wi th a Conti nuum
o f Inte rsecti on Poi nts
3
Jean-Jacques Her ings z
and Dolf Talman x
3
Thisres earch ispartofth eVF-program\Comp e titionandCo op e ration".
z
P.J.J.Herings, Department ofEc onomet ric s and CentER,Tilbur gUnivers ity, P.O.B ox 90153, 5000
LE Tilbur g, TheNe therlands. Theauthor is nancially sup p or tedby the Co op e ration Centre Tilbur g
andEin dhovenU niver sit ie s,TheNethe rlan ds.
x
In all existing intersectio n theo rems conditions a re g iven under which a certain subs et of
a co llection o f sets ha s a non -empty intersection. In th is paper co nditions are formula ted
underwhichtheintersectio nisacontinuumofp ointssatisfyings omeinterestin gtopolo gica l
properties. Inthissen setheintersection theoremsconsideredinthispaperb elongtoanew
cla ss. Theinters ectio ntheoremsa refo rmulatedon theunitcub eanditisshowntha tboth
the vector o f zeroes a nd the vector of o nes lie inthe sa me comp o nent of the intersectio n.
This is interesting fo r some speci c applications . The theorems give a genera liza tio n of
the well- know nlemma sofKna ster,Kura towski,a nd Mazurkiewicz,ofSperner,ofSha pley,
and o f Ichiischi. Moreover theresults can beused to s harp en the usua l formula tion o f the
1 Introduct io n
In intersection theo rems co nditions are givenun der which a certain su bset of a collection
of sets ha s a no n-empty intersection . Well-known intersection theo rems on th e unit
sim-plex are g iveninKnas ter,Kura towski, and Mazurkiewicz(1 92 9) (KKML emma),S p erner
(19 28 ) and Sca rf (19 67 ) (Sp erner L emma),Sha pley (19 73 ) (KKMS Lemma ), Gale (19 84)
(Ga le L emma),and Ichiischi (198 8)(IchiischiLemma). Inters ectio ntheo rems ca nb eus ed
to provetheexistenceo fsolutionsto mathema ticalpro grammingproblems,eco nomicequ
i-librium existence p roblems,and so lutions to g ametheoreticpro blems. T he KKMLemma
and the Sp erner Lemma ca n be u sed to prove Brouwer's xed point theo rem, an d also
to s how the exis tence of an equilib riumin a n exch ange eco nomy with or withou t p ro
duc-tion. Both the KKMS L emma a nd the Ichiishi L emma are very useful when s howing the
non -emptiness o f the core of acoop erative g ame, see S hapley (1 973 ), Ichiishi(1 988 ), and
Sha pley an d Vohra (19 91). In Ga le (1 98 4) a n intersectio n theorem is us ed to show the
existence ofa nequilibriuminan econo my with indivisibleco mmodities. In orderto prove
the exis tence of a Na sh equilibrium in a non-coopera tive ga me it is useful to formula te
intersection theorems on the cub e o r even more g eneral the s implo tope, see for example
van der La an a nd Ta lman (199 3).
It ispossibleto g enera lizethea b ovementioned intersectiontheoremsand to formula te
intersectiontheo remso nap olytope. Inforexa mp leIchiishian dId zik(199 1)a nintersection
theorem o n a p o lyto p e is derived which generalizes b o th th e KKM Lemma a nd the Gale
Lemma . In va nder L aa n, Talma n, and Yan g (1 994 ) a generaltheorem on the p olytope is
sta ted. Mos t o fthe res ultssta ted a b ovecan b e derivedfromthis theorem. Moreover, this
theoremmakesitposs ibleto formu la tea nalogs ofthe KKM, Sp erner,KKMS,and Ichiishi
Lemma on the p o lytope.
Ina llth eintersectiontheoremssta teda b oveco nditionsa regivenunderwh ichacerta in
subs et o fa co llection o f setshas a non-empty intersectio n. The sets inthe collection fo rm
a clo sed covering of a simplex, a simplotope, or a p o lytope. In this paper intersection
theorems arefo rmulatedwithaco ntinuumo finters ectio np o ints. Hence theseintersection
theorems b elong to a new clas s. L et I
n
denote the s et of integ ers f1;:::;n g for s ome
natu ral numb er n: Let Q n = fq 2 IR n j 0 q j 1; 8j 2 I n
g denote the n-dimensiona l
unit cub e. Conditio ns a re given o n a collection of s ubsets covering th e cube such that
certa in s ubsets of this collectio n have an intersection cons isting o f a continuum o f p oints.
Moreover, th e inters ectio n has so me interesting topological properties. It w ill be shown
that it ha s a component, i.e., a maximally connected subset, containing both the vertex
b eing thevectoro f zeroesand thevertex b eingthe vecto rof ones. Fina lly,theintersection
theorems fo rmula ted in th is paper generalize the KKM, S p erner, KKMS, and Ichiischi
the Sp erner Lemma on the cube. Th ere is a close rela tio nship b etween the intersection
theorems of this paper a nd the equilibrium existence pro blem in econo mies with p rice
rigid ities as introduced in Dreze (1 975 ). The inters ectio n theorems o f this paper ca n be
used to show the existence o f a co ntinuumofequilib ria in thes e econo mies.
In Sectio n 2 s ome ma thema tical prelimina ries are g iven. A co rres p on dence s atis fying
the con ditions of a tota l exces s demand correspondence in an economy with p rice rigid
i-ties is introduced a nd using Brow der's xed p o int theoremso me interesting properties of
this corresp o ndence are derived . These results are used in Section 3 to fo rmula te s evera l
intersection theo remson the cub e,a mo ng whichthe ana lo gs o f the KKML emmaa nd the
Sperner Lemma . Con ditions a reg iven u nderw hich acerta in sub seto f a collection o f
sub-sets covering the cube ha s a non-empty intersection, containing , a mo ng other p o ints, the
vector o f zeroes and the vecto r of o nes . Mo reover, th e vector of zeroes a nd the vector of
ones lieinthe same co mponentof theinters ection. Using on eo f the intersection theo rems
of Sectio n 3 it is p os sible to s treng then the usua l fo rmulation o f the Sp erner Lemma on
the cub e. In Section 3 also intersection theoremswith a co ntinuumof inters ectio n points
generalizing theKKMSL emmaandthe IchiischiL emmaaregiven. T heproo fsofall
inter-section theo rems o f Sectio ns3 are derivedfrom the co rres p ond ence introducedin Section
2. Therefo re this corresp o ndence un i es th ese inters ection theorems. In Sectio n 4
atten-tion is focused on s ome well-known intersectio n theo rems on th e unit simplex, where the
existence of a non- empty intersection (in g eneral no t a co ntinuum of points ) of a certa in
subs eto f acollectio nof sets covering the simplex isg uaranteed . Both theS p ernerLemma
and th e KKMLemma, which a rein some sense ea ch others dua l, will b e derived froman
intersection theoremof Section 3. It is a ls o pos sible to derive the KKMS L emma and its
dua l, the IchiischiLemma, froma notherintersectio n theoremgiven inSection 3.
