DISCRETE MATHEMATICS 8 (1974)281-294. - PDF document

DISCRETE MATHEMATICS 8 (1974)281-294. € North-Holland Publishing CompanyEXTREMAL PROBLEMS AMONG SUBSETS OF A SETPaul ERDOSHungarian Academy of Sciences, Budapest, HungaryTechnion, Haifa, IsraelandDaniel J. KLEITMANDepartment of Mathematics, Massachusetts Institute of Technology,Cambridge, Mass. 02139, USAAbstract. This paper is a survey of open problems and results involving extremal size of collec-tions of subsets of a finite set subject to various restrictions, typically on intersections of mem-bers.0. IntroductionThe subsets of a (finite) set form a lattice and in fact a Boolean algebra.The following concepts are natural to them.(A) Intersection.(B) Union.(C) Disjointness.(D) Complement.(E) Containment.(F) Rank (size).In this paper we survey the present status of a number of problems in-volving maximal or minimal sized families of subsets subject to restric-tions involving these concepts.Problems of this kind arise in a large number of contexts in manyareas of mathematics. For example, the divisors of a square free numbercorrespond to the subsets of the prime divisors, so that certain numbertheoretic problems involving divisors of numbers are of this form. Effi-cient error correcting codes and block designs can be considered as ex-tremal collections of subsets satisfying restrictions of this kind.Since the concept of set is as basic in mathematics as the concept ofnumber, one can also investigate the properties considered here for their 282\tP.Erds,D.J. Kleitman, Extremal problems among subsets of a setown sake as one considers similar problems in number theory. Thus wemight ask: "What sort of limitations are imposed upon families of sub-jects of a set by simple restrictions on intersection, union, rank and/orcontainment among members of the family?"Questions of this kind have one additional value. Since the conceptsinvolved are all easily understood by non-mathematicians, results andelegant proofs in this area have tutorial value as illustrations of the powerof mathematical method that are accessible to the layman.To facilitate reference, we divide the problems considered here intofive areas. These are:1. Non-intersection.2. Size limited intersection.3. Intersection and rank limitations.4. Containment limitations.5. Union and intersection restrictions.6. Miscellany.Problems and results in these areas are described in the correspondingsection below.1. Non-disjoint familiesLet S be a finite set havingnelements (I S I =n).Among the simplestrestrictions that can be placed on families of subsets of S is that no twoare disjoint. Thus ifF = {Ail,i=1,...,A withAiCS,we may require thatAin A, 4-0for alli,j.With this one restriction there are several questions that can be raised.Among these are:(a) How large canFbe?(b) IfF is"maximal" in that no subset of S can be added to it withoutviolating the restriction, how small canFbe?(c) How many maximal F's are there of any given size?(d) How manyF'sare there of any size?These four kinds of questions can be raised not only about families ofsubsets restricted as isFabove, but also about families satisfying variantsof the restriction.Among possible variant restrictions of the same general kind are:1.1. LetFbe as defined above, and let G consists of the minimal mem-bers ofFthat is the members ofFnot contained in others. P. Erds, D.J. Kleitman, Extremal problems among subsets of a set\t2831.2.LetG2kbe the union ofkfamilies each restricted as wasFabove.1.3. LetG3kbe a family containing nokmembers that are pairwisedisjoint.1.4. LetG4kbe a family such that the intersection of everykmembersis non-empty.We now describe some results.No collection of non-disjoint subsets can contain a set and its comple-ment. Thus our familyFcan have at most half of the subsets IFI2n-lA maximal familyFcontains every set containing any member. Sinceevery set disjoint fromAis contained in A's complement, ifAcannot beadded to a maximal family F, A is already in it. Thus all maximal familiesconsist of exactly2n-1subsets, exactly one ofAorAfor eachA.Thus questions (a) and (b) are easily answered for families satisfyingthe non-disjointness restriction satisfied byFabove. The number ofmaximal families satisfying this restriction on the other hand has not asyet been determined very well.There exist several levels of inaccuracy in estimates of quantities ofthis kind. Some of these are listed here. One can have:(1) An exact formula.(2)A convergent formula (convergent for largento the exact result).