Non-degenerate Hilbert Cubes in Random Sets

Transcription

1 Journal de Théore des Nombres de Bordeaux 00 (XXXX), Non-degenerate Hlbert Cubes n Random Sets par Csaba Sándor Résumé. Une légère modfcaton de la démonstraton du lemme des cubes de Szemeréd donne le résultat plus précs suvant: s une parte S de {1,..., n} vérfe S n, alors S content un cube de Hlbert non dégénéré de dmenson log log n 3. Dans cet artcle nous montrons que dans un ensemble aléatore avec les probabltés Pr{s S} = 1/ ndépendantes pour 1 s n, la plus grande dmenson d un cube de Hlbert non dégénéré est proche de log log n+log log log n presque sûrement et nous détermnons la foncton seul pour avor un k-cube non dégénéré. Abstract. A slght modfcaton of the proof of Szemeréd s cube lemma gves that f a set S [1, n] satsfes S n, then S must contan a non-degenerate Hlbert cube of dmenson log log n 3. In ths paper we prove that n a random set S determned by Pr{s S} = 1 for 1 s n, the maxmal dmenson of non-degenerate Hlbert cubes s a.e. nearly log log n + log log log n and determne the threshold functon for a non-degenerate k-cube. 1. Introducton Throughout ths paper we use the followng notatons: let [1, n] denote the frst n postve ntegers. The coordnates of the vector A = (a 0, a 1,..., a k ) are selected from the postve ntegers such that k =0 a n. The vectors B, A are nterpreted smlarly. The set f(n) S n s a subset of [1, n]. The notatons f(n) = o(g(n)) means lm g(n) = 0. An arthmetc progresson of length k s denoted by AP k. The rank of a matrx A over the feld F s denoted by r F (A). Let R denote the set of real numbers and F for the fnte feld of order. Let n be a postve nteger, 0 p n 1. The random set S(n, p n ) s the random varable takng ts values n the set of subsets of [1, n] wth the law determned by the ndependence of the events {k S(n, p n )}, 1 k n wth the probablty Pr{k S(n, p n )} = p n. Ths model s often used for provng the exstence of certan sequences. Gven any combnatoral number theoretc property P, there s a probablty that S(n, p n ) satsfes P, whch we wrte Pr{S(n, p n ) = P }. The functon r(n) s called a threshold functon for a combnatoral number theoretc property P f () When p n = o(r(n)), lm Pr{S(n, p n ) = P } = 0, Supported by Hungaran Natonal Foundaton for Scentfc Research, Grant No. T and

2 Csaba Sándor () When r(n) = o(p(n)), lm Pr{S(n, p n ) = P } = 1, or vsa versa. It s clear that threshold functons are not unque. However, threshold functons are unque wthn factors m(n), 0 < lm nf m(n) lm sup m(n) <, that s f p n s a threshold functon for P then p n s also a threshold functon ff p n = O(p n) and p n = O(p n ). In ths sense we can speak of the threshold functon of a property. We call H [1, n] a Hlbert cube of dmenson k or, smply, a k-cube f there s a vector A such that k H = H A = {a 0 + ɛ a : ɛ {0, 1}}. The postve ntegers a 1,..., a k are called the generatng elements of the Hlbert cube. The k-cube s non-degenerate f H = k.e. the vertces of the cube are dstnct, otherwse t s called degenerate. The maxmal dmenson of a non-degenerate Hlbert cube n S n s denoted by H max (S n ),.e. H max (S n ) s the largest nteger l such that there exsts a vector A (l,n) for whch the non-degenerate Hlbert cube H A (l,n) S n. Hlbert orgnally proved that f the postve ntegers are colored wth fntely many colors then one color class contans a k-cube. The densty verson of theorem was proved by Szemeréd and has snce become known as Szemeréd s cube lemma. The best known result s due to Gunderson and Rödl (see [3]): Theorem 1.1 (Szemeréd). For every d 3 there exsts n 0 ( d / ln ) so that, for every n n 0, f A [1, n] satsfes A n 1 1 d 1, then A contans a d-cube. =1 A drect consequence s the followng: Corollary 1.. Every subset S n such that S n n contans a log log n -cube. A slght modfcaton of the proof gves that the above set S n must contan a non-degenerate log log n 3 -cube. Obvously, a sequence S has the Sdon property (that s the sums s + s, s s, s, s S are dstnct) ff S contans no -cube. Godbole, Janson, Locantore and Rapoport studed the threshold functon for the Sdon property and gave the exact probablty dstrbuton n 1999 (see []): Theorem 1.3 (Godbole, Janson, Locantore and Rapoport). Let c > 0 be arbtrary. Let P denote the Sdon property. Then wth p n = cn 3/4, lm Pr{S(n, p n) = P } = e c 4 1. Clearly, a subset H [1, n] s a degenerate -cube ff t s an AP 3. Moreover, an easy argument gves that the threshold functon for the event AP 3 -free s p n = n /3. Hence Corollary 1.4. Let c > 0 be arbtrary. Then wth p n = cn 3/4, lm Pr{S(n, p n) contans no non-degenerate -cube} = e c 4 1.

