A k-signed r-set on[n] = {1, ..., n} is an ordered pair (A, f), where A is an r-subset of [n] and f is a function from A to [k]. Families A1, ..., Ap are said to be cross-intersecting if any set in any family Ai intersects any set in any other family Aj. Hilton proved a sharp bound for the sum of sizes of cross-intersecting families of r-subsets of [n]. Our aim is to generalise Hilton's bound to one for families of k-signed r-sets on [n]. The main tool developed is an extension of Katona's cyclic permutation argument.peer-reviewe

Borg, Peter

Leader, Imre

Abstract A k-signed r-set on [n] = {1, ..., n} is an ordered pair (A, f ), where A is an rsubset of [n] and f is a function from A to [k]. Families A 1 , ..., A p are said to be cross-intersecting if any set in any family A i intersects any set in any other family A j . Hilton proved a sharp bound for the sum of sizes of cross-intersecting families of r-subsets of [n]. Our aim is to generalise Hilton&apos;s bound to one for families of k-signed r-sets on [n]. The main tool developed is an extension of Katona&apos;s cyclic permutation argument

Peter Borg

Imre Leader

CiteSeerX

Multiple cross-intersecting families of signed sets
Peter Borg
Department of Mathematics, University of Malta
Msida MSD 2080, Malta
p.borg.02@cantab.net
Imre Leader
Department of Pure Mathematics and Mathematical Statistics,
Centre for Mathematical Sciences, University of Cambridge,
Wilberforce Road, Cambridge CB3 0WB, United Kingdom
i.leader@dpmms.cam.ac.uk
4th December 2009
Abstract
A k-signed r-set on [n] = {1, ..., n} is an ordered pair (A, f), where A is an r-
subset of [n] and f is a function from A to [k]. Families A1, ...,Ap are said to be
cross-intersecting if any set in any family Ai intersects any set in any other family
Aj . Hilton proved a sharp bound for the sum of sizes of cross-intersecting families of
r-subsets of [n]. Our aim is to generalise Hilton's bound to one for families of k-signed
r-sets on [n]. The main tool developed is an extension of Katona's cyclic permutation
argument.
Keywords: Extremal set theory; Erd®sKoRado theorem; cross-intersecting fami-
lies; signed sets; cyclic permutations.
1 Introduction
For an integer n, the n-set {1, ..., n} is denoted by [n]. The power set {A : A ⊆ X} of a set
X is denoted by 2X , and the uniform sub-family {Y ⊆ X : |Y | = r} of 2X is denoted by(
X
r
)
.
If F is a family of sets and x is an element of the union of all sets in F , then we call
the sub-family of F consisting of those sets that contain x a star of F with centre x.
1
A family A is said to be intersecting if any two sets in A intersect. Note that a star of
a family is trivially intersecting.
The classical Erd®sKoRado (EKR) Theorem [13] says that if r ≤ n/2, then an inter-
secting sub-family A of
(
[n]
r
)
has size at most
(
n−1
r−1
)
, i.e. the size of a star of
(
[n]
r
)
; if r < n/2,
then A attains the bound if and only if A is a star of
(
[n]
r
)
(see [13, 20]). Two alternative
short and beautiful proofs of the EKR Theorem were obtained by Katona [21] and Daykin
[8]. In his proof, Katona introduced a very elegant averaging technique called the cycle
method. Daykin's proof is based on a fundamental result known as the Kruskal-Katona
Theorem [22, 23]. The EKR Theorem inspired a wealth of results and continues to do so;
the survey papers [9, 15] are recommended.
Families A1, . . . ,Ap are said to be cross-intersecting if for any distinct i and j in [p],
any set in Ai intersects any set in Aj.
Hilton [19] established the following best possible cross-intersection result.
Theorem 1.1 (Hilton [19]). Let r ≤ n/2 and p ≥ 2. Let A1, . . . ,Ap be cross-intersecting
sub-families of
(
[n]
r
)
. Then
p∑
i=1
|Ai| ≤

