A general 2-part Erd\H os-Ko-Rado theorem

TL;DR

This paper extends the Erd extH{o}s-Ko-Rado theorem to a two-part setting, establishing bounds on intersecting families within partitioned sets based on intersection properties.

Contribution

It introduces a generalized two-part Erd extH{o}s-Ko-Rado theorem for families with specified intersection sizes across partitions.

Findings

Proved a bound on the size of intersecting families in a two-part set.

Established conditions under which the maximum family size is achieved.

Generalized the classical theorem to a partitioned set context.

Abstract

A two-part extension of the famous Erd\H{o}s-Ko-Rado Theorem is proved. The underlying set is partitioned into $X_1$ and $X_2$. Some positive integers $k_i, \ell_i (1\leq i\leq m)$ are given. We prove that if ${\cal F}$ is an intersecting family containing members $F$ such that $|F\cap X_1|=k_i, |F\cap X_2|=\ell_i$ holds for one of the values $i (1\leq i\leq m)$ then $|{\cal F}|$ cannot exceed the size of the largest subfamily containing one element.

Equations

$$|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right).$$

$$\left(\genfrac{}{}{0ex}{}{n_1-1}{k-1}\right)\left(\genfrac{}{}{0ex}{}{n_2}{\ell }\right),$$

$$\left(\genfrac{}{}{0ex}{}{n_1}{k}\right)\left(\genfrac{}{}{0ex}{}{n_2-1}{\ell -1}\right).$$

$$|\mathcal{F}|\le \max \left\{\left(\genfrac{}{}{0ex}{}{n_1-1}{k-1}\right)\left(\genfrac{}{}{0ex}{}{n_2}{\ell }\right),\left(\genfrac{}{}{0ex}{}{n_1}{k}\right)\left(\genfrac{}{}{0ex}{}{n_2-1}{\ell -1}\right)\right\}.$$

$$\bigcup _{i=1}^m\left(\genfrac{}{}{0ex}{}{X_1,X_2}{k_i,{\ell }_i}\right).$$

$$\bigcup _{i=1}^m\left(\genfrac{}{}{0ex}{}{X_1,X_2}{k_i,{\ell }_i}\right)$$

$$|\mathcal{R}|\le |{\mathcal{R}}^1|\cdot |{\mathcal{R}}^2|$$

$$R_1=I_1\times J_0,R_2=I_2\times J_0,d(I_1,I_2)\ge b+1$$

$$R_3=I_0\times J_3,R_4=I_0\times J_4,d(J_3,J_4)\ge b+1.$$

$$|{\mathcal{R}}^2|\le 3\ell -2<3b.$$

$$|\mathcal{R}|\le 4b^2+(\ell -1)n_1.$$

$$\sum _{B\in {\mathcal{R}}_{\mathrm{b}}^2}\mu (B)\le (\ell -1)n_1.$$

$$\sum _{J\in {\mathcal{R}}_{\mathrm{s}}^2}\mu (J)\le (2\ell -2)(k+b)\le (2b)(2b)=4b^2.$$

$$|\mathcal{R}|=\sum _{J\in {\mathcal{R}}^2}\mu (J)=\sum _{B\in {\mathcal{R}}_{\mathrm{b}}^2}\mu (B)+\sum _{J\in {\mathcal{R}}_{\mathrm{s}}^2}\mu (J)\le 4b^2+(\ell -1)n_1.$$

$$|\mathcal{R}|<9b^2,$$

$$|\mathcal{R}|\le 4b^2+(\ell -1)n_1,$$

$$|\mathcal{R}|\le \ell n_1,$$

$$|\mathcal{R}|\le 4b^2+(k-1)n_2,$$

$$|\mathcal{R}|\le k{n}_2.$$

$$I_0\cap I_2=\varnothing .$$

$$J_0\cap I_4=\varnothing .$$

$$|{\mathcal{R}}_i|<9b^2$$

$$|{\mathcal{R}}_i|\le 4b^2+({\ell }_i-1)n_1$$

$$|{\mathcal{R}}_i|\le {\ell }_i{n}_1$$

$$|{\mathcal{R}}_i|\le 4b^2+(k_i-1)n_2$$

$$|{\mathcal{R}}_i|\le k_i{n}_2.$$

$$|{\mathcal{R}}_i|\le {\ell }_i{n}_1$$

$$|{\mathcal{R}}_i|\le k_i{n}_2.$$

$$\sum _{i=1}^m{\lambda }_i|{\mathcal{R}}_i|\le \max \left\{n_1\sum _{i=1}^m{\lambda }_i{\ell }_i,n_2\sum _{i=1}^m{\lambda }_i{k}_i\right\}$$

