Structural results for conditionally intersecting families and some applications
aa r X i v : . [ m a t h . C O ] M a y Structural results for conditionally intersecting families andsome applications
Xizhi Liu ∗ May 18, 2020
Abstract
Let k ≥ d ≥ F be a k -uniform family on [ n ]. Then F is ( d, s )-conditionally intersecting if it does not contain d sets with union of size at most s and empty intersection. Answering a question of Frankl, we present some structuralresults for families that are ( d, s )-conditionally intersecting with s ≥ k + d −
3, andfamilies that are ( k, k )-conditionally intersecting. As applications of our structuralresults we present some new proofs to the upper bounds for the size of the following k -uniform families on [ n ].(a) ( d, k + d − n ≥ k .(b) ( k, k )-conditionally intersecting families with n ≥ k / ( k − , k )-conditionally intersecting families with n ≥ k (cid:0) kk (cid:1) .Our results for ( c ) confirms a conjecture of Mammoliti and Britz for the case d = 3. Let V be a set, and let S, T be two subsets of V . Then we use S − T to denote the set S \ T , and use (cid:0) Vk (cid:1) to denote the collection of all k -subsets of V . Let [ n ] denote the set { , . . . , n } . A d -cluster of k -sets is a collection of d different k -subsets A , . . . , A d of [ n ]such that | A ∪ · · · ∪ A d | ≤ k, and | A ∩ · · · ∩ A d | = 0 . Let F be a k -uniform family on [ n ]. Then F is ( d, s ) -conditionally intersecting if it doesnot contain d sets with union of size at most s and empty intersection. In particular,a family F is ( d, k )-conditionally intersecting if it does not contain d -clusters. We use h ( n, k, d, s ) to denote the maximum size of a ( d, s )-conditionally intersecting family F .Note that a k -uniform family is (2 , k )-conditionally intersecting if and only if it is in-tersecting. The celebrated Erd˝os-Ko-Rado theorem [4] states that h ( n, k, , k ) ≤ (cid:0) n − k − (cid:1) for all n ≥ k , and when n > k equality holds only if F is a star , i.e. a collection of k -sets that contain a fixed vertex. In [5], Frankl showed that the same conclusion holdsfor n ≥ dk/ ( d −
1) when the intersecting condition is replaced by the d -wise intersecting condition, i.e. every d sets of F have nonempty intersection. ∗ Department of Mathematics, Statistics, and Computer Science, University of Illinois, Chicago, IL,60607 USA. Email: [email protected] heorem 1.1 (Frankl, [5]) . Let k ≥ d ≥ be fixed and n ≥ dk/ ( d − . If F ⊂ (cid:0) [ n ] k (cid:1) is a d -wise intersecting family, then |F | ≤ (cid:0) n − k − (cid:1) , with equality only if F is a star. Later, Frankl and F¨uredi [7] extended Theorem 1.1 and proved that h ( n, k, , k ) ≤ (cid:0) n − k − (cid:1) for all n ≥ k + 3 k , and they conjectured that the same inequality holds for all n ≥ k/ Conjecture 1.2 (Mubayi, [11]) . Let k ≥ d ≥ and n ≥ dk/ ( d − . Suppose that F ⊂ (cid:0) [ n ] k (cid:1) is a ( d, k ) -conditional intersecting family. Then |F | ≤ (cid:0) n − k − (cid:1) , with equality only if F is astar. Conjecture 1.2 has been intensively studied in the past decade. Mubayi [12] proved thisconjecture for the case d = 4 with n sufficiently large. Later, Mubayi and Ramadurai[13], and independently, F¨uredi and ¨Ozkahya [8] settled this conjecture for all d ≥ n sufficiently large. In [2], Chen, Liu and Wang confirmed this conjecture for the case d = k , and they also showed that h ( n, k, d, ( d + 1) k/ ≤ (cid:0) n − k − (cid:1) for all n ≥ dk/ ( d − , Definition 1.3.
Let
H ⊂ [ n ] , and let H ∈ H . A subset G ⊂ H is called unique if thereis no other set in H containing G . The following result of Bollob´as [1] gives an upper bound for the size of a family in whichevery set has a unique subset.