(
n
r
)
if p ≤ n
r
;
p
(
n− 1
r − 1
)
if p ≥ n
r
.
If equality holds and A1 6= ∅, then, unless p = 2 = n/r, one of the following holds:
(i) p < n/r, A1 =
(
[n]
r
)
and A2 = · · · = Ap = ∅;
(ii) p > n/r and |A1| = · · · = |Ap| =
(
n−1
r−1
)
;
(iii) p = n/r and A1, . . . ,Ap are as in (i) or (ii).
The EKR Theorem follows from this result: set p > n/r and A1 = · · · = Ap. Note
that if r > n/2, then it is trivial that the maximum sum of sizes is p
(
n
r
)
because any two
r-subsets of [n] intersect.
We mention that other authors have considered the the maximum product problem (see
[25], [28]); the main result in [25] implies that for any r ≤ n/2 and k ≥ 2, the product of
sizes of k cross-intersecting sub-families of
(
[n]
r
)
is a maximum if they are all the same star
of
(
[n]
r
)
. In this paper, we are interested in the maximum sum problem.
For any family A, we dene A∗ to be the sub-family of A consisting of those sets in A
that intersect each set in A, and we set A′ = A\A∗. So A′ consists of those sets in A that
do not intersect all sets in A.
In [5], the following extension of the EKR Theorem is proved and shown to immediately
yield Theorem 1.1 (it is also shown that in case (ii) of Theorem 1.1, A1 = · · · = Ap and A1
is a star of
(
[n]
r
)
).
Theorem 1.2 (Borg [5]). Let r ≤ n/2, and let A ⊆
(
[n]
r
)
. Then
|A∗|+ r
n
|A′| ≤
(
n− 1
r − 1
)
,
2
and if n > 2r then equality holds if and only if either A′ =
(
[n]
r
)
and A∗ = ∅ or A′ = ∅ and
A∗ is a star of
(
[n]
r
)
.
The proof was obtained by extending Daykin's proof of the EKR Theorem. It will be
easy to see from the proof of our main result (Theorem 1.4 below) how Theorem 1.1 follows
from Theorem 1.2.
As explained below, in this paper we provide an analogue of Theorem 1.2 for signed
sets and use it to obtain an analogue of Theorem 1.1 also for signed sets.
For r ∈ [n] and a positive integer k, let Sn,r,k be the family of k-signed r-sets on [n]
given by
Sn,r,k =
{
{(x1, s1), . . . , (xr, sr)} : {x1, . . . , xr} ∈
(
[n]
r
)
, s1, . . . , sr ∈ [k]
}
.
A well-known analogue of the EKR Theorem for signed sets was rst stated by Meyer
[26] and proved in dierent ways by Deza and Frankl [9] and Bollobás and Leader [4].
Theorem 1.3 (Deza, Frankl [9], Bollobás, Leader [4]). Let r ≤ n and k ≥ 2. Let A be an
intersecting sub-family of Sn,r,k. Then |A| ≤
(
n−1
r−1
)
kr−1, and if kn > 2r then equality holds
if and only if A is a star of Sn,r,k.
The proof of Deza and Frankl is based on the well-known shifting technique (see [15]),
whereas the proof of Bollobás and Leader is based on Katona's cycle method. There are
several other papers in the general area, for example [1, 2, 3, 7, 10, 11, 12, 14, 16, 17, 18,
24, 27].
This brings us to our analogue of Theorem 1.1 for signed sets.
Theorem 1.4. Let r ≤ n, k ≥ 2, p ≥ 2. Let A1, . . . ,Ap be cross-intersecting sub-families
of Sn,r,k. Then
p∑
i=1
|Ai| ≤

