# Fractional L-intersecting families

Let L = {a_1/b_1, ... , a_s/b_s}, where for every i ∈ [s], a_i/b_i∈ [0,1) is an irreducible fraction. Let F = {A_1, ... , A_m} be a family of subsets of [n]. We say F is a fractional L-intersecting family if for every distinct i,j ∈ [m], there exists an a/b∈ L such that |A_i ∩ A_j| ∈{a/b|A_i|, a/b |A_j|}. In this paper, we introduce and study the notion of fractional L-intersecting families.

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09/29/2019

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## 1 Introduction

Let denote and let be a set of non-negative integers. A family of subsets of is -intersecting if for every , , . In 1975, it was shown by Ray-Chaudhuri and Wilson in [13] that if is -uniform, then . Setting , the family is a tight example to the above bound, where denotes the set of all -sized subsets of . In the non-uniform case, it was shown by Frankl and Wilson in the year 1981 (see [7]) that if we don’t put any restrictions on the cardinalities of the sets in , then . This bound is tight as demonstrated by the set of all subsets of of size at most with . The proof of this bound was using the method of higher incidence matrices. Later, in 1991, Alon, Babai, and Suzuki in [2] gave an elegant linear algebraic proof to this bound. They showed that if the cardinalities of the sets in belong to the set of integers with every , then is at most . The collection of all the subsets of of size at least and at most with and forms a tight example to this bound. In 2002, this result was extended by Grolmusz and Sudakov [8] to -wise -intersecting families. In 2003, Snevily showed in [14] that if is a collection of positive integers then . See [11] for a survey on -intersecting families and their variants.

In this paper, we introduce a new variant of -intersecting families called the fractional -intersecting families. Let , where for every , is an irreducible fraction. Let be a family of subsets of . We say is a fractional -intersecting family if for every distinct , there exists an such that . When is -uniform, it is an -intersecting family where and therefore (using the result in [13]), . A tight example to this bound is given by the family where . So what is interesting is finding a good upper bound for in the non-uniform case. Unlike in the case of the classical -intersecting families, it is clear from the above definition that if and are two sets in a fractional -intersecting family, then the cardinality of their intersection is a function of or (or both).

In Section 2.1, we prove the following theorem which gives an upper bound for the cardinality of a fractional -intersecting family in the general case. We follow the convention that is , when .

###### Theorem 1.

Let be a positive integer. Let , where for every , is an irreducible fraction. Let be a fractional -intersecting family of subsets of . Then, , where and . Further,

1. if , then , and

2. if , where is a positive integer constant, then .

Consider the following examples for a fractional -intersecting family.

###### Example 1.

Let , where we omit fractions, like , which are not irreducible. The collection of all the non-empty subsets of is a fractional -intersecting family of cardinality . Here, . Since , we can apply Statement (b) of Theorem 1 to get an upper bound of which is asymptotically tight. In general, when , where is a constant, the set of all the non-empty subsets of of cardinality at most is an example which demonstrates that the bound given in Statement (b) of Theorem 1 is asymptotically tight.

###### Example 2.

Let us now consider another example where is a constant. Let . The collection of all the -sized subsets of is a fractional -intersecting family of cardinality . In this case, the bound given by Theorem 1 is asymptotically tight up to a factor of . We believe that if is a fractional -intersecting family of maximum cardinality, where () is a constant, then .

Coming back to the classical -intersecting families, it is known that when is an -intersecting family where , the Fisher’s Inequality (see Theorem 7.5 in [9]) yields . Study of such intersecting families was initiated by Ronald Fisher in 1940 (see [5]). This fundamental result of design theory is among the first results in the field of -intersecting families. Analogously, consider the scenario when is a singleton set. Can we get a tighter (compared to Theorem 1) bound in this case? We show in Theorem 2 that if is a constant prime we do have a tighter bound.

###### Theorem 2.

Let be a positive integer. Let be a fractional -intersecting families of subsets of , where , , and is a prime. Then, .

Assuming , Examples 3 and 4 in Section 3 give fractional -intersecting families on of cardinality thereby implying that the bound obtained in Theorem 2 is asymptotically tight up to a factor of when is a constant prime. We believe that the cardinality of such families is at most , where is a constant.

The rest of the paper is organized in the following way: In Section 2.1, we give the proof of Theorem 1 after introducing some necessary lemmas in the beginning. In Theorem 6 in Section 2.2, we give an upper bound of for fractional -intersecting families on whose member sets are ‘large enough’. In Section 3, we consider the case when is a singleton set and give the proof of Theorem 2. Later in this section, in Theorem 8, we consider the case when the cardinalities of the sets in the fractional -intersecting family are restricted. Finally, we conclude with some remarks, some open questions, and a conjecture.

## 2 The general case

### 2.1 Proof of Theorem 1

Before we move to the proof of Theorem 1, we introduce a few lemmas that will be used in the proof.

#### 2.1.1 Few auxiliary lemmas

The following lemma is popularly known as the ‘Independence Criterion’ or ‘Triangular Criterion’.

