The Erdős-Szemerédi problem on sum set and product set

Transcription

1 Annals of Mathematics, 157 (003), The Erdős-Szemerédi problem on sum set and product set By Mei-Chu Chang* Summary The basic theme of this paper is the fact that if A is a finite set of integers, then the sum and product sets cannot both be small A precise formulation of this fact is Conecture 1 below due to Erdős-Szemerédi [E-S] (see also [El], [T], and [K-T] for related aspects) Only much weaker results or very special cases of this conecture are presently known One approach consists of assuming the sum set A + A small and then deriving that the product set AA is large (using Freiman s structure theorem) (cf [N-T], [Na3]) We follow the reverse route and prove that if AA <c A, then A + A >c A (see Theorem 1) A quantitative version of this phenomenon combined with the Plünnecke type of inequality (due to Ruzsa) permit us to settle completely a related conecture in [E-S] on the growth in k If g(k) min{ A[1] + A{1} } over all sets A Z of cardinality A = k and where A[1] (respectively, A{1}) refers to the simple sum (resp, product) of elements of A (See (06), (07)) It was conectured in [E-S] that g(k) grows faster than any power of k for k We will prove here that ln g(k) result of this paper () ln Introduction (see Theorem ) which is the main Let A, B be finite sets of an abelian group The sum set of A, B is (01) A + B {a + b a A, b B} We denote by (0) ha A + + A (h fold) the h-fold sum of A *Partially supported by NSA

2 940 MEI-CHU CHANG (03) (04) Similarly we can define the product set of A, B and h-fold product of A AB {ab a A, b B}, A h A A (h fold) If B = {b}, asingleton, we denote AB by b A In 1983, Erdős and Szemerédi [E-S] conectured that for subsets of integers, the sum set and the product set cannot both be small Precisely, they made the following conecture Conecture 1 (Erdős-Szemerédi) For any ε>0 and any h N there is k 0 = k 0 (ε) such that for any A N with A k 0, (05) ha A h A h ε ( ) A + h 1 We note that there is an obvious upper bound ha A h h Another related conecture requires the following notation of simple sum and simple product { k } (06) A[1] ε i a i a i A, ε i =0 or 1, (07) i=1 { k } A{1} a ε i i a i A, ε i =0 or 1 i=1 For the rest of the introduction, we only consider A N Conecture (Erdős-Szemerédi) Let g(k) min A =k { A[1] + A{1} } Then for any t, there is k 0 = k 0 (t) such that for any k k 0,g(k) >k t Toward Conecture 1, all work has been done so far, are for the case h = Erdős and Szemerédi [E-S] got the first bound: Theorem (Erdős-Szemerédi) Let f(k) min A =k A A Then there are constants c 1,c, such that (08) k 1+c 1 <f(k) <k e c ln Nathanson showed that f(k) >ck 3 31, with c =00008 At this point, the best bound is (09) A A >c A 5/4 obtained by Elekes [El] using the Szemerédi-Trotter theorem on line-incidences in the plane (see [S-T])

3 THE ERDŐS-SZEMERÉDI PROBLEM 941 On the other hand, Nathanson and Tenenbaum [N-T] concluded something stronger by assuming the sum set is small They showed Theorem (Nathanson-Tenenbaum) If (010) A 3 A 4, then (011) A = ( ) A ln A Very recently, Elekes and Ruzsa [El-R] again using the Szemerédi-Trotter theorem, established the following general inequality Theorem (Elekes-Ruzsa) If A R is a finite set, then (01) A + A 4 AA ln A > A 6 In particular, their result implies that if (013) A <c A, then (014) A = A c ln A For further result in this direction, see [C] Related to Conecture, Erdős and Szemerédi [E-S] have an upper bound: Theorem (Erdős-Szemerédi) Let g(k) min A =k { A[1] + A{1} } There is a constant c such that () c (015) g(k) <e ln Our first theorem is to show that the h-fold sum is big, if the product is small Theorem 1 Let A N be a finite set If A <α A, then (016) A > 36 α A, and (017) ha >c h (α) A h Here (018) c h (α) =(h h) hα

