 su-free subsets I [n] with I = and I \ O n δ. Proposition 4.2 clearly iplies Proposition 4.1 in the case C n log n. Furtherore, Proposition 4.1 is trivial if O ( ( n, since then the claied upper bound is greater than n. Thus, we need only consider the case C n C n log n. Recall the following well-known result, which is an easy corollary of Janson s inequality (see [5, 33], cobined with Pittel s inequality (see [33]. We refer the reader to [3, Section 5] for a proof. Lea 4.3 (Hypergeoetric Janson Inequality. Suppose that {U i } i J is a faily of subsets of an n-eleent set X and let {0,..., n}. Let µ = i J (/n U i and = i j (/n U i U j, where the second su is over ordered pairs (i, j such that i j and U i U j. Let R be a uniforly chosen rando -subset of X. Then P ( U i R for all i J { } C ax e µ/2, e µ2 /2, for soe absolute constant C > 0. We now turn to the proof of Proposition 4.1. Let C > 0 be a sufficiently large constant, and recall that we ay assue that C n C n log n. We begin by proving the following clai.

10 10 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ Clai. For soe constant c > 0, there are at ost ( { } ( C ax e ck2 /n, e c k k su-free -sets I [n] with I \ O n = k δ. Proof of clai. Let k δ and let S be an arbitrary k-subset of [n] \ O n. Let {U i } i J be the collection of pairs {x, y} O n such that either x + y = z or x y = z for soe z S. In order to bound the nuber of su-free -sets I with I \ O n = S, we shall apply the Hypergeoetric Janson Inequality to the collection {U i } i J and the set X = O n, with R a uniforly chosen rando ( k-subset of X. Note that if S R is su-free, then U i R for all i J. Let µ and be the quantities defined in the stateent of Lea 4.3, and observe that for every even nuber z, there are either at least n/10 pairs {x, y} O n with x + y = z (if z, or at least n/5 such pairs with x y = z (if z. Thus nk/20 J nk, since each pair can be counted at ost twice. Observe that each vertex x O n lies in at ost 2k of the U i. Hence µ nk 20 ( k2 n 2 k2 30n and (2k 2 ( J 2k ( 3 2k2 3. n n 2 By the Hypergeoetric Janson Inequality, if c = 10 4 then there are at ost { } ( C ax e ck2 /n, e c k sets R O n of size k such that S R is su-free. Suing over choices of S, we obtain the claied bound. Now, by (2 and since C n log n n/6, if k n/ then (6 is at ost ( ( k ( ( k ( 3e C e c C e c k 2k assuing δ > 0 is sufficiently sall. However, if k n/ then (6 is at ost ( k ( ( 3e /n C 2 O(n/. 2k e c2 ( To see the final inequality, observe that (since xe x/n n we have e c2 /n n/c, and use the fact that k (a/k k is axiized when k = a/e. This copletes the proof of Proposition Partitions and susets In this section we shall prove Theores 1.3 and 1.4. Recall that p l(k = # { partitions of k into l distinct parts }. We shall use the following easy upper bound on p l (k in the proofs of Theores 1.1 and 1.2., (6

11 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 11 Lea 5.1. For every k, l N, p l(k ( e 2 l k. l 2 Proof. Note that p l (k = 0 if k < ( l+1 2, and that the result is trivial if l 3. So assue that l 4, and consider putting k identical balls into l labelled boxes. There are ( k+l 1 = l k+l ( k+l l l 1 ways to do so, and each partition of k into l distinct parts is counted exactly l! ties. Using the bound l! 2πl ( l l, ( e it follows that, if k l+1 2 and l 4, then p l(k 1 ( l! l k + l 2 ( l ( l ( k + l l l e e(k + l e 2 l k, 2πl l l l 2 as required, since ( k+l l k < e 2 < 6 2π. In order to otivate the proofs of Theores 1.3 and 1.4, we shall first sketch an easy proof of a weaker bound and an incorrect proof of a sharper one; we will use