Families of sets with no matchings of sizes 3 and 4
FFamilies of sets with no matchings of sizes 3 and 4
Peter Frankl, Andrey Kupavskii * Abstract
In this paper, we study the following classical question of extremal set theory: whatis the maximum size of a family of subsets of [ π ] such that no π sets from the family arepairwise disjoint? This problem was first posed by ErdΛos and resolved for π β‘ , β π )by Kleitman in the 60s. Very little progress was made on the problem until recently. Theonly result was a very lengthy resolution of the case π = 3 , π β‘ π = 4 , π β‘ π β‘ β π ) for π β₯ Let [ π ] := { , , . . . , π } be the standard π -element set and 2 [ π ] its power set. A subset β± β [ π ] is called a family . For 0 β€ π β€ π , let (οΈ [ π ] π )οΈ denote the family of all π -subsets of [ π ].For a family β± , let π ( β± ) denote the maximum number of pairwise disjoint members of β± . Note that π ( β± ) β€ π holds unless β β β± . The fundamental parameter π ( β± ) is called the independence number or matching number .Denote the size of the largest family β± β [ π ] with π ( β± ) < π by π ( π, π ). The followingclassical result was obtained by Kleitman. Kleitmanβs Theorem ([7]) Let π β₯ , π β₯ π = π π β , we have π ( π, π ) = βοΈ π β€ π‘ β€ π (οΈ ππ‘ )οΈ , (1)for π = π π, we have π ( π, π ) = π β π (οΈ ππ )οΈ + βοΈ π +1 β€ π‘ β€ π (οΈ ππ‘ )οΈ . (2)The value π ( ππ β , π ) is attained on the family of all sets of size greater than or equal to π .The following matching example for (2) was proposed by Kleitman: {οΈ πΎ β [ π π ] : | πΎ | β₯ π + 1 }οΈ βͺ (οΈ [ π π β π )οΈ . * Moscow Institute of Physics and Technology, Ecole Polytechnique FΒ΄edΒ΄erale de Lausanne; Email: [email protected]
Research supported by the grant RNF 16-11-10014. a r X i v : . [ m a t h . C O ] A ug Note that (οΈ π π β π )οΈ = π β π (οΈ π ππ )οΈ .) Let us mention that for π = 2 both bounds (1) and (2) reduce to π ( π,
2) = 2 π β . This easy statement was proved already by ErdΛos, Ko and Rado [1].Although (1) and (2) are beautiful results, for π β₯ π ΜΈβ‘ , β π ) . For π = 3, the only remaining case was solved by Quinn [8]. However, hisargument is very lengthy and was never published in a refereed journal. In this paper, wereprove his result, as well as extend it to the case π = 4 π + 2 , π = 4. Theorem 1.
Fix an integer π β₯ . Then for π = 3 , and π = π π + π β we have π ( π, π ) = (οΈ π β π β )οΈ + βοΈ π +1 β€ π‘ β€ π (οΈ ππ‘ )οΈ . (3)The following π -matching-free family shows that β β₯ β holds in the equality above for any π β₯ π = π π + π β {οΈ πΏ β [ π ] : | πΏ | β₯ π + 1 }οΈ βͺ {οΈ πΏ β (οΈ [ π ] π )οΈ : 1 β πΏ }οΈ . Theorem 1 bridges the gap that was left between Quinnβs result and the result of the paper[2], where we verified the same statement for π = π π + π β , π β₯
5. Contrary to the intuition,the problem gets easier as π becomes larger, and thus the proof for π = 3 , each such configuration cannot be too large and thenaverage over all such configurations. However, the configuration is quite complicated, the setsin the configuration actually have weights, and, in order to bound the weighted intersection ofthe family with each configuration, we use some kind of discharging method.The method we develop here has proved to be very useful and was already used in severalpapers. In a recent paper [4], we applied it to completely resolve the following problem studiedby Kleitman: what is the maximum cardinality of a family β± β [ π ] that does not contain twodisjoint sets πΉ , πΉ , along with their union πΉ βͺ πΉ ? We refer the reader to the papers [2], [4] fora more detailed introduction to the topic and, in particular, to [2] for the discussion of the caseof general π, π . See also [5], where the method we developed was applied.We note that (1) and (2), along with more general statements, are proved using a simplerversion of our technique in [3]. Recall that β± is called an up-set if for any πΉ β β± all sets that contain πΉ are also in β± . Sincewe aim to upper bound the sizes of families β± with π ( β± ) < π , we may restrict our attention tothe families that are up-sets, which we assume for the rest of the paper.We are going to use the following inequality in the proofs: π β βοΈ π =1 (οΈ π π + π β π )οΈ β€ π β (οΈ π π + π β π )οΈ for any π β€ π, π β₯ . (4)2ndeed, we have ( π π + π β π β π )( π π + π β π β π β ) = π π + π β β π + π +1 π β π β₯ π β π β₯ π β€ π , so by the formulafor the summation of a geometric progression, π β βοΈ π =1 (οΈ π π + π β π )οΈ β€ π β β π β (οΈ π π + π β π )οΈ = 1 π β (οΈ π π + π β π )οΈ . π = 3 We first prove the theorem for π β₯
