Three layer Q 2 -free families in the Boolean lattice
aa r X i v : . [ m a t h . C O ] A ug Three Layer Q -Free Families in the BooleanLattice Jacob Manske ∗ Texas State UniversitySan Marcos, TX, 78666 [email protected]
Jian ShenTexas State UniversitySan Marcos, TX 78666 [email protected]
November 9, 2018
Abstract
We prove that the largest Q -free family of subsets of [ n ] which contains sets ofat most three different sizes has at most (cid:0) √ (cid:1) N/ o ( N ) ≈ . N + o ( N )members, where N = (cid:0) n ⌊ n/ ⌋ (cid:1) . This improves an earlier bound of 2 . N + o ( N ) byAxenovich, Manske, and Martin. Let Q n be the n -dimensional Boolean lattice corresponding to subsets of an n -elementset ordered by inclusion. A poset P = ( X, ≤ ) is a subposet of Q = ( Y, ≤ ′ ) if there is aninjective map f : X → Y such that for x , x ∈ X , x ≤ x implies f ( x ) ≤ ′ f ( x ). Fora poset P , we say that a set of elements F ⊆ [ n ] is P -free if ( F , ⊆ ) does not contain P as a subposet. Let ex ( n, P ) be the size of the largest P -free family of subsets of [ n ]. Wesay that the set of all i -element subsets of [ n ], (cid:0) [ n ] i (cid:1) , is the i th layer of Q n . Finally, let N ( n ) = N = (cid:0) n ⌊ n/ ⌋ (cid:1) ; i.e., N is the size of the largest layer of the Boolean lattice.The first result in this area is Sperner’s Theorem [11], which states that ex ( n, Q ) = N .He also showed that the largest Q -free family is the largest layer in the Boolean lattice.Many largest P -free families are simply unions of the largest layers in Q n . For instance,the largest Q -free family is simply the largest layer in the Boolean lattice. In [5], Erd˝osgeneralized Sperner’s result, showing that the size of the family of subsets of [ n ] whichdoes not contain a chain with k elements, P k , is equal to the number of elements in the ∗ Corresponding author. − Q n . He also showed that the largest P k -free family is the union ofthe ( k −
1) largest layers in the Boolean lattice.De Bonis, Katona and Swanepoel show in [4] that ex ( n, ) = (cid:0) n ⌊ n/ ⌋ (cid:1) + (cid:0) n ⌊ n/ ⌋ +1 (cid:1) , where is a subposet of Q n consisting of distinct sets a, b, c, d such that a, b ⊂ c, d . They alsoshowed that if n = 3 or n ≥
5, the only -free family which achieves this size is the unionof the two largest layers in the Boolean lattice. When n = 4, there is another construction;take all subsets of size 2 together with { } , { } , { , , } , and { , , } .When an exact result is not known, often the asymptotic bounds for ex ( n, P ) areexpressed in terms of N . De Bonis and Katona [3] and independently Thanh [12] showedthat ex ( n, V r +1 ) = N + o ( N ), where V r +1 is a subposet of Q n with distinct elements f, g i , i = 1 , . . . , r , f ⊂ g i for i = 1 , . . . , r . For a poset K s,t , with distinct elements f , . . . , f s ⊂ g , . . . , g t , and a poset P k ( s ), with distinct elements f ⊂ · · · ⊂ f k ⊂ g , g , . . . , g s , Katonaand Tarjan [10] and later De Bonis and Katona [3] proved that ex ( n, K s,t ) = 2 N + o ( N )and ex ( n, P k ( s )) = kN + o ( N ). Griggs and Katona proved in [6] that ex ( n, N ) = N + o ( N ),where N is the poset with distinct elements a, b, c, d , such that a ⊂ c, d , and b ⊂ c . Griggsand Lu [7] proved that ex ( n, P k ( s, t )) = ( k − N + o ( N ), where P k ( s, t ) is a poset withdistinct elements f , f . . . , f s ⊂ g ⊂ g ⊂ · · · ⊂ g k − ⊂ h , . . . , h t , k ≥
