Sharp results concerning disjoint cross-intersecting families

Peter Frankl; Andrey Kupavskii

arXiv:1905.08123·math.CO·May 21, 2019

Sharp results concerning disjoint cross-intersecting families

Peter Frankl, Andrey Kupavskii

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TL;DR

This paper establishes sharp bounds on the maximum sizes of disjoint cross-intersecting families of k-subsets from an n-element set, revealing precise conditions under which these bounds hold.

Contribution

It provides exact formulas and conditions for the maximum sizes of disjoint cross-intersecting families, extending previous results with sharp thresholds.

Findings

01

f(n,k) = floor(1/2 * binomial(n-1, k-1)) for n > c k^2

02

f*(n,k) = floor(1/2 * (binomial(n-1, k-1) - binomial(n-2k, k-1))) + 1 for n ≥ k^3 and k ≥ 5

03

Sharp thresholds for n relative to k for the bounds to hold

Abstract

For an $n$ -element set $X$ let $(k X)$ be the collection of all its $k$ -subsets. Two families of sets $A$ and $B$ are called cross-intersecting if $A \cap B \neq = \emptyset$ holds for all $A \in A$ , $B \in B$ . Let $f (n, k)$ denote the maximum of $min {∣ A ∣, ∣ B ∣}$ where the maximum is taken over all pairs of {\em disjoint}, cross-intersecting families $A, B \subset (k [ n ])$ . Let $c = lo g_{2} e$ . We prove that $f (n, k) = ⌊ \frac{1}{2} (k - 1 n - 1) ⌋$ essentially iff $n > c k^{2}$ (cf. Theorem~1.4 for the exact statement). Let $f^{*} (n, k)$ denote the same maximum under the additional restriction that the intersection of all members of both $A$ and $B$ are empty. For $k \geq 5$ and $n \geq k^{3}$ we show that…

Equations139

∣ F ∣ \leq (k - 1 n - 1) .

∣ F ∣ \leq (k - 1 n - 1) .

∣ F ∣ \leq (k - 1 n - 1) - (k - 1 n - k - 1) + 1.

∣ F ∣ \leq (k - 1 n - 1) - (k - 1 n - k - 1) + 1.

∣ A ∣ + ∣ B ∣ \leq 2^{n}

∣ A ∣ + ∣ B ∣ \leq 2^{n}

∣ A ∣ \cdot ∣ B ∣ \leq (k - 1 n - 1)^{2} .

∣ A ∣ \cdot ∣ B ∣ \leq (k - 1 n - 1)^{2} .

f (n) < \frac{3 - 5}{2} \cdot 2^{n}

f (n) < \frac{3 - 5}{2} \cdot 2^{n}

f (n, k) = ⌊ \frac{1}{2} (k - 1 n - 1) ⌋ .

f (n, k) = ⌊ \frac{1}{2} (k - 1 n - 1) ⌋ .

G <_{L} H iff min {i : i \in G ∖ H} < min {j : j \in H ∖ G} .

G <_{L} H iff min {i : i \in G ∖ H} < min {j : j \in H ∖ G} .

A \cap σ^{(a)} (B^{c}) = \emptyset,

A \cap σ^{(a)} (B^{c}) = \emptyset,

∣ H ∣ \leq k - 1.

∣ H ∣ \leq k - 1.

min {∣ A ∣, ∣ B ∣} \leq (k - 1 n - 1) - (k - 1 n - k - 1) + 1.

min {∣ A ∣, ∣ B ∣} \leq (k - 1 n - 1) - (k - 1 n - k - 1) + 1.

f^{*} (n, k) \geq ⌊ \frac{1}{2} ((k - 1 n - 1) - (k - 1 n - 2 k)) ⌋ + 1.

f^{*} (n, k) \geq ⌊ \frac{1}{2} ((k - 1 n - 1) - (k - 1 n - 2 k)) ⌋ + 1.

∣ A ∣ \leq (k - 1 n - 1) - (k - 1 n - k - 1) .

∣ A ∣ \leq (k - 1 n - 1) - (k - 1 n - k - 1) .

(k - 1 n - 1) - (k - 1 n - k - 1) + 1 \leq \frac{1}{2} (k - 1 n - 1)

(k - 1 n - 1) - (k - 1 n - k - 1) + 1 \leq \frac{1}{2} (k - 1 n - 1)

(k - 1 n - k - 1) \geq \frac{1}{2} (k - 1 n - 1) + 1.

(k - 1 n - k - 1) \geq \frac{1}{2} (k - 1 n - 1) + 1.