Theorem 1.4 (Bollob´as, [1]) . Suppose that for every member H of the family
H ⊂ [ n ] the set G ( H ) ⊂ H is a unique subset. Then X H ∈H (cid:0) n −| H − G ( H ) || G ( H ) | (cid:1) ≤ . Frankl [6] proved the following structural result for (3 , Theorem 1.5 (Frankl, [6]) . Suppose that
F ⊂ (cid:0) [ n ]3 (cid:1) is a (3 , -conditionally intersectingfamily. Then F can be partitioned into two families H and B , and the ground set [ n ] canbe partitioned into two disjoint subsets Y and Z such that the following statements hold.(a) H ⊂ (cid:0) Y (cid:1) and every set H ∈ H contains a unique -subset.(b) B ⊂ (cid:0) Z (cid:1) and B is the vertex disjoint union of | Z | / copies of complete -graphs on vertices. First, let us show how to use Theorem 1.5 to get an upper bound for |F | . Let
F ⊂ (cid:0) [ n ]3 (cid:1) be a (3 , Y, Z, B and H be given by Theorem1.5. Since every set in H contains a unique 2-subset, it follows from Theorem 1.4 that |H| ≤ (cid:0) | Y |− (cid:1) . On the other hand, it is easy to see that |B| = | Z | . Therefore, |F | = |H| + |F | ≤ (cid:18) | Y | − (cid:19) + | Z | ≤ (cid:18) n − (cid:19) , Z = ∅ .In [6], Frankl also asked for a structural result for a (3 , k )-conditionally intersecting family F ⊂ (cid:0) [ n ] k (cid:1) which can imply the (cid:0) n − k − (cid:1) bound for |F | . Here we consider a more generalquestion, namely the structures of ( d, k + d − k ≥ d ≥
3, and we obtain the following result.Let L k denote the collection of all k -graphs on at most 2 k vertices. Theorem 1.6.
Let k ≥ d ≥ be fixed. Suppose that F ⊂ (cid:0) [ n ] k (cid:1) is a ( d, k + d − -conditionally intersecting family. Then F can be partitioned into three families H , B and S , and the ground set [ n ] can be partitioned into two subsets Y and Z such that thefollowing statements hold.(a) H ⊂ (cid:0) Yk (cid:1) and every set H ∈ H contains a unique ( k − -subset.(b) Z has a partition V ∪ · · ·∪ V t with each V i of size at most k such that B ⊂ S ti =1 (cid:0) V i k (cid:1) ,i.e., the family B is the vertex disjoint union of copies of k -graphs in L k (c) S ⊂ (cid:0) [ n ] k (cid:1) − (cid:0) Yk (cid:1) , and for every set S ∈ S and every V i ⊂ Z the size of S ∩ V i is either or at least d . Note that the constraint on | S ∩ V i | in ( c ) for S ∈ S and V i ⊂ Z implies that the family S is actually very sparse. Therefore, the term |S| contributes very little to |F | .Our next result gives a structure for ( k, k )-intersecting families for all k ≥ Theorem 1.7.
Let k ≥ be fixed. Suppose that F ⊂ (cid:0) [ n ] k (cid:1) is a ( k, k ) -conditionallyintersecting family. Then F can be partitioned into two families H and B , and the groundset [ n ] can be partitioned into two subsets Y and Z such that the following statements hold.(a) H ⊂ (cid:0) Yk (cid:1) and every set H ∈ H contains a unique ( k − -subset.(b) B ⊂ (cid:0) Zk (cid:1) and B is the vertex disjoint union of | Z | k +1 copies of complete k -graphs on ( k + 1) vertices. Applying the structural results above we are able to give some new proofs to the followingtheorems.
Theorem 1.8.
Let k ≥ d ≥ be fixed and n ≥ k . Suppose that F ⊂ (cid:0) [ n ] k (cid:1) is a ( d, k + d − -conditionally intersecting family. Then |F | ≤ (cid:0) n − k − (cid:1) . Note that Theorem 1.8 is true for every n ≥ k/ n ≥ k to keep the calculations simple. Theorem 1.9.
Let k ≥ be fixed and n ≥ k / ( k − . Suppose that F ⊂ (cid:0) [ n ] k (cid:1) is a ( k, k ) -conditionally intersecting family. Then |F | ≤ (cid:0) n − k − (cid:1) . Theorem 1.10.
Let k ≥ be fixed and n ≥ k (cid:0) kk (cid:1) . Let F ⊂ (cid:0) [ n ] k (cid:1) be a family that is (3 , k ) -conditionally intersecting but not intersecting. Then |F | ≤ (cid:0) n − k − k − (cid:1) + 1 . Theorem 1.10 shows that Mammoliti and Britz’s conjecture (Conjecture 4.1 in [10]) istrue for the case d = 3. Note that in [9] the author considered Mammoliti and Britz’s3onjecture for all d ≥
3, and showed that their conjecture is true for d = 3, but false for all d ≥
4. However, the method we used here is completely different from the method usedin [9].The remaining part of this paper is organized as follows. In Section 2, we prove Theorems1.6 and 1.7. In Section 3, we prove Theorems 1.8, 1.9, and 1.10.
Let F be a k -uniform family on [ n ] and B ∈ F . We say B is bad if it does not containany unique ( k − B = { b , . . . , b k } is a bad set in F , then thereexist k distinct sets C , . . . , C k in F such that B ∩ C i = B − { b i } for all i ∈ [ k ]. Let V B = B ∪ C · · · ∪ C k and H B = { B, C , . . . , C k } . First let us prove Theorem 1.7. Proof of Theorem 1.7.
