Mixed pairwise cross intersecting families (I)^†^†thanks: This work is supported by NSFC (Grant No. 11931002). E-mail addresses: [email protected] (Yang Huang), [email protected] (Yuejian Peng, corresponding author).

Yang Huang, Yuejian Peng^†
School of Mathematics, Hunan University
Changsha, Hunan, 410082, P.R. China

Abstract

Families $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting if $A\cap B\neq\emptyset$ for any $A\in\mathcal{A}$ and $B\in\mathcal{B}$ . If $n<k+l$ , all families $\mathcal{A}\subseteq{[n]\choose k}$ and $\mathcal{B}\subseteq{[n]\choose l}$ are cross intersecting and we say that $\mathcal{A}$ and $\mathcal{B}$ are cross intersecting freely. An $(n,k_{1},\dots,k_{t})$ -cross intersecting system is a set of non-empty pairwise cross-intersecting families $\mathcal{F}_{1}\subset{[n]\choose k_{1}},\mathcal{F}_{2}\subset{[n]\choose k_{2}},\dots,\mathcal{F}_{t}\subset{[n]\choose k_{t}}$ with $t\geq 2$ and $k_{1}\geq k_{2}\geq\cdots\geq k_{t}$ . If an $(n,k_{1},\dots,k_{t})$ -cross intersecting system contains at least two families which are cross intersecting freely and at least two families which are cross intersecting but not freely, then we say that the cross intersecting system is of mixed type. All previous studies are on non-mixed type, i.e, under the condition that $n\geq k_{1}+k_{2}$ . In this paper, we study for the first interesting mixed type, an $(n,k_{1},\dots,k_{t})$ -cross intersecting system with $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , i.e., families $\mathcal{F}_{i}\subseteq{[n]\choose k_{i}}$ and $\mathcal{F}_{j}\subseteq{[n]\choose k_{j}}$ are cross intersecting freely if and only if $\{i,j\}=\{1,2\}$ . Let $M(n,k_{1},\dots,k_{t})$ denote the maximum sum of sizes of families in an $(n,k_{1},\dots,k_{t})$ -cross intersecting system. We determine $M(n,k_{1},\dots,k_{t})$ and characterize all extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting systems for $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ .

A result of Kruskal-Katona allows us to consider only families $\mathcal{F}_{i}$ whose elements are the first $|\mathcal{F}_{i}|$ elements in lexicographic order (we call them L-initial families). Since $n<k_{1}+k_{2}$ , $\mathcal{F}_{1}\subseteq{[n]\choose k_{1}}$ and $\mathcal{F}_{2}\subseteq{[n]\choose k_{2}}$ are cross intersecting freely. Thus, when we try to bound $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}$ by a function, there are two free variables $I_{1}$ (the last element of $\mathcal{F}_{1}$ ) and $I_{2}$ (the last element of $\mathcal{F}_{2}$ ). This causes more difficulty to analyze properties of the corresponding function comparing to non-mixed type problems (single-variable function). To overcome this difficulty, we introduce new concepts ‘ $k$ -partner’ and ‘parity’, develop some rules to determine whether two L-initial cross intersecting families are maximal to each other, and prove one crucial property that in an extremal L-initial $(n,k_{1},\dots,k_{t})$ -cross intersecting system ( $\mathcal{F}_{1}$ , $\mathcal{F}_{2}$ , $\cdots$ , $\mathcal{F}_{t}$ ), the last element of $\mathcal{F}_{1}$ and the last element of $\mathcal{F}_{2}$ are ‘parities’ (see Section 2 for the definition) to each other. This discovery allows us to bound $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}$ by a single variable function $g(I_{2})$ , where $I_{2}$ is the last element of $\mathcal{F}_{2}$ . Another crucial and challenge part is to verify that $-g(I_{2})$ has unimodality. Comparing to the non-mixed type, we need to overcome more difficulties in showing the unimodality of function $-g(I_{2})$ since there are more terms to be taken care of. We think that the characterization of maximal cross intersecting L-initial families and the unimodality of functions in this paper are interesting in their own, in addition to the extremal result. The most general condition on $n$ is that $n\geq k_{1}+k_{t}$ (If $n<k_{1}+k_{t}$ , then $\mathcal{F}_{1}$ is cross intersecting freely with any other $\mathcal{F}_{i}$ , and consequently $\mathcal{F}_{1}={[n]\choose k_{1}}$ in an extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting system and we can remove $\mathcal{F}_{1}$ .). This paper provides foundation work for the solution to the most general condition $n\geq k_{1}+k_{t}$ .

Key words: Cross intersecting families; Extremal finite sets

2010 Mathematics Subject Classification. 05D05, 05C65, 05D15.

1 Introduction

Let $[n]=\{1,2,\dots,n\}$ . For $0\leq k\leq n$ , let ${[n]\choose k}$ denote the family of all $k$ -subsets of $[n]$ . A family $\mathcal{A}$ is $k$ -uniform if $\mathcal{A}\subset{[n]\choose k}$ . A family $\mathcal{A}$ is intersecting if $A\cap B\neq\emptyset$ for any $A$ and $B\in\mathcal{A}$ . Many researches in extremal set theory are inspired by the foundational result of Erdős–Ko–Rado [1] showing that a maximum $k$ -uniform intersecting family is a full star. This result of Erdős–Ko–Rado has many interesting generalizations. Two families $\mathcal{A}$ and $\mathcal{B}$ are cross-intersecting if $A\cap B\neq\emptyset$ for any $A\in\mathcal{A}$ and $B\in\mathcal{B}$ . Note that $\mathcal{A}$ and $\mathcal{A}$ are cross-intersecting is equivalent to that $\mathcal{A}$ is intersecting. We call $t$ $(t\geq 2)$ families $\mathcal{A}_{1},\mathcal{A}_{2},\dots,\mathcal{A}_{t}$ pairwise cross-intersecting families if $\mathcal{A}_{i}$ and $\mathcal{A}_{j}$ are cross-intersecting when $1\leq i<j\leq t$ . Additionally, if $\mathcal{A}_{j}\neq\emptyset$ for each $j\in[t]$ , then we say that $\mathcal{A}_{1},\mathcal{A}_{2},\dots,\mathcal{A}_{t}$ are non-empty pairwise cross-intersecting. For given integers $n,t,k_{t},\dots,k_{t}$ , if families $\mathcal{A}_{1}\subseteq{[n]\choose k_{1}},\dots,\mathcal{A}_{t}\subseteq{[n]\choose k_{t}}$ are non-empty pairwise cross intersecting and $k_{1}\geq k_{2}\dots\geq k_{t}$ , then we call $(\mathcal{A}_{1},\dots,\mathcal{A}_{t})$ an $(n,k_{1},\dots,k_{t})$ -cross intersecting system.

Define

	$\displaystyle M(n,k_{1},\dots,k_{t})$	$\displaystyle=\max\bigg{\{}\sum_{i=1}^{t}\|\mathcal{A}_{i}\|:(\mathcal{A}_{1},\dots,\mathcal{A}_{t})\text{ is an }(n,k_{1},\dots,k_{t})\text{-cross}$
		$\displaystyle\quad\quad\quad\quad\text{intersecting system }\bigg{\}}.$		(1)

We say that an $(n,k_{1},\dots,k_{t})$ -cross intersecting system $(\mathcal{A}_{1},\dots,\mathcal{A}_{t})$ is extremal if $\sum_{i=1}^{t}|\mathcal{A}_{i}|=M(n,k_{1},\dots,k_{t})$ . In [5] (Theorem 1.1), the authors determined $M(n,k_{1},\\ \dots,k_{t})$ for $n\geq k_{1}+k_{2}$ and characterized the extremal families attaining the bound.

Theorem 1.1 (Huang-Peng [5]).

Let $\mathcal{A}_{1}\subset{[n]\choose k_{1}},\mathcal{A}_{2}\subset{[n]\choose k_{2}},\dots,\mathcal{A}_{t}\subset{[n]\choose k_{t}}$ be non-empty pairwise cross intersecting families with $t\geq 2$ , $k_{1}\geq k_{2}\geq\cdots\geq k_{t}$ , and $n\geq k_{1}+k_{2}$ . Then

\sum_{i=1}^{t}{|\mathcal{A}_{i}|}\leq\textup{max}\left\{{n\choose k_{1}}-{n-k_{t}\choose k_{1}}+\sum_{i=2}^{t}{{n-k_{t}\choose k_{i}-k_{t}}},\,\,\sum_{i=1}^{t}{n-1\choose k_{i}-1}\right\}.

The equality holds if and only if one of the following holds.
(i) ${n\choose k_{1}}-{n-k_{t}\choose k_{1}}+\sum_{i=2}^{t}{{n-k_{t}\choose k_{i}-k_{t}}}>\sum_{i=1}^{t}{n-1\choose k_{i}-1}$ , and there is some $k_{t}$ -element set $T\subset[n]$ such that $\mathcal{A}_{1}=\{F\in{[n]\choose k_{1}}:F\cap T\neq\emptyset\}$ and $\mathcal{A}_{j}=\{F\in{[n]\choose k_{j}}:T\subset F\}$ for each $j\in[2,t]$ ;
(ii) ${n\choose k_{1}}-{n-k_{t}\choose k_{1}}+\sum_{i=2}^{t}{{n-k_{t}\choose k_{i}-k_{t}}}\leq\sum_{i=1}^{t}{n-1\choose k_{i}-1}$ , there are some $i\neq j$ such that $n>k_{i}+k_{j}$ , and there is some $a\in[n]$ such that $\mathcal{A}_{j}=\{F\in{[n]\choose k_{j}}:a\in F\}$ for each $j\in[t]$ ;
(iii) $t=2,n=k_{1}+k_{2}$ , $\mathcal{A}_{1}\subset{[n]\choose k_{1}}$ and $\mathcal{A}_{2}={[n]\choose k_{2}}\setminus\overline{\mathcal{A}_{1}}$ ;
(iv) $t\geq 3,k_{1}=k_{2}=\cdots=k_{t}=k,n=2k$ , fix some $i\in[t]$ , $\mathcal{A}_{j}=\mathcal{A}$ for all $j\in[t]\setminus\{i\}$ , where $\mathcal{A}$ is an intersecting family with size ${n-1\choose k-1}$ , and $\mathcal{A}_{i}={[n]\choose k}\setminus\overline{\mathcal{A}}$ .

Theorem 1.1 generalized the results of Hilton-Milner in [8], Frankl-Tokushige in [4] and Shi-Qian-Frankl in [13]. In Theorem 1.1, taking $t=2$ and $k_{1}=k_{2}$ , we obtain the result of Hilton-Milner in [8]. Taking $t=2$ , we obtain the result of Frankl-Tokushige in [4]. Taking $k_{1}=k_{2}=\dots=k_{t}$ , we obtain the result of Shi-Qian-Frankl in [13]. Note that if $n<k+\ell$ , any family $\mathcal{A}\subseteq{[n]\choose k}$ and any family $\mathcal{B}\subseteq{[n]\choose\ell}$ are cross intersecting. In this case, we say that families $\mathcal{A}\subseteq{[n]\choose k}$ and $\mathcal{B}\subseteq{[n]\choose\ell}$ are cross intersecting freely. If an $(n,k_{1},\dots,k_{t})$ -cross intersecting system contains at least two families which are cross intersecting freely and at least two families which are cross intersecting but not freely, then we say that the cross intersecting system is of mixed type. Otherwise, we say that it is of non-mixed type. All previous studies are of non-mixed type, i.e., $n\geq k_{1}+k_{2}$ . It would be interesting to study for mixed type. The first interesting case is to study an $(n,k_{1},\dots,k_{t})$ -cross intersecting system $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ for $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ . In this case, $\mathcal{F}_{i}\subseteq{[n]\choose k_{i}}$ and $\mathcal{F}_{j}\subseteq{[n]\choose k_{j}}$ are cross intersecting freely if and only if $\{i,j\}=\{1,2\}$ . In this paper, we focus on this case, we determine $M(n,k_{1},\dots,k_{t})$ and characterize all extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting systems. Let us look at the following $(n,k_{1},\dots,k_{t})$ -cross intersecting systems.

Construction 1.2.

Let $k_{1}\geq k_{2}\geq\dots\geq k_{t}$ and $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ . For each $i\in[t]$ , we denote

\mathcal{G}_{i}=\left\{G\in{[n]\choose k_{i}}:1\in G\right\}.

Note that

\sum_{i=1}^{t}|\mathcal{G}_{i}|=\sum_{i=1}^{t}{n-1\choose k_{i}-1}=:\lambda_{1}.

Construction 1.3.

Let $k_{1}\geq k_{2}\geq\dots\geq k_{t}$ and $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ . For $i=1$ or $2$ , let

\mathcal{H}_{i}=\left\{H\in{[n]\choose k_{i}}:H\cap[k_{t}]\neq\emptyset\right\},

and for $i\in[3,t]$ , let

\mathcal{H}_{i}=\left\{H\in{[n]\choose k_{i}}:[k_{t}]\subseteq H\right\}.

Note that

\sum_{i=1}^{t}|\mathcal{H}_{i}|=\sum_{i=1}^{2}\left({n\choose k_{i}}-{n-k_{t}\choose k_{i}}\right)+\sum_{i=3}^{t}{{n-k_{t}\choose k_{i}-k_{t}}}=:\lambda_{2}.

Our main result is that an extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting system must be isomorphic to Construction 1.2 or Construction 1.3 if $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ .

Theorem 1.4.

Let $\mathcal{F}_{1}\subset{[n]\choose k_{1}},\mathcal{F}_{2}\subset{[n]\choose k_{2}},\dots,\mathcal{F}_{t}\subset{[n]\choose k_{t}}$ be non-empty pairwise cross intersecting families with $t\geq 3$ , $k_{1}\geq k_{2}\geq\cdots\geq k_{t}$ and $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ . Then

\sum_{i=1}^{t}{|\mathcal{F}_{i}|}\leq\textup{max}\left\{\sum_{i=1}^{t}{n-1\choose k_{i}-1},\,\,\sum_{i=1}^{2}\left({n\choose k_{i}}-{n-k_{t}\choose k_{i}}\right)+\sum_{i=3}^{t}{{n-k_{t}\choose k_{i}-k_{t}}}\right\}.

The equality holds if and only if $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is isomorphic to $(\mathcal{G}_{1},\dots,\mathcal{G}_{t})$ in Construction 1.2 or $(\mathcal{H}_{1},\dots,\mathcal{H}_{t})$ in Construction 1.3.

In proving Theorem 1.1 in [5], the authors applied a result of Kruskal-Katona (Theorem 2.1) allowing us to consider only families $\mathcal{F}_{i}$ whose elements are the first $|\mathcal{F}_{i}|$ elements in lexicographic order (we call them L-initial families). We bounded $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}$ by a function $f(R)$ of the last element $R$ in the lexicographic order of an L-initial family $\mathcal{F}_{1}$ ( $R$ is called the ID of $\mathcal{F}_{1}$ ), and showed that $-f(R)$ has unimodality. To prove our main result (Theorem 1.4) in this paper, by the result of Kruskal-Katona (Theorem 2.1), we can still consider only pairwise cross-intersecting non-empty L-initial families $\mathcal{F}_{1}\subset{[n]\choose k_{1}},\mathcal{F}_{2}\subset{[n]\choose k_{2}},\dots,\mathcal{F}_{t}\subset{[n]\choose k_{t}}$ . However, in this paper, the condition on $n$ is relaxed to $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , so $\mathcal{F}_{1}\subseteq{[n]\choose k_{1}}$ and $\mathcal{F}_{2}\subseteq{[n]\choose k_{2}}$ are cross intersecting freely. When we try to bound $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}$ by a function, there are two free variables $I_{1}$ (the ID of $\mathcal{F}_{1}$ ) and $I_{2}$ (the ID of $\mathcal{F}_{2}$ ). This causes more difficulty to analyze properties of the corresponding function, comparing to the problem in [5]. To overcome this difficulty, we introduce new concepts ‘ $k$ -partner’ and ‘parity’, develop some rules to determine whether a pair of L-initial cross intersecting families are maximal to each other (see precise definition), and prove one crucial property that for an extremal L-initial $(n,k_{1},\dots,k_{t})$ -cross intersecting system ( $\mathcal{F}_{1}$ , $\mathcal{F}_{2}$ , $\cdots$ , $\mathcal{F}_{t}$ ), the ID $I_{1}$ of $\mathcal{F}_{1}$ and the ID $I_{2}$ of $\mathcal{F}_{2}$ are ‘parities’ to each other (Lemma 3.8). This discovery allows us to bound $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}$ by a single variable function $g(I_{2})$ . Another crucial and challenge part is to verify the unimodality of $-g(I_{2})$ (Lemmas 3.12, 3.13 and 3.14). Comparing to the function in [5], we need to overcome more difficulties in dealing with function $-g(I_{2})$ since there are more ‘mysterious’ terms to be taken care of. We take advantage of some properties of function $f(R)$ obtained in [5] and come up with some new strategies in estimating the change $g(I^{\prime}_{2})-g(I_{2})$ as the ID of $\mathcal{F}_{2}$ increases from $I_{2}$ to $I^{\prime}_{2}$ (Sections 6 and 7).

The most general condition on $n$ is that $n\geq k_{1}+k_{t}$ . If $n<k_{1}+k_{t}$ , then $\mathcal{F}_{1}$ is cross intersecting freely with any other $\mathcal{F}_{i}$ , and consequently $\mathcal{F}_{1}={[n]\choose k_{1}}$ in an extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting system and $\mathcal{F}_{1}$ can be removed. The most general case is more complex and we will deal with it in a forthcoming paper [7]. The work in this paper provides important foundation for the solution to the most general condition $n\geq k_{1}+k_{t}$ .

To obtain the relationship between the ID of $\mathcal{F}_{1}$ and the ID of $\mathcal{F}_{2}$ , we build some foundation work in Section 2, for example, we come up with new concepts ‘ $k$ -partners’ and ‘parity’, and give a necessary and sufficient condition on maximal cross intersecting L-initial families. In Section 3, we give the proof of Theorem 1.4 by assuming the truth of Lemmas 3.8, 3.12, 3.13 and 3.14. In Section 4, we list some results obtained for non-mixed type in [5] which we will apply. In Sections 6 and 7, we give the proofs of Lemmas 3.8, 3.12, 3.13 and 3.14.

2 Partner and Parity

In this section, we introduce new concepts ‘ $k$ -partner’ and ‘parity’, develop some rules to determine whether a pair of L-initial cross intersecting families are maximal to each other (precise definitions will be given in this section). We prove some properties which are foundation for the proof of our main results.

When we write a set $A=\{a_{1},a_{2},\ldots,a_{s}\}\subset[n]$ , we always assume that $a_{1}<a_{2}<\ldots<a_{s}$ throughout the paper. Let $\max A$ denote the maximum element of $A$ , let $\min A$ denote the minimum element of $A$ and $(A)_{i}$ denote the $i$ -th element of $A$ . Let us introduce the lexicographic (lex for short) order of subsets of positive integers. Let $A$ and $B$ be finite subsets of the set of positive integers $\mathbb{Z}_{>0}$ . We say that $A\prec B$ if either $A\supset B$ or $\min(A\setminus B)<\min(B\setminus A)$ . In particular, $A\prec A$ . Let $A$ and $B$ in ${[n]\choose k}$ with $A\precneqq B$ . We write $A<B$ if there is no other $C\in{[n]\choose k}$ such that $A\precneqq C\precneqq B$ .

Let $\mathcal{L}(n,r,k)$ denote the first $r$ subsets in ${[n]\choose k}$ in the lex order. Given a set $R$ , we denote $\mathcal{L}(n,R,k)=\{F\in{[n]\choose k}:F\prec R\}$ . Whenever the underlying set $[n]$ is clear, we shall ignore it and write $\mathcal{L}(R,k)$ , $\mathcal{L}(r,k)$ for short. Let $\mathcal{F}\subset{[n]\choose k}$ be a family, we say $\mathcal{F}$ is L-initial if $\mathcal{F}=\mathcal{L}(R,k)$ for some $k$ -set $R$ . We call $R$ the ID of $\mathcal{F}$ .

The well-known Kruskal-Katona theorem [11, 12] will allow us to consider only L-initial families. An equivalent formulation of this result was given in [3, 9] as follows.

Theorem 2.1 (Kruskal-Katona theorem [3, 9]).

For $\mathcal{A}\subset{[n]\choose k}$ and $\mathcal{B}\subset{[n]\choose l}$ , if $\mathcal{A}$ and $\mathcal{B}$ are cross intersecting, then $\mathcal{L}(|\mathcal{A}|,k)$ and $\mathcal{L}(|\mathcal{B}|,l)$ are cross intersecting as well.

In [5], we proved the following important result.

Proposition 2.2.

[5] Let $a,b,n$ are positive integers and $a+b\leq n$ . For $P\subset[n]$ with $|P|\leq a$ , let $Q$ be the partner of $P$ . Then $\mathcal{L}(Q,b)$ is the maximum L-initial $b$ -uniform family that are cross intersecting to $\mathcal{L}(P,a)$ .

In [5], we worked on non-mixed type: Let $t\geq 2$ , $k_{1}\geq k_{2}\geq\cdots\geq k_{t}$ and $n\geq k_{1}+k_{2}$ and families $\mathcal{A}_{1}\subset{[n]\choose k_{1}},\mathcal{A}_{2}\subset{[n]\choose k_{2}},\dots,\mathcal{A}_{t}\subset{[n]\choose k_{t}}$ be non-empty pairwise cross-intersecting (not freely). Let $R$ be the ID of $\mathcal{A}_{1}$ , and $T$ be the partner of $R$ . In the proof of Theorem 1.1, one important ingredient is that by Proposition 2.2, $\sum_{i=1}^{t}{|\mathcal{A}_{i}|}$ can be bounded by a function of $R$ as following.

\displaystyle f(R)=\sum_{j=1}^{t}|\mathcal{A}_{j}|\leq|\mathcal{L}(R,k_{1})|+\sum_{j=2}^{t}|\mathcal{L}(T,k_{j})|.

(2)

By Theorem 2.1, to prove the quantitative part of Theorem 1.4 we may also assume that $\mathcal{F}_{i}$ is L-initial, that is, $\mathcal{F}_{i}=\mathcal{L}(|\mathcal{F}_{i}|,k_{i})$ for each $i\in[t]$ . In the rest of this chapter, we assume that $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is an extremal non-empty $(n,k_{1},\dots,k_{t})$ -cross intersecting system with $t\geq 3$ , $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , and $\mathcal{F}_{j}$ is L-initial for each $j\in[t]$ with ID $I_{j}$ .

However, the condition on $n$ is relaxed to $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , so $\mathcal{F}_{1}\subseteq{[n]\choose k_{1}}$ and $\mathcal{F}_{2}\subseteq{[n]\choose k_{2}}$ are cross intersecting freely. When we try to bound $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}$ by a function, there are two free variables $I_{1}$ (the ID of $\mathcal{F}_{1}$ ) and $I_{2}$ (the ID of $\mathcal{F}_{2}$ ). This causes more difficulty to analyze properties of the corresponding function, comparing to the problem in [5]. To overcome this difficulty, we introduce new concepts ’ $k$ -partner’ and ’parity’, develop some rules to determine whether a pair of L-initial cross intersecting families are maximal to each other (see precise definition).