\dbinom{n-1}{k-1}\!\Big{/}\!\dbinom{n\!-\!k\!-\!1}{k-1}=\!\prod_{1\leq i\leq k-1}\!1+\frac{k}{n-k-i}<\left(\!1+\frac{k}{ck(k\!-\!1)}\!\right)^{\!k-1}\!<e^{\frac{1}{c}}=2.

\dbinom{n-1}{k-1}\!\Big{/}\!\dbinom{n\!-\!k\!-\!1}{k-1}=\!\prod_{1\leq i\leq k-1}\!1+\frac{k}{n-k-i}<\left(\!1+\frac{k}{ck(k\!-\!1)}\!\right)^{\!k-1}\!<e^{\frac{1}{c}}=2.

(k - 1 n - 1) - (k - 1 n - k) > \frac{1}{2} (k - 1 n - 1) .

(k - 1 n - 1) - (k - 1 n - k) > \frac{1}{2} (k - 1 n - 1) .

(k - 1 n - k) < \frac{1}{2} (k - 1 n - 1) .

(k - 1 n - k) < \frac{1}{2} (k - 1 n - 1) .

\frac{( k - 1 n - k )}{( k - 1 n - 1 )} < (\frac{n - k}{n - 1})^{k - 1} = (1 - \frac{k - 1}{n - 1})^{k - 1} \leq (1 - \frac{k - 1}{c ( k - 1 ) ^{2}})^{k - 1}

\frac{( k - 1 n - k )}{( k - 1 n - 1 )} < (\frac{n - k}{n - 1})^{k - 1} = (1 - \frac{k - 1}{n - 1})^{k - 1} \leq (1 - \frac{k - 1}{c ( k - 1 ) ^{2}})^{k - 1}

= (1 - \frac{1}{c ( k - 1 )})^{k - 1} < e^{- \frac{1}{c}} = \frac{1}{2} .

= (1 - \frac{1}{c ( k - 1 )})^{k - 1} < e^{- \frac{1}{c}} = \frac{1}{2} .

F

F

G

∣ F ∣ + ∣ G ∣ = (k - 1 n - 1) + n - k

∣ F ∣ + ∣ G ∣ = (k - 1 n - 1) + n - k

A = {A \in (k [ n ]) : 1 \in A, A \cap Q \neq = \emptyset} \cup {P},

A = {A \in (k [ n ]) : 1 \in A, A \cap Q \neq = \emptyset} \cup {P},

B = {B \in (k [ n ]) : 1 \in B, B \cap P \neq = \emptyset} \cup {Q} .

B = {B \in (k [ n ]) : 1 \in B, B \cap P \neq = \emptyset} \cup {Q} .

∣ A_{0} ∣ = ⌊ \frac{1}{2} ((k - 1 n - 1) - (k - 1 n - 2 k)) ⌋ + 1,

∣ A_{0} ∣ = ⌊ \frac{1}{2} ((k - 1 n - 1) - (k - 1 n - 2 k)) ⌋ + 1,

∣ B_{0} ∣ = ⌈ \frac{1}{2} ((k - 1 n - 1) - (k - 1 n - 2 k)) ⌉ + 1.

∣ B_{0} ∣ = ⌈ \frac{1}{2} ((k - 1 n - 1) - (k - 1 n - 2 k)) ⌉ + 1.

f (n, k) = ⌊ \frac{1}{2} (k - 1 n - 1) ⌋ if \frac{1}{2} (k - 1 n - 1) \geq (k - 1 n - 1) - (k - 1 n - k - 1)

f (n, k) = ⌊ \frac{1}{2} (k - 1 n - 1) ⌋ if \frac{1}{2} (k - 1 n - 1) \geq (k - 1 n - 1) - (k - 1 n - k - 1)

f (n, k) > \frac{1}{2} (k - 1 n - 1) if \frac{1}{2} (k - 1 n - 1) \leq (k - 1 n - 1) - (k - 1 n - k) .

f (n, k) > \frac{1}{2} (k - 1 n - 1) if \frac{1}{2} (k - 1 n - 1) \leq (k - 1 n - 1) - (k - 1 n - k) .

(k - 1 n - 1) - (k - 1 n - k) < \frac{1}{2} (k - 1 n - 1) < (k - 1 n - 1) - (k - 1 n - k - 1)

(k - 1 n - 1) - (k - 1 n - k) < \frac{1}{2} (k - 1 n - 1) < (k - 1 n - 1) - (k - 1 n - k - 1)

(k - 1 n - k - 1) < \frac{1}{2} (k - 1 n - 1) < (k - 1 n - k) .

