This paper establishes tight bounds on the sizes of two families of subsets of [n] with fractional intersection conditions, generalizing classical intersecting family results and characterizing extremal cases.
Contribution
It introduces the concept of fractional cross intersecting families and provides tight bounds and characterizations for their maximum product sizes, including special cases like half-intersecting families.
Findings
01
Derived tight upper bounds for the product of family sizes.
02
Characterized extremal families when the bounds are achieved.
03
Extended classical intersection results to fractional intersection scenarios.
Abstract
Let A={A1,...,Ap} and B={B1,...,Bq} be two families of subsets of [n] such that for every i∈[p] and j∈[q], ∣Ai∩Bj∣=dc∣Bj∣, where dc∈[0,1] is an irreducible fraction. We call such families "dc-cross intersecting families". In this paper, we find a tight upper bound for the product ∣A∣∣B∣ and characterize the cases when this bound is achieved for dc=21. Also, we find a tight upper bound on ∣A∣∣B∣ when B is k-uniform and characterize, for all dc, the cases when this bound is achieved.
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Full text
Fractional cross intersecting families
Rogers Mathew
Department of Computer Science and Engineering,
Indian Institute of Technology Kharagpur, Kharagpur 721302, India,
Let A={A1,...,Ap} and B={B1,...,Bq} be two families of subsets of [n] such that for every i∈[p] and j∈[q], ∣Ai∩Bj∣=dc∣Bj∣, where dc∈[0,1] is an irreducible fraction. We call such families dc-cross intersecting families. In this paper, we find a tight upper bound for the product ∣A∣∣B∣ and characterize the cases when this bound is achieved for dc=21. Also, we find a tight upper bound on ∣A∣∣B∣ when B is k-uniform and characterize, for all dc, the cases when this bound is achieved.
1 Introduction
Let [n] denote {1,...,n} and let 2[n] denote the power set of [n].We shall use (k[n]) to denote the set of all k-sized subsets of [n]. Let F⊆2[n]. The family F is an intersecting family if every two sets in F intersect with each other. The famous Erdős-Ko-Rado Theorem [1] states that ∣F∣≤(k−1n−1) if F is a k-uniform intersecting family, where 2k≤n. Several variants of the notion of intersecting families have been extensively studied in the literature. Given a set L={l1,…,ls} of non-negative integers, a family F⊆2[n] is L-intersecting if for all Fi,Fj∈F,Fi=Fj,∣Fi∩Fj∣∈L. Ray-Chaudhuri and Wilson in [2] showed that if F is k-uniform and L-intersecting, then ∣F∣≤(sn) and the bound is tight. Frankl and Wilson in [3] showed a tight upper bound of (sn)+(s−1n)+⋯+(0n) if the restriction on the cardinalities of the sets of an L-intersecting family is relaxed. Further, if L is a singleton set, then Fisher inequality [4] gives an upper bound of ∣F∣≤n for the cardinality of an L-intersecting family F. Recently, in [5], Balachandran et al. introduced a fractional variant of the classical L-intersecting families. For a survey on intersecting families, see [6].
Two families A,B⊆2[n] are cross-intersecting if ∣A∩B∣>0, ∀A∈A,B∈B. Pyber in [7] showed that if n≥2k, and A,B⊆(k[n]) is a cross-intersecting pair of families, then ∣A∣∣B∣≤(k−1n−1)2. Frankl et al. in [8] showed that if A,B⊂(k[n]) such that ∣A∩B∣≥t for all A∈A and B∈B, then for all n≥(t+1)(k−t+1), ∣A∣∣B∣≤(k−tn−t)2, the cross-intersecting version of the Erdős-Ko-Rado Theorem. A cross-intersecting pair of families A,B⊆2[n] is said to be l-cross-intersecting if ∀A∈A,\leavevmodeB∈B, ∣A∩B∣=l, for some positive integer l.
Ahlswede, Cai and Zhang showed in [9], for all n≥2l, a simple construction of an l-cross-intersecting pair (A,B) of families of subsets of [n] with ∣A∣∣B∣=(l2l)2n−2l=Θ(l2n). Later Alon and Lubetzky in [10] showed that the Θ(l2n) bound is tight and characterized the cases when the bound is achieved.
