Cross-intersecting families of labeled sets Huajun Zhang∗ Department of Mathematics, Zhejiang Normal University, Jinhua 321004, P.R. China
[email protected] Submitted: Jan 30, 2012; Accepted: Jan 14, 2013; Published: Jan 21, 2013 Mathematics Subject Classifications: 05D05, 06A07
Abstract For two positive integers n and p, let Lp be the family of labeled n-sets given by Lp = {(1, `1 ), (2, `2 ), . . . , (n, `n )} : `i ∈ [p], i = 1, 2 . . . , n . Families A and B are said to be cross-intersecting if A ∩ B 6= ∅ for all A ∈ A and B ∈ B. In this paper, we will prove that for p > 4, if A and B are cross-intersecting subfamilies of Lp , then |A||B| 6 p2n−2 , and equality holds if and only if A and B are an identical largest intersecting subfamily of Lp . Keywords: EKR theorem; Intersecting family; cross-intersecting family; labeled set
1
Introduction
For a positive integer n, let [n] denote the set {1, 2, . . . , n}. Given a set X, by Xk we denote the set of all k-subsets of X, and let 2X denote the set of all subsets of X. A family A of sets is said to be t-intersecting if |A ∩ B| > t for every pair A, B ∈ A. Usually, A is called intersecting if t = 1. The Erd˝os-Ko-Rado Theorem [15] says that if A is an intersecting subfamily of [n] k where n > 2k, then |A| 6 n−1 . This theorem is a central result in extremal set thek−1 ory and inspires abundant fruits in this field, for an excellent introduction to this we recommend the survey paper [13]. This theorem has many generalizations, analogs and variations. First, finite sets are analogous to finite vector spaces ([17, 18, 20]), permutations ([11, 12, 27]) and labeled sets (signed sets [4, 6] or colored sets [22]), etc. Second, the intersection condition was ∗ The author is supported by the National Natural Science Foundation of China (No.11001249 and No.11171310) and the Natural Science Foundation of Zhejiang Province (No.LY12A01006.)
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generalized to t-intersection and cross-intersection. Here, families A1 , A2 , . . . , Am are said to be cross-intersecting ifPA ∩ B 6= ∅ for any A ∈ Ai and B ∈ Aj , i 6= j. Many authors studied the bound of m i=1 |Ai | ([19, 5, 6, 7, 8, 9, 10, 29, 30]), and Pyber [25] first considered the bound of |A||B| for cross-intersecting families A and B. His result was slightly refined by Matsumoto and Tokushige [24] and Bey [3] as follows. Theorem 1. If A ⊆ [n] and B ⊆ [n] are cross-intersecting with n > max{2k, 2`}, then k ` n−1 n−1 |A||B| 6 . k−1 `−1 [n] Moreover, the equality holds if and only if A = {A ∈ [n] : i ∈ A} and B = {B ∈ : k ` i ∈ B} for some i ∈ [n], unless n = 2k = 2`. Tokushige [26] and Ellis, Friedgut and Pilpel [14] generalized the above result to crosst-intersecting families of finite sets and cross-t-intersecting subfamilies of the symmetric group Sn , respectively. This paper provides an analogue of Theorem 1 for families whose sets we refer to as labeled sets, following [5]. For an n-tuple p = (p1 , p2 , . . . , pn ) such that p1 , p2 , . . . , pn are positive integers with p1 6 p2 6 · · · 6 pn , we define the family Lp of labeled sets by Lp = {(1, `1 ), (2, `2 ), . . . , (n, `n )} : `i ∈ [pi ], i = 1, 2 . . . , n . Berge [2] determined the maximum size of intersecting families of labeled n-sets, Livingston [23] characterized partial optimal intersecting families and Borg [5] completely solved it by using the shift operator in an inductive argument. Theorem 2 (Berge, Livingston, Borg). If A is an intersecting subfamily of Lp , then |A| 6 p2 p3 · · · pn . When p1 > 3, equality holds if and only if A = {(1, `1 ), (2, `2 ), . . . , (n, `n )} : `i = j , where pi = p1 and j ∈ [p1 ]. P In [5], Borg also determined the upper