2 Some Preliminaries
In th e follow ing 0 n
will denote a n n-dimensiona l vector containing only zeroes and 1 n
an
n-dimens io nal vecto rof o nes . Th e closure of a subset S of s ometo p o lo gica l sp ace w ill be
denoted by cl(S). The convex hull o f a subset S o f s omeEuclidean spa ce will b e deno ted
by co(S). Fo r the remainder of the paper it will b e usefu l to con sider a correspondence
:Q n
!IR n
sa tisfyingthe fo llowin g a ssumptio n.
Ass umption A T heco rres p on dence :Q n
!IR n
sa tis es:
1. is a corresp o ndence with a clo sed g raph satisfying (q) is non-empty and
convex for every q 2Q n
and [
q 2Q n
2. 8q2Q n
; 9z 2(q)suchtha t foreveryj 2I
n ;q j =0impliesz j 0;a nd q j =1 imp lies z j 0 ; 3. 8q2Q n ; 8z 2(q ); 9p2IR n ++ suchthat p1z =0:
Assumptio nA.1guara nteestha t the co rrespond ence isupp er semi- co ntinuous.
Assump-tion A.2 s p eci es a bounda ry co ndition fo r a nd Assumptio n A.3 is equivalent to the
conditio nth at if z 2(q) fo r so me q 2 Q n ; then z j > 0 for any j 2 I n implies z k < 0 for so me k 2 I n ; an d z j < 0 fo r any j 2 I n implies z k > 0 fo r so me k 2 I n : If n = 1; a
cor-resp o ndence satisfying As sumptio n A necessa rily equa ls the corresp o nden ce a ssociating
witheveryelementinQ 1
thes etf0g:Ap ictureofacorresp o ndence :Q 2
!IR 2
s atis fying
Assumptio n A is drawn inFig ure 1 . InFigure 1 only the z
1
-a xis ha s b een draw n,which
Figu re 1 : A corresp o ndence :Q 2
!IR 2
satisfying Assumptio nA
iss ucientto determinethezeropointso f byAssumptionA.3. For every q2Q 2
theset
(q ) inFigure1con sis tso f o neelement,except when q
1
=1o r q
2
=0 : No ticetha t the set
of zero p oints o f isgivenby fq2Q 2 jq 1 =1 or q 2 =0 g:
Models o f econo mies with price rig idities as introduced in Dreze (19 75) yield excess
dema nd co rres p on dences satisfying Assu mption A a s is show n in Herin gs (19 92). In
The-orem 2.7 interes ting properties o f the set o f zero points of a correspondence s atis fying
allprovedusingTheorem2 .7thisshows tha tthereisaclos erelations hipb etweentheequ
i-librium exis tence problem in econ omies with price rig idities a nd the class of intersection
theorems to be cons idered in this pa p er. In fact, it is poss ibleto sh ow th e existence of a
continuumof equilibria in eco nomies with price rigidities using the inters ectio n theo rems
of this paper.
The followingde nitions , whichca nbefoun dinfo r instanceArms trong(19 83 ),willbe
usefulla ter on.
De nition 2.1 (Connectedness)
A topol og ical space X is con nec ted if it isn ot the union of two non -e mpt y disjoint, cl osed
set s.
A su bset of a topolog ical s pace is con nected if it b eco mes a connected spa ce when g iven
the induced top o lo gy. Intuitively, aconnected setis aset which iso f one piece.
De nition 2.2 (Component)
The compon ent of an el eme nt x in a t opological space X is the un ion of all con nect ed
subse ts of X con tainin g x:
Itisnot dicultto showthatacomp o nentisco nnecteda nd cons idering De nitio n2.1 the
comp o nent of a n element x in a topological space X is the la rgest connecteds ubset of X
containing x: The collection of a ll comp o nentso f a top o lo gica l space X partition sX:
L et a corresp o ndence s atisfying Ass ump tion A be given. Then the set Z
o f zero
p o ints of isde n ed as fo llows, Z
=fq2Q n
j0 n
2(q )g:Theorem 2.7 states tha t there
existsa co nnected s etof zero pointsZ c
Z
containingthe twoextreme p oints0 n
and 1 n
of Q n
: In o rder to s how Th eo rem 2 .7somepreliminarywo rk shou ld be done rs t.
It will be us eful in the proof of Theorem 2.7 to extend a correspondence s atis fying
Assumptio nAsuchtha titisde nedonIR n :Foreveryq2Q n choos eanelement b z(q )2(q ) sa tisfying b z j (q) 0 if q j = 0 and b z j (q ) 0 if q j
= 1: As sumptio n A on gua rantees
that b
z(q ) can b e chos en in this way for every q 2 Q n
: For a non- empty compact set
S IR n
d e ne the correspondence 5
S : IR
n
! S a s the ortho go nal projection o n S; so
5 S (x)=fs 0 2S js 0 2argmin s2S kx0sk 2 g; 8x2IR n
:Itisnotdiculttoshowtha t5
S isa
continuo usfunctionif S isco nvex,s eefo rexa mp leMas- Colell(19 85 ). L et b :IR n !IR n be
the corresp o ndencew ithg raphinIR n
2IR n
g ivenby the setcl(f(x; b z(5 Q n (x)))jx2IR n g):
Notice tha t comp o nent j 2 I
n
of the p rojectio n function 5
Q n is given by (5 Q n(x)) j = maxf0;minfx j ;1gg; 8x2IR n :
Lemma 2.3
Let a corre spon de nce : Q n
! IR n
sat isfying Assumption A be give n. The n the
corre-spon dence b
is n on-e mpty val ued an d upper semi- con tinu ou s. For e ve ryx 2IR n ; z 2 b (x) implie s z 2 (5 Q n(x)):
Moreov er, for e ve ry x 2 IR n ; z 2 b (x) and x j <0 implie s z j 0 ; and z 2 b (x) an d x j >1 impl ies z j 0: Proof Clea rly, b
is a non -empty valu ed correspondence. It follow s from the boundedness o f the
set [
q 2Q n
(q) that the co rrespondence b
is uppersemi-co ntinuous. T hesecond sta tement
in Lemma 2.3 follows immediately us ing the de nitio n of b
a nd the upp er semi-continuity
of : L et z 2 b (x) fo r s ome x 2 IR n with x j < 0 fo r so me j 2 I n
: Then there exists a
sequen ce (x r ;z r ) r 2IN such tha t z r = b z(5 Q n(x r )) and lim r !1 (x r ;z r ) =(x;z): Since x j <0 there exists r 0 2 IN s uch th at r r 0 imp lies x r j < 0 ; hence (5 Q n (x r )) j = 0; a nd z r j 0 : Co nsequently z j
0: It can b e show n in a similar way that z 2 b (x) an d x j > 1 imply z j 0 : Q.E.D.
