Largest families without a r-for Aalisa De Bois Uiversity of Salero Salero, Italy debois@math.it Gyula O.H. Katoa Réyi Istitute Budapest, Hugary ohatoa@reyi.hu 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-ow Sperer theorem ([8]) was the first such theorem. 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 what 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 distict sets with mutual iclusios, F F... F is called a chai of leght. It will be simply deoted by P. Let La(, P ) deote the largest family F without a chai of leght. The wor of the secod author was supported by the Hugaria Natioal Foudatio for Scietific Research grat umbers NK063, AT0886, the Bulgaria Natioal Sciece Fud uder Grat IO-03/005 ad the projects of the Europea Commuity: INTAS 0-77-77, COMBSTRU HPRN-CT-00-00078, FIST MTKD-CT-00-003006.
Theorem. [3] La(, P + ) is equal to the sum of the largest bimomial coefficiets of order. Let V r deote the r-for, that is the followig family of distict sets: F G, F G,... F G r. The quatity La(, V r ), that is, the largest family o elemets cotaiig o V r was first (asymptotically) determied for r =. Theorem.3 [5] ( ) ( + ( )) + O La(, V ) ( ) ( + ). [9] has correctly determied the mai term for V r (i a somewhat more geeral form), provig the followig theorem. Theorem. [9] ( ) ( + r ( )) + O La(V r+ ) ( ) ( + r + o ( )). The mai aim of the preset paper (Sectio ) is to improve the (upper estimate of the) secod term. For the sae of completeess, we repeat the proof of the lower estimate, too. Theorem.5 ( ) ( + r ( )) + O La(V r+ ) ( ) ( + r ( )) + O. I Sectio we show how the upper boud i the geeral form of Tra s theorem [9] ca be attaced by the usage of Theorem.5. I most cases a better secod term is obtaied. I Sectio 3 we give estimates for the maximum size of a family F [] cotaiig o r + s distict members satisfyig A,..., A s B,..., B r.
A auxiliary iequality If F [] is a family let f i deote the umber of its i-elemet members. The Sperer theorem, has the followig sharpeig, ow as the YBLMiequality ([0], [], [6], [7]). Theorem. Let F be a family of subsets of [] without iclusio. If f i = {F : F F, F = i} the holds. f i ( i) (.) The mai igrediet of our Theorem.5 is give below. Theorem. Suppose that the family F cotais o r + -for (0 < r) ad [] F. The ( f ( i i) r ). (.) i Proof. A chai is a family C = {C 0, C,..., C }, C 0 C... C where C i = i (0 i ). We say that a chai C goes through a family F, if C F. Let C(F )(F F) be the set of all chais goig through F. We have C(F ) = F!( F )!. Similarly, let C(F, F ) deote the set of those chais which go through both F, F. This set is empty uless oe of them icludes the other oe. The followig easy lemma, which is actually a primitive sieve, will be applied for chais. Lemma.3 If X,..., X u are subsets of a set X, the X + + X u X + i<j X i X j (.3) holds. Proof. A elemet of X which is outside of all X i is ot couted o the left had side, but it is couted o the right had side. A elemet which belogs to exactly oe X i is couted exactly oce o both sides. Fially, if 3
a elemet belogs to exactly v of X i s the it is couted v times o the left had side ad + ( v ) times o the right had side. The obvious iequality v + ( v ) ( v) completes the proof. L Let X be the set of all chais i [], while the X i s be the chais goig through a give member of F. (.3) becomes C(F ) C(F, F )!. (.) F,F F,F F Itroduce the otatio U(F ) = {G : G F, F G}. Rewrite (.) usig this otatio. C(F ) C(F, G)!. (.5) G U(F ) Here C(F, G) = F! G F!( G )!, (.5) ca be writte i the followig form. F!( F )! F! G F!( G )!!. Divide it by!. ( F ) G U(F ) ( )( F F G U(F ) G F ) = Observe that ( ) F G F ( ) F ( ) F G U(F ) ( F ). G F (.6) sice G F < F. Moreover, sice F cotais o V r, the iequality U(F ) r must hold. Substitutig these facts i (.6) we obtai ( ) r ( ). F F To fiish the proof we oly have itroduce f i. T