(
n
r
)
kr if p ≤ kn
r
;
p
(
n− 1
r − 1
)
kr−1 if p ≥ kn
r
.
Suppose equality holds and A1 6= ∅:
(i) if p < kn
r
then A1 = Sn,r,k and A2 = · · · = Ap = ∅;
(ii) if p > kn
r
then A1 = · · · = Ap and A1 is a star of Sn,r,k;
(iii) if p = kn
r
> 2 then A1, . . . ,Ap are as in (i) or (ii).
Theorem 1.3 follows from this result: set p > n/r and A1 = · · · = Ap. Theorem 1.1 can
be viewed as the cross-intersection result for 1-signed r-sets on [n]. We remark that the
case r = n in Theorem 1.4 is a special case of [6, Theorem 1.5], which employs a method
that is dierent from the one used here.
We will show that Theorem 1.4 is a consequence of the following analogue of Theo-
rem 1.2.
3
Theorem 1.5. Let r ≤ n and k ≥ 2. Let A ⊆ Sn,r,k. Then
|A∗|+ r
kn
|A′| ≤
(
n− 1
r − 1
)
kr−1,
and if kn > 2r then equality holds if and only if either A′ = Sn,r,k and A∗ = ∅ or A′ = ∅
and A∗ is a star of Sn,r,k.
We prove this extension of Theorem 1.3 by rening Katona's cycle method and em-
ploying the idea of good cyclic orderings (see the proof of Theorem 1.5) in the proof of
Theorem 1.3 by Bollobás and Leader [4]. The renement of the cycle method is given by
Lemma 2.1 in the next section, and it is inspired by Theorem 1.2. It is important to point
out that, as is clear from the proof of Theorem 1.5, Lemma 2.1 also leads to a Katona-type
proof of Theorem 1.2.
2 Proofs
The key idea in our work is to extend Katona's method [21] to give a result for any family,
whether intersecting or not; this is achieved in Lemma 2.1 below. The other key point is
to decide what kind of object should play the role of the cyclic orderings; this will become
clear at the very beginning of the proof of Theorem 1.5.
If σ is a cyclic ordering of the elements of a set X and the elements of a subset A of X
are consecutive in σ, then we say that A meets σ.
Lemma 2.1. Let m ≥ 2r, and let X be a set of size m. Let σ be a cyclic ordering of X.
Let C = {C ∈
(
X
r
)
: C meets σ}. For any B ⊆ C we have
|B∗|+ r
m
|B′| ≤ r,
and if m > 2r then equality holds if and only if either B′ = C and B∗ = ∅ or B′ = ∅ and
|B∗| = r.
Proof. Clearly there are m r-subsets of X that meet σ, i.e. |C| = m. So the result
is straightforward if B∗ = ∅. Suppose B∗ 6= ∅. Let B∗ ∈ B∗, and let x1, ..., xr be the
consecutive points in σ such that B∗ = {x1, ..., xr}. For i ∈ [r], let Ci be the r-set
in C beginning with xi in σ, and let C ′i be the r-set in C ending with xi in σ. Let D =
{C1, . . . , Cr}∪{C ′1, . . . , C ′r}. Note that B∗ = C1 = C ′r and hence D = {B∗}∪{C2, . . . , Cr}∪
{C ′1, . . . , C ′r−1}. By the denitions of B∗ and B′, we have B∗ ∪ B′ ⊆ D (because B∗ ∈ B∗)
and, since r ≤ m/2, C ′j−1 /∈ B∗∪B′ for any j ∈ [2, r] such that Cj ∈ B∗. It follows that there
are at least |B∗|−1 sets in D\(B∗∪B′), and hence |B′| ≤ |D|−|B∗|−(|B∗|−1) = 2r−2|B∗|.
So
|B∗|+ r
m
|B′| ≤ |B∗|+ 1
2
|B′| ≤ |B∗|+ 1
2
(2r − 2|B∗|) = r,
and it is immediate from this expression that if r
m
< 1
2
then equality holds throughout if
and only if |B∗| = r and B′ = ∅. Hence the result.
4
Katona [21] proved the above result for intersecting sub-families of C. Our result applies
to any sub-family.
Lemma 2.2. Let r ≤ n and k ≥ 2 such that kn > 2r. Suppose ∅ 6= F ⊆ Sn,r,k such that
for any A ∈ F and B ∈ {S ∈ Sn,r,k : A ∩ S = ∅}, B ∈ F . Then F = Sn,r,k.
Proof. We are given that F contains some set F . For simplicity, we may assume that
F = {(1, 1), (2, 1), . . . , (r, 1)}. Let A be an arbitrary set in Sn,r,k other than F . We are
required to show that A ∈ F .
Suppose k ≥ 3. Then there exist integers s1, . . . , sr ∈ [k] such that the set B =
{(1, s1), . . . , (r, sr)} is disjoint from both F and A. By the conditions of the lemma, we get
B ∈ F , which in turn implies A ∈ F .
Now suppose k = 2. Then n ≥ r + 1 as kn > 2r. We rst show that Sr,r,2 ⊂ F .
Let A1 be an arbitrary set in Sr,r,2 that intersects F on exactly r − 1 elements, say
A1∩F = {(1, 1), . . . , (r−1, 1)}. Then the set A′1 = {(1, 2), . . . , (r−1, 2), (r+1, 2)} ∈ Sn,r,2
is disjoint from both F and A1. So A
′
1 ∈ F , which in turn implies A1 ∈ F . Therefore
F contains A1 = {S ∈ Sr,r,2 : |S ∩ F | = r − 1}. Suppose r ≥ 2. Clearly, for any set
A2 ∈ A2 = {S ∈ Sr,r,2 : |S ∩ F | = r − 2}, there exist S ∈ A1 and A′2 ∈ Sn,r,k such that
|S ∩ A2| = r − 1 and A′2 is disjoint from both S and A2. So A′2 ∈ F and hence A2 ∈ F .
Thus A2 ⊂ F . We can keep on repeating this step until we obtain Sr,r,2 ⊂ F . Finally, if
A /∈ Sr,r,2, then there exists a set in Sr,r,2 that is disjoint from A, and hence A ∈ F .
Proof of Theorem 1.5. Let X = [n] × [k]. Let S = Sn,r,k. For a cyclic ordering σ
