Topology and its Applications 156 (2009) 13711373

Contents lists available at ScienceDirect

O

A

Un

a

ArReRe

Mprse

KeFinChSuTo

1.

soofofprn

wbe

exth

2.

ch

01doTopology and its Applications

www.elsevier.com/locate/topol

n a combinatorial property of choice functions of nite sets

arno Hohti

iversity of Helsinki, Department of Mathematics and Statistics. PO Box 68 (Gustaf Hllstrmin katu 2b), FI-00014 University of Helsinki, Finland

r t i c l e i n f o a b s t r a c t

ticle history:ceived 28 February 2007ceived in revised form 17 July 2008

SC:imary 03E20condary 05C20, 05C65

ywords:ite setoice functionbseturnament

We consider choice functions [X]k X , where X is a nite set and [X]k denotes the setof all k-subsets of X . We dene a property of domination for such maps generalizingthe classical case k = 2 (tournaments) and prove the existence of a dominating elementgeneralizing the existence of a 2-root (king) in the classical case.

2008 Elsevier B.V. All rights reserved.

Introduction

The problem considered in this paper has a natural geometric origin. Let A be a collection of subsets of the plane, or ofme Rn , with the property that for any three members A1, A2, A3 A, there are two Ai, A j such that the -neighbourhoodAi A j contains the third member. Is there a universal pair Ai, A j of elements of A such that the k-neighbourhoodAi A j contains all elements of A, where k is some constant independent of A? A combinatorial formulation of thisoblem reads as follows. Given a function : [n]3 [n] such that (A) A for all A [n]3 (i.e., is a choice function),d a,b [n] such that for every c [n] there are a1,b1, . . . ,ak(3),bk(3) with

({ai,bi,ai+1}

) = ai+1,({ai,bi,bi+1}

) = bi+1,here a1 = a, b1 = b and ak(3) = c. We say that the pair {a,b} k(3)-dominates the element c. This problem can immediatelygeneralized by replacing 3 in the above formulation by an arbitrary positive integer m.In case m = 2, the problem is reduced to complete directed graphs, i.e., tournaments. While tournaments have beentended to uniform hypergraphs by considering hypergraphs with ordered k-edges [2] (hypertournaments), it seems thate other possible extension to choice functions has remained unstudied in this respect.

A notion of domination

Let us now dene a notion of domination connected with our problem. Let X be a nite set and let : [X]3 X be aoice function. If {a,b, c} = c, or if c {a,b}, then we say that the pair {a,b} dominates c, and we write

c {a,b}

E-mail address: hohti@cc.helsinki..

66-8641/$ see front matter 2008 Elsevier B.V. All rights reserved.i:10.1016/j.topol.2008.12.022

1372 A. Hohti / Topology and its Applications 156 (2009) 13711373

(ow

an{an

(mof

3.

foeqsube

ThA

Pri.ech

Le

Puan

Aswisamco

TheaX

th

Siin

Ob

dothbyob

Qur simply c {a,b} if the map is understood from the context). In particular, each pair dominates its elements. Similarly,e dene that {a,b} 2-dominates c with respect to , written c 2 {a,b} if there are a,b X such that a,b {a,b}d c {a,b}. In general, if k > 1, we say that {a,b} k-dominates c, written c k {a,b}, if there are a,b X such that,b} (k 1)-dominates c and {a,b} dominates both a and b . Our result (for triples) gets the following form: if X is aite set and : [X]3 X is a choice function, then there is a pair {a,b} which 3-dominates each element of X .We can obviously generalize the notion of domination for choice functions : [X]m X , where m > 1. We say that an 1)-set A [X]m1 dominates b X , if (A {b}) = b or if b A. Thus, in particular each (m 1)-set dominates eachits elements. As in the preceding paragraph, we dene that A [X]m1 k-dominates b if there is A [X]m1 such that

a k1 A for all a A and a A.The main result

Let us now formulate and prove the main result of this section. It should be noted that a degenerate form of the result,rmulated for choice functions : [X]2 X , has a particularly simple proof. We also note that such choice functions areuivalent to directed complete nite graphs (tournaments) or to reexive and connected relations R on nite sets X : forch a relation R there is x X such that yR2x for all y X (every tournament has a 2-king [1]). The proof presentedlow extends a proof for this degenerate case.

