Monochromatic Infinite Sumsets

Transcription

1 Monochromaic Infinie Sumses Imre Leader Paul A. Russell July 25, 2017 Absrac WeshowhahereisaraionalvecorspaceV suchha,whenever V is finiely coloured, here is an infinie se X whose sumse X+X is monochromaic. Our example is he raional vecor space of dimension sup{ℵ 0,2 ℵ 0,2 2ℵ0,... This complemens a resul of Hindman, Leader and Srauss, who showed ha he resul does no hold for dimension below ℵ ω. So our resul is bes possible under GCH. 1 Inroducion I is a well-known consequence of Ramsey s heorem ha, whenever he naurals are finiely coloured, here is an infinie se X such ha all pairwise sumsofdisincelemensofxhavehesamecolour. Ifoneasksforasronger conclusion, ha he enire sumse X +X = {x+y : x,y X is monochromaic, hen he answer is no: his is because such a sumse auomaically conains wo numbers wih one roughly wice he oher, and his can easily be ruled ou by a suiable 3-colouring (see e.g. [3]). We menion in passing ha i is, surprisingly, unknown as o wheher or no his can be achieved wih a 2-colouring: his is called Owings problem [5]. For background on his, and oher resuls menioned in his inroducion, see [4] alhough we menion ha his paper is self-conained and does no rely on any resuls from [4]. Wha happens if one passes o a larger ambien space, for example he raionals? Here again, he answer is no: here is a finie colouring of Q wih no infinie sumse monochromaic (see e.g. [4]). Wha abou for he reals? Cenre for Mahemaical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK, I.Leader@dpmms.cam.ac.uk Churchill College, Cambridge CB3 0DS, UK, P.A.Russell@dpmms.cam.ac.uk 1

2 Hindman, Leader and Srauss [4] showed ha, for every raional vecor space of dimension smaller ha ℵ ω, here is a finie colouring wihou an infinie monochromaic sumse. Noe ha his esablishes he answer for he reals if we assume CH. (I is sill unknown if he reals have such a bad colouring if we do no make exra se-heoreic assumpions.) However, hey were unable o find a vecor space wih he posiive propery (of having no bad colourings). Our aim in his paper is o show ha such a vecor space does exis. We show ha his is he case for any dimension ha is a leas ℶ ω (read beh-omega ), which is defined o be sup{ℵ 0,2 ℵ 0,2 2ℵ 0,... Noe ha if we assume GCH hen his is exacly ℵ ω, which would be bes possible in ligh of he resul of [4]. We do no know if he vecor space of dimension ℵ ω has his propery if we do no assume GCH. We also prove a similar resul for muliple sums such as X + X + X and so on. The proof involves a perhaps unexpeced use of he Hales-Jewe heorem. For a finie or infinie cardinal κ, we wrie Q κ o denoe he vecor space of dimension κ over Q. Tha is, Q κ is he direc sum of κ copies of Q, no he direc produc. We shall ake Q κ o come equipped wih a basis e 0, e 1, e 2,... ha is well-ordered by he smalles ordinal of cardinaliy κ. 2 Main Resul Consider Q ℶω, he ℶ ω -dimensional vecor space over Q. As remarked above, we shall consider Q ℶω o come equipped wih a well-ordered basis B whose elemens we shall denoe by e 0, e 1, e 2,... Suppose x Q ℶω wih x 0. We may wrie x in erms of he basis B and delee all zero enries o obain a finie lis of non-zero raionals. We call his lis he paern of x. More formally, given a non-zero x Q ℶω, here is a unique way o express x in he form x = n i=1 x ie αi where n is a posiive ineger, each x i is a non-zero raional and α 1 < α 2 < < α n are ordinals. The paern of x is (x 1,x 2,...,x n ). We shall ofen denoe he paern (x 1,x 2,...,x n ) simply by x 1 x 2...x n. We say ha he paern x 1 x 2...x n has lengh n and wrie l(x 1 x 2...x n ) = n. Given a finie colouring of Q ℶω, we seek an infinie se X Q ℶω wih X +X monochromaic. There are wo sages o he proof. We firs show (Lemma 1) ha, given a finie se Π of paerns, here is a large subspace of Q ℶω on which he colour of an x wih paern in Π depends only on he paern. The subspace produced is spanned by a subse of he original basis B of Q ℶω. This par of he proof is a fairly sandard applicaion 2