In this paper, we introduce a fractional variant of the l-cross-intersecting families.
Let A={A1,...,Ap} and B={B1,...,Bq} be two families of subsets of [n] such that for every i∈[p] and j∈[q], ∣Ai∩Bj∣=dc∣Bj∣, where dc∈[0,1] is an irreducible fraction. We call such an (A,B) pair a dc-cross-intersecting pair of families. Given c, d, and n, let Mdc(n) denote the maximum value of ∣A∣∣B∣ where (A,B) is a dc-cross intersecting pair of families of subsets of [n].
We have the following results:
Theorem 1.1**.**
\thlabel
thm:1.1
Mdc(n)=2n
When dc=0, A=2[n], B={∅} is a maximal pair. In fact, A=2[k], B=P(S), where P(S) is the power set of S={k+1,…,n}, are the only maximal pairs up to a relabelling of the elements, 0≤k≤n.
When dc=1, A={[n]} and B=2[n] is a maximal pair. In fact, B=2[k], A={A:A=[k]∪T, where T∈P(S)}, where P(S) is the power set of S={k+1,…,n}, are the only maximal pairs up to a relabelling of the elements, 0≤k≤n.
In \threfthm:1.2, we characterize all maximal pairs when dc=21.
Theorem 1.2**.**
\thlabel
thm:1.2
Let (A,B) be a 21-cross intersecting pair of families of subsets of [n] with ∣A∣∣B∣=2n. Then (A,B) is one of the following ⌊2n⌋+1 pairs of families (Ak,Bk), 0≤k≤⌊2n⌋, up to isomorphism.
Bk={B∈2[n]:∣B∩{2i−1,2i}∣∈{0,2}\leavevmode\leavevmode∀i,1≤i≤k* and ∀j>2k, j∈/B},*
where 1≤k≤⌊2n⌋.
It would be interesting to show a characterization theorem for any dc∈[0,1]. We do have such a general characterization theorem (along with a new tight upper bound) in \threfthm:1.3 for the case when B is k-uniform. The proof is a direct application of Theorem 1.1 in [10].
Theorem 1.3**.**
\thlabel
thm:1.3
Let (A,B) be a dc-cross intersecting pair of families of subsets of [n]. Let B be k-uniform. Then, there exists some k0>0, such that for k>k0 we have
∣A∣∣B∣≤(dckd2ck)2n−d2ck**
and the bound is tight if and only if, either (a) or (b) hold:
(a)
When dc=1, A={{1,…,κ}}×2Y, B=(k[κ]) where Y={κ+1,…,n} and κ∈{2k−1,2k} up to a relabelling of the elements of [n].
2. (b)
When dc=1:
(i)
If k is even, c=1, d=2, dck=⌈2k⌉,
2. (ii)
If k is odd, c=2k+1, d=k, dck=⌈2k⌉,
and for both the cases((i) and (ii)),
there exists some τ such that,
k+τ≤n
and up to a relabelling of the elements of [n],
B={L∪{τ+1,…,k}:L⊂{1,…,τ,k+1,…,k+τ},∣L∩{i,k+i}∣=1* for all i∈[τ]}.*
2 Notations and definitions
Given any S⊆[n], we shall use χ(S) to denote the characteristic vector of S which is a 0−1 vector of size n having its ith entry equal to 1 if and only if i∈S. The weight of a vector is the number of non-zero entries it has, and hence weight of χ(S) is the same as ∣S∣.
For any family A⊆2[n], we shall (ab)use A to denote the collection of characteristic vectors of the members of A as well. The meaning will be clearly stated if not clear from the context.
Let V be a collection of vectors in F2n. Then, we define the following:
span(V): The collection of all the vectors that can be expressed as a linear combination in F2 of the vectors of V. We know that span(V) is a vector space over F2.
2. 2.
basis(V): We use basis(V) to denote the basis of span(V).
3. 3.
dim(V): dim(V)=∣basis(V)∣
Definition 1**.**
V⊆F2n* is a linear code if V=span(V).*
Definition 2**.**
Given a linear code C⊆F2n, the dual codeC⊥ is defined as,
C⊥={x∈F2n∣⟨x,c⟩=0,∀c∈C}**
where ⟨x,y⟩ is the standard inner product over F2.