bound of 16i6m |Ai | for cross-intersecting subfamilies A1 , A2 , . . . , Am of Lp . In this paper, we consider a special case: p1 = p2 = · · · = pn = p. In this case, we write Lp as Lp . The main result in this paper is the following theorem. Theorem 3. Let n and p be two positive integers with p > 4. If A and B are crossintersecting families in Lp , then |A||B| 6 p2n−2 , and equality holds if and only if A = B = {(1, `1 ), (2, `2 ), . . . , (n, `n )} : `i = j for some i ∈ [n] and j ∈ [p]. We will present some preliminary results in the next section, and complete the proof of the above theorem in Section 3. the electronic journal of combinatorics 20(1) (2013), #P17
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2
Preliminary Results
For the labeled set Lp , we can construct a simple graph, whose vertex set is Lp , and A, B ∈ Lp are adjacent if and only if A ∩ B = ∅. For convenience, this graph is also denoted by Lp . Set Γ = Sn o Sp = {(f, g1 , g2 , . . . , gn ) : f ∈ Sn and g1 , g2 , . . . , gn ∈ Sp }, the wreath product of the symmetric groups on [n] and [p]. For σ = (f, g1 , g2 , . . . , gn ) ∈ Γ and {(1, `1 ), (2, `2 ), . . . , (n, `n )} ∈ Lp , define σ({(1, `1 ), . . . , (n, `n )}) = {(f (1), g1 (`1 )), . . . , (f (n), gn (`n ))}. Then Γ acts transitively on Lp . In other words, the graph Lp is vertex-transitive. Moreover, every intersecting subfamily of the labeled set Lp corresponds to an independent set of the graph Lp . In the sequel we shall alternatively use the terms “set” and “graph” when referring to Lp . For a graph G, let α(G) denote the independence number of G. Given a subset A of V (G), we define NG (A) = {b ∈ V (G) : {a, b} ∈ E(G) for some a ∈ A} N G (A) = V (G) − NG (A). If G is clear from the context, for simplicity, we will omit the index G. For B ⊆ V (G), by G[B] we denote the induced subgraph of G. For short, we abbreviate α(G[B]) to α(B). For the labeled set Lp we construct another graph Lbp , whose vertex set is the set {(A, B) ∈ Lp × Lp : A ∩ B 6= ∅}, and (A1 , B1 ) and (A2 , B2 ) are non-adjacent if and only if A1 ∩ B2 6= ∅ and B1 ∩ A2 6= ∅. By definition it is easy to see that if A and B are cross-intersecting subfamilies of Lp , then A × B is an independent set of Lbp . Therefore, |A||B| 6 α(Lbp ). To complete the proof of Theorem 3, it suffices to determine the size and structure of the maximum independent sets in Lbp . Note that the action of Γ on Lp induces an action on the graph Lbp defined by σ(A, B) = (σ(A), σ(B)) for σ ∈ Γ and (A, B) ∈ Lbp . For 1 6 i 6 n, set Lbp,i = {(A, B) ∈ Lp × Lp : |A ∩ B| = i}. Clearly, |A ∩ B| = |σ(A) ∩ σ(B)| holds for all σ ∈ Γ and A, B ∈ Lp , and it is easy to verify that Lbp,1 , Lbp,2 , . . . , Lbp,n are all orbits of Γ on Lbp . In other words, every induced subgraph Lbp,i is vertex-transitive. In the context of vertex-transitive graphs, the following result named the “no-homomorphism lemma” is useful to get bounds on the size of independent sets. Lemma 4 (Albertson and Collins [1]). Let G and G0 be two graphs such that G is vertex0) transitive and there exists a homomorphism φ : G0 7→ G. Then |Vα(G) 6 |Vα(G , and the (G)| (G0 )| equality holds if and only if for any independent set I of cardinality α(G) in G, φ−1 (I) is an independent set of cardinality α(G0 ) in G0 . The following Lemma is a variation of the above. the electronic journal of combinatorics 20(1) (2013), #P17
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Lemma 5. (see [11, Theorem 3]) Let G be a vertex-transitive graph, and Ω a transitive subgroup of Aut(G). Let I be an independent set of G, and let B ⊆ V (G), then |V |I| 6 (G)| α(B) . |B|
Equality holds if and only if |I ∩ σ(B)| = α(B) holds for all σ ∈ Ω.
