Intersecting families, cross-intersecting families, and a proof of a conjecture of Feghali, Johnson and Thomas

TL;DR

This paper extends the Erd ext{o}s-Ko-Rado theorem to cross-intersecting families with weighted sums and proves a conjecture about the structure of maximum intersecting families in certain graphs, specifically depth-two claws.

Contribution

It generalizes the Erd ext{o}s-Ko-Rado theorem to weighted sums under specific conditions and confirms a conjecture on maximum intersecting families in depth-two claw graphs.

Findings

Extended Erd ext{o}s-Ko-Rado theorem to weighted sums for cross-intersecting families.

Proved the full conjecture of Feghali, Johnson, and Thomas for depth-two claw graphs.

Established the existence of a vertex with a largest intersecting family in the graph.

Abstract

A family $\mathcal{A}$ of sets is said to be intersecting if every two sets in $\mathcal{A}$ intersect. Two families $\mathcal{A}$ and $\mathcal{B}$ are said to be cross-intersecting if each set in $\mathcal{A}$ intersects each set in $\mathcal{B}$. For a positive integer $n$, let $[n] = \{1, \dots, n\}$ and $\mathcal{S}_n = \{A \subseteq [n] \colon 1 \in A\}$. In this note, we extend the Erd\H{o}s-Ko-Rado Theorem by showing that if $\mathcal{A}$ and $\mathcal{B}$ are non-empty cross-intersecting families of subsets of $[n]$, $\mathcal{A}$ is intersecting, and $a_0, a_1, \dots, a_n, b_0, b_1, \dots, b_n$ are non-negative real numbers such that $a_i + b_i \geq a_{n-i} + b_{n-i}$ and $a_{n-i} \geq b_i$ for each $i \leq n/2$, then \[\sum_{A \in \mathcal{A}} a_{|A|} + \sum_{B \in \mathcal{B}} b_{|B|} \leq \sum_{A \in \mathcal{S}_n} a_{|A|} + \sum_{B \in \mathcal{S}_n} b_{|B|}.\] For a graph $G$ and an integer $r$, let ${\mathcal{I}_G}^{(r)}$ denote the family of $r$-element independent sets of $G$. Inspired by a problem of Holroyd and Talbot, Feghali, Johnson and Thomas conjectured that if $r < n$ and $G$ is a depth-two claw with $n$ leaves, then $G$ has a vertex $v$ such that $\{A \in {\mathcal{I}_G}^{(r)} \colon v \in A\}$ is a largest intersecting subfamily of ${\mathcal{I}_G}^{(r)}$. They proved this for $r \leq \frac{n+1}{2}$. We use the result above to prove the full conjecture.

Equations

$$\sum _{A\in \mathcal{A}}{a}_{|A|}+\sum _{B\in \mathcal{B}}{b}_{|B|}\le \sum _{A\in {\mathcal{S}}_n}{a}_{|A|}+\sum _{B\in {\mathcal{S}}_n}{b}_{|B|}.$$

$${\mathcal{L}}_{n,k}=\left\{\{(i_1,j_1),\dots ,(i_r,j_r)\}:r\in [n],\{i_1,\dots ,i_r\}\in \left(\genfrac{}{}{0ex}{}{[n]}{r}\right),j_1,\dots ,j_r\in [k]\right\}.$$

$${{\mathcal{I}}_{T_n}}^{(r)}={{\mathcal{L}}_{n,2}}^{(r)}\cup \left\{A\cup \{x_0\}:A\in \left(\genfrac{}{}{0ex}{}{X_n}{r-1}\right)\right\}.$$

$$|{\mathcal{A}}^{(n-r)}|$$

$$=\left(\genfrac{}{}{0ex}{}{n}{r}\right)-|{\mathcal{A}}^{(r)}|-|{\mathcal{B}}^{(r)}\backslash {\mathcal{A}}^{(r)}|.$$