of X, a family F ⊆ S and a set S ∈ S, let Fσ = {F ∈ F : F meets σ} and
Φ(σ, S) =
1 if S meets σ;0 otherwise.
Note that
(A∗)σ ∪ (A′)σ = (Aσ)∗ ∪ (Aσ)′ and (A∗)σ ⊆ (Aσ)∗. (1)
We call a cyclic ordering σ of X good if any n elements (x1, s1), . . . , (xn, sn) of X ap-
pearing consecutively in σ are such that x1, . . . , xn are distinct and hence {x1, . . . , xn} =
[n]. This means that if the elements of X are listed in the order they appear in a
good cyclic ordering starting from an arbitrary element, then the list takes the form
(x1, s11), . . . , (xn, s1n), (x1, s21), . . . , (xn, s2n), . . . , (x1, sk1), . . . , (xn, skn); note that for any
i ∈ [n], {s1i, . . . , ski} = [k]. Let N be the set of all good cyclic orderings of X. The size of
N is h = n!(k!)n/|X| = (n − 1)!(k − 1)!(k!)n−1 (note that the division by |X| comes from
the fact that we are regarding any cyclic ordering and any rotation of it as the same). Any
set in S meets l = r!(n− r)!((k − 1)!)r(k!)n−r cyclic orderings in N . Thus we have
5
l
(
|A∗|+ r
kn
|A′|
)
=
( ∑
A∗∈A∗
l
)
+
r
kn
(∑
A′∈A′
l
)
=
∑
A∗∈A∗
∑
σ∈N
Φ(σ, A∗) +
r
kn
∑
A′∈A′
∑
σ∈N
Φ(σ, A′)
=
∑
σ∈N
( ∑
A∗∈A∗
Φ(σ, A∗) +
r
kn
∑
A′∈A′
Φ(σ, A′)
)
=
∑
σ∈N
(
|(A∗)σ|+
r
kn
|(A′)σ|
)
≤
∑
σ∈N
(
|(Aσ)∗|+
r
kn
|(Aσ)′|
)
(by (1)) (2)
≤
∑
σ∈N
r (by Lemma 2.1) (3)
= rh,
which yields the bound in the theorem.
Suppose A′ = ∅. Then the above immediately gives us |A∗| ≤
(
n−1
r−1
)
kr−1. This is in
fact Theorem 1.3, which also tells us that the inequality is attained only by stars of S if
kn > 2r.
Now suppose kn > 2r, A′ 6= ∅ and we have equality in the theorem. So we have equality
in (2) and (3). By (1) and the equality in (2), we clearly have
(A∗)σ = (Aσ)∗ and (A′)σ = (Aσ)′. (4)
The equality in (3) and Lemma 2.1 give us that for any σ ∈ N , if (Aσ)′ 6= ∅ then (Aσ)′ = Sσ
(and (Aσ)∗ = ∅). Thus, by (4),
for any σ ∈ N , if (A′)σ 6= ∅ then (A′)σ = Sσ. (5)
Let A be an arbitrary set {(x1, p1), . . . , (xr, pr)} inA′. Let B be an arbitrary set {(y1, q1), . . . ,
(yr, qr)} in {S ∈ S : A ∩ S = ∅}. Let X = {x1, . . . , xr}, Y = {y1, . . . , yr}, m = |X ∩ Y |.
As we now show, there exists σA,B ∈ N such that both A and B meet σA,B. If
m = r (i.e. X = Y ), then this is straightforward since A ∩ B = ∅. If m = 0 (i.e. X ∩
Y = ∅, and so 2r ≤ n), then clearly there exist members of N in which the elements
(x1, p1), . . . , (xr, pr), (y1, q1), . . . , (yr, qr) of A ∪ B appear consecutively in the given order.
Now suppose 1 ≤ m ≤ r−1. Let z1, . . . , zm be the elements of X ∩Y . We may re-label the
elements of A as (u1, p
′
1), . . . , (ur−m, p
′
r−m), (z1, p
′
r−m+1), . . . , (zm, p
′
r) and the elements of B
as (z1, q
′
1), . . . , (zm, q
′
m), (v1, q
′
m+1), . . . , (vr−m, q
′
r), and it is clear from the order in which we
listed the elements that there exist members σ of N such that both A and B meet σ.
Finally, since A ∈ (A′)σA,B and B ∈ SσA,B , we have B ∈ (A′)σA,B by (5). So B ∈ A′.
Therefore A′ = S by Lemma 2.2. Hence the result.
6
Proof of Theorem 1.4. Let A =
⋃p
i=1Ai. Clearly A∗ =
⋃p
i=1A∗i and A′ =
⋃p
i=1A′i.
Suppose A′i ∩ A′j 6= ∅, i 6= j. Let A ∈ A′i ∩ A′j. Then there exists Ai ∈ A′i such that
A ∩Ai = ∅, which is a contradiction because A ∈ Aj. So A′i ∩A′j = ∅ for i 6= j, and hence
|A′| =
∑p
i=1 |A′i|. Applying Theorem 1.5, we therefore get
p∑
i=1
|Ai| =
p∑
i=1
|A′i|+
p∑
i=1
|A∗i | ≤ |A′|+ p|A∗| ≤
(
n
r
)
kr +
(
p− kn
r
)
|A∗|. (6)
Suppose p < kn
r
. Then
∑p
i=1 |Ai| ≤
(
n
r
)
kr, and equality holds if and only if A∗ = ∅ and
A = A′ = Sn,r,k. If A ∈ A1 and B is a set in Sn,r,k\A1 that does not intersect A, then
B /∈ Ai, i = 2, . . . , p, and hence B ∈ Sn,r,k\A. Thus, if A = Sn,r,k then the conditions of
Lemma 2.2 hold for A1 (recall that A1 6= ∅), and therefore A1 = A = Sn,r,k. Hence (i).
Next, suppose p > kn
r
. Then, by (6) and Theorem 1.5,
p∑
i=1
|Ai| ≤
(
n
r
)
kr +
(
p− kn
r
)(
n− 1
r − 1
)
kr−1 = p
(
n− 1
r − 1
)
kr−1,
and equality holds if and only if A∗1 = · · · = A∗k = A∗ and |A∗| =
(
n−1
r−1
)
kr−1 = |A|, in which
case A is a star of Sn,r,k by Theorem 1.3. Hence (ii).
Finally, suppose p = kn
r
. Then, by (6),
∑p
i=1 |Ai| ≤ |A′| +
kn
r
|A∗| ≤
(
n
r
)
kr. Suppose
p > 2, i.e. kn
r
> 2. If A∗ = ∅ then A is as in the case p < kn
r
, and it is immediate from
Theorem 1.5 that if A∗ 6= ∅ then A∗ is as in the case p > kn
r
. Hence (iii).
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in Math., Vol. 411, Springer, Berlin, 1974, 127139.
[27] A. Moon, An analogue of the Erd®sKoRado theorem for the Hamming schemes
H(n, q), J. Combin. Theory Ser. A 32 (1982), 386390.
[28] L. Pyber, A new generalization of the Erd®sKoRado theorem, J. Combin. Theory
Ser. A 43 (1986), 8590.
9


