AbstractWe survey results concerning the maximum size of a family F of subsets of an n-element set such that a certain configuration is avoided. When F avoids a chain of size two, this is just Sperner's theorem. Here we give bounds on how large F can be such that no four distinct sets A,B,C,D∈F satisfy A⊂B, C⊂B, C⊂D. In this case, the maximum size satisfies (n⌊n2⌋)(1+1n+Ω(1n2))⩽|F|⩽(n⌊n2⌋)(1+2n+O(1n2)), which is very similar to the best-known bounds for the more restrictive problem of F avoiding three sets B,C,D such that C⊂B, C⊂D

Griggs, Jerrold R.

Katona, Gyula

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No four subsets forming an NJerrold R. Griggs∗University of South CarolinaColumbia, SC 29208 USAgriggs@math.sc.eduGyula O.H. Katona†Re´nyi InstituteBudapest, Hungaryohkatona@renyi.huMarch 2, 20071 IntroductionLet [n] = {1, 2, . . . , n} be a finite set, F ⊂ 2[n] a family of its subsets. Inthe present paper max |F| will be investigated under certain conditions onthe family F . The well-known Sperner’s Theorem ([7]) was the first suchdiscovery.Theorem 1.1 If F is a family of subsets of [n] without inclusion (F,G ∈ Fimplies F 6⊂ G) then|F| ≤(nbn2c)holds, and this estimate is sharp as the family of all bn2c-element subsetsshows.There is a very large number of generalizations and analogues of thistheorem. Here we will mention only some results when the condition on Fexcludes certain configurations that can be expressed by inclusion only. That∗The first author was supported in part by NSF grant DMS-0302307.†The second author was supported by the Hungarian National Foundation for ScientificResearch grant numbers NK0621321, AT048826, the Bulgarian National Science Fundunder Grant IO-03/2005 and the projects of the European Community: INTAS 04-77-7171, COMBSTRU–HPRN-CT-2002-000278, FIST–MTKD-CT-2004-003006.1is, no intersections, unions, etc. are involved. The first such generalizationwas obtained by Erdo˝s [3]. The family of k distinct sets with mutual inclu-sions, F1 ⊂ F2 ⊂ . . . Fk is called a chain of length k, which we denote simplyby Pk. For any family of sets F , with specified inclusions between pairs ofsets, let La(n, F ) denote the size of the largest family F of subsets of [n]without any F . Erdo˝s extended Sperner’s Theorem as follows:Theorem 1.2 [3] La(n, Pk+1) is equal to the sum of the k largest binomialcoefficients of order n.Now consider families other than chains. Let Vr denote the r-fork, whichis a family of r+ 1 distinct sets: F ⊂ G1, F ⊂ G2, . . . , F ⊂ Gr. The quantityLa(n, Vr) was first (asymptotically) determined for r = 2.Theorem 1.3 [5](nbn2c)(1 +1n+O(1n2))≤ La(n, V2) ≤(nbn2c)(1 +2n).This was recently generalized:Theorem 1.4 [1], cf. [8](nbn2c)(1 +rn+O(1n2))≤ La(n, Vr+1) ≤(nbn2c)(1 + 2rn+O(1n2)).Four distinct subsets satisfying A ⊂ C,A ⊂ D,B ⊂ C,B ⊂ D are calleda butterfly and are denoted by B.Theorem 1.5 [2] Let n ≥ 3. Then La(n,B) = ( nbn/2c)+ ( nbn/2c+1).Let us compare Theorems 1.3 and 1.5. When V2 is excluded, then thelargest family has size equal to that of the largest level plus between 1nand 2ntimes the next level. On the other hand, if the butterfly is excluded then thesize of two levels can be achieved. It is a natural question, what happens if aconfiguration between these two is excluded. Namely, let the configuration offour distinct subsets satisfying A ⊂ C,A ⊂ D,B ⊂ C be called and denotedby N . It is somewhat surprising that the result is basically the same (at leastin the first two terms) as in the case of V2. The total “jump” is between Nand B. The goal of the present paper is to prove the following theorem.2Theorem 1.6(nbn2c)(1 +1n+O(1n2))≤ La(n,N) ≤(nbn2c)(1 +2n+O(1n2))holds.2 Notation and