$$c_r+c_{n-r}$$

$$+\left(\left(\genfrac{}{}{0ex}{}{n}{r}\right)-|{\mathcal{A}}^{(r)}|-|{\mathcal{B}}^{(r)}\backslash {\mathcal{A}}^{(r)}|\right)a_{n-r}+\left(\left(\genfrac{}{}{0ex}{}{n}{r}\right)-|{\mathcal{A}}^{(r)}|\right)b_{n-r}$$

$$=|{\mathcal{A}}^{(r)}|(a_r+b_r-a_{n-r}-b_{n-r})-|{\mathcal{B}}^{(r)}\backslash {\mathcal{A}}^{(r)}|(a_{n-r}-b_r)+\left(\genfrac{}{}{0ex}{}{n}{r}\right)(a_{n-r}+b_{n-r})$$

$$\le \left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)(a_r+b_r-a_{n-r}-b_{n-r})+\left(\genfrac{}{}{0ex}{}{n}{r}\right)(a_{n-r}+b_{n-r})$$

$$\sum _{A\in \mathcal{A}}{a}_{|A|}+\sum _{B\in \mathcal{B}}{b}_{|B|}$$

$$=c_0+c_n+\sum _{i=1}^{\frac{n-1}{2}}(c_i+c_{n-i})$$

$$\le a_n+b_n+\sum _{i=1}^{\frac{n-1}{2}}\left(\left(\genfrac{}{}{0ex}{}{n-1}{i-1}\right)(a_i+b_i)+\left(\genfrac{}{}{0ex}{}{n-1}{n-i-1}\right)(a_{n-i}+b_{n-i})\right)$$

$$=\sum _{A\in {\mathcal{S}}_n}{a}_{|A|}+\sum _{B\in {\mathcal{S}}_n}{b}_{|B|}.$$

$$\sum _{A\in \mathcal{A}}{a}_{|A|}+\sum _{B\in \mathcal{B}}{b}_{|B|}$$

$$\le a_n+b_n+\left(\genfrac{}{}{0ex}{}{n-1}{n/2-1}\right)(a_{n/2}+b_{n/2})$$

$$+\sum _{i=1}^{\frac{n}{2}-1}\left(\left(\genfrac{}{}{0ex}{}{n-1}{i-1}\right)(a_i+b_i)+\left(\genfrac{}{}{0ex}{}{n-1}{n-i-1}\right)(a_{n-i}+b_{n-i})\right)$$

$$=\sum _{A\in {\mathcal{S}}_n}{a}_{|A|}+\sum _{B\in {\mathcal{S}}_n}{b}_{|B|},$$

$$\delta_{i,j}(A)=\left\{\begin{array}[]{ll}(A\backslash\{(i,j)\})\cup\{(i,1)\}&\mbox{if $(i,j)\in A$};\\ A&\mbox{otherwise,}\end{array}\right.$$

$${\Delta }_{i,j}(\mathcal{A})=\{{\delta }_{i,j}(A):A\in \mathcal{A}\}\cup \{A\in \mathcal{A}:{\delta }_{i,j}(A)\in \mathcal{A}\}.$$

$${\mathcal{A}}^{\ast }={\Delta }_{n,k}\circ \cdots \circ {\Delta }_{n,2}\circ \cdots \circ {\Delta }_{1,k}\circ \cdots \circ {\Delta }_{1,2}(\mathcal{A}),$$

$$\gamma(A):=\left\{\begin{array}[]{ll}(A\backslash\{x_{0}\})\cup\{y_{1}\}&\mbox{if $x_{0}\in A,y_{1}\notin A$ and $(A\backslash\{x_{0}\})\cup\{y_{1}\}\in\mathcal{R}$;}\\ A&\mbox{otherwise,}\end{array}\right.$$

$$\Gamma (\mathcal{A}):=\{\gamma (A):A\in \mathcal{A}\}\cup \{A\in \mathcal{A}:\gamma (A)\in \mathcal{A}\}.$$