Fina lly, de ne the co rrespond ence : IR n ! IR n by (x) = (x); 8x 2 Q n ; and (x) = co( b (x)); 8x2IR n nQ n
:The following lemma,shown inTo dd (197 6, p. 56 ,Theorem1.4 ),
is usefulto d erive so me pro p erties of :
Lemma 2.4
Let 8 1
:S !T and 8 2
: S!T be upper se mi- contin uouscorresponden ces,where S IR k
and T IR m
:
1. Let 8 : S ! T be de ned by 8(s) = 8 1
(s)[ 8 2
(s); 8s 2 S: The n 8 is uppe r
semi- con tinuous.
2. Let S be c losed and le t e S IR k be such that S e S: De n e 8 : e S ! T by 8(s) = 8 1 (s); 8s2S; and 8 (s)=;; 8s 2 e
SnS: Then 8 is uppe r semi-con tinuous.
3. De ne 8:S !T b y 8(s)=co (8 1
(s)); 8s2S : The n 8 is upper se mi-contin uous.
Lemma 2.5
Let be g ive n a correspon den ce : Q n
!IR n
satisfyin g Assumption A. Then is an uppe r
semi- con tinuou s corre spon de nce , and (q ) is n on-e mpty and conve x for ev ery q 2Q n : For ev ery x2IR n ; z 2(x)implies z2(5 Q
n(x)):Moreove r, for e very x2IR n ; z 2(x) and x j <0 impl ies z j 0; and z 2(x) an d x j >1 impl ies z j 0: Proof
Theco rres p o ndenceisuppersemi-continu oussincetheextensiontoIR n
o fth eupper
semi-continuo uscorrespondence de nedontheclos edsetQ n
;obta inedbya ssigning theempty
n
extension of with the co rres p o nden ce b
is upper semi-continuo us by Lemma 2.4.1, and
a co nvexi ed upper semi-continuo us correspondence is upp er semi- co ntinuous by Lemma
2.4.3. C learly, (q ) is non-empty a nd convex for every q 2 Q n
: T he other sta tements in
Lemma 2.5 follow immediately fro mL emma 2.3. Q.E.D.
The fo llowing lemmais asp ecia l ca se o fTh eo rem 3inMa s-C olell(19 74) in the s ens e that
in the theoremco nvex- valued co rres p on dences a re co nsidered,while Mas -Co lell treats the
moregeneralcaseo fcontractiblevaluedcorresp o ndences. Itisag en era liza tio nofTheorem
2 in Browder(1 960 , p. 18 6) w herethe cas ewith co ntinuous function sis considered .
Lemma 2.6
Let S IR n
bea n on- empty,compact, conv ex set ,let the correspon den ce9:S2[0;1]!S
be uppe r semi-con tinuous, and let 9(s;t) be non -empty an d conv ex , 8(s;t) 2 S 2[0;1]:
Then the se t F 9 = f(s;t) 2 S 2[0;1] j s 2 9(s;t)g contain s a compone nt F c 9 such that (S2f0g)\F c 9 6=; an d (S2f1g)\F c 9 6=;:
Theorem2.7 will b eproved as a na pplicatio no f Lemma 2.6.
Theorem 2.7
Let a correspon de nce : Q n
! IR n
satisfying Assumption A be g ive n. Then the se t Z
contain s a compon en t Z c contain ing 0 n and 1 n : Proof
Let Z b e a compact, convex s et co nta ining [
q 2Q n (q ): De ne the s et R = fr 2 IR n j P n j=1 r j = 0; r j 01; 8j 2 I n
g: Clea rly, the set R is non- empty, compact, and convex.
De nethe corresp o ndence :Z !R by
(z)=fr2R jr1z r1z; 8r2R g; 8z 2Z :
Using the ma ximum theo rem, see fo r example Hildenbrand (19 74, p. 30), it fo llows
im-mediatelyth at the corresp o ndence is upp er s emi- co ntinuous . D e n ethe correspondence
:R2[10n;2]!Z by
(r;t)=(r+t1 n
); 8(r;t)2R2[10n;2 ]:
Then is u pp ersemi-continuo us becau se of the upp er semi- co ntinuity of a nd the co
nti-nuity of the functio n assigning r+t1 n
to (r;t)2R2[10n;2]:De nethe correspondence
9:Z 2R2[10n;2 ]!Z2R by
9(z;r;t)= (r;t)2 (z); 8(z;r;t)2Z 2R2[10n;2]:
non-Using theconvexity o fR and thelin ea rityinr ofr1z itfo llows tha tthe set (z)isco nvex
for every z 2Z: Usinga straightforward g enera liza tio nof Lemma 2.6 itfollows that there
exists a comp o nent F c 9 of the s et F 9 = f(z ;r;t) 2Z 2R2[10n ;2 ] j (z;r ) 2 9(z;r;t)g suchthatF c 9 \(Z2R2f10n g)6=;a ndF c 9 \(Z2R2f2g)6=;:Clearly,(z 3 ;r 3 ;t 3 )2F c 9 implies (z 3 ;r 3 )29(z 3 ;r 3 ;t 3 )=(r 3 +t 3 1 n )2 (z 3 ): Suppose max j2I n z 3 j > 0 : Since z 3 2 (5 Q n(r 3 +t 3 1 n
)) there exists by Ass ump tion A.3
so me p 3 2 IR n ++ su ch th at p 3 1z 3
= 0 and therefo re there exists a k 2 I
n
with z 3
k
< 0 : It
is ea sily veri ed that r 2 (z) for any z 2 Z with z
j > z j 0 ; j ;j 0 2 I n ; implies r j 0 = 01 : Co nsequently r 3 k =01 : Ift 3 <1; then r 3 k +t 3 <0 a nd s incez 3 2(r 3 +t 3 1 n )th is implies z 3 k
0 ; a contradiction. Co nsider th e cas e t 3
1: By de nition of th ere exists j 0 2 I n such th at z 3 j 0 =max j2I n z 3 j >0 and r 3 j 0 >0: Hence r 3 j 0 +t 3 >1 and since z 3 2(r 3 +t 3 1 n ) this implies z 3 j 0
0; a co ntra dictio n. Co nsequently, we have that max
j2I n z 3 j 0: Since p 3 2IR n + + and p 3 1z 3 =0th is impliesz 3 =0 n :
Co nsid ertheco ntinuousfunctio nf :Z2R2[10n;2]!Q n de nedbyf(z;r;t)=5 Q n(r+ t1 n
); 8(z;r;t)2Z2R2[10n;2 ]: Du e to the facttha t the ima geo f aconnectedsetby a
continuo usfunctionisco nnecteditho ldsthatf(F c 9 )Q n iscon nected. Ifq 3 2f(F c 9 );then q 3 =5 Q n (r 3 +t 3 1 n )fo r s ome(z 3 ;r 3 ;t 3 )2F c 9 : Hence 0 n =z 3 2(r 3 +t 3 1 n )(5 Q n(r 3 + t 3 1 n ))= (q 3 ): Co nsequently, f(F c 9 ) Z
: Next, consider the p o ints (0 n ;r 1 ;10n) 2 F c 9 and (0 n ;r 2 ;2)2F c 9 : By de nition, f(0 n ;r 1 ;10n)=5 Q n(r 1 +(10n)1 n ):Sincer 1 2 R it
holds for every j 2I
n th at r 1 j n01 ; an d con sequently 5 Q n(r 1 +(10n)1 n )=0 n : Since r 2 2 R it holds fo r every j 2 I n that r 2 j
01 ; and cons equently 5
Q n(r 2 +211 n ) = 1 n : Hence 0 n ;1 n 2 f(F c 9
): Therefore, the set Z
conta ins a component Z c containing both 0 n and 1 n : Q.E.D.