Suppose that F is a ( k, k )-conditionally intersecting family, andsuppose that B = { b , . . . , b k } is a bad set in F . Let C , · · · , C k , V B , H B be defined asabove. Since | V B | ≤ k , by assumption we have C ∩· · ·∩ C k = ∅ . It follows that | V B | = k +1and, hence, the family H B is a complete k -graph on V B . Let b k +1 denote the vertex in V B − B , and let F ∈ F − H B . Then we claim that F ∩ V B = ∅ . Indeed, suppose that F ∩ V B = ∅ . We may assume that F ∩ V B = { b , . . . , b ℓ } for some ℓ ∈ [ k − H B as B i = V B − b i for all i ∈ [ k + 1]. Since | F ∪ B ∪ · · · ∪ B k − | ≤ k and F ∩ B ∩ · · · ∩ B k − = ∅ , the k sets F, B , . . . , B k − form a k -cluster in F , a contradiction.Therefore, F ∩ V B = ∅ . To finish the proof we just let B be the collection of all bad setsin F , and let H = F − B .Before proving Theorem 1.6 let us present a useful lemma. Let s = 2 k + d − Lemma 2.1.
Suppose that F is a ( d, s ) -conditionally intersecting family and B is a badset in F . Then for every F ∈ F either | F ∩ V B | = 0 or | F ∩ V B | ≥ d .Proof. Let B is a bad set in F and let V B be the set as we defined before. Suppose that F ∈ F has nonempty intersection with V B . It suffices to show that | F ∩ V B | ≥ d . Forcontradiction, suppose that | F ∩ B | = x , | F ∩ ( V B − B ) | = y and x + y ≤ d −
1. Supposethat F ∩ B = { b m , . . . , b m x } and F ∩ ( V B − B ) = { c n , . . . , c n y } .If x = d −
1, then y = 0 and, hence, the d sets F, C m , . . . , C m d − satisfy | F ∪ C m ∪· · · ∪ C m d − | ≤ k and F ∩ C m ∩ · · · ∩ C m d − = ∅ , a contradiction. If x = d −
2, thenthe d sets F, B, C m , . . . , C m d − satisfy | F ∪ B ∪ C m ∪ · · · ∪ C m d − | ≤ k and F ∩ B ∩ C m ∩ · · · ∩ C m d − = ∅ , a contradiction. Therefore, we may assume that x ≤ d −
3. Let p = d − ( x +2). Choose p sets C q , . . . , C q p from { C , . . . , C k }−{ C m , . . . , C m x } . Then the d sets F, B, C m , . . . , C m x , C q , . . . , C q p satisfy | F ∪ B ∪ C m ∪· · ·∪ C m x ∪ C q ∪· · ·∪ C q p | ≤ k + p and F ∩ B ∩ C m ∩ · · · ∩ C m x ∩ C q ∩ · · · ∩ C q p = ∅ . By assumption we have 2 k + p ≥ s and, hence, x = 0 and y ≥ p ′ = d − ( y + 2), and choose p ′ sets C q , . . . , C q p ′ from { C , . . . , C k } − { C n , . . . , C n y } .Then the d sets F, B, C n , . . . , C n y , C q , . . . , C q p ′ satisfy | F ∪ B ∪ C n ∪ · · · ∪ C n y ∪ C q ∪· · · ∪ C q p ′ | ≤ k + p ′ ≤ s and F ∩ B ∩ C n ∩ · · · ∩ C n y ∩ C q ∩ · · · ∩ C q p ′ = ∅ , a contradiction.Therefore, we have | F ∩ V b | ≥ d .Now we are ready to prove Theorem 1.6. 4 roof of Theorem 1.6. Let F be a ( d, s )-conditionally intersecting family. Choose a col-lection of bad sets { B , . . . , B t } for some t from F such that the sets V B , . . . , V B t arepairwise disjoint, and any other bad set in F has nonempty intersection with V B i for some i ∈ [ t ]. Note that this can be done by greedy choosing each B i from F such that B i isdisjoint from S j
Proof of Theorem 1.9.
Suppose that F is a ( k, k )-conditionally intersecting family on[ n ]. Let Y, Z, B and H be given by Theorem 1.7. By Theorem 1.4, H ≤ (cid:0) | Y |− k − (cid:1) . Onthe other hand, it is easy to see that |B| = ( k + 1) × | Z | / ( k + 1) = | Z | . Therefore, |F | = |H| + |B| ≤ (cid:0) | Y |− k − (cid:1) + | Z | ≤ (cid:0) n − k − (cid:1) , and equality holds only if Z = ∅ .Now we apply Theorem 1.6 to prove Theorem 1.8. Proof of Theorem 1.8.