(3) An asymptotic formula (ratio to exact result is convergent),(4) An asymptotic formula for the logarithm.In addition, one can obtain bounds upon any of these levels, one aswell as any others.We can easily find a level 4 expression for the total number of familiesF;it isn(F)= exp2[2n-1(1 +o(1))].1The argument will be described below.The analogous result for the number of maximal families is probablyexp2[(tn~21)(1+o(l))/2]but this has not been proven. It is, however,a lower bound, and an upper bound of exp2[([nn]) (1 + 0(1))]is easilyobtained.To illustrate the kind of reasoning that can be employed to obtainestimates of this kind, we sketch the argument here.A maximalFcan be characterized by its minimal members. That is,we can defineG(F)to be the family consisting of those members ofFwhich contain no others, andG(F)determines F. The familyG(F) isthen what is sometimes called a "Sperner family" or an "antichain"; no1For typographical convenience, 2x Will be denoted exp2x. 284\tRErds,D.J. Kleitman, Extremal problems among subsets of a setAmember of G contains another. (We discuss Sperner families in Section4.)Some information is available about the number of Sperner families,from this an upper bound to the number of maximalF'scan be obtained,the bound being exp2[([n%2])(1 +0(1))].To obtain a lower bound we divide theinelement subsets ofSintothose containing a given element ao,and the rest (the rest here are thecomplements of the members of the former collection). There areexp2[2([n%2])]collections Q made up ofznelement subsets containingao.Each of these determines a collectionF,withFconsisting of all setswith more than2nelements, thoseznelement sets containing aoin Qand complements of theznelement sets containing aonot in Q. The ar-gument fornodd is similar.We expect that the kind of argument used to yield the estimateexp2[([n/21)(1 +0(1))]for the number of Sperner families can be ap-plied to show that the number of Sperner families which contain no dis-joint members is exp2[([n/721)(1 +0(1))/2].Any maximal family having2n-1members has 22n-'subsets. The total member of subsets of allmaximal families, hence the total number ofF's, isno more thanexp2[2n-1+([n%21) (1 +0(1))]which is of the form exp2[2n-1(1+0(1))]as stated above.The other restrictions (1. 1,..., 1.4) have not all been investigated in asmuch detail. We first present the existent results on all these problems.Open problems are then listed.1.1. The properties ofG(F)'sare essentially the properties of maximalF's.They range in size from 1 to([nnI),the number of them can beestimated as discussed above. They are all maximal.1.2. The number of members in the union ofk F'shas been shown tobe no more than 2n- 2n-k(see[ 18]).This bound can be achieved byletting thekfamilies be all subsets containingajfor 1C j'k.1.3. Bounds on the size ofG3k(n),a family containing nokdisjointmembers, have been obtained (see[211). Forn = mk-1,these boundsare realizable; for other values, they seem to be slightly higher than thebest possible results. These results can be obtained by noticing that forany partition ofSintokblocks, at least one block must be outside ofanyG3k(n).Thisfact, for any given set of block sizes, leads to limitationon the number of members ofG3k(n)of these sizes. Manipulation ofthe limiting identities yields the results mentioned above. P.Erds,D.J. Kleitman, Extremal problems among subsets of a set\t285Smallest size of a maximalG3k(n) isno more than2n-2n-k.Thismight be conjectured to be the exact result.1.4. Among the maximal F's are families consisting of all subsets con-taining some single element. Such families have the property that all in-tersections are non-empty. Thus the restriction (onG4k(n))that everykmembers have non-vanishing intersection does not reduce the maximalsize ofG4k(n)below2n-1.There are two natural questions which arisehere. What is the maximal size ofG4k(n)'sin which there exist(k + 1)members whose intersection vanishes? Also what is the minimal size ofa maximalG4k(n)?Milner [33, 34] has some results on the first ofthese questions. The second is open.We now list some open problems in this area.(1) What is the number of maximal families no two members of whichare disjoint?(2) How small can a family be that is maximal with respect to theproperty that is the union ofkdifferent maximal families no two mem-bers of which are disjoint? It is asymptotic to2n-1for largenc.(3) How many such families are there?(4) Does the smallest maximalG3khave2n-2n-kmembers?(5) What are the exact upper bounds onG3k(n)?(6) What is the minimal size of a maximalG4k(n)?