Diamond-free Families

Transcription

1 Diamod-free Families Jerrold R. Griggs Wei-Tia Li Liyua Lu arxiv: v3 [math.co] 6 Sep 011 August 8, 011 Abstract Give a fiite poset P, we cosider the largest size La,P of a family of subsets of [] := {1,...,} that cotais o weak subposet P. This problem has bee studied itesively i recet years, ad it is cojectured that πp := lim La,P/ exists for geeral posets P, ad, moreover, it is a iteger. For k let D k deote the k-diamod poset {A < B 1,...,B k < C}. We study the average umber of times a radom full chai meets a P-free family, called the Lubell fuctio, ad use it for P = D k to determie πd k for ifiitely may values k. A stubbor ope problem is to show that πd = ; here we make progress by provig πd 3 11 if it exists. 1 Itroductio We are iterested i how large a family of subsets of the -set [] := {1,...,} there is that avoids a give weak subposet P. The foudatioal result of this sort, Sperer s Theorem from 198 [16], solves this problem for families that cotai o two-elemet chai that is, for atichais, determiig that the maximum size is. For other excluded subposets, it is iterestig to compare the maximum size of a P-free family to. We give backgroud ad our ew results for this study i the ext sectio. Oe small forbidde poset that cotiues to stymie all iterested researchers is the diamod poset o four elemets. We preset a better ew upper boud o the size of diamod-free families. For k-diamod-free families for geeral k, we provide bouds that, surprisigly, tur out to be best-possible for ifiitely may values of k. Departmet of Mathematics, Uiversity of South Carolia, Columbia, SC 908 USA griggs@math.sc.edu. Departmet of Mathematics, Uiversity of South Carolia, Columbia SC 908 USA li37@mailbox.sc.edu. Departmet of Mathematics, Uiversity of South Carolia, Columbia, SC 908 USA lu@math.sc.edu. This author was supported i part by NSF grat DMS

2 I Sectio 3 we itroduce our method for this subject, the Lubell fuctio of a family, which gives the average umber of times a radom full chai meets the family. The Lubell fuctio yields a upper boud o the size of a family. For diamod-free families, we observe that the maximum possible Lubell fuctio decreases with, ad a calculatio gives our boud. For excludig k-diamods, our ew idea is to partitio the set of full chais, obtaiig bouds o each block of the partitio. The Lubell ad full chai partitio methods hold promise for other families of forbidde subposets. Sectio 4 cotais our detailed proofs, except for the log proof of Theorem.4, which is give i Sectio 5. The paper cocludes with our ideas for advacig the project. Backgroud ad Mai Results For posets P = P, ad P = P,, we say P is a weak subposet of P if there exists a ijectio f: P P that preserves the partial orderig, meaig that wheever u v i P, we have fu fv i P [17]. Throughout the paper, whe we say subposet, we mea weak subposet. We say the height hp of poset P is the maximum size of ay chai i P. Let the Boolea lattice B deote the poset [],. We cosider collectios F []. The F ca be viewed as a subposet of B. If F cotais o subposet P, we say F is P-free. We are iterested i determiig the largest size of a P-free family of subsets of [], deoted La,P. Ithisotatio,Sperer stheorem[16]givesthatla,p =,wherepk deotes the path poset o k poits, usually called a chai of size k. Moreover, Sperer determied that the largest atichais i B are the middle level, [] / for eve ad either of the two middle levels, [] 1/ or [] +1/ for odd, where for a set S, S i deotes the collectio of i-subsets of S. More geerally, Erdős solved the case of P k -free families. Let us deote by Σ,k the sum of the k middle biomial coefficiets i, ad let B,k deote the collectio of subsets of [] of the k middle sizes, that is, the sizes k+1/,..., +k 1/ or else the sizes k+1/,..., +k 1/. So there are either oe or two possible families B,k, depedig o the parities of ad k, ad regardless, B,k = Σ,k. The we have Theorem.1 [6] For k 1 1, La,P k = Σ,k 1. Moreover, the P k -free families of maximum size i B are give by B,k. It follows that for fixed k, La,P k k 1 as. Katoa ad his collaborators promoted the problem of ivestigatig La, P for other posets P, especially its asymptotic behavior for large. Cosider the r-fork poset V r, which has elemets A < B 1,...,B r, r. I 1981 he ad Tarjá [13] obtaied bouds o La,V that he ad DeBois [4] exteded i 007 to geeral V r, r, provig that 1+ r 1 +Ω 1 La,V r 1+ r 1 +O 1.