3. They alsoshowed that ex ( n, O k ) = N + o ( N ), ex ( n, O k − ) ≤ (1 + √ / N + o ( N ), where O i is aposet of height two which is a cycle of length i as an undirected graph. More generally,they proved that if G = ( V, E ) is a graph and P is a poset with elements V ∪ E , with v < e if v ∈ V , e ∈ E and v incident to e , then ex ( n, P ) ≤ (cid:16) p − / ( χ ( G ) − (cid:17) N + o ( N ),where χ ( G ) is the chromatic number of G . Bukh [2] proved that ex ( n, T ) = kN + o ( N ),where T is a poset whose Hasse diagram is a tree and k is the integer which is one less thanthe height of T . For a more complete survey on the subject, see [9] and [8] for alternateproofs of some of the results listed above.The smallest poset for which even an asymptotic result is not known is Q . In [1],Axenovich, Martin, and the first author show that ex ( n, Q ) ≤ . N + o ( N ) andin the special case where if F is a family of subsets of [ n ] with at most 3 different sizesand which is Q -free, then |F | ≤ . N . More recently, Griggs, Li, and Lu were ableto show in [9] that lim n →∞ ex ( n, Q ) N ≤ ex ( n, Q ) ≤ N + o ( N ) and thus reducing the leading coefficient in the bound from[1] by about . F contains sets ofat most 3 sizes; we state the result below as Theorem 1. Theorem 1
Let n be a positive integer. If F ⊂ Q n is a Q -free family, F = S ∪ T ∪ U ,where S is a collection of minimal elements of F , U is a collection of maximal elementsof F and T = F \ ( S ∪ U ) such that for any T ∈ T , S ∈ S , U ∈ U , | T | = k , | U | > k , | S | < k , then |F | ≤ (cid:0) √ (cid:1) N/ o ( N ) ≈ . N + o ( N ) . In particular, if F is a -free subset of three layers of Q n , then |F | ≤ . N + o ( N ) . Following the argument in [1], it suffices to prove Theorem 1 in the case where F containssets of size k , ( k − k + 1).For two functions A ( n ) and B ( n ), by A ( n ) . B ( n ) (or B ( n ) & A ( n )) we meaneither A ( n ) ≤ B ( n ) or lim n →∞ A ( n ) B ( n ) = 1 . Suppose F is a Q -free family from three layers, L , L , L , of the Boolean lattice Q n ,where L = (cid:0) [ n ] k − (cid:1) , L = (cid:0) [ n ] k (cid:1) , L = (cid:0) [ n ] k +1 (cid:1) , and by Lemma 1 from [1], we may assume n/ − n / ≤ k ≤ n/ n / . Let S = F ∩ L , T = F ∩ L , U = F ∩ L . For X ∈ L , Y ∈ L , and Z ∈ L , we define f ( X ) = |{ T ∈ T : X ⊂ T }| ; g ( Z ) = |{ T ∈ T : Z ⊃ T }| ;˘ f ( Y ) = |{ S ∈ S : S ⊂ Y }| ; ˘ g ( Y ) = |{ U ∈ U : U ⊃ Y }| ;Note that X X ∈S f ( X ) = X Y ∈T ˘ f ( Y ) and X Z ∈U g ( Z ) = X Y ∈T ˘ g ( Y ) . From [1], we have |U | + |T | + |S| ≤ N − |T | + 1 k X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) , (1)where N = (cid:0) n ⌊ n/ ⌋ (cid:1) ≈ (cid:0) nk (cid:1) . We start with a few lemmas involving some counting arguments. Lemma 1
For any X ∈ S and Y ∈ T with X ⊂ Y , f ( X ) + ˘ g ( Y ) ≤ n − k + 1 . k. For any Y ∈ T and Z ∈ U with Y ⊂ Z , g ( Z ) + ˘ f ( Y ) ≤ k + 1 . k. Proof.