$$E\cap F\cap X_n\ne \varnothing$$

$$a_0>\cdots >a_r>a_{r+1}=\cdots =a_n=0.$$

$$\sum _{A\in \mathcal{A}}{a}_{|A|}+\sum _{B\in \mathcal{B}}{b}_{|B|}\le \sum _{A\in {\mathcal{X}}_n}{a}_{|A|}+\sum _{B\in {\mathcal{X}}_n}{b}_{|B|},$$

$$|{\mathcal{E}}_0|$$

$$\le \sum _{i=0}^r\left|\left\{E\in {{\mathcal{L}}_{n,2}}^{(r)}:E\cap X_n\in {\mathcal{A}}^{(i)}\right\}\right|=\sum _{i=0}^r|{\mathcal{A}}^{(i)}|a_i=\sum _{A\in \mathcal{A}}{a}_{|A|}$$

$$|\mathcal{E}|=|{\mathcal{E}}_0|+|{\mathcal{E}}_1|\le \sum _{A\in \mathcal{A}}{a}_{|A|}+\sum _{B\in \mathcal{B}}{b}_{|B|}\le \sum _{A\in {\mathcal{X}}_n}{a}_{|A|}+\sum _{B\in {\mathcal{X}}_n}{b}_{|B|}=|{\mathcal{F}}_0|+|{\mathcal{F}}_1|=|\mathcal{F}|.$$

Full text

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For a graph $G$ and an integer $r$, let ${\mathcal{I}_{G}}^{(r)}$ denote the family of $r$-element independent sets of $G$. Inspired by a problem of Holroyd and Talbot, Feghali, Johnson and Thomas conjectured that if $r<n$ and $G$ is a depth-two claw with $n$ leaves, then $G$ has a vertex $v$ such that $\{A\in{\mathcal{I}_{G}}^{(r)}\colon v\in A\}$ is a largest intersecting subfamily of ${\mathcal{I}_{G}}^{(r)}$. They proved this for $r\leq\frac{n+1}{2}$. We use the result above to prove the full conjecture.

1 Introduction

Unless otherwise stated, we shall use small letters such as $x$ to denote non-negative integers or elements of a set, capital letters such as $X$ to denote sets, and calligraphic letters such as $\mathcal{F}$ to denote families (that is, sets whose members are sets themselves). It is to be assumed that arbitrary sets and families are finite. We call a set $A$ an $r$-element set if its size $|A|$ is $r$, that is, if it contains exactly $r$ elements (also called members).

The set $\{1,2,\dots\}$ of positive integers is denoted by $\mathbb{N}$. For any integer $n\geq 0$, the set $\{i\in\mathbb{N}\colon i\leq n\}$ is denoted by $[n]$. Note that $[0]$ is the empty set $\emptyset$. For a set $X$, the power set of $X$ (that is, $\{A\colon A\subseteq X\}$) is denoted by $2^{X}$. The family of $r$-element subsets of $X$ is denoted by $X\choose r$. The family of $r$-element sets in a family $\mathcal{F}$ is denoted by $\mathcal{F}^{(r)}$. If $\mathcal{F}\subseteq 2^{X}$ and $x\in X$, then we denote the family $\{F\in\mathcal{F}\colon x\in F\}$ by $\mathcal{F}(x)$. We call $\mathcal{F}(x)$ a star of $\mathcal{F}$ if $\mathcal{F}(x)\neq\emptyset$.

We say that a set $A$ intersects a set $B$ if $A$ and $B$ have at least one common element (that is, $A\cap B\neq\emptyset$). A family $\mathcal{A}$ is said to be intersecting if for every $A,B\in\mathcal{A}$, $A$ and $B$ intersect. The stars of a family $\mathcal{F}$ (with $|\bigcup_{F\in\mathcal{F}}F|\geq 1$) are the simplest intersecting subfamilies of $\mathcal{F}$. We say that $\mathcal{F}$ has the star property if at least one of the largest intersecting subfamilies of $\mathcal{F}$ is a star of $\mathcal{F}$.

One of the most popular endeavours in extremal set theory is that of determining the size of a largest intersecting subfamily of a given family $\mathcal{F}$. This started in [12], which features the following classical result, known as the Erdős-Ko-Rado (EKR) Theorem.