ideas fro both in the actual proof. We begin with the weaker bound: given c, δ > 0, let C = C(δ, c > 0 be sufficiently large, and suppose that l 3 Ck. We clai that there are at ost 2 δl ( cek l 2 l (7 sets S N with S = l, a S a = k, and S + S ck/l. Note that this is weaker than Theore 1.3 by a factor of (3/2 l. We shall count good sequences (a 1,..., a l of length l, that is, sequences such that the underlying set S = {a 1,..., a l } satisfies S = l, a S a = k and S + S ck/l. Note that each such set S will appear as a sequence exactly l! ties. Set S j = {a 1,..., a j } and observe that ( ( S j + a j+1 \ Sj + S j δ Sj for at ost δl indices j [l], since otherwise S + S δ(δl/2 2 > ck/l, where the last inequality follows because l 3 Ck. We now ake a siple but key observation: that, for every set S N, there are at ost (1 δ 1 S + S eleents y N such that ( S + y \ ( S + S δ S. (8 To prove this, observe that there are S S + S pairs (a, b with a S and b S + S, and that if (8 holds then a + y = b for at least (1 δ S pairs (a, b S (S + S. For each pair (a, b there is at ost one such y, and so there are at ost S S+S eleents y which (1 δ S satisfy (8, as claied. Thus, as in the proof of Lea 5.1, the nuber of good sequences (a 1,..., a l is at ost ( ( ( (1 δl l k + δl S + S ( e ( δl δl ( δl 2ek 1 e O(δl S + S l δl δl 1 δ δ δl S + S ( 2e 2 δl ( l ( l ck e O(δl 2 O( δl ck, (9 δ 2 c l l

12 12 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ assuing δ > 0 is sufficiently sall. Dividing by l!, we obtain (7, as claied. Next, define the span of a set S N to be ax(s in(s and observe that, for any set S, we have 2 span(s = span(s + S. Let B j denote the set of eleents y (as above for which (S j + y \ (S j + S j δ S j. Intuitively, one would expect that B j + S j S j + S j, which iplies that span(b j + span(s j = span(b j + S j span(s j + S j = 2 span(s j, (10 and hence span(b j span(s j. This would iply that in(s j + B j and ax(s j + B j are (alost disjoint, since ax ( in(s j + B j = in(sj + ax(b j ax(s j + in(b j = in ( ax(s j + B j. Hence, if B j + S j S j + S j, then it follows fro the arguent above that B j S j + S j, 2 which would win us a factor of roughly 2 l in (9. Unfortunately, (10 does not hold in general; for exaple, if S j = [k] X, where X is a rando subset of {k + 1,..., 2k} of size δk, then S j + S j [4k] and B j [3k], and so span(s j 2k but span(b j 3k. Nevertheless, we shall be able to prove an approxiate version of (10 (see (14, below by considering an interval J on which S j is sufficiently dense close to the extreal values, ax(j and in(j. Proof of Theore 1.3. Let c c 0 > 0 and δ > 0, and note that without loss of generality we ay assue that δ = δ(c 0 is sufficiently sall. Let C = C(δ, c 0 > 0 be sufficiently large; with foresight, we reark that C = 1/δ 13 will suffice. Note also that if c 3e/2, then the theore follows iediately fro Lea 5.1; we shall therefore assue that c < 3e/2. Given S N with S = l, a S a = k and S + S ck/l, let S = ax { δk/l, in(s } and S = in { k/δl, ax(s }, so δk/l S S k/δl. Moreover, define { t = in t : [ ] } S (1 + δ t S, (1 + δ t+1 S δ 3 l { and t = in t : [ S (1 δ t+1 S, (1 δ t S ] } δ 3 l (or if no such t exists, and set J = [J, J ] = [ (1 + δ t S, (1 δ t S ]. Note that the interval J has dense ends, that is, the sets [(1 δj, J ] and [J, (1 + δj ] each contain ore than δ 3 l eleents of S. We first deal with an easy special case. Case 1: ax { t, t } 1/δ 2. Suppose first that t 1/δ 2. Note that S contains at ost δl eleents greater than S, since a S a = k, and hence at ost 2δl eleents of S greater than (1 δ 1/δ2 S e 1/δ k δl δk l.