The following is a well-known fact that is easy to verify.
Lemma 2.1**.**
If C⊆F2n is a linear code, then C⊥ is also a linear code.
Definition 3**.**
Self orthogonal and self dual codes:
A code C is self orthogonal if C⊆C⊥
and it is self dual if C=C⊥.
3 Bounding Mdc(n)
Let (A,B) be a dc-cross-intersecting pair of families of subsets of [n], where dc∈[0,1] is an irreducible fraction. We shall (ab)use A,B to denote the set of characteristic vectors of the sets in A,B respectively.
For any a∈A,b∈B, we observe that ⟨a,b⟩≡∣A∩B∣\leavevmode(mod\leavevmode2), where a=χ(A), b=χ(B).
Partition the family B into two parts as,
[TABLE]
As all the sets B∈B have their cardinality ∣B∣ divisible by d, {B1,B2} is a valid partition of B.
Therefore ∀a∈A , b∈B, using the dc intersection property, we have:
[TABLE]
Construction 1**.**
Construct a set B1′, by appending a [math] to the left of every vector in B1, and a set B2′ by appending a 1 to the left of every vector in B2. Let B′=B1′∪B2′.
Construct a set A′ by appending a 1 to the left of every vector in A.
We now have, the value of
⟨a,b⟩=0\leavevmode\leavevmode∀a∈A′, b∈B′
So, (span(A′),span(B′)) is a pair of mutually orthogonal subspaces of F2n+1 over F2. We thus have,
dim(span(A′))+dim(span(B′))≤n+1
So, it follows that
[TABLE]
Lemma 3.1**.**
\thlabel
lem:3.1
If the elements of a linear code C⊆F2n are arranged as rows of a matrix MC with n columns, then for each column, one of the following holds,
(i)
All the entries in that column are [math]
2. (ii)
Exactly half the entries in that column are [math], and the rest are 1.
Proof.
As C is a linear code, if we pick any a∈C, and consider the set S={a+x∣x∈C} where a+x is the vector addition in F2n, then by the definition of a linear code S=C. Let MS be a matrix whose rows are the vectors of S, taken in any order. MS and MC have the same set of rows (only their order may differ).
Let j∈[n]. Column j in MC and MS have the same number of 1’s( and [math]’s). Suppose (i) does not hold for column j in MC. Then, some row, say a, in MC has its jth entry as 1. Let S, and thereby MS, be defined according to this vector a. From the definition of S, it is clear that the number of 1’s in the jth column of MS is equal to the number of 1’s in the jth column of MC. Since adding a to any {0,1} vector flips the jth coordinate of v, we conclude that (ii) holds for Mc.
∎
Corollary 3.2**.**
\thlabel
cor:3.2
∣span(A′)∣≥2∣A′∣
Proof.
The leftmost column of MA′ does not contain any [math]. As span(A′) is a linear code and A′⊆span(A′), by condition (ii) of \threflem:3.1 above, span(A′) must have at least ∣A′∣ more elements having their leftmost entry as [math].
∎
Now we prove the main result of this section which is \threfthm:1.1.
Statement of \threfthm:1.1:Mdc(n)=2n
Proof.
A=2[n], B={∅} is a trivial example of a dc cross-intersecting pair of families having ∣A∣∣B∣=2n. Thus, Mdc(n)≥2n.
The proof of the upper bound for Mdc(n) follows from Inequality (3) and \threfcor:3.2. Let (A,B) be a dc cross-intersecting pair of families of subsets of [n]. Let A′, B′ be constructed from A, B, respectively, as explained in the beginning of this section. Note that ∣A′∣=∣A∣ and ∣B′∣=∣B∣ by construction.
[TABLE]
∎
4 Characterization of maximal pairs when dc=21
Definition 4**.**
Cross bisecting pair of families: A pair of families of subsets of [n] is called a cross-bisecting pair if it is a 21 cross-intersecting pair. (A,B) is called a maximal cross bisecting or simply a maximal pair, if it is a cross bisecting pair and ∣A∣∣B∣=2n.
For example, A=2[n] and B={∅} is a trivial maximal pair. In this section, we characterize all maximal pairs.
Let (A,B) be a cross bisecting pair and let (A′,B′) be the associated pair constructed by appending bits as defined in the previous section.