(7) What are more exact estimates on the number of families of eachtype indicated?2. Sizelimited intersectionIn the problems described so far, the basic restriction was that inter-sections do not vanish. Such restrictions can be replaced by size limita-tion on intersections. Thus we could instead require that noAiandAjin F satisfyIA,.nAiI?k,IAiuAiI-IAinAi�I_k,IA,nAiIk,IAiuA/I-IAinAiICk,IAIn AEIIAl.nA~Ik,k. 286P.Erds,D.J. Kleitman, Extremal problems among subsets of a setnrfq(~)(k)The entire range of problems considered above can be raised aboutfamilies defined by each of these restrictions. The generalization whichmost retains the flavor of Section 1 is the first. A maximal sized familyFk(n)restricted by it, consists of all subsets having2(n + k + 1)or moreelements, with((n+k-1) L2)other sets ifn + k isodd. That this is the,largest possible size forN'k(n)was proven byKatona[12]. Few of theother problems have been examined under this restriction.The opposite restriction that subsets do not intersect "too much" isvaguely related to packing and coding problems. The number of membersof size?kof a family restricted so that no two members satisfyI Ain Af�I 1 is at most(k)and is achieved by choosing all subsets ofsizek.If we letfqbe the number of members of such a family havingqelements. We obtainas a size restriction.The coding problem can be described as the study of families limitedby the restriction that the "symmetric difference" between any twomembers be no less thank.The symmetric difference betweenAiandAiisAiUAj-Ain A,.There are many results on the maximal size of codesunder these restrictions and on constructions of optimal codes. Many ofthese are described in, for example, [4].Another problem of this general kind is: How large can a family ofsubsets of S be if the symmetric difference between members is alwaysFor evenq,it has been show that maximal size families consistof any set a and all other whose symmetric difference with it is2q.For oddq,(qnI )of the subsets differing from a by2(q + 1)may alsobe included see[ 19]).3. Intersection and rank limitationsAnother important class of problems involve families of subsets of agiven size subject to intersection restrictions of the kinds already dis-cussed.Erds, Ko and Rado [7] showed that the maximal size of a family ofsubsets of S satisfying(i)all subsets are of size kzn(withI SI =n);and P. Erd‚s, D.J. Kleitman, Extremal problems among subsets of a set\t287(ƒ) no two are disjoint, no one contains another,is(k=i ),the optimum being achieved by choosing all k elements setswhich contain a given element.If2k =n,there are a large number exp2([n/21) of such families. If2k n,however, the maximal sized family is unique up to permutationof the elements.The minimal size of maximal family here may or may not be(2kk1).Among the questions that have been raised in this area are:(1) What is the largest family if one excludes families all of whosemembers contain some element?(2)Given two families such that the members of one all intersect themembers of the other, and subject to the member size limitation de-scribed above; what can be said about their sizes?The following somewhat more general result has been obtained in thisdirection [23].LetFand G be two families of subsets of S, with the members of Fhavingkelements and the members of Gqelements. Letk+ qbe nobigger thann;ifkis no more thani(n +1), thenFcan havekor fewermembers as long as no member contains another; the same possibilityfor G is allowed. Then, eitherIFI(k-1)\tor\tIGI(1).Milner[33, 34]has certain results on the first problem above.For sufficiently largenand givenk,the family consisting of allkelement subsets including one particular element is far larger (of theorder of cnk/k!as opposed toc'nk-1/(k-1)!)than any other. Underthese circumstances, it is easy to answer many of the related questionsthat arise here.Thus, for sufficiently largenfor fixedkandq,we can show the fol-lowing:(A) The number of members in the union ofqsets of k-element non-disjoint subsets of S withI SI = nis no greater than(k-1)+(k-2)+... +(k-q).(B) The number of members of a set of k element subsets of S underthe restriction that no(q + 1)are pairwise disjoint is bounded in thesame way. 1288\tRErds,D.J. Kleitman, Extremal problems among subsets of a set(C) The number of members of a set of k element subsets of S underthe restriction that the intersection of each pair has at leastqelementsin it is at most(k q).One might conjecture that similar results hold so long as2k'n -q + 1for (A) and(B),and that the best result for (C) is the maximum over mofk-m-q\tn+q-2mp=0(2Mm +p)(k-m-p-1)Results of this kind have not yet been obtained.A related problem, also as yet unsolved, is due to Kneser [281. Howmany families of k-element subsets of S, each consisting of subsets whichare not disjoint from one another, are necessary to cover all k-elementsubsets? The answer appears to ben -2k + 1(if this number is at leastone).Restrictions of the kindsubset size = r,size of intersectionrepresent packing problems, or coding problems involving words of"fixed weight". Problems of the formsubset size = k,intersection size= qdescribe such structures as projective planes(q= 2), Steiner systems anddesigns. There exists a vast literature on such questions. Neither class ofproblems will