Let $\mathcal{F}\subseteq{[n]\choose f}$ and $\mathcal{G}\subseteq{[n]\choose g}$ be cross intersecting families. We say that $(\mathcal{F},\mathcal{G})$ is maximal or $\mathcal{F}$ and $\mathcal{G}$ are maximal cross intersecting families if whenever $\mathcal{F}^{\prime}\subseteq{[n]\choose f}$ and $\mathcal{G}^{\prime}\subseteq{[n]\choose g}$ are cross intersecting with $\mathcal{F}\subseteq\mathcal{F}^{\prime}$ and $\mathcal{G}\subseteq\mathcal{G}^{\prime}$ , then $\mathcal{F}=\mathcal{F}^{\prime}$ and $\mathcal{G}=\mathcal{G}^{\prime}$ .

Let $F$ and $G$ be two subsets of $[n]$ . We say $(F,G)$ is maximal if there are two L-initial families $\mathcal{F}\subseteq{[n]\choose|F|}$ and $\mathcal{G}\subseteq{[n]\choose|G|}$ with IDs $F$ and $G$ respectively such that $(\mathcal{F},\mathcal{G})$ is maximal. We say two families $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are maximal pair families if $|\mathcal{A}_{1}|=|\mathcal{A}_{2}|$ and for every $A_{1}\in\mathcal{A}_{1}$ , there is a unique $A_{2}\in\mathcal{A}_{2}$ such that $(A_{1},A_{2})$ is maximal.

Let $F=\{x_{1},x_{2},\dots,x_{k}\}\subseteq[n]$ . We denote

\ell(F)=\begin{cases}\max\{x:[n-x+1,n]\subseteq F\},&\text{if $\max F=n$};\\ 0,&\text{if $\max F<n$}.\end{cases}

Let $F\subseteq[n]$ be a set. We denote

F^{\mathrm{t}}=\begin{cases}\emptyset,&\text{if $\ell(F)=0$};\\ [n-\ell(F)+1,n],&\text{if $\ell(F)\geq 1$}.\end{cases}

Let $F$ and $H$ be two subsets of $[n]$ with size $|F|=f$ and $|H|=h$ . We say that $F$ and $H$ strongly intersect at their last element if there is an element $q$ such that $F\cap H=\{q\}$ and $F\cup H=[q]$ . We also say $F$ is the partner of $H$ , or $H$ is the partner of $F$ . Let $k\leq n-f$ be an integer, we define the $k$ -partner $K$ of $F$ as follows. For $k=h$ , let $K=H$ . If $k>h$ , then let $K=H\cup\{n-k+h+1,\dots,n\}$ . We can see that $|K|=k$ . Indeed, since $F$ and $H$ intersect at their last element, $n^{\prime}=\max H=f+h-1<n-k+h+1$ , so $|K|=|H\cup\{n-k+h+1,\dots,n\}|=k$ . If $k<h$ , then let $K$ be the last $k$ -set in ${[n]\choose k}$ such that $K\prec H$ , in other words, there is no $k$ -set $K^{\prime}$ satisfying $K\precneqq K^{\prime}\precneqq H$ . By the definition of $k$ -partner, we have the following remark.

Remark 2.3.

Let $F\subseteq[n]$ with $|F|=f$ and $k\leq n-f$ . Suppose that $H$ is the partner of $F$ , and $K$ is the k-partner of $F$ , then we have $\mathcal{L}(H,k)=\mathcal{L}(K,k)$ .

Fact 2.4.

Let $F\subseteq[n]$ with $|F|=f$ and $k\leq n-f$ . Then the $k$ -partner of $F$ is the same as the $k$ -partner of $F\setminus F^{\mathrm{t}}$ .

Proof.

If $\ell(F)=0$ , then we are fine. Suppose $\ell(F)>0$ and $F=\{x_{1},\dots,x_{y}\}\cup\{n-\ell(F)+1,\dots,n\}$ . Let $H$ and $H^{\prime}$ be the partners of $F$ and $F\setminus\{n-\ell(F)+1,\dots,n\}$ respectively. Then $|H|>n-f$ , consequently $k<|H|$ , $H=H\cap[x_{y}-1]\cup[x_{y}+1,n-\ell(F)]\cup\{n\}$ and $H^{\prime}=H\cap[x_{y}-1]\cup\{x_{y}\}$ . Suppose that $K$ and $K^{\prime}$ are the $k$ -partners of $F$ and $F\setminus\{n-\ell(F)+1,\dots,n\}$ respectively. By the definition of $k$ -partner, we can see that if $k\leq|H\cap[x_{y}-1]|$ , then $K=K^{\prime}$ , as desired; if $k=|H^{\prime}|$ , then $K^{\prime}=H^{\prime}$ and $K=H\cap[x_{y}-1]\cup\{x_{y}\}=H^{\prime}$ , as desired; if $k>|H^{\prime}|$ , then $K^{\prime}=H^{\prime}\cup[n-k+|H^{\prime}|+1,n]$ and $K=H\cap[x_{y}-1]\cup\{x_{y}\}\cup[n-k+|H^{\prime}|+1,n]=K^{\prime}$ , as desired. ∎

By Remark 2.3 and Proposition 2.2, we have the following fact.

Fact 2.5.

Let $a,b,n$ be positive integers and $n\geq a+b$ . For $A\subset[n]$ with $|A|=a$ , let $B$ be the $b$ -partner of $A$ , then $\mathcal{L}(B,b)$ is the maximum L-initial $b$ -uniform family that are cross intersecting to $\mathcal{L}(A,a)$ . We also say that $\mathcal{L}(B,b)$ is maximal to $\mathcal{L}(A,a)$ , or say $B$ is maximal to $A$ .

Remark 2.6.

Note that families $\mathcal{L}(A,a)$ and $\mathcal{L}(B,b)$ which mentioned in Fact 2.5 may not be maximal cross intersecting, since we don’t know whether $\mathcal{L}(A,a)$ is maximal to $\mathcal{L}(B,b)$ . For example, let $n=9$ , $a=3$ , $b=4$ and $A=\{2,4,7\}$ . Then the $b$ -partner of $A$ is $\{1,3,4,9\}$ . Although $\mathcal{L}(\{1,3,4,9\},4)$ is maximal to $\mathcal{L}(\{2,4,7\},3)$ , $\mathcal{L}(\{1,3,4,9\},4)$ and $\mathcal{L}(\{2,4,7\},3)$ are not maximal cross intersecting families since $\mathcal{L}(\{2,4,7\},3)\subsetneq\mathcal{L}(\{2,4,9\},3)$ , and $\mathcal{L}(\{2,4,9\},3)$ and $\mathcal{L}(\{1,3,4,9\},4)$ are cross intersecting families.

Frankl-Kupavskii [2] gave a sufficient condition for a pair of maximal cross intersecting families, and a necessary condition for a pair of maximal cross intersecting families as stated below.

Proposition 2.7 (Frankl-Kupavskii [2]).

Let $a,b\in\mathbb{Z}_{>0},a+b\leq n$ . Let $P$ and $Q$ be non-empty subsets of $[n]$ with $|P|\leq a$ and $|Q|\leq b$ . If $Q$ is the partner of $P$ , then $\mathcal{L}(P,a)$ and $\mathcal{L}(Q,b)$ are maximal cross intersecting families. Inversely, if $\mathcal{L}(A,a)$ and $\mathcal{L}(B,a)$ are maximal cross intersecting families, let $j$ be the smallest element of $A\cap B$ , $P=A\cap[j]$ and $Q=B\cap[j]$ . Then $\mathcal{L}(P,a)=\mathcal{L}(A,a)$ , $\mathcal{L}(Q,b)=\mathcal{L}(B,b)$ and $P$ , $Q$ satisfy the following conditions: $|P|\leq a$ , $|Q|\leq b$ , and $Q$ is the partner of $P$ .

Based on Proposition 2.7, we point out a necessary and sufficient condition for a pair of maximal cross intersecting families in terms of their IDs.

Fact 2.8.

Let $A$ and $B$ be nonempty subsets of $[n]$ with $|A|+|B|\leq n$ . Let $A^{\prime}=A\setminus A^{\mathrm{t}}$ and $B^{\prime}=B\setminus B^{\mathrm{t}}$ . Then $(A,B)$ is maximal if and only if $A^{\prime}$ is the partner of $B^{\prime}$ .

Proof.

Let $|A|=a$ and $|B|=b$ . From the definitions of $A^{\prime}$ and $B^{\prime}$ , we can see that $\mathcal{L}(A,a)=\mathcal{L}(A^{\prime},a)$ and $\mathcal{L}(B,b)=\mathcal{L}(B^{\prime},b)$ . First we show the sufficiency. Suppose that $A^{\prime}$ is the partner of $B^{\prime}$ . Since $|A^{\prime}|\leq|A|$ and $|B^{\prime}|\leq|B|$ , by Proposition 2.7, $\mathcal{L}(A^{\prime},a)$ and $\mathcal{L}(B^{\prime},b)$ are maximal cross intersecting families. Thus $\mathcal{L}(A,a)$ and $\mathcal{L}(B,b)$ are maximal cross intersecting families, in other words, $(A,B)$ is maximal. Next, we show the necessity. Suppose that $(A,B)$ is maximal. Let $j$ be the smallest element of $A\cap B$ , $P=A\cap[j]$ and $Q=B\cap[j]$ . By Proposition 2.7, $\mathcal{L}(A,a)=\mathcal{L}(P,a)$ , $\mathcal{L}(B,b)=\mathcal{L}(Q,b)$ ; $|P|\leq a$ , $|Q|\leq b$ and $P$ is the partner of $Q$ . Since $\mathcal{L}(A,a)=\mathcal{L}(P,a)$ and $P\subseteq A$ , then we $A=P\cup\{n-a+|P|+1,\dots,n\}$ . Similarly, $B=Q\cup\{n-b+|Q|+1,\dots,n\}$ . By the definitions of $A^{\prime}$ and $B^{\prime}$ , we have $A^{\prime}=P$ and $B^{\prime}=Q$ . Since $P$ is the partner of $Q$ , $A^{\prime}$ is the partner of $B^{\prime}$ . ∎

By Fact 2.8 and the definition of the $k$ -partner, we have the following property.

Fact 2.9.

Let $a,b,k,n$ be integers with $n\geq\max\{a+b,a+k\}$ . Let $A$ be an $a$ -subset of $[n]$ . Suppose that $K$ is the $k$ -partner of $A\setminus A^{\mathrm{t}}$ and there exists a $b$ -set $B$ such that $(A,B)$ is maximal and let $b^{\prime}=|B\setminus B^{\mathrm{t}}|$ . Then $(A,K)$ is maximal if and only if $k\geq b^{\prime}$ .

Fact 2.10.

Let $a,b,n$ be positive integers and $n\geq a+b$ . Suppose that $A$ is an $a$ -subset of $[n]$ , and $B$ is the $b$ -partner of $A$ . Let $A^{\prime}$ be the $a$ -partner of $B$ , then $(A^{\prime},B)$ is maximal. Moreover, if $A^{\prime}\neq A$ , then $A\precneqq A^{\prime}$ .

Proof.

If $(A,B)$ is maximal, then $A$ is the $a$ -partner of $B$ and $A^{\prime}=A$ , we are fine. Suppose that $(A,B)$ is not maximal. By Fact 2.5, $B$ is maximal to $A$ , so $A$ is not maximal to $B$ . Let $A^{\prime}$ be the $a$ -partner of $B$ . By Fact 2.5 again, $A^{\prime}$ is maximal to $B$ and $A\precneqq A^{\prime}$ . Since $B$ is maximal to $A$ , for any $b$ -set $B^{\prime}$ satisfying $B\precneqq B^{\prime}$ , we have $\mathcal{L}(B^{\prime},b)$ and $\mathcal{L}(A^{\prime},a)$ are not cross intersecting. So $B$ is maximal to $A^{\prime}$ . Hence $(A^{\prime},B)$ is maximal. ∎

Definition 2.11.

Let $h_{1}\leq h_{2}$ be positive integers, $H_{1}$ and $H_{2}$ be subsets of $[n]$ with sizes $h_{1}$ and $h_{2}$ respectively. We say $H_{1}$ is the $h_{1}$ -parity of $H_{2}$ , or $H_{2}$ is the $h_{2}$ -parity of $H_{1}$ if $H_{1}\setminus H_{1}^{\rm t}=H_{2}\setminus H_{2}^{\rm t}$ and $\ell(H_{2})-\ell(H_{1})=h_{2}-h_{1}$ .

Remark 2.12.

Let $d\leq f\leq h$ be positive integers and $F\subseteq[n]$ with $|F|=f$ . Then $F$ has an $h$ -parity if and only if $h-f\leq n-\ell(F)-\max(F\setminus F^{\rm t})-1$ , and $F$ has an $d$ -parity if and only if $d\geq f-\ell(F)$ .

Note that for a given integer $k$ and a subset $A\subseteq[n]$ , if $A$ has a $k$ -parity, then it has the unique one. The following fact is derived from the above definition directly.

Fact 2.13.

Let $h_{1}\leq h_{2}\leq h_{3}$ and $H_{i}$ be an $h_{i}$ -set for $i\in[3]$ . If $H_{1}$ is the $h_{1}$ -parity of $H_{2}$ and $H_{2}$ is the $h_{2}$ -parity of $H_{3}$ , then $H_{1}$ is the $h_{1}$ -parity of $H_{3}$ . Also, if $H_{3}$ is the $h_{3}$ -parity of $H_{1}$ and $H_{2}$ is the $h_{2}$ -parity of $H_{1}$ , then $H_{3}$ is the $h_{3}$ -parity of $H_{2}$ .

Fact 2.14.

Let $a,b,k,n$ be positive integers and $n\geq\max\{a+k,b+k\}$ . Let $A$ and $B$ be two subsets of $[n]$ with sizes $|A|=a$ and $|B|=b$ . Suppose that $K_{a}$ and $K_{b}$ are the $k$ -partners of $A$ and $B$ respectively. If $A\prec B$ , then $K_{b}\prec K_{a}$ . In particular, if $A$ is the $a$ -parity of $B$ , then $K_{b}=K_{a}$ .

Definition 2.15.

Let $h_{1}\leq h_{2}$ . For two families $\mathcal{H}_{1}\subseteq{[n]\choose h_{1}}$ and $\mathcal{H}_{2}\subseteq{[n]\choose h_{2}}$ , we say that $\mathcal{H}_{1}$ is the $h_{1}$ -parity of $\mathcal{H}_{2}$ , or $\mathcal{H}_{2}$ is the $h_{2}$ -parity of $\mathcal{H}_{1}$ if

(i) for any $H_{1}\in\mathcal{H}_{1}$ , the $h_{2}$ -parity of $H_{1}$ exists and must be in $\mathcal{H}_{2}$ ;

(ii) for any $H_{2}\in\mathcal{H}_{2}$ , either $H_{2}$ has no $h_{1}$ -parity or it’s $h_{1}$ -parity is in $\mathcal{H}_{1}$ .

Proposition 2.16.

Let $f,g,h,n$ be positive integers with $f\geq g$ and $n\geq f+h$ . Let

	$\displaystyle\mathcal{F}=\left\{F\in{[n]\choose f}:\text{ there exists }H\in{[n]\choose h}\text{ such that }(F,H)\text{ is maximal }\right\},$
	$\displaystyle\mathcal{G}=\left\{G\in{[n]\choose g}:\text{ there exists }H\in{[n]\choose h}\text{ such that }(G,H)\text{ is maximal }\right\}.$

Let $\mathcal{H}_{\mathcal{F}}\subseteq{[n]\choose h}$ and $\mathcal{H}_{\mathcal{G}}\subseteq{[n]\choose h}$ be the families such that $\mathcal{F}$ and $\mathcal{H}_{\mathcal{F}}$ are maximal pair families and $\mathcal{G}$ and $\mathcal{H}_{\mathcal{G}}$ are maximal pair families. Then $\mathcal{F}$ is the $f$ -parity of $\mathcal{G}$ ; $\mathcal{H}_{\mathcal{G}}\subseteq\mathcal{H}_{\mathcal{F}}$ ; and for any $G\in\mathcal{G}$ , let $F\in\mathcal{F}$ be the $f$ -parity of $G$ and $H\in\mathcal{H}$ such that $(F,H)$ is maximal, then $(G,H)$ is maximal.

Proof.

If $f=g$ , then $\mathcal{F}=\mathcal{G}$ and $\mathcal{H}_{\mathcal{G}}=\mathcal{H}_{\mathcal{F}}$ . So we may assume that $f=g+s$ for some $s\geq 1$ . For any $G\in\mathcal{G}$ , there is the unique $H\in\mathcal{H}$ such that $(G,H)$ is maximal. Let $G^{\prime}=G\setminus G^{\rm t}=G\setminus\{n-\ell(G)+1,\dots,n\}$ and $H^{\prime}=H\setminus H^{\rm t}=H\setminus\{n-\ell(H)+1,\dots,n\}$ . Then, by Fact 2.8, $G^{\prime}$ and $H^{\prime}$ are partners of each other . So $\max G^{\prime}\leq g+h-1\leq f+h-s-1$ . Let $F=G^{\prime}\cup\{n-\ell(G)+1-s,\dots,n\}$ . Then $G\subseteq F$ , $|F|=f$ and $\ell(F)-\ell(G)=f-g=s$ . Then, by Definition 2.11, $F$ is the $f$ -parity of $G$ . Let $F^{\prime}=F\setminus F^{\rm t}$ . So $F^{\prime}=G^{\prime}$ . Furthermore, $F^{\prime}$ and $H^{\prime}$ are partners of each other. By Fact 2.8 again, we can see that $(F,H)$ is maximal. So $F\in\mathcal{F}$ , and $\mathcal{F}$ is the $f$ -parity of $\mathcal{G}$ , as desired. For any $G\in\mathcal{G}$ , let $F$ be the $f$ -parity of $G$ and $H\in\mathcal{H}$ be the set such that $(F,H)$ is maximal. By Fact 2.14, $(G,H)$ is maximal, as desired. And this implies $\mathcal{H}_{\mathcal{G}}\subseteq\mathcal{H}_{\mathcal{F}}$ . The proof is complete. ∎

Fact 2.17.

Let $a,b,c,n$ be positive integers, $a\geq b$ and $n\geq a+c$ , and let $C$ be a $c$ -subset of $[n]$ . Suppose that $A$ is the $a$ -partner of $C$ and $B$ is the $b$ -partner of $C$ , then $B\prec A$ or $A$ is the $a$ -parity of $B$ .

Proof.

If $a=b$ , then $A=B$ , we are done. Assume $a>b$ . Let $T$ be the partner of $C$ and let $|T|=c^{\prime}$ . If $c^{\prime}\leq b$ , then $c^{\prime}<a$ . By the definitions of $a$ -partner and $b$ -partner and $n\geq a+c>b+c$ , we can see that $A$ is the $a$ -parity of $B$ . If $b<c^{\prime}\leq a$ , then we have $\min B\setminus A<\min A\setminus B$ , so $B\prec A$ . At last, suppose $b<a<c^{\prime}$ . Let $C^{\prime}=\{x_{1},\dots,x_{b},\dots,x_{a}\}\subseteq C$ be the first $a$ elements of $C$ . If $x_{i+1}=x_{i}+1$ for all $i\in[b,a-1]$ , then $A$ is the $a$ -parity of $B$ . Otherwise, we have $B\prec A$ . ∎

3 Proofs of Theorem 1.4

Recall that $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is an extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting system, each $\mathcal{F}_{i}$ is L-initial, and $I_{1},I_{2},\dots,I_{t}$ are the IDs of $\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t}$ respectively throughout the paper. From Constructions 1.2 and 1.3, we have

\sum_{i=1}^{t}|\mathcal{F}_{i}|\geq\textup{max}\{\lambda_{1},\,\,\lambda_{2}\}.

(3)

We are going to prove that $\sum_{i=1}^{t}|\mathcal{F}_{i}|\leq\textup{max}\{\lambda_{1},\lambda_{2}\}$ .

Claim 3.1.

We may assume that $k_{t}\geq 2$ .

Proof.

We consider the case $k_{t}=1$ . Suppose that $|\mathcal{F}_{t}|=s$ . Then $\mathcal{F}_{t}=\{\{1\},\dots,\{s\}\}$ . Since $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is a cross intersecting system, then for any $i\in[t-1]$ and $F\in\mathcal{F}_{i}$ , we have $[s]\subseteq F$ . Thus,

\sum_{i=1}^{t}|\mathcal{F}_{i}|=\sum_{i=1}^{t-1}{n-s\choose k_{i}-s}+s\leq\lambda_{1}.

So we may assume that $k_{t}\geq 2$ . ∎

From now on, $k_{t}\geq 2$ .

For the case $k_{1}=k_{2}$ , the authors in [6] (Corollary 1.12) have already given a positive answer. Although we can prove for this case in this paper, for simplicity, we assume that

k_{1}>k_{2}.

In [6], the authors proved the following result.

Theorem 3.2.

[6] Let $n$ , $t\geq 2$ , $k_{1},k_{2},\dots,k_{t}$ be positive integers and $d_{1},d_{2},\dots,d_{t}$ be positive numbers. Let $\mathcal{A}_{1}\subset{[n]\choose k_{1}},\mathcal{A}_{2}\subset{[n]\choose k_{2}},\dots,\mathcal{A}_{t}\subset{[n]\choose k_{t}}$ be non-empty cross-intersecting families with $|\mathcal{A}_{i}|\geq{n-1\choose k_{i}-1}$ for some $i\in[t]$ . Let $m_{i}$ be the minimum integer among $k_{j}$ , where $j\in[t]\setminus\{i\}$ . If $n\geq k_{i}+k_{j}$ for all $j\in[t]\setminus\{i\}$ , then

\sum_{1=j}^{t}d_{j}|\mathcal{A}_{j}|\leq\max\left\{d_{i}{n\choose k_{i}}-d_{i}{n-m_{i}\choose k_{i}}+\sum_{j=1,j\neq i}^{t}d_{j}{n-m_{i}\choose k_{j}-m_{i}},\,\,\sum_{j=1}^{t}d_{j}{n-1\choose k_{j}-1}\right\}.