4 94 MEI-CHU CHANG Our approach is to show that there is a constant c such that (019) eπimx h dx<c A h m A by applying an easy result of Freiman s theorem (see the paragraph after Proposition 10) to obtain { ( ) a 1 ( ) } s a1 as (00) A P 0 i <l i b b 1 and carefully analyzing the corresponding trigonometric polynomials (see Proposition 8) These are estimates in the spirit of Rudin [R] The constant c here depends, of course, on s and h In order to have a good universal bound c, weintroduce the concept of multiplicative dimension of a finite set of integers, and derive some basic properties of it (see Propositions 10 and 11) We expect more applications coming out of it Another application of our method together with a Plünnecke type of inequality (due to Ruzsa) gives a complete answer to Conecture Theorem Let g(k) min A =k { A[1] + A{1} } Then there is ε>0 such that (1+ε) (01) k ln b s >g(k) >k ( 1 ε) 8 ln Remark 1 (Ruzsa) The lower bound can be improved to k ( 1 ε) ln We will give more detail after the proof of Theorem Using a result of Laczkovich and Rusza, we obtain the following result related to a conecture in [E-S] on undirected graphs Theorem 3 Let G A A satisfy G >δ A Denote the restricted sum and product sets by (0) (03) A G + A = {a + a (a, a ) G} A G A = {aa (a, a ) G} If (04) A G A <c A, then (05) A + G A >C(δ, c) A

5 THE ERDŐS-SZEMERÉDI PROBLEM 943 The paper is organized as follows: In Section 1, we prove Theorem 1 and introduce the concept of multiplicative dimension In Section, we show the lower bound of Theorem and Theorem 3 In Section 3, we repeat Erdős-Szemerédi s upper bound of Theorem Notation We denote by a the greatest integer a, and by A the cardinality of a set A Acknowledgement The author would like to thank J Bourgain for various advice, and I Ruzsa and the referee for many helpful comments 1 Proof of Theorem 1 Let A N be afinite set of positive integers, and let Γ h,a (n) bethe number of representatives of n by the sum of h (ordered) elements in A, ie, (11) Γ h,a (n) {(a 1,,a h ) a i = n, a i A} The two standard lemmas below provide our starting point Lemma 3 Let A N be finite and let h N If there is a constant c such that (1) Γ h,a (n) <c A h, then n ha (13) ha > 1 c A h Proof Cauchy-Schwartz inequality and the hypothesis give A h = ( ) 1/ Γ h,a (n) ha 1/ Γ h,a (n) n ha n ha < ha 1/ c 1/ A h/ Lemma 4 The following equality holds: ( Γ h,a (n) = ) h e πimx h m A n ha

6 944 MEI-CHU CHANG Proof e πimx m A h h h = e πimx dx = m A ( ) h e πimx m A dx The last equality is Parseval equality ( ) = Γ h,a (n)e πinx n ha = Γ h,a (n) n ha From Lemmas 3 and 4, it is clear that to prove Theorem 1, we want to find a constant c such that ( ) (14) e πimx <c A m A h In fact, we will prove something more general to be used in the inductive argument Proposition 5 Let A N be a finite set with A <α A Then for any {d a } a A R +, (15) d a e πiax <c d a a A h for some constant c depending on h and α only For a precise constant c, see Proposition 9 The following proposition takes care of the special case of (15) when there exists a prime p such that for every nonnegative integer, p appears in the prime factorization of at most one element in A Itisalso the initial step of our iteration First, for convenience, we use the following: Notation We denote by G +, the set of linear combinations of elements in G with coefficients in R + dx

7 THE ERDŐS-SZEMERÉDI PROBLEM 945 Proposition 6 Let p be a fixed prime, and let { (16) F (x) e πip nx + n N, (n, p) =1} Then (17) F h c h F, where c h =h h h Proof To bound F h dx, weexpand F h as (18) ( F ) h ( F ) h Let (19) F 1 F h F h+1 F h be aterm in the expansion of (18) After rearrangement, we may assume 1 h, and h+1 h When (19) is expressed as a linear combination of trignometric functions, atypical term is of the form (110) ne πix(p 1n 1 + +p hn h p h+1n h+1 p hn h ) We note that the integral of (110) is 0, if the expression in the parenthesis in (110) is nonzero In particular, independent of the n i s, the integral of (110) is 0, if (111) 1 h+1, or 1 h+1 min{, h+ }, or h+1 h+ 1 Therefore, if any of the statements in (111) is true, then the integral of (19) is 0 We now consider the integral of (19) where the index set { 1,, h } does not satisfy any of the conditions in (111) For the case 1 = h+1,wesee that in an ordered set of h elements coming from the expansion of (18) (before the rearrangement), there are exactly ( h ) choices for the positions of 1, On the other hand, if F 1 F is factored out, the rest is symmetric with respect to 3,, h, and h+1,, h, ie, all the terms involving 1 = h+1 are simplified to (11) ( ) h (F ) F k k With the same reasoning for the other two cases, we conclude that h

