Fractional cross intersecting families

TL;DR

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

Derived tight upper bounds for the product of family sizes.

Characterized extremal families when the bounds are achieved.

Extended classical intersection results to fractional intersection scenarios.

Abstract

Let $\mathcal{A}=\{A_{1},...,A_{p}\}$ and $\mathcal{B}=\{B_{1},...,B_{q}\}$ be two families of subsets of $[n]$ such that for every $i\in [p]$ and $j\in [q]$, $|A_{i}\cap B_{j}|= \frac{c}{d}|B_{j}|$, where $\frac{c}{d}\in [0,1]$ is an irreducible fraction. We call such families "$\frac{c}{d}$-cross intersecting families". In this paper, we find a tight upper bound for the product $|\mathcal{A}||\mathcal{B}|$ and characterize the cases when this bound is achieved for $\frac{c}{d}=\frac{1}{2}$. Also, we find a tight upper bound on $|\mathcal{A}||\mathcal{B}|$ when $\mathcal{B}$ is $k$-uniform and characterize, for all $\frac{c}{d}$, the cases when this bound is achieved.

Equations

$${\mathcal{B}}_1=\{B\in \mathcal{B}:|B|\equiv 0\text{leavevmode}(\mathrm{mod}\text{leavevmode}2d)\}$$

$${\mathcal{B}}_2=\{B\in \mathcal{B}:|B|\equiv d\text{leavevmode}(\mathrm{mod}\text{leavevmode}2d)\}$$

$$⟨a,b⟩=\{\begin{array}{l}1,ifb\in {\mathcal{B}}_{2}andcisodd\\ 0,otherwise\end{array}$$

$$\begin{array}{rl}|\text{span}({\mathcal{A}}^{{}^{\prime }})|\cdot |\text{span}({\mathcal{B}}^{{}^{\prime }})|& =2^{\text{dim}(span({\mathcal{A}}^{{}^{\prime }}))}\cdot 2^{\text{dim}((span({\mathcal{B}}^{{}^{\prime }}))}\\ & =2^{dim(span({\mathcal{A}}^{{}^{\prime }}))+dim(span({\mathcal{B}}^{{}^{\prime }}))}\\ & \le 2^{n+1}\end{array}$$

$$2^{n+1}$$

$$\ge 2\cdot |{\mathcal{A}}^{{}^{\prime }}|\cdot |\text{span}({\mathcal{B}}^{{}^{\prime }})|$$

$$\ge 2\cdot |{\mathcal{A}}^{{}^{\prime }}|\cdot |{\mathcal{B}}^{{}^{\prime }}|$$

$$=2\cdot |\mathcal{A}|\cdot |\mathcal{B}|$$

$$|\text{span}({\mathcal{A}}^{{}^{\prime }})|$$

$$|\text{span}({\mathcal{B}}^{{}^{\prime }})|$$

$$\text{dim}(span({\mathcal{A}}^{{}^{\prime }}))+\text{dim}(span({\mathcal{B}}^{{}^{\prime }}))$$

$$|B_3|=|B_1\Delta B_2|=|B_1|+|B_2|-2|B_1\cap B_2|$$

$$|B_1|+|B_2|-2|B_1\cap B_2|\equiv 0\text{leavevmode}(\mathrm{mod}\text{leavevmode}4)$$

$$|B_1|+|B_2|-2|B_1\cap B_2|\equiv |B_3|\equiv 2\text{leavevmode}(\mathrm{mod}\text{leavevmode}4)$$

$$\text{dim}(C)+\text{dim}(C^{\perp })=n$$

$$\text{dim}(C)\le \text{dim}(C^{\perp })$$

$$n=\text{dim}(C)+\text{dim}(C^{\perp })\ge 2\text{dim}(C)$$

$$\text{Therefore, }\text{dim}(C)\le \frac{n}{2}$$

$$|\mathcal{B}|\le 2^{\lfloor \frac{n}{2}\rfloor }$$

$$|A\cap (B_1△B_2)|$$

$$\Rightarrow |A\cap ((B_1\backslash B_2)\cup (B_2\backslash B_1))|$$

$$\Rightarrow |A\cap (B_1\backslash B_2)|+|A\cap (B_2\backslash B_1)|$$

$$\Rightarrow |A\cap B_1|-|A\cap (B_1\cap B_2)|+|A\cap (B_2)|-|A\cap (B_1\cap B_2)|=\frac{|B_1|}{2}+\frac{|B_2|}{2}-|B_1\cap B_2|$$

$$\Rightarrow \frac{|B_1|}{2}+\frac{|B_2|}{2}-2|A\cap (B_1\cap B_2)|$$

$$\Rightarrow 2|A\cap (B_1\cap B_2)|$$

$$\Rightarrow |A\cap (B_1\cap B_2)|$$

$$B_1=\bigcap _{\begin{array}{c}1\in B,\\ B\in \mathcal{B}\end{array}}B$$

$$B_2=\bigcap _{\begin{array}{c}2i_1+1\in B,\\ B\in \mathcal{B}\end{array}}B$$

$$B_3=\bigcap _{\begin{array}{c}2i_1+2i_2+1\in B,\\ B\in \mathcal{B}\end{array}}B$$