The equality holds if and only if one of the following holds.
(1) If $d_{i}{n\choose k_{i}}-d_{i}{n-k_{t}\choose k_{i}}+\sum_{j\neq i}^{t}d_{j}{n-m_{i}\choose k_{j}-m_{i}}\geq\sum_{j=1}^{t}d_{j}{n-1\choose k_{j}-1}$ , then there is some $m_{i}$ -element set $T\subset[n]$ such that $\mathcal{A}_{i}=\{F\in{[n]\choose k_{i}}:F\cap T\neq\emptyset\}$ and $\mathcal{A}_{j}=\{F\in{[n]\choose k_{j}}:T\subset F\}$ for each $j\in[t]\setminus\{i\}$ .
(2) If $d_{i}{n\choose k_{i}}-d_{i}{n-k_{t}\choose k_{i}}+\sum_{j\neq i}^{t}d_{j}{n-m_{i}\choose k_{j}-m_{i}}\leq\sum_{j=1}^{t}d_{j}{n-1\choose k_{j}-1}$ , then there is some $a\in[n]$ such that $\mathcal{A}_{j}=\{F\in{[n]\choose k_{j}}:a\in F\}$ for each $j\in[t]$ .
(3) If $t=2$ and $n=k_{i}+k_{3-i}$ . If $d_{i}\leq d_{3-i}$ , then $\mathcal{A}_{3-i}\subseteq{[n]\choose k_{3-i}}$ with $|\mathcal{A}_{3-i}|={n-1\choose k_{3-i}-1}$ and $\mathcal{A}_{i}={[n]\choose k_{i}}\setminus\overline{\mathcal{A}_{3-i}}$ .
(4) If $n=k_{i}+k_{j}$ holds for every $j\in[t]\setminus\{i\}$ and $\sum_{j\neq i}d_{j}=d_{i}$ , then $\mathcal{A}_{j}=\mathcal{A}$ for all $j\in[t]\setminus\{i\}$ , where $\mathcal{A}\subseteq{[n]\choose k}$ is an intersecting family with size $|\mathcal{A}|={n-1\choose k-1}$ , and $\mathcal{A}_{i}={[n]\choose k_{i}}\setminus\overline{\mathcal{A}}$ .

Claim 3.3.

$|\mathcal{F}_{1}|\geq{n-1\choose k_{1}-1}$ and $|\mathcal{F}_{2}|\geq{n-1\choose k_{2}-1}$ , in other words, $\{1,n-k_{1}+2,\dots,n\}\prec I_{1}$ and $\{1,n-k_{2}+2,\dots,n\}\prec I_{2}$ .

Proof.

If $|\mathcal{F}_{i}|<{n-1\choose k_{i}-1}$ for each $i\in[t]$ , then $\sum_{i=1}^{t}{|\mathcal{F}_{i}|}<\lambda_{1}$ , a contradiction to (3). So there is $i\in[t]$ such that $|\mathcal{F}_{i}|\geq{n-1\choose k_{i}-1}$ .

Suppose that $i\in[3,t]$ firstly. Since $k_{1}\geq k_{2}$ , $n\geq k_{1}+k_{3}$ and therefore $n\geq k_{i}+k_{j}$ for all $j\in[t]\setminus\{i\}$ . Let $m_{i}=\min\{k_{j},j\in[t]\setminus\{i\}\}$ . Taking $d_{1}=d_{2}=\dots=d_{t}$ in Theorem 3.2, we obtain

\sum_{j=1}^{t}|\mathcal{F}_{j}|\leq\textup{max}\left\{\sum_{j=1}^{t}{n-1\choose k_{j}-1},\,\,{n\choose k_{i}}-{n-k_{t}\choose k_{i}}+\sum_{j=1,j\neq i}^{t}{n-m_{i}\choose k_{j}-m_{i}}\right\}.

In [6] (Proposition 2.20 [6]), we have shown that

{n\choose k_{i}}-{n-k_{t}\choose k_{i}}+\sum_{j=1,j\neq i}^{t}{n-m_{i}\choose k_{j}-m_{i}}\leq{n\choose k_{1}}-{n-k_{t}\choose k_{1}}+\sum_{j=2}^{t}{n-k_{t}\choose k_{j}-k_{t}}.

	$\displaystyle\sum_{j=1}^{t}\|\mathcal{F}_{j}\|$	$\displaystyle\leq\textup{max}\left\{\sum_{j=1}^{t}{n-1\choose k_{j}-1},{n\choose k_{1}}-{n-k_{t}\choose k_{1}}+\sum_{j=2}^{t}{n-k_{t}\choose k_{j}-k_{t}}\right\}$
		$\displaystyle\leq\textup{max}\{\lambda_{1},\,\,\lambda_{2}\},$

where the last inequality holds by $k_{t}\geq 2$ and ${n\choose k_{2}}-{n-k_{t}\choose k_{2}}>{n-k_{t}\choose k_{2}-k_{t}}$ . Thus if $\max\{\lambda_{1},\,\,\lambda_{2}\}=\lambda_{2}$ , then the last inequality holds strictly, it makes a a contradiction to (3). Suppose that $\max\{\lambda_{1},\,\,\lambda_{2}\}=\lambda_{2}$ . Then Claim 3.3 follows from Theorem 3.2 (see the item (2) of Theorem 3.2). So $i\in[2]$ . Without loss of generality, let $i=1$ . Since $n<k_{1}+k_{2}$ , then any two families $\mathcal{G}\subseteq{[n]\choose k_{1}}$ and $\mathcal{H}\subseteq{[n]\choose k_{2}}$ are cross intersecting freely. Since $|\mathcal{F}_{1}|\geq{n-1\choose k_{1}-1}$ and $\mathcal{F}_{1}$ is L-initial, $\{1,n-k_{1}+2,\dots,n\}\prec I_{1}$ . Note that $n>k_{1}+k_{j}$ for each $j\in[3,t]$ and $\mathcal{F}_{1}$ is cross intersecting with $\mathcal{F}_{j}$ but not freely for each $j\in[3,t]$ , every member of $\mathcal{F}_{j}$ contains 1. Since $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal, all $k_{2}$ -subsets contianing 1 are contained in $\mathcal{F}_{2}$ , so $\{1,n-k_{2}+2,\dots,n\}\prec I_{2}$ , this implies $|\mathcal{F}_{2}|\geq{n-1\choose k_{2}-1}$ . This completes the proof of Claim 3.3. ∎

The following observation is simple and frequently used in this paper.

Remark 3.4.

Let $d\geq 2$ be an integer, we consider $d$ L-initial families $\mathcal{L}(A_{1},a_{1})$ , $\dots$ , $\mathcal{L}(A_{d},a_{d})$ . For $i\in[d]$ and let $S\subseteq[d]\setminus\{i\}$ be the set of all $j\in[d]\setminus\{i\}$ satisfying that $n\geq a_{i}+a_{j}$ . Suppose that $\{1,n-a_{i}+2,\dots,n\}\preceq A_{i}$ . If $\mathcal{L}(A_{i},a_{i})$ and $\mathcal{L}(A_{j},a_{j})$ are cross intersecting for each $j\in S$ , then $\mathcal{L}(A_{j},a_{j})$ , $j\in S$ are pairwise cross intersecting families since for any $j\in S$ , every member of $\mathcal{L}(A_{j},a_{j})$ contains $1$ .

Combining Claim 3.3, Proposition 2.2 and $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal, we have that for $i\in[3,t]$ , $I_{i}$ is $k_{i}$ -partner of $I_{1}$ (if $I_{2}\prec I_{1}$ ) or $I_{2}$ (if $I_{1}\prec I_{2}$ ).

Since $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal, $I_{i}$ (ID of $\mathcal{F}_{i}$ ) must be contained in $\mathcal{R}_{i}$ , which are defined as below.

	$\displaystyle\mathcal{R}_{1}=\{R\in{[n]\choose k_{1}}:\{1,n-k_{1}+2,\dots,n\}\prec R\prec\{k_{t},n-k_{1}+2,\dots,n\}\}.$
	$\displaystyle\mathcal{R}_{2}=\{R\in{[n]\choose k_{2}}:\{1,n-k_{2}+2,\dots,n\}\prec R\prec\{k_{t},n-k_{2}+2,\dots,n\}\}.$

For $i\in[3,t-1]$ , let

	$\displaystyle\mathcal{R}_{i}$	$\displaystyle=\{R\in{[n]\choose k_{i}}:[k_{t}]\cup[n-k_{i}+k_{t}+1,n]\prec R\prec\{1,n-k_{i}+2,\dots,n\}\},$
	$\displaystyle\mathcal{R}_{t}$	$\displaystyle=\{R\in{[n]\choose k_{t}}:\{1,2,\dots,k_{t}\}\prec R\prec\{1,n-k_{t}+2,\dots,n\}\}.$

For a family $\mathcal{A}_{1}\subseteq{[n]\choose k_{1}}$ with ID $A_{1}$ , let

	$\displaystyle m(n,A_{1})=$	$\displaystyle\max\big{\{}\sum_{j\in[t]\setminus\{1\}}\|\mathcal{A}_{j}\|:\mathcal{A}_{j}\subseteq{[n]\choose k_{j}}\text{ is L-initial and }(\mathcal{A}_{1},\dots,\mathcal{A}_{t})\text{ is an }$
		$\displaystyle(n,k_{1},\dots,k_{t})\text{-cross intersecting system, where }k_{1}+k_{3}\leq n<k_{1}+k_{2}\big{\}}.$

For L-initial cross intersecting families $\mathcal{A}_{1}\subseteq{[n]\choose k_{1}}$ and $\mathcal{A}_{2}\subseteq{[n]\choose k_{2}}$ with IDs $A_{1}$ and $A_{2}$ respectively, let

	$\displaystyle m(n,A_{1},A_{2})=$	$\displaystyle\max\big{\{}\sum_{j\in[t]\setminus\{1,2\}}\|\mathcal{A}_{j}\|:\mathcal{A}_{j}\subseteq{[n]\choose k_{j}}\text{ is L-initial and }(\mathcal{A}_{1},\dots,\mathcal{A}_{t})\text{ is an }$
		$\displaystyle(n,k_{1},\dots,k_{t})\text{-cross intersecting system, where }k_{1}+k_{3}\leq n<k_{1}+k_{2}\big{\}}.$

Proposition 3.5.

If $I_{1}=\{1,n-k_{1}+2,\dots,n\}$ or $I_{1}=\{k_{t},n-k_{1}+2,\dots,n\}$ (or $I_{2}=\{1,n-k_{2}+2,\dots,n\}$ or $I_{2}=\{k_{t},n-k_{2}+2,\dots,n\}$ ), then $\sum_{i=1}^{t}|\mathcal{F}_{i}|=\textup{max}\{\lambda_{1},\,\,\lambda_{2}\}$ .

Proof.

The proofs for $I_{1}$ and $I_{2}$ are similar, so we only prove for $I_{1}$ . Suppose that $I_{1}=\{1,n-k_{1}+2,\dots,n\}$ firstly. Since $\{1,n-k_{2}+2,\dots,n\}\prec I_{2}$ , $I_{1}\prec I_{2}$ . By Fact 2.14, we have

m(n,I_{1})=M(n,k_{2},k_{3},\dots,k_{t}).

(4)

Since $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal,

	$\displaystyle\sum_{i=1}^{t}\|\mathcal{F}_{i}\|={n-1\choose k_{1}-1}+m(n,I_{1})$
	$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\overset{(\ref{newdef2})}{=}{n-1\choose k_{1}-1}+M(n,k_{2},k_{3},\dots,k_{t})$
	$\displaystyle~{}~{}~{}~{}~{}~{}\overset{Theorem\ref{HP}}{=}{n-1\choose k_{1}-1}+\textup{max}\left\{\sum_{i=2}^{t}{n-1\choose k_{i}-1},\,{n\choose k_{2}}-{n-k_{t}\choose k_{2}}+\sum_{i=3}^{t}{n-k_{t}\choose k_{i}-k_{t}}\right\}.$

Since $k_{t}\geq 2$ ,

{n-1\choose k_{1}-1}+{n\choose k_{2}}-{n-k_{t}\choose k_{2}}+\sum_{i=3}^{t}{n-k_{t}\choose k_{i}-k_{t}}<\lambda_{2}.

By (3), $\sum_{i=1}^{t}|\mathcal{F}_{i}|=\lambda_{1}$ , as desired.

Next we consider $I_{1}=\{k_{t},n-k_{1}+2,\dots,n\}$ . For each $i\in[3,t]$ , since $\mathcal{F}_{i}$ and $\mathcal{F}_{1}$ are cross intersecting and $n>k_{1}+k_{3}\geq k_{1}+k_{i}$ , every element of $\mathcal{F}_{i}$ must contain $[k_{t}]$ . Since $\mathcal{F}_{i}$ and $\mathcal{F}_{1}$ are cross intersecting for every $i\in[3,t]$ , $k_{2}+k_{3}\leq n<k_{1}+k_{2}$ and $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal, we can see that $I_{2}=\{k_{t},n-k_{2}+2,\dots,n\}$ , that is the last element of $\mathcal{R}_{2}$ . Therefore, $\sum_{i=1}^{t}|\mathcal{F}_{i}|=\lambda_{2}$ , as desired. ∎

By Proposition 3.5, the proof of the quantitative part of Theorem 1.4 will follow from the following proposition.

Proposition 3.6.

If $\{1,n-k_{1}+2,\dots,n\}\precneqq I_{1}\precneqq\{k_{t},n-k_{1}+2,\dots,n\}$ and $\{1,n-k_{2}+2,\dots,n\}\precneqq I_{2}\precneqq\{k_{t},n-k_{2}+2,\dots,n\}$ , then $\sum_{i=1}^{t}|\mathcal{F}_{i}|<\max\{\lambda_{1},\lambda_{2}\}$ .

In order to prove Proposition 3.6, we need some preparations.

Definition 3.7.

Let $\mathcal{F}_{2,3}\subseteq\mathcal{R}_{2}$ and $\mathcal{F}_{3}^{2}\subseteq\mathcal{R}_{3}$ be such that $\mathcal{F}_{2,3}$ and $\mathcal{F}_{3}^{2}$ are maximal pair families.

The following lemma whose proof will be given in Section 5 is crucial. It tells us that for an extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting system $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ with $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , the ID $I_{1}$ of $\mathcal{F}_{1}$ is the parity of the ID $I_{2}$ of $\mathcal{F}_{2}$ .

Lemma 3.8.

Let $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ be an extremal L-initial $(n,k_{1},\dots,k_{t})$ -cross intersecting system with IDs $I_{1},I_{2},\dots,I_{t}$ of $\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t}$ respectively. Then $I_{1}$ and $I_{2}$ satisfy

I_{2}\in\mathcal{F}_{2,3}\,\text{ and }\,I_{1}\text{ is the }k_{1}\text{-parity of }I_{2}.

(5)

Recall that $k_{1}>k_{2}\geq\dots\geq k_{t}$ , and $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ . Let $G_{2}\in\mathcal{F}_{2,3}$ . Let $G_{1}$ be the $k_{1}$ -parity of $G_{2}$ . Then from Fact 2.14, we have the following claim.

Remark 3.9.

For any $j\in[3,t]$ , $G_{1}$ and $G_{2}$ have the same $k_{j}$ -partner.

Fixing any $G_{2}\in\mathcal{F}_{2,3}$ , let $G_{1}$ be the $k_{1}$ -parity of $G_{2}$ , and for any $i\in[3,t]$ , let $T_{i}$ by the $k_{i}$ -partner of $G_{2}$ . By Claim 3.3 and Remark 3.4, we can see that $\mathcal{L}(T_{3},k_{3})$ , …, $\mathcal{L}(T_{t},k_{t})$ are pariwise cross-intersecting. Furthermore, combining Fact 2.5 and Remark 3.9, we conclude that

\displaystyle m(n,G_{1},G_{2})

\displaystyle=\sum_{i=3}^{t}|\mathcal{L}(T_{i},k_{i})|.

(6)

Now we are ready to express $M(n,k_{1},\dots,k_{t})$ in terms of a function of $G_{2}$ .

Remark 3.10.

Let $G_{2}\in\mathcal{F}_{2,3}$ , let $G_{1}$ be the $k_{1}$ -parity of $G_{2}$ and for any $i\in[3,t]$ , let $T_{i}$ be the $k_{i}$ -partner of $G_{2}$ . Define

g(G_{2})=\sum_{i=1}^{2}|\mathcal{L}(G_{i},k_{i})|+\sum_{i=3}^{t}|\mathcal{L}(T_{i},k_{i})|.

(7)

Then by Lemma 3.8 and (6),

M(n,k_{1},\dots,k_{t})=\max_{G_{2}\in\mathcal{F}_{2,3}}\{g(G_{2})\}.

(8)

Definition 3.11.

For each $j\in[k_{2}-1]$ , let $\mathcal{R}_{2,j}=\{R\in\mathcal{R}_{2}:[n-j+1,n]\subset R\}$ and $\mathcal{R}_{2}(j)=\{R\setminus[n-j+1,n]:R\in\mathcal{R}_{2,j}\}$ . Denote $\mathcal{F}_{2,3}(j)=\{R\setminus[n-j+1,n]:R\in\mathcal{F}_{2,3}\cap\mathcal{R}_{2,j}\}$ . For any $j\in[k_{2}-1]$ and any $R\in\mathcal{F}_{2,3}\cap\mathcal{R}_{2,j}$ , we define $g(R\setminus[n-j+1,n])=g(R)$ .

We will prove several key lemmas to show the ‘local unimodality’ of $g(G_{2})$ in Section 5. Before stating these crucial lemmas, we need to introduce some definitions.

Let $\mathcal{A}\subset{[n]\choose k}$ be a family and $c\in[k]$ . We say that $\mathcal{A}$ is $c$ -sequential if there are $A\subset[n]$ with $|A|=k-c$ and $a\geq\max A$ (For a set $A\subset[n]$ , denote $\max A=\max\{x:x\in A\}$ and $\min A=\min\{x:x\in A\}$ ) such that $\mathcal{A}=\{A\sqcup\{a+1,\dots,a+c\},A\sqcup\{a+2,\dots,a+c+1\},\dots,A\sqcup\{b-c+1,\dots,b\}\}$ , then we say that $\mathcal{A}$ is $c$ -sequential, write $A_{1}\overset{c}{\prec}A_{2}\overset{c}{\prec}\cdots\overset{c}{\prec}A_{b-a-c+1}$ , where $A_{1}=A\sqcup\{a+1,\dots,a+c\}$ , $A_{2}=A\sqcup\{a+2,\dots,a+c+1\}$ , $\dots$ , $A_{b-a-c+1}=A\sqcup\{b-c+1,\dots,b\}$ , we also say that $A_{i}$ and $A_{j}$ are $c$ -sequential. In particular, if $l_{2}=l_{1}+1$ , we write $A_{l_{1}}\overset{c}{\prec}A_{l_{2}}$ ; if $\max A_{l_{2}}=n$ , write $A_{l_{1}}\overset{c}{\longrightarrow}A_{l_{2}}$ . Note that if $|\mathcal{A}|=1$ , then $\mathcal{A}$ is $c$ -sequential for any $c\in[k]$ . Let $\mathcal{F}$ be a family and $F_{1},F_{2}\in\mathcal{F}$ . If $F_{1}\precneqq F_{2}$ and there is no $F^{\prime}\in\mathcal{F}$ such that $F_{1}\precneqq F^{\prime}\precneqq F_{2}$ , then we say that $F_{1}<F_{2}$ in $\mathcal{F}$ , or $F_{1}<F_{2}$ simply if there is no confusion.

We will prove the following crucial lemmas in Sections 6 and 7. They show that function $g(R)$ has local unimodality.

Lemma 3.12.

For any $j\in[0,k_{2}-1]$ , let $1\leq c\leq k_{2}-j$ and $F_{2},G_{2},H_{2}\in\mathcal{F}_{2,3}(j)$ with $F_{2}\overset{c}{\prec}G_{2}\overset{c}{\prec}H_{2}$ . Then $g(G_{2})\geq g(F_{2})$ implies $g(H_{2})>g(G_{2})$ .

Lemma 3.13.

Let $4\leq j\leq k_{2}+1$ and $F_{2},G_{2},H_{2}\in\mathcal{F}_{2,3}$ with $[2,j]=F_{2}\setminus[n-\ell(F_{2})+1,n]$ , $[2,j-1]=G_{2}\setminus[n-\ell(G_{2})+1,n]$ and $[2,j-2]=H_{2}\setminus[n-\ell(H_{2})+1,n]$ . Then $g([2,j-1])\geq g([2,j])$ implies $g([2,j-2])>g([2,j-1])$ .

Lemma 3.14.

Let $k_{t}+2\leq j\leq k_{t}+k_{2}-1$ and $F_{2},G_{2},H_{2}\in\mathcal{F}_{2,3}$ with $[k_{t},j]=F_{2}\setminus[n-\ell(F_{2})+1,n]$ , $[k_{t},j-1]=G_{2}\setminus[n-\ell(G_{2})+1,n]$ and $[k_{t},j-2]=H_{2}\setminus[n-\ell(H_{2})+1,n]$ . Then $g([k_{t},j-1])\geq g([k_{t},j])$ implies $g([k_{t},j-2])>g([k_{t},j-1])$ .

Claim 3.15.

Let $A$ be a $k_{2}$ -subset of $[n]$ with $\ell(A)=p$ . Suppose $A=\{x_{1},\dots,x_{k_{2}-p},n-p+1,\dots,n\}$ and $A^{\prime}=A\setminus\{x_{k_{2}-p}\}\cup[n-p,n]$ . If $A\in\mathcal{F}_{2,3}$ , then $A^{\prime}\in\mathcal{F}_{2,3}$ .

Proof.

Let $T$ and $T^{\prime}$ be the partners of $\{x_{1},\dots,x_{k_{2}-p}\}=A\setminus A^{\rm t}$ and $\{x_{1},\dots,x_{k_{2}-p-1}\}=A^{\prime}\setminus A^{\prime\rm t}$ respectively. Since $A\in\mathcal{F}_{2,3}$ , $|T|\leq k_{3}$ . Thus, $|T^{\prime}|\leq|T|\leq k_{3}$ . Let $B=T^{\prime}\cup\{n-k_{3}+|T^{\prime}|+1,\dots,n\}$ if $|T^{\prime}|<k_{3}$ , otherwise, let $B=T^{\prime}$ . Then $B$ is the $k_{3}$ -partner of $\{x_{1},\dots,x_{k_{2}-p-1}\}$ . Fact 2.9 implies that $(A^{\prime},B)$ is maximal, thus $A^{\prime}\in\mathcal{F}_{2,3}$ . ∎

Claim 3.16.

Let $A$ be a $k_{2}$ -subset of $[n]$ with $\ell(A)=p\geq 1$ . Suppose that $A=\{x_{1},\dots,x_{k_{2}-p},n-p+1,\dots,n\}$ and $A^{\prime}=A\setminus\{n-p+1\}\cup\{x_{k_{2}-p}+1\}\cup[n-p+2,n]$ (if $p=1$ , then $[n-p+2,n]=\emptyset$ ). If $A\in\mathcal{F}_{2,3}$ and $\{1,n-k_{2}+2,\dots,n\}\precneqq A$ , then $A^{\prime}\in\mathcal{F}_{2,3}$ .

Proof.