3 While the lower boud is strictly greater tha, we see that La,Vr. Earlier, Thah [18] had ivestigated the more geeral class of broom-like posets. Griggs ad Lu [11] studied the eve more geeral class of bato posets. These posets metioed so far have Hasse diagrams that are trees. I [5] it is show that for the butterfly poset B, with elemets A,B both less tha C,D, oe ca give a exact aswer, La,B = Σ,, for 3, which is asymptotic to. More geerally, for ay s,t, the complete bipartite poset Ks,t with elemets A 1,...,A s all less tha B 1,...,B t, satisfies La,K s,t [4]. The N-poset, with elemets A,B,C,D such that A < B,C < B,C < D, is itermediate betwee V ad B. It is show i [8] that La,N. Based o the examples for which La,P was kow, Griggs ad Lu [11] proposed the cojecture that was certaily apparet to Katoa et al.: Cojecture. For every fiite poset P, the limit πp := lim La,P iteger. exists ad is All of the examples above agree with the cojecture, ad Griggs ad Lu verified it for additioal examples, icludig tree posets of height. For the crow O k, k, which is the poset of height that is a cycle of legth k as a udirected graph, they exteded the butterfly result above ad proved that La,O k for all eve k 4. For odd k 3, it remais a dautig problem to determie the asymptotic behavior of La,O k. At least, Griggs ad Lu ca show La,O k / is asymptotically at most 1+ 1, which is less tha. Whe Griggs lectured o this work o forbidde subposets i 008, Mike Saks ad Peter Wikler observed a patter i all of the examples where πp was determied, which we describe as follows. For poset P defie ep to be the maximum m such that for all, the uio of the m middle levels B,m does ot cotai P as a subposet. Their observatio was πp = ep. For istace, the middle two levels B, cotai o butterfly B, sice o two sets of the same size k cotai the same two subsets of size k 1. Oe gets that eb =, which is πb. I geeral, it is clear that whe it exists, πp must be at least ep. Impressive progress i the developmet of the theory is the result of Bukh [] that for ay tree poset T meaig that the Hasse diagram is a tree, πt = et, so that the cojecture ad observatio are satisfied. It is easily verified that et = ht 1. Is there a coectio for geeral P betwee πp ad the height hp? The result of DeBois ad Katoa for complete bipartite posets K s,t implies that for ay poset P of height, πp, whe it exists. However, there is o such boud for taller posets, as observed by Jiag ad Lu see [11]. Let the k-diamod poset D k, k, cosist of k+ elemets A < B 1,...,B k < C. The hd k = 3 for geeral k, while for k = r 1, the middle r+1 levels B,r+1 caot cotai D k, sice a iterval i B,r+1 with a elemet i the lowest level ad a elemet i the highest level has at most r elemets a subposet B r, ad so at most r elemets i the middle. Hece it is D k -free. 3

4 The diamod D is the most challegig poset o at most four elemets i this theory. It is also the Boolea lattice B. It is easily see that ed =. O the other had, it is a subposet of the path P 4. So, if πd exists which has still ot bee show, it would have to be i [,3]; Its cojectured value is. As a illustratio of the Lubell fuctio method itroduced i the ext sectio, a short applicatio is give that reduces the upper boud o πd from 3 to.5. A refiemet of the Lubell fuctio method, which ivolves partitioig the set of full chais i a appropriate way, gives our first improvemet o the.5 boud: Propositio.3 For all sufficietly large, La,D / <.96. We display thisboud, ot ourbest oe, sice itsproofissimpler thaourbest boud, ad sice its proof gives us further isight ito the Lubell fuctio for D. Some time after we had aouced our boud above, Axeovich, Maske, ad Marti [1] came up with a ew approach which improves the upper boud to.83. Now usig our methods with a much more careful aalysis of diamod-free families for 1, we ca provide a further slight improvemet, which is the best-kow upper boud: Theorem.4 For all sufficietly large, La,D / < o 1 <.73. Cosequetly, if it exists, πd [, 3 11 ]. Because this ew boud requires cosiderably more care, its proof is give i its ow sectio followig the proofs of our other results. We shall see diamod-free families i the proof for which the Lubell fuctio method caot improve the upper boud o πd below.5. Therefore, ew ideas are required to brig the upper boud dow to the cojectured value of. Likewise, it appears that the methods of [1] caot move below.5. See the fial sectio of the paper for more discussio of how we ca do better. Give the great effort that has goe ito improvig the upper boud o πd, it is the quite surprisig that we ca solve the π problem for may of the geeral diamods D k with k >. This ca be regarded as our mai result. Theorem.5 Let k, ad defie m := log k +. 1 If k [ m 1 1, m m m 1], the La,D k = Σ,m. Hece, πd k = ed k = m. Moreover, if F attais the boud La,D k, the F = B,m. If k [ m m m, m ], the, Σ,m La,D k m+1 m k 1 m m. 4

