No four subsets forming an N

Transcription

1 No four subsets formig a N Jerrold R. Griggs Uiversity of South Carolia Columbia, SC 908 USA griggs@math.sc.edu Gyula O.H. Katoa Réyi Istitute Budapest, Hugary ohkatoa@reyi.hu Rev., May 3, 007 Abstract We survey results cocerig the maximum size of a family F of subsets of a -elemet set such that a certai cofiguratio is avoided. Whe F avoids a chai of size two, this is just Sperer s Theorem. Here we give bouds o how large F ca be such that o four distict sets A, B, C, D F satisfy A B, C B, C D. I this case, the maximum size satisfies + + Ω F + + O, which is very similar to the best-kow bouds for the more restrictive problem of F avoidig three sets B, C, D such that C B, C D. Itroductio Let [] = {,,..., } be a fiite set, F [] a family of its subsets. I the preset paper max F will be ivestigated uder certai coditios o the family F. The well-kow Sperer s Theorem [7] was the first such discovery. The first author was supported i part by NSF grat DMS The secod author was supported by the Hugaria Natioal Foudatio for Scietific Research grat umbers NK063, AT04886, the Bulgaria Natioal Sciece Fud uder Grat IO-03/005 ad the projects of the Europea Commuity: INTAS , COMBSTRU HPRN-CT , FIST MTKD-CT

2 Theorem. If F is a family of subsets of [] without iclusio F, G F implies F G the F holds, ad this estimate is sharp as the family of all -elemet subsets shows. There is a very large umber of geeralizatios ad aalogues of this theorem. Here we will metio oly some results whe the coditio o F excludes certai cofiguratios that ca be expressed by iclusio oly. That is, o itersectios, uios, etc. are ivolved. The first such geeralizatio was obtaied by Erdős [3]. The family of k distict sets with mutual iclusios, F F F k is called a chai of legth k, which we deote simply by P k. For ay family of sets F, with specified iclusios betwee pairs of sets, let La, F deote the size of the largest family F of subsets of [] without ay subfamily havig the iclusios specified by F. Erdős exteded Sperer s Theorem as follows: Theorem. [3] La, P k+ is equal to the sum of the k largest biomial coefficiets of order. Now cosider families other tha chais. Let V r deote the r-fork, which is a family of r + distict sets: F G, F G,..., F G r. The quatity La, V r was first asymptotically determied for r =. Theorem.3 [5] + + Ω La, V This was recetly geeralized: Theorem.4 [], cf. [8] + r + Ω La, V r r + O. Four distict subsets satisfyig A C, A D, B C, B D are called a butterfly ad are deoted by B.

3 Theorem.5 [] Let 3. The La, B = / + / +. Let us compare Theorems.3 ad.5. Whe V is excluded, the the largest family has size equal to that of the largest level plus betwee ad times the ext level. O the other had, if the butterfly is excluded the the size of two levels ca be achieved. It is a atural questio, what happes if a cofiguratio betwee these two is excluded. Namely, let the cofiguratio of four distict subsets satisfyig A C, A D, B C be called ad deoted by N. It is somewhat surprisig that the result is basically the same at least i the first two terms as i the case of V. The total jump is betwee N ad B. The goal of the preset paper is to prove the followig theorem. Theorem Ω La, N holds. + + O Notatio ad defiitios A partially ordered set or poset P is a pair P = X, where X is a set i our case always fiite ad is a relatio o X which is reflexive x x holds for every x X, atisymmetric if both x y ad x y hold for x, y X the x = y ad trasitive x y ad y z always implies x z. It is easy to see that if X = [] ad the is defied as, the these coditios are satisfied, that is, the family of all subsets of a -elemet set ordered by iclusio form a poset. We will call this poset the Boolea lattice ad deote it by B. The defiitio of a subposet is obvious: R = Y, is a subposet of P = X, if ad oly if there is a ijectio α of Y ito X i such a way that y, y Y, y y implies αy αy. O the other had, R is a iduced subposet of P whe αy αy holds if ad oly if y y. If P = X, is a poset ad Y X the the poset spaed by Y i P is defied as Y, where agrees with, for all the pairs take from Y. Give a small poset R, let La, R deote the maximum size Y for Y [] that is, the maximum umber of subsets of [] such that R is ot a subposet of the poset spaed by Y i B. 3