(k - 1 n - k - 1) < \frac{1}{2} (k - 1 n - 1) < (k - 1 n - k) .

f (n, k) \leq \frac{1}{2} ((k - 1 n - 1) + n - k - 1) .

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Full text

Sharp results concerning disjoint cross-intersecting families

Peter Frankl111Rényi Institute, Budapest, Hungary, Andrey Kupavskii222University of Oxford and Moscow Institute of Physics and Technology; Email: [email protected] The research was supported by the Advanced Postdoc.Mobility grant no. P300P2_177839 of the Swiss National Science Foundation.

Abstract

For an $n$ -element set $X$ let $\binom{X}{k}$ be the collection of all its $k$ -subsets. Two families of sets $\mathcal{A}$ and $\mathcal{B}$ are called cross-intersecting if $A\cap B\neq\emptyset$ holds for all $A\in\mathcal{A}$ , $B\in\mathcal{B}$ . Let $f(n,k)$ denote the maximum of $\min\{|\mathcal{A}|,|\mathcal{B}|\}$ where the maximum is taken over all pairs of disjoint, cross-intersecting families $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ . Let $c=\log_{2}e$ . We prove that $f(n,k)=\left\lfloor\frac{1}{2}\binom{n-1}{k-1}\right\rfloor$ essentially iff $n>ck^{2}$ (cf. Theorem 1.4 for the exact statement). Let $f^{*}(n,k)$ denote the same maximum under the additional restriction that the intersection of all members of both $\mathcal{A}$ and $\mathcal{B}$ are empty. For $k\geq 5$ and $n\geq k^{3}$ we show that $f^{*}(n,k)=\left\lfloor\frac{1}{2}\left(\binom{n-1}{k-1}-\binom{n-2k}{k-1}\right)\right\rfloor+1$ and the restriction on $n$ is essentially sharp (cf. Theorem 5.4).

1 Introduction

Let $[n]=\{1,2,3,\dots,n\}$ be the standard $n$ -element set and let $2^{[n]}$ denote its power set. A subset $\mathcal{F}\subset 2^{[n]}$ is called a family. For $0\leq k\leq n$ let $\binom{[n]}{k}=\{F\subset[n]:\,|F|=k\}$ . Subsets of $\binom{[n]}{k}$ are called $k$ -uniform. A family $\mathcal{F}$ is called intersecting if $F\cap F^{\prime}\neq\emptyset$ for all $F,F^{\prime}\in\mathcal{F}$ . Let us state one of the central results in extremal set theory

Erdős–Ko–Rado Theorem ([EKR]). Suppose that $\mathcal{F}\subset 2^{[n]}$ is intersecting. Then (i) and (ii) hold.

(i)

$|\mathcal{F}|\leq 2^{n-1}$ .

(ii)

Assuming that $\mathcal{F}$ is $k$ -uniform and $n\geq 2k$ one has

[TABLE]

We should mention that (i) is a trivial consequence of the fact that $F\in\mathcal{F}$ implies $([n]\setminus F)\notin\mathcal{F}$ . There are many different proofs for (1.1). E.g., [D], [Ka2], [Py], [FF1], [HK], to mention a few.

Definition 1.1.

If for some $x\in[n]$ , $x\in F$ for all $F\in\mathcal{F}$ then $\mathcal{F}$ is called a star.

The full star $\mathcal{S}_{x}=\{S\in\binom{[n]}{k}:\,x\in S\}$ shows that (1.1) is sharp.

Hilton–Milner Theorem ([HM]). Suppose that $n>2k$ , $\mathcal{F}\subset\binom{[n]}{k}$ is intersecting and $\mathcal{F}$ is not a star. Then

[TABLE]

There are many known proofs for this important result as well. E.g. [FF2], [F87], [KZ], to mention a few.

Trying to prove results about intersecting families one arrives naturally at the following notion.

Definition 1.2.

Two families $\mathcal{A}$ , $\mathcal{B}$ are called cross-intersecting if $A\cap B\neq\emptyset$ for all $A\in\mathcal{A}$ , $B\in\mathcal{B}$ .

Noting that for $\mathcal{A}=\mathcal{B}$ the cross-intersecting property reduces to $\mathcal{A}$ being intersecting, one can generalize (i) to:

[TABLE]

whenever $\mathcal{A},\mathcal{B}\subset 2^{[n]}$ are cross-intersecting. Although the bound $|\mathcal{A}|+|\mathcal{B}|\leq\binom{n}{k}$ is true, for $n>2k$ it is not sufficient to derive (1.1). However, considering products does the job.

Pyber Theorem [Py]. Suppose that $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ are cross-intersecting, $n\geq 2k$ . Then

[TABLE]

Let us mention [FK1] where a short proof of (1.1) is given.