5 Hece, if πd k exists, the m = ed k πd k m+1 m k 1 m m < m+1. For πd this ew theorem gives a upper boud of.5, ot as good as the theorem before. However, this ew result determies πd k for most values of k, i that for the m 1 valuesofk itherage[ m 1 1, m ], case1appliestoallbut m m C m /m 1/ of them. Moreover, we are able to give La,D k exactly, ot just asymptotically for large, for such values of k. The poset D k ca be viewed as the suspesio of a idepedet set of size k, where we mea that a maximum ad a miimum elemet are added to it. We ca cosider a more geeral suspesio of disjoit paths chais. For k 1 let l 1 l k 3, ad defie the harp poset Hl 1,...,l k to cosist of paths P l1,...,p lk with their top elemets idetified ad their bottom elemets idetified. For istace, i this otatio we have D k is the harp H3,...,3 where there are k 3 s. Theorem.6 If l 1 > > l k 3, the La,Hl 1,...,l k = Σ,l 1 1. Hece, for such harps, π = e = l 1 1. Moreover, for such harps, if F is a harp-free family of subsets of [] of maximum size, the F is B,l 1 1. The theorem above oly determies πh for harps H that have strictly decreasig path legths. However, for the geeral case i which path legths ca be equal there is o boud idepedet of k, sice we have see that for D k, which is a harp, πd k is arbitrarily large as k grows. It is the remarkable that we ca completely solve the problem of maximizig La, H for harps with distict path legths. Aother ovel aspect of this result is that for k the harps it cocers are ot raked posets. 3 The Lubell Fuctio For ow let us fix some family F []. Let C := C deote the collectio of all! full maximal chais {i 1 } {i 1,i } [] i the Boolea lattice B. A method used by Katoa et al. ivolves coutig the umber of full chais that meet F. Here we collect iformatio about the average umber of times chais C C meet F, which ca be used to give a upper boud o F. Recall that the height of F, viewed as a poset, is hf := max F C. C C We cosider what we call the Lubell fuctio of F, which is hf = h F := ave F C. C C 5

6 This is the expected value E F C over a radom full chai C i B. The hf is essetially the fuctio of F at the heart of Lubell s elegat proof of Sperer s Theorem [15], cf. [7] with the observatio. Lemma 3.1 Let F be a collectio of subsets of []. The hf = F F 1/ F. Proof: We have that hf = E F C, where C is picked at radom from C. This expected value is, i tur, the sum over F F of the probability that a radom C cotais F. Sice C meets the k subsets of cardiality k with equal probability, it meas that each set F cotributes 1/ F to the sum. Lubell s proof uses the simple facts that A C 1 for ay atichai A ad that k is maximized by takig k =, to derive Sperer s Theorem that A. By similar reasoig for geeral families F we obtai a geeral upper boud. Lemma 3. Let F be a collectio of subsets of []. If hf m, for real umber m > 0, the F m. Moreover, if m is a iteger, the F Σ,m, ad equality holds if ad oly if F = B,m whe +m is odd, or if F = B,m 1 together with ay m/ subsets of sizes m/ or +m/ whe +m is eve. Proof: We use the symmetry ad strict uimodality of the sequece of biomial coefficiets k, 0 k. If hf m, the F = A F 1 A F / A m. Now assume m > 0 is a iteger. We costruct a family F of maximum size, subject to hf m, by selectig subsets A that cotribute the least to hf, which meas that we miimize 1/ A. Essetially, we are solvig the liear program of maximizig i x i i subject to i x i m, 0 x i 1 for all i. We maximize F by selectig F to be the m middle levels, B, m. Further, if F = Σ,m, it must be that F is B,m whe +m is odd. If +m is eve, the subsets of sizes m/ ad +m/ will tie for the m-th largest size, ad we ca freely choose ay m/ subsets of the two sizes so that F = Σ,m. Weseethatupper boudsotheaverageitersectio size F C leadtoupperbouds o the ratio of particular iterest i this paper, F /. Hece, we get upper bouds o πp, whe it exists, from upper bouds o hf for P-free families F. To illustrate how this ca be useful, we ow give a short proof that, if it exists, πd.5. Cosider a diamod-free family F []. No full chai C C meets F four times, or else F cotais P 4, which has D as a subposet. If o chai meets F three times, we immediately get hf hf. Else, cosider ay three elemets of F X Y Z, ad let Y be ay set ot equal to Y such that X Y Z. Let σ be a permutatio of [] that fixes X ad Z ad seds Y to Y. The σ seds full chais C C through X,Y,Z to full chais C C through X,Y,Z. These chais C meet F oly twice, as F is diamod-free. These chais C are distict. We fid the that hf = E F C.5. 6