$$\sum _{i=1}^m{\lambda }_i|{\mathcal{R}}_i|\le n_1\sum _{i=1}^m{\lambda }_i{\ell }_i$$

Full text

1 Introduction

Let $X$ be a finite set of $n$ elements. A family ${\cal F}\subset 2^{X}$ is called intersecting if $F,G\in{\cal F}$ implies $F\cap G\not=\emptyset$. The family of all $k$-element subsets of $X$ is denoted by ${X\choose k}$. The celebrated theorem of Erdős, Ko and Rado is the following.

Theorem 1

[1]* Suppose that an integer $k\leq{n\over 2}$ is given and ${\cal F}\subset{X\choose k}$ is intersecting. Then*

[TABLE]

The family of all $k$-element subsets containing a fixed element $x\in X$ shows that the estimate is sharp.

The goal of our paper is to consider the problem when the underlying set is partitioned into two parts $X_{1},X_{2}$ and the sets $F\in{\cal F}$ have fixed sizes in both parts. More precisely let $X_{1}$ and $X_{2}$ be disjoint sets of $n_{1}$, respectively $n_{2}$ elements. [2] considered such subsets of $X=X_{1}\cup X_{2}$ which had $k$ elements in $X_{1}$ and $\ell$ elements in $X_{2}$. The family of all such sets is denoted by ${X_{1},X_{2}\choose k,\ell}$. The construction above, taking all possible sets containing a fixed element also works here. If the fixed element is in $X_{1}$ then the number of these sets is

[TABLE]

otherwise it is

[TABLE]

The following theorem of Frankl [2] claims that the larger one of these is the best.

Theorem 2

Let $X_{1},X_{2}$ be two disjoint sets of $n_{1}$ and $n_{2}$ elements, respectively. The positive integers $k,\ell$ satisfy the inequalities $2k\leq n_{1},2\ell\leq n_{2}$. If ${\cal F}$ is an intersecting subfamily of ${X_{1},X_{2}\choose k,\ell}$ then

[TABLE]

The goal of the present paper is to generalize Theorem 2 for the case when other sizes are also allowed that is the family consists of sets satisfying $|F\cap X_{1}|=k_{i},|F\cap X_{2}|=\ell_{i}$ for certain pairs of integers. Using the notation above, we will consider subfamilies of

[TABLE]

The generalization is however a little weaker at one point. In Theorem 2 the thresholds $2k\leq n_{1},2\ell\leq n_{2}$ for validity are natural. If either $n_{1}$ or $n_{2}$ is smaller then the problem becomes trivial, all such sets can be selected in ${\cal F}$. In the generalization below there is no such natural threshold. There will be another difference in the formulation. We give the construction of the extremal family rather than the maximum number of sets. A family is called trivially intersecting if there is an element contained in every member.

Theorem 3

Let $X_{1},X_{2}$ be two disjoint sets of $n_{1}$ and $n_{2}$ elements, respectively. Some positive integers $k_{i},\ell_{i}(1\leq i\leq m)$ are given. Define $b=\max_{i}\{k_{i},\ell_{i}\}$. Suppose that $9b^{2}\leq n_{1},n_{2}$. If ${\cal F}$ is an intersecting subfamily of

[TABLE]

then $|{\cal F}|$ cannot exceed the size of the largest trivially intersecting family satisfying the conditions.

Section 2 gives the proof of Theorem 3, using the method of “cyclic permutations”. Section 3 contains some open questions, while Section 4 shows some similar results from the past as motivations.