8 946 MEI-CHU CHANG F h The right-hand side is [ ( )] h h + h ( ) h = F h F k h F k dx k k + h F h 1 F k F k dx k k ( ) h + F h F k h F k dx k k F F k k h h (h h) F h F k h h 1 k =(h h) F h h F k h k The last inequality is Hölder inequality Now, the next lemma concludes the proof of Proposition 6 Lemma 7 Let F k {e πimkx m k Z} + Then (113) F k F k, for any h Proof h F k dx k = k F k + k k< F k k h F k + F k F k + F k F k + F k dx k k< k k< k k< F k k k F k h dx = F k h k h dx

9 THE ERDŐS-SZEMERÉDI PROBLEM 947 The inequality holds because the coefficients of the trignometric functions (as in (110)) in the expansion are all positive Remark 71 theory This is a special case of a general theorem in martingale Proposition 8 Let p 1,,p t be distinct primes, and let { (114) F 1,, t (x) e πip 1 1 p t t nx + n N, (n, p 1 p t )=1} Then (115) 1,, t F 1,, t h c t h F 1,, t h, where c h =h h 1,, t Proof We do induction on t The left-hand side of (115) becomes F 1,, t c h F 1,, t c h c t 1 h F 1,, t, 1,, t 1,, t 1,, t which is the right-hand side Proposition 5 is proved, if we can find a small t such that the Fourier transform of F 1,, t is supported at one point and such t is bounded by α So we introduce the following notion Definition Let A be afinite set of positive rational numbers in lowest terms (cf (00)) Let q 1,,q l be all the prime factors in the obvious prime factorization of elements in A For a A, let a = q 1 q l l be the prime factorization of a Then the map ν : A R l by sending a to ( 1,, l )is one-to-one The multiplicative dimension of A is the dimension of the smallest (affine) linear space in R l containing ν(a) We note that for any nonzero rational number q, q A and A have the same multiplicative dimension, since ν(q A) is a translation of ν(a) The following proposition is a more precise version of Lemma 5 Proposition 9 Let A N be finite with multdim(a) =m Then ( ) (116) d a e πiax <c m h d a, where c h =h h a A h Proof To use (115) in Proposition 8, we want to show that there are primes q 1,,q m such that a term of the trigonometric polynomial in the

10 948 MEI-CHU CHANG left-hand side of (115), when expressed in terms of the notation in (114), is F 1,, m = d a e πiq 1 1 qm m nx In other words, we want to show that among the prime factors q 1,,q l of elements in A, there are m of them, say q 1,,q m such that ( ) for all ( 1,, m ) Z m, there is at most one a A such that q 1 1 qm m is part of the prime factorization of a This is equivalent to ( ) π ν is inective, where ν is as in the definition of multiplicative dimension and π : R l R m is the proection to the first m coordinates Since dim ν(a) =m, ( ) isclear after some permutation of the q i s Proposition 10 Let A N be finite with multdim A = m Then (117) Γ h,a (n) <cmh h A h, where c h =h h n ha Proof This is a consequence of Lemma 4 and Proposition 9 (with d a = 1) The hypothesis of Theorem 1 gives a universal bound on the multiplicative dimension of A by applying Freiman s theorem (cf [Fr1], [Fr], [Fr3], [Bi], [C1], [Na1]) In fact, we do not need the full content of the Freiman s theorem, but a much easier result by Freiman A small modification (over Q instead of over R) of Lemma 43 in [Bi] is sufficient (As Ruzsa pointed out it is also Lemma 114 in [Fr1]) Theorem (Freiman) Let G R be asubgroup and A 1 G be finite If there is a constant α, α< A 1, such that A 1 <α A 1, then there is an integer s α such that A 1 is contained in an s-dimensional proper progression P 1 ; ie, there exist β,α 1,,α s G and J 1,,J s N such that and P 1 = J 1 J s A 1 P 1 = {β + 1 α s α s 0 i <J i }, Note that if A 1 > α α+1 ( α+1 α), then s α 1 Recall that the full Freiman theorem also permits one to state a bound J 1 J s <c(α) A 1 However this additional information will not be used in what follows We would like to work on a sum set instead of a product set So we define (118) A 1 ln A = {ln a a A} Note that ln is an isomorphism between the two groups (Q +, ) and (ln Q +, +)