13 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 13 Let s 2δl denote the nuber of eleents of S greater than δk/l, and note that ( a l s ( e 2 a l s, l since e(l s l. By Lea 5.1, it follows that there are at ost ( e 2 s ( ( k δk/l e 2 s ( k e 2 l s ( δk l 2 s ( e 2 l ( l δk δk =, (11 s 2 l s s 2 l 2 δs 2 l 2 l 2 such sets S, where the last inequality follows since δ > 0 is sufficiently sall and s l/3. Suing over s, it follows that there are at ost l ( δk l l sets S with t 1/δ 2. Since 2 δ = δ(c 0 was chosen sufficiently sall, the required bound follows. Siilarly, if t 1/δ 2, then S contains at ost 2δl eleents greater than δk/l (at ost δl in the range [δk/l, k/δl] [δk/l, (1 + δ 1/δ2 δk/l], and at ost δl greater than k/δl. Thus l ( δk l l is also an upper bound on the nuber of sets S with 2 t 1/δ 2. We shall now apply the arguent which failed to work above to the set S J := S J. Case 2: ax { t, t } 1/δ 2. Since t, t 1/δ 2, there are at ost 3δl eleents 6 of S outside the set [0, δk/l] J. Moreover, we ay assue that S has at least l/3 eleents larger than δk/l, since the inequalities in (11 hold for every s l/3. It follows that S J l/4, and therefore, using the trivial bound k = a S a S J J, we deduce that J 4k/l. Set J 0 = { a S : a δk/l }, let b = J 0 2l/3 and set r = S \ (J J 0 3δl. Suppose first that span(j < ck/8l. Then, by Lea 5.1, the nuber of choices for S is at ost 7 2l/3 3δl ( ( δk/l e 2 k b b=0 r=0 J,J r 2 r ( ck/8l l b r = k 2 l 2 ax b2l/3, r3δl k 2 l 2 ( 2δl bc ( b ( eδk e 2 r ( k cek ax lb r 2 2l 2 b ( 2el 2 r ( l ( cek 2 δl cek cr 2 2l 2 2l 2 b2l/3, r3δl l b r if δ = δ(c 0 > 0 is sufficiently sall, since the axiu occurs at r = 3δl and b = 2δl/(ec. Thus we ay assue that span(j ck/8l, fro which it follows that span(j = J J δ ( J + J, (12 since J 4k/l, and so J J + ck/8l ( 1 + c/32 J. Let us count sequences a = (a 1,..., a l of distinct eleents such that S = {a 1,..., a l } satisfies a S a = k and S + S ck/l. Given such a sequence a, for each j [l], set and define B j = S j = { a 1,..., a j } J { y N : ( S j + y \ ( } S j + S j δ 6 S j. (13 6 At ost (t + t δ 3 l eleents in the range [δk/l, k/δl], and at ost δl eleents greater than k/δl. 7 Here, as usual, 0 0 = 1. l

14 14 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ We ake the following key clai. Clai. Suppose that the intervals [J, (1 + δj ] and [(1 δj, J ] contain ore than δ 5 l eleents of { } a 1,..., a δl each. Then ( 2 B j 3 + δ S + S for every δl j l. Proof of clai. Fix j with δl j l. Recall fro (13 that ( S j + y \ ( S j + S j δ 6 l for every y B j, and that 2 span(s j = span(s j + S j. We clai that ( 1 δ span(j + span(bj span(s j + S j. (14 Indeed, since [(1 δj, J ] contains ore than δ 5 l eleents of { a 1,..., a δl }, and hence of S j, it follows that x + ax(b j S j + S j for soe x S j [(1 δj, J ]. Siilarly, our assuption on [J, (1 + δj ] { a 1,..., a δl } iplies that x + in(b j S j + S j for soe x S j [J, (1 + δj ], and therefore span(s j + S j span(b j + x x span(b j + span(j δ(j + J. But by (12 we have span(j δ ( J + J, so (14 follows. Now S j J, and so 2 span(j span(s j + S j, which, together with (14, iplies that span(b j 1 + δ span(s j + S j = ( 1 + δ span(s j. (15 2 Next, define the set { A = a S j : ( ( ( } a + B j Sj + S j 1 δ Bj. Note that S j \ A δ B j { (a, y S j B j : a + y S j + S j } δ 6 S j B j, and hence A (1 δ 5 S j. Let A = in A and A = ax A, and consider the set Subclai: D 3 B j /2. D = ( A + B j ( A + B j. Proof of subclai. Since [J, (1+δJ ] and [(1 δj, J ] each contain ore than δ 5 l eleents of { a 1,..., a δl } and A (1 δ 5 S j, we have span(a span(s j δ ( J + J ( 1 2 δ span(s j, where the second inequality follows by (12, and the fact that 2 span(s j span(j. Thus, by (15, span(b j ( 1 + δ span(s j ( δ span(a < 2 span(a, and so ( A + B j ( A + B j Bj /2, which easily iplies the subclai.