$$|\mathcal{A}||\mathcal{B}|$$

Full text

1 Introduction

Let $[n]$ denote $\{1,...,n\}$ and let $2^{[n]}$ denote the power set of $[n]$.We shall use $\binom{[n]}{k}$ to denote the set of all $k$-sized subsets of $[n]$. Let $\mathcal{F}\subseteq 2^{[n]}$. The family $\mathcal{F}$ is an intersecting family if every two sets in $\mathcal{F}$ intersect with each other. The famous Erdős-Ko-Rado Theorem [1] states that $|\mathcal{F}|\leq\binom{n-1}{k-1}$ if $\mathcal{F}$ is a $k$-uniform intersecting family, where $2k\leq n$. Several variants of the notion of intersecting families have been extensively studied in the literature. Given a set $L=\{l_{1},\ldots,l_{s}\}$ of non-negative integers, a family $\mathcal{F}\subseteq 2^{[n]}$ is $L$-intersecting if for all $F_{i},F_{j}\in\mathcal{F},F_{i}\neq F_{j},|F_{i}\cap F_{j}|\in L$. Ray-Chaudhuri and Wilson in [2] showed that if $\mathcal{F}$ is $k$-uniform and $L$-intersecting, then $|\mathcal{F}|\leq\binom{n}{s}$ and the bound is tight. Frankl and Wilson in [3] showed a tight upper bound of $\binom{n}{s}+\binom{n}{s-1}+\cdots+\binom{n}{0}$ 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 $|\mathcal{F}|\leq n$ for the cardinality of an $L$-intersecting family $\mathcal{F}$. 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 $\mathcal{A},\mathcal{B}\subseteq 2^{[n]}$ are cross-intersecting if $|A\cap B|>0$, $\forall$ $A\in\mathcal{A}$,$B\in\mathcal{B}$. Pyber in [7] showed that if $n\geq 2k$, and $\mathcal{A},\mathcal{B}\subseteq\binom{[n]}{k}$ is a cross-intersecting pair of families, then $|\mathcal{A}||\mathcal{B}|\leq\binom{n-1}{k-1}^{2}$. Frankl et al. in [8] showed that if $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ such that $|A\cap B|\geq t$ for all $A\in\mathcal{A}$ and $B\in\mathcal{B}$, then for all $n\geq(t+1)(k-t+1)$, $|\mathcal{A}||\mathcal{B}|\leq\binom{n-t}{k-t}^{2}$, the cross-intersecting version of the Erdős-Ko-Rado Theorem. A cross-intersecting pair of families $\mathcal{A},\mathcal{B}\subseteq 2^{[n]}$ is said to be $l$-cross-intersecting if $\forall A\in\mathcal{A},\leavevmode\nobreak\ B\in\mathcal{B}$, $|A\cap B|=l$, for some positive integer $l$. Ahlswede, Cai and Zhang showed in [9], for all $n\geq 2l$, a simple construction of an $l$-cross-intersecting pair $(\mathcal{A},\mathcal{B})$ of families of subsets of $[n]$ with $|\mathcal{A}||\mathcal{B}|=\binom{2l}{l}2^{n-2l}=\Theta(\frac{2^{n}}{\sqrt{l}})$. Later Alon and Lubetzky in [10] showed that the $\Theta(\frac{2^{n}}{\sqrt{l}})$ 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 $\mathcal{A}=\{A_{1},...,A_{p}\}$ and $\mathcal{B}=\{B_{1},...,B_{q}\}$ be two families of subsets of $[n]$ such that for every $i\in[p]$ and $j\in[q]$, $|A_{i}\cap B_{j}|=\frac{c}{d}|B_{j}|$, where $\frac{c}{d}\in[0,1]$ is an irreducible fraction. We call such an $(\mathcal{A},\mathcal{B})$ pair a $\frac{c}{d}$-cross-intersecting pair of families. Given $c$, $d$, and $n$, let $\mathcal{M}_{\frac{c}{d}}(n)$ denote the maximum value of $|\mathcal{A}||\mathcal{B}|$ where $(\mathcal{A},\mathcal{B})$ is a $\frac{c}{d}$-cross intersecting pair of families of subsets of $[n]$. We have the following results:

Theorem 1.1**.**

\thlabel

thm:1.1 $\mathcal{M}_{\frac{c}{d}}(n)=2^{n}$

When $\frac{c}{d}=0$, $\mathcal{A}=2^{[n]}$, $\mathcal{B}=\{\emptyset\}$ is a maximal pair. In fact, $\mathcal{A}=2^{[k]}$, $\mathcal{B}=\mathcal{P}(S)$, where $\mathcal{P}(S)$ is the power set of $S=\{k+1,\ldots,n\}$, are the only maximal pairs up to a relabelling of the elements, $0\leq k\leq n$. When $\frac{c}{d}=1$, $\mathcal{A}=\{[n]\}$ and $\mathcal{B}=2^{[n]}$ is a maximal pair. In fact, $\mathcal{B}=2^{[k]}$, $\mathcal{A}=\{A:A=[k]\cup T$, where $T\in\mathcal{P}(S)\}$, where $\mathcal{P}(S)$ is the power set of $S=\{k+1,\ldots,n\}$, are the only maximal pairs up to a relabelling of the elements, $0\leq k\leq n$. In \threfthm:1.2, we characterize all maximal pairs when $\frac{c}{d}=\frac{1}{2}$.

Theorem 1.2**.**

\thlabel

thm:1.2 Let $(\mathcal{A},\mathcal{B})$ be a $\frac{1}{2}$-cross intersecting pair of families of subsets of $[n]$ with $|\mathcal{A}||\mathcal{B}|=2^{n}$. Then $(\mathcal{A},\mathcal{B})$ is one of the following $\lfloor\frac{n}{2}\rfloor+1$ pairs of families $(\mathcal{A}_{k},\mathcal{B}_{k})$, $0\leq k\leq\lfloor\frac{n}{2}\rfloor$, up to isomorphism.

$\mathcal{A}_{0}=2^{[n]}$* and $\mathcal{B}_{0}=\{\emptyset\}$*

$\mathcal{A}_{k}=\{A\in 2^{[n]}:|A\cap\{2i-1,2i\}|=1\leavevmode\nobreak\ \leavevmode\nobreak\ \forall i,1\leq i\leq k\}$**

$\mathcal{B}_{k}=\{B\in 2^{[n]}:|B\cap\{2i-1,2i\}|\in\{0,2\}\leavevmode\nobreak\ \leavevmode\nobreak\ \forall i,1\leq i\leq k$* and $\forall j>2k$, $j\notin B\}$,*

where $1\leq k\leq\lfloor\frac{n}{2}\rfloor$.

It would be interesting to show a characterization theorem for any $\frac{c}{d}\in[0,1]$. We do have such a general characterization theorem (along with a new tight upper bound) in \threfthm:1.3 for the case when $\mathcal{B}$ is $k$-uniform. The proof is a direct application of Theorem 1.1 in [10].