3. Suppose that π β₯ π := 3 π + 1 for this section.Consider a family β± β [ π ] with π ( β± ) <
3. Take an arbitrary cyclic permutation π (assumed inwhat follows to be the identity permutation for simplicity) and fix three disjoint π -element setsthat form arcs in that permutation. This is what we call a triple . For π₯ β [3 π + 1], the π₯ -triple is the triple of π -sets that do not contain the element π₯ . It is clear that there is a one-to-onecorrespondence between the π₯ βs and the triples. For each triple, we define three groups of setsof sizes 1 , . . . , π + 3 and assign them weights. We call this ensemble of sets an π₯ -family . Notethat the arithmetic operations in the definitions of the sets are performed modulo π .We define three groups of sets, indexed by π = 0 , ,
2. In what follows, we define group π .The π -th π -set π» ( π ) π ( π₯ ) in the π₯ -family has the form { ππ + π₯ + 1 , . . . , ππ + π₯ + π } . The set π» ( π β π ) π ( π₯ ) of size π β π for π < π has the form { ππ + π + π₯ + 1 , . . . , ππ + π₯ + π } . That is,it consists of the last π β π elements of the π -set, if seen in the clockwise order. The sets β β π» (1) π ( π₯ ) β π» (2) π ( π₯ ) β Β· Β· Β· β π» ( π ) π ( π₯ ) form a full chain. The definition of the sets of size β₯ π + 1 is less straightforward. Each of the ( β₯ π + 1)-sets in the π -th group contains thecorresponding π -set. The ( π + 1)-set π» ( π +1) π ( π₯ ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { π₯ } in group π is called central . Note that the extra element it has is the element that was left outby the π -sets and so π» ( π +1) π ( π₯ ; π₯ ) is disjoint of π» ( π ) π ( π₯ ) for π ΜΈ = π . The two others π» ( π +1) π ( π ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { ππ + π₯ + 1 } for π β { , , } β { π } are called lateral and are disjoint of the corresponding π» ( π β π ( π₯ ) and the remaining π -set π» ( π ) π ( π₯ ), where { π , π, π } = { , , } . For each π β { , , } β { π } , we define two ( π + 2)-elementsets: the central set π» ( π +2) π ( π₯, π ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { ππ + π₯ + 1 , π₯ } and the lateral set π» ( π +2) π ( π, π ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { ππ + π₯ + 2 , ππ + π₯ + 3 } . The former ones are disjoint of π» ( π β π ( π₯ ) and the π -set from the remaining π -th group, where { π , π, π } = { , , } , while the latter ones are disjoint of π» ( π β π ( π₯ ). Note that π» ( π +2) π ( π₯, π ; π₯ )and π» ( π +2) π ( π, π ; π₯ ) are disjoint for { π , π, π } = { , , } . Finally, we have one ( π + 3)-set in eachgroup: π» ( π +3) π ( π₯ ) := π» ( π ) π ( π₯ ) βͺ { π₯, π₯ + 1 , π + π₯ + 1 , π + π₯ + 1 } . π’ ( π ) modulo rotation. The elements of the ground set are rep-resented by dots, and the red dots represent the elements that are contained in the set. Insidethe circles we specify the length of any interval of points that simultaneously either belong ordo not belong to the set (excluding the intervals of length 1 and 2). The weights of π -sets arespecified and divided by (οΈ ππ )οΈ for shorthand. For ( π + 1)- and ( π + 2)-sets we also mention, howmany times does a given set appear as a central and lateral set.It is disjoint of the ( π β π€ π the weight of the π -element sets, withpossible superscripts π, π depending on whether the set is lateral or central, respectively. Put πΌ := (3 π + 2) π π + 3)(2 π + 1) , πΌ β² := 1 β πΌ. (5)Note that πΌ β² > πΌ . The weights are as follows ( π β₯ π€ π β π := (οΈ ππ β π )οΈ ; π€ ππ +1 := πΌ (οΈ ππ + 1 )οΈ ; π€ ππ +1 := πΌ β² (οΈ ππ + 1 )οΈ ; π€ π +3 := (οΈ ππ + 3 )οΈ ; π€ ππ +2 := 18 (οΈ ππ + 2 )οΈ ; π€ ππ +2 := 38 (οΈ ππ + 2 )οΈ . (6)Thus, the total weight of all sets in an π₯ -family is 3 βοΈ π +3 π =1 (οΈ ππ )οΈ . Each set πΉ that appears inseveral π₯ -families accumulates all the weight π€ ( πΉ ) that it was assigned. On Fig. 1, we listed all4he types of sets that are assigned non-zero weights, together with the corresponding weights.We recommend the reader to verify Fig. 1, since we shall use the information provided on thefigure later in the proof! The elements of the ground set are placed on the circle and the sets arerepresented modulo rotation. We denote by π’ ( π ) the family of all sets that got non-zero weightfor a given permutation π . Note that, for each π = 1 , . . . , π + 3, we have βοΈ πΊ βπ’ ( π ) β© ( [ π ] π ) π€ ( πΊ ) = 3 π (οΈ ππ )οΈ . Claim 2.
To prove the theorem for π = 3 , it is sufficient to show that for any π we have βοΈ πΉ ββ±β©π’ ( π ) π€ ( πΉ ) β€ π (οΈ(οΈ π β π β )οΈ + π +3 βοΈ π = π +1 (οΈ ππ )οΈ)οΈ = 3 π (οΈ ππ )οΈ + 3 π π +3 βοΈ π = π +1 (οΈ ππ )οΈ . (7) Proof.
