On the size of $k$-cross-free families

TL;DR

This paper improves the upper bound on the size of families of subsets with no $k$ pairwise crossing sets from $O_k(n \, \log n)$ to $O_k(n \log^{*} n)$, refining a longstanding conjecture in combinatorics.

Contribution

The authors establish a tighter upper bound of $O_k(n \log^{*} n)$ for the size of $k$-cross-free families, advancing previous bounds and addressing a 40-year-old conjecture.

Findings

Bound improved from $O_k(n \log n)$ to $O_k(n \log^{*} n)$

Confirms the conjecture for larger $k$ with a nearly optimal bound

Uses novel combinatorial techniques to refine existing bounds

Abstract

Two subsets $A,B$ of an $n$-element ground set $X$ are said to be \emph{crossing}, if none of the four sets $A\cap B$, $A\setminus B$, $B\setminus A$ and $X\setminus(A\cup B)$ are empty. It was conjectured by Karzanov and Lomonosov forty years ago that if a family $\mathcal{F}$ of subsets of $X$ does not contain $k$ pairwise crossing elements, then $|\mathcal{F}|=O_{k}(n)$. For $k=2$ and $3$, the conjecture is true, but for larger values of $k$ the best known upper bound, due to Lomonosov, is $|\mathcal{F}|=O_{k}(n\log n)$. In this paper, we improve this bound by showing that $|\mathcal{F}|=O_{k}(n\log^{*} n)$ holds, where $\log^{*}$ denotes the iterated logarithm function.

Equations

$$F^{\prime }=\{A\in F:|A|<n/2\}\cup \{A\in F:|A|=n/2\mathrm{and}1\in A\}$$

$$|F^{\prime }|=|F|/2\le 1+\sum _{s=1}^{n/2}(k-1)n/s=O_k(n\log n).$$

$$|F|\le 4(k-1)n-2\left(\genfrac{}{}{0ex}{}{2k-1}{2}\right),$$

$$\left(\genfrac{}{}{0ex}{}{x}{k}\right)=\{\begin{array}{ll}\frac{x(x-1)\dots (x-k+1)}{k!}& \text{mbox}ifx\ge k-1\\ 0& \text{mbox}ifx<k-1.\end{array}$$

$${\mathcal{F}}^{\prime }=\{A\in {\mathcal{F}}^{\prime }:1\notin A\}\cup \{[n]\setminus A:A\in {\mathcal{F}}^{\prime },1\in A\}.$$

$${\mathcal{F}}_i:=\{X\in \mathcal{F}:2^i<|X|\le 2^{i+1}\}.$$

$$|\mathcal{A}|\le \frac{(k-1)n}{2^i}.$$

$$(k-1)n\ge \sum _{A\in \mathcal{A}}|A|\ge |\mathcal{A}|2^i.$$

$$\sum _{C\in {\Gamma }_i}|C|\ge \frac{|{\mathcal{F}}_i|}{k-1}.$$

$$|{\Gamma }_i|\le \frac{(k-1)n}{2^i}.$$

$$\sum _{y\in [n]}d(y)=\sum _{C\in {\Gamma }_{a,b}}|Y(C)|=\sum _{C\in {\Gamma }_{a,b}}|C|\ge \frac{|{\mathcal{F}}_{a,b}|}{k-1},$$

$$g(y)\ge \left(\genfrac{}{}{0ex}{}{d(y)/(k-1{)}^2}{k}\right).$$

$$\left(\genfrac{}{}{0ex}{}{r}{k}\right)\ge \left(\genfrac{}{}{0ex}{}{d(y)/(k-1{)}^2}{k}\right)$$

$$N<\frac{2(k-1{)}^{k}n}{2^a}\left(\genfrac{}{}{0ex}{}{b}{k-1}\right).$$

$$|{\Gamma }_{a,b}|=\sum _{a<i\le b}|{\Gamma }_i|\le \sum _{a<i\le b}\frac{(k-1)n}{2^i}<\frac{2(k-1)n}{2^a};$$

$$\frac{2(k-1{)}^{k}n}{2^a}\left(\genfrac{}{}{0ex}{}{b}{k-1}\right).$$

$$\mathcal{H}:=\big{\{}C_{i,k+1-i}:i\in[k]\big{\}}$$

$$M\le (k-1)N<\frac{2(k-1{)}^{k+1}n}{2^a}\left(\genfrac{}{}{0ex}{}{b}{k-1}\right)\le \frac{2n(k-1{)}^{k+1}{b}^{k-1}}{2^a(k-1)!}.$$