3 Non-degenerate Hlbert Cubes n Random Sets 3 In Theorem 1.5 we extend the prevous Corollary. Theorem 1.5. For any real number c > 0 and any nteger k, f p n = cn k+1 k then lm Pr{S(n, p n) contans no non-degenerate k-cube} = e c k (k+1)!k!. In the followng we shall fnd bounds on the maxmal dmenson of non-degenerate Hlbert cubes n the random subset S(n, 1 ). Let D n (ɛ) = log log n + log log log n + (1 ɛ) log log log n log log log n and E n (ɛ) = log log n + log log log n + (1 + ɛ) log log log n. log log log n The next theorem mples that for almost all n, H max (S(n, 1 )) concentrates on a sngle value because for every ɛ > 0, D n (ɛ) = E n (ɛ) except for a sequence of zero densty. Theorem 1.6. For every ɛ > 0 lm Pr{D n(ɛ) H max (S(n, 1 )) E n(ɛ)} = 1.. Proofs In order to prove the theorems we need some lemmas. Lemma.1. For k n = o( log n log log n ) the number of non-degenerate k n-cubes n [1, n] s (1 + o(1)) ( ) n 1 k n+1 k n!, as n. Proof. All vectors A (k n,n) are n 1-1 correspondence wth all vectors (v 0, v 1,..., v kn ) wth 1 v 1 < v < < v kn n n R kn+1 accordng to the formulas (a 0, a 1,..., a kn ) (v 0, v 1,..., v kn ) = (a 0, a 0 +a 1,..., a 0 +a 1 + +a kn ); and (v 0, v 1,..., v kn ) (a 0, a 1,..., a kn ) = (v 0, v 1 v 0,..., v kn v kn 1). Consequently, ( ) n = {A (kn,n) : H k n + 1 A (kn,n) s non-degenerate} + {A (kn,n) : H A (kn,n) s degenerate}. By the defnton of a non-degenerate cube we have {A (k n,n) : H A (kn,n) s non-degenerate} = k n! {non-degenerate k n -cubes n [1, n]}, because permutatons of a 1,..., a k gve the same k n -cube. It remans to verfy that the number of vectors A (kn,n) whch generate degenerate k n -cubes s o( ( n k n +1) ). Let A (k n,n) be a vector for whch H A (kn,n) s a degenerate k n -cube. Then there exst ntegers 1 u 1 < u <... < u s k n, 1 v 1 < v <... < v t k n such that a 0 + a u a us = a 0 + a v a vt, where we may assume that the ndces are dstnct, therefore s + t k n. Then the equaton x 1 + x x s x s+1... x s+t = 0

4 4 Csaba Sándor can be solved over the set {a 1, a..., a kn }. The above equaton has at most n s+t 1 n k n 1 solutons over [1, n]. Snce we have at most kn possbltes for (s, t) and at most n possbltes for a 0, therefore the number of vectors A (k n,n) for whch H A (kn,n) s degenerate s at most knn kn = o( ( n k n+1) ). In the remanng part of ths secton the Hlbert cubes are non-degenerate. The proofs of Theorem 1.5 and 1.6 wll be based on the followng defnton. For two ntersectng k-cubes H A, H B let H A H B = {c 1,..., c m } wth c 1 <... < c m, where k k c d = a 0 + α d,l a l = b 0 + β d,l b l, α d,l, β d,l {0, 1} for 1 d m and 1 l k. The rank of the ntersecton of two k-cubes H A, H B s defned as follows: we say that, H B f for the matrces A = (α d,l ) m k, B = (β d,l ) m k we have r R (A) = s and r R (B) = t. The matrces A and B are called matrces of the common vertces of H A, H B. Lemma.. The condton, H B ) = (s, t) mples that mn{s,t}. H B Proof. We may assume that s t. The nequalty H B s s obvously true for s = k. Let us suppose that s < k and the number of common vertces s greater than s. Then the correspondng (0 1)-matrces A and B have more than s dfferent rows, therefore r F (A) > s, but we know from elementary lnear algebra that for an arbtrary (0 1)-matrx M we have r F (M) r R (M), whch s a contradcton. Lemma.3. Let us suppose that the sequences A and B generate non-degenerate k-cubes. Then (1) {(A, B ( ) :, H B ) = (s, t)} k n ) k+1 n k+1 max{s,t} for all 0 s, t k; () {(A, B ) :, H B ) = (r, r), H B = r ( } k n ) k+1 n k r for all 0 r < k; (3) {(A, B ) :, H B ) = (k, k), H B > k 1 } k +k ( n k+1). Proof. (1): We may assume that s t. In ths case we have to prove that the number of correspondng pars (A, B ) s at most ( n k+1) k n k+1 t. We have already seen n the proof of Lemma 1 that the number of vectors A s at most ( n k+1). Fx a vector A and count the sutable vectors B. Then the matrx B has t lnearly ndependent rows, namely r R ((β d,l) t k ) = t, for some 1 d 1 < < d t m, where k k a 0 + α d,la l = b 0 + β d,lb l, α d,l, β d,l {0, 1} for 1 t.