15 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 15 Finally, observe that D \ (S j + S j 2δ B j by the definition of A. Hence ( 3 2 2δ B j S j + S j S + S, as required. Now, recall that (by definition, the intervals [J, (1+δJ ] and [(1 δj, J ] each contain at least δ 3 l eleents of S. Let I(S denote the collection of orderings of the eleents of S such that the intervals [J, (1 + δj ] and [(1 δj, J ] each contain ore than δ 5 l eleents of { } ( a 1,..., a δl, and write a I if a I {a1,..., a l }. Observe that, given a rando ordering a = (a 1,..., a l of the eleents of S, the probability that a I(S is at least 1/2. Thus there are at least l!/2 orderings a I. In order to count sequences a I, recall that J 0 = { a S : a δk/l } and b = J 0, let Ĵ 0 = { } j [l] : a j J 0, and set Q = { 1,..., δl } { } j [l] : a j B j 1 and j Ĵ0. We clai that if a I, then Q 5δl. To see this, recall that r = S \ (J J 0 3δl, and suppose that there are at least δl values of j δl with a j+1 J \ B j. Then each such j adds at least δ 6 S j δ 11 l new eleents to S J + S J, since a I iplies that S δl δ 5 l. But then S + S S J + S J δl δ 11 l > ck/l, since l 3 Ck = k/δ 13 > ck/δ 12, which contradicts our assuption. Thus, by the Clai, and setting q = Q, the nuber of choices for S is at ost 2l/3 5δl b=0 q=0 2 l! ( l q ( k + q 1 q 1 ( l b ( δk l b (( δ l q b ck. (16 l Indeed, for each choice of the sets Ĵ0 and Q, and the values of a j for each j Ĵ0 Q, there are at ost B j ( 2 + δ S + S choices for each reaining eleent a 3 j+1. Recall that q 5δl, by the observations above, and that each set S is counted at least l!/2 ties as a sequence a I. Since S + S ck/l, (16 follows. Finally, note that the suand in (16 is bounded above by ( 2 el l! q 2k q ( q 3l el 2ck b δk b (( l ( l 3l 2 ck 2ck 3 + δ = 2O(δl 3el 2 q ( b ( l 3eδl 2ck l l! cq 2 2cb 3l which is axiized with b = 3δl/2c and q = 5δl, and so is at ost ( l 2cek 2 γl, 3l 2 where γ = γ(δ, c 0 0 as δ 0 for any fixed c 0 > 0. Since we chose δ = δ(c 0 to be sufficiently sall, the theore follows.

16 16 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ The proof of Theore 1.4 is alost identical to the proof of Theore 1.3 given above; we need only to add the following observations: that S j B j, and that a j+1 S j. Proof of Theore 1.4. Let δ > 0, and note that without loss of generality we ay assue that δ is sufficiently sall. Suppose that l N is sufficiently large and that ( l 2 k l 2 /δ, and set c 0 = 2l 2 /k. Let C = C(δ 3, c 0 > 0 be given by Theore 1.3, and note that since δ l 2 /k 2, C depends only on δ. Let λ 2 and set c = λl 2 /k, and note that λl = ck/l and l 3 (δlk Ck. We are therefore in the setting of Theore 1.3, and hence we can repeat the proof above up to (16, except replacing δ everywhere by δ 3. Using the observations that S j B j and a j+1 S j, we deduce that the nuber of choices for S is at ost l 5δ 3 l ( ( ( ( 2 l k + q 1 l δ 3 b k (( 2 l! q q 1 b l 3 + δ3 S + S S j, (17 b=0 q=0 where Z = { } j > δ 3 l : a j+1 B j, so in particular Z = l q b. Recalling that S j j q b and S + S λl, and applying the AM-GM inequality, we see that j Z (( δ3 S + S S j and hence (17 is at ost l 5δ 3 l b=0 q=0 2 O(δ2 l l! ( (2 j Z 3 + δ3 λl + q + b 1 Z l q b j, ( e 3 q ( kl δ 3 b ( ek 2λl q 2 b 3 l l q b 3(q + b +. ( Since ( l 2 k l 2 /δ and λ 2, the suand in (18 is axiized with q = 5δ 3 l and b 3δ 2 l/c, and hence (18 is at ost as required. 2 δl ( (4λ 3e 6 We end this section by briefly discussing the factor (4/3 l which separates the bound in Theore 1.3 fro that in Conjecture 1.5. Although it ay see obvious that we lost this factor in the subclai, where one ight hope that D (2 δ B j, we reark that this is in fact not the case. Indeed, let x y z, and consider the set S j = (T + z (T + 2z, where T = U W is coposed of U = [y, y + x] and W, a rando subset of [0, 2y] of density ε. Note that if δz > 2y, then this set is dense near its extrees, and oreover B j [z, 2y + z] [2z, 2y + 2z] and A j (U + z (U + 2z. Thus A j + B j [ y + z, x + 3y + z ] [ y + 2z, x + 3y + 2z ] [ y + 3z, x + 3y + 3z ] l, has size roughly 3 B j /2, and so in this case the subclai is sharp. j Z