The following natural question was first considered in [DF].

Determine or estimate $f(n)\stackrel{{\scriptstyle\mathrm{def}}}{{=}}\min\{|\mathcal{A}|,|\mathcal{B}|\}$ where $\mathcal{A},\mathcal{B}\subset 2^{[n]}$ are cross-intersecting and $\mathcal{A}\cap\mathcal{B}=\emptyset$ . The following, rather surprising result was proved in [DF].

[TABLE]

and the constant $\frac{3-\sqrt{5}}{2}$ is optimal.

The second author was the first to consider the corresponding function $f(n,k)$ for $k$ -uniform families.

Conjecture 1.3 ([K17]).

Suppose that $n>2k\geq 4$ . Let $f(n,k)=$ $\max\min\{|\mathcal{A}|,|\mathcal{B}|\}$ where the maximum is over all disjoint and cross-intersecting families $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ . Then

[TABLE]

A few months later Huang and independently the present authors disproved (1.5) for $n<k^{2}$ (cf. [H]). On the positive side Huang [H] proved (1.5) for $n>2k^{2}$ .

In the present paper we determine the range where (1.5) holds almost completely. Set $c=\log_{2}e$ and note that $c<3/2$ .

Theorem 1.4.

(i)

If $n\geq ck^{2}+(2-c)k$ then (1.5) is true.

(ii)

If $n\leq ck^{2}-2ck+1$ then (1.5) fails.

2 Tools of proofs

For a family $\mathcal{F}$ and a positive integer $l$ define the $l$ -shadow $\sigma^{(l)}(\mathcal{F})=\{G:\,|G|=l,\ \exists F\in\mathcal{F},\,G\subset F\}$ . In other words, $\sigma^{(l)}(\mathcal{F})=\bigcup\limits_{F\in\mathcal{F}}\binom{F}{l}$ .

One of the most important results in extremal set theory is the Kruskal–Katona Theorem. For given positive integers $m,k,l$ , $k>l$ it determines the minimum of $|\sigma^{(l)}(\mathcal{F})|$ where $\mathcal{F}$ is $k$ -uniform and $|\mathcal{F}|=m$ .

It was Daykin’s proof [D] of the Erdős–Ko–Rado Theorem that established the connection between these two important theorems. To state this connection in the more general setting of pairs of cross-intersecting families let us define the lexicographic order, $<_{L}$ on $\binom{[n]}{t}$ , $1\leq t\leq n$ . For $G,H\in\binom{[n]}{t}$ ,

[TABLE]

That is, $\{1,\,2,\,77\}<_{L}\{1,\,3,\,4\}$ .

For fixed $t$ and $m$ , $1\leq m<\binom{n}{t}$ let $\mathcal{L}(n,t,m)$ denote the initial segment, the first $m$ subsets of size $t$ in the lexicographic order.

Let now $a,b$ be positive integers, $a+b<n$ . Hilton [Hi] observed that $\mathcal{A}\subset\binom{[n]}{a}$ and $\mathcal{B}\subset\binom{[n]}{b}$ are cross-intersecting iff

[TABLE]

where $\mathcal{B}^{c}=\{[n]\setminus B:\,B\in\mathcal{B}\}$ is the family of complements. This permits the following equivalent formulation of the Kruskal–Katona Theorem.

Kruskal–Katona Theorem ([Ka1], [Kr]). Let $a,b$ be positive integers, $n>a+b$ . Suppose that $\mathcal{A}\subset\binom{[n]}{a}$ and $\mathcal{B}\subset\binom{[n]}{b}$ are cross-intersecting. Then $\mathcal{L}(n,a,|\mathcal{A}|)$ and $\mathcal{L}(n,b,|\mathcal{B}|)$ are cross-intersecting too.

To get the reader familiar with this formulation let us show an easy consequence that we need later.

Corollary 2.1.

Let $n>2k>0$ and suppose that $\mathcal{G}\subset\binom{[n-1]}{k-1}$ and $\mathcal{H}\subset\binom{[n-1]}{k}$ are cross-intersecting and $|\mathcal{G}|>\binom{n-1}{k-1}-\binom{n-k}{k-1}$ . Then

[TABLE]

Proof.