Theorem 2.7 will turn out to b e a very useful tool for proving a numb er of intersection
theoremsinthenexts ectio n. Sin ceTheorem2.7isu sedinthepro o fofa lltheseintersection
theorems , T heo rem2 .7 can be seen as a unifying theorem.
3 Intersect io n T heorems with a Cont inuum of
Inter-section Po ints
In the following, for q 2 Q n , I 0 (q) = fj 2 I n j q j = 0g; I 1 (q ) = fj 2 I n j q j = 1g; s 0 (q )
denotes the number of elements in the set I 0
(q ), s 1
(q ) d eno tes the numb er of elements in
the setI 1 (q); and s(q)= s 0 (q)+s 1
(q ): The j- th unit vector in IR n
w ill b e denoted by e j
:
In Theorem3 .1itw illbea ssumedthat if anindexj 2I
n
istaken , thenj+1=1if j =n;
Theorem 3.1 Let C 1 ;:::;C n be cl osed subse ts of Q n satisfying [ n j=1 C j = Q n : Moreove r, if for q 2 Q n ; q j =0 or q j+ 1 =1; then q2C j
: The n the re ex ists a con nec ted se t S in Q n such that 0 n ; 1 n 2S an d S \ n j=1 C j : Proof
De ne for every q 2 Q n
the s et J(q ) =fj 2 I
n
jq 2 C j
g: Let t(q) den ote the number of
elements inthe s et J(q ): De nethe correspondence :Q n !IR n by (q )=co e j 0 1 n 1 n j 2J(q ) :
Firs t we verify that sa tis es Assumptio n A. From L emma 2 .4 if follow s immediately,
usingthe closedness ofthe s etsC j
; 8j 2I
n
; tha t isu pp ers emi-co ntinuous . C learly, (q )
is non- empty a nd co nvex fo r every q2 Q n
a nd [
q 2Q
n(q ) is b o unded. Hence Ass umption
A.1 is satis ed by : If,forso mej 2I n ; q j =0 ;thenq 2C j an dhencee j 0 1 n 1 n
2(q):If, forsomej 2I
n ; q j = 1;then q 2C j0 1 and hencee j01 0 1 n 1 n
2(q ):Three cases have to b e considered.
1. If q = 1 n
; then con sider z 2 (q) with z = P j2I n 1 n (e j 0 1 n 1 n ) = 0 n : So it follows that z j 0for every j 2I n : 2. If 8j 2 I n ; 0 q j < 1 ; then de n e z = P j2J(q ) 1 t(q ) (e j 0 1 n 1 n
): It follows immed iately
that z 2(q) an d q j =0 implies z j = 1 t(q ) 0 1 n 0: 3. If 9j 1 2I n with q j 1 =1 and 9j 2 2I n with q j 2 <1 ; th en choose so me j 0 2I n s atis fying q j 0 = 1 a nd q j 0 01 6= 1: De ne z = P j2I 0 (q ) 1 n e j +(10 s 0 (q ) n )e j 0 01 0 1 n 1 n
: Then it is eas ily
veri ed that z 2 (q ): Moreover, q
j = 0 implies z j 1 n 0 1 n = 0 a nd q j = 1 implies z j =0 1 n <0 :
The Ca ses 1 ,2 , and 3 showthat sa tis es Ass ump tion A.2.
Since 8q2Q n
; 8z 2(q ); 1 n
1z =0;also As sumptio n A.3is satis edby :
If 0 n 2 (q 3 ) for some q 3 2 Q n ; or equiva lently q 3 2Z
; then obviou sly q 3 2C j for every j 2 I n ; so q 3 2 \ j2In C j
: By Th eo rem 2.7 there is a co nnected set o f p o ints Z c Z sa tisfying 0 n 2Z c and 1 n 2Z c
:C onsequently, S can b eta kenequ al to Z c
: Q.E.D.