2 The proof of the main theorem

We will use the method of cyclic permutations [5] giving a simple proof of the EKR theorem. There the analogous problem is solved for intervals along a cyclic permutation and then a double counting easily finishes the proof. Here we need a pair of cyclic permutations: one for $X_{1}$ and one for $X_{2}$. A cycle of size $n_{i}$ will be represented by the integers $\mod n_{i}$. The usual notation is $\mathbb{Z}_{n_{i}}$. Hence the pair of cycles will be $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. The direct product of the intervals of length $k$ and $\ell$, in $\mathbb{Z}_{n_{1}}$ and $\mathbb{Z}_{n_{2}}$, respectively, will be a $k\times\ell$ rectangle in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. Problems analogous to our Theorem 3 will be considered for such rectangles.

The proof will be divided into lemmas. If only one cycle is involved we use the notation $\mathbb{Z}_{n}$ or even forget about this notation.

The distance $d(u,v)$ of the elements $u,v\in\mathbb{Z}_{n}$ is the smaller distance along the cycle.

Lemma 1

(Folklore)* Suppose $2\leq 2k<n$ and let $G_{n}^{k}=(\mathbb{Z}_{n},E)$ be a graph where two vertices are adjacent if their distance (mod $n$) is at most $k-1$. Then the largest click (complete subgraph) in $G_{n}^{k}$ has $k$ vertices. If $k$ vertices form a click then these vertices are consecutive (mod $n$).*

Proof. $k$ consecutive elements obviously form a click. In order to prove that there is no larger click, choose a vertex of a click. By symmetry one can suppose that [math] is this vertex. The potential other vertices of the click are $-(k-1),-(k-2),\ldots,-1,1,\ldots,k-2,k-1$. By the assumption $2k<n$ all these vertices are distinct. The pairs $(-(k-1),1),(-(k-2),2),\ldots,(-1,k-1)$ are not adjacent, therefore the click can contain only one of them, proving that the click contains at most $1+(k-1)=k$ vertices.

To prove the second statement of the lemma suppose that $U\subset\mathbb{Z}_{n}$ spans a click, moreover $|U|=k,0\in U$ hold. Let $i\in U(1<i\leq k-1)$ be in the click. Its “pair” cannot be chosen: $-(k-i)\not\in U$. However the distance of $i$ and $-(k-i-1)$ is also at least $n-(k+1)\geq 2k+1-(k+1)=k$. Therefore $-(k-i-1)\not\in U$ also holds. Its “pair” $i-1$ must be in $U$. Let $j$ be the largest element of $U$. We have seen that $j-1,j-2,\ldots,1$ are in $U$. On the other hand, since $j+1,\ldots,k-1\not\in U$, their “pairs” must be in $U$. It is fully determined: $U=\{-(k-j-1),-(k-j-2),\ldots,0,1,\ldots,j\}$. $\Box$

An interval of length $a$ in $\mathbb{Z}$ is a set of form $\{i+1,i+2,\ldots,i+a\}$ (mod $n$). The left end of this interval is $i+1$. The distance of two intervals is the minimum distance between two elements, one from each intervals.

Lemma 2

Let $k,b,n$ be positive integers satisfying $2(k+b)\leq n$. If $k+b+1$ distinct intervals of length $k$ are given in $\mathbb{Z}_{n}$ then there is a pair among them whose distance is at least $b+1$.

Proof. Apply Lemma 1 with $k+b$. The left ends of our intervals cannot form a click in $G_{n}^{k+b}$ therefore two of them must have a distance at least $k+b$. Consequently the distance of these intervals is at least $b+1$. $\Box$

Let $I$ be an interval of length $k$ in $\mathbb{Z}_{n_{1}}$ while $J$ is an interval of length $\ell$ in $\mathbb{Z}_{n_{2}}$. The direct product $I\times J$ is a $k\times\ell$ rectangle in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. We say that the rectangles $I_{1}\times J_{1}$ and $I_{2}\times J_{2}$ are proj-intersecting if either $I_{1}\cap I_{2}$ or $J_{1}\cap J_{2}$ is non-empty. A family of rectangles is proj-intersecting if any two rectangles in the family are proj-intersecting. If ${\cal R}$ is a family of rectangles, let ${\cal R}^{1}$ denote the set of intervals obtained by projecting the members of ${\cal R}$ on $\mathbb{Z}_{n_{1}}$. The family ${\cal R}^{2}$ is defined similarly. The inequality

[TABLE]

is obvious.