Theorem 1.3**.**

\thlabel

thm:1.3 Let $(\mathcal{A}$,$\mathcal{B})$ be a $\frac{c}{d}$-cross intersecting pair of families of subsets of $[n]$. Let $\mathcal{B}$ be $k$-uniform. Then, there exists some $k_{0}>0$, such that for $k>k_{0}$ we have

$|\mathcal{A}||\mathcal{B}|\leq\binom{\frac{2ck}{d}}{\frac{ck}{d}}2^{n-\frac{2ck}{d}}$**

and the bound is tight if and only if, either $(a)$ or $(b)$ hold:

1. (a)

When $\frac{c}{d}=1$, $\mathcal{A}=\{\{1,\ldots,\kappa\}\}\times 2^{Y}$, $\mathcal{B}=\binom{[\kappa]}{k}$ where $Y=\{\kappa+1,\ldots,n\}$ and $\kappa\in\{2k-1,2k\}$ up to a relabelling of the elements of $[n]$. 2. (b)

When $\frac{c}{d}\neq 1$:

1. (i)

If $k$ is even, $c=1$, $d=2$, $\frac{ck}{d}=\lceil\frac{k}{2}\rceil$, 2. (ii)

If $k$ is odd, $c=\frac{k+1}{2}$, $d=k$, $\frac{ck}{d}=\lceil\frac{k}{2}\rceil$,

and for both the cases($(i)$ and $(ii)$), there exists some $\tau$ such that, $k+\tau\leq n$ and up to a relabelling of the elements of $[n]$,

$\mathcal{A}=\{\cup_{T\in J}\,{T}:J\subset\{\{1,k+1\},\ldots,\{\tau,k+\tau\},\{\tau+1\},\ldots,\{k\}\},|J|=\lceil\frac{k}{2}\rceil\}\times 2^{X}$**

where $X=\{k+\tau+1,\ldots,n\}$ and

$\mathcal{B}=\{L\cup\{\tau+1,\ldots,k\}:L\subset\{1,\ldots,\tau,k+1,\ldots,k+\tau\},|L\cap\{i,k+i\}|=1$* for all $i\in[\tau]\}$.*

2 Notations and definitions

Given any $S\subseteq[n]$, we shall use $\chi(S)$ to denote the characteristic vector of $S$ which is a $0-1$ vector of size $n$ having its $i^{th}$ entry equal to $1$ if and only if $i\in S$. The weight of a vector is the number of non-zero entries it has, and hence weight of $\chi(S)$ is the same as $|S|$.

For any family $\mathcal{A}\subseteq 2^{[n]}$, we shall (ab)use $\mathcal{A}$ to denote the collection of characteristic vectors of the members of $\mathcal{A}$ as well. The meaning will be clearly stated if not clear from the context.

Let $V$ be a collection of vectors in $\mathbb{F}^{n}_{2}$. Then, we define the following:

$span(V)$: The collection of all the vectors that can be expressed as a linear combination in $\mathbb{F}_{2}$ of the vectors of $V$. We know that $span(V)$ is a vector space over $\mathbb{F}_{2}$. 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**.**

$V\subseteq\mathbb{F}^{n}_{2}$* is a linear code if $V=span(V)$.*

Definition 2**.**

Given a linear code $C\subseteq\mathbb{F}^{n}_{2}$, the dual code $C^{\perp}$ is defined as,

$C^{\perp}=\{x\in\mathbb{F}^{n}_{2}|\langle x,c\rangle=0,\forall c\in C\}$**

where $\langle x,y\rangle$ is the standard inner product over $\mathbb{F}_{2}$.

The following is a well-known fact that is easy to verify.

Lemma 2.1**.**

If $C\subseteq\mathbb{F}^{n}_{2}$ is a linear code, then $C^{\perp}$ is also a linear code.

Definition 3**.**

Self orthogonal and self dual codes: A code $C$ is self orthogonal if $C\subseteq C^{\perp}$ and it is self dual if $C=C^{\perp}$.

3 Bounding $\mathcal{M}_{\frac{c}{d}}(n)$

Let $(\mathcal{A},\mathcal{B})$ be a $\frac{c}{d}$-cross-intersecting pair of families of subsets of $[n]$, where $\frac{c}{d}\in[0,1]$ is an irreducible fraction. We shall (ab)use $\mathcal{A},\mathcal{B}$ to denote the set of characteristic vectors of the sets in $\mathcal{A},\mathcal{B}$ respectively. For any $a\in\mathcal{A},b\in\mathcal{B}$, we observe that $\langle a,b\rangle\equiv|A\cap B|\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 2)$, where $a=\chi(A)$, $b=\chi(B)$.

Partition the family $\mathcal{B}$ into two parts as,

[TABLE]

As all the sets $B\in\mathcal{B}$ have their cardinality $|B|$ divisible by $d$, $\{\mathcal{B}_{1}$,$\mathcal{B}_{2}\}$ is a valid partition of $\mathcal{B}$. Therefore $\forall a\in\mathcal{A}$ , $b\in\mathcal{B}$, using the $\frac{c}{d}$ intersection property, we have:

[TABLE]

Construction 1**.**

Construct a set $\mathcal{B}^{{}^{\prime}}_{1}$, by appending a [math] to the left of every vector in $\mathcal{B}_{1}$, and a set $\mathcal{B}^{{}^{\prime}}_{2}$ by appending a 1 to the left of every vector in $\mathcal{B}_{2}$. Let $\mathcal{B}^{{}^{\prime}}=\mathcal{B}^{{}^{\prime}}_{1}\cup\mathcal{B}^{{}^{\prime}}_{2}$. Construct a set $\mathcal{A}^{{}^{\prime}}$ by appending a $1$ to the left of every vector in $\mathcal{A}$.

We now have, the value of

$\langle a,b\rangle=0\leavevmode\nobreak\ \leavevmode\nobreak\ \forall a\in\mathcal{A}^{{}^{\prime}}$, $b\in\mathcal{B}^{{}^{\prime}}$

So, $(span(\mathcal{A}^{{}^{\prime}}),span(\mathcal{B}^{{}^{\prime}}))$ is a pair of mutually orthogonal subspaces of $\mathbb{F}^{n+1}_{2}$ over $\mathbb{F}_{2}$. We thus have,

$dim(span(\mathcal{A}^{{}^{\prime}}))+dim(span(\mathcal{B}^{{}^{\prime}}))\leq n+1$

So, it follows that

[TABLE]

Lemma 3.1**.**

\thlabel

lem:3.1 If the elements of a linear code $C\subseteq\mathbb{F}^{n}_{2}$ are arranged as rows of a matrix $M_{C}$ with $n$ columns, then for each column, one of the following holds,

1. (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 $a\in C$, and consider the set $S=\{a+x|x\in C\}$ where $a+x$ is the vector addition in $\mathbb{F}^{n}_{2}$, then by the definition of a linear code $S=C$. Let $M_{S}$ be a matrix whose rows are the vectors of $S$, taken in any order. $M_{S}$ and $M_{C}$ have the same set of rows (only their order may differ).