4 Redefie our small cofiguratios i terms of posets. The chai P k cotais k elemets: a,..., a k where a <... < a k. The r-fork cotais r + elemets: a, b,..., b r where a < b,..., a < b r. The butterfly B cotais 4 elemets: a, b, c, d with a < c, a < d, b < c, b < d. Fially, we have the followig relatios i N: a < c, a < d, b < c. It is easy to see that the defiitios of La, P k, La, V r, La, B ad La, N i Sectios ad agree. I the rest of the paper we will use both sets of termiology. A poset is coected if for ay pair z 0, z k of its elemets there is a sequece z,..., z k such that either z i < z i+ or z i > z i+ holds for 0 i < k. If the poset is ot coected, maximal coected subposets are called its coected compoets. Give a family F of subsets of [], it spas a poset i B. We will cosider its coected compoets i two differet ways. First as posets themselves, secodly as they are represeted i B. I the latter case the sizes of the sets are also idicated. A full chai i B is a family of sets A 0 A... A where A i = i. Let us metio that the umber of full chais i B is!. We say that a full chai goes through a family subposet F if they itersect, that is, if the chai goes through at least oe member of the family. 3 Proof of Theorem.6 The lower estimate is obtaied from Theorem.3, sice La, V La, N. The upper estimate uses a idea geeralizig the proof of Sperer s Theorem give by Lubell [6], which is based o coutig the umber of full chais passig through a family. Let F be a family of subsets of [] cotaiig o four distict members formig a N. Cosider the poset P F spaed by F i B. Its coected compoets are deoted by F,..., F K. Let cf i deote the umber of full chais goig through F i. Observe that a full chai caot go through two distict compoets. Otherwise, they would be the same compoet. Therefore, the followig iequality holds: cf i!. What ca these compoets be? A compoet might be a P 3, but oe ca check by cases that o compoet ca cotai a P 3 as a proper subposet, sice addig oe more elemet to P 3 creates a N, o matter which elemet 4

5 of P 3 is related to the ew elemet. Hece, if a < b are two elemets of a compoet of size at least four, the a ad b caot be both comparable with the same other elemet of the compoet which would create P 3, though oe of a, b ca be comparable with may other elemets i the compoet. Therefore, the oly possible compoets are these: a < b < c, a < b i i r where r 0, 3 a > b i i r where r. 4 These are deoted by P 3, V r, Λr i this order. The elemets b i are urelated here. Notice that the umber of full chais goig through a poset of these types depeds oly o the sizes of the elemets of the poset, that is, the sizes of the members of the family F i. To idicate this size iformatio, we itroduce the otatio P 3; u, v, w for posets of type P 3 where a = u < b = v < c = w. The aalogous otatios for the other poset types above are are V r; u, u,..., u r u < u,..., u r ad Λr; v, v,..., v r v > v,..., v r. All compoets F i i are of this form. Pla of the proof. A good upper boud is sought for F = F i 5 where P 3 = 3, V r = Λr = r + are obvious. This upper boud will be determied etirely o the basis of. Deote the umbers of F i s of types P 3, V r, Λr by φ, νr, λr respectively. The 5 ca be writte i the form 3φ + r + νr + r + λr. 6 r=0 If the miima or good lower bouds mi u,v,w cp 3; u, v, w, mi cv r; u, u,..., u r, mi cλr; v, v,..., v r u,u,...,u r v,v,...,v r are determied the leads to a liear combiatio of φ, νr, ad λr. That is, oe liear combiatio, amely 5, has to be maximized uder the coditio that aother combiatio is bouded from above. This will be a easy task. Therefore, our mai problem ow is to determie the miima i the display above. This will be doe i the followig two lemmas. 5 r=