###### Lemma 3 (Lemma 13.11 in [9], Proposition 2.5 in [3]).

For let be functions and elements such that

1. for all ;

2. for all .

Then are linearly independent members of the space .

###### Lemma 4.

Let be a prime; . Let and let . For any , let denote its -

incidence vector and let

. Assume , for every . Then, the set of functions is linearly independent in the vector space over .

###### Proof.

Arrange every subset of of cardinality less than in a linear order, say , such that implies . For any two distinct sets and , we know that when , where denote the evaluation of the function at . Suppose has a non-trivial solution. Then, identify the first set, say , in the linear order for which is non-zero. Evaluate the functions on either side of the above equation at to get which is a contradiction to our assumption. ∎

The following lemma is from [3] (see Lemma 5.38).

###### Lemma 5 (Lemma 5.38 in [3]).

Let be a prime; . Let be defined as . For any , let denote its - incidence vector and let . Assume and . Then, the set of functions is linearly independent in the vector space over .

#### 2.1.2 The proof

###### Proof.

Let be a prime and let . We partition into parts, namely , where .

#### Estimating |Fi|, when i>0.

Let and let denote their corresponding - incidence vectors. Define functions to , where each , in the following way.

 fj(x)=(⟨Vj,x⟩−a1b1i)(⟨Vj,x⟩−a2b2i)⋯(⟨Vj,x⟩−asbsi).

Note that since , . Since , for every , unless . So,

 fj(x){≠0, if x=Vj=0, otherwise. (1)

So, ’s are linearly independent in the vector space over (by Lemma 3). Since , , for any positive integer . Each is thus an appropriate linear combination of distinct monomials of degree at most . Therefore, . We can improve this bound by using the “swallowing trick” in a way similar to the way it is used in the proof of Theorem 1.1 in [2]. Let be defined as . From Lemma 4, we know that the set of functions is linearly independent in the vector space over .

###### Claim 5.1.

is a collection of functions that is linearly independent in the vector space over .

In order to prove the claim, assume for some . Evaluating at , all terms in the second sum vanish (since ) and by Equation 1, only the term with subscript remains of the first sum. We infer that , for every . It then follows from Lemma 4 that every is zero thus proving the claim.

Since each function in the collection of functions in Claim 5.1 can be obtained as a linear combination of distinct monomials of degree at most , we can infer that . We thus have

 |Fi|≤{(ns)+(ni), if i

Observe that . We will shortly see that the prime we choose is always at most , where . So if , the condition (here stands for the symbol in Lemma 5) given in Lemma 5 is satisfied and therefore the more powerful Lemma 5 can be used instead of Lemma 4 while applying the swallowing trick. We can then claim that (proof of this claim is similar to the proof of Claim 5.1 and is therefore omitted) (, where ) is a collection of functions that is linearly independent in the vector space over which can be obtained as a linear combination of distinct monomials of degree at most . It then follows that .

In the rest of the proof, we shall assume the general bound for given by Inequality 2. (Using the upper bound for in place of Inequality 2 when in the rest of the proof will yield the tighter bound for given in Statement (a) in the theorem.)

Observe that we still do not have an estimate of

since when . To overcome this problem, consider the collection of smallest primes with ( denotes the -th prime; and so on) such that for every , there exists a prime with . Note that if we repeat the steps done above for each , we obtain the following upper bound.

 |F| ≤ (pq+1+⋯+pr−(r−q))(ns)+(r−q)s−1∑j=1(nj) < (r−q)(pr(ns)+s−1∑j=1(nj))

To obtain a small cardinality set of the desired requirement, we choose the minimum such that . If , for some positive integer constant , then satisfies the desired requirements of . We thus have,

 |F| <⎧⎪⎨⎪⎩c1(pr(ns)+∑s−1j=1(nj)), {\color[rgb]{0,0,0} if t>n−c1 (here c1 is a% positive integer constant)}r(pr(ns)+∑s−1j=1(nj)), otherwise% (3)

The product of the first primes is the primorial function and it is known that . Given a natural number , let denote the product of all the primes less than or equal to (some call this the primorial function). It is known that . Since , setting , we get, . Using the prime number theorem, the th prime is at most . Thus, we have . Substituting for and in Inequality 3 gives the theorem. ∎

### 2.2 When the sets in F are ‘large enough’

In the following theorem, we show that when the sets in a fractional -intersecting are ‘large enough’, then is at most .

###### Theorem 6.

Let be a positive integer. Let , where for every , is an irreducible fraction. Let . Let be a fractional -intersecting family of subsets of such that for every , , where . Then, .

###### Proof.

Let . For every , we define its -incidence vector as:

 XAi(j)={+1, if j∈Ai−1, if j∉Ai. (4)

We prove the theorem by proving the following claim.

###### Claim 6.1.

are linearly independent in the vector space over .