11 THE ERDŐS-SZEMERÉDI PROBLEM 949 Applying the theorem to A 1 ln Q +, then pushing back by (ln) 1,we have { a (119) A P b (a 1 ) 1 ( a } s ) s 0 i <J i Q +, b 1 b s where a, b, a i,b i,j i N, and (a, b) =1, (a i,b i )=1 Moreover, s α 1 and different ordered sets ( 1,, s ) represent different rational numbers Clearly, (10) mult dim A multdim P = dim E s α 1, where E is the vector space generated by ν( a 1 b 1 ),,ν( as b s ) Therefore, we have Proposition 11 Let A N be a finite set If A <α A for some constant α, α< A 1/, then multdim A α Furthermore, if A > α α+1 ( α+1 α), then multdim A α 1 Putting Propositions 10 and 11 together, we have Proposition 1 Let A N be finite If A <α A for some constant α, α< A 1/, then Γ h,a (n) <cαh h A h, where c h =h h n ha Now, Theorem 1 follows from Proposition 1 and Lemma 3 Simple sums and products In this section we will prove the lower bound in Theorem Let A N be finite We define (1) g(a) A[1] + A{1}, where A[1] and A{1} are the simple sum and simple product of A (See (06), (07) for precise definitions) We will show that for any ε and any A N with A = k 0, () g(a) >k ( 1 ε) 8 ln For those who like precise bounds, we show: For 0 <ε 1,ε < 1, (3) g(a) >e 3 k 1 ε ( 1 4 ε 1 ) ln,

12 950 MEI-CHU CHANG if A = k is large enough such that (4) ln >, 8ε 1 and (5) ln > ε Proposition 13 Let B N be finite with multdim B = m Then for any h 1 N, [ ] B h1 (6) h 1 B B[1] > (h 1 h 1) m+1 Proof Since h 1 B B[1] is the set of simple sums with exactly h 1 summands, we have ( ) B (7) Γ h1,b(n) Therefore, h 1 n h 1 B B[1] ( ) B h1 < (h 1 B B[1]) 1/ h 1 n h 1 B Γ h 1,B (n) 1/ (h 1 B B[1]) 1/ [(h 1 h 1 ) mh 1 B h 1 ] 1/ The first inequality is because of the Cauchy-Schwartz inequality and the fact that h 1 B B[1] h 1 B The second inequality is Proposition 10 Remark 131 Clearly, from our proof, the denominator in (6) can be replaced by (h 1 h 1) m h 1 Proposition 14 Let B N with B k and multdim B = m For any 0 <ε 1 < 1, if ( 1 (8) m +1 4 ε ) 1 ln, then (9) g(b) >k ε 1 ln k Proof Inequality (8) is equivalent to (10) () m+ k 1/ ε 1

13 THE ERDŐS-SZEMERÉDI PROBLEM 951 In Proposition 13, we take h 1 = This gives (11) h 1 () Combining (11), (10) and (6), we have ( k 1/ g(b) > h 1 B B[1] > k 1/ ε 1 ) = k ε 1 Remark 141 Let A N with A = k, k 0 (see (4)) The set B in Proposition 14 will be taken as a subset of A Then the bound in (9) is bigger than that in (), and our proof is done Therefore for the rest of the section, we assume (1) multdimb ( 1 4 ε ) ln, for any B A with B > k We need the following: Notation We denote B ν(b) for any B A, where ν = A Z l is as in the definition of multiplicative dimension Note that (13) B [1] = B{1} We will use the following Plünnecke type of inequality due to Ruzsa Ruzsa s Inequality [Ru] For any h, l N: If M + N ρ M, then hn ln ρ h+l M Proof of () We divide A into k pieces B 1,B,, each of cardinality at least kfor0<ε < 1, let (14) ρ =1+k 1/+ε, and let (15) A s There are two cases: s B i i=1 (i) For all s, (A s Bs+1 ) [1] >ρ A s[1] Iterating gives (16) A [1] = (B 1 B ) [1] >ρ k k