Let F be a ( d, k + d − n ≥ k vertices. Let Y, Z, B , H and S be given by Theorem 1.6. Let v i = | V i | for i ∈ [ t ]. Let Y = Y and Y i = Y i − ∪ V i for i ∈ [ t ] and let y i = | Y i | for 0 ≤ i ≤ t . Define H i = F ∩ (cid:0) Y i k (cid:1) and let h i = |H i | . By Lemma 2.1, every set H ∈ H i is either disjoint from V i or hasan intersection of size at least d with V i . Therefore, |H i | ≤ |H i − | + P kℓ = d (cid:0) v i ℓ (cid:1)(cid:0) y i − k − ℓ (cid:1) .Inductively, we obtain |F | ≤ |H| + t − X i =0 k X ℓ = d (cid:18) v i +1 ℓ (cid:19)(cid:18) y i k − ℓ (cid:19) ≤ (cid:18) y − k − (cid:19) + t − X i =0 k X ℓ = d (cid:18) kℓ (cid:19)(cid:18) n − k − k − ℓ (cid:19) . Since (cid:0) kℓ (cid:1)(cid:0) n − k − k − ℓ (cid:1) ≥ (cid:0) kℓ +1 (cid:1)(cid:0) n − k − k − ℓ − (cid:1) , we obtain |F | ≤ (cid:18) y − k − (cid:19) + t − X i =0 ( k − d ) (cid:18) kd (cid:19)(cid:18) n − k − k − d (cid:19) ≤ (cid:18) y − k − (cid:19) + ( k − d ) (cid:18) kd (cid:19)(cid:18) n − k − k − d (cid:19) n − y k + 1 ≤ (cid:18) y − k − (cid:19) + (cid:18) k (cid:19)(cid:18) n − k − k − (cid:19) ( n − y ) . Now let δ = (cid:16) (cid:0) k (cid:1)(cid:17) − . If n − y ≤ δn , then |F | < (cid:18) n − k − (cid:19) − k (cid:18) n − k − k − (cid:19) + n (cid:18) n − k − k − (cid:19) < (cid:18) n − k − (cid:19) , y ≤ (1 − δ ) n . Then |F | ≤ − (cid:0) k (cid:1) ! (cid:18) n − k − (cid:19) + (cid:18) n − k − k − (cid:19) n ≤ (cid:18) n − k − (cid:19) , and this completes the proof of Theorem 1.8.The remaining part of this section is devoted to prove Theorem 1.10. We will use thefollowing lemma in our proof.The shadow ∂ H of a family H ⊂ (cid:0) [ n ] k (cid:1) is defined as follows: ∂ H = (cid:26) G ∈ (cid:18) [ n ] k − (cid:19) : ∃ H ∈ H such that G ⊂ H (cid:27) . Lemma 3.1.
Suppose that
H ⊂ (cid:0) [ n ] k (cid:1) , and every set H ∈ H has a unique ( k − -subset G ( H ) ⊂ H . Then |H| ≤ n − k + 1 n | ∂ H| . Proof.
Consider a weight function ω ( G, H ) for all pairs G ⊂ H ∈ F with | G | = k −
1. Forevery G ∈ ∂ H and every H ∈ H assign weight 1 to ( G, H ) if G = G ( H ) and ( n − k + 1) − if G = G ( H ). Then an easy double counting gives (cid:18) k − n − k + 1 (cid:19) |H| = X ( G,H ) ω ( G, H ) ≤ | ∂ H| , which implies |H| ≤ ( n − k + 1) | ∂ H| /n . Definition 3.2.
Let
F ⊂ (cid:0) [ n ] k (cid:1) and S ⊂ [ n ] . Then F is a full star on S if it is the collectionof all k -subsets of S that contain a fixed vertex v , and F is a star if it is a subfamily ofsome full star on S . In either case, we call v the core of F . Now we prove Theorem 1.10.
Proof of Theorem 1.10.