We only prove f ( X ) + ˘ g ( Y ) ≤ n − k + 1 . k , and the other inequality followssimilarly. By definition, ˘ g ( Y ) = |{ U ∈ U : U ⊃ Y }| ≤ n − | Y | = n − k . So we maysuppose without loss of generality that f ( X ) ≥
2. For any Y ′ ∈ T with X ⊂ Y ′ = Y , wehave | Y ∪ Y ′ | = | Y | + | Y ′ | − | Y ∩ Y ′ | = | Y | + | Y ′ | − | X | = 2 k − ( k −
1) = k + 1 and thus3 ∪ Y ′ ∈ L . Since F is Q -free, we have Y ∪ Y ′ ∈ ( L − U ). Further Y ∪ Y ′ = Y ∪ Y ′′ for any other Y ′′ with X ⊂ Y ′′ ∈ ( T − {
Y, Y ′ } ) . Therefore,˘ g ( Y ) ≤ |{ U ∈ U : U ⊃ Y and U = Y ∪ Y ′ with X ⊂ Y ′ ∈ ( T − { Y } ) }| = |{ U ∈ U : U ⊃ Y }| − |{ Y ′ ∈ T : X ⊂ Y ′ ∈ ( T − { Y } ) }| = n − k − f ( X ) + 1 . ✷ Lemma 2 |S| & X Y ∈T ˘ f ( Y ) k − ˘ g ( Y ) , |U | & X Y ∈T ˘ g ( Y ) k − ˘ f ( Y ) . Proof.
We use double counting to only prove the first inequality, and the otherinequality follows symmetrically. First X ( X,Y ): S∋ X ⊂ Y ∈T f ( X ) = X X ∈S X X ⊂ Y ∈T f ( X ) = X X ∈S |S| . Second, by Lemma 1, X ( X,Y ): S∋ X ⊂ Y ∈T f ( X ) = X Y ∈T X S∋ X ⊂ Y f ( X ) & X Y ∈T X S∋ X ⊂ Y k − ˘ g ( Y ) = X Y ∈T ˘ f ( Y ) k − ˘ g ( Y ) . ✷ Lemma 3
For any non-negative reals x and y with x < k , y < k , and x + y ≥ k , xk − y + yk − x ≥ x + 2 y k − x − y . Proof. x ( k − x )(2 k − x − y )+ y ( k − y )(2 k − x − y ) − (2 x + 2 y )( k − x )( k − y ) = ( x − y ) ( x + y − k ) ≥ . ✷ Define T := { Y ∈ T : ˘ f ( Y ) + ˘ g ( Y ) ≥ k } and T := { Y ∈ T : ˘ f ( Y ) + ˘ g ( Y ) < k } .4 emma 4 X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) ≤ k |T | , X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) . k |T | ( |S| + |U | ) |S| + |U | + 2 |T | . Proof.
By the definition of T , we have P Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) ≤ P Y ∈T k = k |T | . We now prove the second inequality of the lemma. Recall that T ∪ T forms a disjointunion of T . By Lemma 2 and Lemma 3 (with x = ˘ f ( Y ) and y = ˘ g ( Y )), |S| + |U | & X Y ∈T ˘ f ( Y ) k − ˘ g ( Y ) + ˘ g ( Y ) k − ˘ f ( Y ) ! ≥ X Y ∈T f ( Y ) + 2˘ g ( Y )2 k − ˘ f ( Y ) − ˘ g ( Y )= X Y ∈T k k − ˘ f ( Y ) − ˘ g ( Y ) − ! = X Y ∈T k k − ˘ f ( Y ) − ˘ g ( Y ) − |T | . Then, by the Cauchy-Schwarz inequality,( |S| + |U | + 2 |T | ) (cid:16) k |T | − P Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17)(cid:17) & X Y ∈T k k − ˘ f ( Y ) − ˘ g ( Y ) X Y ∈T (cid:16) k − ˘ f ( Y ) − ˘ g ( Y ) (cid:17) ≥ k |T | , which is equivalent to X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) . k |T | − k |T | |S| + |U | + 2 |T | = 2 k |T | ( |S| + |U | ) |S| + |U | + 2 |T | . ✷ We are now ready to prove Theorem 1.