For an event π΄ , denote by πΌ [ π΄ ] its indicator random variable. Denote ππ the identitypermutation. Indeed, if we take a permutation π uniformly at random, then, for each π =1 , . . . , π + 3, we haveE π [οΈ βοΈ πΉ ββ±β©π’ ( π ) β© ( [ π ] π ) π€ ( πΉ ) ]οΈ = βοΈ πΉ ββ±β© ( [ π ] π ) E π [οΈ βοΈ πΊ βπ’ ( ππ ) β© ( [ π ] π ) πΌ [ π ( πΉ ) = πΊ ] π€ ( πΊ ) ]οΈ == βοΈ πΉ ββ±β© ( [ π ] π ) βοΈ πΊ βπ’ ( ππ ) β© ( [ π ] π ) Pr[ π ( πΉ ) = πΊ ] π€ ( πΊ ) = βοΈ πΉ ββ±β© ( [ π ] π ) βοΈ πΊ βπ’ ( ππ ) β© ( [ π ] π ) π€ ( πΊ ) (οΈ ππ )οΈ == βοΈ πΉ ββ±β© ( [ π ] π ) 3 π = 3 π βββ β± β© (οΈ [ π ] π )οΈβββ . Therefore, (7) implies that π +3 βοΈ π =1 βββ β± β© (οΈ [ π ] π )οΈβββ = 13 π E π [οΈ βοΈ πΉ ββ±β©π’ ( π ) π€ ( πΉ ) ]οΈ β€ (οΈ π β π β )οΈ + π +3 βοΈ π = π +1 (οΈ ππ )οΈ , which implies the statement of the theorem for π = 3.Our strategy to prove (7) is as follows. For a set πΉ β π’ ( π ), we define the charge π ( πΉ ) tobe equal to π€ ( πΉ ) if πΉ β β± , and to be 0 otherwise. Clearly, βοΈ πΉ βπ’ ( π ) π ( πΉ ) = βοΈ πΉ ββ±β©π’ ( π ) π€ ( πΉ ).If among the ( β€ π β π’ ( π ) there are no sets from β± , as well as there are at most ππ -sets, then we are done since each π -set appears in exactly three π₯ -families. Otherwise, certain( β₯ π + 1)-sets do not appear in β± . Then we transfer (a part of) the charge of the ( β€ π )-setsto the ( β₯ π + 1)-sets that have zero charge. We show that the total charge transferred to each( β₯ π + 1)-set is at most its weight. As a result of this procedure, the ( β€ π β π -sets will have total charge at most 3 π (οΈ ππ )οΈ , and each ( β₯ π + 1)-set willhave a charge not greater than its own weight. This will obviously conclude the proof of thetheorem.Next, we design a charging scheme that satisfies the above requirements. For the sets of sizeat most π β
1, we transfer their charge within each π₯ -family, assuring that the charge that we5 ( π β π ( π₯ ) π» ( π +2) π ( π , π ; π₯ ) π» ( π +2) π ( π₯, π ; π₯ ) π» ( π β π ( π₯ ) π» ( π β π ( π₯ ) π» ( π +3) π ( π₯ ) π» ( π β π ( π₯ ) π» ( π ) π ( π₯ ) π» ( π +2) π ( π₯, π ; π₯ ) π» ( π β π ( π₯ ) π» ( π +1) π ( π₯ ; π₯ ) π» ( π +1) π ( π ; π₯ ) π» ( π ) π ( π₯ ) π» ( π ) π ( π₯ ) π» ( π +1) π ( π₯ ; π₯ )Table 1: The list of all types of triples of pairwise disjoint sets that we employ in the proof. Weassume that { π , π , π } = { , , } . The triples are listed in the order they appear in the proof.transferred to a bigger set in one π₯ -family is smaller than the weight that this bigger set got from this π₯ -family . See Table 1 for all the triples of pairwise disjoint sets we use in the proof.The reader is welcome to verify that all the triples are actually disjoint. Stage 1. Transferring charge from the ( β€ m β ) -sets to ( m + ) -sets .Assume that, for some π₯ β [ π ] and π β { , , } , the set π» ( π β π ( π₯ ) is in the family. Choose π , π such that { π , π , π } = { , , } . Then at least one set from each of the two pairs (οΈ π» ( π +2) π ( π, π ; π₯ ) ,π» ( π +2) π ( π₯, π ; π₯ ) )οΈ , (οΈ π» ( π +2) π ( π, π ; π₯ ) , π» ( π +2) π ( π₯, π ; π₯ ) )οΈ is missing from β± . We transfer of the chargeof the subsets π» ( π ) π ( π₯ ), π β€ π β
3, which is at most βοΈ π β π =1 π€ π , to some two of these missingsets. We again remark that, for each π₯ , we transfer only the part of the charge of the sets π» ( π ) π ( π₯ )that they got as the member of the π₯ -family. We have12 π β βοΈ π =1 π€ π = 12 π β βοΈ π =1 (οΈ ππ )οΈ (4) β€ (οΈ ππ β )οΈ = 12 βοΈ π =0 π + 2 β π π + π (οΈ ππ + 2 )οΈ < (οΈ ππ + 2 )οΈ . (We could have put instead of , but it does not matter for the calculations.) Since each( π + 2)-set in the π₯ -family may get this charge from each of the two groups to which it doesnot belong, the total charge transferred in that way is at most (οΈ ππ +2 )οΈ (6) = π€ ππ +2 . The lateral( π + 2)-sets are not going to get any more charge. As for the central ( π + 2)-sets, we have tomake sure that they will get not more than π€ ππ +2 β π€ ππ +2 (6) = (οΈ ππ +2 )οΈ additional charge.We note that the ( π β π β Stage 2. Transferring charge from pairs of ( m β ) -sets to ( m + ) -sets .Due to the fact that β± is an up-set, from now on we may assume that the charge of any ( β€ π β π₯ and π, π , π , where { π , π , π } = { , , } , both π» ( π β π ( π₯ ) and π» ( π β π ( π₯ ) belong to β± . Then the set π» ( π +3) π ( π₯ ) is not in β± and, consequently, has zero charge.Transfer the charge of the sets π» ( π ) π ( π₯ ) and π» ( π ) π ( π₯ ), π = π β , π β