Let n ≥ k (cid:0) kk (cid:1) and let F be a family on [ n ] such that F is(3 , k )-conditionally intersecting but not intersecting. Suppose that B ∈ F is a bad set.Let V B , H B be as defined at the beginning of this section and let F ′ = F ∩ (cid:0) [ n ] − V B k (cid:1) . Since F ′ is also (3 , k )-intersecting, by result in [11], |F ′ | ≤ (cid:0) n −| V B |− k − (cid:1) | ≤ (cid:0) n − k − k − (cid:1) . Then byLemma 2.1, |F | ≤ |F ′ | + k X i =3 (cid:18) ki (cid:19)(cid:18) n − k − k − i (cid:19) ≤ (cid:18) n − k − k − (cid:19) + k (cid:18) k (cid:19)(cid:18) n − k − k − (cid:19) = (cid:18) n − k − k − (cid:19) − (cid:18)(cid:18) n − k − k − (cid:19) − k (cid:18) k (cid:19)(cid:18) n − k − k − (cid:19)(cid:19) < (cid:18) n − k − k − (cid:19) + 1 , and we are done. So we may assume that every F ∈ F has a unique ( k − G ( F ).6ince F is not intersecting, there exist two disjoint sets A, B in F . Assume that A = { a , . . . , a k } and B = { b , . . . , b k } . Let I = { a , . . . , a k , b , . . . , b k } and let U = [ n ] − I .For every set C ⊂ U of size at most k − F ( C ) on I as follows: F ( C ) = { F − C : F ∈ F and F ∩ U = C } . For every i ∈ { , , ..., k } let F i = { F ∈ F : | F ∩ I | = i } . First notice that F k = { A, B } , since any extra edge in F k together with A, B would forma 3-cluster in F . Next, we will prove ℓ X i =0 |F i | ≤ ℓ X i =1 (cid:18) n − kk − i (cid:19)(cid:18) k − i − (cid:19) . (1)for all ℓ ∈ [ k ]. Suppose that (1) is true, then by letting ℓ = k we obtain |F | = k X i =0 |F i | ≤ k − X i =1 (cid:18) n − kk − i (cid:19)(cid:18) k − i − (cid:19) + 2 = (cid:18) n − k − k − (cid:19) + 1 , and this will complete the proof of Theorem 1.10. One could compare (1) with a similarinequality in [11], which is |F | ≤ k X ℓ =1 (cid:18) n − tkk − ℓ (cid:19)(cid:18) tk − ℓ − (cid:19) = (cid:18) n − k − (cid:19) , (2)where t is the maximum number of pairwise disjoint sets in F . For the case t = 2, thesummand in (2) is (cid:0) n − kk − ℓ (cid:1)(cid:0) k − ℓ − (cid:1) , but the summand in (1) is (cid:0) n − kk − ℓ (cid:1)(cid:0) k − ℓ − (cid:1) , which is smallerwhen ℓ ≥ Claim 3.3.
Let F ∈ F . Then the set F ∩ U is a unique ( k − -subset of F in F .Proof of Claim 3.3. Without lose of generality, we may assume that F = { a , f , . . . , f k − } ,where f , . . . , f k − are contained in U . Suppose that there is another edge F ′ ∈ F con-taining { f , . . . , f k − } . Then the three sets A, F, F ′ form a 3-cluster in F , a contradiction.Therefore, F ∩ U = { f , . . . , f k − } is a unique ( k − F in F .Now we prove (1) for ℓ = 1. Let us consider the family F ∪ F . Define M = (cid:26) G ∈ (cid:18) Uk − (cid:19) : ∃ F ∈ F ∪ F such that G ⊂ F (cid:27) . By assumption, every set F ∈ F ∪ F has a unique ( k − G ( F ), and by Claim 3.3,we may assume that G ( F ) ⊂ U . Let G = { G ( F ) : F ∈ F } . For every set F ∈ F , theset G ( F ) cannot be contained in ∂ F , since otherwise one could easily find a 3-cluster.Therefore, G and ∂ F are disjoint. Since |G| = |F | , by Lemma 3.1, we have | U || U | − k + 1 |F | + |F | ≤ |M| ≤ (cid:18) n − kk − (cid:19) , and hence |F | + |F | ≤ (cid:0) n − kk − (cid:1) . 7o prove (1) for ℓ ≥
2, we need to give an upper bound for |F i | for every 2 ≤ i ≤ k − |F i | = P C ∈ ( Uk − i ) |F ( C ) | , it suffices to give an upper bound for |F ( C ) | for every C ∈ (cid:0) Uk − i (cid:1) . Unfortunately, the inequality |F ( C ) | ≤ (cid:0) k − i − (cid:1) is not true in general. So, in ourproof, we will build a relationship between F i and S j
2. We say C is perfect if the family F ( C ) is a fullstar on either A or B . Let D ⊂ U be a set of size k −
1. We say D is perfect if there existsa set F in F that contains D .For every i ∈ [ k −
1] let P i be the collection of all perfect sets in (cid:0) Uk − i (cid:1) , and let N i be thecollection of non-perfect sets in (cid:0) Uk − i (cid:1) . Let p i = |P i | and n i = |N i | for i ∈ [ k −
1] and noticethat p i + n i = (cid:0) | U | k − i (cid:1) .For every i ∈ { , . . . , k − } let P ′ i denote the collection of all sets C ∈ (cid:0) Uk − i (cid:1) such that C iscontained in a perfect set in (cid:0) Uk − i +1 (cid:1) , and let N ′ i denote the collection all of sets D ∈ (cid:0) Uk − i (cid:1) such that D is not contained in any perfect set in (cid:0) Uk − i +1 (cid:1) . Let p ′ i = |P ′ i | and n ′ i = |N ′ i | for i ∈ { , . . . , k − } . Let G i = N i ∩ P ′ i and B i = N i ∩ N ′ i , and let g i = |G i | and b i = |B i | for i ∈ { , . . . , k − } . Let G = N , and let g = n , b = 0. Note that by definition, b i + g i = n i and n ′ i ≥ b i for i ∈ [ k − |F ( C ) | = (cid:0) k − i − (cid:1) for all C ∈ P i . Later we will show that |F ( C ) | < (cid:0) k − i − (cid:1) for all C ∈ G i . For every C ∈ B i it could be true that |F ( C ) | > (cid:0) k − i − (cid:1) .However, for every C ∈ B i there are either many sets in G i − containing C , which meansthat there are many sets D ∈ (cid:0) Uk − i +1 (cid:1) with |F ( D ) | smaller than its expected value, orthere are many sets in B i − , in which case we turn to consider sets in (cid:0) Uk − i +2 (cid:1) and repeatthis argument until we end up with many sets P in (cid:0) Uk − (cid:1) with |F ( P ) | smaller than itsexpected value.The next claim gives a relation between n i and b i +1 . Claim 3.4.