7 Ufortuately, the behavior of hf does ot match that of F / asymptotically there ca be a gap. We shall see examples of this for diamod-free families. Noetheless, i may cases we ca obtai πp from hf. Besides that, it is iterestig i its ow right to maximize hf for P-free families F, though obtaiig a good boud o hf ca be difficult. We have discovered that a partitio method ca be fruitful. Specifically, we partitio the set C of full chais ito blocks Ci ad the, for each i separately, we boud the average size F C over full chais C Ci. The priciple is that the average size F C over all full chais C is at most the maximum over i of the average over block Ci. A aalogy to baseball is helpful for some readers: A hitter s average over a whole seaso is ever more tha his maximum mothly average over the moths i the seaso. We illustrate the partitio method by sketchig a derivatio of La,B. Let F be a butterfly-free family of subsets of [], 3. Oe ca check that if F cotais or [], the F < Σ, although, oe may have hf >. Else, suppose,[] / F. We show hf. Defie the collectio M of subsets M F of [] for which there exists a chai C C passig through A,M,B F with A M B. Notice that sice F cotais o butterfly B, it cotais o P 4, ad so the collectio M is a atichai. Now partitio the set of full chais C as follows: For M M, C M cosists of all full chais meetig M, while C cotais all full chais that do ot meet M. By defiitio of M, o chai i C meets F three times, ad so ave C C F C. For M M, similar to the argumet above for D -free families, for ay chai C C M meetig F three times, it must meet F i A,M,B, ad there is a correspodig chai i C M meetig F oly at M ad avoidig A,B, so that ave C CM F C. Hece, we have partitioed C ito blocks such that F meets chais i each block at most twice, o average, ad hece at most twice, o average, over all of C. Thus, hf, ad it follows that La,B = Σ,, sice B, is butterfly-free. Regardig the extremal butterfly-free families as far as achievig La, B, Lemma 3. above applies. I fact, it is kow that F must be B, for 5, though it is ot true for = 4: Cosider F cosistig of {{1},{},{1,3,4},{,3,4}} ad all six -subsets. However, i some cases we ca show that La,P is attaied oly by F = B,k: Lemma 3.3 Suppose that for poset P, ep = m, a iteger. Suppose that for all, all P-free families F [] satisfy hf m. The for all, La,P = Σ,m, ad if F is a extremal family, the F = B,m. Proof: Let F be a P-freefamily with size La,P. Accordig to Lemma 3., La,P = Σ,m, sice hf m. Further, F = B,m whe + m is odd. Hece, suppose + m is eve, so that F = B,m 1 together with ay m/ subsets of sizes m/ or + m/. Suppose for cotradictio that F cotais subsets of both sizes m/ ad + m/. By the atural geeralizatio of Sperer s proof of Sperer s Theorem or by usig the ormalized matchig property o the raks [] m/ ad [] +m/ [7], we ca fid subsets A,B F with A B, A = m/ ad B = +m/. The the iterval [A,B] F. This iterval is a Boolea lattice, B m. 7

8 However, h m B m = m + 1 > m, so that by hypothesis, B m must cotai subposet P, which cotradicts our assumptio that F is P-free. Hece, F oly cotais oe of the two sizes m/ ad +m/. We saw that the coclusio of the lemma above fails for the butterfly P = B but oly for small. The reaso we could ot apply this lemma to P = B is that the hypothesis fails for = : The full Boolea lattice B, which has oe more elemet tha B,, is butterfly-free. 4 Proofs of Results.3,.5,.6 We ow illustrate our partitio method to brig the boud for D -free families below.5. Our best boud is derived i the ext sectio. Proof of Propositio.3: Let F be a D -free family of subsets of [] with maximum Lubell fuctio value hf, ad let d deote this value. We claim that d is oicreasig for. By easy direct case study we get that d =.5 ad d 3 = d 4 = 7/3.33. For 5, if both ad [] are i F, the we have oly oe more subset i F, ad hf We will later give examples of families with Lubell fuctio >., so F caot satisfy this coditio. The we may assume by symmetry that [] / F. We partitio the set C of full chais ito the blocks C,i, where the chais C C,i pass through set []\{i}. A radom full chai i C is equally likely to belog to each C,i, ad hf is simply the average over i of the values E F C take over C C,i, viewed as the Lubell fuctio for the subsets of []\{i}. That is, hf is the average of terms, each of which is at most d 1. Hece, d = hf d 1. Returig to the calculatios, for d 5 we ote that sice hf is a sum of terms, each 1 or 1/5 or 1/10, d 5 is a multiple of 1/10, ad hece at most.3 sice it is at most 7/3. The d 7, which is similarly a multiple of 1/105, must be less tha.3, ad hece at most 41/105 <.953, ad so by Lemma 3., for 7, F <.953, which implies the theorem. Next is the result for D k -free families for geeral k. Proof of Theorem.5: Let,k, ad defie m := log k +. For the lower bouds, cosider F = B,m. We have B A m 1 for ay two subsets A B i F. There are at most m 1 subsets S satisfyig A S B. Hece F is D k -free. So m ed k πd k. Now we derive the upper bouds. Let F be a largest D k -free family i B. We take what we call the mi-max partitio of the set C of full chais i B accordig to F: For subsets A B [] with A,B F, the block C A,B cosists of the full chais C such that the smallest ad the largest subsets i F C are A ad B, respectively. We deote by C the block of full chais that do ot meet F at all. For C C, we have F C = 0. 8