Let $T$ and $T^{\prime}$ be the partners of $\{x_{1},\dots,x_{k_{2}-p}\}=A\setminus A^{\rm t}$ and $\{x_{1},\dots,x_{k_{2}-p},\\ x_{k_{2}-p}+1\}=A^{\prime}\setminus A^{\prime\rm t}$ respectively. Then $|T|=|T^{\prime}|$ and $\max T^{\prime}-\max T=1$ . Since $A\in\mathcal{F}_{2,3}$ , by Definition 3.7, there is $B\in\mathcal{F}_{3}^{2}$ such that $(A,B)$ is maximal. By Fact 2.8, $T=B\setminus\{n-\ell(B)+1,\dots,n\}$ , $B=T\cup\{n-k_{3}+|T|+1,\dots,n\}$ and $\max T<n-k_{3}+|T|$ . We claim that $\max T<n-k_{3}+|T|-1$ . Since otherwise $\max T=n-k_{3}+|T|-1$ , this implies that $|\{x_{1},\dots,x_{k_{2}-p}\}|+|T|=n-k_{3}+|T|$ and then

|A|+|B|\geq|\{x_{1},\dots,x_{k_{2}-p}\}|+1+|T|+|\{n-k_{3}+|T|+1,\dots,n\}|=n+1,

this is a contradiction to $n\geq k_{2}+k_{3}\,\,(=|A|+|B|)$ . Let $B^{\prime}=T^{\prime}\cup\{n-k_{3}+|T^{\prime}|+1,\dots,n\}$ if $|T^{\prime}|<k_{3}$ , and $B^{\prime}=T^{\prime}$ otherwise. Thus,

\max T^{\prime}=\max T+1<n-k_{3}+|T|=n-k_{3}+|T^{\prime}|.

Therefore, $T^{\prime}=B^{\prime}\setminus\{n-\ell(B^{\prime})+1,\dots,n\}$ . Trivially, $\{x_{1},\dots,x_{k_{2}-p},x_{k_{2}-p}+1\}=A^{\prime}\setminus\{n-\ell(A^{\prime})+1,\dots,n\}$ . By Fact 2.8 again, $(A^{\prime},B^{\prime})$ is maximal and since $\{1,n-k_{2}+2,\dots,n\}\precneqq A$ , we have $A^{\prime}\in\mathcal{F}_{2,3}$ , as desired. ∎

Definition 3.17.

Let $\mathcal{F}_{1,3}\subseteq\mathcal{R}_{1}$ and $\mathcal{F}_{3}^{1}\subseteq\mathcal{R}_{3}$ such that $\mathcal{F}_{1,3}$ and $\mathcal{F}_{3}^{1}$ are maximal pair families. By Proposition 2.16, $\mathcal{F}_{3}^{2}\subseteq\mathcal{F}_{3}^{1}$ . Let $\mathcal{F}^{\prime}_{1,3}$ be the subfamily of $\mathcal{F}_{1,3}$ such that $\mathcal{F}^{\prime}_{1,3}$ and $\mathcal{F}_{3}^{2}$ are maximal pair families.

Observation 3.18.

Since $k_{3}\geq k_{t}\geq 2$ and $n\geq k_{1}+k_{3}$ , by Fact 2.8, it is easy to see that $\{1,n-k_{2}+1,\dots,n\},\{2,\dots,k_{2}+1\},\{k_{t},k_{t}+1,\dots,k_{t}+k_{2}-1\},\{k_{t},n-k_{2}+1,\dots,n\}$ are in $\mathcal{F}_{2,3}$ . As a consequence of Claim 3.15 and Claim 3.16, we have the following observation. $\{2,\dots,k_{2},n\},\{2,\dots,k_{2}-1,n-1,n\},\dots,\{2,n-k_{2}+2,\dots,n\},\{k_{t},k_{t}+1,\\ \dots,k_{t}+k_{2}-2,n\},\{k_{t},k_{t}+1,\dots,k_{t}+k_{2}-3,n-1,n\},\dots,\{k_{t},n-k_{2}+2,\dots,n\}$ are in $\mathcal{F}_{2,3}$ as well. $\mathcal{F}^{\prime}_{1,3}$ contains $\{1,n-k_{1}+2,\dots,n\},\{2,\dots,k_{1}+1\},\{2,\dots,k_{1},n\},\{2,\\ \dots,k_{1}-1,n-1,n\},\dots,\{2,n-k_{1}+2,\dots,n\},\{k_{t},k_{t}+1,\dots,k_{t}+k_{1}-1\},\{k_{t},k_{t}+1,\dots,k_{t}+k_{1}-2,n\},\{k_{t},k_{t}+1,\dots,k_{t}+k_{1}-3,n-1,n\},\dots,\{k_{t},n-k_{1}+2,\dots,n\}$ . Note that $\mathcal{F}^{\prime}_{1,3}\subseteq\mathcal{L}(\{k_{t},n-k_{1}+2,\dots,n\},k_{1})$ since we require that $(n,k_{1},\dots,k_{t})$ -cross intersecting systems is non-empty.

Assuming that Lemmas 3.8, 3.12, 3.13 and 3.14 hold, we are going to complete the proof of Proposition 3.6. We will prove Lemmas 3.8, 3.12, 3.13 and 3.14 in Sections 5, 6 and 7.

Definition 3.19.

For a family $\mathcal{G}\subseteq\mathcal{F}_{2,3}$ , denote $g(\mathcal{G})=\max\{g(G):G\in\mathcal{G}\}$ .

Proof of Proposition 3.6.

By Observation 3.18 and Lemma 3.12, we have

g(\mathcal{F}_{2,3})=\max\{g([2,k_{2}+1]),g([k_{t},k_{t}+1,k_{t}+k_{2}-1]),g(\mathcal{F}_{2,3}(1))\}.

(9)

Applying Lemma 3.12 and Observation 3.18 repeatedly, we have

		$\displaystyle g(\mathcal{F}_{2,3}(1))=\max\{g([2,k_{2}]),g([k_{t},k_{t}+k_{2}-2]),g(\mathcal{F}_{2,3}(2))\},$
		$\displaystyle g(\mathcal{F}_{2,3}(2))=\max\{g([2,k_{2}-1]),g([k_{t},k+k_{2}-3]),g(\mathcal{F}_{2,3}(3))\},$
		$\displaystyle\quad\vdots$
		$\displaystyle g(\mathcal{F}_{2,3}(k_{2}-1))<\max\{g(\{1\}),g(\{k_{t}\})\}.$		(10)

By Lemma 3.13, we have

		$\displaystyle\max\{g([2,k_{2}+1]),g([2,k_{2}]),\dots,g(\{2,3\}),g(\{2\})\}$
		$\displaystyle=\max\{g([2,k_{2}+1]),g(\{2\})\},$		(11)

and

g(\{2\})<\max\{g(\{1\}),g(\{k_{t}\})\}.

(12)

By Lemma 3.14, we have

		$\displaystyle\max\{g([k_{t},k_{t}+k_{2}-1]),g([k_{t},k_{t}+k_{2}-2]),\dots,g(\{k_{t},k_{t}+1\})\}$
		$\displaystyle=\max\{g([k_{t},k_{t}+k_{2}-1]),g(\{k_{t},k_{t}+1\})\},$		(13)

and

g(\{k_{t},k_{t}+1\})<\max\{g([k_{t},k_{t}+k_{2}-1]),g(\{k_{t}\})\}.

(14)

In Section 7, we will prove the following proposition.

Proposition 3.20.

	$\displaystyle g([2,k_{2}+1])$	$\displaystyle<\max\{g(\{1\}),g([2,k_{2}])\}$		(15)
	$\displaystyle g([k_{t},k_{t}+k_{2}-1])$	$\displaystyle<\max\{g(\{k_{t}-1\}),g([k_{t},k_{t}+k_{2}-2])\}.$		(16)

Combining (15), (3) and (12), we have

g([2,k_{2}+1])<\max\{g(\{1\}),g(\{k_{t}\})\}.

(17)

Combining (16), (3) and (14), we have

g([k_{t},k_{t}+k_{2}-1])<\max\{g(\{k_{t}\}),g(\{k_{t}-1\})\}\leq\max\{g(\{1\}),g(\{k_{t}\})\}.

(18)

Combining (9), (10), (3), (12), (3), (14),(17) and (18), we have

g(\mathcal{F}_{2,3})<\max\{g(\{1\}),g(\{k_{t}\})\}.

Recall (8), we obtain

\displaystyle M(n,k_{1},\dots,k_{t})

\displaystyle=\max_{G_{2}\in\mathcal{F}_{2,3}}g(G_{2})<\max\{g(\{1\}),g(\{k_{t}\})\}.

(19)

This completes the proof of Proposition 3.6. ∎

We apply Propositions 3.5 and 3.6 to prove Theorem 1.4.

Proof of Theorem 1.4.

Propositions 3.5 and 3.6 imply the quantitative part of Theorem 1.4. Now we show that extremal $(n,k_{1},\dots,k_{t})$ -cross intersecting systems with $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ must be isomorphic to Constructions 1.2 or 1.3. By Claim 3.3, Propositions 3.5 and 3.6, and Theorem 2.1, we conclude that: if $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},\dots,k_{t})$ -cross intersecting system with $\sum_{i=1}^{t}|\mathcal{F}_{i}|=\max\{\lambda_{1},\lambda_{2}\}$ , then either for each $i\in[t]$ , $|\mathcal{F}_{i}|={n-1\choose k_{i}-1}$ or $|\mathcal{F}_{1}|={n\choose k_{1}}-{n-k_{t}\choose k_{1}}$ , $|\mathcal{F}_{2}|={n\choose k_{2}}-{n-k_{t}\choose k_{2}}$ and $|\mathcal{F}_{i}|={n-k_{t}\choose k_{i}-k_{t}}$ holds for each $i\in[3,t]$ . If the previous happens, then $(\mathcal{F}_{1},\mathcal{F}_{3},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},k_{3},\dots,k_{t})$ -cross intersecting system with $\sum_{i=1,i\neq 2}^{t}|\mathcal{F}_{i}|=\sum_{i=1,i\neq 2}^{t}{n-1\choose k_{i}-1}$ and $(\mathcal{F}_{2},\mathcal{F}_{3},\dots,\mathcal{F}_{t})$ is an $(n,k_{2},k_{3},\dots,k_{t})$ -cross intersecting system with $\sum_{i=2}^{t}|\mathcal{F}_{i}|=\sum_{i=2}^{t}{n-1\choose k_{i}-1}$ . By Theorem 1.1, $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is isomorphic to $(\mathcal{G}_{1},\dots,\mathcal{G}_{t})$ which is defined in Construction 1.2. If the later happens, then $(\mathcal{F}_{1},\mathcal{F}_{3},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},k_{3},\dots,\\ k_{t})$ -cross intersecting system with $\sum_{i=1,i\neq 2}^{t}|\mathcal{F}_{i}|={n\choose k_{1}}-{n-k_{t}\choose k_{1}}+\sum_{i=3}^{t}{n-k_{t}\choose k_{i}-k_{t}}$ and $(\mathcal{F}_{2},\mathcal{F}_{3},\dots,\mathcal{F}_{t})$ is an $(n,k_{2},k_{3},\dots,k_{t})$ -cross intersecting system with $\sum_{i=2}^{t}|\mathcal{F}_{i}|={n\choose k_{2}}-{n-k_{t}\choose k_{2}}+\sum_{i=3}^{t}{n-k_{t}\choose k_{i}-k_{t}}$ . By Theorem 1.1, $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is isomorphic to $(\mathcal{H}_{1},\dots,\mathcal{H}_{t})$ which is defined in Construction 1.3. This completes the proof of Theorem 1.4. ∎

We owe the proofs of Lemmas 3.8, 3.12, 3.13, 3.14 and Proposition 3.20. Before giving their proofs, we list some results obtained in [5] for non-mixed type in the next section. Figure 1 is the flow chart of the proofs of the main theorem and lemmas.

Refer to caption — Figure 1: Flow chart of the proofs

4 Results of non-mixed type

Lemma 4.1.

[5] Let $k_{1}\geq k_{2}\geq\dots\geq k_{t}$ , $n\geq k_{1}+k_{2}$ and $(\mathcal{A}_{1},\mathcal{A}_{2},\dots,\mathcal{A}_{t})$ be a non-empty L-initial $(n,k_{1},k_{2},\dots,k_{t})$ -cross intersecting system with $|\mathcal{A}_{1}|\geq{n-1\choose k_{1}-1}$ . Let $R$ be the ID of $\mathcal{A}_{1}$ and $T$ be the partner of $R$ . Then

\displaystyle\sum_{j=1}^{t}|\mathcal{A}_{j}|\leq|\mathcal{L}(R,k_{1})|+\sum_{j=2}^{t}|\mathcal{L}(T,k_{j})|=:f(R).

(20)

Another crucial part in the proof of Theorem 1.1 is to show local unimodality of $f(R)$ . Let $R$ and $R^{\prime}$ be $k_{1}$ -subsets of $[n]$ satisfying $R\prec R^{\prime}$ . In order to analyze $f(R^{\prime})-f(R)$ , two related functions are defined in [5] as follows.

Definition 4.2.

Let $t\geq 2$ , $k_{1},k_{2},\dots,k_{t}$ be positive integers with $k_{1}\geq k_{2}\geq\dots\geq k_{t}$ and $n\geq k_{1}+k_{2}$ . Let $R$ and $R^{\prime}$ be $k_{1}$ -subsets of $[n]$ satisfying $R\prec R^{\prime}$ and let $T$ and $T^{\prime}$ be the partners of $R$ and $R^{\prime}$ respectively. We define

		$\displaystyle\alpha(R,R^{\prime}):=\|\mathcal{L}(R^{\prime},k_{1})\|-\|\mathcal{L}(R,k_{1})\|,$		(21)
		$\displaystyle\beta(R,R^{\prime}):=\sum_{j=2}^{t}\big{(}\|\mathcal{L}(T,k_{j})\|-\|\mathcal{L}(T^{\prime},k_{j})\|\big{)}.$		(22)

It is easy to see that

\displaystyle f(R^{\prime})-f(R)=\alpha(R,R^{\prime})-\beta(R,R^{\prime}).

Let

\mathcal{R}_{1}=\{R\in{[n]\choose k_{1}}:\{1,n-k_{1}+2,\dots,n\}\prec R\prec\{k_{t},n-k_{1}+2,\dots,n\}\}.

In order to show local unimodality of $f(R)$ , we proved the following results in [5]. These results give us some foundation in showing local unimodality of $g(G_{2})$ (recall (7)).

Proposition 4.3.

[5] Let $F<G\in\mathcal{R}_{1}$ and $\max G=q$ . Then $\beta(F,G)=\sum_{j=2}^{t}{n-q\choose k_{j}-(q-k_{1})}$ . If $n=k_{1}+k_{j}$ holds for any $j\in[t]\setminus\{1\}$ , then $\beta(F,G)=\sum_{j=2}^{t}1=t-1$ ; otherwise, we have $\beta(F,G)\geq 0$ and $\beta(F,G)$ decrease as $q$ increase.

Lemma 4.4.

[5] Let $c\in[k_{1}]$ and $F,G,F^{\prime},G^{\prime}\in\mathcal{R}_{1}$ . If $F,G$ are $c$ -sequential, $F^{\prime},G^{\prime}$ are $c$ -sequential and $\max F=\max F^{\prime},\max G=\max G^{\prime}$ , then $\alpha(F,G)=\alpha(F^{\prime},G^{\prime})$ and $\beta(F,G)=\beta(F^{\prime},G^{\prime})$ .

Definition 4.5.

For $k\in[k_{1}-1]$ , let $\mathcal{R}_{1,k}=\{R\in\mathcal{R}_{1}:[n-k+1,n]\subset R\}$ , and $\mathcal{R}_{1}(k)=\{R\setminus[n-k+1,n]:R\in\mathcal{R}_{1,k}\}$ . In addition, we will write $\mathcal{R}_{1}(0)=\mathcal{R}_{1}$ .

Remark 4.6.

When we consider $\mathcal{R}_{i}(k)$ , we regard the ground set as $[n-k]$ . For $R\in\mathcal{R}_{i}(k)$ , we write $f(R)$ simply, in fact, $f(R)=f(R\cup[n-k+1,n])$ .

Lemma 4.7.

[5] Let $1\leq j\leq k_{1}-1$ and $1\leq d\leq k_{1}-j$ . Let $F,H,F^{\prime},H^{\prime}\in\mathcal{R}_{1}(j)$ and $F$ , $H$ are $d$ -sequential, $F^{\prime}$ , $H^{\prime}$ are $d$ -sequential. If $\max F=\max F^{\prime}$ , then $\alpha(F,H)=\alpha(F^{\prime},H^{\prime})$ and $\beta(F,H)=\beta(F^{\prime},H^{\prime})$ .

The following two lemmas confirm that $f(R)$ has local unimodality.

Lemma 4.8.

[5] Suppose that if $t=2$ , then $n>k_{1}+k_{t}$ . For any $j\in[0,k_{1}-1]$ , let $1\leq c\leq k_{1}-j$ and $F$ , $G$ , $H$ be contained in $\mathcal{R}_{1}(j)$ with $F\overset{c}{\prec}G\overset{c}{\prec}H$ . If $f(G)\geq f(F)$ , then $f(H)>f(G)$ .

Lemma 4.9.

[5] Suppose that if $t=2$ , then $n>k_{1}+k_{t}$ . Let $1\leq m\leq k_{t}$ and $m+1\leq j\leq m+k_{1}-1$ . If $f_{1}([m,j])\leq f_{1}([m,j-1])$ , then $f([m,j-1])<f([m,j-2])$ .

5 Verify Parity: Proof of Lemma 3.8

In this section, we will give the proof of Lemma 3.8. The proof is divided into two cases.

Case 1: $I_{1}\prec I_{2}$ .

By Fact 2.14 , Fact 2.5, Remark 3.4 and $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal, we have that for each $i\in[3,t]$ , $I_{i}$ is the $k_{i}$ -partner of $I_{2}$ . By Fact 2.17, for all $i\in[4,t]$ , we have

I_{i}\prec I_{3}\text{ or }I_{3}\text{ is the $k_{3}$-parity of }I_{i}.

(23)

Claim 5.1.

$(I_{2},I_{3})$ is maximal.

Proof.

Assume on the contrary that $(I_{2},I_{3})$ is not maximal. Let $F_{2,i}$ be the $k_{2}$ -partner of $I_{i}$ for each $i\in[3,t]$ , by Fact 2.10, $(I_{2,i},I_{i})$ is maximal. Since $(I_{2},I_{3})$ is not maximal, $I_{2}\precneqq I_{2,3}$ . By (23) and Fact 2.14, we have $I_{2,3}\prec I_{2,j}$ for all $j\in[4,t]$ . This implies that for $i\in[3,t]$ , $\mathcal{L}(I_{2,3},k_{2})$ and $\mathcal{L}(I_{i},k_{i})$ are cross intersecting. Further more, $(\mathcal{F}_{1},\mathcal{L}(I_{2,3},k_{2}),\mathcal{F}_{3},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},\dots,k_{t})$ -cross intersecting system. However, $I_{2}\precneqq I_{2,3}$ implies $\mathcal{F}_{2}\subsetneqq\mathcal{L}(I_{2,3},k_{2})$ , this is a contradiction to the assumption that $(\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t})$ is extremal. ∎

Claim 5.1 tells us that $I_{2}\in\mathcal{F}_{2,3}$ . By Proposition 2.16, $\mathcal{F}_{1,3}$ is the $k_{1}$ -parity of $\mathcal{F}_{2,3}$ . Then there exists $I^{\prime}_{1}\in\mathcal{F}_{1,3}$ such that $I^{\prime}_{1}$ is the $k_{1}$ -parity of $I_{2}$ . If $I_{1}=I^{\prime}_{1}$ , then $I_{1}$ and $I_{2}$ satisfy (5), as desired. So we may assume $I_{1}\neq I^{\prime}_{1}$ . By Proposition 2.16, $(I^{\prime}_{1},I_{3})$ is maximal. Since $\mathcal{F}_{1}$ and $\mathcal{F}_{3}$ are cross intersecting, $I_{1}\precneqq I^{\prime}_{1}$ . Let $I^{\prime}_{i}$ be the $k_{i}$ -partner of $I^{\prime}_{1}$ for each $i\in[3,t]$ . It follows from Fact 2.14 that $I^{\prime}_{i}=I_{i}$ , $i\in[3,t]$ . Therefore, $(\mathcal{L}(I^{\prime}_{1},k_{1}),\mathcal{F}_{2},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},\dots,k_{t})$ -cross intersecting system. However, $I_{1}\precneqq I^{\prime}_{1}$ implies $\mathcal{F}_{1}\subsetneqq\mathcal{L}(I^{\prime}_{1},k_{1})$ , this is a contradiction to the assumption that $(\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t})$ is extremal.

Case 2: $I_{2}\precneqq I_{1}$ .

Using a similar argument to Case 1, we conclude the following three properties:

(a) $I_{i}$ is the $k_{i}$ -partner of $I_{1}$ for each $i\in[3,t]$ ;

(b) for all $i\in[4,t]$ , $I_{i}\prec I_{3}$ or $I_{3}$ is the $k_{3}$ -parity of $I_{i}$ ;

If $(I_{2},I_{3})$ is maximal, then $I_{2}\in\mathcal{F}_{2,3}$ , and Proposition 2.16 implies that $I_{1}$ is the $k_{1}$ -parity of $I_{2}$ . However, $I_{2}\precneqq I_{1}$ implies that $I_{1}$ is not the $k_{1}$ -parity of $I_{2}$ , a contradiction. Therefore $(I_{2},I_{3})$ is not maximal. Since $(\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t})$ is extremal, combining (b), Fact 2.14 and Fact 2.5, we have that $I_{2}$ is the $k_{2}$ -partner of $I_{3}$ . Therefore, Fact 2.10 implies that the $k_{3}$ -partner $I^{\prime}_{3}$ of $I_{2}$ is such that $(I_{2},I^{\prime}_{3})$ is maximal. So $I_{2}\in\mathcal{F}_{2,3}$ . By Proposition 2.16, there exists $I^{\prime}_{1}\in\mathcal{F}_{1,3}$ such that $I^{\prime}_{1}$ is the $k_{1}$ -parity of $I_{2}$ . For $i\in[3,t]$ , let $I^{\prime}_{i}$ be the $k_{i}$ -partner of $I_{2}$ . It follows from Facts 2.14 and 2.5 that $I^{\prime}_{i}$ is maximal to $I^{\prime}_{1}$ for all $i\in[3,t]$ . Combining with Remark 3.4, we conclude that $(\mathcal{L}(I^{\prime}_{1},k_{1}),\mathcal{F}_{2},\mathcal{L}(I^{\prime}_{3},k_{3}),\dots,\mathcal{L}(I^{\prime}_{t},k_{t}))$ is an $(n,k_{1},\dots,k_{t})$ -cross intersecting system. Since $(\mathcal{F}_{1},\dots,\mathcal{F}_{t})$ is extremal,

|\mathcal{L}(I^{\prime}_{1},k_{1})|+|\mathcal{F}_{2}|+\sum_{i=3}^{t}|\mathcal{L}(I^{\prime}_{i},k_{i})|\leq\sum_{i=1}^{t}|\mathcal{F}_{i}|.