be considered here. Erds, Ko and Rado [7] conjecturedthat if I S I = 4k andFconsists of subsets of size 2k of S which overlapby at least two, then max IFI=((zk) -(2k)2)/2.4. Containment restrictionIn this section we consider families of subsets that are subject to con-tainment restrictions. The prototype of such restrictions is that satisfiedby a "Sperner family" or antichain, no member contains another. Sperner[37] in 1927 showed that such a family could have at most([n"/2])mem-bers. Lubell [30] in 1959 and independently Meshalkin [32] in 1963 ob-tained a somewhat stronger restriction. Iffkis the number of k-element P.Erds,D.J. Kleitman, Extremal problems among subsets of a set\t289members of a Sperner family of subsets of S with ISI =n,then the in-equalitynfkl(k)1k=0holds. Equality can only occur iffk= 1for some value ofk.Sperner'sresult is a corollary of this inequality since it is trivial thatn\tnLJfkl([n12])\tfkl(k)k=0\tk=0Lubell's argument is so simple that we repeat it here. A maximal chainis a set ofn+ 1subsets ofStotally ordered by inclusion. Eachkelementsubset occurs in the same proportion (1/(k)) of maximal chains. Since nochain can contain more than one member of a Sperner family, the sumof the proportion of maximal chains containing each member cannot ex-ceed one, which is the Lubell-Meshalkin inequality.The same argument implies that the maximal number of members ina family which has at most q members in common with any chain is thesum of the largestqbinomial coefficients. This result follows from theinequalitynfkl(k)qk=0which must be satisfied by such a family. Lubell's argument can be ap-plied in many other contexts. Thus, by its use, along with certain addi-tional arguments, the following generalization has been obtained[27].Letfbe any function defined in the members of any partial order andletFbe a family which has at mostkmembers in common with anychain in the partial order. Let G be a permutation group defined on thepartial order which preservesf(forginG,f(gA) =f(A))and is a sym-metry of the partial order(ACBif and only ifgAgBfor everyginG). Then the maximum value of the sum offover the members ofF isachieved for someFwhich is the union of orbits under G. That is, thereis anFsuch thatEf(A)'rf(A)AEF\tAEF 290\tP.Erds,D.J. Kleitman, Extremal problems among subsets „f a setwith F the union of complete orbits under G. Lubell [31] has obtainedstill further generalizations of his result.The following questions have also been raised about Sperner families.LetFbe a Sperner family, let G+be the family connecting of all subsetswhich contain at least one member of F, and letGIbe the family of allsubsets ordered by inclusion with respect to at least one ofF.How large can IFIbe, given IG+I? GivenI GII ?If I�F I ([nl2 1),howmany pairsA, BwithABmust there be inF?The following results along these lines have been obtained:(1) If IFI�(k)fork in,then IG+I� Eg_a(g) (see[241).(2)IFI/IGII(~n~J)/2n(see[17]).(3) The number of containment pairs is minimized if F consists ofall subsets having[n12],[n/2] + 1, [n/2] -1, [n/2]+ 2,... elements andof the remaining members ofFall have a number of elements given bythe next entry on this list[20].The minimal number of "containment triples" has not been found asyet, although one could guess the same conclusion.Sperner's conclusion can be obtained, when the restriction of non-containment is relaxed considerably. Suppose, for example, that S isthe union of two disjoint sets Tiand T2(see [13, 17]),S= T,UT2,\tT,nT2=0,and suppose thatFis restricted such that ifADBforA,BEF,thenA -B ~ TlandA -BqT2.ThenF I([n~21),that is, Sperner's boundstill applies with these weakened requirements onF.An interesting un-solved problem is the analogue of this for S = TlUT2UT3all T's dis-joint; under these circumstances the analogous restriction onFis notsufficient to get the same bound on IF I.One can ask: What is the bestbound? Also: What are the weakest additional restrictions necessary toimpose uponFto get back to the Sperner bound in this case? One canalso ask: What analogue of Lubell's inequality can be obtained for theS = TlUT2problem?Katona[15],Sch‚nheim [36] and Erds[11]have obtained furthergeneralizations of Sperner's theorem.The number of Sperner families of subsets of S has been investigatedby many authors beginning with Dedekind. The best recent result [251is that this number is greater than exp2[([nn2])(1 + cn-1/2logn)].Katona[14] and Kruskal[291 have considered a related question.Given an f member familyFof k-element subsets of S. Let G consist ofthe(k + 1)element subsets which contain one or more members ofF. P.Erds,D.J. Kleitman, Extremal problems among subsets of a set\t291How small can IGI be, givenf?His result is an exact one:fcan be uni-quely expressed as(k-I ) +(k-2)+(k33)+... +(krmm)with rl�r2�...