definitionsA partially ordered set (or poset) P is a pair P = (X,≤) where X is a set (inour case always finite) and ≤ is a relation on X which is reflexive (x ≤ x holdsfor every x ∈ X), antisymmetric (if both x ≤ y and x ≥ y hold for x, y ∈ Xthen x = y) and transitive (x ≤ y and y ≤ z always implies x ≤ z). It iseasy to see that if X = 2[n] and the ≤ is defined as ⊆, then these conditionsare satisfied, that is, the family of all subsets of an n-element set ordered byinclusion form a poset. We will call this poset the Boolean lattice and denoteit by Bn.The definition of a subposet is obvious: R = (Y,≤2) is a subposet ofP = (X,≤1) if and only if there is an injection α of Y into X in such a waythat y1, y2 ∈ Y, y1 ≤2 y2 implies α(y1) ≤1 α(y2). On the other hand, R isan induced subposet of P when α(y1) ≤1 α(y2) holds if and only if y1 ≤2 y2.If P = (X,≤) is a poset and Y ⊂ X then the poset spanned by Y in Pis defined as (Y,≤∗) where ≤∗ agreers with ≤, for all the pairs taken fromY . Given a “small” poset R, let La(n,R) denote the maximum size |Y | forY ⊂ 2[n] (that is, the maximum number of subsets of [n]) such that R is nota subposet of the poset spanned by Y in Bn.Redefine our “small” configurations in terms of posets. The chain Pkcontains k elements: a1, . . . , ak where a1 < . . . < ak. The r-fork containsr+1 elements: a, b1, . . . , br where a < b1, . . . a < br. The butterfly B contains4 elements: a, b, c, d with a < c, a < d, b < c, b < d. Finally, we have thefollowing relations in N : a < c, a < d, b < c. It is easy to see that thedefinitions of La(n, Pk), La(n, Vr), La(n,B) and La(n,N) in Sections 1 and2 agree. In the rest of the paper we will use both sets of terminology.A poset is connected if for any pair (z0, zk) of its elements there is asequence z1, . . . , zk−1 such that either zi < zi+1 or zi > zi+1 holds for 0 ≤ i <k. If the poset is not connected, maximal connected subposets are called itsconnected components. Given a family F of subsets of [n], it spans a poset inBn. We will consider its connected components in two different ways. First3as posets themselves, secondly as they are represented in Bn. In the lattercase the sizes of the sets are also indicated. A full chain in Bn is a family ofsets A0 ⊂ A1 ⊂ . . . ⊂ An where |Ai| = i. Let us mention that the numberof full chains in Bn is n!. We say that a (full) chain goes through a family(subposet) F if they intersect, that is, if the chain “goes through” at leastone member of the family.3 Proof of Theorem 1.6The lower estimate is obtained from Theorem 1.3, since La(n, V2) ≤ La(n,N).The upper estimate uses an idea generalizing the proof of Sperner’s Theo-rem given by Lubell [6], which is based on counting the number of full chainspassing through a family. Let F be a family of subsets of [n] containing nofour distinct members forming an N . Consider the poset P (F) spanned byF in Bn. Its connected components are denoted by F1, . . . ,FK . Let c(Fi)denote the number of full chains going through Fi. Observe that a full chaincannot go through two distinct components. Otherwise, they would be thesame component. Therefore, the following inequality holds:K∑i=1c(Fi) ≤ n!. (1)What can these components be? A component might be a P3, but one cancheck by cases that no component can contain a P3 as a proper subposet,since adding one more element to P3 creates an N , no matter which elementof P3 is related to the new element. Hence, if a < b are two elements of acomponent of size at least four, then a and b cannot be both comparable withthe same other element of the component (which would create P3), thoughone of a, b can be comparable with many other elements in the component.Therefore, the