17 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 17 It therefore sees that in order to prove Conjecture 1.5, one would have to use soe structural properties of S j. Indeed, since B j B j+1 for each j [l] and a typical S j is a rando-like subset of B 1... B j 1, one ight hope to bound the probability that (i.e., the nuber of S j for which in the subclai we have D < (2 δ B j. However, since the bounds in Theores 1.3 and 1.4 will suffice to prove Theores 1.1 and 1.2, we shall not pursue this atter here. 6. The proof of Theore 1.1 We are now ready to prove Theore 1.1, which generalizes the Caeron-Erdős Conjecture to su-free sets of size, and its structural analogue, Theore 1.2. Both theores will follow fro essentially the sae proof; we shall first prove Theore 1.1, and then (in Section 7 point out how the proof can be adapted to deduce Theore 1.2. As noted earlier, we shall for siplicity assue throughout that n is even. The proof is fairly long and technical so, in order to aid the reader, we shall start by giving a brief sketch. The arguent is broken into a series of six clais, each relying on the earlier ones; the first five being relatively straightforward, and the last being soewhat ore involved. We begin, in Clai 1, by using the Hypergeoetric Janson Inequality to give a general upper bound on the nuber of su-free -sets I [n] with S = I [] fixed. In Clai 2, we use this bound, together with Propositions 2.3 and 4.2, to prove the theore in the case (1/2 δn. Then, in Clai 3, we use Clai 2, Propositions 2.3 and 4.2, and induction on n, to deal with the case S δ. Writing l = S and k = a S ( a, in Clais 4 and 5 we use Clais 1, 2 and 3 and Lea 5.1 to deal with the (easy cases l = O(n/ and k l 2 n/. Finally, in Clai 6, we deal with the reaining (harder cases; however, since we now have l 3 Ck, we ay apply Theore 1.3 in place of Lea 5.1. In fact, it turns out that when = Θ(n the bound in Theore 1.3 is not quite strong enough for our purposes, but in this case we have S + S = O( S, and so ay instead use Theore 1.4. Each of the clais is stated in such a way as to facilitate the deduction of Theore 1.2, which follows by exactly the sae arguent, with a couple of inor tweaks. Proof of Theore 1.1. Fix δ > 0 sufficiently sall, let C = C(δ > 0 be a sufficiently large constant, and let n N and 1. We shall show that there are at ost 2 Cn/( su-free -subsets of [n]. We shall use induction on n, and so we will assue that the result holds for all saller even values of n. Note that the result is trivial if C 1/3 n, since in that case 2 Cn/( ( n. We shall therefore assue that C 1/3 n, and hence (via our choice of C that n is sufficiently large. For any set I [n], let S(I = { x I : x } denote the collection of eleents of I which are at ost, as in the stateent of Theore 1.2. Moreover, given a set S [], let S = { x S : x > n/4 }. We begin by giving a general bound on the nuber of su-free -sets I [n] with S(I = S, for each S []. For each l [] and k N, let S(k, l denote the collection