The first $\binom{n-1}{k-1}-\binom{n-k}{k-1}=\binom{n-2}{k-2}+\dots+\binom{n-k}{k-2}$ subsets $G\in\binom{[n-1]}{k-1}$ are $\mathcal{L}(n-1,\,k-1,\,\binom{n-1}{k-1}-\binom{n-k}{k-1})=\{G\in\binom{[n-1]}{k-1}:\,G\cap[k-1]\neq\emptyset\}$ . The next $(k-1)$ -set is $[k,\,2k-2]$ . The only $k$ -subsets intersecting all of them are $[k-1]\cup\{j\},\,k\leq j\leq 2k-2$ .∎

The proof of Theorem 1.4 (i) is based on the following result extending the Hilton–Milner Theorem to two families.

Mörs Theorem ([M]). Suppose that $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ are cross-intersecting, $n>2k>0$ and neither $\mathcal{A}$ nor $\mathcal{B}$ is a star. Then

[TABLE]

Let us note that for $k=1$ all the problems considered so far are trivial. For $k=2$ , and neither $\mathcal{A}$ nor $\mathcal{B}$ being a star, either $\mathcal{A}=\mathcal{B}=\binom{T}{2}$ for a 3-element set $T$ , or one of $\mathcal{A}$ and $\mathcal{B}$ consists of two pairwise disjoint 2-sets and the other is a subset of the four 2-sets intersecting both. Therefore from now on we are going to assume $k\geq 3$ .

In a recent work [FK2] we extended the Mörs Theorem using the notion of diversity (cf. the definition and statement in Section 4). This result appears to be essential in establishing the exact value of $f^{*}(n,k):=\min\bigl{\{}|\mathcal{A}|,|\mathcal{B}|:\,\mathcal{A},\mathcal{B}\subset\binom{[n]}{k},\ \mathcal{A}\cap\mathcal{B}=\emptyset$ , $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting, and neither of them is a star $\bigr{\}}$ .

Proposition 2.2.

For $n>2k$ one has

[TABLE]

3 The proof of Theorem 1.4 and Proposition 2.2

Let $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ be disjoint and cross-intersecting. If one of them (say $\mathcal{A}$ ) is a star then there are two cases. For definiteness suppose $1\in A$ for all $A\in\mathcal{A}$ . Should $1\in B$ hold for all $B\in\mathcal{B}$ then $|\mathcal{A}\cup\mathcal{B}|\leq\binom{n-1}{k-1}$ implies (1.5).

If there is some $B_{0}\in\mathcal{B}$ with $1\notin B_{0}$ then the cross-intersecting property implies

[TABLE]

On the other hand, if neither of $\mathcal{A}$ and $\mathcal{B}$ is a star, we may apply the Mörs Theorem and get the bound (2.2). Consequently, if

[TABLE]

then (1.5) holds. Equivalently

[TABLE]

Recall the inequality $1+x<e^{x}$ valid for all $x$ . Suppose that $n\geq c(k^{2}-k)+2k-1$ . Then

[TABLE]

This proves (3.2) for the corresonding range thereby establishing Theorem 1.4 (i).

Let us now prove that for $n\leq c(k-1)^{2}+1$ ,

[TABLE]

Equivalently,

[TABLE]

We derive (3.4) from the following chain of equalities and inequalities:

[TABLE]

To use (3.3) let us define two families $\mathcal{F}$ and $\mathcal{G}$ .

[TABLE]

Then

[TABLE]

and the families $\mathcal{F},\mathcal{G}$ are cross-intersecting while $\mathcal{G}$ is intersecting. In view of (3.3), $|\mathcal{G}|>\frac{1}{2}\binom{n-1}{k-1}$ .

If $|\mathcal{F}|>\frac{1}{2}\binom{n-1}{k-1}$ then we are done.

Suppose $|\mathcal{F}|\leq\frac{1}{2}\binom{n-1}{k-1}$ and let $\mathcal{G}_{0}\subset\mathcal{G}$ be an arbitrary subfamily satisfying $|\mathcal{G}_{0}|=\lfloor\frac{1}{2}\binom{n-1}{k-1}\rfloor+1-|\mathcal{F}|$ . Set $\mathcal{A}=\mathcal{F}\cup\mathcal{G}_{0}$ , $\mathcal{B}=\mathcal{G}\setminus\mathcal{G}_{0}$ . By definition $|\mathcal{A}|=\lfloor\frac{1}{2}\binom{n-1}{k-1}\rfloor+1$ and (3.5) implies $|\mathcal{B}|>\frac{1}{2}\binom{n-1}{k-1}$ as well. This completes the proof of Theorem 1.4. $\square$

Let us turn to the proof of Proposition 2.2.

Let $P,Q\in\binom{[2n]}{k}$ satisfy $P\cap Q=\{2\}$ . Define

[TABLE]

It is easy to see that $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting and neither of them is a star. However, they are not disjoint.