6 Lemma 3. cp 3; u, v, w u < v < w takes its miimum for the values u =, v =, w = +, that is,!! + cp 3; u, v, w. Proof. The umber of full chais goig through P 3; u, v, w is cp 3; u, v, w = u! u! + v! v! + w! w! u!v u! v! v!w v! w! u!w u! w! 7 +u!v u!w v! w! by the ordiary sieve. Sice this expressio has the same value whe we replace u, v, w by w, v, u, respectively, v ca be supposed. Aother useful form of 7 is cp 3; u, v, w! = + u v = + u v v + v u w + w v v u w w v w w u v w + w u v u w w v v u. 8 Here w is a decreasig fuctio of w i the iterval [v, ] with fixed u, v. O the other had, w u ad w v are icreasig fuctios. Therefore, 8 is smallest for w = v + : Oe has to see that the coefficiets of / w ad / w v are oegative, but this is a easy task if 0 < u, ad, otherwise, 8 is equal to, sice every full chai goes through the empty set. Replacig w by v + i 7: cp 3; u, v, v + = u! u!+v! v! u!v u! v! u. 9 Aother useful form of 9 is cp 3; u, v, v +! = u + v u v. 0 u Here v is a decreasig fuctio of v i the iterval [, ], while is icreasig. Therefore, if we wat to miimize 0, v ca be chose v u 6

7 as the smallest iteger with v > u. Two cases will be distiguished: v = u + ad v =. Case. v = u +. I this case 9 becomes cp 3; v, v, v + = v! v +! + v! v! v! v! v + = v! v!v v + =! v v +. v It is easy to see that both factors attai their respective miima at v =. The lemma is proved for this case. Case. v =. Rewrite 0: cp 3; u, v, v + = +!. u v u v u is a icreasig fuctio of u i the iterval [0, ] while u decreasig. The best choice for u is. Lemma 3. Suppose 6, r. The u! u! + ru!u u! cv r; u, u,...,u r u < u,...,u r v is L holds where u = u = if is eve, u = if is odd ad r, while u = 3 if is odd ad < r. Proof.. Oe ca easily show by usig the sieve that cv r; u, u,..., u r = r u! u! + u i! u i! This will actually be used i the form r u!u i u! u i!. cv r; u, u,..., u r = r r u! u! + u i! u i! u!u i u! u i!. 7

8 Dividig oe term by!, two useful forms are obtaied for the summad i : r + ui = u u i u i r + ui u u u i u ad r + u = u u i u + u i u i u r u u i. 3. First we will show that -3 attais its miimum for some pair u, u i = u +. If u, fix u ad cosider chagig u i i. Here, u i is a decreasig fuctio of u i i the iterval [, ], while u i u is icreasig. Therefore, oe ca suppose that u i = u +, ad we are doe. Else, > u ad the method i above leads to u i. Fix this value ad icrease u usig 3. It will ot icrease by movig u to u =. Hece, we obtaied the lower estimate mi u r u! u! + u +! u! u!! u! = mi u r u! u! + u!u u! for -3 ad therefore we have mi u u! u! + ru!u u! cv r; u, u,..., u r 4 This miimum will be determied i the rest of the proof. 3. Suppose ow that r. Take the derivative of f r u = u! u! + ru!u u!, that is, compare f r u at two cosecutive values of u. Whe does the iequality f r u = u! u +! + ru!u u! < f r u = u! u! + ru!u u! 5 hold? It is equivalet to 0 < r u r 3 + r u + r. 8