Definition 5**.**
Let fA:A→A′ be a bijective mapping that maps every vector in A to its corresponding vector in A′, and let gA:A′→A be its inverse. Likewise, define functions fB and gB between B and B′. For any set V⊆A, we shall use, fA(V) to denote {fA(A)∣A∈V} and for any V⊆A′, we use gA(V) to denote {gA(A)∣A∈V}.
Similarly, for any V⊆B, we use, fB(V) to denote {fB(B)∣B∈V} and for any V⊆B′, gB(V) to denote {gB(B)∣B∈V}
Observation 1**.**
\thlabel
*obv
fB(B1)=B1′ and fB(B2)=B2′.
Similarly, gB(B1′)=B1 and gB(B2′)=B2
Suppose (A,B) is a maximal pair. Then from the proof of \threfthm:1.1, we must have :
[TABLE]
Proposition 4.1**.**
\thlabel
lem:4.3
B=span(B). Further, fB is a linear map.
Proof.
This follows from equation (5). Let x1,x2∈B. We show that x3=x1+x2∈B. This would imply B is closed under addition in F2n over F2, and hence B=span(B).
Let x1′=fB(x1) and x2′=fB(x2). From Equation (5), we have, w=x1′+x2′∈B′. Since w and x3 agree on each of the rightmost n bits of x3, we have gB(w)=x3. Since w∈B′, from the definition of the function gB we have x3=gB(w)∈B. Further, observe that fB(x1)+fB(x2)=w=fB(x3)=fB(x1+x2) and hence fB is a linear map.
∎
That B is a linear code from \threflem:4.3 implies closure of the family of subsets B under symmetric difference. In fact, we have the following stronger result.
Proposition 4.2**.**
\thlabel
lem:4.4
Let vectors b1,b2∈B. Then, b1+b2∈B1 if and only if either b1,b2∈B1, or b1,b2∈B2. Otherwise, b1+b2∈B2.
Proof.
We prove the 2-way implication, and rest of the proposition follows from \threflem:4.3. Let b1′=fB(b1),b2′=fB(b2).
•
b1+b2∈B1⇒b1 and b2 are both from B1, or both from B2
Since fB is a linear map, we have (b1+b2∈B1)⇒(fB(b1+b2)=fB(b1)+fB(b2)=b1′+b2′∈B1′). So, the leftmost bit of b1′+b2′ is [math]. This means that the leftmost bit must be the same in b1′ and b2′, which directly implies that either b1′,b2′∈B1′, or b1′,b2′∈B2′.
•
Either b1,b2∈B1, or b1,b2∈B2⇒b1+b2∈B1
Since b1′ and b2′ agree upon the leftmost bit, b1′+b2′ has a [math] in its leftmost bit. So, b1′+b2′∈B1′. From the \threfobv above, we have b1+b2∈B1.
∎
Proposition 4.3**.**
\thlabel
lem:4.5
B is a self-orthogonal code.
Proof.
We prove the proposition by showing that ∀b1,b2∈B, ⟨b1,b2⟩=0. Let B1,B2 be the sets corresponding to the vectors b1,b2, respectively. Since we are operating in the field F2, it is enough to show that ∣B1∩B2∣ is even.
Let b3=b1+b2. We observe that b3 is the characteristic vector of B3=B1ΔB2, the symmetric difference of B1 and B2.
We have,
[TABLE]
As dc=21, ∀B∈B1, we have ∣B∣≡0\leavevmode(mod\leavevmode4).
By \threflem:4.3, B1ΔB2=B3∈B as B is a linear code.
Taking equation (7) modulo 4, if B3∈B1, then
[TABLE]
By \threflem:4.4, both B1 and B2 are either from B1 or from B2. In both cases, ∣B1∣+∣B2∣≡0\leavevmode(mod\leavevmode4)
Therefore, 2∣B1∩B2∣≡0\leavevmode(mod\leavevmode4) or
∣B1∩B2∣≡0\leavevmode(mod\leavevmode2).
If B3∈B2, then
[TABLE]
Again by \threflem:4.4, ∣B1∣+∣B2∣≡2\leavevmode(mod\leavevmode4).