In Figure 2, T heo rem 3.1 is illus tra ted for the case n = 2 : The set C 1
\C 2
cons ists of
four comp o nents, a nd one o f them co ntain s b oth the p o ints (0;0 ) a nd (1;1): It sho uld
b e mentio ned tha t Fig ure 2 illustrates a rather nice ca se in th e sense tha t the sets C 1
and C 2
have afa irly easy structure. Besides the bounda ryconditio na nd the requirement
that these two sets cover Q 2
; the o nly requ irement ma de is tha t the s ets C 1
and C 2
a re
clo sed. Henceingeneralmuchmore co mplicatedsituations mig hta rise. Thea b overema rk
is true for a ll illustra tio ns in the sequel. In S ectio n 4 it w ill b e shown that Theorem 3.1
Figure 2 : Illustration o f Theorem 3.1, case n=2:
It s hould b e no ticed tha t it is possible to repla ce the bounda ry conditio n q
j = 0 or q j+1 = 1 implies q 2 C j
by the mo re general co ndition tha t there exis t permutations
1 = ( 1 1 ;:::; 1 n ) and 2 = ( 2 1 ;:::; 2 n ) o f I n
su ch that for every j 2 I
n ; 1 j 6= 2 j ; and q 1 j =0 or q 2 j =1 implies q 2 C j
: Th eo rem 3 .1 co rres p on ds to the choice 1
= (1;:::;n)
and 2
= (2 ;:::;n;1): Fig ure 3 gives a n easy counterexample fo r the ca se in w hich the
conditio n 1 j 6= 2 j ; 8j 2 I n
; is not sa tis ed. Fig ure 3 corresp o nds to the ca se 1 = (2;1) and 2 =(2;1 ): The s et C 1 \C 2
co nsis ts oftwo comp onents,o ne containin gth e p o int 0 n
and the o ther o ne the point 1 n
: The follow ing theo rem generalizes b o th Theorem3 .1 and
the more general speci ca tio n with the twopermutations 1 and 2 : Theorem 3.2 Let C 1 ;:::;C n be c losed subset s of Q n satisfying [ n j=1 C j = Q n
: Moreove r, for any q 2
Q n ; q j =0 impl ie s q 2C j ; and q j =1 b ut q 6=1 n impl ies q 2C k for some k 2I n nI 1 (q):
Then the re e xists a conn ected se t S in Q n suc h that 0 n ;1 n 2S an d S \ n j=1 C j : Proof
De ne for every q 2 Q n
the set J(q) =fj 2 I
n
j q 2 C j
g a nd de ne the correspondence
:Q n !IR n by (q )=co e j 0 1 n 1 n j 2J(q ) :
Simila r to th e proof of Theorem3 .1 it can be shown that is an upper semi-co ntinuous
corresp o ndence satisfying (q) is non- empty and convex fo r every q 2 Q n
an d [
q 2Q n
Figure 3 : Co unterexa mp le,ca sen =2:
is b o unded. Moreover, 8q 2 Q n
; 8z 2 (q); 1 n
1z =0: So Assumptio ns A.1 a nd A.3 a re
sa tis ed by : Assumptio nA.2 remains to be veri ed. The cas ewith q =1 n
is considered
rst. For every 0 < " 1 a nd for every j 2 I
n it holds that 1 n 0 "e j 2 C j using
the assumptions o f T heo rem 3.2. Since C j
is clo sed fo r every j 2 I
n
; this implies that
1 n 2 \ n j=1 C j a nd hen ce 0 n 2 (1 n
): Next consider the ca se with q 2 Q n n f1 n g and I 0 (q)[I 1 (q )6=;: Letk 2I n nI 1 (q)b e such tha t q2C k : De ne z = X j2I 0 (q ) 1 s(q) e j + s 1 (q) s(q) e k 0 1 n 1 n : Obvious ly, z 2 (q): If q j = 0 ; then z j 1 s(q ) 0 1 n 0 : If q j = 1 ; then z j = 0 1 n < 0 :
Hence As sumptio n A.2 is satis ed by : By Theo rem 2.7 there exists a con nected set Z c containing b o th0 n a nd 1 n :Ifq 3 2Z c ;then0 n 2(q 3
)anditiseas ilys eenthat thisimplies
q 3 2\ n j= 1 C j :So the setZ c
satis esa ll the requirements imposed o n the setS : Q.E.D.
For the ca se n =2 Theorems 3.1 and 3 .2 a reequiva lent. For the ca se n 3 T heo rem3.2
is clearlymore general. Thereforeit isalso p os sib leto derivethe Sperner L emma directly
from Theorem 3.2. By s ymmetry cons iderations the fo llowing du al theo rem fo llows a s a
corollary to T heo rem3 .2 .
Theorem 3.3 Let C 1 ;:::;C n be cl osedsubse ts of Q n satisfying[ n C j =Q n :Moreover, for an y q2Q n ,
q j =1 impl ies q2C j ; an d q j =0 b ut q6=0 n implies q2C k forsome k 2I n nI 0 (q ): Then
the re e xists a conn ected se t S such that0 n ;1 n 2S and S \ n j= 1 C j :
It will b e shown in S ectio n 4 that the well-known KKML emma of Knaster, Kuratows ki,
and Ma zu rkiew icz (192 9) fo llows a lmos t immediately fro m T heo rem 3.3. Since T heo rems
3.2 a nd 3.3 are completely symmetric it sho uld be clea r that the KKM L emma ca n also
b e easily derived from Theorem 3.2. Similarly, the Sp erner Lemma ca n b e derived fro m
Theorem3.3.
So fa rintersectio nth eo remshaveb eencons ideredw hereastatementismadeabout the
intersection o f a ll the sets covering Q n
: In for exa mp le th e KKMS L emma or the Ichiishi
Lemma (see Shap ley (197 3) a nd Ichiis hi (1 98 8), respectively) a statement is made about
the inters ectio n of sets in certa in subs ets o f the co llection of sets covering Q n
: Theorem
3.4 is also an intersection theo rem in this spirit. Moreover, unlike Theorems 3 .1 -3.3 it is
completely symmetric with resp ect to the as sumptio ns ma de o n th e sets in the cover of
Q n : Theorem 3.4 LetC 1 ;:::;C n an dD 1 ;:::;D n bec losedsubse tsofQ n satisfyin g([ n j= 1 C j ) [ ([ n j=1 D j )=Q n :
Moreov er, for e ve ry q 2Q n ; q j =0 impl ies q2C j and q j =1 implie s q 2D j : Then t he re
ex ists a conn ected set S su ch that 0 n ;1 n 2 S and q 3 2 S impl ies q 3 2 C j \D j for some j 2I n ; orq 3 2\ n j= 1 C j ; or q 3 2\ n j= 1 D j : Proof De neJ 0 (q)=fj 2I n jq2C j g an d J 1 (q)=fj 2I n jq 2D j g: No ticethat I 0 (q )J 0 (q ) and I 1 (q )J 1
(q ): De nethe correspondence :Q n !IR n by (q )=co e j 0 1 n 1 n j 2J 0 (q ) [ 1 n 1 n 0e j j 2J 1 (q ) :
Then it fo llows immed ia tely from L emma 2.4, usin g the closedness of the sets C j and D j ; 8j 2 I n
; that is upp er semi-continuou s. Mo reover, (q) is non-empty and co nvex
for every q 2 Q n
: Clea rly [
q2Q n
(q) is bounded. Hence As sumptio n A.1 is sa tis ed by :
Since 8q2C n
; 8z 2(q ); 1 n
1z =0;As sumptio nA.3 is also satis edby :
Co nsid er a n elementq 2Q n sa tisfying I 0 (q )[I 1 (q)6=;: De ne z 2(q ) by z = X j2I 0 (q ) 1 s(q) (e j 0 1 n 1 n )+ X j2I 1 (q ) 1 s(q ) ( 1 n 1 n 0e j ): Ifq j =0; thenz j = 1 s(q ) 0 s 0 (q ) s(q )n + s 1 (q ) s(q )n n0s 0 (q ) s(q )n
0: Similarly itca nb e shown tha t q
j =1 implies z j s 1 (q )0n s (q )n
0 : Hence satis es Assumption A.2.