Proof. Set D = {σ(B) : σ ∈ Ω} and Du = {D ∈ D : u ∈ D} for u ∈ V (G). Note that the action of Ω on V (G) is transitive. The size of Du , denoted by r, is independent of the choice of u. Hence, r|V (G)| = |B||D|. On the other hand, for each D ∈ D, I ∩ D is also an independent set of D, and so |D ∩ I| 6 α(G[B]). Therefore, r|I| 6 α(G[B])|D|. Combining the above two inequalities gives |V |I| 6 α(G[B]) , and equality holds if and only (G)| |B| if |D ∩ I| = α(G[B]) for each D ∈ D. Since all Lbp,i are vertex-transitive, the above lemma can be applied to them. In more b be a subset of Lbp such that K b ∩ Lbp,i 6= ∅ for 1 6 i 6 n. Write K bi = K b ∩ Lbp,i detail, let K for i ∈ [n]. Then, for any independent set Ib of Lbp and i ∈ [n], |Ib ∩ Lbp,i | 6 α(Lbp,i ), and by b Lemma 5, α(Lbp,i ) 6 |Lbp,i | α(|KbKi| ) . Therefore, i
b = |I|
n X
|Ib ∩ Lbp,i | 6
i=1
k X
|Lbp,i |
i=1
bi) α(K , bi| |K
b = α(K b i ) for all and equality holds if and only if |Ib ∩ Lbp,i | = α(Lbp,i ) and |Ib ∩ Lbp,i ∩ σ(K)| i = 1, 2, . . . , n and σ ∈ Γ. Equivalently, for each σ ∈ Γ, b = |Ib ∩ σ(K)|
n X
−1
b ∩K b ∩ Lbp,i | = |σ (I)
n X
i=1
b i ) = α(K). b α(K
i=1
We state it as a lemma as follows. b be a subset of Lbp such that K bi = bi = K b ∩ Lbp,i . Lemma 6. Let K 6 ∅ for 1 6 i 6 n, where K If Ib is an independent set of Lbp , then b 6 |I|
n X i=1
|Lbp,i |
bi) α(K , bi| |K
b = Pn α(K b i ) = α(K) b for each σ ∈ Γ. and equality holds if and only if |Ib ∩ σ(K)| i=1 Arrange the elements (1, 1), (2, 1), . . . , (n, 1), (1, 2), (2, 2), . . . , (n, 2), . . . , (1, p), (2, p), . . . , (n, p) in a cycle. Let Ri denote the ith n-interval {(s, j), (s + 1, j) . . . , (n, j), (1, j + 1), . . . , (s − 1, j + 1)} of this cycle, where i = n(j − 1) + s with 1 6 s 6 n. Set R = {R1 , R2 , . . . , Rnp } b = {(A, B) ∈ R × R : A ∩ B 6= ∅}. Then, R b ⊆ Lbp and R bi = R b ∩ Lbp,i 6= ∅ for each and R 1 6 i 6 n. the electronic journal of combinatorics 20(1) (2013), #P17
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Clearly, Ri ∩ Rj 6= ∅ if and only if |i − j| < n or |i + np − j| < n for Ri , Rj ∈ R, and the subgraph of Lp induced by R, which will also be denoted by R, is isomorphic to the well-known circular graph Circ(n, np). Here, the graph Circ(n, np) has the vertex set [np], and i and j are not adjacent if and only if |i − j| < n or |np + i − j| < n. Hence, α(R) = n, and by the well-known result of Katona [21], the maximum independent sets b is the desired subset. of R are stars. In the following we will prove that R Let A and B be cross-intersecting subfamilies of R. Then, it is obvious that B ⊆ N R (A). For every non-empty A ⊂ V (Circ(n, np))(p > 3), we have proved that if |A| > 2n, N (A) = ∅; if |A| < 2n, |N (A)| + |A| 6 2n, and equality holds if and only if A = {i, i + 1, . . . , i + |A| − 1} for some i (see [16, Lemma 3.1] or [28, Lemma 2.3]). Therefore, if A and B are both non-empty, then |A| + |B| 6 |A| + |N R (A)| 6 2n. Note that |A||B| = 0 if one of A and B is empty. So we have that |A||B| 6 |A|(2n − |A|) 6 n2 , and equality holds if and only if A and B are some identical maximum independent set of R. Therefore, b = n2 . In the following, we give a stronger result. α(R) Lemma 7. Suppose p > 4. Then b = n2 = α(R)
n X
b i ), α(R
i=1