Lemma 3

Suppose that the positive integers $k,\ell,b,n_{1},n_{2}$ satisfy the inequalities $k,\ell\leq b,2(k+b)\leq n_{1},2(\ell+b)\leq n_{2}$. Let ${\cal R}$ be a proj-intersecting family of size $|{\cal R}|\geq 9b^{2}$ of $k\times\ell$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. Then either there are two rectangles $R_{1},R_{2}\in{\cal R}$ such that

[TABLE]

or there are two rectangles $R_{3},R_{4}\in{\cal R}$ such that

[TABLE]

Proof. (1) implies that either $|{\cal R}^{1}|>2b$ or $|{\cal R}^{2}|>2b.$

First suppose that $|{\cal R}^{1}|>2b\geq k+b$. Using Lemma 2 two members of ${\cal R}^{1}$ are obtained such that the distance between these two intervals $I_{5},I_{6}$ is at least $b$. Let $R_{5}$ and $R_{6}$ be two rectangles whose projections to $\mathbb{Z}_{n_{1}}$ are $I_{5},I_{6}$, respectively. Of course $I_{5}\cap I_{6}$ is empty, moreover no interval of length $k$ can intersect both of them, since $k\leq b$. Choose an arbitrary $R_{7}\in{\cal R}$ with projection $I_{7}$. It can intersect at most one of $I_{5}$ and $I_{6}$. Since ${\cal R}$ is proj-intersecting, $J_{5}\cap J_{6}\not=\emptyset$ and one of $J_{7}\cap J_{5},J_{7}\cap J_{6}$ is non-empty, too. The length of $J_{5}\cup J_{6}$ is at most $2\ell-1$. Here $J_{7}$ must meet $J_{5}\cup J_{6}$ therefore there are at most $3\ell-2$ choices for $J_{7}$. Hence we have

[TABLE]

The multiplicity $\mu(J)$ of $J\in{\cal R}^{2}$ is the number of members of ${\cal R}$ with projection $J$ on $\mathbb{Z}_{n_{2}}$. Obviously $\sum_{J\in{\cal R}^{2}}\mu(J)=|{\cal R}|$ holds. $|{\cal R}|\geq 9b^{2}$ and (4) imply that there is a member $J_{0}$ of ${\cal R}^{2}$ having multiplicity at least $3b\geq k+b+1$. Lemma 2 implies again that there are two rectangles $R_{1},R_{2}\in{\cal R}$ such that they both have the same projection $J_{0}$ on $\mathbb{Z}_{n_{2}}$ while their projections $I_{1},I_{2}$ have distance at least $b+1$. A pair of rectangles of form (2) was found.

The other case when $|{\cal R}^{2}|>2b$ is analogous, then a pair of type (3) can be found. $\Box$

We say that the rectangles $R_{1}$ and $R_{2}$ in (2) form a $b$-blocking pair with base $J_{0}$.

Lemma 4

Suppose that the rectangles $R_{1},R_{2}$ form a $b$-blocking pair with base $J_{0}$ and these two and a third rectangle $U\times V$ are pairwise proj-intersecting where $U\subset\mathbb{Z}_{n_{1}},V\subset\mathbb{Z}_{n_{2}}$ are intervals of lengths $1\leq|U|,|V|\leq b$. Then $J_{0}\cap V\not=\emptyset.$

Proof. Since $d(I_{1},I_{2})\geq b+1$, the projection $U$ can meet only one of them. Suppose $U\cap I_{2}=\emptyset.$Then the other projections of $R_{2}$ and $U\times V$ must meet: $J_{0}\cap V\not=\emptyset$ , as stated. $\Box$

Lemma 5

Suppose that the positive integers $k,\ell,b,n_{1},n_{2}$ satisfy the inequalities $k,\ell\leq b,2(k+b)<n_{1},2(\ell+b)<n_{2}$. Let ${\cal R}$ be a proj-intersecting family of $k\times\ell$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. Suppose that ${\cal R}$ contains $\ell$ pieces of $b$-blocking pairs of form (2) with distinct bases. Then there is a $\beta\in\mathbb{Z}_{n_{2}}$ such that the projection of every member of ${\cal R}$ contains it.