So, we have 2∣B1∩B2∣≡0\leavevmode(mod\leavevmode4) or
∣B1∩B2∣≡0\leavevmode(mod\leavevmode2).
Thus in both cases, ∣B1∩B2∣
is even, so B is a self-othogonal code.
∎
Lemma 4.4**.**
\thlabel
lem:4.6
Let (A,B) be a maximal pair, then ∣B∣≤2⌊2n⌋
Proof.
It is a known result (see [11]) that for a linear code C⊆F2n and its dual code C⊥,
Since B is a self-orthogonal code (\threflem:4.5), we get dim(B) ≤2n. Hence,
[TABLE]
∎
Proposition 4.5**.**
\thlabel
lem:4.7
If a set A bisects B1, B2 and B1ΔB2, then A also bisects B1∩B2.
Proof.
[TABLE]
[TABLE]
[TABLE]
∎
Proposition 4.6**.**
\thlabel
cor:4.8
B is closed under intersection.
Proof.
Let B1,B2∈B. We show that B1∩B2∈B. By \threflem:4.3, b1+b2∈B i.e., B1ΔB2∈B. Let A be any arbitrary member of A.
Now, A bisects B1,B2 and B1ΔB2 as (A,B) is a cross bisecting pair. By \threflem:4.7, A bisects B1∩B2. Since (A,B) is a maximal pair, we conclude that B1∩B2∈B.
∎
Now, we prove the main result of this section,\threfthm:1.2, the characterization of maximal pairs.
**Statement of \threfthm:1.2: **
Let (A,B) be a 21-cross intersecting pair of families of subsets of [n] with ∣A∣∣B∣=2n. Then (A,B) is one of the following ⌊2n⌋+1 pairs of families (Ak,Bk), 0≤k≤⌊2n⌋, up to isomorphism.
Bk={B∈2[n]:∣B∩{2i−1,2i}∣∈{0,2}\leavevmode\leavevmode∀i,1≤i≤k* and ∀j>2k, j∈/B},*
*where 1≤k≤⌊2n⌋.
*
By isomorphism, it is meant that for any maximal pair (A,B), ∃ a bijective mapping
f:[n]→[n]
such that if every A∈A is replaced by Af={f(i)∣i∈A} and every B∈B is replaced by Bf={f(i)∣i∈B} then the families (Af,Bf), where Af={Af∣A∈A} and Bf={Bf∣B∈B}, is a maximal pair which is one of (Ak,Bk) , 0≤k≤⌊2n⌋.
Proof.
Consider a maximal pair (A,B) where B={∅}. We write the elements of B as rows of a 0−1 matrix M0. Suppose n0 columns have only [math] entries in all the rows(n0 may be [math]). As the characterization is up to isomorphism, we may assume that these are the rightmost n0 columns of the matrix M0.
In each of the remaining n−n0 columns, from \threflem:3.1, there are exactly 2∣B∣1’s and 2∣B∣ [math]’s as B is a linear code. (by \threflem:4.3)
Define
[TABLE]
We write the 2∣B∣ rows containing 1 in the leftmost column of M0 as the top 2∣B∣ rows to obtain a new matrix M1 from M0. And B1 is one of these rows according to \threfcor:4.8.
Moreover, as all intersections are of even cardinality (\threflem:4.5), ∣B1∣ is even.
Let ∣B1∣=2i1, i1≥1.
So, there are 2i1−1 elements in B1 other than the element 1. Due to isomorphism, we may assume them to be 2,3,…,2i1.
If 2i1+1≤n−n0, then define the set B2 as:
[TABLE]
Claim 4.7**.**
\thlabel
lem:4.ten
1∈/B2
Proof.
Assume for the sake of contradiction, 1∈B2. This implies that for all the 2∣B∣ sets which contain the element 2i1+1 also contain the element 1. From \threflem:3.1, (number of sets in B that contain the element 1) = (number of sets in B that contain the element 2i1+1) = 2∣B∣. Hence, for any B∈B, 1∈B⟺2i1+1∈B. This implies that 2i1+1∈B1, which is a contradiction. Hence, 1∈/B2 and therefore B2 does not belong to the top 2∣B∣ rows of M1.
∎
Claim 4.8**.**
\thlabel
lem:4.eleven
B1∩B2=∅
Proof.