9 We ow boud the expected size of F C for a radom chai C i C A,B. If B A m 1, the this is at most m immediately. For the remaider, assume B A m. We use the Lubell fuctio Lemma 3.1 to calculate E F C by addig the cotributios of each subset S F [A,B], which is 1/ B A S A. Sice F is Dk -free ad cotais both A ad B, there are at most k 1 subsets S F [A,B] besides A ad B. The E F C is maximized if we take the k 1 terms with largest cotributio, i.e., with miimum B A S A, which meas the sets S closest to the eds A or B, so with S A equal to 1 or B A 1, the or B A, ad so o. The cotributio from each full level we iclude is the oe. For the case 1, where k 1 m m, we see that for B A = m, the k 1 m terms are at most eough to accout for all subsets S [A,B] with S A ot equal to m/, that is, we get Lubell fuctio at most m whe we iclude the terms for A ad B. For B A > m, sice the levels workig up from A or dow from B are larger, the k 1 terms are o loger sufficiet to cover as may full levels, ad the Lubell fuctio is strictly less tha m. Sice every block i our partitio has expected value at most m, we coclude that hf m. Lemma 3. gives us F Σ,m. Furthermore, we also have ed k m. Hece m = ed k = πd k hf m. By Lemma 3.3 the extremal family F must i fact be B,m. For the case, where k 1 > m m, we see that for B A = m, the m largest sum of k 1 terms leads to Lubell fuctio at most m+1 m k 1/ m. m As i case 1, if B A > m, the sice the levels workig up from the bottom or dow from the top i [A,B] are larger, the Lubell fuctio is strictly less tha this boud. The boud holds for every block C A,B of the mi-max partitio. Therefore, if πd k exists, πd k hf m+1 m k 1/ m m. Now we use the mi-max partitio of the set of full chais to prove the Harp Theorem. Proof of Theorem.6: We argue that hf l 1 1 for ay Hl 1,...,l k -free F usig iductio o k. The case k = 1 cocers a family F that cotais o chai of height l 1, for which we get immediately that hf = E F C max C F C l 1 1 which implies Erdős s Theorem.1. Let k, ad assume the boud o h for harps with k 1 paths. Let F be a H-free familyofsubsets of[], where H = Hl 1,...,l k, adcosider ablockc A,B ithemi-max partitio of the set of full chais C iduced by F. Let t be the largest height of ay chai i F [A,B]. If t < l 1 we get that for full chais C i this block, E F C l 1 1. Otherwise, t l 1. Cosider a largest chai Z i F A,B, say S 1 S t, where A S 1 ad S t B. Let F be F [A,B] with the sets i Z removed. The Z ad F are disjoit ad EF C for radom full chais C i this block is the sum of EZ C ad EF C. For the Z term, by Lemma 3.1 we get i 1/ B A S i A t / B A < 1. For the other term, we observe that F is Hl,...,l k -free i the Boolealatticeof subsets of[a,b]. Byiductio o k, hf l 1 l 1. So C meets F o average at most l 1 times. Combiig terms, we fid that C meets F at most 9