Let $j\in[n]$. Column $j$ in $M_{C}$ and $M_{S}$ have the same number of $1$’s( and [math]’s). Suppose (i) does not hold for column $j$ in $M_{C}$. Then, some row, say $a$, in $M_{C}$ has its $j^{th}$ entry as 1. Let $S$, and thereby $M_{S}$, be defined according to this vector $a$. From the definition of $S$, it is clear that the number of $1$’s in the $j^{th}$ column of $M_{S}$ is equal to the number of $1$’s in the $j^{th}$ column of $M_{C}$. Since adding $a$ to any $\{0,1\}$ vector flips the $j^{th}$ coordinate of $v$, we conclude that $(ii)$ holds for $M_{c}$. ∎

Corollary 3.2**.**

\thlabel

cor:3.2 $|span(\mathcal{A}^{{}^{\prime}})|\geq 2|\mathcal{A}^{{}^{\prime}}|$

Proof.

The leftmost column of $\mathcal{M}_{\mathcal{A}^{{}^{\prime}}}$ does not contain any [math]. As $span(\mathcal{A}^{{}^{\prime}})$ is a linear code and $\mathcal{A}^{{}^{\prime}}\subseteq span(\mathcal{A}^{{}^{\prime}})$, by condition (ii) of \threflem:3.1 above, $span(\mathcal{A}^{{}^{\prime}})$ must have at least $|\mathcal{A}^{{}^{\prime}}|$ 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: $\mathcal{M}_{\frac{c}{d}}(n)=2^{n}$

Proof.

$\mathcal{A}=2^{[n]}$, $\mathcal{B}=\{\emptyset\}$ is a trivial example of a $\frac{c}{d}$ cross-intersecting pair of families having $|\mathcal{A}||\mathcal{B}|=2^{n}$. Thus, $\mathcal{M}_{\frac{c}{d}}(n)\geq 2^{n}$. The proof of the upper bound for $\mathcal{M}_{\frac{c}{d}}(n)$ follows from Inequality (3) and \threfcor:3.2. Let $(\mathcal{A},\mathcal{B})$ be a $\frac{c}{d}$ cross-intersecting pair of families of subsets of $[n]$. Let $\mathcal{A}^{{}^{\prime}}$, $\mathcal{B}^{{}^{\prime}}$ be constructed from $\mathcal{A}$, $\mathcal{B}$, respectively, as explained in the beginning of this section. Note that $|\mathcal{A}^{{}^{\prime}}|=|\mathcal{A}|$ and $|\mathcal{B}^{{}^{\prime}}|=|\mathcal{B}|$ by construction.

[TABLE]

∎

4 Characterization of maximal pairs when $\frac{c}{d}=\frac{1}{2}$

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 $\frac{1}{2}$ cross-intersecting pair. $(\mathcal{A},\mathcal{B})$ is called a maximal cross bisecting or simply a maximal pair, if it is a cross bisecting pair and $|\mathcal{A}||\mathcal{B}|=2^{n}$.

For example, $\mathcal{A}=2^{[n]}$ and $\mathcal{B}=\{\emptyset\}$ is a trivial maximal pair. In this section, we characterize all maximal pairs. Let $(\mathcal{A},\mathcal{B})$ be a cross bisecting pair and let $(\mathcal{A}^{{}^{\prime}},\mathcal{B}^{{}^{\prime}})$ be the associated pair constructed by appending bits as defined in the previous section.

Definition 5**.**

Let $f_{\mathcal{A}}:\mathcal{A}\rightarrow\mathcal{A}^{{}^{\prime}}$ be a bijective mapping that maps every vector in $\mathcal{A}$ to its corresponding vector in $\mathcal{A}^{{}^{\prime}}$, and let $g_{\mathcal{A}}:\mathcal{A}^{{}^{\prime}}\rightarrow\mathcal{A}$ be its inverse. Likewise, define functions $f_{\mathcal{B}}$ and $g_{\mathcal{B}}$ between $\mathcal{B}$ and $\mathcal{B}^{{}^{\prime}}$. For any set $V\subseteq\mathcal{A}$, we shall use, $f_{\mathcal{A}}(V)$ to denote $\{f_{\mathcal{A}}(A)|\ A\in V\}$ and for any $V\subseteq\mathcal{A}^{{}^{\prime}}$, we use $g_{\mathcal{A}}(V)$ to denote $\{g_{\mathcal{A}}(A)|\ A\in V\}$. Similarly, for any $V\subseteq\mathcal{B}$, we use, $f_{\mathcal{B}}(V)$ to denote $\{f_{\mathcal{B}}(B)|\ B\in V\}$ and for any $V\subseteq\mathcal{B}^{{}^{\prime}}$, $g_{\mathcal{B}}(V)$ to denote $\{g_{\mathcal{B}}(B)|\ B\in V\}$

Observation 1**.**

\thlabel

*obv $f_{\mathcal{B}}(\mathcal{B}_{1})=\mathcal{B}_{1}^{{}^{\prime}}$ and $f_{\mathcal{B}}(\mathcal{B}_{2})=\mathcal{B}_{2}^{{}^{\prime}}$. Similarly, $g_{\mathcal{B}}(\mathcal{B}_{1}^{{}^{\prime}})=\mathcal{B}_{1}$ and $g_{\mathcal{B}}(\mathcal{B}_{2}^{{}^{\prime}})=\mathcal{B}_{2}$

Suppose $(\mathcal{A},\mathcal{B})$ is a maximal pair. Then from the proof of \threfthm:1.1, we must have :

[TABLE]

Proposition 4.1**.**

\thlabel

lem:4.3 $\mathcal{B}=\text{span}(\mathcal{B})$. Further, $f_{\mathcal{B}}$ is a linear map.

Proof.

This follows from equation (5). Let $x_{1},x_{2}\in\mathcal{B}$. We show that $x_{3}=x_{1}+x_{2}\in\mathcal{B}$. This would imply $\mathcal{B}$ is closed under addition in $\mathbb{F}^{n}_{2}$ over $\mathbb{F}_{2}$, and hence $\mathcal{B}=\text{span}(\mathcal{B})$.