By Theorem 2.7 there is a co nnected set o f p o ints Z c sa tisfying 0 n ;1 n 2 Z c a nd q 3 2 Z c
implies 0 n 2(q 3 ):L et 0 n b e a n element of (q 3 ) for some q 3 2Q n
:Then there exis ts, for
everyj 2I n ; j 0 a nd j 0 suchthat n X j=1 j (e j 0 1 n 1 n )+ n X j=1 j ( 1 n 1 n 0e j )=0 n ; where j =0ifq62C j a nd j =0ifq 62D j ,a nd P n j= 1 j + P n j=1 j =1:De ne= P n j= 1 j and= P n j=1 j :Foreveryj 2I n itho ldsthat j 0 1 n = j 0 1 n :Hence j 0 j = 1 n (0);
b eing indep endent of j :T hreeposs ibilities ca noccur.
If> ;then8j 2I n ; j 0 j >0;h ence8j 2I n ; j
>0;and cons equentlyq 3 2\ n j=1 C j : If = ; then 8j 2 I n ; j = j : Since fo r some k 2 I n ; k > 0 or k >0 ; it ho lds that q 3 2C k \D k forso mek 2I n : If <; then q 3 2\ n j= 1 D j : Q.E.D.
Theorem 3 .4 is illus trated in Figure 4 for the case n =2: It is eas ily veri ed tha t the set
Figure 4 : Illustration o f Theorem 3.4, case n=2:
(C 1 \D 1 )[(C 2 \D 2 )[(C 1 \C 2 )[(D 1 \D 2
) co nsists o f two co mp onents, on eo f them
containing the points(0;0) and (1 ;1):
We w ill show that at least one p ointin the s et S lies in the intersection of C k
a nd D k
for so me index k2I
n
:It is evenpos sible to showtha t (\ n j=1 C j )\D k 6=;fo r some k 2I n and (\ n j= 1 D j )\C k 6=; fo r so me k 2I n : Theorem 3.5 LetC 1 ;:::;C n an dD 1 ;:::;D n bec losedsubse tsofQ n satisfyin g([ n j= 1 C j ) [ ([ n j=1 D j )=Q n :
Moreov er, for e ve ry q 2Q n ; q =0 impl ies q2C j and q =1 implie s q 2D j : Then t he re
ex ists a con nect ed set S wit h the propertie s giv en in Theore m 3.4 and points s 1 ;s 2 2 S such that s 1 2[ n k=1 ((\ n j=1 C j )\D k ) and s 2 2[ n k = 1 ((\ n j=1 D j )\C k ): Proof
By Theorem 3.4 there exists a connected set S such that 0 n ;1 n 2 S ; and q 3 2 S implies q 3 2 C j \ D j for some j 2 I n ; o r q 3 2 \ n j=1 C j ; o r q 3 2 \ n j= 1 D j
: Clea rly, S can be
cho sen such tha t it is a clos ed set. De ne J 0
(q ) a nd J 1
(q ) as in the pro o f of Theorem
3.4 a nd let j 0
(q ) a nd j 1
(q ) denote the numb er of elements in these sets, respectively.
De ne the corresp o ndence 8 0 : S ! IR by 8 0 (q) = fj 0 (q)g if j 0 (q) > 0 a nd 8 0 (q) = ; if j 0
(q )=0; 8q2S:De nethecorresp o ndence8 1 :S !IRby8 1 (q )=f0j 1 (q)gifj 1 (q )>0 and 8 1 (q ) = ; if j 1
(q) = 0; 8q 2 S: Finally, de ne the co rres p ond ence 8 : S ! IR by
8(q) = co(8 0
(q )[8 1
(q )); 8q 2 S: Usin g L emma 2 .4 and Theorem 3.4 it follows eas ily
that 8isan upper semi-continuo us co rres p ond encea nd 8 (q ) isnon- empty and convex for
everyq2Q n
:It w illb e s hownth at8 (S)isaconnectedsubs et ofIRa ndhence aninterval.
Suppose 8(S) is not connected, then it can bepa rtitioned in twono n-empty disjoint s ets
T 1
andT 2
;b o thb eing clo sedin 8(S): By Propositio n1 o fHildenbrand (19 74 ,p.22 ) both
8 0 1 (T 1 ) and 8 0 1 (T 2
) are clo sed in S : Supp o se 8 0 1 (T 1 )\8 0 1 (T 2
) 6= ; then there exists
an element q 2 S and points t 1 a nd t 2 such tha t t 1 2 8(q )\ T 1 and t 2 2 8 (q )\T 2 :
Since 8 is co nvex- valu ed it holds that (10)t 1
+ t 2
2 8(q ) for every 2 [0 ;1 ]: Hence
there exists a continuo us function f : [0;1 ] ! 8(S) such tha t f(0) = t 1 a nd f(1) = t 2 : So t 1 and t 2
are an element o f the s ame co mp onent o f 8(S); a contradiction. Th erefore,
8 0 1 (T 1 )\ 8 01 (T 2 ) = ;: Clea rly 8 0 1 (T 1 )[8 0 1 (T 2
) = S: Hence S is not co nnected, a
contradiction. Cons equently 8(S)is co nnected.
It holds that 0 n 2 S; n 2 8(0 n ); 1 n 2 S; an d 0n 2 8(1 n ): T herefore 8 (S) = [0n;n]: Suppose th atfo r every k 2I n ; (\ n j= 1 C j )\D k \S =;:Then 8((\ n j=1 C j )\S)=fng and 8(Sn\ n j= 1 C j )[0n ;n01];s o 8(S) [0n;n01][fng; a contradiction . Co nsequently there is a p oint s 1 2S such tha t s 1 2 (\ n j= 1 C j )\D k fo r some k 2I n : Similarly it can be
show nthat thereis ap o ints 2 2S such tha t s 2 2(\ n j=1 D j )\C k fo r somek 2I n : Q.E.D.