Proof. Lemma 4 implies that the bases must pairwise intersect. Let the bases be $B_{0},B_{1},\ldots,B_{\ell-1}$ . By Lemma 1 $B_{0},B_{1},\ldots,B_{\ell-1}$ are $\ell$ consecutive intervals: all intervals of length $\ell$ in an interval $A$ of length $2\ell-1$. All these bases contain the middle element of $A$, this will play the role of $\beta$.

Apply Lemma 4 for an arbitrary member of ${\cal R}$ (that is $|U|=k,|V|=\ell.$) By this lemma $V$ intersects each $B_{i}$. It is easy to see that $V$ must be equal to one of them. Hence it contains $\beta$. $\Box$

Corollary 1

$|{\cal R}|\leq\ell n_{1}$* holds under the conditions of Lemma 5.*

Proof. $\ell n_{1}$ is the total number of $k\times\ell$ rectangles whose projection on $\mathbb{Z}_{n_{2}}$ contains a fixed element $\beta$. $\Box$

Lemma 6

Suppose that the positive integers $k,\ell,b,n_{1},n_{2}$ satisfy the inequalities $k,\ell\leq b,2(k+b)<n_{1},2(\ell+b)<n_{2}$. Let ${\cal R}$ be a proj-intersecting family of $k\times\ell$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. Suppose that the number of $b$-blocking pairs of form (2) with distinct bases is at least one and at most $\ell-1$. Then

[TABLE]

Proof. Define a partition of ${\cal R}^{2}={\cal R}_{\rm b}^{2}\cup{\cal R}_{\rm s}^{2}$ where ${\cal R}_{\rm b}^{2}$ denotes the set of members of ${\cal R}^{2}$ with multiplicity at least $k+b+1$. If $B\in{\cal R}_{\rm b}^{2}$ then, by Lemma 2 there is a $b$-blocking pair with basis $B$. One of the conditions of the lemma is that $|{\cal R}_{\rm b}^{2}|\leq\ell-1$. These members of ${\cal R}_{\rm b}^{2}$ have only the trivial bound on the multiplicities: $n_{1}$. We have

[TABLE]

Choose one member $B_{0}\in{\cal R}_{\rm b}^{2}$. Lemma 4 implies that $B_{0}$ must meet every member of ${\cal R}_{\rm s}^{2}$. There are at most $2\ell-2$ intervals of length $\ell$ meeting $B_{0}$. This results in $|{\cal R}_{\rm s}^{2}|\leq 2\ell-2.$ Suppose $J\in{\cal R}_{\rm s}^{2}$. By Lemma 2, if $\mu(J)\geq k+b+1$ then there is a $b$-blocking pair with basis $J$. Therefore $\mu(J)\leq k+b$ holds. We obtained

[TABLE]

We can finish the proof using (5) and (6):

[TABLE]

$\Box$

The statements of Lemmas 3, 5 and 6 can be summarized in the following Corollary.

Corollary 2

Suppose that the positive integers $k,\ell,b,n_{1},n_{2}$ satisfy the inequalities $k,\ell\leq b,2(k+b)<n_{1},2(\ell+b)<n_{2}$. Let ${\cal R}$ be a proj-intersecting family of $k\times\ell$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$. Then one of the followings hold.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

This corollary, however is not sufficient for our final goal when there are rectangles of different sizes. The reason is that the statements of Corollary 2 cannot be independently used for different sizes, because they strongly interact. See the lemma below.

Lemma 7

Suppose that the positive integers $k_{1},k_{2},\ell_{1},\ell_{2},b,n_{1},n_{2}$ satisfy the inequalities $k_{1},k_{2},\ell_{1},\ell_{2}\leq b,4b<n_{1},4b<n_{2}$. Let ${\cal R}_{i}(i=1,2)$ be a family of $k_{i}\times\ell_{i}$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}$ and suppose that ${\cal R}={\cal R}_{1}\cup{\cal R}_{2}$ is proj-intersecting. Then ${\cal R}$ cannot simultaneously contain a $b$-blocking pair with basis in $\mathbb{Z}_{n_{1}}$ (of the form (2)) and another one with basis in $\mathbb{Z}_{n_{2}}$ (of the form (3)).