Proof.
We will show that |S| + |U | + |T | . √ N ≈ . N.
5y (1) and Lemma 4, |U | + |T | + |S| . N − |T | + 1 k X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) = 2 N − |T | − |T | + 1 k X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) + 1 k X Y ∈T (cid:16) ˘ f ( Y ) + ˘ g ( Y ) (cid:17) . N − |T | + 2 |T | ( |S| + |U | ) |S| + |U | + 2 |T | = 2 N + |T | ( |S| + |U | − |T | ) |S| + |U | + 2 |T | . N + ( |U | + |T | + |S| ) · f ( x ) , where f ( x ) = x − x + 2)( x + 1)with x = |S| + |U ||T | >
0. Note that the function f ( x ) has a unique critical point at x = (cid:0) √ (cid:1) on the interval [0 , ∞ ). The function f ( x ) achieves its maximum value at x = (cid:0) √ (cid:1) and thus |U | + |T | + |S| . N +( |U | + |T | + |S| ) · f (cid:16) √ (cid:17) = 2 N + (cid:16) − √ (cid:17) ( |U | + |T | + |S| ) , from which we have |U | + |T | + |S| . (cid:0) √ (cid:1) N/ ≈ . N . ✷ There are two ways of extending the argument to get a general bound on ex ( n, Q ). Oneis to adapt the counting argument above to work with a family of subsets of [ n ] withmore than 3 sizes. Another way is to show that if F is a family of size ex ( n, Q ), then F contains sets of at most 3 different sizes. If the latter is true, then Theorem 1 shows that ex ( n, Q ) ≤ . N .We may also investigate π ( Q ) = lim n →∞ ex ( n, Q ) N , as the authors in [9] do. It is notknown if π ( Q ) exists, although it is conjectured in [7] that π ( P ) exists and is an integerfor any finite poset P . If true, then π ( Q ) = 2 and ex ( n, Q ) = 2 N + o ( N ). Acknowledgments.
Shen’s research was partially supported by NSF (CNS 0835834,DMS 1005206) and Texas Higher Education Coordinating Board (ARP 003615-0039-2007).6 eferences [1] M. Axenovich, J. Manske, and R. Martin. Q -free families of the Boolean lattice. accepted by Order , 2011.[2] B. Bukh. Set families with a forbidden subposet. Electron. J. Combin. , 16(1):Researchpaper 142, 11, 2009.[3] A. De Bonis and G. O. H. Katona. Largest families without an r -fork. Order ,24(3):181–191, 2007.[4] A. De Bonis, G. O. H. Katona, and K. J. Swanepoel. Largest family without A ∪ B ⊆ C ∩ D . J. Combin. Theory Ser. A , 111(2):331–336, 2005.[5] P. Erd˝os. On a lemma of Littlewood and Offord.
Bull. Amer. Math. Soc. , 51:898–902,1945.[6] J. R. Griggs and G. O. H. Katona. No four subsets forming an N . J. Combin. TheorySer. A , 115(4):677–685, 2008.[7] J. R. Griggs and L. Lu. On families of subsets with a forbidden subposet.
Combina-torics, Probability and Computing , 18:731–748, 2009.[8] J.R. Griggs and W.-T. Li. The partition method for poset-free families. preprint ,2011.[9] J.R. Griggs, W.-T. Li, and L. Lu. Diamond-free families. accepted by J. CombinatorialTheory (Ser. A) , 2011.[10] G. O. H. Katona and T. G. Tarj´an. Extremal problems with excluded subgraphs inthe n -cube. In Graph theory ( Lag´ow, 1981) , volume 1018 of
Lecture Notes in Math. ,pages 84–93. Springer, Berlin, 1983.[11] E. Sperner. Ein Satz ¨uber Untermengen einer endlichen Menge.
Math. Z. , 27(1):544–548, 1928.[12] H. T. Thanh. An extremal problem with excluded subposet in the Boolean lattice.