eorem 3.1. There is for each n > 1 a number k = k(n) such that for any nite set X and any choice function : [X]n X there is [X]n1 with b k A for all b X. We can take k(n) = n.

oof. We shall proceed by induction on n. The claim is true for n = 2. Suppose that n > 2 and the claim is valid for n 1,., there is k(n 1) such that the claim of the theorem is true for k(n 1). Let X be a nite set and let : [X]n X be aoice function. Obviously we can assume that |X | n + 1. Let {a1, . . . ,an1} [X]n1 be arbitrary, and dene

E1 ={b X: b {a1, . . . ,an1}

}.

t F1 = X \ E1. Write A1 = {a1, . . . ,an1}. If F1 = , choose an F1 and deneE2 =

{b F1: i [n 1]

(b (A1 \ {ai}

) {an})}

.

t F2 = F1 \ E2. In general, assuming that Fi = has been dened, let ai+n1 Fi be arbitrary, let Ai = {a1, . . . ,ai+n2}d let

Ei+1 ={b Fi: S [Ai]n2

(b S {ai+n1}

)}.

before, let Fi+1 = Fi \ Ei+1. Since X is nite, after nitely many steps we arrive at m such that Em+1 = Fm . For each x Xe dene a mapping x : [X \{x}]n1 X by setting x(A) = (A{x}). We claim that the restriction of am+n1 to [Am]n1a choice function. Indeed, let A [Am]n1. We have to show that am+n1(A) A. But if (A {am+n1}) = am+n1, then+n1 would be dominated by an (n 1)-set in Am , which would imply am+n1 E1 E2 Em , which would be antradiction, because

am+n1 Fm = Fm1 \ Em X \ (E1 E2 Em).erefore, by the inductive hypothesis, there is k(n 1) and an (n 2)-set A [Am](n2) such that A k(n 1)-dominatesch element of Am with respect to am+n1 . We claim that A

= (A {am+n1}) (k(n 1) + 1)-dominates every element ofwith respect to . Too see this, let c X . First assume that c / Em+1. Because

X = E1 E2 Em+1,ere is l

A. Hohti / Topology and its Applications 156 (2009) 13711373 1373

4. Applications

To obtain the geometric application mentioned in Introduction, let A be a nite family of subsets of R2 with the propertyconsidered above. Dene a mapping : [A]3 A by choosing for each triple {A1, A2, A3} an element ({A1, A2, A3})contained in the -neighbourhood of the two other elements. By the theorem there is a 3-dominating pair {Ai, A j} [A]2.The reader will observe that the relation of being contained in an -neighbourhood is compatible with the relation ofdomination in the sense that for each p, Ak p {Ai, A j} implies Ak Bp(Ai A j). Thus, the 3-neighbourhood of Ai A jcontains all elements of A.

To continue with another geometric example, let X be a nite subset of R2 with the property that for any three pointsx, y, z X , one is contained in the vertical -stripe determined by the other two points. (The vertical -stripe determinedby p,q R2 is the union of all those vertical translates of the segment Lpq joining p and q which are -close to Lpq .) Againwe note that the relation of being contained in the vertical -stripe is compatible with the relation of domination. Thus,there is a pair x, y X such that the 3-stripe determined by x, y contains X .

References

[1] H. Landau, On dominance relations and the structure of animal societies, III: The condition for score structure, Bull. Math. Biophys. 15 (2) (1953)143148.

[2] R. Assous, Enchainabilite et seuil de monomorphie des tournois n-aires, Discrete Math. 62 (1986) 119125.

On a combinatorial property of choice functions of finite setsIntroductionA notion of dominationThe main resultApplicationsReferences