Noting $|\mathcal{A}|=|\mathcal{B}|$ and $|\mathcal{A}\cup\mathcal{B}|=\binom{n-1}{k-1}-\binom{n-2k}{k-1}+2$ , one can remove equitably the members of $\mathcal{A}\cap\mathcal{B}$ from exactly one of the two families to obtain $\mathcal{A}_{0}\subset\mathcal{A}$ , $\mathcal{B}_{0}\subset\mathcal{B}$ ,

[TABLE]

Obviously, $\mathcal{A}_{0}$ and $\mathcal{B}_{0}$ are cross-intersecting and for $k\geq 2$ neither of them is a star. This concludes the proof of (2.3). $\square$

4 In the grey zone

In the previous section we proved Theorem 1.4. To be more exact, we proved that

[TABLE]

and also

[TABLE]

Let us try and say something about the “grey zone”, about the narrow range where

[TABLE]

or equivalently

[TABLE]

Let us prove that even if (1.5) fails it is almost true.

Proposition 4.1.

Suppose that (4.1) hold for the pair $(n,k)$ , $n>2k$ . Then

[TABLE]

Proof.

Let again $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ be disjoint and cross-intersecting, moreover, $\min\{|\mathcal{A}|,|\mathcal{B}|\}\geq\frac{1}{2}\binom{n-1}{k-1}$ . If both are stars then (1.5) holds. Suppose now that $\mathcal{A}$ is a star where $n\in A$ for all $A\in\mathcal{A}$ . In view of (4.1) we may apply Corollary 2.1 with $\mathcal{G}=\mathcal{A}(n)=\big{\{}A\setminus\{n\}:\,A\in\mathcal{A}\big{\}}$ , $\mathcal{H}=\mathcal{B}(\bar{n})=\{B\in\mathcal{B}:\,n\notin B\}$ . This gives $|\mathcal{H}|\leq k-1$ . Except for $\mathcal{H}$ , all members of $\mathcal{A}\cup\mathcal{B}$ contain $n$ . Thus we infer

[TABLE]

Since $\mathcal{A}$ and $\mathcal{B}$ are disjoint, (4.3) follows.

The case that remains is when neither $\mathcal{A}$ nor $\mathcal{B}$ is a star. To deal with this case we need the notion of diversity and the extension of Mörs Theorem.

Definition 4.2.

For a family $\mathcal{F}\in 2^{[n]}$ define its diversity $\gamma(\mathcal{F})$ by $\gamma(\mathcal{F})=\min\{|\mathcal{F}(\bar{i})|:\,i\in[n]\}$ .

It should be clear that $\mathcal{F}$ is a star iff $\gamma(\mathcal{F})=0$ .

Theorem 4.3 ([FK2]).

Suppose that $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ are cross-intersecting, $n>2k$ , $k>3$ . Let $u$ be an integer, $3\leq u\leq k$ and suppose that

[TABLE]

Then either equality holds in (4.4) for both $\mathcal{A}$ and $\mathcal{B}$ or

[TABLE]

moreover both families share the same (unique) element of maximum degree.

Let us return to the proof of Proposition 4.1. We apply Theorem 4.3 with $u=k-1$ . If

[TABLE]

then via (4.1)

[TABLE]

follows.

In the opposite case from (4.4) we derive

[TABLE]

By symmetry we may assume

[TABLE]

By disjointness of $\mathcal{A}$ and $\mathcal{B}$ we infer

[TABLE]

Thus

[TABLE]

follows. ∎

5 Determining $f^{*}(n,k)$ for $n\geq k(k+5)$ , $k\geq 5$

Throughout this section let $k\geq 5$ , $n\geq k(k+5)$ and let $\mathcal{A},\mathcal{B}\subset\binom{[n]}{k}$ be disjoint, cross-intersecting. Moreover we assume that neither of $\mathcal{A}$ and $\mathcal{B}$ is a star.

Since we are trying to determine $f^{*}(n,k)$ , in view of Proposition 2.2 we may assume that

[TABLE]

First we show that the above conditions on $n$ and $k$ guarantee that Theorem 4.3 can be applied with some $3\leq u\leq k$ .

Claim 5.1.

[TABLE]

Proof.