Multiple cross-intersecting families of signed sets

OAR@UM

Journal of Combinatorial Theory, Series A 117 (2010) 583–588Contents lists available at ScienceDirectJournal of Combinatorial Theory,Series Awww.elsevier.com/locate/jctaNoteMultiple cross-intersecting families of signed setsPeter Borg a, Imre Leader ba Department of Mathematics, University of Malta, Msida MSD 2080, Maltab Department of Pure Mathematics and Mathematical Statistics, Centre for Mathematical Sciences, University of Cambridge,Wilberforce Road, Cambridge CB3 0WB, United Kingdoma r t i c l e i n f o a b s t r a c tArticle history:Received 15 July 2008Available online 8 January 2010Keywords:Extremal set theoryErdo˝s–Ko–Rado TheoremCross-intersecting familiesSigned setsCyclic permutationsA k-signed r-set on [n] = {1, . . . ,n} is an ordered pair (A, f ), whereA is an r-subset of [n] and f is a function from A to [k]. FamiliesA1, . . . ,Ap are said to be cross-intersecting if any set in any familyAi intersects any set in any other family A j . Hilton proved asharp bound for the sum of sizes of cross-intersecting families ofr-subsets of [n]. Our aim is to generalise Hilton’s bound to one forfamilies of k-signed r-sets on [n]. The main tool developed is anextension of Katona’s cyclic permutation argument.© 2009 Elsevier Inc. All rights reserved.1. IntroductionFor an integer n, the n-set {1, . . . ,n} is denoted by [n]. The power set {A: A ⊆ X} of a set X isdenoted by 2X , and the uniform sub-family {Y ⊆ X: |Y | = r} of 2X is denoted by (Xr ).If F is a family of sets and x is an element of the union of all sets in F , then we call the sub-family of F consisting of those sets that contain x a star of F with centre x.A family A is said to be intersecting if any two sets in A intersect. Note that a star of a family istrivially intersecting.The classical Erdo˝s–Ko–Rado (EKR) Theorem [13] says that if r  n/2, then an intersecting sub-family A of ([n]r ) has size at most (n−1r−1), i.e. the size of a star of ([n]r ); if r < n/2, then A attains thebound if and only if A is a star of ([n]r ) (see [13,20]). Two alternative short and beautiful proofs of theEKR Theorem were obtained by Katona [21] and Daykin [8]. In his proof, Katona introduced a veryelegant averaging technique called the cycle method. Daykin’s proof is based on a fundamental resultknown as the Kruskal–Katona Theorem [22,23]. The EKR Theorem inspired a wealth of results andcontinues to do so; the survey papers [9,15] are recommended.E-mail addresses: p.borg.02@cantab.net (P. Borg), i.leader@dpmms.cam.ac.uk (I. Leader).0097-3165/$ – see front matter © 2009 Elsevier Inc. All rights reserved.doi:10.1016/j.jcta.2009.11.010584 P. Borg, I. Leader / Journal of Combinatorial Theory, Series A 117 (2010) 583–588Families A1, . . . ,Ap are said to be cross-intersecting if for any distinct i and j in [p], any set in Aiintersects any set in A j .Hilton [19] established the following best possible cross-intersection result.Theorem 1.1. (See Hilton [19].) Let r  n/2 and p  2. Let A1, . . . ,Ap be cross-intersecting sub-familiesof([n]r). Thenp∑i=1|Ai|{(nr)if p  nr ;p(n−1r−1)if p  nr .If equality holds and A1 = ∅, then, unless p = 2 = n/r, one of the following holds:(i) p < n/r, A1 =([n]r)and A2 = · · · = Ap = ∅;(ii) p > n/r and |A1| = · · · = |Ap| =(n−1r−1);(iii) p = n/r and A1, . . . ,Ap are as in (i) or (ii).The EKR Theorem follows from this result: set p > n/r and A1 = · · · = Ap . Note that if r > n/2,then it is trivial that the maximum sum of sizes is p(nr)because any two r-subsets of [n] intersect.We mention that other authors have considered the maximum product problem (see [25,28]); themain result in [25] implies that for any r  n/2 and k 2, the product of sizes of k cross-intersectingsub-families of([n]r)is a maximum if they are all the same star of([n]r). In this paper, we are interestedin the maximum sum problem.For any family A, we deﬁne A∗ to be the sub-family of A consisting of those sets in A thatintersect each set in A, and we set A′ = A \ A∗ . So A′ consists of those sets in A that do notintersect all sets in A.In [5], the following extension of the EKR Theorem is proved and shown to immediately yieldTheorem 1.1 (it is also shown that in case (ii) of Theorem 1.1, A1 = · · · = Ap and A1 is a star of([n]r)).Theorem 1.2. (See Borg [5].) Let r  n/2, and let A ⊆ ([n]r ). Then∣∣A∗∣∣+ rn∣∣A′∣∣ (n − 1r − 1),and if n > 2r then equality holds if and only if either A′ = ([n]r ) and A∗ = ∅ or A′ = ∅ and A∗ is a star of ([n]r ).The proof was obtained by extending Daykin’s proof of the EKR Theorem. It will be easy to seefrom the proof of our main result (Theorem 1.4 below) how Theorem 1.1 follows from Theorem 1.2.As explained below, in this paper we provide an analogue of Theorem 1.2 for signed sets and use itto obtain an analogue