$$M\le \frac{c_1(k)n{b}^{k-1}}{2^a}.$$

$$M=\sum _{y\in [n]}g(y),$$

$$\sum _{y\in [n]}g(y)\ge \sum _{y\in [n]}\left(\genfrac{}{}{0ex}{}{d(y)/(k-1{)}^2}{k}\right).$$

$$n\left(\genfrac{}{}{0ex}{}{{\sum }_{y\in [n]}d(y)/(k-1{)}^{2}n}{k}\right).$$

$$M\ge n\left(\genfrac{}{}{0ex}{}{|F_{a,b}|/(k-1{)}^{3}n}{k}\right).$$

$$\left(\genfrac{}{}{0ex}{}{|F_{a,b}|/(k-1{)}^{3}n}{k}\right)>{\left(\frac{|{\mathcal{F}}_{a,b}|}{2(k-1{)}^{3}n}\right)}^k\frac{1}{k!}.$$

$$M>\frac{c_2(k)|F_{a,b}{|}^k}{n^{k-1}}.$$

$$\frac{c_1(k)n{b}^{k-1}}{2^a}>\frac{c_2(k)|F_{a,b}{|}^k}{n^{k-1}},$$

$$|{\mathcal{F}}_{a,b}|<n{\left(\frac{c_1(k)}{c_2(k)}\right)}^{1/k}\frac{b^{(k-1)/k}}{2^{a/k}}.$$

$$|{\mathcal{F}}_{a,b}|<\max \left\{2k(k-1{)}^{3}n,\frac{c_3(k)n{b}^{(k-1)/k}}{2^{a/k}}\right\}$$

$$|{\mathcal{F}}_{a_0,a_1}|=|{\mathcal{F}}_{0,k^2}|\le \sum _{l=1}^{2^{k^2}}\frac{(k-1)n}{l}=O_k(n),$$

$$|{\mathcal{F}}_{a_i,a_{i+1}}|<\max \left\{2k(k-1{)}^{3}n,\frac{c_3(k)n{a}_{i+1}^{(k-1)/k}}{2^{a_i/k}}\right\}=\max \{2k(k-1{)}^3,c_3(k)\}n.$$

Full text

1 Introduction

As usual, denote $[n]:=\{1,\ldots,n\}$ and let $2^{[n]}$ be the family of all subsets of $[n]$. Two sets $A,B\in 2^{[n]}$ are said to be crossing, if $A\setminus B$, $B\setminus A$, $A\cap B$ and $[n]\setminus(A\cup B)$ are all non-empty.

We say that a family $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ is $k$-cross-free if it does not contain $k$ pairwise crossing sets. The following conjecture was made by Karzanov and Lomonosov [12], [11] and later by Pevzner [14]; see also Conjecture 3 in  [4], Section 9.

Conjecture 1**.**

Let $k\geq 2$ and $n$ be positive integers, and let $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ be a $k$-cross-free family. Then $|\operatorname*{\mathcal{F}}|=O_{k}(n)$.

Here and in the rest of this paper, $f(n)=O_{k}(n)$ means that $f(n)\leq c_{k}n$ for a suitable constant $c_{k}>0$, which may depend on the parameter $k$.

It was shown by Edmonds and Giles [8] that every $2$-cross-free family $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ has at most $4n-2$ members. Pevzner [14] proved that every $3$-cross-free family on an $n$-element underlying set has at most $6n$ elements, and Fleiner [9] established the weaker bound $10n$, using a simpler argument. For $k>3$, Conjecture 1 remains open. The best known general upper bound for the size of a $k$-cross-free family is $O_{k}(n\log n)$, which can be obtained by the following elegant argument, due to Lomonosov.