Theorem3.5 s treng thensthe usua lfo rmula tio nofthea nalogofthe SpernerL emmaon the
cub e (see Freund (19 86) a nd van der La an, Talman, and Yang (199 4)), which is given in
Co rollary3 .6 . Corollary 3.6 LetC 1 ;:::;C n an dD 1 ;:::;D n bec losedsubse tsofQ n satisfyin g([ n j= 1 C j ) [ ([ n j=1 D j )=Q n :
Moreov er, for e ve ry q 2Q n ; q j =0 impl ies q2C j and q j =1 implie s q 2D j : Then t he re is an inde x k2I n such thatC k \D k 6=;:
Lemma a nd the Ichiishi L emma w ill b e con sidered. In order to do this we rst give a
de nition o f a balanced co llectio n of sets. De ne fo r every non -empty s ubset T of I
n the vector e T by e T j = 1 jTj if j 2 T an d e T j
= 0 if j 62 T; where jTj denotes the number of
elements in the setT:De neth evecto re ; bye ; j = 1 n ; 8j 2I n
: D eno te the collectio n ofall
subs ets of I n by T n : No ticethat ;2T n : De nition 3.7
Let B bea n on -e mpt ycollect ion of el emen tsof T
n
; sayB=fT 1
;:::;T m
g:The coll ectionB
isbalancedifthereex istpositivenumbe rs 1 ;:::; m suc ht hat P m i= 1 i =1an d P m i= 1 i e T i = 1 n 1 n :
De nitio n 3.7 is slightly more general than the usu al d e nitio n of ba la ncedness s ince the
empty set is not excluded a s an element of a balanced collection of sets. If only
non-empty subs ets of I
n
a re co nsidered, then D e nition 3 .7 redu ces to the usua l de nitio n of
balan cedness. In Sectio n 4 it will be shown tha t the next theoremg eneralizes the KKMS
Lemma an d the IshiishiL emma.
Theorem 3.8
Let fC T
j T 2 T
n
g be a coll ection of cl osed sub sets of Q n sat isfying [ T2Tn C T = Q n :
Moreov er, for e very q 2 Q n w ith ; 6= I 0 (q ) 6= I n ; q 2 C T for a se t T 2 T n satisfying I 0
(q)T; and for e very q 2 Q n w it h ; 6=I 1 (q ) 6=I n ; q 2C T for a set T 2T n satisfying I 1 (q) I n
nT: Then t here ex ists a conn ec ted se t S suc h that 0 n
;1 n
2 S and for ev ery
q 3
2S t he re is a bal anced collec tion fT 1 ;:::;T m g of set s in T n such that q 3 2\ m i=1 C T i : Proof For every q 2 Q n de ne th e set J(q ) = fT 2 T n j q 2 C T
g: De ne the correspondence
:Q n !IR n by (q )=co e T 0 1 n 1 n T 2J(q ) :
Using L emma 2.4 a nd the closedness of th e sets C T
; 8T 2 T
n
; it fo llows that is upper
semi-continu ous. Clea rly (q )is no n-empty and convex foreveryq2Q n
;an d [
q 2Q
n(q)is
b o unded, so Ass umption A.1 is satis edby :Ass ump tion A.3is trivia lly sa tis ed by :
Co nsid er thepoint q=0 n
:Dueto the b ou ndarycon dition itholds forevery j 2I
n
and for
every" 2(0;1 ]thatthep oint0 n +"e j b elo ngsto C Innfjg o rto C In :Hence,sinceea chC T is clo sed,0 n 2\ j2In C I n nfjg or 0 n 2C I n
:Clearly, both the collection fI
n nfjg2 T n jj 2I n g
and the collection fI
n
g are ba la nced a nd therefore 0 n
2 (0 n
): Simila rly, since 1 n 0"e j b elong s to C fjg or to C ;
fo r every " 2 (0;1 ] it holds that 1 n 2 \ j2In C fjg or 1 n 2 C ; : Hence 0 n 2 (1 n
) s ince b o ththe collection ffjg2 T
n
j j 2 I
n
g and the co llection f;g a re
Co nsid er a p o int q 2Q n nf0 n ;1 n g w ith I 0 (q)[I 1 (q )6=;: L etT 0 2T n
b e a sets atis fying
I 0 (q)T 0 a nd q2C T 0 ;and letT 1 2T n beas etsatisfying I 1 (q )I n nT 1 a nd q2C T 1 :If T 0 =; o rT 1 =;; then clearly 0 n
2(q ) and As sumptio n A.2 is sa tis ed. Hence co nsider
the ca se T 0 6= ; and T 1 6= ;: De ne z = jT 0 j n e T 0 + n0 jT 0 j n e T 1 0 1 n 1 n : Clearly, z 2 (q): For j 2 I 0 (q ) it holds that z j jT 0 j n 1 jT 0 j 0 1 n =0 a nd fo r j 2I 1 (q) tha t z j jT 0 j n 1 jT 0 j 0 1 n =0 :
Co nsequently,As sumptio n A.2is satis ed.
By Theorem 2.7 there is a co nnected set o f p o ints Z c sa tisfying 0 n ;1 n 2 Z c a nd q 3 2 Z c implies 0 n 2(q 3 ):Notice tha t 0 n 2 (q 3
) if and o nly if there exists a balanced collection
fT 1 ;:::;T m g of s ets in T n sa tisfying tha t q 3 2 \ m i=1 C T i
; i.e., there exis ts 31 ;:::; 3m >0 with P m i=1 3i =1such tha tq 3 2C T i fo reveryi2I m and0 n = P m i=1 3i ( 1 n 1 n 0e T i );hence P m i=1 3i e T i = 1 n 1 n : Q.E.D.
Since the bounda ry con dition in T heo rem 3.8 is no t sp eci ed for q = 0 n a nd q = 1 n ; it is possible that C ; = ; o r C I n
= ;: It sho uld b e no ticed that the b o undary co ndition
speci ed in Theorem 3 .8 is weaker tha n the conditio n that for every q 2 Q n nf0 n ;1 n g with ; 6= I 0 (q)[I 1 (q ); q 2 C T fo r a set T 2T n satisfyin g I 0 (q) T a nd I 1 (q) I n nT:
Theorem3.8isillus tra tedinFig ure5. Intheillustrationnequals2:Inthislow-dimensiona l
Figure 5 : Illustration o f Theorem 3.8, case n=2:
case the on ly di eren ce with Theorem 3.1 o r Theorem 3.2 is the possibility o f non- empty
sets C ;
or C f1;2g
: In th e ca se n = 2 the minima l b ala nced co llections of s ets are g iven
by fC ; g; fC f1;2g g; a nd fC f1g ;C f2g
g: It is easily veri ed tha t in Figure 5 the union over
all ba la nced co llections of sets B of th e intersection of the sets in B consists of three
comp o nents, witho necomponent containingb o th thepoints0 n
and 1 n
situation may bemuch mo reco mplica ted thanin Th eo rems 3 .1and 3.2.