�rm.Then�IGI_(r2)+(r3)+...+(rmk\tk-2\tk-Meshalkin[32]has obtained a result on families of partitions ofn-element sets into k labelled blocks restricted so that no blocks properlycontains a block with the same label. The result, the largest k-nomialcoefficient, is really a corollary of the Lubell-Meshalkin identity.5. Union and intersection restrictionsThere are a number of problems that have been studied which involveintersection restriction involving three or more subsets. The followingset of limitations have been considered.(a) F1is limited in that no three membersA, B, CsatisfyAUB = C(A n B = Cwould be equivalent).(b)F2obeys the restriction that no four membersA, B, C, DsatisfyAuB=CAnB=D.(c) No three members ofA, B, CofF3satisfyAUB = CorA n B = C.(d) No three members ofF4satisfyAUBDC (equivalently,A n BCC).(e) No three members of FSsatisfyAUBCC.(f) No2kmembers ofFkform a Boolean algebra under union andintersection.(g) Given anykmembersA1...Akof.F7k,the intersectionA1n A2n A3n..,n Akis nonempty and the same restriction holds if any or allAd'sare replaced by their complements.(h) Given two disjoint members ofF8,their union is a nonmemberAuB=C,AnB=0is excluded.Results on these areas have been as follows:(a) The restrictionAUB Cwould seem to limit F1to(,n%2)(1+cn-1)members. The best limitation[26]obtained has been(n~2~) ~(1+c/n-1/2 ).(b) Under the restrictionAUB CorA n B4,D,2can havec2-nn-1/4members. Upper and lower bounds of this form have been obtained; theymay or may not be equal [ 8 ].(c) The restriction stated above probably requires thatF3can have atmost([n'/21)+1 members forneven. Clements (private communication)has found examples having this many members. 292\tP.Erds,D.J.Kleitman, Extremal problems among subsets of a set(d) The number of members of Fqis exponentially small comparedto2n.Little is known about this limitation.(e) UnderAUBqC,the size of FScannot exceed([n~21)(1 + c/n)which bound can be achieved.(f) Little is known beyond case (b) above for this restriction,(g) This problem has been considered by Joel Spencer (private commu-nication). Fork = 2,it is resolved that the bound is(nniX21).Fork = 3,upper and lower bounds of the formCnwith 1- c2have been ob-tained. They are not close to one another. This restriction includes thatof(d),namelyA1n A2A3fork ?3.(h) Roughly speaking, under these restrictions, the family G can con-tain all sets having'3nto'3nelements. Best results have been obtained forn = 3k +1.Forn = 3k, 3k + 2,there is a slight gap between the best boundand the best existing results.Another set of related problems are due to Erds and Moser[9].Re-wards for their solution are available from the former author."Find bounds forf(n) =the least number of subsets of a setAofnelements such that every subset ofAis the union of two of thef(n)sub-sets. It is easy to prove that.\/2…2n()We offer525.00deciding (with proof) whetherf(2n)'2n;is or (1.75)2for sufficiently largen.""Find bounds onf(n) =the largest number of subsets A1,A2,... , Af(n)of a set ofnelements such that the(f zn))setsAiv Ai,1 iC f(n),aredistinct. We can prove that for largen,(I+E1)n()(),where0E1E2 l, and offer525.00for findingE1, E2with62161 .01."6. MiscellanyAnother kind of problem involves families of sets of a specified sizeout of a not necessarily specified set.Two problems of this kind are:(1) Suppose that no three subsets have pairwise the same intersection,and they are of sizek.How many can there be?(2)Suppose that any subset which interests all members of the familycontains at least one member. How few members can the family have? P.Erds,D.J. Kleitman, Extremal problems among subsets of a set\t293The property mentioned in(2),called "Property 13", has been exten-sively studied. Forn = 3,one can find a 7 member family with this prop-erty. Forn = 4,the smallest family size is unknown but probably around20.Erds [5, 6] has an upper bound ofcn22nand Schmidt[35]has alower bound of2n(1 + 41n)-l.These results have recently been improvedslightly by Herzog and Sch‚nheim (private communication).The best bound for problem (1) here is probably of the formck.Thebest result obtained so far for an upper bound has been of the formk! ck[3, 10].References[11H.L. Abbott, Some remarks on a combinatorial theorem of Erds and Rado, Can. Math.Bull. 9 (2) (1966) 155-160.[2] H.L. Abbott and B. Gardner, On a combinatorial theorem of Erds and Rado, to appear.[3] H.L. Abbott, D. Hanson and N. Sauer, Intersection theorems for systems of sets, to ap-pear.[4] E. Berlekamp, Algebraic Coding Theory (McGraw-Hill, New York, 1968).[5] P. Erds, On a combinatorial problem 11, Acta. Math. Acad. Sci. Hungar. 15 (1964) 445-449.[6] P. Erds, On a combinatorial problem 111, Can. Math. Bull. 12 (1969) 413-416.[7] P. Erds, Chao Ko and R. 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