Let $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ be a maximal $k$-cross-free family. Notice that for any set $A\in\operatorname*{\mathcal{F}}$, the complement of $A$ also belongs to $\operatorname*{\mathcal{F}}$. Thus, the subfamily

[TABLE]

contains precisely half of the members of $\operatorname*{\mathcal{F}}$. For every $s,\,1\leq s\leq n/2$, any two $s$-element members of $\operatorname*{\mathcal{F}}^{\prime}$ that have a point in common, are crossing. Since $\operatorname*{\mathcal{F}}^{\prime}$ has no $k$ pairwise crossing members, every element of $[n]$ is contained in at most $k-1$ members of $\operatorname*{\mathcal{F}}^{\prime}$ of size $s$. Thus, the number of $s$-element members is at most $(k-1)n/s$, and

[TABLE]

The main result of the present note represents the first improvement on this 40 years old bound. Let $\log_{(i)}n$ denote the function $\log\ldots\log n$, where the $\log$ is iterated $i$ times, and let $\log^{*}n$ denote the iterated logarithm of $n$, that is, the largest positive integer $i$ such that $\log_{(i)}n>1$.

Theorem 2**.**

Let $k\geq 2$ and $n$ be positive integers, and let $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ be a $k$-cross-free family. Then $|\operatorname*{\mathcal{F}}|=O_{k}(n\log^{*}n).$

Conjecture 1 has been proved in the following special case. Let $\operatorname*{\mathcal{F}}$ be a $k$-cross-free family consisting of contiguous subintervals of the cyclic sequence $1,2,\ldots,n$. It was shown by Capoyleas and Pach [5] that in this case

[TABLE]

provided that $n\geq 2k-1$. This bound cannot be improved.

A geometric graph $G$ is a graph drawn in the plane so that its vertices are represented by points in general position in the plane and its edges are represented by (possibly crossing) straight-line segments between these points. Two edges of $G$ are said to be crossing if the segments representing them have a point in common.

Conjecture 3**.**

Let $k\geq 2$ and $n$ be positive integers, and let $G$ be a geometric graph with $n$ vertices, containing no $k$ pairwise crossing edges. Then the number of edges of $G$ is $O_{k}(n)$.

The result of Capoyleas and Pach mentioned above implies that Conjecture 3 holds for geometric graphs $G$, where the points representing the vertices of $G$ form the vertex set of a convex $n$-gon. It is also known to be true for $k\leq 4$; see [3], [1], [2]. For $k>4$, it was proved by Valtr [16] that if a geometric graph on $n$ vertices contains no $k$ pairwise crossing edges then it its number of edges is $O_{k}(n\log n)$ edges.

A bipartite variant of Conjecture 1 was proved by Suk [15]. He showed that if $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ does not contain $2k$ sets $A_{1},\ldots,A_{k}$ and $B_{1},\ldots,B_{k}$ such that $A_{i}$ and $B_{j}$ are crossing for all $i,j\in[k]$, then $|\operatorname*{\mathcal{F}}|\leq(2k-1)^{2}n$.

The notion of $k$-cross-free families was first introduced by Karzanov [11] in the context of multicommodity flow problems. Let $G=(V,E)$ be a graph, $X\subset V$. A multiflow $f$ is a fractional packing of paths in $G$. We say that $f$ locks a subset $A\subset X$ in $G$ if the total value of all paths between $A$ and $X\setminus A$ is equal to the minimum number of edges separating $A$ from $X\setminus A$ in $G$. A family $\operatorname*{\mathcal{F}}$ of subsets of $X$ is called lockable if for every graph $G$ with the above property there exists a multiflow $f$ that locks every member $A\in\operatorname*{\mathcal{F}}$. The celebrated locking theorem of Karzanov and Lomonosov [12] states that a set family is lockable if and only if it is $3$-cross-free. This is a useful extension of the Ford-Fulkerson theorem for network flows, and it generalizes some previous results of Cherkasky [6] and Lovász [13]; see also [10].

2 The proof of Theorem 2

In this section, we prove our main theorem. Throughout the proof, floors and ceilings are omitted whenever they are not crucial, and $\log$ stands for the base $2$ logarithm. Also, for convenience, we shall use the following extended definition of binomial coefficients: if $x$ is a real number and $k$ is a positive integer,

[TABLE]

Let us remark that the function $f(x)=\binom{x}{k}$ is monotone increasing and convex.

A pair of sets, $A,B\in 2^{[n]},$ are said to be weakly crossing, if $A\setminus B$, $B\setminus A$ and $A\cap B$ are all non-empty. Clearly, if $A$ and $B$ are crossing, then $A$ and $B$ are weakly crossing as well. We call a set family $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ weakly $k$-cross-free if it does not contain $k$ pairwise weakly crossing sets.