1, to π» ( π +3) π ( π₯ ). The chargetransferred is at most 2 π€ π β + 2 π€ π β , which is2 (οΈ ππ β )οΈ + 2 (οΈ ππ β )οΈ β€ (οΈ ππ )οΈ = 3( π + 1)2(2 π + 1) (οΈ ππ + 1 )οΈ β€ (οΈ ππ + 1 )οΈ β€ (οΈ ππ + 3 )οΈ (6) = π€ π +3 , since π β₯
3. The ( π + 3)-sets are not going to get any more charge.6 tage 3. Transferring charge from ( m β ) -sets paired with m-sets to central ( m + ) -sets .Assume that, for some π₯ β [ π ] and π, π , π , where { π , π , π } = { , , } , both π» ( π β π ( π₯ ) and π» ( π ) π ( π₯ ) belong to β± . Then the central ( π +2)-set π» ( π +2) π ( π₯, π ; π₯ ) is not in β± and, consequently,received at most π€ ππ +2 charge within the π₯ -family (because of the charge possibly transferredon Stage 1). Transfer the charge of the sets π» ( π ) π ( π₯ ), π = π β , π β
1, to π» ( π +2) π ( π₯, π ; π₯ ). Thecharge transferred is at most π€ π β + π€ π β , which is (οΈ ππ β )οΈ + (οΈ ππ β )οΈ = 3 π + 22 π + 3 (οΈ ππ β )οΈ = 3 π + 22 π + 3 βοΈ π =0 π + π π β π + 2 (οΈ ππ + 2 )οΈ β€ (οΈ ππ + 2 )οΈ . The last inequality is valid for any π β₯ π€ ππ +2 β π€ ππ +2 , and so the total charge of the central ( π + 2)-sets is at most π€ ππ +2 in each π₯ -family. The ( π + 2)-sets are not going to get any more charge. Stage 4. Transferring charge from single ( m β ) -sets to ( m + ) -sets .After the above redistribution of charges, for each π₯ β [ π ], there are no ( β€ π β π β π β π₯ -family, moreover, we cannothave an ( π β π -sets with non-zero charges in the π₯ -family.Assume that for some π₯ β [ π ] and π, π , π , where { π , π , π } = { , , } , the set π» ( π β π ( π₯ )belongs to β± . Then one set from each of the two pairs (οΈ π» ( π +1) π ( π₯ ; π₯ ) , π» ( π +1) π ( π ; π₯ ) )οΈ and (οΈ π» ( π +1) π ( π ; π₯ ) , π» ( π +1) π ( π₯ ; π₯ ) )οΈ is missing from β± . We transfer of the charge of the subsets π» ( π ) π ( π₯ ), π = π β , π β
1, which is at most ( π€ π β + π€ π β ), to each of these missing sets. Wehave12 ( π€ π β + π€ π β ) = 12 (οΈ ππ β )οΈ + 12 (οΈ ππ β )οΈ = (3 π + 2) π π + 3)(2 π + 1) (οΈ ππ + 1 )οΈ (6) = π€ ππ +1 . (8)Recall that πΌ β² > πΌ , and, therefore, π€ ππ +1 > π€ ππ +1 , which means that no ( π + 1)-set gets morecharge than its weight up to this stage. Stage 5. Transferring charge from pairs of m-sets to central ( m + ) -sets .Denote the number of π -sets that have non-zero charge (that is, that are contained in β± β© π’ ( π ))by π . If π β€ π , then we are clearly done since each π -set appears in exactly three π₯ -families.Assume that π > π . On the one hand, it makes an extra contribution 3( π β π ) (οΈ ππ )οΈ to theleft hand side of (7). On the other hand, the number of triples with two π -sets belonging to β± β© π’ ( π ) is non-zero. Indeed, if, for π β€
2, we denote by π§ π the number of triples with π π -setsin the family, then, since each π -set participates in three triples, we have π§ + 2 π§ = 3 π . Since π§ + π§ + π§ = π , we have π§ β₯ π β π . Assume that for some π₯ β [ π ] and π, π , π , where { π , π , π } = { , , } , both π» ( π ) π ( π₯ ) and π» ( π ) π ( π₯ ) belong to β± . Then the central ( π + 1)-set π» ( π +1) π ( π₯ ; π₯ ) is not in the family. Moreover, no charge was transferred to it from the π₯ -family,since we could not have had two π -sets and an ( π β π β π ) π§ (οΈ ππ )οΈ charge to π» ( π +1) π ( π₯ ; π₯ ) from the π -sets.First note that we have transferred π§ π β π ) π§ (οΈ ππ )οΈ charge from the π -sets to the central ( π +1)-sets, which results in π -sets having total charge of 3 π (οΈ ππ )οΈ . This is precisely what we neededto have, and we are only left to verify that we did not overcharge the central ( π + 1)-sets.7nfortunately, we can run into problems in this situation, so we have to consider two cases.First, assume that π β₯ π + 2. Then the charge on each central set in each π₯ -family is at most3( π β π ) π§ (οΈ ππ )οΈ β€ π β π )3 π β π (οΈ ππ )οΈ β€ (οΈ ππ )οΈ = 6( π + 1)5(2 π + 1) (οΈ ππ + 1 )οΈ . (9)The second inequality holds due to the fact that for π β₯ π + 1 the function π β π )3 π β π decreases as π grows. Therefore, if πΌ β² = (1 β πΌ ) β₯ π +1)5(2 π +1) , then we are done. Let us verify that this inequalityis implied by (5). Adding 2 πΌ to the right hand side of the inequality, we get6( π + 1)5(2 π + 1) + (3 π + 2) π π + 3)(2 π + 1) = 12( π + 1)(2 π + 3) + 5(3 π + 2) π π + 3)(2 π + 1) == 39 π + 70 π + 3640 π + 80 π + 30 < , where the last inequality holds for any π β₯