of Theorem 1.1 also for signed sets.For r ∈ [n] and a positive integer k, let Sn,r,k be the family of k-signed r-sets on [n] given bySn,r,k ={{(x1, s1), . . . , (xr, sr)}: {x1, . . . , xr} ∈([n]r), s1, . . . , sr ∈ [k]}.A well-known analogue of the EKR Theorem for signed sets was ﬁrst stated by Meyer [26] andproved in different ways by Deza and Frankl [9] and Bollobás and Leader [4].Theorem 1.3. (See Deza and Frankl [9], Bollobás and Leader [4].) Let r  n and k 2. Let A be an intersectingsub-family of Sn,r,k. Then |A|(n−1r−1)kr−1 , and if kn > 2r then equality holds if and only if A is a star of Sn,r,k.The proof of Deza and Frankl is based on the well-known shifting technique (see [15]), whereasthe proof of Bollobás and Leader is based on Katona’s cycle method. There are several other papers inthe general area, for example [1–3,7,10–12,14,16–18,24,27].This brings us to our analogue of Theorem 1.1 for signed sets.P. Borg, I. Leader / Journal of Combinatorial Theory, Series A 117 (2010) 583–588 585Theorem 1.4. Let r  n, k 2, p  2. Let A1, . . . ,Ap be cross-intersecting sub-families of Sn,r,k. Thenp∑i=1|Ai|{(nr)kr if p  knr ;p(n−1r−1)kr−1 if p  knr .Suppose equality holds and A1 = ∅:(i) if p < knr then A1 = Sn,r,k and A2 = · · · = Ap = ∅;(ii) if p > knr then A1 = · · · = Ap and A1 is a star of Sn,r,k;(iii) if p = knr > 2 then A1, . . . ,Ap are as in (i) or (ii).Theorem 1.3 follows from this result: set p > n/r and A1 = · · · = Ap . Theorem 1.1 can be viewed asthe cross-intersection result for 1-signed r-sets on [n]. We remark that the case r = n in Theorem 1.4is a special case of [6, Theorem 1.5], which employs a method that is different from the one usedhere.We will show that Theorem 1.4 is a consequence of the following analogue of Theorem 1.2.Theorem 1.5. Let r  n and k 2. Let A ⊆ Sn,r,k. Then∣∣A∗∣∣+ rkn∣∣A′∣∣ (n − 1r − 1)kr−1,and if kn > 2r then equality holds if and only if either A′ = Sn,r,k and A∗ = ∅ or A′ = ∅ and A∗ is a starof Sn,r,k.We prove this extension of Theorem 1.3 by reﬁning Katona’s cycle method and employing the ideaof good cyclic orderings (see the proof of Theorem 1.5) in the proof of Theorem 1.3 by Bollobás andLeader [4]. The reﬁnement of the cycle method is given by Lemma 2.1 in the next section, and it isinspired by Theorem 1.2. It is important to point out that, as is clear from the proof of Theorem 1.5,Lemma 2.1 also leads to a Katona-type proof of Theorem 1.2.2. ProofsThe key idea in our work is to extend Katona’s method [21] to give a result for any family, whetherintersecting or not; this is achieved in Lemma 2.1 below. The other key point is to decide what kindof object should play the role of the cyclic orderings; this will become clear at the very beginning ofthe proof of Theorem 1.5.If σ is a cyclic ordering of the elements of a set X and the elements of a subset A of X areconsecutive in σ , then we say that A meets σ .Lemma 2.1. Let m  2r, and let X be a set of size m. Let σ be a cyclic ordering of X . Let C ={C ∈ (Xr ): C meets σ }. For any B ⊆ C we have∣∣B∗∣∣+ rm∣∣B′∣∣ r,and if m > 2r then equality holds if and only if either B′ = C and B∗ = ∅ or B′ = ∅ and |B∗| = r.Proof. Clearly there are m r-subsets of X that meet σ , i.e. |C| = m. So the result is straightforwardif B∗ = ∅. Suppose B∗ = ∅. Let B∗ ∈ B∗ , and let x1, . . . , xr be the consecutive points in σ such thatB∗ = {x1, . . . , xr}. For i ∈ [r], let Ci be the r-set in C beginning with xi in σ , and let C ′i be the r-set in C ending with xi in σ . Let D = {C1, . . . ,Cr} ∪ {C ′1, . . . ,C ′r}. Note that B∗ = C1 = C ′r and henceD = {B∗}∪{C2, . . . ,Cr}∪{C ′1, . . . ,C ′r−1}. By the deﬁnitions of B∗ and B′ , we have B∗∪B′ ⊆ D (because586 P. Borg, I. Leader / Journal of Combinatorial Theory, Series A 117 (2010) 583–588B∗ ∈ B∗) and, since r m/2, C ′j−1 /∈ B∗ ∪ B′ for any j ∈ [2, r] such that C j ∈ B∗ . It follows that thereare at least |B∗| − 1 sets in D \ (B∗ ∪ B′), and hence |B′| |D| − |B∗| − (|B∗| − 1) = 2r − 2|B∗|. So∣∣B∗∣∣+ rm∣∣B′∣∣ ∣∣B∗∣∣+ 12∣∣B′∣∣ ∣∣B∗∣∣+ 12(2r − 2∣∣B∗∣∣)= r,and it is immediate from this expression that if rm <12 then equality holds throughout if and only if|B∗| = r and B′ = ∅. Hence the result. Katona [21] proved the above result for intersecting sub-families of C . Our result applies to anysub-family.Lemma 2.2. Let r  n and k  2 such that kn > 2r. Suppose ∅ = F ⊆ Sn,r,k such that for any A ∈ F andB ∈ {S ∈ Sn,r,k: A ∩ S = ∅}, B ∈ F . Then F = Sn,r,k.Proof. We are given that F contains some set F . For simplicity, we may assume that F ={(1,1), (2,1), . . . , (r,1)}. Let A be an arbitrary set in Sn,r,k other than F . We are required to showthat A ∈ F .Suppose k  3. Then there exist integers s1, . . . , sr ∈ [k] such that the set B = {(1, s1), . . . , (r, sr)}is disjoint from both F and A. By the conditions of the lemma, we get