As our first step of the proof, we show that if $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ is a $k$-cross-free family, then we can pass to a weakly $k$-cross-free family $\operatorname*{\mathcal{F}}^{\prime}\subset 2^{[n]}$ by losing a factor of at most $2$ in the cardinality.

Lemma 4**.**

Let $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ be a $k$-cross-free family. Then there exists a weakly $k$-cross-free family $\operatorname*{\mathcal{F}}^{\prime}\subset 2^{[n]}$ such that $|\operatorname*{\mathcal{F}}^{\prime}|\geq|\operatorname*{\mathcal{F}}|/2$.

Proof.

Let

[TABLE]

Clearly, we have $|\operatorname*{\mathcal{F}}^{\prime}|\geq|\operatorname*{\mathcal{F}}|/2$.

Note that two sets $A,B\in[n]$ are crossing if and only if $A$ and $[n]\setminus B$ are crossing. Hence, $\mathcal{F}^{\prime}$ does not contain $k$ pairwise crossing sets. But no set in $\mathcal{F}^{\prime}$ contains $1$, so we cannot have $A\cup B=[n]$ for any $A,B\in\operatorname*{\mathcal{F}}^{\prime}$. Thus, $A,B\in\operatorname*{\mathcal{F}}^{\prime}$ are crossing if and only if $A$ and $B$ are weakly crossing. Hence, $\operatorname*{\mathcal{F}}^{\prime}$ satisfies the conditions of the lemma. ∎

Now Theorem 2 follows trivially from the combination of Lemma 4 and the following theorem.

Theorem 5**.**

Let $k\geq 2$ and $n$ be positive integers and let $\operatorname*{\mathcal{F}}\subset 2^{[n]}$ be a weakly $k$-cross-free family. Then $|\operatorname*{\mathcal{F}}|=O_{k}(n\log^{*}n).$

The rest of this section is devoted to the proof of this theorem. Let us briefly sketch the idea of the proof while introducing some of the main notation.

Let $\operatorname*{\mathcal{F}}$ be a weakly $k$-cross-free family. First, we shall divide the elements of $\operatorname*{\mathcal{F}}$ into $\log n$ parts according to their sizes: for $i=0,\ldots,\log n$, let $\mathcal{F}_{i}:=\{X\in\mathcal{F}:2^{i}<|X|\leq 2^{i+1}\}$. We might refer to the families $\operatorname*{\mathcal{F}}_{i}$ as blocks. Next, we show that, as the block $\operatorname*{\mathcal{F}}_{i}$ is weakly $k$-cross-free, it must have the following property: a positive proportion of $\operatorname*{\mathcal{F}}_{i}$ can be covered by a collection of chains $\Gamma_{i}$ with the maximal elements of these chains forming an antichain. These chains are going to be the objects of main interest in our proof.

We show that if $\operatorname*{\mathcal{F}}$ has too many elements, then we can find $k$ chains $\operatorname*{\mathcal{C}}_{1}\subset\Gamma_{i_{1}},\ldots,\operatorname*{\mathcal{C}}_{k}\subset\Gamma_{i_{k}}$ for some $i_{1}<\ldots<i_{k}$, and $k$ elements $C_{j,1}\subset\ldots\subset C_{j,k}$ in each chain $\operatorname*{\mathcal{C}}_{j}$ such that $C_{j,l}\subset C_{j^{\prime},l^{\prime}}$ if $j\leq j^{\prime}$ and $l\leq l^{\prime}$, and $C_{j,l}$ and $C_{j^{\prime},l^{\prime}}$ are weakly crossing otherwise. But then we arrive to a contradiction since the $k$ sets $C_{1,k},C_{2,k-1},\ldots,C_{k,1}$ are pairwise weakly crossing.

Now let us show how to execute this argument precisely.

Proof of Theorem 5..

Without loss of generality, we can assume that $\operatorname*{\mathcal{F}}$ does not contain the empty set and $1$-element sets, since by deleting them we decrease the size of $\operatorname*{\mathcal{F}}$ by at most $n+1$.