1. Thus, we fulfilled all the requirements on thecharging scheme and we are done in the case π β₯ π + 2.In the case π = π +1, however, we run into trouble: the inequality πΌ β² β₯ π β π )( π +1) π§ (2 π +1) = π +1) π§ (2 π +1) may not hold. Recall that π§ β₯ π β π = 2. We are still fine if π§ β₯ π§ = 2, which weassume until the end of this section. The equation π§ = 2 means that there are exactly twotriples with two π -sets from β± . This is the only part of the proof when we are not going tocompare the amount of charge passed to the ( π + 1)-sets to the portion of its weight inside the π₯ -family. Instead, we compare the charge to the full weight of the ( π + 1)-set.We have two possible configurations with π§ = 2. One possibility is that we have two π -setsfrom β± forming an interval of length 2 π on the circle, and then the two triples contributing to π§ share the same two π -sets. In this case, the central ( π + 1)-set that we forbid is the samein both triples, and it is of type 1a (see Fig. 1). Recall that this set has weight 2( πΌ + πΌ β² ) (οΈ ππ +1 )οΈ .The other possibility is that we have two pairs of π -sets, with each pair separated on the twosides by a third π -set forming a triple with the pair, and by the element missing from the triple,respectively. In this case, in each of the corresponding two π₯ -families we forbid a central ( π + 1)-set of type 1b. Each of these two sets (that are clearly different) has weight ( πΌ + πΌ β² ) (οΈ ππ +1 )οΈ . Ineither case, we need to transfer π +1)2 π +1 (οΈ ππ +1 )οΈ amount of charge from the π -sets to some of the( π + 1)-sets. We transfer this weight to the central ( π + 1)-set(s), possibly overcharging it.Assume first that we do not have any ( π β β± β© π’ ( π ). Then, in either of thepossibilities described above, the central ( π + 1)-sets have zero charge before Stage 5. Therefore,we are good if the weight of these (one or two) ( π + 1)-sets is greater than the amount of chargewe transfer to them from the pairs of π -sets. Namely, we are good if2( πΌ + πΌ β² ) = 2 β πΌ β₯ π + 1)2 π + 1 β β₯ π + 1)2 π + 1 + (3 π + 2) π π + 3)(2 π + 1) . (10)We have 3( π + 1)2 π + 1 + (3 π + 2) π π + 3)(2 π + 1) = 15 π + 32 π + 188 π + 16 π + 6 < , where the last inequality holds for π β₯
3. Thus, this case is covered.8igure 2: In each of the two cases the ( π β π + 1)-sets, which form a matching with the ( π β π β πΉ β β± β© π’ ( π ). Then, asone can see from Fig. 2, it forbids at least one ( π + 1)-set of each of the types 1a and 1b toappear. Denote them by π , π . We have seen two paragraphs above that in either case of thearrangement of the π -sets we forbid sets of the same type (either one of type 1a, or two of type1b). Therefore, at least one of π , π that did not get any charge at Stage 5. We assume thatit is a set π of type 1b (the other case is easier and is treated similarly). The set π appearsin two π₯ -families and got some charge only at Stage 4. Moreover, π got at most πΌ (οΈ ππ +1 )οΈ chargefrom each of the two π₯ -families. Thus, the charge of π after all five stages is at most 2 πΌ (οΈ ππ +1 )οΈ .This, in turn, means that it has extra capacity of at least ( πΌ β² β πΌ ) (οΈ ππ +1 )οΈ . We redistribute somepart of the charge from the two ( π + 1)-sets (that appeared in the π₯ -families with the pairs of π -sets from β± ) to π . In order to be able to fulfil the requirements on the charging scheme, weneed the total capacity of these ( π + 1)-sets to be greater than the charge we transfer. Moreprecisely, it is sufficient if the following inequality holds:(3 πΌ β² β πΌ ) β₯ π + 1)2 π + 1 β β πΌ β₯ π + 1)2 π + 1 . Note that we replaced the capacity of the (one or two) missing central ( π + 1)-set(s) by 2 πΌ β² sincethe πΌ -part of the charge may have been already used up by the ( π β β πΌ with 2 β πΌ .One can easily see (cf. (5)) that πΌ β€ , so we have 3 β πΌ > β πΌ . The case π = π + 1 isexamined in its entirety, and the proof of the theorem in the case π = 3 , π β₯ π β€ , π = 3 In the argument above, we assumed that π β₯
3. However, we want to prove the theorem for π β₯
1, which leaves us with two cases: π = 1 and π = 2. If π = 1, then we have π = 4,and we have to show that at least four sets, including the empty set, are missing from a family β± β [4] with π ( β± ) β€