Assume for the sake of contradiction, x∈B1∩B2. Then x is present in the 2∣B∣ rows of the matrix M1 whose intersection yields B1. Since x∈B2 and B2 does not belong to these 2∣B∣ rows of M1 (by \threflem:4.ten). Thus, we have the element x present in at least 2∣B∣+1 rows of M1, contradicting \threflem:3.1.
∎
We take the rows corresponding to the sets containing the (2i1+1)th element that are not among the first 2∣B∣ rows in M1 and arrange them below
the top 2∣B∣ rows to create a matrix called M2 from M1. Again from \threflem:4.5, ∣B2∣ is even, say 2i2. Due to isomorphism and \threflem:4.eleven, we may assume that 2i1+1,…,2i1+2i2 are these 2i2 elements.
If 2i1+2i2+1≤n−n0, then define,
[TABLE]
Claim 4.9**.**
\thlabel
lem:4.twelve
1∈/B3 and 2i1+1∈/B3.
The proof is similar to that of \threflem:4.ten
Claim 4.10**.**
\thlabel
lem:4.thirteen
B1∩B3=∅ and B2∩B3=∅.
The proof is again similar to that of \threflem:4.eleven.
We take the rows corresponding to the sets containing the (2i1+2i2+1)th element that are not among the first r rows (r>2∣B∣) in M2 which contain the elements 1 or
2i1+1 and arrange them below the top r rows of M2 to create a
matrix called M3 from M2. From \threflem:4.5 and the definition
of B3, we have ∣B3∣=2i3, i3≥1. Due to isomorphism and
\threflem:4.thirteen, we may assume that
2i1+2i2+1,…,2i1+2i2+2i3 are these 2i3 elements.
We continue in this manner for k steps by constructing sets
B1,…,Bk and matrices M1,…,Mk, where k≥1, until we have 2i1+⋯+2ik=n−n0.
Observe that B1,…,Bk and P={n−n0+1,…,n} is a partition of [n].
Claim 4.11**.**
\thlabel
lem:4.fourteen
For any set B∈B, j∈[k], we have B∩Bj∈{∅,Bj}. Further, B∩P=∅.
Proof.
From the definition of P, we have B∩P=∅. Let j∈[k]. Since Bj is equal to the intersection of some 2∣B∣ sets in B, we have Bj present as a subset of all these 2∣B∣ sets. Applying \threflem:3.1, we can say that no element of Bj is present in any set in B other than these 2∣B∣ sets. Hence, the claim.
∎
From \threflem:4.fourteen, observe that S={B1,…,Bk} forms a basis of the row space of the matrix Mk.
The advantage of such a
“disjoint basis” is that the bisection in one part is independent of another.
Claim 4.12**.**
\thlabel
lem:4.fifteen
A set A∈A bisects every set in B if and only if it bisects every set in the basis S of B.
Proof.
The forward direction is straightforward as S⊆B. For the opposite direction, let A∈A be a set that bisects every member of S. Since the sets corresponding to the members in S are disjoint, any B∈B can be written as a union of some of these sets.
Since each set A∈A bisects the sets B1,…,Bk and P, from \threflem:4.fifteen, the set A may contain any of the 2n0 subsets of P, and ∣A∩B1∣=i1,…,∣A∩Bk∣=ik. Since dim(B)=k, by \threflem:4.3, we have ∣B∣=2k.
[TABLE]
Recall that j=1∑k2ij=n−n0. Right hand side of Equation (9), is equal to 2n if and only if ij=1, ∀j∈[k].
Thus, if B={∅}, then (Ak,Bk),k≥1, defined in the statement of the theorem are the only maximal pairs. This completes the proof of \threfthm:1.2.
∎
5 Tight upper bound on Mdc(n) when B is k-uniform and characterization of the cases when the bound is achieved
Let (A,B) be a dc cross-intersecting pair of families of subsets of [n], where dc∈[0,1] is an irreducible fraction. In this section, we deal with the scenario when B is k-uniform, where 0<k≤n.
Since B is k-uniform, for any A∈A and any B∈B, ∣A∩B∣=dck=l. Since c is relatively prime with d, and ∣A∩B∣ is an integer, we have k divisible by d. Therefore, we have a uniformly cross intersecting pair of families.