Let us remind the reader of the definition of blocks: for $i=0,1,\ldots,\log n$, we have

[TABLE]

The next claim gives an upper bound on the size of an antichain in $\operatorname*{\mathcal{F}}_{i}$.

Claim 6**.**

If $\mathcal{A}\subset\mathcal{F}_{i}$ is an antichain, then

[TABLE]

Proof.

Suppose that there exists $x\in[n]$ such that $x$ is contained in $k$ sets from $\mathcal{A}$. Then these $k$ sets are pairwise weakly crossing. Hence, every element of $[n]$ is contained in at most $k-1$ of the sets in $\mathcal{A}$, which implies that

[TABLE]

∎

In the next claim, we show that a positive proportion of $\operatorname*{\mathcal{F}}_{i}$ can be covered by chains whose maximal elements form an antichain. We shall use the following notation concerning chains. If $\operatorname*{\mathcal{C}}$ is a chain of size $l$, denote its elements by $\operatorname*{\mathcal{C}}(1)\subset\ldots\subset\operatorname*{\mathcal{C}}(l)$. Accordingly, let $\min\operatorname*{\mathcal{C}}=\operatorname*{\mathcal{C}}(1)$ and $\max\operatorname*{\mathcal{C}}=\operatorname*{\mathcal{C}}(l)$.

Claim 7**.**

For every $i\geq 0$, there exists a collection $\Gamma_{i}$ of chains in $\operatorname*{\mathcal{F}}_{i}$ such that $\{\max\operatorname*{\mathcal{C}}:\operatorname*{\mathcal{C}}\in\Gamma_{i}\}$ is an antichain and

[TABLE]

Proof.

Let $\operatorname*{\mathcal{M}}$ be the family of maximal elements of $\operatorname*{\mathcal{F}}_{i}$ with respect to containment. For each $M\in\operatorname*{\mathcal{M}}$, let $\mathcal{H}_{M}\subset\operatorname*{\mathcal{F}}_{i}$ be a family of sets contained in $M$ such that the system $\{\mathcal{H}_{M}\}_{M\in\operatorname*{\mathcal{M}}}$ forms a partition of $\operatorname*{\mathcal{F}}_{i}$.

Note that any two sets in $\mathcal{H}_{M}$ have a nontrivial intersection, as every $A\in\mathcal{H}_{M}$ satisfies $A\subset M$ and $|A|>|M|/2$. Hence, $\mathcal{H}_{M}$ cannot contain an antichain of size $k$, otherwise, these $k$ sets would be pairwise weakly crossing. Therefore, by Dilworth’s theorem [7], $\mathcal{H}_{M}$ contains a chain $\operatorname*{\mathcal{C}}_{M}$ of size at least $|\mathcal{H}_{M}|/(k-1)$. The collection $\Gamma_{i}=\{\operatorname*{\mathcal{C}}_{M}:M\in\operatorname*{\mathcal{M}}\}$ meets the requirements of the Claim. ∎

Let $\Gamma_{i}$ be a collection of chains in $\operatorname*{\mathcal{F}}_{i}$ satisfying the conditions in Claim 7. As the maximal elements of the chains in $\Gamma_{i}$ form an antichain, Claim 6 gives the following upper bound on the size of $\Gamma_{i}$:

[TABLE]

From now on, fix some positive real numbers $a,b$ with $a\leq b\leq\log n$ and consider the union of blocks $\operatorname*{\mathcal{F}}_{a,b}=\bigcup_{a<i\leq b}\operatorname*{\mathcal{F}}_{i}$. Analogously, let $\Gamma_{a,b}=\bigcup_{a<i\leq b}\Gamma_{i}$. Allowing $a$ and $b$ to be not necessarily integers will serve as a slight convenience. In what follows, we bound the size of $\operatorname*{\mathcal{F}}_{a,b}$.

For each chain $\operatorname*{\mathcal{C}}\in\Gamma_{a,b}$, define a set $Y(\operatorname*{\mathcal{C}})$ by picking an arbitrary element from each of the difference sets ${\operatorname*{\mathcal{C}}(j+1)\setminus\operatorname*{\mathcal{C}}(j)}$ for $j=1,\ldots,|\operatorname*{\mathcal{C}}|-1$, and from $\operatorname*{\mathcal{C}}_{1}$, as well. Clearly, we have $|Y(\operatorname*{\mathcal{C}})|=|\operatorname*{\mathcal{C}}|$. For every $y\in[n]$, let $d(y)$ be the number of chains $\operatorname*{\mathcal{C}}$ in $\Gamma_{a,b}$ such that $y\in Y(\operatorname*{\mathcal{C}})$. Note that

[TABLE]

where the last inequality holds by Claim 7.