Assume for contradiction that are linearly dependent in the vector space over . Then, we have some reals where not all of them are zeroes such that

 λA1XA1+⋯+λAmXAm=0. (5)

It is given that, for every , . Let be the all ones vector. Then, , for every . Therefore, if all non-zero s in Equation (5) are of the same sign, say positive, then the inner product of with the L.H.S of Equation (5) would be non-zero which is a contradiction. Hence, we can assume that not all s are of the same sign. We rewrite Equation (5) by moving all negative s to the R.H.S. Without loss of generality, assume are non-negative and the rest are negative. Thus, we have

 v=λA1XA1+⋯+λAkXAk=−(λAk+1XAk+1+⋯+λAmXAm),

where is a non-zero vector.

For any two distinct sets , such that

 ⟨XA,XB⟩=⎧⎪⎨⎪⎩n−2|A|+4ai−2bibi|B|, if |A∩B|=aibi|B|, n−2|B|+4ai−2bibi|A|, otherwise (that is, if |A∩B|=aibi|A|). (6)

Since , we have or . Applying the fact that the cardinality of every set in satisfies , where , we get . This implies that which is a contradiction. This proves the claim and thereby the theorem. ∎

## 3 L is a singleton set

As explained in Section 1, the Fisher’s Inequality is a special case of the classical -intersecting families, where . In this section, we study fractional -intersecting families with ; a fractional variant of the Fisher’s inequality.

### 3.1 Proof of Theorem 2

Statement of Theorem 2: Let be a positive integer. Let be a fractional -intersecting families of subsets of , where , , and is a prime. Then, .

###### Proof.

It is easy to see that if , then with the set of all singleton subsets of forming a tight example to this bound. So assume . Let , where . From the definition of a fractional -intersecting family it is clear that . The rest of the proof is to show that . We do this by partitioning into parts and then showing that each part is of size at most . We define as

 Fji={A∈F||A|≡j (mod i)}.

Since divides , for every , under this definition can be partitioned into families , where and . We show that, for every and for every , .

In order to estimate , for each , create a vector as follows:

 XA(j)={1√bk−2, if j∈A;0, otherwise.

Note that, for

 ⟨XA,XB⟩≡{b (mod b2), if A=B, ai (mod b), if A≠B, (7)

Let . Let denote the matrix formed by taking s as rows for each . Then, can be proved by considering and showing that (, where is the all 1 matrix, ) has full rank; determinant of is non-zero since the only term not divisible by the prime in the expansion of its determinant comes from the product of all the diagonals (note that , , and since is a prime, we have ). ∎

We shall call a bisection closed family if is a fractional -intersecting family where . We have two different constructions of families that are bisection closed and are of cardinality on .

###### Example 3.

Let be an even positive integer. Let denote the collection of 2-sized sets that contain only 1 as a common element in any two sets, i.e. ; and let denote collection of 4-sized sets that contain only as common elements, i.e. . It is not hard to see that is indeed bisection closed.

###### Example 4.

The second example of a bisection closed family of cardinality comes from Recursive Hadamard matrices. A Recursive Hadamard matrix of size can be obtained from of size as follows

 H(k)=[H(k−1)H(k−1)H(k−1)−H(k−1)],

where . Now consider the matrix:

 M(k)=⎡⎢⎣H(k−1)H(k−1)H(k−1)−H(k−1)H(k−1)J(k−1)⎤⎥⎦, where J(k−1) denotes the 2k−1×2k−1 % all 1s' matrix.

Let be the matrix obtained from by removing the first and the th rows and replacing the -1’s by 1’s and 1’s by 0’s. is clearly bisection closed and has cardinality , where .

### 3.2 Restricting the cardinalities of the sets in F

When , where is a prime, Theorem 2 yields an upper bound of for . However, we believe that when , the cardinality of any fractional -intersecting family on would be at most , where is a constant. To this end, we show in Theorem 8 that when the sizes of the sets in are restricted, we can achieve this.

The following lemma is crucial to the proof of Theorem 8.

###### Lemma 7.

[1, 4] Let be an real symmetric matrix with and for all . Let denote the trace of , i.e., the sum of the diagonal entries of A. Let denote the rank of . Then,

 rk(A)≥(tr(A))2tr(A2)≥m1+(m−1)ϵ2.
###### Proof.

Let

denote the eigenvalues of

. Since only eigenvalues of are non-zero, , where the inequality follows from the Cauchy-Schwartz Inequality. Thus, . Substituting and in the above inequality proves the theorem. ∎

###### Theorem 8.

Let be a positive integer and let . Let be a fractional -intersecting family of subsets of , where , is an irreducible fraction and for every , in an integer in the range . Then, .

###### Proof.

For any , let be a vector defined as:

 YA(j)=⎧⎨⎩+1√n, if j∈A−1√n, if j∉A.

Clearly, . For any two distinct sets , we have

 ⟨YA,YB⟩=⎧⎪ ⎪⎨⎪ ⎪⎩n−2|A|+4a−2bb|B|n, if |A∩B|=ab|B|, n−2|B|+4a−2bb|A|n, otherwise (that is, if |A∩B|=ab|A|). (8)

Suppose . Let be the matrix with as its rows. Then, from Equation 8, it follows that is an real symmetric matrix with the diagonal entries being and the absolute value of any other entry being at most . Applying Lemma 7, we have