2. If there is at most one singleton in β± , then we are done. If there are atleast two singletons, say, { } and { } , then β± β© { , } is empty, which gives 4 missing sets. The9ase π = 1 is covered.If π = 2 then π = 7, and we have to show that at least 1 + (οΈ )οΈ + (οΈ )οΈ = 23 sets are missingfrom β± . If there is at least one singleton in β± , say { } , then β± β© [2 , is intersecting, and so, bythe ErdΛos-Ko-Rado theorem, a half of the sets are missing from it. This gives 32 missing sets.Thus, we may assume that there are no sets of size smaller than 2 in β± . Now we may slightlymodify the proof for the case π β₯ π β₯
2. Namely, among the ( π + 1)-sets,we give weights only to the central ( π + 1)-sets (the weights on other layers stay the same).Claim 2 stays true in this case. Since we do not have sets of size smaller than π , we can goto Stage 5 of the analysis, where we want to show that with the new weights (9) holds for any π β₯ π + 1 for π = 2 , π = 7:3( π β π )3 π β π (οΈ ππ )οΈ β€ (οΈ ππ )οΈ β€ (οΈ ππ + 1 )οΈ β (οΈ )οΈ = 632 β€ (οΈ )οΈ = 35 . The last inequality obviously holds. Thus, for π = 2 we may terminate the proof right after (9).The proof is complete. π = 4 We first prove the theorem for π β₯
3. We put π := 4 π + 2 for some π β₯ π = 3 π + 1, andthe proof is in a sense even simpler. We present it somewhat more concisely.We fix an arbitrary permutation π of the ground set. For simplicity, we assume that π is theidentity permutation. Quite predictably, define four groups of sets, indexed by π = 0 , , , π₯ -family . The four π -sets π» ( π ) π ( π₯ ) in an π₯ -family are disjoint and form an intervalof length 4 π , leaving two contiguous elements π₯ β , π₯ out (thus, the π₯ -family is indexed bythe last of the two missing elements in the clockwise order). In what follows, we define the π -thgroup. The sets in the π -th group of size π β π, π = 1 , . . . , π , form a full chain together with π» ( π ) π ( π₯ ): π» ( π β π ) π ( π₯ ) := { π₯ + 1 + π + ππ, . . . , π₯ + ( π + 1) π } . We again have both central and lateral ( π + 1)- and ( π + 2)-sets. The ( π + 1)-sets π» ( π +1) π ( π₯ β² ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { π₯ β² } for π₯ β² = π₯, π₯ β π are called central . Note that the extra element in both sets is left out by the π -sets,and so π» ( π +1) π ( π₯ β² ; π₯ ) for both π₯ β² = π₯ β , π₯ is disjoint of the π -set from the π -th group, π ΜΈ = π .The three others π» ( π +1) π ( π ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { ππ + π₯ + 1 } for π β { , . . . , } β { π } are called lateral and are disjoint of the corresponding π» ( π β π ( π₯ ) and of the π -set in the π β² -group, π β² ΜΈ = π, π .For each π β { , . . . , } β { π } , we define two lateral ( π + 2)-element sets: π» ( π +2) π ( π₯ β² , π ; π₯ ) := π» ( π +1) π ( π ; π₯ ) βͺ { π₯ β² } for π₯ β² = π₯, π₯ β , π we define one central set: π» ( π +2) π ( π₯ β , π₯ ; π₯ ) := π» ( π ) π ( π₯ ) βͺ { π₯ β , π₯ } . The former ones are disjoint of the ( π β π and the two π -sets from theremaining groups, while the latter one is disjoint of the three π -sets from the groups π, π ΜΈ = π .Finally, we have one ( π + 5)-element set for each π : π» ( π +5) π ( π₯ ) := π» ( π ) π ( π₯ ) βͺ { π₯ β , π₯, π₯ + 1 , π + π₯ + 1 , π + π₯ + 1 , π + π₯ + 1 } . Each set in each group gets a weight. We denote by π€ π the weight of the π -element sets,with possible superscripts π, π depending on whether the set is lateral or central, respectively.The weights are as follows ( π β₯ π€ π β π := (οΈ ππ β π )οΈ ; π€ ππ +1 := π π + 2) (οΈ ππ + 1 )οΈ ; π€ ππ +1 := 12 (οΈ ππ + 1 )οΈ β π€ ππ +1 ; π€ π +5 := (οΈ ππ + 5 )οΈ ; π€ ππ +2 := 122 (οΈ ππ + 2 )οΈ ; π€ ππ +2 := 811 (οΈ ππ + 2 )οΈ . (11)It is easy to check that, for each π and π β { , . . . , π + 2 } βͺ { π + 5 } , in each group the weightof π -element sets sums up to (οΈ ππ )οΈ for fixed π₯ , and that π€ ππ +1 is positive.As before, each set πΉ that appears in some π₯ -families accumulates all the weight π€ ( πΉ ) thatit was assigned. We denote by π’ ( π ) the family of all sets that got non-zero weight for a givenpermutation π . Analogously to Claim 2, to prove the theorem in this case, it is sufficient toshow that for any π we have βοΈ πΉ ββ±β©π’ ( π ) π€ ( πΉ ) β€ π (οΈ(οΈ π β π β )οΈ + βοΈ π β{ , , } (οΈ ππ + π )οΈ)οΈ = 4 π (οΈ ππ )οΈ + 4 π βοΈ π β{ , , } (οΈ ππ + π )οΈ . (12)For a set πΉ β π’ ( π ), we define the charge π ( πΉ ) to be equal to π€ ( πΉ ) if πΉ β β± , and π ( πΉ ) := 0otherwise. Clearly, βοΈ πΉ βπ’ ( π ) π ( πΉ ) = βοΈ πΉ ββ±β©π’ ( π ) π€ ( πΉ ). We again design a scheme for the transferof (a part of) the charge of the ( β€ π )-sets to the ( β₯ π + 1)-sets