We will bound the size of $\operatorname*{\mathcal{F}}_{a,b}$ by arguing that one cannot have $k$ different elements of $[n]$ appearing in $Y(\operatorname*{\mathcal{C}})$ for many different sets $\operatorname*{\mathcal{C}}\in\Gamma_{a,b}$ without violating the condition that $\operatorname*{\mathcal{F}}$ is weakly $k$-cross-free. Thus, $\sum_{y\in[n]}d(y)$ must be small. For this, we need the following definition.

Definition 8**.**

Let $y\in[n]$. Consider a $k$-tuple of chains $(\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{k})$ in $\Gamma_{a,b}$, where $\operatorname*{\mathcal{C}}_{i}\in\Gamma_{j_{i}}$ for a strictly increasing sequence $j_{1}<\ldots<j_{k}$. We say that $(C_{1},\ldots,C_{k})$ is good for $y$ if

(i)

$y\in Y(\operatorname*{\mathcal{C}}_{i})$* for $i\in[k]$,*

(ii)

if $C_{i}\in\operatorname*{\mathcal{C}}_{i}$ is the smallest set such that $y\in C_{i}$, then $C_{1}\subset\ldots\subset C_{k}$.

Next, we show that if $d(y)$ is large, then $y$ is good for many $k$-tuples of chains. Let $g(y)$ denote the number of good $k$-tuples for $y$.

Claim 9**.**

For every $y\in[n]$, we have

[TABLE]

Proof.

Let $d=d(y)$ and let $\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{d}\in\Gamma_{a,b}$ be the chains such that $y\in Y(\operatorname*{\mathcal{C}}_{i})$. Also, for $i=1,\ldots,d$, let $C_{i}$ be the smallest set in $\operatorname*{\mathcal{C}}_{i}$ containing $y$, and let $\mathcal{H}=\{C_{1},\ldots,C_{d}\}$.

The family $\mathcal{H}$ is intersecting. Therefore, it cannot contain an antichain of size $k$, as any two elements of such an antichain are weakly crossing. Applying Dilworth’s theorem [7], we obtain that $\mathcal{H}$ contains a chain of size at least $s=\lceil d/(k-1)\rceil$. Without loss of generality, let $C_{1}\subset\ldots\subset C_{s}$ be such a chain.

For any $a\leq i\leq b$, $\operatorname*{\mathcal{F}}_{i}$ contains at most $k-1$ members of the sequence $C_{1},\ldots,C_{s}$. Otherwise, if $\operatorname*{\mathcal{C}}_{j_{1}},\ldots,\operatorname*{\mathcal{C}}_{j_{k}}\in\Gamma_{i}$ for some $1\leq j_{1}<\ldots<j_{k}\leq s$, the maximal elements $\max\operatorname*{\mathcal{C}}_{j_{1}},\ldots,\max\operatorname*{\mathcal{C}}_{j_{k}}$ are pairwise weakly crossing, because these sets form an antichain and contain $y$.

This implies that the sets $C_{1},\ldots,C_{s}$ are contained in at least $r=\lceil s/(k-1)\rceil\geq d/(k-1)^{2}$ different blocks. Thus, we can assume that there exist $i_{1}<\ldots<i_{r}$ and $j_{1}<\ldots<j_{r}$ such that $C_{i_{l}}\in\operatorname*{\mathcal{F}}_{j_{l}}$ for $l\in[r]$.

Then any $k$-element subset of $\{\operatorname*{\mathcal{C}}_{i_{1}},\ldots,\operatorname*{\mathcal{C}}_{i_{r}}\}$ is a good $k$-tuple for $y$, resulting in at least

[TABLE]

good $k$-tuples for $y$. ∎

Now we give an upper bound on the total number of $k$-tuples that may be good for some $y\in[n]$. A $k$-tuple of chains in $\Gamma_{a,b}$ is called nice if it is good for some $y\in[n]$. Let $N$ be the number of nice $k$-tuples.