14 95 MEI-CHU CHANG Therefore (17) g(a) > A{1} = A [1] >e ( k )ln ρ+ 1 >e ( k ) 4 5 k 1/+ε + 1 >e 4 5 kε Inequality (5) is equivalent to k ε > () which is certainly stronger than what we need to show () (ii) There exists s such that (A s B s+1 ) [1] ρ A s[1] We use the fact that (A s B s+1 ) [1] = A s[1] + B s+1 [1], and Ruzsa s inequality (with h = h +1,l=1)toobtain (18) (h +1)B s+1[1] B s+1[1] ρ h + A s[1] Let m =multdimb s+1 Foraset B, for h N, denote (19) B[h] { εi x i ε i =0,,h,x i B The left-hand side of (18) is (0) (1) h B s+1[1] B s+1[h ] h m We take h = k 1/ ε Then in the right-hand side of (18), ρ h + (1 + k 1/+ε ) k1/ ε + < (e k 1/+ε ) k1/ ε + <e 3 Therefore, (18), (0) and (1) imply g(a) >g(a s ) > A s[1] >e 3 h m >e 3 k 1/ ε ( 1 4 ε 1 ) ln The last inequality follows from our choice of h and Remark 141 Proof of Remark 1 In Proposition 14, if we take B with B k, then we will replace (8), and (9) by (8 ) m +1 1 (1 ε 1) ln, }

15 THE ERDŐS-SZEMERÉDI PROBLEM 953 and ( k (9 ε 1 ) g(b) > Let m 0 = 1 (1 ε 1) Then (1) will be replaced by ) ln (1 ) multdimb m 0, for any B A with B > k Now we modify the proof of () Since A = k > k, wehave multdima m 0 So there is B 1 A with multdimb 1 = m 0 and B 1 = m 0 +1 Similarly, we have B A B 1 with multdimb = m 0 and B = m 0 +1 We continue this process until k r> (m 0 +1) Wehave A B 1 B r with and With more replacements, (5 ) and multdimb i = m 0, B i = m 0 +1 ln > 3 ε (14 ) ρ =1+k 1+ε, Identical arguments give g(a) >e 3 k 1 ε ( 1 ε 1 ) ln Sketch of Proof of Theorem 3 Let A = N Then the Laczkovich-Ruzsa theorem [L-R] and (04) give A 1 A with () A 1 A 1 <c N, and (3) G (A 1 A 1 ) >δ N The weak Freiman theorem and () imply (4) multdima 1 <c

16 954 MEI-CHU CHANG It follows from Proposition 10 (with h =)and the proof of Lemma 4 that (5) β {(n 1,n,n 3,n 4 ) A 4 1 n 1 n + n 3 n 4 =0} < 36 c N Hence (6) δ N < G {(n 1,n ) A1 n = n 1 + n } < A 1 + A1 1 1 β G n A 1+A1 The first inequality is (3), while the second one is the Cauchy-Schwartz inequality Therefore, (5) and (6) give A G + A A 1 G + A1 (δ ) N 4 β >CN 3 The example In this section for completeness we repeat a family of examples by Erdős- Szemerédi which provide the upper bound in Theorem Precisely, we will show Proposition 15 Given ε 3 > 0, for J so large that ln J (31) ln ln J > 1, ε 3 there is a set A of cardinality A = k J J, such that (1+ε) (3) g(a) < k ln, where (33) ε =3ε 3 + ε 3 The example really comes from the proof of the lower bound of Theorem Let p 1,,p J be the first J primes, and let { } (34) A p 1 1 p J J 0 i <J Then (35) k A = J J We will use the following relations between k and J

17 THE ERDŐS-SZEMERÉDI PROBLEM 955 Lemma 16 Let k, J be as in (35) Then (i) = Jln J (ii) ln =lnj +lnlnj If J and ε 3 satisfy (31), then (iii) ln <(1 + ε 3 )ln J (iv) J<(1 + ε 3 ) ln (v) J < (1 + ε )( ln ), where ε =ε 3 + ε 3 Proof Each one follows immediately from the preceding one implying (iv), we use J = ln J For (iii) Remark 161 The inequality ln J ln ln J > 1 ε 3 clearly implies (36) ln > 1 ε 3 Lemma 17 (i) For all a A, a<() J, (ii) A[1] <k() J, (iii) A{1} < (kj) J Proof (i) For a A, (34) gives ( ) J ( a< p i < iln i i<j i<j ) J < ( J J (ln J) J) J =(Jln J) J = () J The second inequality is by the Prime Number Theorem The last equality is Lemma 16 (i) (ii) follows from (i) (iii) We see that { k s=1 (37) A{1} = p (s) 1 1 p k s=1 (s) J J } 0 (s) i <J Since k s=1 (s) i <kj, (iii) holds