Proof. Let $R_{1},R_{2}\in{\cal R}_{1}$ be a $b$-blocking pair with basis $J_{0}\subset\mathbb{Z}_{n_{2}}$. In other words $R_{1}=I_{1}\times J_{0},R_{2}=I_{2}\times J_{0}$ where $d(I_{1},I_{2})\geq b+1$. On the other hand let $R_{3},R_{4}\in{\cal R}_{2}$ be the other $b$-blocking pair with basis $I_{0}\subset\mathbb{Z}_{n_{2}}$ where $R_{3}=I_{0}\times J_{3},R_{4}=I_{0}\times J_{4}$ with $d(J_{3},J_{4})\geq b+1$.

Here $d(I_{1},I_{2})\geq b+1$ implies that $I_{0}$ can intersect only one of them. Suppose

[TABLE]

On the other hand, using $d(J_{3},J_{4})\geq b+1$, the interval $J_{0}$ can meet at most one of $J_{3}$ and $J_{4}$. Suppose

[TABLE]

(7) and (8) show that $R_{2}$ and $R_{4}$ are not proj-intersecting. This contradiction finishes the proof. $\Box$

Lemma 8

Suppose that the positive integers $k_{i},\ell_{i},b,n_{1},n_{2}$ satisfy the inequalities $k_{i},\ell_{i}\leq b(1\leq i\leq m),4b<n_{1},n_{2}$. Let ${\cal R}_{i}$ be a family of $k_{i}\times\ell_{i}$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}(1\leq i\leq m)$. Suppose that ${\cal R}=\bigcup_{i=1}^{m}{\cal R}_{i}$ is a proj-intersecting family. Assume that there is a $b$-blocking pair say in $R_{1}$ with basis in $\mathbb{Z}_{n_{2}}$. Then either

[TABLE]

or

[TABLE]

or

[TABLE]

holds for all $i(1\leq i\leq m).$

Proof. If (9) does not hold for an $i$ then by Lemma 3 there is a $b$-blocking pair in ${\cal R}_{i}$. By Lemma 7 it must be one with a basis in $\mathbb{Z}_{n_{2}}$. Suppose that the number of distinct bases of $b$-blocking pairs ${\cal R}_{i}$ is $\ell_{i}$. Then by Lemma 5 and Corollary 1 we obtain (11). On the other hand if the number of distinct bases is between 1 and $\ell_{i}-1$ then Lemma 6 implies (10). $\Box$

Remark 1

Of course, if the bases of the $b$-blocking pairs are in $\mathbb{Z}_{n_{1}}$ then (10) and (11) are replaced by

[TABLE]

or

[TABLE]

Remark 2

The bases of $b$-blocking pairs for distinct ${\cal R}_{i}$’s must also intersect. This fact gives a stricter structure for the system of bases, but we will not use this fact.

Now we get rid of the cases, assuming that $n_{1}$ and $n_{2}$ are large enough.

Lemma 9

Suppose that the positive integers $k_{i},\ell_{i},b,n_{1},n_{2}$ satisfy the inequalities $k_{i},\ell_{i}\leq b(1\leq i\leq m),9b^{2}<n_{1},n_{2}$. Let ${\cal R}_{i}$ be a family of $k_{i}\times\ell_{i}$ rectangles in $\mathbb{Z}_{n_{1}}\times\mathbb{Z}_{n_{2}}(1\leq i\leq m)$. Suppose that ${\cal R}=\bigcup_{i=1}^{m}{\cal R}_{i}$ is a proj-intersecting family. Then either

[TABLE]

holds for all $i(1\leq i\leq m)$ or

[TABLE]

Proof. We only have to notice that each of (9) and (10) implies (11) under the condition $9b^{2}<n_{1},n_{2}$, while (12) implies (13). $\Box$

Corollary 3

Let $\lambda_{i}>0(1\leq i\leq m)$ be real numbers. Under the conditions of Lemma 9

[TABLE]

holds.