b if and only if Ib = S × S for some maximum and Ib is a maximum independent set of R independent set of R. b i |. Let S Proof. For any subsets A, B of R and 1 6 i 6 n, set (A, B)i = |(A × B) ∩ R be a fixed maximum independent set of R and write (S, S)i = ai . Clearly, ai does not b = n2 = P depend on the choice of S, and α(R) 16i6n ai . To complete the proof, it suffices b i ) = ai for each 1 6 i 6 n. To do this, we only need to verify that for to prove that α(R b |Ib ∩ R b i | 6 ai for 1 6 i 6 n. every independent set Ib of R, b b Then there exists a pair of cross-intersecting Let I be an independent set of R. subfamilies C and D of R such that Ib ⊆ C × D. Since |C| + |D| 6 2n, we may assume |C| = s 6 n. We first consider the simple case when C consists of consecutive elements of R. Without loss of generality, assume C = {Rn , Rn+1 , . . . , Rn+s−1 }. For 1 6 t 6 n, set Ct = {Rn , Rn+1 , . . . , Rn+t−1 }. Then, D ⊆ N (Cs ). For each 1 6 t < n and 1 6 i 6 n, it is easy to verify that Ct+1 × N (Ct+1 ) = [Ct × N (Ct+1 )] ∪ [{Rn+t } × N (Ct+1 )] = [Ct × N (Ct )] ∪ [{Rn+t } × N (Ct+1 )] − [Ct × {Rt }] and ({Rn+t }, N (Ct+1 ))i > (Ct , {Rt })i , and consequently we have (Ct+1 , N (Ct+1 ))i = (Ct , N (Ct ))i + ({Rn+t }, N (Ct+1 ))i − (Ct , {Rt })i > (Ct , N (Ct ))i . Therefore, for 1 6 i 6 n, (C, D)i 6 (Cs , N (Cs ))i 6 (Cs+1 , N (Cs+1 ))i 6 · · · 6 (Cn , N (Cn ))i = ai the electronic journal of combinatorics 20(1) (2013), #P17
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because Cn = N (Cn ) is a maximum independent set of R. Now we consider the general case. Without loss of generality, assume R2n ∈ D. Then C ⊆ N (D) ⊆ N ({R2n }) = {Rn+1 , Rn+2 , . . . , R3n−1 }. Suppose C = {Ri1 , Ri2 , . . . , Ris }, where n + 1 6 i1 < i2 < · · · < is 6 3n − 1. Noting p > 4, if Rj ∈ N ({Ri1 }) ∩ N ({Ris }), then it follows from definition that |j − i1 | < n and |j − is | < n, that is, is − n + 1 6 j 6 i1 +n−1. Therefore, N ({Ri1 })∩N ({Ris }) = {Ris −n+1 , Ris −n+2 , . . . , Ri1 +n−1 } = N (C). Set C 0 = {Ri1 , Ri1 +1 , . . . , Ris } and D0 = {Ris −n+1 , Ris −n+2 , . . . , Ri1 +n−1 }. Then, C 0 = N (D0 ), and the above argument implies that the inequality (C 0 , D0 )i 6 ai holds for each 1 6 i 6 n. Note that C ⊆ C 0 and D ⊆ D0 . Hence, (C, D)i 6 (C 0 , D0 )i 6 ai . Remark. In the above result, the condition that p > 4 is necessary. For example, assume n = 6 and p = 3, set S = {R1 , R2 , R3 , R4 , R5 , R6 }, C = {R6 , R14 } and D = {R1 , R11 }, it is easy to see that S is a maximum independent set ofP R and C × D is independent set Pan 6 6 b b b b of R, but 2 = (S, S)1 < (C, D)1 = 3 6 α(R1 ), and so i=1 α(Ri ) > i=1 (S, S)i = α(R).
3
Proof of Theorem 3
In this section we complete the proof of Theorem 3. Proof of Theorem 3. Take a maximum independent set S 0 of Lp and set Ib0 = S 0 ×S 0 . Then α(Lp ) |S 0 | = = = p1 . For Ib0 is an independent set of Lbp with |Ib0 | = p2n−2 . Note that α(R) |R| |Lp | |Lp | b = n2 , and each σ ∈ Γ, Lemma 5 implies |S 0 ∩σ(R)| = α(R) = n, that is to say, |Ib0 ∩σ(R)| P b i ) hold by Lemma 7. Then, it follows b = α(R) b = n α(R so the equalities |Ib0 ∩ σ(R)| i=1 P b from Lemma 6 that p2n−2 = |Ib0 | = ni=1 |Lbp,i | α(|RbRi| ) . Therefore, for every independent i Pn b α(Rb i ) 2n−2 b b b set I of Lp , we have |I| 6 . Furthermore, the equality holds bi| = p i=1 |Lp,i | |R b = α(R) b for all σ ∈ Γ. Then, for each σ ∈ Γ, by Lemma 7, if and only if |Ib ∩ σ(R)| b = Sσ × Sσ for some maximum independent set Sσ of σ(R). Set S = ∪σ∈Γ Sσ . Ib ∩ σ(R) Noting that the maximality of Ib implies that Ib = C × D for a pair of cross-intersecting b subfamilies C and D of Lp . Then we have that S is an independent set and S × S ⊆ I. On the other hand, it is easy to see that |S ∩ σ(R)| = α(R) holds for all σ ∈ Γ, so Lemma 5 implies S is a maximum independent set of Lp . Then we obtain Ib = S × S b = p2n−2 = |S × S|. This completes the proof of Theorem 3. since |I|
Acknowledgements The author is greatly indebted to the anonymous referees for giving useful comments and suggestions that have considerably improved the manuscript. He also thanks Jun Wang for giving many valuable suggestions.
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