3 of he Erdős-Rado heorem [1]. The hear of he proof comes in he second sage. The main obsacle o overcome is o deermine how we should proceed following he reducion givenbylemma1. Thaisosay, whichpaernsshouldweconsiderandhow do we force all he elemens of X+X o have he desired paern or paerns? While we are able o work wihin a subspace spanned by a counable subse A B, i is ineresing o noe ha our proof ofen requires his subse A o have an order-ype greaer han ω. We herefore ask he subspace produced in Lemma 1 o have dimension ℵ 1 ; his allows us o always find A as required. We now proceed o he firs of he wo sages deailed above. Firs, we recall he Erdős-Rado heorem. As usual, we denoe by exp r (κ) he r-fold exponenial of κ, i.e. exp 0 (κ) = κ and exp r+1 (κ) = 2 exp r (κ). Erdős-Rado heorem ([1]). Le r be a non-negaive ineger and le κ be an infinie cardinal. Suppose he (r + 1)-elemen subses of a se of cardinaliy exp r (κ) + are coloured wih κ colours. Then here is a subse of cardinaliy κ + all of whose (r+1) elemen subses are he same colour. In paricular, his immediaely implies ha for every posiive ineger r, if he r-elemen subses of a se of cardinaliy ℶ ω are coloured wih finiely many colours hen here is a subse of cardinaliy ℵ 1 all of whose r-elemen subses are he same colour. Lemma 1. Le k be a posiive ineger and suppose Q ℶω is k-coloured. Le Π be a finie se of paerns. Then here is a subse A B of cardinaliy ℵ 1 such ha for each π Π he se is monochromaic. {x Q ℶω : x is in he span of A and has paern π Proof. Le c be he given k-colouring of Q ℶω. Le r be he lengh of he longes paern in Π. Le Π = { {{ π : π Π. r l(π) Wrie Π = {π (1),π (2),...,π (n). We define n k-colourings c 1, c 2,..., c n of he r-elemen subses of B as follows. Given S B wih S = r, wrie S = {e α1,e α2,...,e αr wih α 1 < α 2 < < α r. Then se c i (S) = r j=1 π (i) j e αj. 3

4 Now define a single k n -colouring c of he r-elemen subses of B by c (S) = (c 1 (S),c 2 (S),...,c n (S)). We apply he Erdős-Rado heorem o his final colouring c o obain A B wih all r-subses of A he same colour and A = ℵ 1. Removing he r leas elemens of A, we obain our se A as required. We are now ready o proceed o he main par of he proof. Theorem 2. Le k be a posiive ineger, and suppose Q ℶω is k-coloured. Then here is an infinie se X Q ℶω such ha he sumse X +X is monochromaic. Proof. Le c be he given k-colouring of Q ℶω. For a = 0, 1, 2,..., k, le π a be he paern π a = {{ {{ a 2(k a) and le Π = {π a : 0 a k. By Lemma 1, we can find A B wih A = ℵ 1 and colours c a (0 a k) such ha if x is in he span of A and has paern π a hen c(x) = c a. By he pigeonhole principle, we mus have c a = c b for some a and b wih 0 a < b k. Le C be a subse of A of order-ype α = ω(b a+2) and lis he elemens of C in order as f 0, f 1, f 2,... Now le X = {x i : i < ω, where, for each i < ω, we define a 1 b a x i = f r + f ωr+i + r=0 r=1 2(k b) 1 r=0 1 2 f ω(b a+1)+r. Then for all i, j N, we observe ha x i +x j has paern π a or π b according as i j or i = j. Thus X +X is monochromaic, as claimed. 3 Exensions There are wo obvious direcions in which one migh seek o exend Theorem 2. Firs, wha if insead of simply requiring ha X be infinie, we seek an X of cardinaliy ℵ 1, say, or of some larger specified cardinaliy? This is possible if we sar wih a vecor space of sufficienly large cardinaliy, and requires only a rivial modificaion o he proof of Theorem 2. 4