Theorem 1.1** (EKR Theorem [12])**

If $r\leq n/2$ and $\mathcal{A}$ is an intersecting subfamily of ${[n]\choose r}$, then $|\mathcal{A}|\leq{n-1\choose r-1}$.

This means that ${[n]\choose r}$ has the star property. There are various proofs of the EKR Theorem (see [10, 23, 25]), two of which are particularly short and beautiful: Katona’s [23], which introduced the elegant cycle method, and Daykin’s [10], using the fundamental Kruskal–Katona Theorem [24, 26]. The EKR Theorem gave rise to some of the highlights in extremal set theory [1, 15, 25, 28] and inspired many results that establish how large a system of sets can be under certain intersection conditions; see [4, 11, 14, 16, 17, 20, 21].

If $\mathcal{A}$ and $\mathcal{B}$ are families such that each set in $\mathcal{A}$ intersects each set in $\mathcal{B}$, then $\mathcal{A}$ and $\mathcal{B}$ are said to be cross-intersecting.

For intersecting subfamilies of a given family $\mathcal{F}$, the natural question to ask is how large they can be. A natural variant of this intersection problem is the problem of maximizing the sum or the product of sizes of cross-intersecting subfamilies (not necessarily distinct or non-empty) of $\mathcal{F}$. This has recently attracted much attention. The relation between the original intersection problem, the sum problem and the product problem is studied in [6]. Solutions have been obtained for various families; most of the known results are referenced in [2, 7], which treat the product problem for families of subsets of $[n]$ of size at most $r$.

Here we consider the sum problem for the case where at least one of two cross-intersecting families $\mathcal{A}$ and $\mathcal{B}$ of subsets of $[n]$ is an intersecting family. We actually consider a more general setting of weighted sets, where each set of size $i$ is assigned two non-negative integers $a_{i}$ and $b_{i}$, and the objective is to maximize $\sum_{A\in\mathcal{A}}a_{|A|}+\sum_{B\in\mathcal{B}}b_{|B|}$. Let $\mathcal{S}_{n}$ denote the star $\{A\subseteq[n]\colon 1\in A\}$ of $2^{[n]}$. In Section 2, we prove the following extension of the EKR Theorem.

Theorem 1.2

If $\mathcal{A}$ and $\mathcal{B}$ are non-empty cross-intersecting families of subsets of $[n]$, $\mathcal{A}$ is intersecting, and $a_{0},a_{1},\dots,a_{n},b_{0},b_{1},\dots,b_{n}$ are non-negative real numbers such that $a_{i}+b_{i}\geq a_{n-i}+b_{n-i}$ and $a_{n-i}\geq b_{i}$ for each $i\leq n/2$, then

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The EKR Theorem is obtained by taking $r\leq n/2$, $\mathcal{B}=\mathcal{A}\subseteq{[n]\choose r}$, and $b_{i}=0=a_{i}-1$ for each $i\in\{0\}\cup[n]$.

We use Theorem 1.2 to prove a conjecture of Feghali, Johnson and Thomas [13, Conjecture 2.1]. Before stating the conjecture, we need some further definitions and notation.

A graph $G$ is a pair $(X,\mathcal{Y})$, where $X$ is a set, called the vertex set of $G$, and $\mathcal{Y}$ is a subset of ${X\choose 2}$ and is called the edge set of $G$. The vertex set of $G$ and the edge set of $G$ are denoted by $V(G)$ and $E(G)$, respectively. An element of $V(G)$ is called a vertex of $G$, and an element of $E(G)$ is called an edge of $G$. We may represent an edge $\{v,w\}$ by $vw$. If $vw$ is an edge of $G$, then we say that $v$ is adjacent to $w$ (in $G$). A subset $I$ of $V(G)$ is an independent set of $G$ if $vw\notin E(G)$ for every $v,w\in I$. Let $\mathcal{I}_{G}$ denote the family of independent sets of $G$. An independent set $J$ of $G$ is maximal if $J\nsubseteq I$ for each independent set $I$ of $G$ such that $I\neq J$. The size of a smallest maximal independent set of $G$ is denoted by $\mu(G)$.