This implies that

\sum_{i=1,i\neq 2}^{t}|\mathcal{L}(I^{\prime}_{i},k_{i})|\leq\sum_{i=1,i\neq 2}^{t}|\mathcal{F}_{i}|.

(24)

Let $H_{1}=I_{1}\setminus I_{1}^{\rm t}$ and $g=|H_{1}|$ . Recall that $I_{2}$ is the $k_{2}$ -partner of $I_{3}$ , $(I_{2},I_{3})$ is not maximal and $(I_{1},I_{3})$ is maximal. It follows from Facts 2.9 and 2.4 that

k_{2}<g.

Let $H_{2}=\{x_{1},\dots,x_{i},x_{i+1},\dots,x_{k_{2}}\}$ be the first $k_{2}$ elements of $H_{1}$ . Let $i\in[0,k_{2}-1]$ be the subscript such that $x_{i}+2\leq x_{i+1}$ and $x_{j}+1=x_{j+1}$ for all $j\in[i+1,k_{2}-1]$ , where $i=0$ if $x_{j+1}=x_{j}+1$ for all $j\in[k_{2}-1]$ . Since $I_{2}$ is the $k_{2}$ -partner of $I_{3}$ and $(I_{1},I_{3})$ is maximal, we can see that if $i<k_{2}-1$ , then $I_{2}=\{x_{1},\dots,x_{i},x_{i+1}-1,n-k_{2}+i+2,\dots,n\}$ , if $i=k_{2}-1$ , then $I_{2}=\{x_{1},\dots,x_{k_{2}-1},x_{k_{2}}-1\}$ . Since $I^{\prime}_{1}$ is the $k_{1}$ -parity of $I_{2}$ and $k_{1}>k_{2}$ , we get $\ell(I^{\prime}_{1})>0$ and

I^{\prime}_{1}=\{x_{1},\dots,x_{i},x_{i+1}-1,n-k_{1}+i+2,\dots,n\},\,where\,\,i\leq k_{2}-1.

Claim 5.2.

$I^{\prime}_{1}\precneqq I_{1}\precneqq I^{\prime\prime}_{1}:=\{x_{1},\dots,x_{i+1},n-k_{1}+i+2,\dots,n\}.$

Proof.

Obviously, $I^{\prime}_{1}\precneqq I_{1}$ . We are going to prove $I_{1}\precneqq I^{\prime\prime}_{1}$ . Notice that $\{x_{1},\dots,x_{i},\\ x_{i+1}\}\subset I_{1}$ and $|I^{\prime\prime}_{1}|=k_{1}$ , these imply $I_{1}\prec I^{\prime\prime}_{1}$ . If $I_{1}=I^{\prime\prime}_{1}$ , then $H_{1}=\{x_{1},\dots,x_{i+1}\}$ . So $g=i+1\leq k_{2}$ since $i\leq k_{2}-1$ , this is a contradiction to $k_{2}<g$ . ∎

Since $(\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},k_{2},\dots,k_{t})$ -cross intersecting system with $k_{1}+k_{3}\leq n<k_{1}+k_{2}$ , $(\mathcal{F}_{1},\mathcal{F}_{3},\dots,\mathcal{F}_{t})$ is an $(n,k_{1},k_{3},\dots,k_{t})$ -cross intersecting system with $n\geq k_{1}+k_{3}$ . Denote

	$\displaystyle m^{\prime}$	$\displaystyle=\max\big{\{}\sum_{i=1}^{t}\|\mathcal{G}_{i}\|:(\mathcal{G}_{1},\mathcal{G}_{3},\dots,\mathcal{G}_{t})\text{ is an L-initial }(n,k_{1},k_{3},\dots,k_{t})\text{-cross intersecting}$
		$\displaystyle\quad\quad\quad\quad\text{system with $n\geq k_{1}+k_{3}$, the ID}\ {G}_{1}\ \text{of}\ \mathcal{G}_{1}\ \text{satisfies}\ I^{\prime}_{1}\prec G_{1}\prec I^{\prime\prime}_{1}\big{\}}.$

Suppose that $(\mathcal{L}(G_{1},k_{1}),\mathcal{L}(G_{3},k_{3}),\dots,\mathcal{L}(G_{t},k_{t}))$ is an $(n,k_{1},\dots,k_{t})$ -cross intersecting system with $n\geq k_{1}+k_{3}$ such that $m^{\prime}=\sum_{i=1,i\neq 2}^{t}|\mathcal{G}_{i}|$ and $I^{\prime}_{1}\prec G_{1}\prec I^{\prime\prime}_{1}$ . Since $I^{\prime}_{1}$ is the $k_{1}$ -parity of $I_{2}$ and $I^{\prime}_{1}\prec G_{1}\prec I^{\prime\prime}_{1}$ , we have that either $I_{2}\prec G_{1}$ (if $G_{1}\neq I^{\prime}_{1}$ ) or $G_{1}$ is the $k_{1}$ -parity of $I_{2}$ (if $G_{1}=I^{\prime}_{1}$ ). Thus, $(\mathcal{L}(G_{1},k_{1}),\mathcal{F}_{2},\mathcal{L}(G_{3},k_{3}),\dots,\mathcal{L}(G_{t},k_{t}))$ is an $(n,k_{1},\dots,k_{t})$ -cross intersecting system as well. Since $(\mathcal{F}_{1},\mathcal{F}_{2},\dots,\mathcal{F}_{t})$ is extremal, we have $\sum_{i=1,i\neq 2}^{t}|\mathcal{F}_{i}|\geq\sum_{i=1,i\neq 2}^{t}|\mathcal{G}_{i}|=m^{\prime}$ . Therefore,

m^{\prime}=\sum_{i=1,i\neq 2}^{t}|\mathcal{F}_{i}|.

(25)

To complete the proof of Lemma 3.8, we only need to prove the following claim since it makes a contradiction to (25) and Claim 5.2.

Claim 5.3.

Let $(\mathcal{G}_{1},\mathcal{G}_{3},\dots,\mathcal{G}_{t})$ be an L-initial $(n,k_{1},k_{3},\dots,k_{t})$ -cross intersecting system with $n\geq k_{1}+k_{3}$ and $G_{1}$ be the ID of $\mathcal{G}_{1}$ satisfying $I^{\prime}_{1}\prec G_{1}\prec I^{\prime\prime}_{1}$ . Then $\sum_{i=1,i\neq 2}^{t}|\mathcal{G}_{i}|=m^{\prime}$ if and only if $G_{1}=I^{\prime}_{1}$ or $G_{1}=I^{\prime\prime}_{1}$ .

Proof.

For each $i\in[3,t]$ , let $T_{i}$ be the $k_{i}$ -partner of $G_{1}$ . By Remark 2.3 and Lemma 4.1, we have

\sum_{i=1,i\neq 2}^{t}|\mathcal{G}_{i}|\leq|\mathcal{L}(G_{1},k_{1})|+\sum_{i=3}^{t}|\mathcal{L}(T_{i},k_{i})|=:f_{1}(G_{1}).

By the definition of $m^{\prime}$ ,

m^{\prime}=\max\left\{f_{1}(R):R\in{[n]\choose k_{1}}\ \text{and}\ I^{\prime}_{1}\prec R\prec I^{\prime\prime}_{1}\right\}.

Denote

\mathcal{F}=\{F\in{[n]\choose k_{1}}:I^{\prime}_{1}\prec F\prec I^{\prime\prime}_{1}\}.

Then $m^{\prime}=f_{1}(\mathcal{F})$ (recall that $f_{1}(\mathcal{F})=\max\{f_{1}(F):F\in\mathcal{F}\}$ ). For each $j\in[k_{1}]$ , denote

\mathcal{F}(j)=\{F\setminus[n-j+1,n]:F\in\mathcal{F},[n-j+1,n]\subseteq F\}.

To prove Claim 5.3 is equivalent to prove the following claim.

Claim 5.4.

$f_{1}(\mathcal{F})=\max\{f_{1}(I^{\prime}_{1}),f_{1}(I^{\prime\prime}_{1})\}$ , and for any $F\in\mathcal{F}$ with $I^{\prime}_{1}\precneqq F\precneqq I^{\prime\prime}_{1}$ , we have $f_{1}(F)<f_{1}(\mathcal{F})$ .

Proof.

By the definitions of $I^{\prime}_{1}$ and $I^{\prime\prime}_{1}$ , we can see that $\ell(I^{\prime}_{1})=\ell(I^{\prime\prime}_{1})=:k>0$ , $\mathcal{F}(k)=\{I^{\prime}_{1}\setminus[n-\ell(I^{\prime}_{1})+1,n],\,I^{\prime\prime}_{1}\setminus[n-\ell(I^{\prime\prime}_{1})+1,n]\}$ and for each $F\in\mathcal{F}$ , $\ell(F)\leq k$ . Denote

	$\displaystyle A_{0}$	$\displaystyle=\{x_{1},\dots,x_{i},x_{i+1},x_{i+1}+1,\dots,x_{i+1}+k_{1}-i-1\},$
	$\displaystyle A_{1}$	$\displaystyle=\{x_{1},\dots,x_{i},x_{i+1},x_{i+1}+1,\dots,x_{i+1}+k_{1}-i-2\}\cup\{n\},$
		$\displaystyle\vdots$
	$\displaystyle A_{k-1}$	$\displaystyle=\{x_{1},\dots,x_{i},x_{i+1},x_{i+1}+1\}\cup[n-k_{1}+i+3,n].$

Then $A_{0}$ is the $k_{1}$ -set with $F^{\prime}_{1}<A_{0}$ . Applying Lemma 4.8 repeatedly, we obtain

	$\displaystyle f_{1}(\mathcal{F})$	$\displaystyle=\max\{f_{1}(A_{0}),f_{1}(\mathcal{F}(1))\},$
	$\displaystyle f_{1}(\mathcal{F}(1))$	$\displaystyle=\max\{f_{1}(A_{1}),f_{1}(\mathcal{F}(2))\},$
		$\displaystyle\vdots$
	$\displaystyle f_{1}(\mathcal{F}(k-1))$	$\displaystyle=\max\{f_{1}(A_{k-1}),f_{1}(\mathcal{F}(k))\},$
	$\displaystyle f_{1}(\mathcal{F}(k))$	$\displaystyle=\max\{f_{1}(F^{\prime}_{1}),f_{1}(F^{\prime\prime}_{1})\}.$

Thus

f_{1}(\mathcal{F})=\max\{f_{1}(A_{0}),f_{1}(A_{1}),\dots,f_{1}(A_{k-1}),f_{1}(I^{\prime}_{1}),f_{1}(I^{\prime\prime}_{1})\}.

Since $k_{1}>k_{2}$ and $n\geq k_{2}+k_{t}$ , $n>k_{1}+k_{t}$ . Then by Lemma 4.8 again, we can see that for any $F\in\mathcal{F}\setminus\{A_{0},A_{1},\dots,A_{k-1},I^{\prime}_{1},I^{\prime\prime}_{1}\}$ , we have $f_{1}(F)<f_{1}(\mathcal{F})$ . Combing with the following Claims 5.5 and 5.6, we will get Claim 5.4.

Claim 5.5.

$\max\{f_{1}(A_{0}),f_{1}(A_{1}),\dots,f_{1}(A_{k-1}),f_{1}(I^{\prime\prime}_{1})\}=\max\{f_{1}(A_{0}),f_{1}(I^{\prime\prime}_{1})\}$ .

Proof.

If $k=1$ , then we are fine. We may assume that $k\geq 2$ . Let $I^{\prime\prime}_{1}=A_{k}$ . To prove Claim 5.5, it is sufficient to prove that for any $j\in[0,k-2]$ , if $f_{1}(A_{j})\leq f_{1}(A_{j+1})$ , then $f_{1}(A_{j+1})<f_{1}(A_{j+2})$ . Let $j\in[0,k-2]$ . Clearly, $\ell(A_{j+1})=\ell(A_{j})+1\geq 1$ and $\ell(A_{j+2})=\ell(A_{j+1})+1$ . Let $A^{\prime}_{j}$ and $A^{\prime}_{j+1}$ be the $k_{1}$ -sets such that $A_{j}<A^{\prime}_{j}$ and $A_{j+1}<A^{\prime}_{j+1}$ . Then $A^{\prime}_{j}\overset{\ell(A_{j+1})}{\longrightarrow}A_{j+1}$ . Let $J$ be the $k_{1}$ -set such that $A^{\prime}_{j+1}\overset{\ell(A_{i+1})}{\longrightarrow}J$ . Then $A^{\prime}_{j}$ , $A_{j+1}$ are $\ell(A_{i+1})$ -sequential and $A^{\prime}_{j+1}$ , $J$ are $\ell(A_{i+1})$ -sequential. Clearly, $\max A^{\prime}_{j}=\max A^{\prime}_{j+1}$ and $\max A_{j+1}=\max J=n$ . Applying Lemma 4.4, we have

\alpha(A^{\prime}_{j},A_{j+1})=\alpha(A^{\prime}_{j+1},J)

and

\beta(A^{\prime}_{j},A_{j+1})=\beta(A^{\prime}_{j+1},J)

(26)

If $\ell(A^{\prime}_{j})\geq 1$ , then $A_{j}<A^{\prime}_{j}=A_{j+1}<A_{j+2}=A^{\prime}_{j+1}$ . In this case, $\alpha(A_{j+1},A_{j+2})=1$ . By Proposition 4.3 and $n>k_{1}+k_{t}$ (since $k_{1}>k_{2}$ and $n\geq k_{2}+k_{t}$ ), we have $\beta(A_{j+1},A_{j+2})=0$ . So $f(A_{j+1})<f(A_{j+2})$ , as desired. Next we may assume that $\ell(A^{\prime}_{j})=0$ . Clearly, $\alpha(A_{j},A^{\prime}_{j})=\alpha(A_{j+1},A^{\prime}_{j+1})=1$ . By Proposition 4.3 and $\max A^{\prime}_{j}=\max A^{\prime}_{j+1}$ , we have $\beta(A_{j},A^{\prime}_{j})=\beta(A_{j+1},A^{\prime}_{j+1})$ . Combining with (26), we get

\alpha(A_{j},A_{j+1})=\alpha(A_{j+1},J)

and

\beta(A_{j},A_{j+1})=\beta(A_{j+1},J)

Thus $f(J)\geq f(A_{j+1})$ since $f(A_{j})\leq f(A_{j+1})$ . Note that $\ell(A_{j+1})=\ell(J)$ and $A_{j+1}\setminus A_{j+1}^{\rm t}\in\mathcal{R}_{1}(\ell(A_{j+1}))$ , so $J\setminus J^{\rm t}\in\mathcal{R}_{1}(\ell(A_{j+1}))$ and $A_{j+1}\setminus A_{j+1}^{\rm t}\overset{1}{\prec}J\setminus J^{\rm t}$ . Since $\ell(A_{j+2})=\ell(A_{j+1})+1$ , $A_{j+2}\setminus[n-\ell(A_{j+1})+1,n]\in\mathcal{R}_{1}(\ell(A_{j+1}))$ . And in $\mathcal{R}_{1}(\ell(A_{j+1}))$ , we have

A_{j+1}\setminus A_{j+1}^{\rm t}\overset{1}{\prec}J\setminus J^{\rm t}\overset{1}{\longrightarrow}A_{j+2}\setminus[n-\ell(A_{j+1})+1,n].

Recall that $n>k_{1}+k_{t}$ . By Lemma 4.8 and $f(J)\geq f(A_{j+1})$ , we obtain $f(A_{j+2})>f(J)\geq f(A_{j+1})$ , as required. ∎

Claim 5.6.

If $f_{1}(A_{0})\geq f_{1}(I^{\prime}_{1})$ , then $f_{1}(A_{0})<f_{1}(I^{\prime\prime}_{1})$ .

Proof.

Note that $I^{\prime}_{1}<A_{0}$ in $\mathcal{R}_{1}$ . Then $\alpha(I^{\prime}_{1},A_{0})=1$ . Since $f_{1}(A_{0})\geq f_{1}(I^{\prime}_{1})$ , $\alpha(I^{\prime}_{1},A_{0})\geq\beta(I^{\prime}_{1},A_{0})$ , so $\beta(I^{\prime}_{1},A_{0})\leq 1$ . Consider the family $\mathcal{F^{\prime}}:=\{F:|F|=k_{1},A_{0}\prec F\prec I^{\prime\prime}_{1}\}$ . We can see that for each $F\in\mathcal{F^{\prime}}$ , $\max F\geq\max A_{0}$ . By Proposition 4.3 and $n>k_{1}+k_{t}$ (since $k_{1}>k_{2}$ and $n\geq k_{2}+k_{t}$ ), for each two $k_{1}$ -sets $F$ and $G$ with $A_{0}\prec F<G\prec I^{\prime\prime}_{1}$ , we have that $\beta(F,G)<\beta(I^{\prime}_{1},A_{0})\leq 1$ and $\alpha(F,G)=1$ . Thus, $f_{1}(I^{\prime\prime}_{1})>f_{1}(A_{0})$ . This completes the proof of Claim 5.6 ∎

Since we have shown Claims 5.5 and 5.6, the proof of Claim 5.4 is complete. ∎

Since the proof of Claim 5.4 is complete, Claim 5.3 follows as noted before. ∎

Since the proof of Claim 5.3 is complete, the proof of Lemma 3.8 is complete.

6 Verify Unimodality: Proof of Lemma 3.12

In this section, we are going to prove a more general result Lemma 6.10 which will be applied for the most general case $n\geq k_{1}+k_{t}$ in [7]. Lemma 3.12 will follow from Lemma 6.10 immediately. Before stating the result, we need to make some preparations.

Definition 6.1.

Suppose that $t\geq 2$ is a positive integer, $s\in[t-1]$ , $k_{1}\geq k_{2}\geq\dots\geq k_{t}$ and $k_{1}+k_{s+1}\leq n<k_{s-1}+k_{s}$ . We define $\mathcal{R}_{i}$ for every $i\in[t]$ as follows. For $i\in[s]$ , let

\mathcal{R}_{i}=\{R\in{[n]\choose k_{i}}:\{1,n-k_{i}+2,\dots,n\}\prec R\prec\{k_{t},n-k_{i}+2,\dots,n\}\}.

For $i\in[s+1,t]$ , let

\mathcal{R}_{i}=\{R\in{[n]\choose k_{i}}:[k_{t}]\subseteq R\}.

In Section 3, we defined notations $\mathcal{R}_{i}$ , $1\leq i\leq t$ . Throughout the paper except in this section, $\mathcal{R}_{i}$ follows from the definitions in Section 3. $\mathcal{R}_{i}$ in this section follows from the above definition. When $s=2$ , they are consistent.

Definition 6.2.

Let $R$ and $T$ be two subsets of $[n]$ with different sizes. We write $R\overset{\sim}{<}T$ if $R\prec T$ and there is no other set $R^{\prime}$ such that $|R^{\prime}|=|R|$ and $R\precneqq R^{\prime}\prec T$ .

By the definition of parity, we have the following simple remark.

Remark 6.3.

For any $R\in\mathcal{R}_{1}$ and $i\in[2,s]$ , $R$ has a $k_{i}$ -parity if and only if $|R\setminus R^{\rm t}|\leq k_{i}$ .

From the above observation, for any $R_{1}\in\mathcal{R}_{1}$ and $i\in[2,s]$ , we define the corresponding $k_{i}$ -set of $R_{1}$ as follows.

Definition 6.4.

Let $R_{1}\in\mathcal{R}_{1}$ and $i\in[2,s]$ . If $R$ has a $k_{i}$ -parity, then let $R_{i}$ be the $k_{i}$ -parity of $R_{1}$ ; otherwise, let $R_{i}$ be the $k_{i}$ -set such that $R_{i}\overset{\sim}{<}R_{1}$ . We call $R_{i}$ the corresponding $k_{i}$ -set of $R_{1}$ .

Definition 6.5.

Let $R_{1}\in\mathcal{R}_{1}$ . For each $i\in[2,s]$ , let $R_{i}$ be the corresponding $k_{i}$ -set of $R_{1}$ as in Definition 6.4 and for each $i\in[s+1,t]$ , let $R_{i}$ be the $k_{i}$ -partner of $R_{1}$ . We denote

f(R_{1})=\sum_{i=1}^{t}|\mathcal{L}(R_{i},k_{t})|.

Definition 6.6.

For each $j\in[k_{1}-1]$ , let

	$\displaystyle\mathcal{R}_{1,j}$	$\displaystyle=\{R\in\mathcal{R}_{1}:[n-j+1,n]\subseteq R\},$
	$\displaystyle\mathcal{R}_{1}(j)$	$\displaystyle=\{R\setminus[n-j+1,n]:R\in\mathcal{R}_{1,j}\}.$

For any $j\in[k_{1}-1]$ and any $R\in\mathcal{R}_{1,j}$ , we define $f(R\setminus[n-j+1,n])=f(R)$ . For example, $f(\{1\})=f(\{1,n-k_{1}+2,\dots,n\})$ .

Definition 6.7.

Let $R_{1},R^{\prime}_{1}\in\mathcal{R}_{1}$ with $R_{1}\prec R^{\prime}_{1}$ and for any $i\in[s+1,t]$ , $R_{i},R^{\prime}_{i}$ be the $k_{i}$ -partners of $R_{1},R^{\prime}_{1}$ respectively, and for each $i\in[2,s]$ , let $R_{i},R^{\prime}_{i}$ be the corresponding $k_{i}$ -sets of $R_{1},R^{\prime}_{1}$ respectively as in Definition 6.4. We define the following functions.

For $i\in[s]$ , let

	$\displaystyle\alpha_{i}(R_{1},R^{\prime}_{1})$	$\displaystyle=\|\mathcal{L}(R^{\prime}_{i},k_{1})\|-\|\mathcal{L}(R_{i},k_{1})\|,$
	$\displaystyle\gamma(R_{1},R^{\prime}_{1})$	$\displaystyle=\sum_{i=1}^{s}\alpha_{i}(R_{1},R^{\prime}_{1}),$
	$\displaystyle\delta(R_{1},R^{\prime}_{1})$	$\displaystyle=\sum_{i=s+1}^{t}\big{(}\|\mathcal{L}(R_{i},k_{i})\|-\|\mathcal{L}(R^{\prime}_{i},k_{i})\|\big{)}.$

Definition 6.8.

For any $j\in[k_{1}-1]$ and any $R_{1},R^{\prime}_{1}\in\mathcal{R}_{1,j}$ with $R_{1}\prec R^{\prime}_{1}$ , we define $\alpha_{i}(R_{1}\setminus[n-j+1,n],R^{\prime}_{1}\setminus[n-j+1,n])=\alpha_{i}(R_{1},R^{\prime}_{1})$ for each $i\in[s]$ , $\gamma(R_{1}\setminus[n-j+1,n],R^{\prime}_{1}\setminus[n-j+1,n])=\gamma(R_{1},R^{\prime}_{1})$ and $\delta(R_{1}\setminus[n-j+1,n],R^{\prime}_{1}\setminus[n-j+1,n])=\delta(R_{1},R^{\prime}_{1})$ .