10 l 1 1 times o average for C i this block, ad hece for all radom full chais C C. We have that hf l 1 1. By Lemma 3. we get that La,Hl 1,...,l k Σ,l 1 1. The family B,l 1 1 achieves the upper boud just give, sice it does ot cotai a l 1 -chai, ad is thus Hl 1,...,l k -free. We see that ehl 1,...,l k = πhl 1,...,l k = l 1 1. Moreover, by Lemma 3.3, the oly harp-free family of maximum size is B,l Proof of D Theorem.4 We ivestigate the structure of D -free families with maximum Lubell fuctio, ad use this iformatio to improve our earlier boud. Before provig Theorem.4, we cotiue from the proof of Propositio.3 i the last sectio, assumig all otatio ad facts from that. We adopt the otatio that for ay families F 1,...,F m of sets, F 1 F m deotes the family {F 1 F m if i F i }. Give disjoit sets S,T we defie the followig three costructios: Costructio C 1 S,T: F = { } S 1 S 1 T 1 T. Costructio C S,T: F = { } S T S T 1 S 1 T. Costructio C 3 S,T: F = [] 1 S T S T 1 S 1 T. We will typically partitio [] ito subsets S, T i usig these costructios, ad we write C i s, s for C i [s],[] \ [s], for itegers s, 0 < s <. The families above are D - free ad each hc i s, s = + s s. For, the maximum value over s is 1 + / / >.5, achieved by s = / or /. 1 I our approach the key to provig Theorem.4 is to focus o D -free families F that cotai. Let δ be the maximum value of hf for all such families. Defiitios give that + / / δ 1 d. Eve though we do ot obtai the values of d, we ca obtai δ for up to 1. This techical iformatio icludig the extremal families for δ makes up the followig lemma, which is the hard part i provig the Theorem. Lemma 5.1 The sequece {δ } satisfies the followig properties. 1 It is oicreasig for 4. For 4 1, if F cotais ad h F 3, the up to relabelig elemets 11 of [], F is C 1 s, s for s = or, or C,. Hece, δ = + / /. 1 Proof of Lemma 5.1: To show 1, let F be a D -free family of subsets of [] such that h F = δ. For 5, if both ad [] are i F, the we have oly oe more subset i F, ad h F 10

11 + 1.. Thus [] / F. The similar to the proof of Propositio.3, we partitio the set C of full chais ito the blocks C,i, where the chais C C,i pass through set []\{i}. Agai, hf is the average over i of the values E F C take over C C,i, viewed as the Lubell fuctio for the subsets of []\{i}. That is, h F is the average of terms, each of which is at most δ 1. I other words, for 1 i, let F i = {F F i F}. The each F i is a D -free family i []\{i}. We have h F = ave ave F C 1 i C C,i = 1 h 1 F i. Hece, δ = h F δ 1. We claim the followig two facts which are eeded i showig. Claim 1: The iequality + / / < 3 This ca be verified by a simple computatio. 11 i=1 holds for all 1. Claim : For 4 1, suppose F C 1 s, s with h F 3. The F = 11 C 1 s, s with s = or. Similarly, if F C s, s with h F 3, the 11 F = C, The same as C, by relabelig the elemets. Oe ca calculate that if s s > 1, the h C i s, s < 3. Furthermore, if 11 F C 1 s, s with s = or, the h F + / / 1 1 < 3. Similarly, 11 if F C,, the h F + / / < 3. So Claim holds. 11 Weshowbyiductioo. Whe = 4,itcabedirectlyverifiedbyeumeratio. There are 17 classesup to relabelig of elemets of [] of D -free families cotaiig. The classes C 1, ad C, satisfy hf = 1 while the rest of them have hf at 3 most 1 3 which is less tha Assume 5 ad the statemets are true for 1. Now we cosider a D -free family F [] satisfyig h F 3 ad F. Agai, the full set [] is ot i F. Otherwise, 11 F cotais at most oe more subset other tha ad [], ad h F + 1 < 3. Sice 11 h F 3, there exists i so that h 11 1 F i 3. We may assume h 11 1 F 3. By 11 iductive hypothesis, F is C 1 1, 1, C 1 1, 1, or C 1, 1. We cosider two cases. Case 1: F = C 1 S,T where S = 1 or 1. ItremaistodecidethesubsetsiF\F. Herearetwosubcases depedig owhether {} is i F. Subcase 1a: {} F. SiceF isd -free, itcotaisosubsets offorms{s 1,}, {s 1,s,,...}, {s 1,t 1,,...}, ad {t 1,t,,...} for s i, S ad t i T. Thus, F C 1 S {},T. Sice h F 3, 11 we coclude that S +1 must be or. Thus, by relabelig elemets of [] we have that F = C 1,. Subcase 1b: {} F. 11