For every i ∈ [ k − we have n i ≥ n − kk b i +1 . Proof of Claim 3.4.
Let C ∈ N ′ i +1 , and let u ∈ U − C . By definition C ∪ { u } is a non-perfect set in (cid:0) Uk − i (cid:1) . Therefore, we have ( k − i ) n i ≥ n ′ i +1 ( n − k + i + 1) ≥ b i +1 ( n − k ).It follows that n i ≥ ( n − k ) b i +1 /k . Claim 3.5.
The following statement holds for all ℓ ≥ ( k + 1) / . Suppose that C ⊂ U is aperfect set of size ℓ , and F ( C ) is a full star on A (or on B ) with core v . Then for every ( ℓ − -subset C ′ of C the family F ( C ′ ) is a star on A (or on B ) with core v .Proof of Claim 3.5. Let C ⊂ U such that F ( C ) is a full star on A with core v ∈ A .Without loss of generality we may assume that v = a . Let E ′ ∈ F ( C ′ ). If E ′ ⊂ B , thenchoose a set E from F ( C ), and the three sets E ∪ C, E ′ ∪ C ′ , B form a 3-cluster in F , acontradiction. If E ′ ∩ A = ∅ and E ′ ∩ B = ∅ , then let x = | E ′ ∩ A | and y = | E ′ ∩ B | .8ince x + y = k − ℓ + 1, we have x ≤ k − ℓ and y ≤ k − ℓ . If a E ′ ∩ A , then bythe assumption that ℓ ≥ ( k + 1) / F ( C ) is a full star, there exists a set E ∈ F ( C )such that ( E ′ ∩ A ) ∩ E = ∅ . So the three sets E ′ ∪ C ′ , E ∪ C, A form a 3-cluster in F ,a contradiction. If a ∈ E ′ ∩ A , then by assumption there exists a set E ∈ F ( C ) suchthat E ′ ∩ A ⊂ E . However, the three sets E ∪ C, E ′ ∪ C ′ , B form a 3-cluster in F , acontradiction. Therefore, every set in F ( C ′ ) is completely contained in A .Next, we show that every set E ′ ∈ F ( C ′ ) contains a . Suppose there exists a set E ′ ∈F ( C ′ ) such that a E ′ . By assumption we have k − ℓ + 1 + k − ℓ ≤ k , so there existsa set E ∈ F ( C ) such that E ∩ E ′ = ∅ . However, the three sets E ′ ∪ C ′ , E ∪ C, A form a3-cluster in F , a contradiction. Therefore, the family F ( C ′ ) is a star on A with core a .For every i ∈ [ k −
1] let w i = (cid:0) k − i − (cid:1)(cid:0) n − kk − i (cid:1) and k i = (cid:0) ki (cid:1) − (cid:0) k − i − (cid:1) + 1. Our next claim givesan upper bound for |F i | for 2 ≤ i ≤ ( k + 1) / Claim 3.6.
For every i satisfying ≤ i ≤ ( k + 1) / we have |F i | ≤ w i + k i b i − n i . Proof of Claim 3.6.