that have zero charge. Weshow that the charge transferred to each ( β₯ π + 1)-set is at most its weight. As a result ofthis procedure, the ( β€ π β π -sets will have total charge4 π (οΈ ππ )οΈ , and each ( β₯ π + 1)-set will have charge not greater than its own weight. This willobviously conclude the proof of the theorem.Next we design a charging scheme that satisfies the above requirements. For π = 4 π + 2 itis sufficient in all cases to redistribute the charge within each π₯ -family, assuring that the chargethat we transferred to the larger set in one π₯ -family is smaller than the weight that this biggerset got from this π₯ -family. Stage 1. Transferring charge from triples of ( m β ) -sets to ( m + ) -sets .Assume that, for some π₯ β [ π ] and π , π , π , π , where { π , π , π , π } = { , , , } , the sets π» ( π β π π’ ( π₯ )for π’ β [3], belong to β± . Then π» ( π +5) π ( π₯ ) is missing from β± , and, consequently, has zero charge.Transfer all the charge of the sets π» ( π ) π π’ ( π₯ ), π β€ π β π + 5)-set.The charge transferred is at most 3 βοΈ π β π =1 π€ π , which is3 (οΈ ππ β )οΈ + 3 π β βοΈ π =1 (οΈ ππ )οΈ (4) β€ (οΈ ππ β )οΈ = 9 βοΈ π =0 ( π + π )2 βοΈ π = β (3 π + π ) (οΈ ππ + 5 )οΈ < (οΈ ππ + 5 )οΈ (11) = π€ π +5 , π β₯
2. Note that we apply (4) for π = π β π + 5)-sets are not going to get any morecharge. Stage 2. Transferring charge from pairs of ( m β ) -sets to lateral ( m + ) -sets .Assume that, for some π₯ β [ π ] and π , π , π , π , where { π , π , π , π } = { , , , } , both π» ( π β π ( π₯ )and π» ( π β π ( π₯ ) belong to β± . Then in each of the four pairs (οΈ π» ( π +2) π ( π₯ β² , π β² ; π₯ ) , π» ( π +2) π ( π₯ β²β² , π β²β² ; π₯ ) )οΈ ,where { π β² , π β²β² } = { π , π } , { π₯ β² , π₯ β²β² } = { π₯ β , π₯ } one of the ( π + 2)-sets is missing from β± , and,consequently, has zero charge. Note that all these ( π + 2)-sets are lateral. Transfer one quarterof the charge of the sets π» ( π ) π ( π₯ ) and π» ( π ) π ( π₯ ), π β€ π β
1, to each of these missing sets.The charge transferred to each lateral ( π + 2)-set is at most βοΈ π β π =1 π€ π , which is12 π β βοΈ π =1 (οΈ ππ )οΈ (4) β€ (οΈ ππ β )οΈ = 3 π ( π + 1)( π + 2)4(3 π + 3)(3 π + 2)(3 π + 1) (οΈ ππ + 2 )οΈ < (οΈ ππ + 2 )οΈ (11) = π€ ππ +2 , where the last inequality holds for any π β₯
1. The lateral ( π + 2)-sets are not going to get anymore charge. Stage 3. Transferring charge from single ( m β ) -sets to ( m + ) -sets .After the above redistribution of charges, we have at most one ( π β π₯ -family.Assume that for some π₯ β [ π ] and π, π , π , π , where { π , π , π , π } = { , , , } , the set π» ( π β π ( π₯ ) belongs to β± and still has non-zero charge. Then one set from each of the sixtriples (οΈ π» ( π +1) π ( π ( π ); π₯ ) , π» ( π +1) π ( π ( π₯ β π₯ ) , π» ( π +1) π ( π ( π₯ ); π₯ ) )οΈ , where π is a permutation of theset { π, π₯ β , π₯ } , is missing from β± . It is not difficult to see that it means that at least three outof the listed sets are missing from β± . Note that among the possible missing sets there are bothcentral and lateral ( π + 1)-sets.We transfer of the charge of π» ( π ) π ( π₯ ), π β€ π β
1, to each of the three missing sets. This isat most βοΈ π β π =1 π€ π , which is13 π β βοΈ π =1 (οΈ ππ )οΈ (4) β€ (οΈ ππ β )οΈ = π ( π + 1)2(3 π + 3)(3 π + 2) (οΈ ππ + 1 )οΈ = π π + 2) (οΈ ππ + 1 )οΈ (11) < π€ ππ +1 , We are not going to transfer any more weight to the lateral ( π + 1)-sets. Stage 4. Transferring charge from pairs and triples of m-sets .At this stage only the sets of size greater than or equal to π have non-negative charge. Denotethe number of π -sets that have non-zero charge (that is, that are contained in β± β© π’ ( π )) by π .If π β€ π , then we are clearly done.Assume that π > π . On the one hand, it makes an extra contribution 4( π β π ) (οΈ ππ )οΈ to theleft hand side of (12). On the other hand, the number of quadruples with two or three π -setsbelonging to β± β© π’ ( π ) is non-zero. Indeed, if we denote by π§ π the number of quadruples with ππ -sets in the family, for π β€
3, then we have π§ + 2 π§ + 3 π§ = 4 π . Since π§ + π§ + π§ + π§ = π ,we have π§ + 2 π§ β₯ π β π. (13)12e proceed as follows. (i) Triples of m-sets. Assume that, for some π₯ β [ π ] and π , π , π , π , where { π , π , π , π } = { , , , } , the sets π» ( π ) π π’ ( π₯ ) belong to β± for all π’ = 1 , ,