Holroyd and Talbot introduced the problem of determining whether ${\mathcal{I}_{G}}^{(r)}$ has the star property for a given graph $G$ and an integer $r\geq 1$. The Holroyd–Talbot (HT) Conjecture [21, Conjecture 7] claims that ${\mathcal{I}_{G}}^{(r)}$ has the star property if $\mu(G)\geq 2r$. It is proved in [3] that the conjecture is true if $\mu(G)$ is sufficiently large depending on $r$. By the EKR Theorem, it is true if $G$ has no edges. The HT Conjecture has been verified for several classes of graphs [8, 9, 18, 19, 20, 21, 22, 27, 29]. As demonstrated in [9], for $r>\mu(G)/2$, whether ${\mathcal{I}_{G}}^{(r)}$ has the star property or not depends on $G$ and $r$ (both cases are possible).

A depth-two claw is a graph consisting of $n$ pairwise disjoint edges $x_{1}y_{1},\dots,x_{n}y_{n}$ together with a vertex $x_{0}\notin\{x_{1},\dots,x_{n},y_{1},\dots,y_{n}\}$ that is adjacent to each of $y_{1},\dots,y_{n}$. This graph will be denoted by $T_{n}$. Thus, $T_{n}=(\{x_{0},x_{1},\dots,x_{n},y_{1},\dots,y_{n}\},\{x_{0}y_{1},\dots,x_{0}y_{n},x_{1}y_{1},\dots,x_{n}y_{n}\})$. For each $i\in[n]$, we may take $x_{i}$ and $y_{i}$ to be $(i,1)$ and $(i,2)$, respectively. Let $X_{n}=\{x_{i}\colon i\in[n]\}$ and

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Note that

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The family ${\mathcal{I}_{T_{n}}}^{(r)}$ is empty for $r>n+1$, and consists only of the set $\{x_{0},x_{1},\dots,x_{n}\}$ for $r=n+1$. In [13], Feghali, Johnson and Thomas showed that ${\mathcal{I}_{T_{n}}}^{(r)}$ does not have the star property for $r=n$, and they made the following conjecture.

Conjecture 1.3** ([13])**

If $r\leq n-1$, then ${\mathcal{I}_{T_{n}}}^{(r)}$ has the star property.

They proved the conjecture for $r\leq\frac{n+1}{2}$.

Theorem 1.4** ([13])**

If $n\geq 2r-1$, then ${\mathcal{I}_{T_{n}}}^{(r)}$ has the star property.

In the next section, we settle the full conjecture using Theorem 1.2 for $r>\frac{n+1}{2}$.

Theorem 1.5

Conjecture 1.3 is true.

Our proof for $r\leq\frac{n}{2}+1$ provides an alternative proof of Theorem 1.4.

2 Proofs

In this section, we prove Theorems 1.2 and 1.5.

Proof of Theorem 1.2. For each $i\in\{0\}\cup[n]$, let $c_{i}=|\mathcal{A}^{(i)}|a_{i}+|\mathcal{B}^{(i)}|b_{i}$. Since $\mathcal{A}$ and $\mathcal{B}$ are non-empty and cross-intersecting, we have $\emptyset\notin\mathcal{A}$ and $\emptyset\notin\mathcal{B}$, so $\mathcal{A}^{(0)}=\mathcal{B}^{(0)}=\emptyset$. Thus, $c_{0}=0$. The result is immediate if $n=1$. Suppose $n>1$.