One can rewrite the RHS as

[TABLE]

In view of (5.1) it is sufficient to show that

[TABLE]

Noting that ${\binom{n-i-1}{k-2}}\Bigm{/}{\binom{n-i}{k-2}}=1-\frac{k-2}{n-i}\geq 1-\frac{k-2}{n-8}$ for $2\leq i\leq 8$ , using $n\geq k^{2}+5k>(k-2)(k+7)+8$ we infer that the above ratio is always more than

[TABLE]

Now (5.3) follows from $1+\frac{11}{12}+(\frac{11}{12})^{2}+\dots+(\frac{11}{12})^{7}=12\cdot\big{(}1-(\frac{11}{12})^{8}\big{)}$ and $(\frac{11}{12})^{8}<\frac{1}{2}$ . ∎

Now let $u$ be the maximal integer, $3\leq u\leq k$ such that

[TABLE]

In view of Theorem 4.3 we may assume that

[TABLE]

Since neither $\mathcal{A}$ nor $\mathcal{B}$ is a star, $u<k$ follows.

Our next goal is to show that for $A\in\mathcal{A}(\bar{1})$ and $B\in\mathcal{B}(\bar{1})$ necessarily

[TABLE]

The cross-intersecting property implies that every $C\in(\mathcal{A}(1)\cup\mathcal{B}(1))$ satisfies $D\cap C\neq\emptyset$ . Consequently,

[TABLE]

By (5.4) we have

[TABLE]

Comparing with (5.1) we infer

[TABLE]

However, simple computation shows that this inequality fails for $n\geq k(k+5)$ .

Indeed,

[TABLE]

For $k\geq 5$ and $n\geq k(k+5)$ , $\frac{n-2k-1}{k-3}>\frac{n-2k}{k-2}>k+3\geq 8$ . Also $\frac{2(k-1)}{n-(k-1)}<\frac{2}{k+5}$ and $\left(1-\frac{2}{k+5}\right)^{k-2}>e^{-2}$ . These show that

[TABLE]

contradicting (5.7). Now (5.5) is proved.

Claim 5.2.

If $A,A^{\prime}\in\mathcal{A}(\bar{1})$ and $B,B^{\prime}\in\mathcal{B}(\bar{1})$ then

[TABLE]

Proof.

Set $D=A\cup B$ , $D^{\prime}=A^{\prime}\cup B^{\prime}$ and suppose indirectly $D\neq D^{\prime}$ . As we pointed out before, the cross-intersecting property implies both $C\cap D\neq\emptyset$ and $C\cap D^{\prime}\neq\emptyset$ for all $C\in\mathcal{A}(1)\cup\mathcal{B}(1)$ . It is easy to see that the total number of $(k-1)$ -subsets $\overset{\sim}{C}\subset[2n]$ , intersecting both $D$ and $D^{\prime}$ is largest if $|D\cap D^{\prime}|=2k-2$ .

In that case the number is

[TABLE]

It is only a slight difference with respect to (5.6) and we can get a contradiction in exactly the same way. ∎

Claim 5.3.

$\min\{|\mathcal{A}(\bar{1})|,|\mathcal{B}(\bar{1})|\}=1$ .

Proof.

Suppose the contrary. WLOG let $[2,k+1]\in\mathcal{A}(\bar{1})$ , $[k+1,2k]\in\mathcal{B}(\bar{1})$ . Choose some different $A\in\mathcal{A}(\bar{1})$ and $B\in\mathcal{B}(\bar{1})$ . By Claim 5.2, $A\cup B=[2,2k]$ . By (5.5), $A\cap B=\{j\}$ for some $j\in[2,2k]$ .

In the case $j=k+1$ , $A\neq[2,k+1]$ implies $|A\cap[k+1,2k]|\geq 2$ , contradicting (5.5). Suppose by symmetry $2\leq j\leq k$ . Now (5.5) implies $[2,k+1]\cap B=\{j\}$ whence $B=\{j\}\cup[k+2,2k]$ . From $A\cap B=\{j\}$ , $A=[2,k+1]$ follows which is in contradiction with our choice $A\neq[2,k+1]$ . ∎

By symmetry, let us suppose that $|\mathcal{A}(\bar{1})|=1$ . WLOG let $[2,k+1]$ be the unique member of $\mathcal{A}(\bar{1})$ and let $[k+1,2k]$ be one of the members of $\mathcal{B}(\bar{1})$ . Using Claim 5.2 and (5.5) we infer that

[TABLE]

in particular

[TABLE]

Together with

[TABLE]

this implies

[TABLE]

In the case $|\mathcal{B}(\bar{1})|=1$ we infer

[TABLE]

in accordance with (2.3).

Let us show that for $n\geq k^{3}$ the RHS of the last displayed inequality is the value of $f^{*}(n,k)$ .

Theorem 5.4.

For $n\geq k^{3}$

[TABLE]

Proof.