3 Proof of Theorem.5 Upper boud. (.) i the form f ( i ) i i i r suggests that oe has to fid the maximum of ( ) i b(i) = i i r i i. Lemma 3. Suppose 6r + 3 <. The ( ) ( i i i r ) r holds for 0 i < r. (3.) Proof. Cosider the derivative of the fuctio b(i) (0 i < r), that is, compare two cosecutive values ( i < r): f(i ) =! (i )!( i)!( i + r) < f(i) =! i!( i )!( i r). This is equivalet to i( i r) < ( i)( i + r) ad 0 < i (3 r +)i+ r +. The discrimiat of the correspodig quadratic equatio is (3 r + ) 8( r + ) = + r r r + = ( r) r +. The followig iequalities are obvious for 6r + 3 <. ( r ) < ( r) r + < ( r ) (3.) Let α < α be the roots of the equatio. Usig (3.) we obtai the followig boud for the roots. + < α < + 3 ad r < α < r. 5
This shows that f(i) is growig util ad is decreasig from this poit to r. L Partitio the family F accordig to the sizes of its members: F = {F : F F, F < r}, F = {F : F F, F r}. Apply Theorem. for F ad use Lemma 3.. r ( f ( i i) r ) r = i = ( ) r r f ( i ) i i i r r f i = ( ) r F. f ( i ) r Hece we obtai the right upper boud for F. O the other had, F r ( ) i shows F = O( r ) F + F = F fiishes the proof. ( ) ( ) O. Up Lower boud. (See [9].) For a fixed ad a tae all the subsets {x,..., x } of [] satisfyig x + + x a (mod r ). (xi s are differet.) Suppose that some sets have the same -elemet itersectio, say x + + x. The the equatios x + + x + y a ad x + + x + y a imply y y (mod r ). It is obvious that there are at most r such umbers (mod ). Choose a maximizig the umber of solutios. For this a the umber of solutios is at least ( ) r ( ) r + r ( ) ( ) r r. Tae = +. The the family cosistig of all -elemet sets ad the + -elemet oes costructed above will cotai o r+-for. The umber of sets is as it is give i the theorem. Lo 6
Remar. This costructio is a slight geeralizatio of the case whe r is []. It was show that oe ca fid approximately part of all the sets of size + without a itersectio of size. O the other had, it is trivial that there is a upper boud which is approximately twice as much. It is a old ope problem of codig theory which oe is the right costat. Or either oe? This is why to get rid of the factor i the secod term i Theorems.3,.5,.-. ad 5. seems to be difficult. Aother approach. Earlier we had a more complicated proof for the upper boud. It cotaied a iequality what might have some iterest i its ow right. Theorem 3. Suppose that the family F cotais o r + -for (0 < r) ad the sizes of all members of F are at most m where m < r. The m f ( i ) + r m + r ( m r). (3.3) i Proof. Replacig i by m i the coefficiet, the statemet of Theorem. becomes m f ( i i) r. m The easy iequality r m + r m + r ( m r) fiishes the proof. T The proof of Theorem.5 ca be fiished by choosig m to be somewhat more tha. The Theorem 3. esures that the umber of members of F with size m caot exceed the desired boud. O the other had, the umber of sets i F is at most the sum of the biomial coefficiets from m to what is much less tha the largest biomial coefficiet, if m is chose properly. But this is oly a setch, the details eed some theorems from the asymptotical theory of biomial coefficiets ad tedious calculatios. Problem. What is the maximum of the left had side of (3.3) uder the coditio F F implies F m? If there is o upper boud for the sizes 7
i F the the family cosistig of all i-elemet sets ad [] gives while the family cotais o -for. If [] is excluded, that is, m = the the followig family gives asymptotically +. Suppose is eve ad divide [] ito two equal parts, [/] ad its complemet. Tae all -elemet sets cotaiig [/] ad all -elemet sets ot cotaiig it. This family cotais o -for. Is this the asymptotically best for family without -fors? Tra s theorem ad its partial improvemet A r-for with a -shaft is a family of distict subsets F, F,..., F, G, G,..., G r such that F F... F, F G, F G,..., F G r. It is a combiatio of a path ad a for, it is deoted by V r. Tra s Theorem i its full geerality was the followig. Theorem. [9] Let r, be give itegers. The + i= + La( V r+ ) ( ) ( + i ++ ) ( ( )) r + O ( ) ( + r( ) +r + o ( ) ). We are goig to prove a somewhat stroger upper boud i most cases, amely the followig statemet. Theorem. La( V r+ ) ( ) + + r ( ) + O. Proof. Suppose that the family F cotais o V r. Let F deote the family of those members F of F for which there is o chai of leght below F, that is, there are o distict sets F, F,..., F F such that F F... F F. O the other had, F = F F. It is easy to see that F cotais o chai of legth, therefore F + i= + 8 ( ) i (.)