Claim 10**.**

We have

[TABLE]

Proof.

Let $\operatorname*{\mathcal{C}}\in\Gamma_{a,b}$. Let us count the number of nice $k$-tuples $(\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{k})$ for which $\operatorname*{\mathcal{C}}=\operatorname*{\mathcal{C}}_{1}$. Note that in a nice $k$-tuple $(\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{k})$, the set $\min\operatorname*{\mathcal{C}}_{1}$ is contained in $\max\operatorname*{\mathcal{C}}_{1},\ldots,\max\operatorname*{\mathcal{C}}_{k}$.

But then, for any positive integer $i$ satisfying $a<i\leq b$, there are at most $k-1$ chains in $\Gamma_{i}$ that can belong to a nice $k$-tuple with first element $\operatorname*{\mathcal{C}}$. Indeed, suppose that there exist $k$ chains $\mathcal{D}_{1},\ldots,\mathcal{D}_{k}$ in $\Gamma_{i}$ that all appear in a nice $k$-tuple with their first element being $\operatorname*{\mathcal{C}}$. Then $\{\max\mathcal{D}_{1},\ldots,\max\mathcal{D}_{k}\}$ is an intersecting antichain: it is intersecting because $\max\mathcal{D}_{j}$ contains $\min\operatorname*{\mathcal{C}}$ for $j\in[k]$, and it is an antichain, by the definition of $\Gamma_{i}$. Thus, any two sets among $\max\mathcal{D}_{1},\ldots,\max\mathcal{D}_{k}$ are weakly crossing, a contradiction.

Hence, the number of nice $k$-tuples $(C_{1},\ldots,C_{k})$ for which $\operatorname*{\mathcal{C}}_{1}=\operatorname*{\mathcal{C}}$ is at most $\binom{b}{k-1}(k-1)^{k-1}$, as there are at most $\binom{b}{k-1}$ choices for $j_{2}<\ldots<j_{k}\leq b$ such that $C_{l}\in\Gamma_{j_{l}}$ for $l=2,\ldots,k$, and there are at most $k-1$ further choices for each chain $C_{l}$ in $\Gamma_{j_{l}}$.

Clearly, the number of choices for $\operatorname*{\mathcal{C}}=\operatorname*{\mathcal{C}}_{1}$ is at most the size of $\Gamma_{a,b}$, which is

[TABLE]

see (1) for the first inequality. Hence, the total number of nice $k$-tuples is at most

[TABLE]

∎

The next claim is the key observation in our proof. It tells us that a $k$-tuple of chains cannot be good for $k$ different elements of $[n]$.

Claim 11**.**

There are no $k$ different elements $y_{1},\ldots,y_{k}\in[n]$ and a $k$-tuple $(\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{k})$ such that $(\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{k})$ is good for $y_{1},\ldots,y_{k}$.

Proof.

Suppose that there exist such a $k$-tuple $(\operatorname*{\mathcal{C}}_{1},\ldots,\operatorname*{\mathcal{C}}_{k})$ and $k$ elements $y_{1},\ldots,y_{k}$. For $i,j\in[k]$, let $C_{i,j}$ be the smallest set in $\operatorname*{\mathcal{C}}_{i}$ that contains $y_{j}$. By the definition of a good $k$-tuple, we have $C_{1,j}\subset\ldots\subset C_{k,j}$ for $j\in[k]$. Also, the sets $C_{1,1},\ldots,C_{1,k}$ are distinct elements of the chain $\operatorname*{\mathcal{C}}_{1}$, so, without loss of generality, we can assume that $C_{1,1}\subset C_{1,2}\subset\ldots\subset C_{1,k}$.

First, we show that this assumption forces $C_{i,1}\subset\ldots\subset\operatorname*{\mathcal{C}}_{i,k}$ for all $i\in[k]$, as well. To this end, it is enough to prove that we cannot have $C_{i,j^{\prime}}\subset C_{i,j}$ for some $1\leq j<j^{\prime}\leq k$. Indeed, suppose that $C_{i,j^{\prime}}\subset C_{i,j}$. Then $y_{j}\in C_{i,j}$, but $y_{j}\not\in C_{i,j^{\prime}}$. However, $y_{j}\in C_{1,j}$ and $C_{1,j}\subset C_{1,j^{\prime}}\subset C_{i,j^{\prime}}$, contradiction.