5 Non-degenerate Hlbert Cubes n Random Sets 5 The number of possble b 0 s s at most n. For fxed b 0, α d,l, β d,l let us study the system of equatons k k a 0 + α d,la l = b 0 + β d,lx l, α d,l, β d,l {0, 1} for 1 t. The assumpton r R (β d,l) t k = t mples that the number of solutons over [1, n] s at most n k t. Fnally, we have at most kt possbltes on the left-hand sde for α d,ls and, smlarly, we have at most kt possbltes on the rght-hand sde for β d,ls, therefore the number of possble systems of equatons s at most k (): The number of vectors A s ( n k+1) as n Part 1. Fx a vector A and count the sutable vectors B. It follows from the assumptons, H B ) = (r, r), (k) H B (k) = r that the vectors (α d,1,..., α d,k ), d = 1,..., r and the vectors (β d,1,..., β d,k ), d = 1,..., r, respectvely form r-dmensonal subspaces of F k. Consderng the zero vectors of these subspaces we get a 0 = b 0. The ntegers b 1,..., b k are solutons of the system of equatons k k a 0 + α d,l a l = b 0 + β d,l x l α d,l, β d,l {0, 1} for 1 d r. Smlarly to the prevous part ths system of equaton has at most n k r solutons over [1, n] and the number of choces for the r lnearly ndependent rows s at most k. (3): Fx a vector A. Let us suppose that for a vector B we have, H B ) = (k, k) and H B > k 1. Let the common vertces be a 0 + k α d,l a l = b 0 + k β d,l b l, α d,l, β d,l {0, 1} for 1 d m, where we may assume that the rows d 1,..., d k are lnearly ndependent,.e. the matrx B k = (β d,l) k k s regular. Wrte the rows d 1,..., d k n matrx form as (1) a = b B k b, wth vectors a = (a 0 + k α d,la l ) k 1, 1 = (1) k 1 and b = (b ) k 1. It follows from (1) that b = B 1 k (a b 01) = B 1 k a b 0B 1 k 1. Let B 1 k 1 = (d ) k 1 and B 1 k a = (c ) k 1. Obvously, the number of subsets { 1,... l } {1,..., k} for whch d d l 1 s at least k 1, therefore there exst 1 u 1 <... < u s k and 1 v 1 <... < v t k such that a 0 +a u a us = b 0 +b v b vt, and d v d vt 1. Hence a 0 + a u a us = b 0 + b v b vt = b 0 + c v c vt b 0 (d v d vt ) b 0 = a 0 + a u a us c v1... c vt. 1 (d v d vt ) To conclude the proof we note that the number of sets {u 1,..., u s } and {v 1,..., v t } s at most k and there are at most k choces for B k and a, respectvely. Fnally, for gven B k, a, b 0, 1 u 1 <... < u s k and 1 v 1 <... < v t k, the vector B s determned unquely. In order to prove the theorems we need two lemmas from probablty theory (see e.g. [1] p. 41, ). Let X be the ndcator functon of the event A and S n = X X N. For ndces