From the above definition, we have

f(R^{\prime}_{1})-f(R_{1})=\gamma(R_{1},R^{\prime}_{1})-\delta(R_{1},R^{\prime}_{1}).

Remark 6.9.

Suppose that $A_{1},B_{1},C_{1}\in\mathcal{R}_{1}$ with $A_{1}\prec B_{1}\prec C_{1}$ . Then for any $i\in[s]$ , $\alpha_{i}(A_{1},C_{1})=\alpha_{i}(A_{1},B_{1})+\alpha_{i}(B_{1},C_{1})$ , and $\gamma(A_{1},C_{1})=\gamma(A_{1},B_{1})+\gamma(B_{1},C_{1})$ , $\delta(A_{1},C_{1})=\delta(A_{1},B_{1})+\delta(B_{1},C_{1})$ .

We will prove the following more general lemma, which implies Lemma 3.12.

Lemma 6.10.

Let $k\in[0,k_{1}-1]$ and $F_{1},G_{1},H_{1}\in\mathcal{R}_{1}(k)$ with $F_{1}\overset{c}{\prec}G_{1}\overset{c}{\prec}H_{1}$ for some $c\in[k_{1}]$ . If $f(F_{1})\leq f(G_{1})$ , then $f(G_{1})<f(H_{1})$ .

Let us explain why Lemma 6.10 implies Lemma 3.12. Take $s=2$ (see Definition 6.1). Take $F_{2},G_{2},H_{2}\in\mathcal{F}_{2,3}(j)$ satisfying the condition $F_{2}\overset{c}{\prec}G_{2}\overset{c}{\prec}H_{2}$ (see Lemma 3.12). Let $F^{\prime}_{2}=F_{2}\cup[n-j+1,n],G^{\prime}_{2}=G_{2}\cup[n-j+1,n]$ and $H^{\prime}_{2}=H_{2}\cup[n-j+1,n]$ . Then $F^{\prime}_{2},G^{\prime}_{2},H^{\prime}_{2}\in\mathcal{F}_{2,3}$ and $g(F_{2})=g(F^{\prime}_{2})$ , $g(G_{2})=g(G^{\prime}_{2})$ and $g(H_{2})=g(H^{\prime}_{2})$ (see Definition 3.11). Let $F^{\prime}_{1},G^{\prime}_{1},H^{\prime}_{1}$ be the $k_{1}$ -parities of $F^{\prime}_{2},G^{\prime}_{2},H^{\prime}_{2}$ respectively. (Proposition 2.16) guarantees the $k_{1}$ -parities of $F^{\prime}_{2},G^{\prime}_{2},H^{\prime}_{2}$ exist.) Take $k=j+k_{1}-k_{2}$ in Lemma 6.10. Let $F_{1}=F^{\prime}_{1}\setminus[n-k+1,n]$ , $G_{1}=G^{\prime}_{1}\setminus[n-k+1,n]$ and $H_{1}=H^{\prime}_{1}\setminus[n-k+1,n]$ . Then $F_{1},G_{1},H_{1}\in\mathcal{R}_{1}(k)$ with $F_{1}\overset{c}{\prec}G_{1}\overset{c}{\prec}H_{1}$ and $f(F_{1})=g(F^{\prime}_{1})$ , $f(G_{1})=f(G^{\prime}_{1})$ and $f(H_{1})=f(H^{\prime}_{1})$ (see Definition 6.6). By Fact 2.14, for each $i\in[3,t]$ , $F^{\prime}_{1}$ and $F^{\prime}_{2}$ have the same $k_{i}$ -partner, $G^{\prime}_{1}$ and $G^{\prime}_{2}$ have the same $k_{i}$ -partner and $H^{\prime}_{1}$ and $H^{\prime}_{2}$ have the same $k_{i}$ -partner. Therefore, $g(F_{2})=g(F^{\prime}_{2})=f(F^{\prime}_{1})=f(F_{1})$ , $g(G_{2})=g(G^{\prime}_{2})=f(G^{\prime}_{1})=f(G_{1})$ and $g(H_{2})=g(H^{\prime}_{2})=f(H^{\prime}_{1})=f(H_{1})$ . Now applying Lemma 6.10, we have that if $f(F_{1})=f(F^{\prime}_{1})=g(F_{2})\leq g(G_{2})=f(G^{\prime}_{1})=f(G_{1})$ , then $g(G_{2})=f(G^{\prime}_{1})=f(G_{1})<f(H_{1})=f(H^{\prime}_{1})=g(H_{2})$ .

Before proving Lemma 6.10, we need to make some preparations. Denote

s^{\prime}=\{i:i\in[s],k_{i}=k_{1}\}.

(27)

The following remark is useful.

Remark 6.11.

When $k_{1}=k_{2}=\dots=k_{s}$ , the authors have proved the truth of Theorem 1.4 in [6], see Corollary 1.12 in [6] (taking $k=k_{1}=\dots=k_{s}$ in Corollary 1.12). So we may assume that $s^{\prime}<s$ . So $n>k_{s^{\prime}}+k_{s+1}$ .

Claim 6.12.

Let $R_{1},R^{\prime}_{1}\in\mathcal{R}_{1}$ with $R_{1}<R^{\prime}_{1}$ and $R=R^{\prime}_{1}\setminus{R^{\prime}_{1}}^{\rm t}$ Then for each $i\in[s]$ , we have

\alpha_{i}(R_{1},R^{\prime}_{1})={\ell(R^{\prime}_{1})\choose k_{i}-|R|}.

In particular, $\alpha_{i}(R_{1},R^{\prime}_{1})=0$ if and only if $\ell(R^{\prime}_{1})<k_{1}-k_{i}$ . Furthermore, if $\ell(R^{\prime}_{1})=0$ , then $\gamma(R_{1},R^{\prime}_{1})=s^{\prime}$ .

Proof.

Clearly, for $i\in[s^{\prime}]$ , $\alpha_{i}(R_{1},R^{\prime}_{1})=1={\ell(R^{\prime}_{1})\choose k_{i}-|R|}$ . We next consider for $i\in[s^{\prime}+1,s]$ . In this case, $k_{i}<k_{1}$ . Let $R_{i}$ and $R^{\prime}_{i}$ be the corresponding $k_{i}$ -sets of $R_{1}$ and $R^{\prime}_{1}$ respectively as in Definition 6.4. Then $R_{i}\prec R^{\prime}_{i}$ , moreover $\alpha_{i}(R_{1},R^{\prime}_{1})\geq 0$ and $\alpha_{i}(R_{1},R^{\prime}_{1})=0$ if and only if $R_{i}=R^{\prime}_{i}$ clearly. $R_{i}=R^{\prime}_{i}$ implies $R^{\prime}_{i}\overset{\sim}{<}R^{\prime}_{1}$ , furthermore, $\ell(R^{\prime}_{1})<k_{1}-k_{i}$ . Then we can see that if $\ell(R^{\prime}_{1})=0$ , then $\alpha_{i}(R_{1},R^{\prime}_{1})=0={\ell(R^{\prime}_{1})\choose k_{i}-|R|}$ . Therefore, $\gamma(R_{1},R^{\prime}_{1})=s^{\prime}$ , as required. We next assume that $R^{\prime}_{i}$ is the $k_{i}$ -parirty of $R^{\prime}_{1}$ . So $\ell(R^{\prime}_{1})\geq 1$ . In this case, since $R_{1}<R^{\prime}_{1}$ , $\ell(R_{1})=\ell(R^{\prime}_{1})-1$ . If $R_{i}\overset{\sim}{<}R_{1}$ , then $\ell(R^{\prime}_{1})=k_{i}-|R|$ and $\alpha_{i}(R_{1},R^{\prime}_{1})=1={\ell(R^{\prime}_{1})\choose k_{i}-|R|}$ , as required. Last, we assume that $R_{i}$ is the $k_{i}$ -parirty of $R_{1}$ . So $\ell(R_{1})\geq 1$ . Let $k_{1}-k_{i}=k$ . In this case we have

	$\displaystyle R^{\prime}_{1}$	$\displaystyle=R\cup[n-\ell(R^{\prime}_{1})+1,n];$
	$\displaystyle R_{1}$	$\displaystyle=R\cup\{n-\ell(R^{\prime}_{1})\}\cup[n-\ell(R^{\prime}_{1})+2,n];$
	$\displaystyle R_{i}$	$\displaystyle=R\cup\{n-\ell(R^{\prime}_{1})\}\text{ if $\ell(R_{i})=0$,}$
	$\displaystyle R_{i}$	$\displaystyle=R\cup\{n-\ell(R^{\prime}_{1})\}\cup[n-\ell(R^{\prime}_{1})+2+k,n]\text{ if $\ell(R_{i})\geq 1$};$
	$\displaystyle R^{\prime}_{i}$	$\displaystyle=R\cup[n-\ell(R^{\prime}_{1})+1+k,n].$

Then

\alpha_{i}(R_{1},R^{\prime}_{1})=|\mathcal{L}(R^{\prime}_{i},k_{i})|-|\mathcal{L}(R_{i},k_{i})|={\ell(R^{\prime}_{1})\choose k_{i}-|R|},

as required. ∎

From Definitions 6.7 and 4.2, we have the following observation.

Remark 6.13.

For any $R_{1},R_{1}^{\prime}\in\mathcal{R}_{1}$ with $R_{1}\prec R^{\prime}_{1}$ , we have that $\alpha_{i}(R_{1},R^{\prime}_{1})=\alpha(R_{1},R^{\prime}_{1})$ holds for each $i\in[s^{\prime}]$ , and $\delta(R_{1},R^{\prime}_{1})=\beta(R_{1},R^{\prime}_{1})$ .

Claim 6.14.

Let $j\in[0,k_{1}-1]$ , $d\in[k_{1}-j]$ and $A_{1},B_{1},C_{1},D_{1}\in\mathcal{R}_{1}(j)$ . Suppose that $A_{1},B_{1}$ are $d$ -sequential and $C_{1},D_{1}$ are $d$ -sequential with $\max A_{1}=\max C_{1}$ and $\max B_{1}=\max D_{1}$ . Then $\gamma(A_{1},B_{1})=\gamma(C_{1},D_{1})$ and $\delta(A_{1},B_{1})=\delta(C_{1},D_{1})$ . In particular, if $A_{1}\overset{d}{\longrightarrow}B_{1}$ , $C_{1}\overset{d}{\longrightarrow}D_{1}$ and $\max A_{1}=\max C_{1}$ , then $\gamma(A_{1},B_{1})=\gamma(C_{1},D_{1})$ and $\delta(A_{1},B_{1})=\delta(C_{1},D_{1})$ .

Proof.

By the definitions of $A_{1},B_{1},C_{1},D_{1}$ , from Lemma 4.7 and Remark 6.13, we have $\delta(A_{1},B_{1})=\delta(C_{1},D_{1})$ and $\alpha_{i}(A_{1},B_{1})=\alpha_{i}(C_{1},D_{1})$ holds for each $i\in[s^{\prime}]$ . Next, we aim to show that for each $i\in[s^{\prime}+1,s]$ , $\alpha_{i}(A_{1},B_{1})=\alpha_{i}(C_{1},D_{1})$ . Let $i\in[s^{\prime}+1,s]$ . Denot

	$\displaystyle\mathcal{A}$	$\displaystyle=\{R:A_{1}\cup[n-j+1,n]\prec R\prec B_{1}\cup[n-j+1,n]\text{ and }\|R\|=k_{1}\},$
	$\displaystyle\mathcal{B}$	$\displaystyle=\{T:C_{1}\cup[n-j+1,n]\prec T\prec D_{1}\cup[n-j+1,n]\text{ and }\|T\|=k_{1}\}.$

Since $\alpha_{1}(A_{1},B_{1})=\alpha_{1}(C_{1},D_{1})$ , $|\mathcal{A}|=|\mathcal{B}|=:h$ . Let $\mathcal{A}=\{R_{1},R_{2},\dots,R_{h}\}$ and $\mathcal{B}=\{T_{1},T_{2},\dots,T_{h}\}$ , where $R_{1}\prec R_{2}\prec\dots\prec R_{h}$ and $T_{1}\prec T_{2}\prec\dots\prec T_{h}$ . For any $j\in[h]$ , we have $\ell(R_{j})=\ell(T_{j})$ and $|R_{j}\setminus R_{j}^{\rm t}|=|T_{j}\setminus T_{j}^{\rm t}|$ . Thus, by Claim 6.12, for any $j\in[h-1]$ , $\alpha_{i}(R_{j},R_{j+1})=\alpha_{i}(T_{j},T_{j+1})$ . Furthermore, by Remark 6.9, we conclude that $\alpha_{i}(A_{1},B_{1})=\alpha_{i}(C_{1},D_{1})$ . ∎

Claim 6.15.

Let $A_{1},B_{1},C_{1}\in\mathcal{R}_{1}$ and $a$ be an integer. Suppose $A_{1}\setminus A_{1}^{\rm t}=A\cup\{a,a+1\}$ , $B_{1}=A\cup\{a\}\cup[n-\ell(B_{1})+1,n]$ and $C_{1}=A\cup\{a+1\}\cup[n-\ell(B_{1})+1,n]$ for some $A$ . Then $\delta(A_{1},B_{1})=\delta(B_{1},C_{1})$ . If $a+1\leq n-\ell(A_{1})-2$ , then $\gamma(A_{1},B_{1})=\gamma(B_{1},C_{1})$ . If $a+1=n-\ell(A_{1})-1$ , then $\gamma(A_{1},B_{1})\leq\gamma(B_{1},C_{1})$ , equality holds if and only if $C_{1}$ does not have $k_{s^{\prime}+1}$ -parity (recall that $s^{\prime}$ is the integer such that $k_{1}=\dots=k_{s^{\prime}}>k_{s^{\prime}+1}$ ).

Proof.

Let $A^{\prime}_{1}$ and $B^{\prime}_{1}$ be the $k_{1}$ -sets such that $A_{1}<A^{\prime}_{1}$ and $B_{1}<B^{\prime}_{1}$ . Then $\max A^{\prime}_{1}=\max B^{\prime}_{1}$ , and $A^{\prime}_{1}$ , $B_{1}$ are $(\ell(A_{1})+1)$ -sequential, $B^{\prime}_{1}$ , $C_{1}$ are $(\ell(A_{1})+1)$ -sequential. Note that $\max B_{1}=\max C_{1}=n$ , then applying Lemma 4.4 and Reamrk 6.13, we have that $\alpha_{i}(A^{\prime}_{1},B_{1})=\alpha_{i}(B^{\prime}_{1},C_{1})$ holds for each $i\in[s^{\prime}]$ and $\delta(A^{\prime}_{1},B_{1})=\delta(B^{\prime}_{1},C_{1})$ . Clearly, $\alpha_{1}(A_{1},A^{\prime}_{1})=\alpha_{1}(B_{1},B^{\prime}_{1})=1$ holds for each $i\in[s^{\prime}]$ and using Proposition 4.3 and Remark 6.13, we have $\delta(A_{1},A^{\prime}_{1})=\delta(B_{1},B^{\prime}_{1})$ . Therefore, by Remark 6.9, we have $\delta(A_{1},B_{1})=\delta(B_{1},C_{1})$ . Clearly, for each $i\in[s^{\prime}]$ ,

\alpha_{i}(A_{1},B_{1})=\alpha_{i}(A_{1},A^{\prime}_{1})+\alpha_{i}(A^{\prime}_{1},B_{1})=\alpha_{i}(B_{1},B^{\prime}_{1})+\alpha_{i}(A^{\prime}_{1},C_{1})=\alpha_{i}(B_{1},C_{1}).

(28)

Since $A_{1}\setminus A_{1}^{\rm t}=A\cup\{a,a+1\}$ , $a+1\leq n-\ell(A_{1})-1$ . By the definitions of $A_{1}$ , $B_{1}$ , and $C_{1}$ , if $a+1\leq n-\ell(A_{1})-2$ , then $\ell(B_{1})=\ell(C_{1})=\ell(A_{1})+1$ , and if $a+1=n-\ell(A_{1})-1$ , then $\ell(B_{1})=\ell(A_{1})+1$ and $\ell(C_{1})=\ell(B_{1})+1$ . Using Claim 6.14, for each $j\in[s^{\prime}+1,s]$ , we have

\alpha_{j}(A^{\prime}_{1},B_{1})=\alpha_{j}(B^{\prime}_{1},C_{1}).

(29)

If the previous case happens, then $\ell(A^{\prime}_{1})=\ell(B^{\prime}_{1})=0$ , so Claim 6.12 gives $\alpha_{i}(A_{1},A^{\prime}_{1})=\alpha_{i}(B_{1},B^{\prime}_{1})=0$ for each $i\in[s^{\prime}+1,s]$ . Combing this with (28), (29) and Remark 6.9, we get

	$\displaystyle\gamma(A_{1},B_{1})$	$\displaystyle=\sum_{i=1}^{s}\alpha_{i}(A_{1},B_{1})$
		$\displaystyle=\sum_{i=1}^{s^{\prime}}\alpha_{i}(A_{1},B_{1})+\sum_{i=s^{\prime}+1}^{s}(\alpha_{i}(A_{1},A^{\prime}_{1})+\alpha_{i}(A^{\prime}_{1},B_{1}))$
		$\displaystyle=\sum_{i=1}^{s^{\prime}}\alpha_{1}(B_{1},C_{1})+\sum_{i=s^{\prime}+1}^{s}(\alpha_{i}(B_{1},B^{\prime}_{1})+\alpha_{i}(B^{\prime}_{1},C_{1}))$
		$\displaystyle=\gamma(B_{1},C_{1}),$

as desired. If the later case happens, then $A_{1}<B_{1}<C_{1}$ , by Claim 6.12, since $\ell(B_{1})=\ell(A_{1})+1$ and $\ell(C_{1})=\ell(B_{1})+1$ , then for each $j\in[s^{\prime}+1,s]$ , we have $\alpha_{j}(A_{1},B_{1})=\alpha_{j}(B_{1},C_{1})$ if $k_{j}<|C_{1}\setminus C_{1}^{\rm t}|$ , i.e., $C_{1}$ dest not have $k_{j}$ -parity; and $\alpha_{j}(A_{1},B_{1})<\alpha_{j}(B_{1},C_{1})$ if $k_{j}\geq|C_{1}\setminus C_{1}^{\rm t}|$ , i.e., $C_{1}$ has $k_{j}$ -parity. Note that $k_{s^{\prime}+1}\geq\dots\geq k_{s}$ , so if $C_{1}$ does not have $k_{s^{\prime}+1}$ -parity, then it does not have any $k_{j}$ -parity for $j\in[s^{\prime}+1,s]$ . Therefore, $\sum_{j=1}^{s}\alpha_{j}(A_{1},B_{1})\leq\sum_{j=1}^{s}\alpha_{j}(B_{1},C_{1})$ , and the equality holds if and only if $C_{1}$ has $k_{s^{\prime}+1}$ -parity, that is, $\gamma(A_{1},B_{1})\leq\gamma(B_{1},C_{1})$ , and the equality holds if and only if $C_{1}$ does not have $k_{s^{\prime}+1}$ -parity, as desired. ∎

We are going to prove Lemma 6.10 by induction on $k$ .

6.1 The Base Case

We are going to show that Lemma 6.10 holds for $k=0$ . Assume that $F_{1},G_{1},H_{1}\in\mathcal{R}_{1}$ with $F_{1}\overset{c}{\prec}G_{1}\overset{c}{\prec}H_{1}$ for some $c\in[k_{1}]$ and $f(F_{1})\leq f(G_{1})$ , we are going to show that $f(G_{1})<f(H_{1})$ . Since $F_{1},G_{1},H_{1}\in\mathcal{R}_{1}$ with $F_{1}\overset{c}{\prec}G_{1}\overset{c}{\prec}H_{1}$ , $\max F_{1}\leq n-2$ , $\max G_{1}\leq n-1$ and $\max H_{1}\leq n$ . Let $G^{\prime}_{1}$ and $H^{\prime}_{1}$ be the $k_{1}$ -sets such that $G^{\prime}_{1}<G_{1}$ and $H^{\prime}_{1}<H_{1}$ . we first show the following observation.

Claim 6.16.

Let $F_{m}$ be the $k_{1}$ -set such that $F_{1}\overset{c}{\longrightarrow}F_{m}$ , where $m$ is the number of $k_{1}$ -subsets $R$ satisfying $F_{1}\prec R\prec F_{m}$ . If $\delta(G^{\prime}_{1},G_{1})<s^{\prime}$ , then $f(R)$ , where $R\in\mathcal{F}$ , increases on $\mathcal{F}$ , in particular, $f(G_{1})<f(H_{1})$ .

Proof.

We may assume that $\mathcal{F}=\{F_{i}:1\leq i\leq m\}$ and $F_{1}<F_{2}<\dots<F_{m}$ . Then $\max F_{2}=\max G_{1}$ . Since $F_{1}<F_{2}$ and $G^{\prime}_{1}<G_{1}$ , by Proposition 4.3 and Remark 6.13, $\delta(F_{1},F_{2})=\delta(G^{\prime}_{1},G_{1})<s^{\prime}$ (assumption). Since $\max F_{i}\geq F_{2}$ for any $i\in[2,m]$ , then by Proposition 4.3 and Remark 6.11, $\delta(F_{i},F_{i+1})<s^{\prime}$ holds for all $i\in[m-1]$ . On the other hand, $\gamma(F_{i},F_{i+1})\geq s^{\prime}$ for all $i\in[m-1]$ , thus, $f(R)$ , where $R\in\mathcal{F}$ , increases on $\mathcal{F}$ . Clearly, $G_{1},H_{1}\in\mathcal{F}$ and $G_{1}\precneqq H_{1}$ , therefore, $f(G_{1})<f(H_{1})$ , as required. ∎

By Claim 6.16, next we may assume that

\displaystyle\delta(G^{\prime}_{1},G_{1})\geq s^{\prime}.

(30)

Note that

	$\displaystyle f(G_{1})$	$\displaystyle=f(G^{\prime}_{1})+\gamma(G^{\prime}_{1},G_{1})-\delta(G^{\prime}_{1},G_{1}),$		(31)
	$\displaystyle f(H_{1})$	$\displaystyle=f(H^{\prime}_{1})+\gamma(H^{\prime}_{1},H_{1})-\delta(H^{\prime}_{1},H_{1}).$		(32)

Since $\ell(G_{1})=0$ , by Claim 6.12, for each $i\in[s^{\prime}+1,s]$ , we have

\alpha_{i}(G^{\prime}_{1},G_{1})=0\leq\alpha_{i}(H^{\prime}_{1},H_{1}).