18 18 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ of sets S [] such that S = l and ( n 2 a = k. a S The following clai follows easily fro the Hypergeoetric Janson Inequality. Clai 1. For every k, l N with l /2, and every S S(k, l, there are at ost {( S + S { } ( } in, C ax e k2 /2n 2, e k/8nl l l su-free -sets I [n] such that S(I = S. Proof of Clai 1. Since I is su-free and S I, it follows that I contains no eleent of S + S { + 1,..., n}. Since S(I = S and S = l, the first bound follows. For the second bound, we use the Hypergeoetric Janson Inequality. Define the graph G of forbidden pairs by setting V (G = { + 1,..., n } and E(G = { {x, x + s}: s S }, and observe that I \ S is an independent set in G, and that G has k edges and axiu degree at ost 2l, since S(I = S S(k, l. Let µ and be the quantities defined in the stateent of Lea 4.3 and note that we are applying the lea with X = and R = l. Recalling that l /2, we have µ = k ( l2 ( 2 k2 n 2 and (2l 2 ( k l ( l ( 3 32kl( l3 n 3. Thus µ/2 k 2 /2n 2 and µ 2 /2 k/8nl, and so the claied inequality follows. In the calculation below, we shall on several occasions wish to ake the assuption that n 2 δn. The next clai deals with the copleentary case. ( ( 1 1 Clai 2. If 2 δ n, then there are at ost ( su-free -sets l n 2 I [n] such that S(I = l and I O n. Proof of Clai 2. The result is trivial if l = 0, so let us assue that l 1. Note that the conditions of Propositions 2.3 and 4.2 are satisfied, since δ > 0 is sufficiently sall and n is sufficiently large. Thus, by Proposition 2.3, there exists ε = ε(δ > 0 such that all but 2 ε( su-free -sets I [n] satisfy either I \ On δ, or I \ B δ for soe ( interval B of length. Moreover, by Proposition 4.2 there are at ost 1 n 3 su-free -sets I [n] such that 1 I \ O n δ. It thus suffices to count su-free -sets I [n] such that I \ B δ for soe interval B of length. We shall divide into three cases, depending on the size of in(i. Set a(i := n 2 in(i, and fix an interval B of length such that I \ B δ.

19 Case 1: a(i 4δ. A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 19 We clai that no such su-free set I exists. Indeed, note that a(i I and that ax(b in(b = 1. Since I is su-free, it follows that I contains at ost half of the eleents of the set 8 { in(b,..., in(b + 4δ 1 } { ax(b a(i + 1,..., ax(b a(i + 4δ }. Thus I B 4δ + δ < ( 1 2 δ n, which contradicts our assuption on I. Case 2: 2l a(i 4δ. Set a = a(i and note that, as in Case 1, since a I, it follows that I contains at ost a eleents of the set { + 1,..., + a} {n a + 1,..., n}. Thus I contains at least a l 3a/2 eleents of { + a + 1,..., n a}. The reaining eleents are contained in a set of size 3a, and thus, by (3, there are at ost 2 3a ( 2a 3a/2 = 2 3a ( 2a t a/2 such su-free -sets, where t = δn. Case 3: a(i in{2l, 4δ}. ( 2 3a t t a/2 ( t 1 2 l+1 ( Fix soe set S [ 4δ, ] with S = l, and note that S + S 3l/2, where S = S \ [n/4], since S = S. Hence, by Clai 1 and using (3, the nuber of su-free -sets I [n] such that S(I = S is at ost ( ( ( ( S + S 3l/2 3l/2 l l = t l/2 t t l/2 (, (19 t where again t = δn. Now, since a(i 2l, there are at ost 2 2l choices for the set S(I. Therefore, by (19, there are at ost ( ( S 2 2l + S l/2 ( 16t 1 ( l t t 2 l+1 such su-free -sets I [n] with S(I = l. Finally, suing over the various cases, the clai follows. Fro now on we shall assue that n 2 2δn. Recall that Clai 1 allows us to count su-free sets with at ost δ eleents less than. We shall use the induction hypothesis to count the sets I that have ore than δ eleents in []. Clai 3. There are at ost δ 2 Cn/( su-free -sets I [n] with at least δ eleents less than. 8 For siplicity, we assued here that a(i 4δ. However, if I contains an eleent b < 4δ then we can easily find a atching of pairs {x, x + b} of size 4δ in B.