6 6 Csaba Sándor, wrte f and the events A, A are depandant. We set Γ = Pr{A A } (the sum over ordered pars). Lemma.4. If E(S n ) and Γ = o(e(s n ) ), then S n > 0 a.e. In many nstances, we would lke to bound the probablty that none of the bad events B, I, occur. If the events are mutually ndependent, then Pr{ I B } = I Pr{B }. When the B are mostly ndependent, the Janson s nequalty allows us, sometmes, to say that these two quanttes are nearly equal. Let Ω be a fnte set and R be a random subset of Ω gven by Pr{r R} = p r, these events beng mutually ndependent over r Ω. Let E, I be subsets of Ω, where I a fnte ndex set. Let B be the event E R. Let X be the ndcator random varable for B and X = I X be the number of E s contaned n R. The event I B and X = 0 are then dentcal. For, I, we wrte f and E E. We defne = Pr{B B }, here the sum s over ordered pars. We set M = I Pr{B }. Lemma.5 (Janson s nequalty). Let ε ]0, 1[, let B, I,, M be as above and assume that Pr{B ) ε for all. Then M Pr{ I B ) Me 1 1 ε. Proof of Theorem 1.5. Let H A,..., H 1 A N be the dstnct non-degenerate k-cubes n [1, n]. S(n, cn k+1 k ). Then Pr{B } = c k n (k+1) = o(1) and N = Let B be the event H A (1 + o(1)) ( n k+1) 1 k!. It s enough to prove snce then Janson s nequalty mples = Pr{B B } = o(1) Pr{S(n, cn k+1 k ) does not contan any k-cubes} = Pr{ N =1B } = (1 + o(1))(1 (cn k+1 k ) k ) (1+o(1))( k+1) n 1 k! = (1 + o(1))e c k (k+1)!k!. It remans to verfy that Pr{B B } = o(1). We splt ths sum accordng to the ranks n the followng way k k Pr{B B } = Pr{B B } = k 1 r=0,h A H A k s 1 )=(r,r) = r s=0 t=0,h A k 1 Pr{B B } +,H A r=1 Pr{B B }+,H A H A )=(r,r) < r Pr{B B }+

7 ,H A H A Non-degenerate Hlbert Cubes n Random Sets 7 )=(k,k) k 1 Pr{B B } + The frst sum can be estmated by Lemmas. and.3 (3) k s 1,H A n o(1) k s=1 Pr{B B } k s 1,H A H A )=(k,k) > k 1 ( ( n )n k k+1 s k + 1 k+1 s 1 n k n s = n o(1) (n k+1 k 1 + n k+1 k ) = o(1), Pr{B B }. c n k+1 k ) k t = snce the sequence a s = s 1 k+1 s s decreasng for 1 s k log k (k + 1) + 1 and ncreasng for k log (k + 1) + 1 < s k. To estmate the second sum we apply Lemma.3 () k 1 r=0,h A H A )=(r,r) = r k 1 ( ) n Pr{B B } k n k r c ( ) k + 1 n k+1 k r=0 k r = k 1 n 1+o(1) r=0 The thrd sum can be bounded usng Lemma.3 (1) k+1 r n k n r = n 1+o(1) (n k+1 k + n k+1 (k 1) ) = o(1). k 1 r=1,h A H A )=(r,r) < r k+1 k 1 o(1) n k r=1 k 1 ( ) n Pr{B B } k k + 1 k+1 r n k n r r=1 n k+1 r ( = n o(1) k+1 k (n k+1 k 1 + n k+1 (k 1) ) = o(1). c n k+1 k ) k r +1 Smlarly, for the fourth sum we apply Lemma.3 (1) Pr{B B } n o(1) n k+ (,H A H A )=(k,k) k 1 c n k+1 k ) 1.5 k = o(1).

8 8 Csaba Sándor To estmate the ffth sum we note that H A 3.3 that,h A H A whch completes the proof. )=(k,k) > k 1 Pr{B B } k +k n k+1 ( H A k + 1. It follows from Lemma c n k+1 k ) k +1 = o(1), Proof of Theorem 1.6. Let ɛ > 0 and for smplcty let D n = D n (ɛ) and E n = E n (ɛ). In the proof we use the estmatons () D n and (3) E n+1 log log n+log log log n+ (1 ɛ) log log log n log log log n log log n+log log log n+ (1+ɛ) log log log n log log log n In order to verfy Theorem we have to show that (4) lm Pr{S(n, 1 ) contans a D n-cube} = 1 and = n log log n+(1 ɛ+o(1)) log log log n = n log log n+(1+ɛ+o(1)) log log log n (5) lm Pr{S(n, 1 ) contans an (E n + 1)-cube} = 0. To prove the lmt n (4) let H A,..., H 1 A N n [1, n], B be the event H A X = X X N be the number of H A by Lemma 1 and nequalty () be the dfferent non-degenerate D n -cubes S(n, 1 ), X be the ndcator random varable for B and S(n, 1 ). The lnearty of expectaton gves ( ) n 1 E(X) = NE(X ) = (1 + o(1)) D n + 1 D n! Dn n log log n+(1+o(1)) log log log n n log log n (1 ɛ+o(1)) log log log n = n (ɛ+o(1)) log log log n, therefore E(X), as n. By Lemma.4 t remans to prove that Pr{B B } = o(e(x) ) where means that the events B, B are not ndependent.e. the cubes H A, H A have common vertces. We splt ths sum accordng to the ranks D n D n Pr{B B } = Pr{B B ) s=0 t=0,h A (6),H A )=(0,0) D n Pr{B B ) + s,h A Pr{B B }.