Next, we show that any two sets in the family

[TABLE]

are weakly crossing. Every element of $\mathcal{H}$ contains $C_{1,1}$, so $\mathcal{H}$ is an intersecting family. Our task is reduced to showing that $\mathcal{H}$ is an antichain. Suppose that $C_{i,k+1-i}\subset C_{i^{\prime},k+1-i^{\prime}}$ for some $i,i^{\prime}\in[k]$, $i\neq i^{\prime}$. Then we must have $i<i^{\prime}$. Otherwise, $|C_{i,k+1-i}|>|C_{i^{\prime},k+1-i^{\prime}}|,$ as $C_{i,k+1-i}\in\operatorname*{\mathcal{F}}_{j_{i}}$ and $C_{i^{\prime},k+1-i^{\prime}}\in\operatorname*{\mathcal{F}}_{j_{i^{\prime}}}$ hold for some $j_{i^{\prime}}<j_{i}$. But if $i<i^{\prime}$, we have $y_{k+1-i}\in C_{i,k+1-i}$ and $y_{k+1-i}\not\in C_{i^{\prime},k+1-i^{\prime}}$, so $C_{i,k+1-i}\not\subset C_{i^{\prime},k+1-i^{\prime}}$.

Thus, any two sets of the $k$-element family $\mathcal{H}$ are weakly crossing, which is a contradiction. ∎

Let $M$ be the number of pairs $(y,(C_{1},\ldots,C_{k}))$ such that $(C_{1},\ldots,C_{k})$ is a good $k$-tuple for $y\in[n]$. Let us double count $M$.

On one hand, Claim 11 implies that $M\leq(k-1)N$. Plugging in our upper bound of Claim 10 for $N$, we get

[TABLE]

For simplicity, write $c_{1}(k)=2(k-1)^{k+1}/(k-1)!$, then our inequality becomes

[TABLE]

On the other hand, we have

[TABLE]

where $g(y)$, as before, stands for the number of good $k$-tuples for $y$. Applying Claim 9, we can bound the right-hand side from below, as follows.

[TABLE]

Exploiting the convexity of the function $\binom{x}{k}$, Jensen’s inequality implies that the right-hand side is at least

[TABLE]

Finally, using (2), we obtain

[TABLE]

Suppose that $|\operatorname*{\mathcal{F}}_{a,b}|>2k(k-1)^{3}n$. In this case, we have

[TABLE]

Writing $c_{2}(k)=1/2^{k}(k-1)^{3k}k!$, we can further bound the right-hand side of (4) and arrive at the inequality

[TABLE]

Comparing (3) and (5), we obtain

[TABLE]

which yields the following upper bound for the size of $\operatorname*{\mathcal{F}}_{a,b}$:

[TABLE]

Recall that (5) and the last inequality hold under the assumption that $|\operatorname*{\mathcal{F}}_{a,b}|>2k(k-1)^{3}n$. Hence, writing $c_{3}(k)=(c_{1}(k)/c_{2}(k))^{1/k}$, we get that

[TABLE]

holds without any assumption.

We finish the proof by choosing an appropriate sequence $\{a_{i}\}_{i=0}^{s}$ and applying the bound (6) for the families $\operatorname*{\mathcal{F}}_{a_{i},a_{i+1}}$.

Define the sequence $\{a_{i}\}_{i=0,1,\ldots}$ such that $a_{0}=0$, $a_{1}=k^{2}$ and $a_{i+1}=2^{a_{i}/(k-1)}$ for $i=1,2,\ldots$. Let $s$ be the smallest positive integer such that $a_{s}>\log n$. Clearly, we have $s=O_{k}(\log^{*}(n))$. Also,

[TABLE]

as $\operatorname*{\mathcal{F}}$ has at most $(k-1)n/l$ elements of size $l$ for $l\in[n]$, by the weakly $k$-cross-free property. Finally, for $i=1,\ldots,s-1$, (6) yields that

[TABLE]

The proof of Theorem 5 can be completed by noting that

[TABLE]

∎

Acknowledgement. We are grateful to Peter Frankl for his useful remarks. In particular, he pointed out that with more careful computation Claim 9 can be improved to

[TABLE]