only possible components are these:a < b < c, (2)a < bi(1 ≤ i ≤ r) where r ≥ 0, (3)a > bi(1 ≤ i ≤ r) where r ≥ 2. (4)These are denoted by P (3), V (r),Λ(r) in this order. (The elements bi areunrelated here.) Notice that the number of full chains going through a poset4of these types depends only on the sizes of the elements of the poset, that is,the sizes of the members of the family Fi. To indicate this size information,we introduce the notation P (3;u, v, w) for posets of type P (3) where |a| =u < |b| = v < |c| = w. The analogous notations for the other poset typesabove are are V (r;u, u1, . . . , ur) (u < u1, . . . , ur) and Λ(r; v, v1, . . . , vr) (v >v1, . . . , vr). All components Fi in (1) are of this form.Plan of the proof. A good upper bound is sought for|F| =K∑i=1|Fi| (5)where |P (3)| = 3, |V (r)| = |Λ(r)| = r + 1 are obvious. This upper boundwill be determined entirely on the basis of (1). Denote the numbers of Fis oftypes P (3), V (r),Λ(r) by φ, ν(r), λ(r) respectively. Then (5) can be writtenin the form3φ+∞∑r=0(r + 1)ν(r) +∞∑r=2(r + 1)λ(r). (6)If the minima (or good lower bounds)minu,v,wc(P (3;u, v, w)), minu,u1,...,urc(V (r;u, u1, . . . , ur)), minv,v1,...,vrc(Λ(r; v, v1, . . . , vr))are determined then (1) leads to a linear combination of φ, ν(r), and λ(r).That is, one linear combination, namely (5), has to be maximized under thecondition that another combination is bounded from above. This will be aneasy task. Therefore, our main problem now is to determine the minima inthe display above. This will be done in the following two lemmas.Lemma 3.1 c(P (3;u, v, w)) (u < v < w) takes its minimum for the valuesu = bn2c − 1, v = bn2c, w = bn2c+ 1, that is,(⌊n2⌋− 1)!(⌈n2⌉− 1)!(⌊n2⌋2− n⌊n2⌋+ n2 − 1)≤ c(P (3;u, v, w)).Proof. The number of full chains going through P (3;u, v, w) is5c(P (3;u, v, w)) = u!(n− u)! + v!(n− v)! + w!(n− w)!−u!(v − u)!(n− v)!− v!(w − v)!(n− w)!− u!(w − u)!(n− w)! (7)+u!(v − u)!(w − v)!(n− w)!by the ordinary sieve. Since this expression has the same value when wereplace u, v, w by n − u, n − v, n − w, respectively, n2≤ v can be supposed.Another useful form of (7) isc(P (3;u, v, w))n!=1(nu) + 1(nv) + 1(nw) − 1(nv)(vu) − 1(nw)(wv) − 1(nw)(wu) + 1(nw)(wv)(vu)=1(nu) + 1(nv) − 1(nv)(vu) + 1(nw) (1− 1(wu) − 1(wv) (1− 1(vu))) . (8)Here(nw)is a decreasing function of w in the interval [v, n] (with fixed u, v).On the other hand,(wu)and(wv)are increasing functions. Therefore, (8) issmallest for w = v + 1: One has to see that the coefficients of 1/(nw)and1/(wv)are nonnegative, but this is an easy task if 0 < u, and, otherwise, (8)is equal to 1.w can be replaced by v + 1 in (7):c(P (3;u, v, v+1)) = u!(n−u)!+nv!(n−v−1)!−u!(v−u)!(n−v−1)!(n−u).(9)Another useful form of (9) isc(P (3;u, v, v + 1))n!=1(nu) + 1(n−1v) (1− n− un· 1(vu)) . (10)Here(n−1v)is a decreasing function of v in the interval [bn−12c, n − 1], while(vu)is increasing. Therefore, if we want to minimize (10), v can be chosenas the smallest integer ≥ bn2c with v > u. Two cases will be distinguished:v = u+ 1 and v = bn2c.Case 1. v = u+ 1. In this case (9) becomesc(P (3; v − 1, v, v + 1))= (v − 1)!(n− v + 1)! + nv!(n− v − 1)!− (v − 1)!(n− v − 1)!(n− v + 1)6= (v − 1)!(n− v − 1)!(v2 − nv + n2 − 1) = (n− 2)!(n−2v−1) (v2 − nv + n2 − 1).It is easy to see that both factors attain their respective minima at v = bn2c.The lemma is proved for this case.Case 2. v = bn2c. Rewrite (10):c(P (3;u, v, v + 1))n!=1(n−1v) + 1(nu) (1− 1(n−u−1n−v−1)) .(nu)is an increasing function of u in the interval [0, bn2c] while (n−u−1n−v−1)isdecreasing. The best choice for u is bn2c − 1. LLemma 3.2 Suppose n ≥ 6, r ≥ 1. Thenu∗!