12 Let S = {s S {s,} F} ad T = {t T {t,} F}. Sice F is D -free, F caot have subsets of forms {s 1,s,,...}, {s 1,t 1,,...}, {t 1,t,,...}, ad {t 1,t,t 3,...} for s i S, t i T, ad t T. Equivaletly, S T T\T F F {{}} {{}} {{}}. 1 1 The h F 1+ S + S T + T 1+ S + S T + T + S +f T. + S + T + T T 3 Here f T = T T 3 + T is a quadratic fuctio of T defied o the iteger poits of the iterval [0, T ]. Its maximum is reached at oe of the two eds, amely T = 0 or T = T. Claim 3: If hf 3, the 11 T = T. For = 5, we have S = T =. If T < T, the f T max{f0,f T 1} = 1. Thus, hf 1+ S 10 + S T + T + S +f T < 3 11,whichcotradictsourassumptio, ad so T = T. For = 6, we have either S = ad T = 3, or else S = 3, T =. If T < T, the f T max{f0,f T 1} = 3 for S, T =,3, ad 0 f T max{f0,f T 1} = 1 15 hf 1+ S + S T + T + S for S, T = 3,. By direct computatio, both cases give +f T < 3, which is agai a cotradictio. 11 For 7 1, both f0 ad f T 1 are at most T 1 3. Thus, h F 1+ S + S T + T h C 1 S, T / / 1 + S + T < This cotradictio agai proves T = T, ad completes the proof of Claim 3. Hece, T\T {{}} is a ull family. Namely, F C1 S,T {}. By the coditio h F 3, we coclude by relabelig elemets of [] that F = C 11 1,, which is oe of the listed possibilities i. Case : F = C S,T where S = 1 ad T = 1. 1

13 We determie what are the possible subsets i F \ F. Cosider the two subcases depedig o whether {} F. Subcase a: {} F. Sice F isd -free, F caot cotaisubsets offorms{s 1,s,,...}, {t 1,t,,...}, ad {u,v,w,,...} for s 1,s S, t 1,t T ad u,v,w []. Let S = {s S {s,} F} ad T = {t T {t,} F}. The S T S\S F F {{}} {{}} {{}} T\T {{}} We have h F 1+ S + T + T S + S T S + T + S S T T 3 = h C S +1, T b, where we see that b := 1 + S S T is a biliear fuctio of 3 S, T defied o [0, S ] [0, T ]. To fid the extremal values of b it suffices to check the four corer poits 0,0, S,0, 0, T, S, T. We fid the miimum value of b, at S, T, is S T S Hece, h F + / / 1 + T + S T + S T S T 1 3 < 3 11, which cotradicts our assumptio, so this subcase is impossible. Subcase b: {} F. Similar to subcase a, let S = {s S {s,} F} ad T = {t T {t,} F}. The family F caot have subsets of forms {s,s,,...}, {s,t,,...}, {t,t,,...}, ad {u,v,w,,...} for s S, t T, s S, t T ad u,v,w []. The S T S F F 1 {{}} 1 {{}} 1 T S 1 \ 1 T 1 {{}} S\S T\T {{}} {{}}. 13

14 We have h F 1+ S + T + T S + S T 3 + S T S T + S\S + T\T 3 = h C S +1, T g 3, where S S T g = 3 + = ε 1 T +ε S \S, with + S + T T + S T S \S T \T ε 1 = T 1 5 S \S ad ε = S S 1 T If S, T = S,0 or 0, T, the F C S {},T or F C S,T {}. Sice h F 3, we have, by relabelig elemets of [], F = C 11,, which is aother of the alteratives listed i. Claim 4: Whe S, T S,0 or 0, T, the g 1. Recall that 0 S S = 1 ad 0 T T = 1. Suppose = k, so we have S = k 1 ad T = k, k. Rewrite ε 1 = 3 T \T +5 S /6 ad ε = T \T +3 S +/6. Note that S ad T \T are ot both zero, or are T ad S\S both zero. Oe ca see that ε 1 ad ε are each at least 1/, ad so g 1 uless T = 1 ad S \S = 0, or T = 0 ad S \S = 1. But either pair of coditios icreases the ε s ad still leads to g 1. Else, suppose = k +1, S = T = k ad k. This time rewrite ε 1 = 3 T \T +5 S 1/6 ad ε = T \T +3 S +1/6. Agai it is simple to check that g 1, ad Claim 4 holds. From Claim 4, we have if S, T S,0 or 0, T, the oce agai we get the cotradictio h F + / / 1 This completes the Case ad the proof of the Lemma. Now we are ready to prove our improved boud. 1 3 < Proof of Theorem.4: Let F be a D -free family of subsets of []. Partitio F ito F k ad F such that F k cotais subsets of sizes i [k, k] where k = l, ad F = F \F k. We kow that F / < for large see [11], Lemma 1. 14