5 Theorem 3. Le k be a posiive inegerand le κ be an infiniecardinal. Then here is an infinie cardinal λ such ha whenever he λ-dimensional raional vecor space Q λ is k-coloured, here is a subse X Q λ wih X = κ and X +X monochromaic. Indeed, wih a similar applicaion of he Erdős-Rado heorem as above, we may ake λ = sup{κ,2 κ,2 2κ,... More ineresingly, wha if raher han simply looking for he sumse X + X we seek a monochromaic sum of many copies of X? For example, define he riple sumse of X o be X +X +X = {x+y +z : x,y,z X. If we finiely colour Q ℶω, can we always find an infinie X Q ℶω wih X +X +X monochromaic? Le us firs consider informally how one migh ry o exend he proof of Theorem 2 o deal wih his problem. Previously, we spli our basis vecors ino sreches of lengh ω. Depending on he colouring, we hen defined each x i o eiher ake value 1 or 1 on cerain fixed coordinaes in he srech 2 (a fixed srech ), or we defined each x i o ake value 1 on coordinae i of he srech and 0 elsewhere (a variable srech ). This resuled in x i + x j always having a paern consising of 1 s and 2 s. More precisely, he paern on a given fixed srech is always he same, whereas he paern on a variable srech could be eiher 11 or 2. Now, suppose we consider x h +x i +x j wih a similar definiion of he x i. The variable sreches will now have paern 111 or 21 or 12 or 3. To deal wih his, i urns ou ha we need a somewha unexpeced applicaion of he Hales-Jewe Theorem [2]. Theorem 4. Le k and be posiive inegers and suppose Q ℶω is k-coloured. Then here is an infinie se X Q ℶω such ha X +X + +X {{ is monochromaic. Proof. Le c be he given k-colouring of Q ℶω. LeΠbeheseofallpaernsofheformx 1 x 2...x n wherex 1, x 2,..., x n are posiive inegers summing o. Noe ha Π is finie. Le N be a posiive ineger such ha whenever Π N is k-coloured i conains a monochromaic combinaorial line. (Such N exiss by he Hales-Jewe Theorem.) Le Π be he se of paerns obained by concaenaing N paerns from Π. By Lemma 1, here exis a subse A B wih A = ℵ 1 and colours c π (π Π ) such ha if x is in he span of A and has paern π hen c(x) = c π. 5

6 We induce a colouring of Π N by giving (π 1,...,π N ) Π N he colour of any x in he span of A wih paern π 1 π 2...π N. (Noe ha his does no depend on he choice of x). We may now find a monochromaic combinaorial line L in Π N. Le J be he se of acive coordinaes of L and, for each π Π, le I π be he se of inacive coordinaes where L akes consan value π. (Noe ha we ake our coordinaes o range from 0 o N 1.) Le C be a subse of A of order-ype ωn and lis he elemens of C in order as f 0, f 1, f 2,... Le X = {x i : i < ω where x i = r J f ωr+i + π Π l(π) r I π s=1 Then each elemen of X +X + +X {{ X +X + +X {{ is monochromaic. π s f ωr+s. has paern in L and hus We remark ha, exacly as he proof of Theorem 2 was adaped o yield Theorem 3, we may similarly adap he proof of Theorem 4 o give: Theorem 5. Le k and be posiive inegers and le κ be an infinie cardinal. Then here is an infinie cardinal λ such ha whenever Q λ is k-coloured here is an infinie se X Q λ wih X = κ and X +X + +X {{ monochromaic. As wih Theorem 3, i suffices o ake λ = sup{κ,2 κ,2 2κ,... References [1] Erdős, P., and Rado, R., A pariion calculus in se heory, Bull. Amer. Mah. Soc 62 (1956), [2] Hales, A. W., and Jewe, R. I., Regulariy and posiional games, Trans. Amer. Mah. Soc. 106 (1963), [3] Hindman, N., Pariions and sums of inegers wih repeiion, J. Comb. Theory (A) 27 (1979),

7 [4] Hindman, N., Leader, I., and Srauss, D., Pairwise sums in colourings of he reals, Abh. Mah. Sem. Univ. Hamburg, o appear. [5] Owings, J., Problem E2494, Amer. Mah. Monhly 81 (1974),