(n− u∗)! + ru∗!u∗(n− u∗ − 1)! ≤ c(V (r;u, u1, . . . ,ur))(u < u1, . . . ,ur)holds where u∗ = u∗(n) = n2− 1 if n is even, u∗ = n−12if n is odd andr − 1 ≤ n, while u∗ = n−32if n is odd and n < r − 1.Proof.1. One can easily show by using the sieve thatc(V (r;u, u1, . . . , ur)) =u!(n− u)! +r∑i=1ui!(n− ui)!−r∑i=1u!(ui − u)!(n− ui)!.This will actually be used in the formc(V (r;u, u1, . . . , ur)) =r∑i=1(1ru!(n− u)! + ui!(n− ui)!− u!(ui − u)!(n− ui)!).(11)Dividing one term by n!, two useful forms are obtained for the summandin (11):1r(nu) + 1(nui) − 1(nui)(uiu) = 1r(nu) + 1(nui) (1− 1(uiu)) (12)7and1r(nu) + 1(nui) − 1(nu)(n−un−ui) = 1(nui) + 1(nu) (1r− 1( n−un−ui)) . (13)2. First we will show that (12)-(13) attains its minimum for some pairu, ui = u+ 1.If n2− 1 ≤ u, fix u and consider changing ui in (12). Here,(nui)is adecreasing function of ui in the interval [bn2 c, n], while(uiu)is increasing.Therefore, one can suppose that ui = u+ 1, and we are done.Else, n2− 1 > u and the method in (12) above leads to ui ≤ bn2 c. Fixthis value and increase u using (13). It will not increase by moving u tou = bn2c − 1.Hence, we obtained the lower estimateminu(1ru!(n− u)! + (u+ 1)!(n− u− 1)!− u!1!(n− u− 1)!)=minu(1ru!(n− u)! + u!u(n− u− 1)!)for (12)-(13) and therefore we haveminu(u!(n− u)! + ru!u(n− u− 1)!) ≤ c(V (r;u, u1, . . . , ur)) (14)This minimum will be determined in the rest of the proof.3. Suppose now that 2 ≤ r. Take the “derivative” of fr(u) = u!(n−u)! +ru!u(n−u−1)!, that is, compare fr(u) at two consecutive values of u. Whendoes the inequalityfr(u− 1) = (u− 1)!(n− u+ 1)! + r(u− 1)!(u− 1)(n− u)! <fr(u) = u!(n− u)! + ru!u(n− u− 1)!(15)hold? It is equivalent to0 < 2(r − 1)u2 − (n(r − 3) + r − 1)u− n2 + (r − 1)n.The discriminant of the corresponding quadratic equation in u is(n(r−3)+r−1)2+8(r−1)(n2−(r−1)n) = (r+1)2n2−2(r−1)(3r−1)n+(r−1)2.8The latter expression can be strictly upperestimated by((r + 1)n− (r − 1))2 ,if r + 1 < 3r − 1 holds, that is, if r > 1. Hence, the larger root α2 of thequadratic equation is less thann(r − 3) + r − 1 + (r + 1)n− (r − 1)4(r − 1) =n2.On the other hand, as it is easy to see, (n(r + 1) − 3(r − 1))2 is a lowerbound for the discriminant if r−1 ≤ n holds. Using this estimate, we obtainthat n−12≤ α2 in this case. Substituting this lower estimate into the formulafor the smaller root α1, we obtain α1 ≤ 0 when n ≥ r − 1. Since (15) holdsexactly below α1 and above α2, we can state that fr(u) attains its minimumat u = bα2c. By the inequalities above we can conclude that this is at n2 − 1if n is even and n−12if n is odd. The statement of the lemma is proved in thecase of n ≥ r − 1.Else suppose n < r − 1. The inequality α2 < n−12 can be proved in thesame way as in the previous case. On the other hand, 6 ≤ n implies that(n(r+ 1)− 5(r− 1))2 is a lower estimate on the discriminant, hence we haven2− 1 < α2. This gives that α1 < 32 . The if n is even, bα2c is again n2 − 1,while bα2c = n−32 when n is odd. Although fr(0) < fr(1) is allowed by thisestimate, it is easy to check that fr(0) > fr(1) holds in reality. By (14) theproof is finished for r ≥ 2.The case r = 1 is much easier. The comparison (15) leads to a linearinequality which is an equality for u = n2. The formula f1(u) also has itsminimum at⌊n−12⌋. (But it has the same value at n2− 1 and n2.) LThe case r = 0 has to be treated differently. Then the number of fullchains going through the only (u-element) set is u!(n− u)!. Its minimum isbn2c!dn2e!.Introduce the notation m(n) = u∗!(n − u∗)! + ru∗!u∗(n − u∗ − 1)! whereu∗ = u∗(n) is determined in Lemma 3.2.Proof of the Theorem. We will need the following inequalities in theproof:u∗!u∗(n− u∗ − 1)! < u∗!