In view of (5.10), all we have to show is $|\mathcal{B}(\bar{1})|=1$ . Suppose for contradiction $|\mathcal{B}(\bar{1})|\geq 2$ . WLOG $[k+1,2k]$ and $\{k\}\cup[k+2,2k]$ belong to $|\mathcal{B}(\bar{1})|$ . Then for $A\in\mathcal{A}(1)$ either

[TABLE]

The number of $(k-1)$ -sets $A\subset[2,n]$ satisfying these conditions is $\binom{n-1}{k-1}-\binom{n-k}{k-1}+\binom{n-k-2}{k-3}.$ In view of $|\mathcal{A}(\bar{1})|=1$ , to conclude the proof it is sufficient to show

[TABLE]

Equivalently,

[TABLE]

The last term in brackets is of smaller order of magnitude. E.g., for $n-k>4(k^{2}-4)$ it is smaller than $\frac{1}{2(k+2)}\binom{n-k-1}{k-2}$ .

We break up $\binom{n-k-1}{k-2}$ into $k+1$ equal parts and use $\frac{1}{k+1}\binom{n-k-1}{k-2}>\frac{1}{k+1}\binom{n-k-j}{k-2}$ for $j\geq 2$ . One of these terms we use to compensate for the last term in (5.12). Consequently, instead of (5.12) it suffices to show that

[TABLE]

This inequality follows once we show

[TABLE]

Let us expand the LHS of (5.13) and use the Bernoulli inequality

[TABLE]

Now $n\geq k^{3}$ implies $n-2k\geq k^{3}-2k>(k+2)k(k-2)$ , i.e., the RHS is at least $1-\frac{1}{k+2}=\frac{k+1}{k+2}$ completing the proof. ∎

Let us mention that our argument was essentially sharp, that is for $n<(1-\varepsilon)k^{3}$ the inequality (5.11) would fail completely. That is the difference of the two sides would be much more than $k$ . Consequently, imitating the proof of Proposition 2.2 we can show the following.

Proposition 5.5.

For any $\varepsilon>0$ and $k>k_{0}(\varepsilon)$ in the range $k^{2}+5k<n<(1-\varepsilon)k^{3}$ one has

[TABLE]

$\square$

Bibliography20

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1]
2[D] D. E. Daykin, Erdős–Ko–Rado from Kruskal–Katona, Journal of Combinatorial Theory A 17 (1974), 254–255.
3[DF] D. E. Daykin, P. Frankl, Extremal sets of subsets satisfying conditions induced by a graph, in: Graph theory and combinatorics (Cambridge, 1983), pp. 107–126, Academic Press, London, 1984.
4[F] P. Frankl, A new short proof for the Kruskal–Katona Theorem, Discrete Mathematics 48 (1984), 327–329.
5[F 87] P. Frankl, Erdős–Ko–Rado Theorem with conditions on the maximal degree, Journal of Combinatorial Theory A 46 (1987), N 2, 252–263.
6[FF 1] P. Frankl, Z. Füredi, A new short proof of the EKR theorem, Journal of Combinatorial Theory A 119 (2012), 1388–1390.
7[FF 2] P. Frankl, Z. Füredi, Non-trivial intersecting families, Journal of Combinatorial Theory A 41 (1986), 150–153.
8[FK 1] P. Frankl, A. Kupavskii, A size-sensitive inequality for cross-intersecting families, European Journal of Combinatorics 62 (2017), 263–271.

TL;DR

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Sharp results concerning disjoint cross-intersecting families

Abstract

1 Introduction

Definition 1.1**.**

Definition 1.2**.**

Conjecture 1.3** ([K17]).**

Theorem 1.4**.**

2 Tools of proofs

Corollary 2.1**.**

Proof.

Proposition 2.2**.**

3 The proof of Theorem 1.4 and Proposition 2.2

4 In the grey zone

Proposition 4.1**.**

Proof.

Definition 4.2**.**

Theorem 4.3** ([FK2]).**

5 Determining f∗(n,k)f^{*}(n,k)f∗(n,k) for n≥k(k+5)n\geq k(k+5)n≥k(k+5), k≥5k\geq 5k≥5

Claim 5.1**.**

Proof.

Claim 5.2**.**

Proof.

Claim 5.3**.**

Proof.

Theorem 5.4**.**

Proof.

Proposition 5.5**.**

Definition 1.1.

Definition 1.2.

Conjecture 1.3 ([K17]).

Theorem 1.4.

Corollary 2.1.

Proposition 2.2.

Proposition 4.1.

Definition 4.2.

Theorem 4.3 ([FK2]).

5 Determining $f^{*}(n,k)$ for $n\geq k(k+5)$ , $k\geq 5$

Claim 5.1.

Claim 5.2.

Claim 5.3.

Theorem 5.4.

Proposition 5.5.