Consider any positive integer $r\leq n/2$. Since $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting and $\mathcal{A}$ is intersecting, $[n]\backslash A\notin\mathcal{A}^{(n-r)}$ for each $A\in\mathcal{A}^{(r)}\cup\mathcal{B}^{(r)}$, so

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Similarly, since $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting, $|\mathcal{B}^{(n-r)}|\leq{n\choose r}-|\mathcal{A}^{(r)}|$. Thus,

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by Theorem 1.1 and the given conditions $a_{r}+b_{r}\geq a_{n-r}+b_{n-r}$ and $a_{n-r}\geq b_{r}$. Therefore, $c_{r}+c_{n-r}\leq{n-1\choose r-1}(a_{r}+b_{r})+{n-1\choose r}(a_{n-r}+b_{n-r})={n-1\choose r-1}(a_{r}+b_{r})+{n-1\choose n-r-1}(a_{n-r}+b_{n-r})$. Note that if $r=n/2$, then we have $c_{n/2}+c_{n/2}\leq{n-1\choose n/2-1}(a_{n/2}+b_{n/2})+{n-1\choose n/2-1}(a_{n/2}+b_{n/2})$, and hence $c_{n/2}\leq{n-1\choose n/2-1}(a_{n/2}+b_{n/2})$.

If $n$ is odd, then

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Similarly, if $n$ is even, then

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as required.  $\Box$

For $(i,j)\in[n]\times[2,k]$, let $\delta_{i,j}\colon\mathcal{L}_{n,k}\rightarrow\mathcal{L}_{n,k}$ be defined by

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and let $\Delta_{i,j}\colon 2^{\mathcal{L}_{n,k}}\rightarrow 2^{\mathcal{L}_{n,k}}$ be the compression operation defined by

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It is well known that $|\Delta_{i,j}(\mathcal{A})|=|\mathcal{A}|$ and that, if $\mathcal{A}$ is intersecting, then $\Delta_{i,j}(\mathcal{A})$ is intersecting. Moreover, we have the following special case of [5, Corollary 3.2].

Lemma 2.1

If $\mathcal{A}$ is an intersecting subfamily of $\mathcal{L}_{n,k}$ and

[TABLE]

then $|A\cap B\cap X_{n}|\geq 1$ for any $A,B\in\mathcal{A}^{*}$.

Proof of Theorem 1.5. The result is trivial for $0\leq r\leq 1$, so consider $2\leq r\leq n-1$. Let $\mathcal{R}={\mathcal{I}_{T_{n}}}^{(r)}$. Let $\mathcal{E}$ be an intersecting subfamily of $\mathcal{R}$. Let $\mathcal{F}=\{A\in\mathcal{R}\colon x_{1}\in A\}$. For any $\mathcal{H}\subseteq\mathcal{R}$, let $\mathcal{H}_{0}=\{H\in\mathcal{H}\colon x_{0}\notin H\}$, $\mathcal{H}_{1}=\{H\in\mathcal{H}\colon x_{0}\in H\}$ and $\mathcal{H}_{1}^{\prime}=\{H\backslash\{x_{0}\}\colon H\in\mathcal{H}_{1}\}$; by (1), $\mathcal{H}_{0}\subseteq{\mathcal{L}_{n,2}}^{(r)}$ and $\mathcal{H}_{1}^{\prime}\subseteq{X_{n}\choose r-1}$.

Case 1: $n\geq 2r-2$. Then $r\leq\frac{n}{2}+1$. Let $\gamma\colon\mathcal{R}\rightarrow\mathcal{R}$ be defined by

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and let $\Gamma\colon 2^{\mathcal{R}}\rightarrow 2^{\mathcal{R}}$ be the compression operation defined by