Alon and Lubetzky in [10] found a tight upper bound for the case of uniformly cross intersecting families and fully characterized
the cases when the bound is achieved in the following theorem:
Theorem 5.1**.**
\thlabel
thm:5.1 [Theorem 1.1 in [10]]
There exists some l0>0 such that, for all l≥l0, every l-cross intersecting pair A,B⊂2[n] satisfies:
∣A∣∣B∣≤(l2l)2n−2l**
Furthermore, if ∣A∣∣B∣=(l2l)2n−2l, then there exists some choice of parameters κ,τ,n′:
κ∈{2l−1,2l},τ∈{0,⋯,κ}
κ+τ≤n′≤n**
such that upto a relabelling of the elements of [n] and swapping A,B, the following holds:
B={L∪{τ+1,⋯,κ}:L⊂{1,⋯,τ,κ+1,⋯,κ+τ},∣L∩{i,κ+i}∣=1*
for all i∈[τ]}×2Y*
where X={κ+τ+1,⋯,n′} and Y={n′+1,⋯,n}.
Let (A,B) be a dc cross-intersecting
family where B is k-uniform. From \threfthm:5.1, there
exists a k0>0 such that if dck=l>k0, then
∣A∣∣B∣≤(l2l)2n−2l. Consider the case
when B corresponds to B of \threfthm:5.1. If
∣A∣∣B∣=(l2l)2n−2l, then n′=n,
Y=∅, and k=κ in the statement of \threfthm:5.1.
Since l=dck and k∈{d2ck−1,d2ck}, we have the following two cases:
Case 1:k=d2ck−1. Then, (k+1)d=2ck. Since gcd(c,d)=1 and gcd(k,k+1)=1, we have k∣d∣2k. Thus, d=k or d=2k. We claim that d=2k is an invalid case. This is because, when d=2k, we have c=k+1. Since gcd(c,d)=1, k cannot be odd. And if k is even, then l=dck=2k+1 is not an integer. So, the only valid case is d=k, c=2k+1=l and k is an odd integer.
Case 2:k=d2ck. Then, dc=21, that is (A,B) is a cross bisecting pair. Since l=dck=2k is an integer, k must be even in this case.
If B corresponds to A of \threfthm:5.1, we have X=∅, τ=0, B is k(=l)-uniform, l=dck. Thus, we have dc=1, A={{1,…,κ}}×2Y where Y={κ+1,…,n} and B=(k[κ]), κ∈{2k−1,2k} up to a relabelling of the elements.
This leads us to the main result of this section.
Statement of \threfthm:1.3:Let (A,B) be a dc-cross intersecting pair of families of subsets of [n]. Let B be k-uniform. Then, there exists some k0>0, such that for k>k0 we have
∣A∣∣B∣≤(dckd2ck)2n−d2ck**
and the bound is tight if and only if, either (a) or (b) hold:
(a)
When dc=1, A={{1,…,κ}}×2Y, B=(k[κ]) where Y={κ+1,…,n} and κ∈{2k−1,2k} up to a relabelling of the elements of [n].
2. (b)
When dc=1:
(i)
If k is even, c=1, d=2, dck=⌈2k⌉,
2. (ii)
If k is odd, c=2k+1, d=k, dck=⌈2k⌉,
and for both the cases((i) and (ii)),
there exists some τ such that,
k+τ≤n
and up to a relabelling of the elements of [n],
B={L∪{τ+1,…,k}:L⊂{1,…,τ,k+1,…,k+τ},∣L∩{i,k+i}∣=1* for all i∈[τ]}.*
6 Discussion
What are those pairs of dc-cross intersecting families (A,B) which achieve ∣A∣∣B∣=2n (equal to the upper bound for Mdc(n) proved in \threfthm:1.1)? In the introduction we characterize such families when dc=0 and dc=1. In \threfthm:1.2, we characterize such families when dc=21. From \threfthm:1.3, we see that when B is k-uniform, ∣A∣∣B∣ is maximized when dc is 1 or nearly 21(21 or 21+2k1). For dc∈(0,1), besides the case A=2[n], B={∅}, is ∣A∣∣B∣=2n achieved only when dc is close to 21?
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