3. Then the central ( π + 2)-set π» ( π +2) π ( π₯ β , π₯ ; π₯ ) is not in the family β± . Moreover, it has zero charge. We transfer π β π )4 π β π (οΈ ππ )οΈ charge to this set. We have π β π )4 π β π β€ π β₯ π + 1, since this function for π β₯ π + 1 decreasesas π grows. Therefore, we have8( π β π )4 π β π (οΈ ππ )οΈ β€ (οΈ ππ )οΈ = 4 ( π + 1)( π + 2)(3 π + 2)(3 π + 1) (οΈ ππ + 2 )οΈ β€ (οΈ ππ + 2 )οΈ (11) = π€ ππ +2 . (14)The last inequality holds for π β₯ (ii) Pairs of m-sets. Assume that for some π₯ β [ π ] and π , π , π , π , where { π , π , π , π } = { , , , } , exactly two π -sets π» ( π ) π ( π₯ ) and π» ( π ) π ( π₯ ) from the π₯ -family belong to β± . Then in eachof the two pairs of central ( π + 1)-sets (οΈ π» ( π +1) π ( π₯ β² ; π₯ ) , π» ( π +1) π ( π₯ β²β² ; π₯ ) )οΈ for { π₯ β² , π₯ β²β² } = { π₯ β , π₯ } one of the sets is not in the family. Moreover, each of them has received at most π€ ππ +1 chargewithin this π₯ -family (they could have received charge only in Stage 3). We transfer π β π )4 π β π (οΈ ππ )οΈ charge to each of the two missing central sets. We have to verify that the charge transferred isat most π€ ππ +1 β π€ ππ +1 . Note that π β π )4 π β π β€
1. Therefore, it is enough to verify (οΈ ππ )οΈ β€ π€ ππ +1 β π€ ππ +1 (11) β (οΈ ππ )οΈ = 2 π + 23 π + 2 (οΈ ππ + 1 )οΈ β€ (οΈ ππ + 1 )οΈ β π€ ππ +1 . (15)The last inequality holds (with equality) since by (11) we have 5 π€ ππ +1 = π π +2 (οΈ ππ +1 )οΈ .Now we only have to make sure that we have transferred enough charge. Indeed, we havetransferred a total amount of charge equal to4( π β π )4 π β π (οΈ ππ )οΈ π§ + 8( π β π )4 π β π (οΈ ππ )οΈ π§ = 4( π β π )4 π β π (οΈ ππ )οΈ ( π§ + 2 π§ ) (13) β₯ π β π ) (οΈ ππ )οΈ . Therefore, the total amount of charge that is left on the π -sets is at most 4 π (οΈ ππ )οΈ , moreover, allsets of size not greater than π β π = 4 , π β₯ π β€ , π = 4 In the argument above, we assumed that π β₯
3. However, we wang to prove the theorem for π β₯
1, which leaves us with two cases: π = 1 and π = 2. If π = 1, then we have π = 6, andwe have to show that at least 6 sets, including the empty set, are missing from a family β± β [6] with π ( β± ) β€
3. If there is at most one singleton in β± , then we are done. If there are at leasttwo singletons, say, { } and { } , then β± β© [3 ,
6] is intersecting, and, consequently, at least 8 setsare missing from β± among the sets from 2 [3 , . The case π = 1 is covered.13f π = 2 then π = 10 and we have to show that at least 1 + (οΈ )οΈ + (οΈ )οΈ = 47 sets are missingfrom β± . If there is at least one singleton in β± , say { } , then, applying (2) to β± β© [2 , , we getthat at least 1 + (οΈ )οΈ + (οΈ )οΈ + (οΈ )οΈ sets are missing from β± , which is more than 47.Thus, we may assume that there are no sets of size smaller than 2 in β± . Now we may slightlymodify the proof for the case π β₯ π β₯
2. Namely, among the ( π + 1)-and ( π + 2)-sets we give weights only to the central ( π + 1)- and ( π + 2)-sets (each central( π + 1)-set receives a weight of (οΈ ππ +1 )οΈ , each central ( π + 2)-set receives a weight of (οΈ ππ +2 )οΈ , andthe weights on other layers stay the same). Since β± contains no sets of size smaller than π , wemay go to part 4 of the analysis, where we have to verify the following analogues of (15) and(14) for π = 2 , π = 10:2 π + 23 π + 2 (οΈ ππ + 1 )οΈ β€ (οΈ ππ + 1 )οΈ , π + 1)( π + 2)(3 π + 2)(3 π + 1) (οΈ ππ + 2 )οΈ β€ (οΈ ππ + 2 )οΈ . Both hold for π = 2. The rest of the proof stays the same. The proof is complete. Acknowledgements.
We thank the anonymous referees for their helpful commentson the presentation of the paper.
References [1] P. ErdΛos, C. Ko, R. Rado,
Intersection theorems for systems of finite sets , The QuarterlyJournal of Mathematics, 12 (1961) N1, 313β320.[2] P. Frankl, A. Kupavskii,
Families with no π pairwise disjoint sets , Journal of the LondonMath. Soc. 95 (2017), N3, 875β894.[3] P. Frankl, A. Kupavskii, Two problems on matchings in set families β in the footsteps ofErdΛos and Kleitman , accepted at J. Comb. Th. Ser. B, arXiv:1607.06126[4] P. Frankl, A. Kupavskii,
Partition-free families of sets , accepted at Proceedings of theLondon Mathematical Society, arXiv:1706.00215[5] P. Frankl, A. Kupavskii,
New inequalities for families without π pairwise disjoint members ,J. Comb. Th. Ser. A 157 (2018), 427-434.[6] G. Katona, Intersection theorems for systems of finite sets , Acta Math. Acad. Sci. Hung.15 (1964), 329β337.[7] D.J. Kleitman,
Maximal number of subsets of a finite set no π of which are pairwisedisjointof which are pairwisedisjoint