Let $x^{{}^{\prime}}_{1}=f_{\mathcal{B}}(x_{1})$ and $x^{{}^{\prime}}_{2}=f_{\mathcal{B}}(x_{2})$. From Equation (5), we have, $w=x^{{}^{\prime}}_{1}+x^{{}^{\prime}}_{2}\in\mathcal{B}^{{}^{\prime}}$. Since $w$ and $x_{3}$ agree on each of the rightmost $n$ bits of $x_{3}$, we have $g_{\mathcal{B}}(w)=x_{3}$. Since $w\in\mathcal{B}^{{}^{\prime}}$, from the definition of the function $g_{\mathcal{B}}$ we have $x_{3}=g_{\mathcal{B}}(w)\in\mathcal{B}$. Further, observe that $f_{\mathcal{B}}(x_{1})+f_{\mathcal{B}}(x_{2})=$ $w=f_{\mathcal{B}}(x_{3})=f_{\mathcal{B}}(x_{1}+x_{2})$ and hence $f_{\mathcal{B}}$ is a linear map. ∎

That $\mathcal{B}$ is a linear code from \threflem:4.3 implies closure of the family of subsets $\mathcal{B}$ under symmetric difference. In fact, we have the following stronger result.

Proposition 4.2**.**

\thlabel

lem:4.4 Let vectors $b_{1},b_{2}\in\mathcal{B}$. Then, $b_{1}+b_{2}\in\mathcal{B}_{1}$ if and only if either $b_{1}$,$b_{2}$ $\in\mathcal{B}_{1}$, or $b_{1}$,$b_{2}$ $\in\mathcal{B}_{2}$. Otherwise, $b_{1}+b_{2}\in\mathcal{B}_{2}$.

Proof.

We prove the 2-way implication, and rest of the proposition follows from \threflem:4.3. Let $b^{{}^{\prime}}_{1}=f_{\mathcal{B}}(b_{1}),b^{{}^{\prime}}_{2}=f_{\mathcal{B}}(b_{2})$.

• •

$b_{1}+b_{2}\in\mathcal{B}_{1}\Rightarrow b_{1}\text{ and }b_{2}$ are both from $\mathcal{B}_{1}$, or both from $\mathcal{B}_{2}$

Since $f_{\mathcal{B}}$ is a linear map, we have $(b_{1}+b_{2}\in\mathcal{B}_{1})\Rightarrow(f_{\mathcal{B}}(b_{1}+b_{2})=f_{\mathcal{B}}(b_{1})+f_{\mathcal{B}}(b_{2})=b^{{}^{\prime}}_{1}+b^{{}^{\prime}}_{2}\in\mathcal{B}^{{}^{\prime}}_{1})$. So, the leftmost bit of $b^{{}^{\prime}}_{1}+b^{{}^{\prime}}_{2}$ is [math]. This means that the leftmost bit must be the same in $b^{{}^{\prime}}_{1}$ and $b^{{}^{\prime}}_{2}$, which directly implies that either $b_{1}^{{}^{\prime}}$,$b_{2}^{{}^{\prime}}$ $\in\mathcal{B}_{1}^{{}^{\prime}}$, or $b_{1}^{{}^{\prime}}$,$b_{2}^{{}^{\prime}}$ $\in\mathcal{B}_{2}^{{}^{\prime}}$.

• •

Either $b_{1}$,$b_{2}$ $\in\mathcal{B}_{1}$, or $b_{1}$,$b_{2}$ $\in\mathcal{B}_{2}$ $\Rightarrow b_{1}+b_{2}\in\mathcal{B}_{1}$

Since $b_{1}^{{}^{\prime}}$ and $b_{2}^{{}^{\prime}}$ agree upon the leftmost bit, $b^{{}^{\prime}}_{1}+b^{{}^{\prime}}_{2}$ has a [math] in its leftmost bit. So, $b^{{}^{\prime}}_{1}+b^{{}^{\prime}}_{2}\in\mathcal{B}_{1}^{{}^{\prime}}$. From the \threfobv above, we have $b_{1}+b_{2}\in\mathcal{B}_{1}$.

∎

Proposition 4.3**.**

\thlabel

lem:4.5 $\mathcal{B}$ is a self-orthogonal code.

Proof.

We prove the proposition by showing that $\forall b_{1},b_{2}\in\mathcal{B}$, $\langle b_{1},b_{2}\rangle=0$. Let $B_{1},B_{2}$ be the sets corresponding to the vectors $b_{1},b_{2}$, respectively. Since we are operating in the field $\mathbb{F}_{2}$, it is enough to show that $|B_{1}\cap B_{2}|$ is even.

Let $b_{3}=b_{1}+b_{2}$. We observe that $b_{3}$ is the characteristic vector of $B_{3}=B_{1}\Delta B_{2}$, the symmetric difference of $B_{1}$ and $B_{2}$. We have,

[TABLE]

As $\frac{c}{d}=\frac{1}{2}$, $\forall B\in\mathcal{B}_{1}$, we have $|B|\equiv 0\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 4)$. By \threflem:4.3, $B_{1}\Delta B_{2}=B_{3}\in\mathcal{B}$ as $\mathcal{B}$ is a linear code. Taking equation (7) modulo 4, if $B_{3}\in\mathcal{B}_{1}$, then

[TABLE]

By \threflem:4.4, both $B_{1}$ and $B_{2}$ are either from $\mathcal{B}_{1}$ or from $\mathcal{B}_{2}$. In both cases, $|B_{1}|+|B_{2}|\equiv 0\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 4)$ Therefore, $2|B_{1}\cap B_{2}|\equiv 0\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 4)$ or $|B_{1}\cap B_{2}|\equiv 0\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 2)$. If $B_{3}\in\mathcal{B}_{2}$, then

[TABLE]

Again by \threflem:4.4, $|B_{1}|+|B_{2}|\equiv 2\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 4)$.

So, we have $2|B_{1}\cap B_{2}|\equiv 0\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 4)$ or $|B_{1}\cap B_{2}|\equiv 0\leavevmode\nobreak\ (\mathrm{mod}\leavevmode\nobreak\ 2)$. Thus in both cases, $|B_{1}\cap B_{2}|$ is even, so $\mathcal{B}$ is a self-othogonal code. ∎

Lemma 4.4**.**

\thlabel

lem:4.6 Let $(\mathcal{A},\mathcal{B})$ be a maximal pair, then $|\mathcal{B}|\leq 2^{\lfloor\frac{n}{2}\rfloor}$

Proof.

It is a known result (see [11]) that for a linear code $C\subseteq\mathbb{F}^{n}_{2}$ and its dual code $C^{\perp}$,

[TABLE]

For any self-orthogonal code $C$, $C\subseteq C^{\perp}$. So,

[TABLE]

Applying equation (8) in this inequality, we get

[TABLE]

Since $\mathcal{B}$ is a self-orthogonal code (\threflem:4.5), we get dim($\mathcal{B}$) $\leq\frac{n}{2}$. Hence,

[TABLE]

∎

Proposition 4.5**.**

\thlabel

lem:4.7 If a set $A$ bisects $B_{1}$, $B_{2}$ and $B_{1}\Delta B_{2}$, then $A$ also bisects $B_{1}\cap B_{2}$.