Let us give an upper bound for |F ( C ) | for every C ∈ (cid:0) Uk − i (cid:1) . Firstnotice that by definition |F ( C ) | = (cid:0) k − i − (cid:1) for all C ∈ P i . By Claim 3.5, |F ( C ) | ≤ (cid:0) k − i − (cid:1) − C ∈ G i . On the other hand, it is trivially true that |F ( C ) | ≤ (cid:0) ki (cid:1) for all C ∈ B i .Therefore, |F i | = X C ∈P i |F ( C ) | + X C ∈G i |F ( C ) | + X C ∈B i |F ( C ) |≤ (cid:18) k − i − (cid:19) p i + (cid:18)(cid:18) k − i − (cid:19) − (cid:19) g i + (cid:18) ki (cid:19) b i = (cid:18) k − i − (cid:19)(cid:18) n − kk − i (cid:19) + (cid:18)(cid:18) ki (cid:19) − (cid:18) k − i − (cid:19) + 1 (cid:19) b i − n i = w i + k i b i − n i . Here we used that fact that b i + g i = n i and n i + p i = (cid:0) n − kk − i (cid:1) .Recall that Claim 3.4 says that n i ≥ ( n − k ) b i +1 /k . Since n ≥ k (cid:0) kk (cid:1) and k i +1 < (cid:0) kk (cid:1) , wehave n i / ≥ k i +1 b i +1 . Combining this inequality with Claim 3.6 we obtain the followingclaim. Claim 3.7.
For every ℓ satisfying ≤ ℓ ≤ ( k + 1) / we have ℓ X i =0 |F i | ≤ ℓ X i =1 w i − ℓ X i =1 n i . Proof of Claim 3.7.
The case ℓ = 1 follows from the inequality that |F | + |F | ≤ |M| = (cid:18) n − kk − (cid:19) − n . For ℓ ≥ ℓ X i =0 |F i | ≤ ℓ X i =1 ( w i + k i b i − n i ) = ℓ X i =1 w i − ℓ − X i =1 ( n i − k i +1 b i +1 ) − n ℓ ≤ ℓ X i =1 w i − ℓ X i =1 n i . ℓ > ( k + 1) / Claim 3.8.
Let C ⊂ U be a set of size ℓ ≥ . Suppose that F ( C ) is a full-star on A (oron B ) with core v and there exists a perfect set P ∈ (cid:0) Uk − (cid:1) containing C . Then, for every ( ℓ − -subset C ′ ⊂ C the family F ( C ′ ) is a star on A (or on B ) with core v .Proof of Claim 3.8. Let C ⊂ U be a set of size ℓ such that F ( C ) is a full-star on A withcore v . Without loss of generality we may assume that v = a . Let P ∈ (cid:0) Uk − (cid:1) be a perfectset containing C . By the definition of perfect set there exists a set F ∈ F containing P .Suppose that F = P ∪ { u } , and we want to show that u = a . Suppose that u A . Thenfor every E ∈ F ( C ) the three sets A, F, E ∪ C form a 3-cluster in F , a contradiction.Therefore, u ∈ A .Now suppose for the contrary that u = a . Then by assumption there exists a set E ∈F ( C ) not containing u and, hence, the three sets A, F, E ∪ C form a 3-cluster in F , acontradiction. Therefore, u = a .Let C ′ ⊂ C be a set of size ℓ − E ′ ∈ F ( C ′ ). If E ′ ⊂ B , then for every E ∈ F ( C )the three sets E ∪ C, E ′ ∪ C ′ , B form a 3-cluster in F , a contradiction. If E ′ ∩ A = ∅ and E ′ ∩ B = ∅ , then let x = | E ′ ∩ A | and y = | E ′ ∩ B | . Since x + y = k − ℓ + 1, we have x ≤ k − ℓ and y ≤ k − ℓ . If x ≤ k − ℓ −
1, then by assumption there exists a set E ∈ F ( C )containing E ′ ∩ A . However, the three sets E ∪ C, E ′ ∪ C ′ , B form a 3-cluster in F , acontradiction. Therefore, we may assume that x = k − ℓ . If a ∈ E ′ ∩ A , then there existsa set E ∈ F ( C ) such that E ′ ∩ A = E . However, the three sets E ∪ C, E ′ ∪ C ′ , B form a3-cluster in F , a contradiction. If a E ′ ∩ A , then the three sets A, F, E ′ ∪ C ′ form a3-cluster in F , a contradiction. Therefore, every set in F ( C ′ ) is completely contained in A .Suppose that there is a set E ′ ∈ F ( C ′ ) not containing a , then the three sets A, F, E ′ ∪ C ′ would form a 3-cluster in F , a contradiction. Therefore, every set in F ( C ′ ) contains a ,and this complete the proof of Claim 3.8.Let c = ⌊ ( k + 1) / ⌋ and let m = ⌊ k/ ⌋ , and notice that m + c = k . The next claim showsthat (1) holds for ℓ = c + 1. Claim 3.9.
We have c +1 X i =0 |F i | ≤ c +1 X i =1 w i − c +1 X i =1 n i . Proof of Claim 3.9.