B ∈ F , which in turn impliesA ∈ F .Now suppose k = 2. Then n r+1 as kn > 2r. We ﬁrst show that Sr,r,2 ⊂ F . Let A1 be an arbitraryset in Sr,r,2 that intersects F on exactly r − 1 elements, say A1 ∩ F = {(1,1), . . . , (r − 1,1)}. Then theset A′1 = {(1,2), . . . , (r − 1,2), (r + 1,2)} ∈ Sn,r,2 is disjoint from both F and A1. So A′1 ∈ F , which inturn implies A1 ∈ F . Therefore F contains A1 = {S ∈ Sr,r,2: |S ∩ F | = r − 1}. Suppose r  2. Clearly,for any set A2 ∈ A2 = {S ∈ Sr,r,2: |S ∩ F | = r − 2}, there exist S ∈ A1 and A′2 ∈ Sn,r,k such that|S ∩ A2| = r − 1 and A′2 is disjoint from both S and A2. So A′2 ∈ F and hence A2 ∈ F . Thus A2 ⊂ F .We can keep on repeating this step until we obtain Sr,r,2 ⊂ F . Finally, if A /∈ Sr,r,2, then there existsa set in Sr,r,2 that is disjoint from A, and hence A ∈ F . Proof of Theorem 1.5. Let X = [n] × [k]. Let S = Sn,r,k . For a cyclic ordering σ of X , a family F ⊆ Sand a set S ∈ S , let Fσ = {F ∈ F : F meets σ } andΦ(σ , S) ={1 if S meets σ ;0 otherwise.Note that(A∗)σ∪ (A′)σ= (Aσ )∗ ∪ (Aσ )′ and(A∗)σ⊆ (Aσ )∗. (1)We call a cyclic ordering σ of X good if any n elements (x1, s1), . . . , (xn, sn) of X appearingconsecutively in σ are such that x1, . . . , xn are distinct and hence {x1, . . . , xn} = [n]. This meansthat if the elements of X are listed in the order they appear in a good cyclic ordering startingfrom an arbitrary element, then the list takes the form (x1, s11), . . . , (xn, s1n), (x1, s21), . . . , (xn, s2n),. . . , (x1, sk1), . . . , (xn, skn); note that for any i ∈ [n], {s1i, . . . , ski} = [k]. Let N be the set of all goodcyclic orderings of X . The size of N is h = n!(k!)n/|X | = (n − 1)!(k − 1)!(k!)n−1 (note that the divisionby |X | comes from the fact that we are regarding any cyclic ordering and any rotation of it as thesame). Any set in S meets l = r!(n − r)!((k − 1)!)r(k!)n−r cyclic orderings in N . Thus we havel(∣∣A∗∣∣+ rkn∣∣A′∣∣)= ( ∑A∗∈A∗l)+ rkn( ∑A′∈A′l)=∑A∗∈A∗∑σ∈NΦ(σ , A∗)+ rkn∑A′∈A′∑σ∈NΦ(σ , A′)=∑( ∑∗ ∗Φ(σ , A∗)+ rkn∑′ ′Φ(σ , A′))σ∈N A ∈A A ∈AP. Borg, I. Leader / Journal of Combinatorial Theory, Series A 117 (2010) 583–588 587=∑σ∈N(∣∣(A∗)σ∣∣+ rkn∣∣(A′)σ∣∣)∑σ∈N(∣∣(Aσ )∗∣∣+ rkn∣∣(Aσ )′∣∣) (by (1))(2)∑σ∈Nr (by Lemma 2.1) (3)= rh,which yields the inequality in the theorem.Suppose A′ = ∅. Then the above immediately gives us |A∗| (n−1r−1)kr−1. This is in fact Theorem 1.3,which also tells us that the bound is attained only by stars of S if kn > 2r.Now suppose kn > 2r, A′ = ∅ and we have equality in the theorem. So we have equality in (2)and (3). By (1) and the equality in (2), we clearly have(A∗)σ= (Aσ )∗ and(A′)σ= (Aσ )′. (4)The equality in (3) and Lemma 2.1 give us that for any σ ∈ N , if (Aσ )′ = ∅ then (Aσ )′ = Sσ (and(Aσ )∗ = ∅). Thus, by (4),for any σ ∈ N, if (A′)σ= ∅ then (A′)σ= Sσ . (5)Let A be an arbitrary set {(x1, p1), . . . , (xr, pr)} in A′ . Let B be an arbitrary set {(y1,q1), . . . , (yr,qr)}in {S ∈ S: A ∩ S = ∅}. Let X = {x1, . . . , xr}, Y = {y1, . . . , yr}, m = |X ∩ Y |.As we now show, there exists σA,B ∈ N such that both A and B meet σA,B . If m = r (i.e. X = Y ),then this is straightforward since A ∩ B = ∅. If m = 0 (i.e. X ∩ Y = ∅, and so 2r  n), then clearlythere exist members of N in which the elements (x1, p1), . . . , (xr, pr), (y1,q1), . . . , (yr,qr) of A ∪ Bappear consecutively in the given order. Now suppose 1m r−1. Let z1, . . . , zm be the elements ofX ∩Y . We may re-label the elements of A as (u1, p′1), . . . , (ur−m, p′r−m), (z1, p′r−m+1), . . . , (zm, p′r) andthe elements of B as (z1,q′1), . . . , (zm,q′m), (v1,q′m+1), . . . , (vr−m,q′r), and it is clear from the order inwhich we listed the elements that there exist members σ of N such that both A and B meet σ .Finally, since A ∈ (A′)σA,B and B ∈ SσA,B , we have B ∈ (A′)σA,B by (5). So B ∈ A′ . Therefore A′ = Sby Lemma 2.2. Hence the result. Proof of Theorem 1.4. Let A = ⋃pi=1 Ai . Clearly A∗ = ⋃pi=1 A∗i and A′ = ⋃pi=1 A′i . SupposeA′i ∩ A′j = ∅, i = j. Let A ∈ A′i ∩ A′j . Then there exists Ai ∈ A′i such that A ∩ Ai = ∅, which is acontradiction because A ∈ A j . So A′i ∩ A′j = ∅ for i = j, and hence |A′| =∑pi=1 |A′i |. Applying Theo-rem 1.5, we therefore getp∑i=1|Ai| =p∑i=1∣∣A′i∣∣+p∑i=1∣∣A∗i ∣∣ ∣∣A′∣∣+ p∣∣A∗∣∣(nr)kr +(p − knr)∣∣A∗∣∣. (6)Suppose p < knr . Then∑pi=1 |Ai | (nr)kr , and equality holds if and only if A∗ = ∅ and A = A′ =Sn,r,k . If A ∈ A1 and B is a set in Sn,r,k \ A1 that does not intersect A, then B /∈ Ai , i = 2, . . . , p, andhence B ∈ Sn,r,k \ A. Thus, if A = Sn,r,k then the conditions of Lemma 2.2 hold for A1 (recall thatA1 = ∅), and therefore A1 = A = Sn,r,k . Hence (i).Next, suppose p > knr . Then, by (6) and Theorem 1.5,p∑i=1|Ai|(nr)kr +(p − knr)(n − 1r − 1)kr−1 = p(n − 1r − 1)kr−1,and equality holds if and only if A∗1 = · · · = A∗k = A∗ and |A∗| =(n−1r−1)kr−1 = |A|, in which case A isa star of Sn,r,k by Theorem 1.3. Hence (ii).588 P. Borg, I. Leader / Journal of Combinatorial Theory, Series A 117 (2010) 583–588Finally, suppose p = knr . Then, by (6),∑pi=1 |Ai | |A′| + knr |A∗|(nr)kr . Suppose p > 2, i.e. knr > 2.If A∗ = ∅ then A is as in the case p < knr , and it is immediate from Theorem 1.5 that if A∗ = ∅ thenA∗ is as in the case p > knr . Hence (iii). References[1] R. Ahlswede, L.H. Khachatrian, The diametric theorem in Hamming spaces—Optimal anticodes, Adv. in Appl. Math. 20(1998) 429–449.