(n− u∗)! + ru∗!u∗(n− u∗ − 1)!r + 1(16)holds for each of the 3 possible values of u∗. They easily reduce to n2− 1 <n2+ 1, n−12< n+12and n−32< n+32when u∗ = n2− 1, n−12and n−32, respectively.9A different inequality is needed in the case r = 0:u∗!u∗(n− u∗ − 1)! <⌊n2⌋!⌈n2⌉!. (17)One can verify it separating the 3 cases.Similarlyu∗!u∗(n− u∗ − 1)! ≤13(⌊n2⌋− 1)!(⌈n2⌉− 1)!(⌊n2⌋2− n⌊n2⌋+ n2 − 1)(18)can be obtained for each case.Start the proof by (1):n! ≥K∑i=1c(Fi) =K∑i=1c(Fi)|Fi| |Fi| (19)If Fi is a P (3), then Lemma 3.1 and (18) givec(Fi)|Fi| =c(Fi)3≥ u∗!u∗(n− u∗ − 1)!.If Fi is an V (r) then Lemma 3.2 and (16) provec(Fi)|Fi| =c(Fi)r + 1≥ u∗!u∗(n− u∗ − 1)!for 1 ≤ r and (17) shows its validity when r = 0. By symmetry one can saythe same about Λ(r). Display (19) results inn! ≥K∑i=1c(Fi) =K∑i=1c(Fi)|Fi| |Fi| ≥u∗!u∗(n− u∗ − 1)!K∑i=1|Fi| = u∗!u∗(n− u∗ − 1)!|F|.Hence we obtained|F| ≤ n!u∗!u∗(n− u∗ − 1)! ,10and this is equal to(n⌊n2⌋) n2n2− 1 ,(n⌊n2⌋) n+12n−12,(n⌊n2⌋) n−12n−32in the cases u∗ = n2− 1, n−12and n−32, respectively. These are all equal to=(n⌊n2⌋)(1 + 2n+O( 1n2)).T4 RemarkIt is somewhat disturbing that the constant in the second term in Theorem1.6 (and (1.3)) is not fully determined. Is it 1 or 2? Let us make the situationclear.The construction of a family avoiding V2 is the following: Take all thesets of size bn2c and a family A1, . . . Am of bn2 c+ 1-element sets satisfying thecondition |Ai ∩ Aj| < bn2 c for every pair i < j. It is easy to see that thisfamily contains no V2. We only have to maximize m. Denote this maximumby m(n). Since the bn2c-element subsets of the Ais are all distinct, we havem(⌊n2⌋+ 1)≤(nbn2c).This gives the upper estimatem(n) ≤(nbn2c)2n.There is a very nice construction (see [4]) of such sets Ai withm =(nbn2c+ 1)1n.We obtained (nbn2c)(1n+O(1n2))≤ m(n) ≤(nbn2c)(2n). (20)11It is a longstanding conjecture of coding theory what the right constant ishere, 1 or 2. Does this limit exist at all?It is easy to see that(nbn2c)+m(n) ≤ La(n,N).The upper estimate of our Theorem 1.6 is a (far-reaching) generalization (upto the second term) of the upper estimate in (20). Replacing the constant 2in the upper estimate in Theorem 1.6 by 1 would improve the upper estimatein (20), too, solving the old coding problem. On the other hand, if one couldconstruct a family with a constant 2 rather than 1, this would improve thelower estimate in Theorem 1.6. Summarizing, there is a strong connectionbetween the two problems. Of course, it is possible that the proper constantin m(n) is 1, while the one in La(n,N) is 2.References[1] Annalisa De Bonis, Gyula O.H. Katona, Largest families without anr-fork, submitted.[2] Annalisa De Bonis, Gyula O.H. Katona, Konrad J. Swanepoel, Largestfamily without A ∪ B ⊆ C ∪D, J. Combin. Theory Ser. A 111(2005),331-336.[3] P. Erdo˝s, On a lemma of Littlewood and Offord, Bull. Amer. Math. Soc.51(1945), 898-902.[4] R.L. Graham and N.J.A. Sloane, Lower bounds for constant weightcodes, IEEE IT 26(1980), 37-43.[5] G.O.H. Katona and T.G. Tarja´n, Extremal problems with excluded sub-graphs in the n-cube, in: Borowiecki, M., Kennedy, J.W., Sys lo M.M.(eds.) Graph Theory,  Lago´w, 1981, Lecture Notes in Math., 1018 84-93.Springer, Berlin Heidelberg New York Tokyo, 1983.[6] D. Lubell, A short proof of Sperner’s lemma, J. Combin. Theory 1(1966),299.12[7] E. Sperner, Ein Satz u¨ber Untermegen einer endlichen Menge, Math. Z.27(1928), 544–548.[8] Hai Tran Thanh, An extremal problem with excluded subposets in theBoolean lattice, Order 15(1998), 51-57.13

No four subsets forming an N

Katona, Gyula O.H.

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No four subsets forming an N 

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No four subsets forming an N

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