Proof.

[TABLE]

[TABLE]

[TABLE]

∎

Proposition 4.6**.**

\thlabel

cor:4.8 $\mathcal{B}$ is closed under intersection.

Proof.

Let $B_{1}$,$B_{2}$ $\in\mathcal{B}$. We show that $B_{1}\cap B_{2}\in\mathcal{B}$. By \threflem:4.3, $b_{1}+b_{2}\in\mathcal{B}$ i.e., $B_{1}\Delta B_{2}\in\mathcal{B}$. Let $A$ be any arbitrary member of $\mathcal{A}$. Now, $A$ bisects $B_{1},B_{2}$ and $B_{1}\Delta B_{2}$ as $(\mathcal{A},\mathcal{B})$ is a cross bisecting pair. By \threflem:4.7, $A$ bisects $B_{1}\cap B_{2}$. Since $(\mathcal{A},\mathcal{B})$ is a maximal pair, we conclude that $B_{1}\cap B_{2}\in\mathcal{B}$. ∎

Now, we prove the main result of this section,\threfthm:1.2, the characterization of maximal pairs.

**Statement of \threfthm:1.2: ** Let $(\mathcal{A},\mathcal{B})$ be a $\frac{1}{2}$-cross intersecting pair of families of subsets of $[n]$ with $|\mathcal{A}||\mathcal{B}|=2^{n}$. Then $(\mathcal{A},\mathcal{B})$ is one of the following $\lfloor\frac{n}{2}\rfloor+1$ pairs of families $(\mathcal{A}_{k},\mathcal{B}_{k})$, $0\leq k\leq\lfloor\frac{n}{2}\rfloor$, up to isomorphism.

$\mathcal{A}_{0}=2^{[n]}$* and $\mathcal{B}_{0}=\{\emptyset\}$*

$\mathcal{A}_{k}=\{A\in 2^{[n]}:|A\cap\{2i-1,2i\}|=1\leavevmode\nobreak\ \leavevmode\nobreak\ \forall i,1\leq i\leq k\}$**

$\mathcal{B}_{k}=\{B\in 2^{[n]}:|B\cap\{2i-1,2i\}|\in\{0,2\}\leavevmode\nobreak\ \leavevmode\nobreak\ \forall i,1\leq i\leq k$* and $\forall j>2k$, $j\notin B\}$,*

*where $1\leq k\leq\lfloor\frac{n}{2}\rfloor$. *

By isomorphism, it is meant that for any maximal pair $(\mathcal{A},\mathcal{B})$, $\exists$ a bijective mapping $f:[n]\rightarrow[n]$ such that if every $A\in\mathcal{A}$ is replaced by $A_{f}=\{f(i)|i\in A\}$ and every $B\in\mathcal{B}$ is replaced by $B_{f}=\{f(i)|i\in B\}$ then the families $(\mathcal{A}_{f},\mathcal{B}_{f})$, where $\mathcal{A}_{f}=\{A_{f}|A\in\mathcal{A}\}$ and $\mathcal{B}_{f}=\{B_{f}|B\in\mathcal{B}\}$, is a maximal pair which is one of $(\mathcal{A}_{k},\mathcal{B}_{k})$ , $0\leq k\leq\lfloor\frac{n}{2}\rfloor$.

Proof.

Consider a maximal pair $(\mathcal{A},\mathcal{B})$ where $\mathcal{B}\neq\{\emptyset\}$. We write the elements of $\mathcal{B}$ as rows of a $0-1$ matrix $M_{0}$. Suppose $n_{0}$ columns have only [math] entries in all the rows($n_{0}$ may be [math]). As the characterization is up to isomorphism, we may assume that these are the rightmost $n_{0}$ columns of the matrix $M_{0}$. In each of the remaining $n-n_{0}$ columns, from \threflem:3.1, there are exactly $\frac{|\mathcal{B}|}{2}$ $1$’s and $\frac{|\mathcal{B}|}{2}$ [math]’s as $\mathcal{B}$ is a linear code. (by \threflem:4.3)

Define

[TABLE]

We write the $\frac{|\mathcal{B}|}{2}$ rows containing 1 in the leftmost column of $M_{0}$ as the top $\frac{|\mathcal{B}|}{2}$ rows to obtain a new matrix $M_{1}$ from $M_{0}$. And $B_{1}$ is one of these rows according to \threfcor:4.8. Moreover, as all intersections are of even cardinality (\threflem:4.5), $|B_{1}|$ is even.

Let $|B_{1}|=2i_{1}$, $i_{1}\geq 1$. So, there are $2i_{1}-1$ elements in $B_{1}$ other than the element 1. Due to isomorphism, we may assume them to be $2,3,\ldots,2i_{1}$.

If $2i_{1}+1\leq n-n_{0}$, then define the set $B_{2}$ as:

[TABLE]

Claim 4.7**.**

\thlabel

lem:4.ten $1\notin B_{2}$

Proof.

Assume for the sake of contradiction, $1\in B_{2}$. This implies that for all the $\frac{|\mathcal{B}|}{2}$ sets which contain the element $2i_{1}+1$ also contain the element $1$. From \threflem:3.1, (number of sets in $\mathcal{B}$ that contain the element $1$) = (number of sets in $\mathcal{B}$ that contain the element $2i_{1}+1$) = $\frac{|\mathcal{B}|}{2}$. Hence, for any $B\in\mathcal{B}$, $1\in B$ $\Longleftrightarrow$ $2i_{1}+1\in B$. This implies that $2i_{1}+1\in B_{1}$, which is a contradiction. Hence, $1\notin B_{2}$ and therefore $B_{2}$ does not belong to the top $\frac{|\mathcal{B}|}{2}$ rows of $M_{1}$. ∎

Claim 4.8**.**

\thlabel

lem:4.eleven $B_{1}\cap B_{2}=\emptyset$

Proof.