[2] C. Berge, Nombres de coloration de l’hypergraphe h-parti complet, in: Hypergraph Seminar, Columbus, Ohio, 1972, in:Lecture Notes in Math., vol. 411, Springer, Berlin, 1974, pp. 13–20.[3] C. Bey, An intersection theorem for weighted sets, Discrete Math. 235 (2001) 145–150.[4] B. Bollobás, I. Leader, An Erdo˝s–Ko–Rado theorem for signed sets, Comput. Math. Appl. 34 (1997) 9–13.[5] P. Borg, A short proof of a cross-intersection theorem of Hilton, Discrete Math. 309 (2009) 4750–4753.[6] P. Borg, Intersecting and cross-intersecting families of labeled sets, Electron. J. Combin. 15 (2008), note paper N9.[7] P. Borg, Intersecting systems of signed sets, Electron. J. Combin. 14 (2007), research paper R41.[8] D.E. Daykin, Erdo˝s–Ko–Rado from Kruskal–Katona, J. Combin. Theory Ser. A 17 (1974) 254–255.[9] M. Deza, P. Frankl, The Erdo˝s–Ko–Rado theorem—22 years later, SIAM J. Algebraic Discrete Methods 4 (1983) 419–431.[10] K. Engel, An Erdo˝s–Ko–Rado theorem for the subcubes of a cube, Combinatorica 4 (1984) 133–140.[11] K. Engel, P. Frankl, An Erdo˝s–Ko–Rado theorem for integer sequences of given rank, European J. Combin. 7 (1986) 215–220.[12] P.L. Erdo˝s, U. Faigle, W. Kern, A group-theoretic setting for some intersecting Sperner families, Combin. Probab. Comput. 1(1992) 323–334.[13] P. Erdo˝s, C. Ko, R. Rado, Intersection theorems for systems of ﬁnite sets, Quart. J. Math. Oxford (2) 12 (1961) 313–320.[14] P. Erdo˝s, J. Schönheim, On the set of non pairwise coprime divisors of a number, in: Combinatorial Theory and ItsApplications, Balatonfüred, Hungary, 1969, in: Colloq. Math. Soc. János Bolyai, vol. 4, North-Holland, Amsterdam, 1970,pp. 369–376.[15] P. Frankl, The shifting technique in extremal set theory, in: C. Whitehead (Ed.), Combinatorial Surveys, Cambridge Univ.Press, London–New York, 1987, pp. 81–110.[16] P. Frankl, Z. Füredi, The Erdo˝s–Ko–Rado theorem for integer sequences, SIAM J. Algebraic Discrete Methods 1 (1980) 376–381.[17] P. Frankl, N. Tokushige, The Erdo˝s–Ko–Rado theorem for integer sequences, Combinatorica 19 (1999) 55–63.[18] H.-D.O.F. Gronau, More on the Erdo˝s–Ko–Rado theorem for integer sequences, J. Combin. Theory Ser. A 35 (1983) 279–288.[19] A.J.W. Hilton, An intersection theorem for a collection of families of subsets of a ﬁnite set, J. Lond. Math. Soc. (2) 15 (1977)369–376.[20] A.J.W. Hilton, E.C. Milner, Some intersection theorems for systems of ﬁnite sets, Quart. J. Math. Oxford (2) 18 (1967) 369–384.[21] G.O.H. Katona, A simple proof of the Erdo˝s–Chao Ko–Rado theorem, J. Combin. Theory Ser. B 13 (1972) 183–184.[22] G.O.H. Katona, A theorem of ﬁnite sets, in: Theory of Graphs, in: Proc. Colloq. Tihany, Akadémiai Kiadó, 1968, pp. 187–207.[23] J.B. Kruskal, The number of simplices in a complex, in: Mathematical Optimization Techniques, University of CaliforniaPress, Berkeley, CA, 1963, pp. 251–278.[24] M.L. Livingston, An ordered version of the Erdo˝s–Ko–Rado Theorem, J. Combin. Theory Ser. A 26 (1979) 162–165.[25] M. Matsumoto, N. Tokushige, The exact bound in the Erdo˝s–Ko–Rado theorem for cross-intersecting families, J. Combin.Theory Ser. A 52 (1989) 90–97.[26] J.-C. Meyer, Quelques problèmes concernant les cliques des hypergraphes k-complets et q-parti h-complets, in: HypergraphSeminar, Columbus, Ohio, 1972, in: Lecture Notes in Math., vol. 411, Springer, Berlin, 1974, pp. 127–139.[27] A. Moon, An analogue of the Erdo˝s–Ko–Rado theorem for the Hamming schemes H(n,q), J. Combin. Theory Ser. A 32(1982) 386–390.[28] L. Pyber, A new generalization of the Erdo˝s–Ko–Rado theorem, J. Combin. Theory Ser. A 43 (1986) 85–90.

AbstractA k-signed r-set on [n]={1,…,n} is an ordered pair (A,f), where A is an r-subset of [n] and f is a function from A to [k]. Families A1,…,Ap are said to be cross-intersecting if any set in any family Ai intersects any set in any other family Aj. Hilton proved a sharp bound for the sum of sizes of cross-intersecting families of r-subsets of [n]. Our aim is to generalise Hilton's bound to one for families of k-signed r-sets on [n]. The main tool developed is an extension of Katona's cyclic permutation argument

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Multiple cross-intersecting families of signed sets 

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