9 The discrimiat of the correspodig quadratic equatio i u is r 3+r +8r r = r+ r 3r +r. The latter expressio ca be strictly upperestimated by r + r, if r + < 3r holds, that is, if r >. Hece, the larger root α of the quadratic equatio is less tha r 3 + r + r + r 4r =. O the other had, as it is easy to see, r + 3r is a lower boud for the discrimiat if r holds. Usig this estimate, we obtai that α i this case. Substitutig this lower estimate ito the formula for the smaller root α, we obtai α 0 whe r. Sice 5 holds exactly below α ad above α, we ca state that f r u attais its miimum at u = α. By the iequalities above we ca coclude that this is at if is eve ad if is odd. The statemet of the lemma is proved i the case of r. Else suppose < r. The iequality α < ca be proved i the same way as i the previous case. O the other had, 6 implies that r + 5r is a lower estimate o the discrimiat, hece we have < α. This gives that α < 3. If is eve, α is agai, while α = 3 whe is odd. Although f r 0 < f r is allowed by this estimate, it is easy to check that f r 0 > f r holds i reality. By 4 the proof is fiished for r. The case r = is much easier. The compariso 5 leads to a liear iequality which is a equality for u =. The formula f u also has its miimum at. But it has the same value at ad. L The case r = 0 has to be treated differetly. The the umber of full chais goig through the oly u-elemet set is u! u!. Its miimum is!!. Proof of the Theorem. We will eed the followig iequalities i the proof: u!u u! < u! u! + ru!u u! 6 r + 9

10 holds for each of the 3 possible values of u. They easily reduce to < +, < + ad 3 < +3 whe u =, ad 3, respectively. A differet iequality is eeded i the case r = 0: u!u u! <!!. 7 Oe ca verify it separatig the 3 cases. Similarly u!u u!! 3 ca be obtaied for each case. Start the proof by :! + 8! cf i = If F i is a P 3, the Lemma 3. ad 8 give cf i F i F i 9 cf i F i = cf i 3 u!u u!. If F i is a V r the Lemma 3. ad 6 prove cf i F i = cf i r + u!u u! for r ad 7 shows its validity whe r = 0. By symmetry oe ca say the same about Λr. Display 9 results i! cf i = Hece we obtaied u!u u! F cf i F i F i F i = u!u u! F.! u!u u!, 0

11 ad this is equal to, +, 3 i the cases u =, = ad 3 +, respectively. These are all equal to + O. T 4 Remark It is somewhat disturbig that the costat i the secod term i Theorem.6 ad.3 is ot fully determied. Is it or? Let us make the situatio clear. The costructio of a family avoidig V is the followig: Take all the sets of size ad a family A,..., A m of + -elemet sets satisfyig the coditio A i A j < for every pair i < j. It is easy to see that this family cotais o V. We oly have to maximize m. Deote this maximum by m. Sice the -elemet subsets of the A i s are all distict, we have m +. This gives the upper estimate m. There is a very ice costructio see [4] of such sets A i with m = +. We obtaied + Ω m. 0

12 It is a logstadig cojecture of codig theory what the right costat is here, or. Does this limit exist at all? It is easy to see that + m La, N. The upper estimate of our Theorem.6 is a far-reachig geeralizatio up to the secod term of the upper estimate i 0. Replacig the costat i the upper estimate i Theorem.6 by would improve the upper estimate i 0, too, solvig the old codig problem. O the other had, if oe could costruct a family with a costat rather tha, this would improve the lower estimate i Theorem.6. Summarizig, there is a strog coectio betwee the two problems. Of course, it is possible that the proper costat i m is, while the oe i La, N is. Refereces [] Aalisa De Bois, Gyula O.H. Katoa, Largest families without a r-fork, submitted. [] Aalisa De Bois, Gyula O.H. Katoa, Korad J. Swaepoel, Largest family without A B C D, J. Combi. Theory Ser. A 005, [3] P. Erdős, O a lemma of Littlewood ad Offord, Bull. Amer. Math. Soc. 5945, [4] R.L. Graham ad N.J.A. Sloae, Lower bouds for costat weight codes, IEEE IT 6980, [5] G.O.H. Katoa ad T.G. Tarjá, Extremal problems with excluded subgraphs i the -cube, i: Borowiecki, M., Keedy, J.W., Sys lo M.M. eds. Graph Theory, Lagów, 98, Lecture Notes i Math., Spriger, Berli Heidelberg New York Tokyo, 983. [6] D. Lubell, A short proof of Sperer s lemma, J. Combi. Theory 966, 99.

13 [7] E. Sperer, Ei Satz über Utermege eier edliche Mege, Math. Z. 798, [8] Hai Tra Thah, A extremal problem with excluded subposets i the Boolea lattice, Order 5998,