Assume for the sake of contradiction, $x\in B_{1}\cap B_{2}$. Then $x$ is present in the $\frac{|\mathcal{B}|}{2}$ rows of the matrix $M_{1}$ whose intersection yields $B_{1}$. Since $x\in B_{2}$ and $B_{2}$ does not belong to these $\frac{|\mathcal{B}|}{2}$ rows of $M_{1}$ (by \threflem:4.ten). Thus, we have the element $x$ present in at least $\frac{|\mathcal{B}|}{2}+1$ rows of $M_{1}$, contradicting \threflem:3.1. ∎

We take the rows corresponding to the sets containing the $(2i_{1}+1)^{th}$ element that are not among the first $\frac{|\mathcal{B}|}{2}$ rows in $M_{1}$ and arrange them below the top $\frac{|\mathcal{B}|}{2}$ rows to create a matrix called $M_{2}$ from $M_{1}$. Again from \threflem:4.5, $|B_{2}|$ is even, say $2i_{2}$. Due to isomorphism and \threflem:4.eleven, we may assume that $2i_{1}+1,$…$,2i_{1}+2i_{2}$ are these $2i_{2}$ elements.

If $2i_{1}+2i_{2}+1\leq n-n_{0}$, then define,

[TABLE]

Claim 4.9**.**

\thlabel

lem:4.twelve $1\notin B_{3}$ and $2i_{1}+1\notin B_{3}$.

The proof is similar to that of \threflem:4.ten

Claim 4.10**.**

\thlabel

lem:4.thirteen $B_{1}\cap B_{3}=\emptyset$ and $B_{2}\cap B_{3}=\emptyset$.

The proof is again similar to that of \threflem:4.eleven.

We take the rows corresponding to the sets containing the $(2i_{1}+2i_{2}+1)^{th}$ element that are not among the first $r$ rows $(r>\frac{|\mathcal{B}|}{2})$ in $M_{2}$ which contain the elements $1$ or $2i_{1}+1$ and arrange them below the top $r$ rows of $M_{2}$ to create a matrix called $M_{3}$ from $M_{2}$. From \threflem:4.5 and the definition of $B_{3}$, we have $|B_{3}|=2i_{3}$, $i_{3}\geq 1$. Due to isomorphism and \threflem:4.thirteen, we may assume that $2i_{1}+2i_{2}+1,$…$,2i_{1}+2i_{2}+2i_{3}$ are these $2i_{3}$ elements.

We continue in this manner for $k$ steps by constructing sets $B_{1},\ldots,B_{k}$ and matrices $M_{1},\ldots,M_{k}$, where $k\geq 1$, until we have $2i_{1}+\cdots+2i_{k}=n-n_{0}$. Observe that $B_{1},\ldots,B_{k}$ and $P=\{n-n_{0}+1,\ldots,n\}$ is a partition of $[n]$.

Claim 4.11**.**

\thlabel

lem:4.fourteen For any set $B\in\mathcal{B}$, $j\in[k]$, we have $B\cap B_{j}\in\{\emptyset,B_{j}\}$. Further, $B\cap P=\emptyset$.

Proof.

From the definition of $P$, we have $B\cap P=\emptyset$. Let $j\in[k]$. Since $B_{j}$ is equal to the intersection of some $\frac{|\mathcal{B}|}{2}$ sets in $\mathcal{B}$, we have $B_{j}$ present as a subset of all these $\frac{|\mathcal{B}|}{2}$ sets. Applying \threflem:3.1, we can say that no element of $B_{j}$ is present in any set in $\mathcal{B}$ other than these $\frac{|\mathcal{B}|}{2}$ sets. Hence, the claim. ∎

From \threflem:4.fourteen, observe that $S=\{B_{1},\ldots,B_{k}\}$ forms a basis of the row space of the matrix $M_{k}$. 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 $A\in\mathcal{A}$ bisects every set in $\mathcal{B}$ if and only if it bisects every set in the basis $S$ of $\mathcal{B}$.

Proof.

The forward direction is straightforward as $S\subseteq\mathcal{B}$. For the opposite direction, let $A\in\mathcal{A}$ be a set that bisects every member of $S$. Since the sets corresponding to the members in $S$ are disjoint, any $B\in\mathcal{B}$ can be written as a union of some of these sets.

Let $B=B_{1}\cup\cdots\cup B_{l}$, where $\{B_{1},\ldots,B_{l}\}\subseteq S$. Then,

$|A\cap B|=|A\cap(\bigcup\limits_{j=1}^{l}B_{j})|=\sum\limits_{j=1}^{l}|A\cap B_{j}|=\sum\limits_{j=1}^{l}\frac{|B_{j}|}{2}=\frac{|\bigcup\limits_{j=1}^{l}B_{j}|}{2}=\frac{|B|}{2}$

∎

Since each set $A\in\mathcal{A}$ bisects the sets $B_{1},\ldots,B_{k}$ and $P$, from \threflem:4.fifteen, the set $A$ may contain any of the $2^{n_{0}}$ subsets of $P$, and $|A\cap B_{1}|=i_{1},\ldots,|A\cap B_{k}|=i_{k}$. Since dim$(\mathcal{B})=k$, by \threflem:4.3, we have $|\mathcal{B}|=2^{k}$.

[TABLE]

Recall that $\sum\limits_{j=1}^{k}2i_{j}=n-n_{0}$. Right hand side of Equation (9), is equal to $2^{n}$ if and only if $i_{j}=1$, $\forall j\in[k]$.

Thus, if $\mathcal{B}\neq\{\emptyset\}$, then $(\mathcal{A}_{k},\mathcal{B}_{k}),$ $k\geq 1$, 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 $M_{\frac{c}{d}}(n)$ when $\mathcal{B}$ is $k$-uniform and characterization of the cases when the bound is achieved

Let $(\mathcal{A},\mathcal{B})$ be a $\frac{c}{d}$ cross-intersecting pair of families of subsets of $[n]$, where $\frac{c}{d}\in[0,1]$ is an irreducible fraction. In this section, we deal with the scenario when $\mathcal{B}$ is $k$-uniform, where $0<k\leq n$. Since $\mathcal{B}$ is $k$-uniform, for any $A\in\mathcal{A}$ and any $B\in\mathcal{B}$, $|A\cap B|=\frac{ck}{d}=l$. Since $c$ is relatively prime with $d$, and $|A\cap B|$ 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 $l_{0}>0$ such that, for all $l\geq l_{0}$, every $l$-cross intersecting pair $\mathcal{A},\mathcal{B}\subset 2^{[n]}$ satisfies:

$|\mathcal{A}||\mathcal{B}|\leq\binom{2l}{l}2^{n-2l}$**

Furthermore, if $|\mathcal{A}||\mathcal{B}|=\binom{2l}{l}2^{n-2l}$, then there exists some choice of parameters $\kappa,\tau,n^{{}^{\prime}}$:

$\kappa\in\{2l-1,2l\},\tau\in\{0,\cdots,\kappa\}$

$\kappa+\tau\leq n^{{}^{\prime}}\leq n$**

such that upto a relabelling of the elements of $[n]$ and swapping $\mathcal{A},\mathcal{B}$, the following holds:

$\mathcal{A}=\{\bigcup\limits_{T\in J}\,{T}:J\subset\{\{1,\kappa+1\},\cdots,\{\tau,\kappa+\tau\},\{\tau+1\},\cdots,\{\kappa\}\},|J|=l\}\>\times\>2^{X}$,

$\mathcal{B}=\{L\cup\{\tau+1,\cdots,\kappa\}:L\subset\{1,\cdots,\tau,\kappa+1,\cdots,\kappa+\tau\},|L\cap\{i,\kappa+i\}|=1$* for all $i\in[\tau]\}\>\times\>2^{Y}$*

where $X=\{\kappa+\tau+1,\cdots,n^{{}^{\prime}}\}$ and $Y=\{n^{{}^{\prime}}+1,\cdots,n\}.$

Let $(\mathcal{A},\mathcal{B})$ be a $\frac{c}{d}$ cross-intersecting family where $\mathcal{B}$ is $k$-uniform. From \threfthm:5.1, there exists a $k_{0}>0$ such that if $\frac{ck}{d}=l>k_{0}$, then $|\mathcal{A}||\mathcal{B}|\leq\binom{2l}{l}2^{n-2l}$. Consider the case when $\mathcal{B}$ corresponds to $\mathcal{B}$ of \threfthm:5.1. If $|\mathcal{A}||\mathcal{B}|=\binom{2l}{l}2^{n-2l}$, then $n^{\prime}=n$, $Y=\emptyset$, and $k=\kappa$ in the statement of \threfthm:5.1. Since $l=\frac{ck}{d}$ and $k\in\{\frac{2ck}{d}-1,\frac{2ck}{d}\}$, we have the following two cases:

Case 1: $k=\frac{2ck}{d}-1$. 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 $l=\frac{ck}{d}=\frac{k+1}{2}$ is not an integer. So, the only valid case is $d=k$, $c=\frac{k+1}{2}=l$ and $k$ is an odd integer.

Case 2: $k=\frac{2ck}{d}$. Then, $\frac{c}{d}=\frac{1}{2}$, that is $(\mathcal{A},\mathcal{B})$ is a cross bisecting pair. Since $l=\frac{ck}{d}=\frac{k}{2}$ is an integer, $k$ must be even in this case.

If $\mathcal{B}$ corresponds to $\mathcal{A}$ of \threfthm:5.1, we have $X=\emptyset$, $\tau=0$, $\mathcal{B}$ is $k(=l)$-uniform, $l=\frac{ck}{d}$. Thus, we have $\frac{c}{d}=1$, $\mathcal{A}=\{\{1,\ldots,\kappa\}\}\times 2^{Y}$ where $Y=\{\kappa+1,\ldots,n\}$ and $\mathcal{B}=\binom{[\kappa]}{k}$, $\kappa\in\{2k-1,2k\}$ up to a relabelling of the elements.

This leads us to the main result of this section.

Statement of \threfthm:1.3: Let $(\mathcal{A}$,$\mathcal{B})$ be a $\frac{c}{d}$-cross intersecting pair of families of subsets of $[n]$. Let $\mathcal{B}$ be $k$-uniform. Then, there exists some $k_{0}>0$, such that for $k>k_{0}$ we have

$|\mathcal{A}||\mathcal{B}|\leq\binom{\frac{2ck}{d}}{\frac{ck}{d}}2^{n-\frac{2ck}{d}}$**

and the bound is tight if and only if, either $(a)$ or $(b)$ hold:

1. (a)

When $\frac{c}{d}=1$, $\mathcal{A}=\{\{1,\ldots,\kappa\}\}\times 2^{Y}$, $\mathcal{B}=\binom{[\kappa]}{k}$ where $Y=\{\kappa+1,\ldots,n\}$ and $\kappa\in\{2k-1,2k\}$ up to a relabelling of the elements of $[n]$. 2. (b)

When $\frac{c}{d}\neq 1$:

1. (i)

If $k$ is even, $c=1$, $d=2$, $\frac{ck}{d}=\lceil\frac{k}{2}\rceil$, 2. (ii)

If $k$ is odd, $c=\frac{k+1}{2}$, $d=k$, $\frac{ck}{d}=\lceil\frac{k}{2}\rceil$,

and for both the cases($(i)$ and $(ii)$), there exists some $\tau$ such that, $k+\tau\leq n$ and up to a relabelling of the elements of $[n]$,

$\mathcal{A}=\{\cup_{T\in J}\,{T}:J\subset\{\{1,k+1\},\ldots,\{\tau,k+\tau\},\{\tau+1\},\ldots,\{k\}\},|J|=\lceil\frac{k}{2}\rceil\}\times 2^{X}$**

where $X=\{k+\tau+1,\ldots,n\}$ and

$\mathcal{B}=\{L\cup\{\tau+1,\ldots,k\}:L\subset\{1,\ldots,\tau,k+1,\ldots,k+\tau\},|L\cap\{i,k+i\}|=1$* for all $i\in[\tau]\}$.*

6 Discussion

What are those pairs of $\frac{c}{d}$-cross intersecting families $(\mathcal{A},\mathcal{B})$ which achieve $|\mathcal{A}||\mathcal{B}|=2^{n}$ (equal to the upper bound for $\mathcal{M}_{\frac{c}{d}}(n)$ proved in \threfthm:1.1)? In the introduction we characterize such families when $\frac{c}{d}=0$ and $\frac{c}{d}=1$. In \threfthm:1.2, we characterize such families when $\frac{c}{d}=\frac{1}{2}$. From \threfthm:1.3, we see that when $\mathcal{B}$ is $k$-uniform, $|\mathcal{A}||\mathcal{B}|$ is maximized when $\frac{c}{d}$ is $1$ or nearly $\frac{1}{2}$($\frac{1}{2}$ or $\frac{1}{2}+\frac{1}{2k}$). For $\frac{c}{d}\in(0,1)$, besides the case $\mathcal{A}=2^{[n]}$, $\mathcal{B}=\{\emptyset\}$, is $|\mathcal{A}||\mathcal{B}|=2^{n}$ achieved only when $\frac{c}{d}$ is close to $\frac{1}{2}$?