(33)

Clearly, for $i\in[s^{\prime}]$ , $\alpha_{i}(G^{\prime}_{1},G_{1})=\alpha_{i}(H^{\prime}_{1},H_{1})=1$ . So $\gamma(G^{\prime}_{1},G_{1})\leq\gamma(H^{\prime}_{1},H_{1})$ . Note that $\max G_{1}<\max H_{1}$ , combining Proposition 4.3, Remark 6.11, Remark 6.13 and (30), we obtain $\delta(G^{\prime}_{1},G_{1})>\delta(H^{\prime}_{1},H_{1})$ . Combining with equalities (31) and (32), to show $f(G_{1})<f(H_{1})$ , it is sufficient to show the following claim.

Claim 6.17.

$f(G^{\prime}_{1})\leq f(H^{\prime}_{1})$ .

Proof.

If $c=1$ , then $F_{1}=G^{\prime}_{1}$ and $G_{1}=H^{\prime}_{1}$ . Since $f(F_{1})\leq f(G_{1})$ , $f(G^{\prime}_{1})\leq f(H^{\prime}_{1})$ , as desired. Thus, we have proved for $c=1$ .

Next, we consider $c\geq 2$ . In this case, $F_{1}\overset{c-1}{\longrightarrow}G^{\prime}_{1}$ and $G_{1}\overset{c-1}{\longrightarrow}H^{\prime}_{1}$ . Let $F^{\prime}_{1}$ be the $k_{1}$ -set such that $F_{1}\overset{c-1}{\prec}F^{\prime}_{1}$ . Therefore, $\max F^{\prime}_{1}=\max G_{1}$ and $F^{\prime}_{1}\overset{c-1}{\longrightarrow}G^{\prime}_{1}$ . Note that $\max G^{\prime}_{1}=\max H^{\prime}_{1}=n$ . By Lemma 4.4 and Remark 6.13, we have $\alpha_{1}(F^{\prime}_{1},G^{\prime}_{1})=\alpha_{1}(G_{1},H^{\prime}_{1})$ and $\delta(F^{\prime}_{1},G^{\prime}_{1})=\delta(G_{1},H^{\prime}_{1})$ . Let $F^{\prime\prime}_{1}$ be the $k_{1}$ -set such that $F^{\prime\prime}_{1}<F^{\prime}_{1}$ . Thus, $F^{\prime\prime}_{1}$ , $G^{\prime}_{1}$ and $H^{\prime}_{1}$ satisfy the condition of Claim 6.15. So we conclude that $\gamma(F^{\prime\prime}_{1},G^{\prime}_{1})\leq\gamma(G^{\prime}_{1},H^{\prime}_{1})$ and $\delta(F^{\prime\prime}_{1},G^{\prime}_{1})=\delta(G^{\prime}_{1},H^{\prime}_{1})$ . Note that

	$\displaystyle f(G^{\prime}_{1})=f(F^{\prime\prime}_{1})+\gamma(F^{\prime\prime}_{1},G^{\prime}_{1})-\delta(F^{\prime\prime}_{1},G^{\prime}_{1}),$
	$\displaystyle f(H^{\prime}_{1})=f(G^{\prime}_{1})+\gamma(G^{\prime}_{1},H^{\prime}_{1})-\delta(G^{\prime}_{1},H^{\prime}_{1}).$

Thus, to show $f(G^{\prime}_{1})\leq f(H^{\prime}_{1})$ is sufficient to show the following claim.

Claim 6.18.

$f(F^{\prime\prime}_{1})\leq f(G^{\prime}_{1})$ .

Proof.

Suppose on the contrary that $f(F^{\prime\prime}_{1})>f(G^{\prime}_{1})$ . Since $\alpha_{i}(G^{\prime}_{1},G_{1})=1$ holds for each $i\in[s^{\prime}]$ and $\alpha_{j}(G^{\prime}_{1},G_{1})=0$ holds for each $j\in[s^{\prime}+1,s]$ (see (33)), $\gamma(G^{\prime}_{1},G_{1})=s^{\prime}$ . Since $\delta(G^{\prime}_{1},G_{1})\geq s^{\prime}$ and $f(G_{1})=f(G^{\prime}_{1})+\gamma(G^{\prime}_{1},G_{1})-\delta(G^{\prime}_{1},G_{1})$ , $f(G^{\prime}_{1})\geq f(G_{1})$ . Therefore, $f(G^{\prime}_{1})\geq f(F_{1})$ since $f(G_{1})\geq f(F_{1})$ . Hence, $f(F^{\prime\prime}_{1})>f(F_{1})$ since $f(F^{\prime\prime}_{1})>f(G^{\prime}_{1})$ . This implies $F_{1}\neq F^{\prime\prime}_{1}$ . Therefore, $c\geq 3$ and $F_{1}\overset{c-2}{\longrightarrow}F^{\prime\prime}_{1}$ . Since $F_{1}\overset{c}{\prec}G_{1}$ , we may assume that there exists some $(k_{1}-c)$ -set $F$ and some integer $x$ such that $F_{1}=F\cup[x+1,x+c]$ and $\max F\leq x$ if $F\neq\emptyset$ . Then $F^{\prime\prime}_{1}=F\cup\{x+1,x+2\}\cup[n-c+3,n]$ .

The forthcoming claim will be used to make a contradiction to $f(F^{\prime\prime}_{1})>f(G^{\prime}_{1})$ , thereby ending the proof of Claim 6.18. We will explain this after stating the claim.

Claim 6.19.

Let $d$ be a positive integer, $B_{1}=B\cup[y+1,y+d]$ for some set $B$ with $\max B\leq y$ and $y+d<n$ . Suppose $i\in[d]$ and $C_{1}$ is the $k_{1}$ -set such that $B_{1}\overset{i}{\longrightarrow}C_{1}$ . Let $p=n-y-d$ . We define $k_{1}$ -sets $D_{1},D_{2},\dots,D_{p}$ as follows. If $i=1$ , then let $D_{1}=B_{1}$ and $D_{j}=B\cup[y+1,y+d-1]\cup\{y+d+j-1\}$ for each $j\in[2,p]$ , when $d=1$ , we regard $[y+1,y+d-1]$ as an empty set. If $i\geq 2$ , then let $D_{j}=B\cup[y+1,y+d-i]\cup\{y+d+j-i\}\cup[n-i+2,n]$ for each $j\in[p]$ , when $d=i$ , we regard $[y+1,y+d-i]$ as an empty set. If $f(B_{1})\leq f(C_{1})$ , then for each $j\in[p]$ , $f(D_{j})\leq f(C_{1})$ .

To finish the proof of Claim 6.18, we only need to prove Claim 6.19. Assume that Claim 6.19 holds, taking $B_{1}=F_{1}$ , $y=x$ , $d=c$ and $i=c-1$ in Claim 6.19, we can see that $C_{1}=G^{\prime}_{1}$ and $D_{1}=F^{\prime\prime}_{1}$ . Since $f(F_{1})\leq f(G^{\prime}_{1})$ , then Claim 6.19 gives $f(F^{\prime\prime}_{1})\leq f(G^{\prime}_{1})$ , which is a contradiction to the assumption that $f(F^{\prime\prime}_{1})>f(G^{\prime}_{1})$ . This completes the proof of Claim 6.18 by assuming the truth of Claim 6.19. ∎

Proof for Claim 6.19.

We are going to prove Claim 6.19 by induction on $i$ .

We first consider the case $i=1$ . In this case, $B_{1}=D_{1}\overset{1}{\prec}D_{2}\overset{1}{\prec}\dots\overset{1}{\prec}D_{p}\overset{1}{\prec}C_{1}$ . If $p=1$ , then we have nothing to say. We may assume $p\geq 2$ . If $f(D_{2})>f(C_{1})$ , then $f(D_{1})<f(D_{2})$ since $f(B_{1})=f(D_{1})\leq f(C_{1})$ . Note that we have already proved Lemma 6.10 when $c=1$ . By taking $c=1$ in Lemma 6.10, we have $f(D_{2})<f(D_{3})<\dots<f(D_{p})<f(C_{1})$ , a contradiction. Using the same argument, we can see that for each $j\in[2,p]$ , $f(D_{j})<f(C_{1})$ , as required.

We next consider the case $i\geq 2$ and assume that Claim 6.19 holds for $i-1$ . If $f(D_{1})\geq f(C_{1})$ , then $f(D_{1})\geq f(B_{1})$ since $f(B_{1})\leq f(C_{1})$ . Note that $B_{1}\overset{i-1}{\longrightarrow}D_{1}$ . By replacing $C_{1}$ with $D_{1}$ and $i$ with $i-1$ in Claim 6.19, we define $E_{1},E_{2},\dots,E_{p}$ as definitions of $D_{1},D_{2},\dots,D_{p}$ . Then by induction hypothesis, we obtain

f(E_{j})\leq f(D_{1})

holds for each

j\in[p]

(34)

For convenience, we denote $C_{1}=D_{p+1}$ . We have the following claim.

Claim 6.20.

For each $j\in[p]$ , $\delta(E_{j},D_{1})=\delta(D_{j},D_{j+1})$ . For each $j\in[p-1]$ , $\gamma(E_{j},D_{1})=\gamma(D_{j},D_{j+1})$ and $\gamma(E_{p},D_{1})\leq\gamma(D_{p},D_{p+1})$ .

Proof.

Note that $E_{p}<D_{1}$ , $D_{p}<D_{p+1}$ , $\ell(D_{1})=\ell(D_{p+1})-1$ and $|D_{1}\setminus D_{1}^{\rm t}|=|D_{p+1}\setminus D_{p+1}^{\rm t}|+1$ . By Claim 6.12, for each $j\in[s^{\prime}+1,s]$ , $\alpha_{j}(E_{p},D_{1})\leq\alpha_{j}(D_{p},D_{p+1})$ . Trivially, for each $j\in[s^{\prime}]$ , $\alpha_{i}(E_{p},D_{1})\leq\alpha_{i}(D_{p},D_{p+1})=1$ , therefore, $\gamma(E_{p},D_{1})\leq\gamma(D_{p},D_{p+1})$ , as required.

For each $j\in[2,p]$ , let $F_{j}$ and $H_{j}$ be the $k_{1}$ -sets such that $D_{j-1}<F_{j}$ and $E_{j-1}<H_{j}$ . Thus, for each $j\in[2,p]$ , $F_{j}\overset{i-1}{\longrightarrow}D_{j}$ and we have $B_{1}\overset{i-1}{\prec}H_{2}\overset{i-1}{\prec}H_{3}\overset{i-1}{\prec}\dots\overset{i-1}{\prec}H_{p}\overset{i-1}{\prec}D_{1}$ and $B_{1}\overset{i}{\prec}F_{2}\overset{i}{\prec}F_{3}\overset{i}{\prec}\dots\overset{i}{\prec}F_{p}\overset{i}{\prec}C_{1}.$ Then for each $j\in[2,p]$ , we have $\max H_{j}=\max F_{j}$ . By Claim 6.14, $\gamma(H_{j},D_{1})=\gamma(F_{j},D_{j})$ and $\delta(H_{j},D_{1})=\delta(F_{j},D_{j})$ . Note that for each $j\in[2,p]$ , $D_{j-1}<F_{j}$ and $E_{j-1}<H_{j}$ , hence Claim 6.12 gives $\gamma(D_{j-1},F_{j})=\gamma(E_{j-1},H_{j})$ , Proposition 4.3 and Remark 6.13 give $\delta(D_{j-1},F_{j})=\delta(E_{j-1},H_{j})$ . So for each $j\in[2,p]$ , we have

	$\displaystyle\gamma(E_{j-1},D_{1})$	$\displaystyle=\gamma(E_{j-1},H_{j})+\gamma(H_{j},D_{1})$
		$\displaystyle=\gamma(D_{j-1},F_{j})+\gamma(F_{j},D_{j})$
		$\displaystyle=\gamma(D_{j-1},D_{j}),$

and

	$\displaystyle\delta(E_{j-1},D_{1})$	$\displaystyle=\delta(E_{j-1},H_{j})+\delta(H_{j},D_{1})$
		$\displaystyle=\delta(D_{j-1},F_{j})+\delta(F_{j},D_{j})$
		$\displaystyle=\delta(D_{j-1},D_{j}).$

Since $E_{p}<D_{1}$ , $D_{p}<D_{p+1}$ and $\max D_{1}=\max D_{p+1}=\max C_{1}=n$ . By Proposition 4.3 and Remark 6.13, $\delta(E_{p},D_{1})=\beta(E_{p},D_{1})=\beta(D_{p},D_{p+1})=\delta(D_{p},D_{p+1})$ . This completes the proof of Claim 6.20. ∎

Let us continue the proof of Claim 6.19. Note that for each $j\in[p]$ , $f(D_{j+1})=f(D_{j})+\gamma(D_{j},D_{j+1})-\delta(D_{j},D_{j+1})$ and $f(D_{1})=f(E_{j})+\gamma(E_{j},D_{1})-\delta(E_{j},D_{1})$ . By Claim 6.20 and (34), we conclude that

f(D_{1})\leq f(D_{2})\leq\dots\leq f(D_{p+1})=f(C_{1}).

This completes the proof of Claim 6.19. ∎

Since we have shown that Claim 6.19 holds, Claim 6.17 holds. ∎

This completes the base case of Lemma 6.10. We next to consider the induction step.

6.2 The Induction Step

Recall that $\mathcal{R}_{1,k}=:\{R\in\mathbb{R}_{1}:[n-k+1,n]\subset R\},\mathcal{R}_{1}(k)=:\{R\setminus[n-k+1,n]:R\in\mathcal{R}_{1,k}\}$ for $k\in[k_{1}-1]$ . The authors have shown the following result in [5].

Claim 6.21.

[5] Let $j\in[0,k_{1}-j]$ , $F_{1}<F^{\prime}_{1},G_{1}<G^{\prime}_{1}$ in $\mathcal{R}_{1}(j)$ , and $\max F^{\prime}_{1}=\max G^{\prime}_{1}$ . Then $\alpha(F_{1},F^{\prime}_{1})=\alpha(G_{1},G^{\prime}_{1})$ and $\beta(F_{1},F^{\prime}_{1})=\beta(G_{1},G^{\prime}_{1})$ .

We have the following claim.

Claim 6.22.

Let $j\in[0,k_{1}-1]$ , $F_{1}<F^{\prime}_{1},G_{1}<G^{\prime}_{1}$ and $F_{1}\prec F^{\prime}_{1}\prec G_{1}\prec G^{\prime}_{1}$ in $\mathcal{R}_{1}(j)$ , and $\max F^{\prime}_{1}=\max G^{\prime}_{1}$ with $\ell(F^{\prime}_{1})=\ell(G^{\prime}_{1})$ in $\mathcal{R}_{1}(j)$ . Then $\gamma(F_{1},F^{\prime}_{1})=\gamma(G_{1},G^{\prime}_{1})$ and $\delta(F_{1},F^{\prime}_{1})=\delta(G_{1},G^{\prime}_{1})$ .

Proof.

If $j=0$ , then we are fine (recall Claim 6.12 and Proposition 4.3). Assume $j\geq 1$ . Let $A_{1}=F_{1}\cup[n-j+1,n]$ , $A^{\prime}_{1}=F^{\prime}_{1}\cup[n-j+1,n]$ , $B_{1}=G_{1}\cup[n-j+1,n]$ and $B^{\prime}_{1}=G^{\prime}_{1}\cup[n-j+1,n]$ . Let $A^{\prime\prime}_{1}$ and $B^{\prime\prime}_{1}$ be the $k_{1}$ -sets such that $A_{1}<A^{\prime\prime}_{1}$ and $B_{1}<B^{\prime\prime}_{1}$ . Then by the definitions of $F_{1},F^{\prime}_{1},G_{1},G^{\prime}_{1}$ , we have $\max A^{\prime\prime}_{1}=\max B^{\prime\prime}_{1}$ , $A^{\prime\prime}_{1}\overset{j}{\longrightarrow}A^{\prime}_{1}$ and $B^{\prime\prime}_{1}\overset{j}{\longrightarrow}B^{\prime}_{1}$ . By Claim 6.14, $\gamma(A^{\prime\prime}_{1},A^{\prime}_{1})=\gamma(B^{\prime\prime}_{1},B^{\prime}_{1})$ and $\delta(A^{\prime\prime}_{1},A^{\prime\prime}_{1})=\delta(B^{\prime\prime}_{1},B^{\prime}_{1})$ . By Claim 6.12 and $\ell(F^{\prime}_{1})=\ell(G^{\prime}_{1})$ in $\mathcal{R}_{1}(j)$ , $\gamma(A_{1},A^{\prime}_{1})=\gamma(B_{1},B^{\prime\prime}_{1})$ and $\delta(A_{1},A^{\prime\prime}_{1})=\delta(B_{1},B^{\prime\prime}_{1})$ . Thus, $\gamma(A_{1},A^{\prime}_{1})=\gamma(B_{1},B^{\prime}_{1})$ and $\delta(A_{1},A^{\prime}_{1})=\delta(B_{1},B^{\prime}_{1})$ , that is $\gamma(F_{1},F^{\prime}_{1})=\gamma(G_{1},G^{\prime}_{1})$ and $\delta(F_{1},F^{\prime}_{1})=\delta(G_{1},G^{\prime}_{1})$ , as desired. ∎

We are ready to give the induction step of Lemma 6.10.

Proof of Lemma 6.10.

By induction on $k$ . We have shown that it holds for $k=0$ in Section 6.1. Suppose it holds for $k\in[0,k_{1}-2]$ , we are going to prove it holds for $k+1$ . Let $F_{1},G_{1},H_{1}\in\mathcal{R}_{1}(k+1)$ with $F_{1}\overset{c}{\prec}G_{1}\overset{c}{\prec}H_{1}$ and $f(G_{1})\geq f(F_{1})$ , i.e., $\gamma(F_{1},G_{1})\geq\delta(F_{1},G_{1})$ . We are going to apply induction hypothesis to show $f(H_{1})>f(G_{1})$ , i.e., $\gamma(G_{1},H_{1})>\delta(G_{1},H_{1})$ . Let $F^{\prime}_{1}=F_{1}\cup\{\max F+1\},G^{\prime}_{1}=G_{1}\cup\{\max G_{1}+1\}$ and $H^{\prime}_{1}=H_{1}\cup\{\max H_{1}+1\}$ . Then $F^{\prime}_{1},G^{\prime}_{1},H^{\prime}_{1}\in\mathcal{R}_{1}(k)$ . Moreover, $F^{\prime}_{1}\overset{c+1}{\prec}G^{\prime}_{1}\overset{c+1}{\prec}H^{\prime}_{1}$ in $\mathcal{R}_{1}(k)$ .

Let $A,B,C,D,E$ be the $(k_{1}-k)$ -sets satisfying $C<G^{\prime}_{1}<D,E<H^{\prime}_{1},F^{\prime}_{1}\overset{c}{\prec}A$ and $F^{\prime}_{1}<B$ in $\mathcal{R}_{1}(k)$ . Let $\widetilde{F}_{1}=F_{1}\sqcup\{n-k\},\widetilde{G}_{1}=G_{1}\sqcup\{n-k\}$ , $\widetilde{H}_{1}=H_{1}\sqcup\{n-k\}$ . Then $\widetilde{F}_{1},\widetilde{G}_{1},\widetilde{H}_{1}\in\mathcal{R}_{1}(k)$ . We can see that if $c\geq 2$ , then

F^{\prime}_{1}<B\overset{1}{\longrightarrow}\widetilde{F}_{1}\precneqq A\overset{c}{\longrightarrow}C<G^{\prime}_{1}<D\overset{1}{\longrightarrow}\widetilde{G}_{1}\,\,\text{and}\,\,G^{\prime}_{1}\overset{c}{\longrightarrow}E<H^{\prime}_{1}.

(35)

If $c=1$ , then

F^{\prime}_{1}<A=B\overset{1}{\longrightarrow}\widetilde{F}_{1}=C<G^{\prime}_{1}<D\overset{1}{\longrightarrow}\widetilde{G}_{1}\,\,\text{and}\,\,G^{\prime}_{1}\overset{1}{\longrightarrow}E<H^{\prime}_{1}.

(36)

Claim 6.23.

$\gamma(A,C)>\delta(A,C).$

Proof.

Suppose on the contrary that $\gamma(A,C)\leq\delta(A,C).$ We first consider the case $c\geq 2$ . By (35),

	$\displaystyle\gamma(\widetilde{F}_{1},\widetilde{G}_{1})=\gamma(\widetilde{F}_{1},A)+\gamma(A,C)+\gamma(C,G^{\prime}_{1})+\gamma(G^{\prime}_{1},\widetilde{G}_{1}),$
	$\displaystyle\delta(\widetilde{F}_{1},\widetilde{G}_{1})=\delta(\widetilde{F}_{1},A)+\delta(A,C)+\delta(C,G^{\prime}_{1})+\delta(G^{\prime}_{1},\widetilde{G}_{1}).$

Note that $f(F_{1})=f(\widetilde{F}_{1})$ and $f(G_{1})=f(\widetilde{G}_{1})$ , therefore, $\gamma(F_{1},G_{1})\geq\delta(F_{1},G_{1})$ implies $\gamma(\widetilde{F}_{1},\widetilde{G}_{1})\geq\delta(\widetilde{F}_{1},\widetilde{G}_{1})$ . Since $\gamma(A,C)\leq\delta(A,C)$ , then

\gamma(\widetilde{F}_{1},A)+\gamma(C,G^{\prime}_{1})+\gamma(G^{\prime}_{1},\widetilde{G}_{1})\geq\delta(\widetilde{F}_{1},A)+\delta(C,G^{\prime}_{1})+\delta(G^{\prime}_{1},\widetilde{G}_{1}).

(37)

Note that $\max B=\max G^{\prime}_{1}$ with $\ell(B)=\ell(G^{\prime}_{1})=0$ in $\mathcal{R}_{1}(j)$ (i.e., $\max B=\max G^{\prime}_{1}<n-k$ ). By Claim 6.22, we have $\delta(F^{\prime}_{1},B)=\delta(C,G^{\prime}_{1})$ and $\gamma(F^{\prime}_{1},B)=\gamma(C,G^{\prime}_{1})$ . Note that $B\overset{1}{\longrightarrow}\widetilde{F}_{1},G^{\prime}_{1}\overset{1}{\longrightarrow}\widetilde{G}_{1}$ , $\max B=\max G^{\prime}_{1}$ and $\max\widetilde{F}_{1}=\max\widetilde{G}_{1}$ , it follows from Claim 6.14 that $\gamma(B,\widetilde{F}_{1})=\gamma(G^{\prime}_{1},\widetilde{G}_{1})$ and $\delta(B,\widetilde{F}_{1})=\delta(G^{\prime}_{1},\widetilde{G}_{1}).$ Then

\displaystyle\gamma(\widetilde{F}_{1},A)+\gamma(C,G^{\prime}_{1})+\gamma(G^{\prime}_{1},\widetilde{G}_{1})

\displaystyle=\gamma(\widetilde{F}_{1},A)+\gamma(F^{\prime}_{1},B)+\gamma(B,\widetilde{F}_{1})=\gamma(F^{\prime}_{1},A).

Similarly, we have

\delta(\widetilde{F}_{1},A)+\delta(C,G^{\prime}_{1})+\delta(G^{\prime}_{1},\widetilde{G}_{1})=\delta(F^{\prime}_{1},A).