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Note that $\gamma(A)\neq A$ if and only if $x_{0}\in A$ and $x_{1},y_{1}\notin A$. Let $\mathcal{G}=\Gamma(\mathcal{E})$. Then $\mathcal{G}\subseteq\mathcal{R}$ and $|\mathcal{G}|=|\mathcal{E}|$. By [9, Lemma 2.1], $\mathcal{G}_{0}$ and $\mathcal{G}_{1}^{\prime}$ are intersecting. Since $r-1\leq\frac{n}{2}$ and $\mathcal{G}_{1}^{\prime}\subseteq{X_{n}\choose r-1}$, Theorem 1.1 gives us $|\mathcal{G}_{1}^{\prime}|\leq|\{A\in{X_{n}\choose r-1}\colon x_{1}\in A\}|=|\mathcal{F}_{1}^{\prime}|$. It is well known that, for any $k\geq 2$, ${\mathcal{L}_{n,k}}^{(r)}$ has the star property [11, Theorem 5.2]. Thus, $|\mathcal{G}_{0}|\leq|\{A\in{\mathcal{L}_{n,2}}^{(r)}\colon x_{1}\in A\}|=|\mathcal{F}_{0}|$. Therefore, we have $|\mathcal{E}|=|\mathcal{G}|=|\mathcal{G}_{0}|+|\mathcal{G}_{1}^{\prime}|\leq|\mathcal{F}_{0}|+|\mathcal{F}_{1}^{\prime}|=|\mathcal{F}|$.

Case 2: $n\leq 2r-3$. Then $r>\frac{n}{2}+1$. Let $\mathcal{E}_{0}^{\prime}=\Delta_{n,2}\circ\dots\circ\Delta_{1,2}(\mathcal{E}_{0})$. Thus, $\mathcal{E}_{0}^{\prime}\subseteq{\mathcal{L}_{n,2}}^{(r)}$. By Lemma 2.1,

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For any $E\in\mathcal{E}_{0}$ and any $F\in\mathcal{E}_{1}$, we have $\emptyset\neq E\cap F\subseteq X_{n}$ and $E\cap X_{n}\subseteq\delta_{i,2}(E)$ for any $i\in[n]$. It clearly follows that

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Let $\mathcal{A}=\{E\cap X_{n}\colon E\in\mathcal{E}_{0}^{\prime}\}$ and $\mathcal{B}=\mathcal{E}_{1}^{\prime}$. By (2), $\mathcal{A}$ is intersecting. By (3), $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting. Let $a_{i}={n-i\choose r-i}$ for each $i\in\{0\}\cup[r]$, and let $a_{i}=0$ for each $i\in[r+1,n]$. Since $r\leq n-1$, we have

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Let $b_{r-1}=1$ and let $b_{i}=0$ for each $i\in(\{0\}\cup[n])\backslash\{r-1\}$.

Consider any $i\leq\frac{n}{2}$. Then $i<r-1$, so $b_{i}=0\leq a_{n-i}$. If $i<n/2$, then $a_{i}>a_{n-i}$ (by (4) as $i<\frac{n}{2}<r$), so $a_{i}+b_{i}\geq a_{n-i}+b_{n-i}$. If $i=n/2$, then $n-i=i$, so $a_{i}+b_{i}=a_{n-i}+b_{n-i}$. (Note that if we ignore Case 1 and rely solely on Theorem 1.4, then we need to prove Conjecture 1.3 for $n\leq 2r-2$, that is, $r\geq\frac{n}{2}+1$. In this case, we may have $i=r-1$. Suppose $i=r-1$. Since $i\leq\frac{n}{2}\leq r-1=i$, $i=\frac{n}{2}$. Thus, $b_{i}=1=b_{n-i}$ and $a_{i}=a_{n-i}\geq 1$, and hence again $a_{i}+b_{i}=a_{n-i}+b_{n-i}$ and $a_{n-i}\geq b_{i}$.)

We have shown that $a_{i}+b_{i}\geq a_{n-i}+b_{n-i}$ and $a_{n-i}\geq b_{i}$ for each $i\leq n/2$. By Theorem 1.2,

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where $\mathcal{X}_{n}=\{A\subseteq X_{n}\colon x_{1}\in A\}$. Now

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and $|\mathcal{E}_{1}|=|\mathcal{B}|=\sum_{B\in\mathcal{B}}b_{r-1}=\sum_{B\in\mathcal{B}}b_{|B|}$. Clearly, $|\mathcal{F}_{0}|=\sum_{A\in\mathcal{X}_{n}}a_{|A|}$ and $|\mathcal{F}_{1}|=\sum_{B\in\mathcal{X}_{n}}b_{|B|}$. We have

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Hence the result.  $\Box$