20 20 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ Proof of Clai 3. Recall that C 1/3 n and that C = C(δ > 0 is sufficiently large. Thus, by Proposition 2.3, it follows that there exists ε = ε(δ > 0 such that all but 2 ε( su-free -sets I [n] satisfy either I \ O n δ 3, or I \ B δ 3 for soe interval B of length. Moreover, since δ > 0 is sufficiently sall, by Proposition 4.1 there are at ost 2 C( su-free -sets I [n] with I \ On δ 3. We ay therefore restrict our attention to the collection X of su-free -sets I [n] that satisfy I \ B δ 3 for soe interval B of length. First, we shall show that there are only few sets in X which contain ore than δ 3 eleents less than 2δ 2 n. Indeed, such a set contains at ost δ 3 eleents of the interval {n 2δ 2 n + 1,..., n} and hence, by the induction hypothesis and (3, there are at ost 2 C(n δ2 n/( s ( ( δ 2 n 2δ 2 n s s 2 Cn/( 1 2δ 2 ( s 2 n 2 s ( 2eδ 2 n s s ( such sets with s δ 3 eleents greater than n 2δ 2 n. Suing over s, and recalling that n 2 δn, it follows that there are at ost δ 3 ( s ( ( δ 2 Cn/ e 2δ2 ( s 2 δn 2eδ2 n δ 3 e δ2 4e 3 ( 2 Cn/ (20 s δ 2 s=0 such sets, which is at ost δ 2 2 Cn/(, since δ > 0 was chosen sufficiently sall and C 1/3 is sufficiently large. It only reains to count the sets in X which contain at least δ δ 3 > δ/2 eleents of the interval { 2δ 2 n,..., }, and at ost δ 3 eleents less than 2δ 2 n. Note that S + S 2 S 1 for every S Z and so, by Clai 1, there are at ost δ 3 a a=0 b=δ/2 ( ( 2δ 2 n + 1 a b ( δ a b such su-free sets. Note that 5δn (else there are no such sets, and so by (3 we have ( ( a b ( a+b ( ( a+b ( δ δ 3. a b n Hence (21 is at ost δ 3 a ( a ( 3e 7eδ 2 2a b a=0 b=δ/2 b ( so this copletes the proof of the clai. ( δ 3e 3 ( ( 2 δ/2 14eδ 2δ 3 (21 ( e δ, Since δ > 0 was arbitrary, we ay fro now on restrict our attention to those su-free subsets I [n] for which S(I δ 3. The reainder of the proof involves soe careful counting using Theores 1.3 and 1.4 and Lea 5.1. We shall break up the calculation into three clais. In the first two, which are fairly straightforward, we count the sets I for

21 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 21 which S(I is sall (Clai 4 or S(I is not sall but a S(I ( a is large (Clai 5. Finally in Clai 6, which is uch ore delicate, we count the reaining sets. Clai 4. If l (2Cδn/, then there are at ost 2 C( su-free -sets I [n] with S(I = l. Proof of Clai 4. Let k N and fix a set S S(k, l. By Clai 1, there are at ost { } ( C ax e k2 /2n 2, e k/8nl l su-free -sets I [n] with S(I = S. Recall that l δ 3 and n 2 δn, and suppose first that e k2 /2n 2 e k/8nl, i.e., that l n/4. By Lea 5.1, there are at ost ( e 2 k l l 2 choices for S, and hence, using (2, we can bound the nuber of sets I as follows: ( C e k2 /2n 2 ( e 2 l ( l ( k C e k2 /2n 2 2. (23 l l 2 n 2 k k S S(k,l Now, using (4 to bound the su over k, this is at ost ( e C 2 2 l 2 l ( 2n 2 l+1 ( ( 3 ( l ( l Cn 4en 2 δn e 2 δl ( 2 Cn/3, where in the last two steps we used the bound l n/4. Suppose next that e k2 /2n 2 e k/8nl, i.e., that n/4 l (2Cδn/. The calculation is alost the sae: k S S(k,l C e k/8nl ( l ( Cn 3 ( 4e δ k l ( ( e 2 k C l 2 l e k/8nl ( 2 n 2 ( (4e/δ ( 3δ Cn/ l ( (22 ( 2 Cn/3, (24 where we again used Lea 5.1, (2, (4 and the bound l (2Cδn/. Clai 5. If l n/4, then there are at ost e l( su-free -sets I [n] such that S(I S(k, l for soe k l 2 n/δ. Proof of Clai 5. The calculation is siilar to that in the previous clai. Indeed, note that (22 still holds, and that e k2 /2n 2 e k/8nl, since l n/4. Thus, by Lea 5.1 and (2, and since n 2 δn, ( e k/8nl ( e 2 k l l 2 l ( e k/8nl (25 2 δn kl 2 n/δ S S(k,l kl 2 n/δ is an upper bound on the nuber of su-free -sets I [n] such that S(I S(k, l for soe k l 2 n/δ.