holds by Theorem.. O the other had, F cotais o V r, therefore ( ) F ( + r ( )) + O. (.) (.) ad (.) imply F = F + F + i= + ( ) ( ) + ( i + r ( )) + O. (.3) ( Comparig ) the lower ( estimate i Theorem. ad (.3) we see that + is replaced by ). Let us study their ratio ( ) ( + ) = K + i (.) i where K is a otatio for +. Observe that K = whe both ad are eve, it is if is eve, is odd, it is whe is odd, is eve, ad it is + i the last case whe ad are both odd. Oe factor of (.) ca be rewritte i the form + i = + K i (.5) i where K deotes + ad is actually equal to,, ad followig the order at K. Sice the depedece o should be avoided, tae the trivial upper bouds K + ad K. It is easy to see that (.5) ca be upperbouded by + = + i + O( ). Usig this boud for all factors i (.), the followig upper boud is obtaied for the ratio: ( K + ) + O( ) = + + ( ) + O. 9
Therefore ( ) ca be replaced i (.3) by ( ) ( + + + ( = + ) + ( ) ) + O ( ) + O. Let us see ow that ( ) + r + r r holds for 3 r. The right had side is ot icreased by replacig r by 3 i the biomial coefficiet. The remaiig iequality ( ) + + r r is really easy to prove. That is, our upper boud is stroger wheever 3. However, this is ot true i geeral for r =,. However we strogly believe that ca be completely deleted from the the secod term of the upper estimate. Cojecture La( V r+ ) ( ) ( + r ( )) + O. 5 Complete two level posets I this sectio we are tryig to maximize the size of a family F cotaiig o r +s distict members satisfyig A,..., A s B,..., B r. Let T r,s deote the poset with two levels, s elemet o the lower, r elemets o the upper level, every lower oe is i relatio with every upper oe. It is easy to see that our coditio ca be formulated i the way that we are looig for the maximum umber of the elemets i the Boolea lattice of subsets of [] (defied by iclusio) without cotaiig T r,s as a subposet. Let the maximum be deoted by La(, P ) for a arbitrary poset P. 0
Theorem 5. Suppose that s, 3 r ad s r hold. The ( ) ( ) ( ) ( + + r + s ( )) + O La(, T r,s ) Proof. Upper estimate. ( ) ( + r + s 3 ( )) + O. Lemma 5. If a poset P cotais o T r,s as a subposet, the it ca be partitioed ito posets P ad P so that P cotais o T r, ad P cotais o T,s. Proof. Let P be the set of those elemets a of P for which the umber of elemets b P satisfyig a < b is at most r. The it is obvious that the poset iduced by theses elemets cotais o T r,. Let P = P P. Suppose that i cotrast to our statemet, P cotais a T,s as a subposet: a,... a s, b P, a < b,... a s < b. Sice b is ot i P, there are some c,..., c r i P such that b < c,..., c r < b holds. It is easy to see that all these elemets, a,... a s, b, c,..., c r form a T r,s i P L Apply the lemma for the poset spaed by the family cotaiig o sets formig a T r,s. The two families obtaied are deoted by F ad F. Theorem.5 ca be directly used for F. Sice complemetatio preserves iclusio, Theorem.5 ca also be used for F. Addig up the two upper estimates, the upper estimate of Theorem 5. is obtaied. Up Lower estimate. Let F be a family of sets of size such that the size of the uio of ay s members of F is at least, ad let F be a family of sets of size + such that the size of the itersectio of ay r of them is at most. We deote by F the family cotaiig all members of F ad F alog with all sets of size ad all those of size. The family F cotais o r + s distict members A,..., A s, B,..., B r satisfyig A,..., A s B,..., B r. Ideed, it is possible to see that the size of the uio of ay s members of F is at least whereas the itersectio of ay r members of F has size at most. Let A,..., A s F. If at least two sets amog A,..., A s have size at least the the uio of these two sets has a size at least. Otherwise, if at most oe set amog A,..., A s has size larger tha or equal
to the at least s sets amog A,..., A s belog to F. By costructio the size of the uio of these s sets is at least. O the other had let us cosider r members B,..., B r of F. If at least two sets amog B,..., B r have size at most the the itersectio of these two sets is at most. If at most oe set amog B,..., B r has size at most, the at least r sets amog B,..., B r belog to F ad cosequetly their itersectio is at most. By usig the costructio i the proof of Theorem.5 F ca be made as large as ( ) ( ( )) r + O. By the same costructio we obtai ( ) ( s ( )) + O sets of size + such that the itersectio of ay s of them is at most. The family F is obtaied by taig the complemet of each of those sets. Addig up the sizes of the four families, the lower estimate of the theorem is obtaied. Up Remars.. I the lower estimate the mai term is the sum of the two largest biomial coefficiets of order. I the upper estimate it is the double of the largest oe. They are the same if is odd, but differet for eve s. If oe of the umbers ( ( ) is replaced by the secod largest biomial coefficiet ) the we loose a little. The r + s 3 should be replaced by r + s.. The proofs wor for r = s = as well. However the upper estimate is too wee. It has bee proved [] that La(, T, ) is the sum of the two largest biomial coefficiets. 3. We believe that the lower estimate is the (asymptotically) correct oe up to the secod term. Refereces [] B. Bollobás, O geeralized graphs, Acta. Math. Acad. Sci. Hugar., 6(965) 7-5.
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