Proof. Indeed, summing up (14) for $i$ in Lemma 9

[TABLE]

is obtained while (15) leads to

[TABLE]

Since one of them must hold, the statement of the lemma follows. $\Box$

Proof of Theorem 3 Define the families

[TABLE]

We use double counting for the sum

[TABLE]

where ${\cal C}_{j}$ is a cyclic permutation of $\mathbb{Z}_{n_{j}}(j=1,2)$, $F\in{\cal F}$ and it forms a rectangle for the product of these two cyclic permutations and the weight $s(F)$ is defined in the following way:

[TABLE]

For a fixed set $F\in{\cal F}_{i}$ there are $k_{i}!(n_{1}-k_{i})!\ell_{i}!(n_{2}-\ell_{i})!$ pairs of cyclic permutations $({\cal C}_{1},{\cal C}_{2})$ in which $F$ is a rectangle. That is (16) is equal to

[TABLE]

[TABLE]

Now fix the permutations in (16):

[TABLE]

where ${\cal R}_{i}$ denotes the set of rectangles obtained from ${\cal F}_{i}$ in these fixed cyclic permutations. ${\cal F}$ is an intersecting family. It is easy to see that this implies that ${\cal R}=\cup{\cal R}_{i}$ is proj-intersecting. Corollary 3 can be applied with $\lambda_{i}=s_{i}(F)$ since its conditions are satisfied.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Taking into account that the number of pairs of cyclic permutations ${\cal C}_{1},{\cal C}_{2}$ is $(n_{1}-1)!(n_{2}-1)!$ and using (20) we obtain an upper estimate on (19):

[TABLE]

Since both (18) and (19) are equal to (16), we arrived to the inequality

[TABLE]

The quantities in the $\max$ are the numbers of all sets containing a fixed element of $X_{2}$ and $X_{1}$, respectively. $\Box$

3 Open problems

The upper bound in the Erdős-Ko-Rado theorem is reached for the family of all $k$-element sets containing a fixed element. What happens if we exclude this construction? Hilton and Milner found the largest intersecting, but not trivially intersecting family.

Theorem 4

[3]* If ${\cal F}$ is an intersecting but not a trivially intersecting family, ${\cal F}\subset{[n]\choose k}(2k\leq n)$ then*

[TABLE]

The construction giving equality is the following. Fix an element $x\in X$ and a $k$-element set $K$ such that $x\not\in K,K\subset X$. The extremal family will consist of all $k$-element sets containing $x$ and intersecting $K$.

This construction can be imitated for two parts, but the size of the family depends on weather $x$ and $K$ are in $X_{1}$ or $X_{2}$. Suppose that $x\in X_{1},K\subset X_{1}$. Then the construction of the family is the following:

[TABLE]

The case when $x$ and $K$ are in $X_{2}$ is analogous. We conjecture that one of these constructions is the best.

Conjecture 1

If ${\cal F}$ is a non-trivially intersecting subfamily ${X_{1},X_{2}\choose k,\ell}$ then

[TABLE]

[TABLE]

The constructions in this conjecture are non-trivially intersecting, but the “intersections happen” in one side. Our next question is what happens if one side is not enough for satisfying the intersection conditions. The families satisfying the following conditions are called two-sided intersecting: there are members $F_{11},F_{12},F_{21},F_{22}\in{\cal F}$ such that $F_{11}\cap F_{12}\cap X_{1}=\emptyset$ and $F_{21}\cap F_{22}\cap X_{2}=\emptyset$ hold.

To better understand our “best” two-sided intersecting construction one more notion and one more statement are needed. The families ${\cal F},{\cal G}\subset{[n]\choose k}$ are cross-intersecting if $F\cap G$ is non-empty for every pair of members $F\in{\cal F},G\in{\cal G}$. Here the total number $|{\cal F}|+|{\cal G}|$ of members should be maximized. But only for non-empty families, otherwise it is trivial and uninteresting.

Theorem 5

[3]* If ${\cal F},{\cal G}\subset{[n]\choose k}$ are non-empty cross-intersecting families then $|{\cal F}|+|{\cal G}|\leq 1+{n\choose k}-{n-k\choose k}$.*

This estimate is sharp: let $K$ be a $k$-element subset, let ${\cal F}_{1}(n)=\{K\}$ and let ${\cal G}_{1}(n)$ consist of all $k$-element subsets intersecting $K$.