9 Non-degenerate Hlbert Cubes n Random Sets 9 The condton thus by Lemma 3.,H A o,h A, H A ) = (0, 0) mples that H A )=(0,0) Pr{B B } D n ( (( ) ) ) n 1 D n + 1 D n! Dn = o(e(x) ). H A = Dn+1 1, ( ) n n Dn D N = D n + 1 In the lght of Lemmas. and.3 (1) the second term n (6) can be estmated as D n s D n s ( ) n Pr{B B } D nn Dn+1 s D n+ t = D n + 1 (( ) ) n 1 D n D n + 1 D n! Dn n o(1) s=1 s t=0 (( ) ) t n 1 n s = D n D n + 1 D n! Dn n o(1) s n s s=1 Fnally, the functon f(x) = x n decreases on (, log x log n log log ] and ncreases on [log log n log log, ), therefore by () whch proves the lmt n (4). D n s=1 s n r = no(1) ( 4 n + Dn n D n ) = n 1+o(1), In order to prove the lmt n (5) let H C (En +1,n),..., H 1 C (En +1,n) be the dstnct (E n +1)-cubes K n [1, n] and let F be the event H C (En +1,n) S(n, 1 ). By (3) we have K Pr{S n contans an (E n + 1)-cube} = Pr{ K =1F } Pr{F } ( n ) En+1 nlog log n+(1+o(1)) log log log n E n + n log log n+(1+ɛ+o(1)) log log log n = o(1), whch completes the proof. =1 3. Concludng remarks The am of ths paper s to study non-degenerate Hlbert cubes n a random sequence. A natural problem would be to gve analogous theorems for Hlbert cubes, where degenerate cubes are allowed. In ths stuaton the domnant terms may come from arthmetc progressons. An AP k+1 forms a k-cube. One can prove by the Janson nequalty (see Lemma.5) that for a fxed k lm Pr{S(n, cn k+1 ) contans no APk+1 } = e ck+1 k. An easy argument shows (usng Janson s nequalty agan) that for all c > 0, wth p n = cn /5 lm Pr{S(n, p n) contans no 4-cubes} = e c 5 8.

10 10 Csaba Sándor Conecture 3.1. For k 4 lm Pr{S(n, cn k+1 ) contans no k-cubes} = e ck+1 k. A smple calculaton mples that n the random subset S(n, 1/) the length of the longest arthmetc progresson s a.e. nearly log n, therefore t contans a Hlbert cube of dmenson ( ε) log n. Conecture 3.. For every ε > 0 lm Pr{the maxmal dmenson of Hlbert cubes n S(n, 1 ) s < ( + ε)log n} = 1. N. Hegyvár (see [5]) studed the specal case where the generatng elements of Hlbert cubes are dstnct. He proved that n ths stuaton the maxmal dmenson of Hlbert cubes s a.e. between c 1 log n and c log n log log n. In ths problem the lower bound seems to be the correct magntude. References [1] N. Alon, and J. Spencer, The Probablstc Method, Wley-Interscence, Seres n Dscrete Math. and Optmzaton, 199. [] A. Godbole, S. Janson, N. Locantore and R. Rapoport, Random Sdon Seqence, J. Number Theory, 75 (1999), no. 1, 7-. [3] D. S. Gunderson and V. Rödl, Extremal problems for Affne Cubes of Integers, Combn. Probab. Comput., 7 (1998), no. 1, [4] R. L. Graham, B. L. Rothchld and J. Spencer, Ramsey Theory, Wley-Interscence, Seres n Dscrete Math. and Optmzaton, [5] N. Hegyvár, On the dmenson of the Hlbert cubes, J. Number Theory, 77 (1999), no., Csaba Sándor Department of Stochastcs Budapest Unversty of Technology and Economcs Egry J. u. 1, 1111 Budapest Hungary E-mal : csandor@math.bme.hu