Similar to the proof of Claim 3.6, for every C ∈ P c +1 we have |F ( C ) | = (cid:0) k − c (cid:1) , and for every C ∈ B c +1 we have |F ( C ) | ≤ (cid:0) kc +1 (cid:1) .For every perfect set D ∈ (cid:0) Um (cid:1) we say that D is a good container if D itself is contained ina perfect ( k − D is a bad container . Let S be the collectionof all sets in G c +1 that are contained in a good container. Let T be the collection of allsets in G c +1 that are not contained in any good container. Let s = |S| and t = |T | . Sinceevery bad container in (cid:0) Um (cid:1) has m subsets of size m −
1, the number of bad containers in (cid:0) Um (cid:1) is at least t/m .Let D ∈ (cid:0) Um (cid:1) be a bad container. Then for every E ∈ (cid:0) U − Dk − m − (cid:1) the set D ∪ E is non-perfectin (cid:0) Uk − (cid:1) . Therefore, n ≥ (cid:0) n − k − mc − (cid:1) t/ (cid:16) m (cid:0) k − m (cid:1)(cid:17) . By definition, every set C ∈ G c +1 is10ontained in a perfect set D ∈ (cid:0) Um (cid:1) . If C ∈ S , then by Claim 3.8, |F ( C ) | ≤ (cid:0) k − c (cid:1) −
1. If C ∈ T , then it is trivially true that |F ( C ) | ≤ (cid:0) kc +1 (cid:1) . Therefore, |F c +1 | = X C ∈P c +1 |F ( C ) | + X C ∈B c +1 |F ( C ) | + X C ∈S |F ( C ) | + X C ∈T |F ( C ) |≤ (cid:18) k − c (cid:19) p c +1 + (cid:18) kc + 1 (cid:19) b c +1 + (cid:18)(cid:18) k − c (cid:19) − (cid:19) s + (cid:18) kc + 1 (cid:19) t = w c +1 + k c +1 b c +1 + k c +1 t − n c +1 . Here we used the fact that s + t = g c +1 , g c +1 + b c +1 = n c +1 and n c +1 + p c +1 = (cid:0) n − kk − c − (cid:1) .Combining the inequality above with Claim 3.6, we obtain c +1 X i =0 |F i | ≤ c +1 X i =1 ( w i + k i b i − n i ) + k c +1 t. Since n / ≥ k c +1 t and n i / ≥ k i +1 b i +1 , c +1 X i =0 |F i | ≤ c +1 X i =1 w i − c +1 X i =1 n i . Claim 3.10.
Every set C ⊂ U of size at most k − c is contained in a perfect ( k − -set.Proof of Claim 3.10. Let C ⊂ U be a set of size ℓ ≤ k − c . Suppose that C is not containedin any perfect ( k − S ∈ (cid:0) U − Ck − ℓ − (cid:1) the set C ∪ S is non-perfect andof size k −
1. Therefore, we have n ≥ (cid:0) n − k − ℓk − ℓ − (cid:1) / (cid:0) k − ℓ (cid:1) ≥ (cid:0) n − k − ℓc − (cid:1) / (cid:0) k − ℓ (cid:1) . On the otherhand, we have P k − i = c +2 |F i | ≤ P k − i = c +2 (cid:0) ki (cid:1)(cid:0) n − kk − i (cid:1) . Since n ≥ k (cid:0) kk (cid:1) , n / > P k − i = c +2 |F i | .Therefore, by Claim 3.9, k − X i =0 |F i | = c +1 X i =1 |F i | + k − X i = c +2 |F i | ≤ c +1 X i =1 w i − c +1 X i =1 n i k − X i = c +2 (cid:18) ki (cid:19)(cid:18) n − kk − c − (cid:19) < k − X i =1 w i , and we are done. So we may assume that C is contained in a perfect ( k − Claim 3.11.
The inequality |F i | ≤ w i + t i b i − n i holds for all i ≥ c + 1 .Proof. By Claim 3.10, every set C ⊂ U of size at most k − c is contained in a perfect( k − |F i | = X C ∈P i |F ( C ) | + X C ∈G i |F ( C ) | + X C ∈B i |F ( C ) |≤ (cid:18) k − i − (cid:19) p i + (cid:18)(cid:18) k − i − (cid:19) − (cid:19) g i + (cid:18) ki (cid:19) b i . By Claims 3.4, 3.6, and 3.11, k − X i =0 |F i | ≤ k − X i =1 ( w i + t i b i − n i ) = k − X i =1 w i − k − X i =1 ( n i − t i +1 b i +1 ) − n k − ≤ k − X i =1 w i − k − X i =1 n i , C is perfect for every C ∈ (cid:0) Ui (cid:1) and forevery i ∈ [ k − F is the disjoint union of a k -set and a full star. We are very grateful to Dhruv Mubayi for his guidance, expertise, fruitful discussions thatgreatly assisted this research, and suggestions that greatly improved the presentation ofthis paper. We are also very grateful to the referee for a careful reading of this manuscriptand several helpful suggestions.
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