18 956 MEI-CHU CHANG (38) Proof of Proposition 15 Lemma 17 (ii) and Lemma 16 (v) give A[1] <k() (1+ε )( ln ) = e +(1+ε () ) ln = e (1+(1+ε ) ln ) (1+ε) <e ln (1+ε) = k ln Here ε = ε + ε 3 =3ε 3 + ε 3 Weuse (36) for the last inequality (39) Lemmas 17 (iii), 16 (iv), and (35) give The last inequality is again by (36) A{1} <k J k = k J+1 <k (1+ε 3) ln +1 <k (1+ε 3) ln (1+ε) Putting (38) and (39) together, we have g(a) < k ln University of California, Riverside, CA address: mcc@mathucredu References [Bi] Y Bilu, Structure of sets with small sumset, in Structure Theory of Set Addition, Astérisque 58 (1999), [C1] M-C Chang, Apolynomial bound in Freiman s theorem, Duke Math J 113 (00), [C],Factorization in generalized arithmetic progressions and applications to the Erdös-Szemerédi sum-product problems, GAFA, toappear [El] G Elekes, Onthe number of sums and products, Acta Arith 81 (1997), [El-R] [E] G Elekes and I Ruzsa, Product sets are very large if sumsets are very small, preprint P Erdős, Problems and results on combinatorial number theory III, in Number Theory Day, 43 7 (Proc Conf Rockefeller Univ, New York, 1976), Lecture Notes in Math 66, Springer-Verlag, New York, 1977 [E-S] P Erdős and E Szemerédi, On sums and products of integers, Studies in Pure Mathematics, Birkhäuser, Basel, 1983, [Fr1] G A Freiman, Foundations of a Structural Theory of Set Addition, Transl of Math Monographs 37, A M S, Providence, RI, 1973 [Fr], Onthe addition of finite sets I, Izv Vysh Ucheb Zaved Matematika 13 (1959), 0 13 [Fr3], Inverse problems of additive number theory VI On the addition of finite sets III, Izv Vysh Ucheb Zaved Matematika 8 (196), [H-T] R R Hall and G Tenenbaum, Divisors, Cambridge Tracts in Math 90, Cambridge Univ Press, Cambridge, 1988 [K-T] N Katz and T Tao, Some connections between Falconer s distance set conecture and sets of Furstenberg type, New York J Math 7 (001),

19 THE ERDŐS-SZEMERÉDI PROBLEM 957 [L-R] M Laczkovich and I Z Ruzsa, The number of homothetic subsets, in The Mathematics of P Erdős, II (R L Graham and J Nesetril, eds), Springer-Verlag, New York, 1977, [Na1] M B Nathanson, Additive Number Theory Inverse Problems and the Geometry of Sumsets, Grad Texts in Math 165, Springer-Verlag, New York, 1996 [Na], The simplest inverse problems in additive number theory, in Number Theory with an Emphasis on the Markloff Spectrum (Provo, UT, 1991), , Marcel Dekker, New York, 1993 [Na3], Onsums and products of integers, Proc Amer Math Soc 15 (1997), 9 16 [N-T] M Nathanson and G Tenenbaum, Inverse theorems and the number of sums and products, in Structure Theory of Set Addition, Astérisque 58 (1999), [P] H Plünnecke, Eine zahlentheoretische Anwendung der Graphtheorie, J Reine Angew Math 43 (1970), [R] W Rudin, Trigonometric series with gaps, J Math Mech 9 (1960), 03 7 [Ru] I Z Ruzsa, Generalized arithmetical progressions and sumsets, Acta Math Hungar 65 (1994), [Ru], Sums of finite sets, in Number Theory (New York, ), Springer- Verlag, New York, 1996 [S-T] E Szemerédi and W Trotter, Extremal problems in discrete geometry, Combinatorica 3 (1983), [T] T Tao, From rotating needles to stability of waves: emerging connections between combinatorics, analysis, and PDE, Notices Amer Math Soc 48 (001), (Received November 7, 001)