15 Now cocetrate o the family of sets ear the middle, F k. We take what we call the mi partitio of the set C of full chais i B : Let C be the block cotaiig the full chais that do ot meet F k at all. For each subset A F k, let C A be the block cotaiig all full chais C havig A as the miimal elemet i F k C. We see that the average umber of times a chai i C A meets F k is obtaied by cosiderig oly the subsets i F k that cotai A ad viewig them after removig A from each as a diamod-free family of subsets of []\A cotaiig. We deduce that ave C CA F k C δ A. For large eough, we have A > l 1. Sice the δ are oicreasig, we have for large that hf k δ 1 = 3. It follows that for sufficietly large, all D 11 -free families F i B satisfy F = Fk + F Cosequetly, if it exists, the limit πd Further Research Beyod diamods D k, we cotiue to ivestigate why the limit πp exists for geeral posets P. The methods itroduced i this paper have prove to be useful for determiig πp for several other small posets P, which we are collectig separately [9, 10, 14]. Oe example is the subposet J of D cosistig of four elemets A,B,C,D with B < A ad B < C < D. Forbiddig J is more restrictive tha forbiddig D. We show that for 1, La,J = Σ,, ad hece πj =. All kow values of πp satisfy Cojecture.. I order to resolve the asymptotics for diamod-free posets, ad show that πd = as expected, it is ot eough to work with the Lubell fuctio due to families such as those i the costructios C i described above. These examples show that the terms i the sequece δ, which was show to be oicreasig for, are at least.5 for all. We suspect that the limit lim d, which is kow to exist, is.5. This would follow from the cojecture below. Here, the cojugate of F is the family F = { F F F} where F = []\F is the complemet of F. Cojecture 6.1 For every 4, the value hf of ay D -free family F [] satisfies h F + 4 ad equality holds if ad oly if, up to relabellig elemets of [], F 1 or F is C i S,T i = 1,,3 with S T 1. The how might we reduce our upper boud o πd if it exists to below.5? The examples above are owhere ear as large as, yet have very small sets that make large cotributios to the Lubell fuctio. To build large diamod-free families, we should restrict our attetio to families with o small or large sets, say F B,k with k = f <. If we could show that F is at most +o 1, for suitable f, we would have πd = as we expect, sice most of the subsets are cocetrated ear the middle rak. 15

16 Aother idicatio of the challege facig us for D is that we have costructed three D -free families for = 6 of size 36, which is oe more tha Σ6,. This is i cotrast to the values k for which Theorem.5 determies La,D k completely, ad its value is exactly Σ,m. Thus, the solutio for D, ad probably also D k for the usettled values k, is likely goig to be more complicated. Refereces [1] M. Axeovich, J. Maske, ad R. Marti, Q -free families i the Boolea lattice, Order published olie: 15 March 011. [] B. Bukh, Set families with a forbidde poset, Elect. J. Combi , R14, 11p. [3] T. Carroll ad G. O. H. Katoa, Bouds o maximal families of sets ot cotaiig three sets with A B C,A B, Order [4] A. De Bois ad G. O. H. Katoa, Largest families without a r-fork, Order 4 007, [5] A. De Bois, G. O.H. Katoa ad K. J. Swaepoel, Largest family without A B C D, J. Combi. Theory Ser. A , [6] P. Erdős, O a lemma of Littlewood ad Offord, Bull. Amer. Math. Soc , [7] C. Greee ad D. J. Kleitma, Proof techiques i the theory of fiite sets, i: G.C. Rota ed., Studies i Combiatorics, MAA Studies i Mathematics 17, MAA, Providece, 1978, pp. 79. [8] J. R. Griggs ad G. O. H. Katoa, No four subsets formig a N, J. Combiatorial Theory Ser. A , [9] J. R. Griggs ad W.-T. Li, The partitio method for poset-free families, preprit 011. [10] J. R. Griggs ad W.-T. Li, Uiformly L-bouded posets, preprit 011. [11] J. R. Griggs ad L. Lu, O families of subsets with a forbidde subposet, Combiatorics, Probability, ad Computig , [1] Gyula O. H. Katoa, Forbidde iclusio patters i the families of subsets itroducig a method, i Horizos of Combiatorics, Bolyai Society Mathematical Studies, 17, Bolyai Mathematical Society, Budapest ad Spriger-Verlag, 008, pp

17 [13] G. O. H. Katoa ad T. G. Tarjá, Extremal problems with excluded subgraphs i the -cube, i: M. Borowiecki, J. W. Keedy, ad M. M. Sys lo eds. Graph Theory, Lagów, 1981, Lecture Notes i Math., , Spriger, Berli Heidelberg New York Tokyo, [14] W.-T. Li, Extremal Problems o Families of Subsets with Forbidde Subposets, Ph.D. dissertatio, Uiversity of South Carolia, 011. [15] D. Lubell, A short proof of Sperer s lemma, J. Combi. Theory 11966, 99. [16] E. Sperer, Ei Satz über Utermege eier edliche Mege, Math. Z , [17] R. P. Staley, Eumerative Combiatorics Vol. 1, Cambridge Uiversity Press, [18] H. T. Thah, A extremal problem with excluded subposets i the Boolea lattice, Order ,