22 22 NOGA ALON, JÓZSEF BALOGH, ROBERT MORRIS, AND WOJCIECH SAMOTIJ Now, note that k l 2 n/δ > 3l 8nl/ since δ > 0 is sufficiently sall. Therefore ( e 2 k l 2 l e k/8nl 16nl ( 2e 2 l e l/8δ e l, 2 δn δ 2 k=l 2 n/δ since g(x = x a e bx is decreasing on [a/b, and g(x + 1/b < g(x/2 if x > 3a/b. It follows that the right-hand side of (25 is bounded above by e l(, as claied. Since 3 < 2 e, the following clai now copletes the proof of Theore 1.1. Clai 6. If k l 2 n/δ and l (2Cδn/, then there are at ost ( ( l ( 3 2 O(δl 2 + e δ e su-free -sets I [n] with S(I S(k, l. This is the ost difficult case, and we shall have to count ore carefully, using Theore 1.3 (in the case = o(n and Theore 1.4 (in the case = Θ(n. Given a set S [], we write S = S [n/4] and S = S \ S. For siplicity, we shall fix integers k, l N and consider only sets in S = { S S(k, l : S S(k, l }. Note that suing over choices of k and l only costs us a factor of kl = O(l 4, which is absorbed by the error ter 2 O(δl. Set l = l l and k = k k, and observe that l = l l = S [n/4] 4k 4l2 δl, (27 n δ since k l 2 n/δ and l δ 3. We shall use this fact (that l l on nuerous occasions in the calculation below. Proof of Clai 6. We begin by slightly iproving the bound in Clai 1; this will allow us to deal with the case in which S is reasonably large. To be precise, we shall show that if n 2 δn and l δ 3, then there are at ost C S S ax { e δ S 2 /n, e δ } ( S + S l ( + e δ (26 (28 su-free -sets I [n] with S(I S. To prove (28, we shall partition S into two sets by setting S 1 = { S S : S + S δn }, and S 2 = S \ S 1. By Clai 1, and using (3, there are at ost ( ( δn ( 1 2δ ( l ( l ( l 2 en l l δn 2l choices for I with S(I S 1, since n 2 δn and l δ 3. ( e δ,

23 A REFINEMENT OF THE CAMERON-ERDŐS CONJECTURE 23 Suppose now that S S 2. Siilarly as in the proof of Clai 1, in order to bound the nuber of sets I with S(I = S we apply the Hypergeoetric Janson Inequality to the graph G with V (G = { + 1,..., n } \ (S + S and E(G = { {x, x + s}: s S }. Observe that k l n/4, and hence G has at least ( n k (2l S + S 2l 8 S + S l n 8 edges, and axiu degree at ost 2l. Hence, letting µ and to be the quantities defined in the stateent of Lea 4.3, we have µ l n 8 ( l2 l 2 ( 2 3n and n 2 (2l 2 ( 3 l 20(l 2 3. δn n 2 Thus µ/2 δl 2 /n and µ 2 /2 δ, and so (28 follows. To see why the bound in (28 is so useful, observe that ( S } ax {e δ S 2 /n, e δ 2 δl. (29 S Indeed, if l n/ then the left-hand side of (29 is at ost 1, since l δ 3. On the other hand, if l n/ then ( l } ( l ( l ax {e δl 2 /n, e δ = l l /n n/δ e δ2 e n/δ 2 δl, l since e δ2 /n n/δ, as in the proof of Proposition 4.1, and the function x (c/x x is axiized when x = c/e; the last inequality holds since l (2Cδn/. In order to coplete the calculation and obtain (26, we break into cases according to the order of agnitude of. We begin with the case δn. Case 1: C 1/3 n δn. We wish to bound (28 fro above; the key idea is to break up S according to the size of S +S. If it is very sall (at ost δ 3 k /l, then we easily obtain a bound via Theore 1.3, and if it is very large (at least (n/δ k /l, then the trivial bound of Lea 5.1 suffices. In the iddle range, we shall partition into sufficiently sall intervals (see (34, below; inside such an interval we have an upper bound on S + S, which allows us to apply Theore 1.3, and also a lower bound, which is useful when we apply (28. (a S + S δ 3 k /l. Let S a = { S S : S + S δ 3 k /l }. We shall first apply Theore 1.3 to bound the nuber of sets S S a ; we ay do so since (l 3 k l3 2k l δ 2n Cδ2, (30