Now we are ready to construct a large two-sided intersecting family ${\cal F}$. It will be done in two steps. First the “projection” ${\cal F}_{2}$ of our family ${\cal F}$ for $X_{2}$ will be given, formally

[TABLE]

Then we will determine the families ${\cal F}(M)$ “belonging” to the members $M\in{\cal F}_{2}$:

[TABLE]

${\cal F}_{2}$ will be an “almost intersecting” family in which there is only one non-intersecting pair. Start with a family extremal for Theorem 4 in $X_{2}$ where $x$ is a fixed element, $L$ is a fixed $\ell$-element subset, not containing $x$:

[TABLE]

Add another $\ell$-element set $L^{\prime}$ such that $x\in L^{\prime}$ holds. Of course, $L\cap L^{\prime}=\emptyset$. Define the “projection” (22) as

[TABLE]

$L\cap L^{\prime}=\emptyset,$ all other pairs are intersecting. Hence ${\cal F}(M)$ can be chosen to be ${X_{1}\choose k}$ if $M\not=L$ or $L^{\prime}$. However the families ${\cal F}(L)$ and ${\cal F}(L^{\prime})$ must be a pair of non-empty cross-intersecting families. To maximize the sum of their sizes the construction of Theorem 5 should be used. Choose a $k$-element subset $K\subset X_{1}$ and define ${\cal F}(L)=\{K\},{\cal F}(L^{\prime})=\{F\subset X_{1}:\ F\cap K\not=\emptyset\}.$ We believe that either this, or its symmetric version is the largest such family.

Conjecture 2

If ${\cal F}$ is a two-sided intersecting subfamily of ${X_{1},X_{2}\choose k,\ell}$ then

[TABLE]

[TABLE]

4 Related results from the past

$k$-part Sperner theorems. Let ${\cal F}$ be a family of subsets of the $n$-element $X$ satisfying the condition that $F\in{\cal F},G\in{\cal F}$ implies $F\not\subset G$. (In other words ${\cal F}$ is inclusion-free.) The well-known theorem of Sperner [9] states that $|{\cal F}|\leq{n\choose\lfloor n/2\rfloor}$ and this estimate is sharp: take all sets of size $\lfloor n/2\rfloor$.

Let now $X$ be a disjoint union of $X_{1}$ and $X_{2}$ where $|X_{1}|+|X_{2}|=n.$ It was noticed in [4] and [6] that Sperner’s bound remains valid if only such pairs of subsets $F\subset G$ are excluded which are equal in one of the parts, that is either $F\cap X_{1}=G\cap X_{1}$ or $F\cap X_{2}=G\cap X_{2}$ holds.

But this statement does not generalize to three parts. One can give a family ${\cal F}$ containing more than ${n\choose\lfloor n/2\rfloor}$ subsets of $X=X_{1}\cup X_{2}\cup X_{3}$ such that there are no pairs $F\in{\cal F},G\in{\cal F},F\subset G$ with $G-F\subset X_{i}$ for some $i$. It seems to be difficult to determine the largest ${\cal F}$ under this condition. The best estimates can be found in [7].

Non-empty intersections in both parts. Sali [8] considered the following problem. Let

[TABLE]

be a $t+1$-intersecting family with an additional condition: $F\cap G\cap X_{1}\not=\emptyset,F\cap G\cap X_{2}\not=\emptyset$ for all $F,G\in{\cal F}$. The exact maximum size of $|{\cal F}|$ is determined under this condition if both $|X_{1}|$ and $|X_{2}|$ are large enough. In spite of formal similarities this problem is very different from our problem in nature.

Using the two parts differently. The following result is also slightly related. Let

[TABLE]

that is the members of the family either have exactly $k$ elements in the first part, or exactly $\ell$ elements in the whole underlying set. Wang and Zhang [10] determined the maximum size of an intersecting family of this form.

5 Acknowledgement

The author is indebted to Noga Alon, Peter Frankl and Casey Tompkins for helpful discussions and the anonymous referee for careful reading and corrections.