So inequality (37) gives $\gamma(F^{\prime}_{1},A)\geq\delta(F^{\prime}_{1},A)$ . Note that $F^{\prime}_{1}\overset{c}{\prec}A,A\overset{c}{\longrightarrow}C$ in $\mathcal{R}_{1}(k)$ for $c\in[k_{i}-k]$ , by induction hypothesis, $\gamma(A,C)>\delta(A,C)$ . A contradiction to our assumption.

We next consider the case $c=1$ . By (36), we have

	$\displaystyle\gamma(\widetilde{F}_{1},\widetilde{G}_{1})=\gamma(C,G^{\prime}_{1})+\gamma(G^{\prime}_{1},\widetilde{G}_{1}),$
	$\displaystyle\delta(\widetilde{F}_{1},\widetilde{G}_{1})=\delta(C,G^{\prime}_{1})+\delta(G^{\prime}_{1},\widetilde{G}_{1}).$

Note that $\gamma(F_{1},G_{1})\geq\delta(F_{1},G_{1})$ implies $\gamma(\widetilde{F}_{1},\widetilde{G}_{1})\geq\delta(\widetilde{F}_{1},\widetilde{G}_{1})$ and

\gamma(C,G^{\prime}_{1})+\gamma(G^{\prime}_{1},\widetilde{G}_{1})\geq\delta(C,G^{\prime}_{1})+\delta(G^{\prime}_{1},\widetilde{G}_{1}).

(38)

Note that $\max B=\max G^{\prime}_{1}$ and $\ell(B)=\ell(G^{\prime}_{1})=0$ in $\mathcal{R}_{1}(k)$ . By Claim 6.22, we have $\delta(F^{\prime}_{1},B)=\delta(C,G^{\prime}_{1})$ and $\gamma(F^{\prime}_{1},B)=\gamma(C,G^{\prime}_{1})$ . Note that $B\overset{1}{\longrightarrow}\widetilde{F}_{1},G^{\prime}_{1}\overset{1}{\longrightarrow}\widetilde{G}_{1}$ , $\max B=\max G^{\prime}_{1}$ and $\max\widetilde{F}_{1}=\max\widetilde{G}_{1}$ , it follows from Claim 6.14 that $\gamma(B,\widetilde{F}_{1})=\gamma(G^{\prime}_{1},\widetilde{G}_{1})$ and $\delta(B,\widetilde{F}_{1})=\delta(G^{\prime}_{1},\widetilde{G}_{1}).$ Then

\displaystyle\gamma(C,G^{\prime}_{1})+\gamma(G^{\prime}_{1},\widetilde{G}_{1})

\displaystyle=\gamma(F^{\prime}_{1},B)+\gamma(B,\widetilde{F}_{1})=\gamma(F^{\prime}_{1},\widetilde{F}_{1}).

Similarly, we have

\delta(C,G^{\prime}_{1})+\delta(G^{\prime}_{1},\widetilde{G}_{1})=\delta(F^{\prime}_{1},\widetilde{F}_{1}).

So inequality (38) gives $\gamma(F^{\prime}_{1},\widetilde{F}_{1})\geq\delta(F^{\prime}_{1},\widetilde{F}_{1})$ .

In view of (36) that

\gamma(F^{\prime}_{1},A)=\gamma(F^{\prime}_{1},\widetilde{F}_{1})-\gamma(A,C)

and

\delta(F^{\prime}_{1},A)=\delta(F^{\prime}_{1},\widetilde{F}_{1})-\delta(A,C).

Since $\gamma(A,C)\leq\delta(A,C)$ and $\gamma(F^{\prime}_{1},\widetilde{F}_{1})\geq\delta(F^{\prime}_{1},\widetilde{F}_{1})$ , we get $\gamma(F^{\prime}_{1},A)\geq\delta(F^{\prime}_{1},A)$ . Note that $F^{\prime}_{1}\overset{c}{\prec}A,A\overset{c}{\longrightarrow}C\in\mathcal{R}_{1}(k)$ , where $c=1\in[k_{1}-k]$ , by induction hypothesis, $\gamma(A,C)>\delta(A,C)$ . A contradiction to our assumption. This completes the proof of Claim 6.23. ∎

By (35) and (36), we have $G^{\prime}_{1}\overset{c}{\longrightarrow}E,A\overset{c}{\longrightarrow}C,\max G^{\prime}_{1}=\max A,\max E=\max C$ in $\mathcal{R}_{1}(k)$ . By Claim 6.14 and Claim 6.23, we get

\gamma(G^{\prime}_{1},E)>\delta(G^{\prime}_{1},E).

(39)

Claim 6.24.

$\gamma(D,H^{\prime}_{1})>\delta(D,H^{\prime}_{1})$ .

Proof.

Since $G^{\prime}_{1}<D$ , $E<H^{\prime}_{1}$ and $G^{\prime}_{1}\prec D\prec E\prec H^{\prime}_{1}$ in $\mathcal{R}_{1}(k)$ , we will meet the following two cases: $\ell(D)=\ell(H^{\prime}_{1})$ and $\ell(D)<\ell(H^{\prime}_{1})$ . We first consider the case $\ell(D)=\ell(H^{\prime}_{1})$ . Then by Claim 6.22, we have $\gamma(G^{\prime}_{1},D)=\gamma(E,H^{\prime}_{1})$ and $\delta(G^{\prime}_{1},D)=\delta(E,H^{\prime}_{1})$ . Therefore,

	$\displaystyle\gamma(D,H^{\prime}_{1})$	$\displaystyle=\gamma(G^{\prime}_{1},E)-\gamma(G^{\prime}_{1},D)+\gamma(E,H^{\prime}_{1})$
		$\displaystyle=\gamma(G^{\prime}_{1},E)$
		$\displaystyle>\delta(G^{\prime}_{1},E)$
		$\displaystyle=\delta(G^{\prime}_{1},E)-\delta(G^{\prime}_{1},D)+\delta(E,H^{\prime}_{1})$
		$\displaystyle=\delta(D,H^{\prime}_{1}),$

as desired. Next we assume that $\ell(D)<\ell(H^{\prime}_{1})$ . If $c=1$ , then $G^{\prime}_{1}<D=E<H^{\prime}_{1}$ . By Claim 6.12,

\gamma(D,H^{\prime}_{1})\geq\gamma(G^{\prime}_{1},D)\overset{(\ref{hh})}{>}\delta(G^{\prime}_{1},E)=\delta(D,H^{\prime}_{1}),

where the last equality holds by Proposition 4.3, as required. If $c\geq 2$ , then $G^{\prime}_{1}<D=\widetilde{G}_{1}\precneqq E<H^{\prime}_{1}$ , $\max D=\max E=n-k$ and $\ell(E)>\ell(D)\geq 0$ . By Claim 6.12 and Remark 6.9,

\gamma(D,H^{\prime}_{1})\geq\gamma(G^{\prime}_{1},E)\overset{(\ref{hh})}{>}\delta(G^{\prime}_{1},E)=\delta(D,H^{\prime}_{1}),

as required. ∎

Consequently, $f(D)<f(H^{\prime}_{1})$ following from Claim 6.24. Recall that $D\overset{1}{\longrightarrow}\widetilde{G}_{1}$ and $H^{\prime}_{1}\overset{1}{\longrightarrow}\widetilde{H}_{1}$ . Hence, $f(\widetilde{G}_{1})<f(\widetilde{H}_{1})$ by applying Claim 6.14. This implies $\gamma(G_{1},H_{1})>\delta(G_{1},H_{1})$ , as desired.

This completes the proof of Lemma 6.10. ∎

7 Verify Unimodality: The Proofs of Lemmas 3.13, 3.14 and Proposition 3.20

Lemmas 3.13 and 3.14 will follow from the following lemma.

Lemma 7.1.

Let $B_{0}=\{b_{1},\dots,b_{x}\}\cup[y,y+k]$ with $b_{x}<y-1$ , $k\geq 1$ and $y+k<n$ . For $i\in[k]$ , let $B_{i}=\{b_{1},\dots,b_{x}\}\cup[y,y+k-i]\cup[n-i+1,n]$ . Suppose that $B_{i},B_{i+1},B_{i+2}\in\mathcal{R}_{1}$ for some $i\in[0,k-2]$ and $f(B_{i})\leq f(B_{i+1})$ , then $f(B_{i+1})<f(B_{i+2})$ .

Let us explain why Lemma 7.1 implies Lemmas 3.13 and 3.14. Let $F_{2},G_{2},H_{2}$ be as in Lemma 3.13 or Lemma 3.14. Let $F_{1},G_{1},H_{1}$ be the $k_{1}$ -parities of $F_{2},G_{2},H_{2}$ respectively (Proposition 2.16 guarantees that $F_{1},G_{1},H_{1}$ exists).

By Fact 2.14, for each $i\in[3,t]$ , $F_{1}$ and $F_{2}$ have the same $k_{i}$ -partner. Take $s=2$ (see Definition 6.1), we have $g(F_{2})=f(F_{1})$ , $g(G_{2})=f(G_{1})$ and $g(H_{2})=f(H_{1})$ . We may take $B_{i}=F_{1}$ , $B_{i+1}=G_{1}$ and $B_{i+2}=H_{1}$ . So $g(F_{2})\leq g(G_{2})$ implies

f(B_{i})=f(F_{1})=g(F_{2})\leq g(G_{2})=f(G_{1})=f(B_{i+1}).

Then by Lemma 7.1, we have $f(B_{i+2})>f(B_{i+1})$ . So

g(H_{2})=f(H_{1})=f(B_{i+2})>f(B_{i+1})=f(G_{1})=g(G_{2}).

Proof for Lemma 7.1.

Clearly, $\ell(B_{i+1})=\ell(B_{i})+1\geq 1$ and $\ell(B_{i+2})=\ell(B_{i+1})+1$ . Let $B^{\prime}_{i}$ and $B^{\prime}_{i+1}$ be the $k_{1}$ -sets such that $B_{i}<B^{\prime}_{i}$ and $B_{i+1}<B^{\prime}_{i+1}$ . Then $B^{\prime}_{i}\overset{\ell(B_{i+1})}{\longrightarrow}B_{i+1}$ . Let $J$ be the $k_{1}$ -set such that $B^{\prime}_{i+1}\overset{\ell(B_{i+1})}{\longrightarrow}J$ . Then $B^{\prime}_{i},B_{i+1}$ are $\ell(B_{i+1})$ -sequential and $B^{\prime}_{i+1},J$ are $\ell(B_{i+1})$ -sequential. Clearly, $\max B^{\prime}_{i}=\max B^{\prime}_{i+1}$ and $\max B_{i+1}=\max J=n$ . By Claim 6.14,

\gamma(B^{\prime}_{i},B_{i+1})=\gamma(B^{\prime}_{i+1},J)

and

\delta(B^{\prime}_{i},B_{i+1})=\delta(B^{\prime}_{i+1},J)

(40)

If $\ell(B^{\prime}_{i})\geq 1$ , then $B_{i}<B^{\prime}_{i}=B_{i+1}<B_{i+2}=B^{\prime}_{i+1}$ . By Proposition 4.3, $\delta(B_{i},B_{i+1})=\delta(B_{i+1},B_{i+2})$ . By Claim 6.12, we have $\gamma(B_{i},B_{i+1})\leq\gamma(B_{i+1},B_{i+2})$ . So $f(B_{i+1})-f(B_{i})=\gamma(B_{i},B_{i+1})-\delta(B_{i},B_{i+1})\leq\gamma(B_{i+1},B_{i+2})-\delta(B_{i+1},B_{i+2})=f(B_{i+2})-f(B_{i+2})$ . So if $f(B_{i})<f(B_{i+1})$ , then $f(B_{i+1})<f(B_{i+2})$ , as desired. We next assume that $f(B_{i})=f(B_{i+1})$ . Then $\delta(B_{i},B_{i+1})=\gamma(B_{i},B_{i+1})\geq s^{\prime}$ . If $\gamma(B_{i},B_{i+1})=s^{\prime}$ , then $\delta(B_{i},B_{i+1})=s^{\prime}$ . Since $B_{i}<B_{i+1}$ and $\max B_{i+1}=n$ , $\delta(B_{i},B_{i+1})=\beta(B_{i},B_{i+1})=0$ or $1$ , and $\delta(B_{i},B_{i+1})=\beta(B_{i},B_{i+1})=s^{\prime}$ if and only if $n=k_{1}+k_{t}=k_{2}+k_{t-1}=\dots=k_{s}^{\prime}+k_{t-s^{\prime}+1}$ (see Proposition 4.3). This is a contradiction to $n>k_{1}+k_{t}$ (since $s^{\prime}<s$ and $n\geq k_{s}+k_{t}$ ). So $\gamma(B_{i},B_{i+1})>s^{\prime}$ . Consequently, there exists $j\in[s^{\prime}+1,s]$ such that $\alpha_{j}(B_{i},B_{i+1})>0$ . Let $j$ be any integer that satisfies the above condition. By Claim 6.12, $\ell(G_{i+1})\geq k_{1}-k_{j}$ . Since $\ell(B_{i+2})=\ell(B_{i+1})+1$ , $\ell(G_{i+1})>k_{1}-k_{j}$ . By Claim 6.12 again, $\alpha_{j}(B_{i+1},B_{i+2})>\alpha_{j}(B_{i},B_{i+1})$ . By the arbitrariness of $j$ and the definitions of $\gamma(B_{i},B_{i+1})$ and $\gamma(B_{i+1},B_{i})$ (see Definition 6.7), we conclude that $\gamma(B_{i+1},B_{i})<\gamma(B_{i},B_{i+1})$ . Combining with $\delta(B_{i},B_{i+1})=\delta(B_{i+1},B_{i+2})$ , we have $f(B_{i+1})-f(B_{i})=\gamma(B_{i},B_{i+1})-\delta(B_{i},B_{i+1})<\gamma(B_{i+1},B_{i+2})-\delta(B_{i+1},B_{i+2})=f(B_{i+2})-f(B_{i+2})$ . So if $f(B_{i})=f(B_{i+1})$ , then $f(B_{i+1})<f(B_{i+2})$ , as desired.

Next we assume that $\ell(B^{\prime}_{i})=0$ . By Claim 6.12, for each $j\in[s^{\prime}+1,s]$ , $\alpha_{j}(B_{i},B^{\prime}_{i})=\alpha_{j}(B_{i+1},B^{\prime}_{i+1})=0$ . Clearly, for for each $j\in[s^{\prime}]$ , $\alpha_{j}(B_{i},B^{\prime}_{i})=\alpha_{j}(B_{i+1},B^{\prime}_{i+1})=1$ and then $\gamma(B_{i},B^{\prime}_{i})=\gamma(B_{i+1},B^{\prime}_{i+1})$ . Combining with (40), we get

\gamma(B_{i},B_{i+1})=\gamma(B_{i+1},J)

and

\delta(B_{i},B_{i+1})=\delta(B_{i+1},J)

Thus $f(J)\geq f(B_{i+1})$ since $f(B_{i})\leq f(B_{i+1})$ . Note that $\ell(B_{i+1})=\ell(J)$ and $B_{i+1}\setminus B_{i+1}^{\rm t}\in\mathcal{R}_{1}(\ell(B_{i+1}))$ , so $J\setminus J^{\rm t}\in\mathcal{R}_{1}(\ell(B_{i+1}))$ and $B_{i+1}\setminus B_{i+1}^{\rm t}\overset{1}{\prec}J\setminus J^{\rm t}$ . Since $\ell(B_{i+2})=\ell(B_{i+1})+1$ , $B_{i+2}\setminus[n-\ell(B_{i+1})+1,n]\in\mathcal{R}_{1}(\ell(B_{i+1}))$ . And in $\mathcal{R}_{1}(\ell(B_{i+1}))$ , we have

B_{i+1}\setminus B_{i+1}^{\rm t}\overset{1}{\prec}J\setminus J^{\rm t}\overset{1}{\longrightarrow}B_{i+2}\setminus[n-\ell(B_{i+1})+1,n].

By Lemma 6.10 and $f(J)\geq f(B_{i+1})$ , we obtain $f(B_{i+2})>f(J)\geq f(B_{i+1})$ , as required. ∎

Proof for Proposition 3.20.

The proofs of equalities (15) and (16) are quite similar, we prove the previous one only.

Let $G_{2}=[2,k_{2}+1]$ and $G_{1}$ be the $k_{1}$ -parity of $G_{2}$ . Since $k_{1}>k_{2}$ , $\ell(G_{1})\geq 1$ . So $G_{1}=[2,k_{2}+1]\cup[n-k_{1}+k_{2}+1,n]$ . Let $A$ be the $k_{1}$ -parity of $[2,k_{2}]\cup\{n\}$ . So $A=[2,k_{2}]\cup[n-k_{1}+k_{2},n]$ . To prove (15), it is equivalent to prove that $f(G_{1})<\max\{f(\{1\}),f(A)\}$ . Let $k=k_{1}-k_{2}$ . Note that $G_{1}\setminus G_{1}^{\rm t}$ , $\{1,n-k_{1}+2,\dots,n\}\setminus[n-k_{1}+k_{2}+1,n]$ and $A\setminus[n-k_{1}+k_{2}+1,n]$ are contained in $\mathcal{R}_{1}(k)$ and

\{1,n-k_{1}+2,\dots,n\}\setminus[n-k_{1}+k_{2}+1,n]<G_{1}\setminus G_{1}^{\rm t}\overset{1}{\longrightarrow}A\setminus[n-k_{1}+k_{2}+1,n]

in $\mathcal{R}_{1}(k)$ . Let $C_{1}=G_{1}\setminus G_{1}^{\rm t}\cup\{\max(G_{1}\setminus G_{1}^{\rm t})+1\}\cup[n-\ell(G_{1})+2,n]$ (if $\ell(G_{1})=2$ , then $[n-\ell(G_{1})+2,n]=\emptyset$ ). Since $\ell(G_{1})\geq 1$ , $|C_{1}|=k_{1}$ . Let $B\in\mathcal{R}_{1}(k)$ be the set such that $G_{1}\setminus G_{1}^{\rm t}<B$ and $B_{1}=B\cup[n-\ell(G_{1})+1,n]$ . Clearly, $\{1,n-k_{1}+2,\dots,n\}\precneqq C_{1}\precneqq G_{1}\precneqq B_{1}\prec A$ . We may also assume that $f(G_{1})\geq f(C_{1})$ since otherwise, $g(\mathcal{F}_{2,3})>g(G_{1})=f(G_{1})$ , we are done. Based on these, we may apply Lemma 7.1 to $C_{1},G_{1},B_{1}$ . Since $f(G_{1})\geq f(C_{1})$ , Lemma 7.1 gives $f(G_{1})<f(B_{1})$ . Consequently, by Lemma 6.10, $f(A)>f(B)>f(G_{1})$ , as desired. ∎

8 Acknowledgements

This work was supported by NSFC (Grant No. 11931002).

References

[1] P. Erdős, C. Ko, R. Rado, Intersection theorems for systems of finite sets, Quart. J. Math. Oxf. 2(12) (1961) 313–320.
[2] P. Frankl, A. Kupavskii, Erdős-Ko-Rado theorem for {0, ±1}-vectors, J. Comb. Theory Ser. A 155 (2018), 157–179.
[3] P. Frankl, A. Kupavskii, Sharp results concerning disjoint cross intersecting families, Europ J. Combin 86 (2020) 103089.
[4] P. Frankl, N. Tokushige, Some best possible inequalities concerning crossing-intersecting families, J. Combin. Theory Ser. A 61 (1992) 87-97.
[5] Yang Huang, Yuejian Peng, The maximum sum of sizes of non-empty pairwise cross intersecting families. arXiv: 2306.03473.
[6] Yang Huang, Yuejian Peng, Coefficients added non-empty pairwise cross intersecting families. Manuscript.
[7] Yang Huang, Yuejian Peng, Mixed pairwise cross intersecting families (II). In preparation.
[8] A.J.W. Hilton, E.C. Milner, Some intersection theorems for systems of finite sets, Quart. J. Math. Oxf. 2 (18) (1967) 369-384.
[9] A.J.W. Hilton, The Erdős-Ko-Rado theorem with valency conditions, Unpublished Manuscript, 1976.
[10] A.J.W. Hilton, An intersection theorem for a collection of families of subsets of a finite set, J. London Math. Soc. 2 (1977) 369-376.
[11] G.O.H. Katona, A theorem of finite sets, in: Theory of Graphs, Proc. Colloq. Tihany, Akadémai Kiadó, (1968) 187-207.
[12] J.B. Kruskal, The number of simplices in a complex, in: Math. Opt. Techniques, Univ. of Calif. Press, (1963) 251-278.
[13] C. Shi, P. Frankl, J. Qian, On non-empty cross-intersecting families, Combinatorica 42 (2022) 1513–1525

Mixed pairwise cross intersecting families (I)††thanks: This work is supported by NSFC (Grant No. 11931002). E-mail addresses: [email protected] (Yang Huang), [email protected] (Yuejian Peng, corresponding author).

Abstract

1 Introduction

Theorem 1.1 (Huang-Peng [5]).

Construction 1.2.

Construction 1.3.

Theorem 1.4.

2 Partner and Parity

Theorem 2.1 (Kruskal-Katona theorem [3, 9]).

Proposition 2.2.

Remark 2.3.

Fact 2.4.

Proof.

Fact 2.5.

Remark 2.6.

Proposition 2.7 (Frankl-Kupavskii [2]).

Fact 2.8.

Proof.

Fact 2.9.

Fact 2.10.

Proof.

Definition 2.11.

Remark 2.12.

Fact 2.13.

Fact 2.14.

Definition 2.15.

Proposition 2.16.

Proof.

Fact 2.17.

Proof.

3 Proofs of Theorem 1.4

Claim 3.1.

Proof.

Theorem 3.2.

Claim 3.3.

Proof.

Remark 3.4.

Proposition 3.5.

Proof.

Proposition 3.6.

Definition 3.7.

Lemma 3.8.

Remark 3.9.

Remark 3.10.

Definition 3.11.

Lemma 3.12.

Lemma 3.13.

Lemma 3.14.

Claim 3.15.

Proof.

Claim 3.16.

Proof.

Definition 3.17.

Observation 3.18.

Definition 3.19.

Proof of Proposition 3.6.

Proposition 3.20.

Proof of Theorem 1.4.

4 Results of non-mixed type

Lemma 4.1.

Definition 4.2.

Proposition 4.3.

Lemma 4.4.

Definition 4.5.

Remark 4.6.

Lemma 4.7.

Lemma 4.8.

Lemma 4.9.

5 Verify Parity: Proof of Lemma 3.8

Claim 5.1.

Proof.

Claim 5.2.

Proof.

Claim 5.3.

Proof.

Claim 5.4.

Proof.

Claim 5.5.

Proof.

Claim 5.6.

Mixed pairwise cross intersecting families (I)^†^†thanks: This work is supported by NSFC (Grant No. 11931002). E-mail addresses: [email protected] (Yang Huang), [email protected] (Yuejian Peng, corresponding author).