Bysymmetrycon sidera tion sTheorem3.9fo llowsimmediatelyasacoro llarytoTheorem
3.8. It will b e shown in the next sectio n tha t u sing Theorem 3.9 it is ea sy to derive the
KKMS Lemma . Theorem 3.9 Let fC T j T 2 T n
g be a coll ection of cl osed sub sets of Q n sat isfying [ T2T n C T = Q n :
Moreov er, for e very q 2 Q n w ith ; 6= I 0 (q ) 6= I n ; q 2 C T for a se t T 2 T n satisfying I 0 (q) I n nT; an d for ev ery q 2 Q n w ith ; 6= I 1 (q ) 6= I n ; q 2 C T for a se t T 2 T n satisfyin g I 1
(q)T:The nthe reex istsa connec tedsetS suc hthat0 n ;1 n 2S an d forev ery q 3
2S t he re is a bal anced collec tion fT 1 ;:::;T m g of set s in T n such that q 3 2\ m i=1 C T i :
4 Intersect io n Theorems on the Unit Simplex
In this section a numb er of well-known intersection theorems o n the (n01 )-dimensiona l
unit s implex, S n = fp 2 IR n + j P n j= 1 p j
= 1g; will b e shown to follow as corolla ries to
the theorems o f Section 3. Theorem 3.1 leads to the Sperner L emma. This L emma is
illustrated fo r the ca se n = 3 in Figure 6 . In Figure 6 there is exactly one intersection
Figure 6 : Illustration of Sperner L emma,case n =3:
p o int. Inthe proof o f T heo rem 4.1a coverfC 1
;:::;C n
gof S n
s atisfying the co nditions of
Theorem4 .1isextendedinastra ig htfo rwardwayto acover f b C 1 ;:::; b C n g ofQ n s atis fying
thecond itionsofTheorem3.1. Thenitfollows tha tthereexistsaconnectedset b Ssuchthat 0 n ;1 n 2 b S a nd b S \ n b C j
: It will b e show n that th is co nnected set b
intersection with th e un it simplex S n
: Fina lly it is show n that \ n j=1 b C j \S n =\ n j= 1 C j : In
Theorem4 .1 it is a ga inas sumed tha t if an in dex j 2I
n
is taken, thenj +1=1 if j = n;
and j01=n if j =1 :
Theorem 4.1 (Sp erner Lemma)
Let C 1 ;:::;C n be c losed subse ts of S n satisfyin g [ n j=1 C j =S n : Moreove r, p j =0 for some j 2I n implie s p2 C j : Then \ n j=1 C j 6=;: Proof
The cas en =1 is trivial, s ocons ider the cas en2: For every j 2I
n d e ne the s et b C j =fq2Q n j5 S n(q )2C j g[fq2Q n jq j =0 g[fq2Q n j q j+ 1 =1 g:
Itwillb es hown that thesets b C 1 ;:::; b C n
satisfyth eco nditio nso fT heo rem3.1. Itiseas ily
veri edthat,usingtheclosednessofthesetsC 1
;:::;C n
;theco ntinuityofthefu nction5
S n;
andthepro p ertythat[ n j=1 C j =S n ;foreveryj 2I n the set b C j isclos edand[ n j= 1 b C j =Q n : If, for q 2 Q n ; q j = 0 or q j+ 1
=1; then clea rly q 2 b C j : So the sets b C 1 ;:::; b C n sa tis fy the
conditio ns of T heo rem 3 .1 . Hence there exists a co nnected s et b S such tha t 0 n ;1 n 2 b S and b S \ n j=1 b C j : De ne f : b S ! IR by f(s) = P n j= 1 s j ; 8s 2 b S: Sin ce th e imag e of a
connected set und er a co ntinuous functio n is co nnected, f(0 n ) =0 a nd f(1 n ) = n; there exists b s 2 b S such tha t f (b s) = 1 ; o r equiva lently b s 2 b S \S n \ n j= 1 ( b C j \S n ): C learly C j b C j \S n ; 8j 2 I n
: Su pp os e there exists an element b q 2 ( b C k \S n )nC k fo r s ome k 2I n :Then since b q2 S n
imp lies tha t 5
S n (b q )= b q it ho lds tha t b q k =0 or b q k +1 =1: Since b q k = 0 implies b q 2 C k it holds that b q k + 1 = 1 and b q k
> 0; yielding a contrad ictio n since
b
q2S n
withn 2:Cons equently b C j \S n =C j ; 8j 2I n ; a nd b s2\ n j=1 C j : Q.E.D.
Theorem3.3 leadsto the KKML emma. Th is lemma is illustrated inFigure7forthe cas e
n = 3 : In the proof of Theorem 4 .2 a cover fC 1
;:::;C n
g of S n
sa tisfying the cond itions
of T heo rem 4.2 is extend ed in more or les s the same straightforward way a s in th e proof
of Theorem4 .1 to yield a cover f b C 1 ;:::; b C n g of Q n
satisfyin g the conditio ns of Theorem
3.3. Some nota tio n is introduced rs t. For J 2 T
n let Q
n
(J) deno te the set fq 2 Q n j q j = 0; 8j 2 Jg: D e ne S n (J) = fp 2 S n j p j = 0; 8j 2 Jg =S n \Q n (J): No tice that Q n (;)=Q n , S n (;)=S n ; a nd S n
(J) 6=; if and o nly if J is a proper s ubset of I
n
:D eno te
the collection of a ll pro p er subs ets ofI
n by T 0 n ;s o T 0 n =T n nfI n g: Theorem 4.2 (KKM Lemma) Let C 1 ;:::;C n be cl osed su bsets of S n satisfyin g [ n j=1 C j =S n
:Moreove r, forev ery p2S n wit h I 0 (p) 6=; t he re ex ist s some j 2I n nI 0 (p) su ch t hat p2C j : Then \ n j= 1 C j 6=;:
Figure 7: Illus tration o f KKMLemma,cas en =3 :
Proof
For every j 2I
n
de ne the set
b C j =[ J2T 0 n fq 2Q n (J)j5 S n (J) (q)2C j g[fq 2Q n jq j =1g:
It will b esh own th at th esets b C 1 ;:::; b C n
s atisfy the cond itions of Theorem3.3. Using the
clo sednes s of the s ets C 1
;:::;C n
; the continuity o f the fu nction 5
S n
(J)
for every proper
subs et J of I n ; a nd the fa ct that [ n j=1 C j =S n
;it ho lds that for every j 2I
n the set b C j is clo seda nd [ n j=1 b C j =Q n :Clea rly,q 2Q n andq j =1impliesq 2 b C j :Co nsid erq 2Q n nf0 n g with I 0 (q)6= ;: T hen q 2Q n (I 0 (q )) a nd 9k 2I n nI 0 (5 S n (I 0 (q )) (q )) I n nI 0 (q) such that 5 S n (I 0 (q )) (q) 2 C k ; so q 2 b C k
: Co nsequently the sets b C 1 ;:::; b C n
sa tis fy the cond itions of
Theorem3 .3 and there existsa co nnected set b S s uch that 0 n ;1 n 2 b S an d b S \ n j=1 b C j : As
inth eproofofTheorem4 .1itfollow sthatthereexists b s2 b S\S n \ n j= 1 ( b C j \S n ):C learly C j b C j \S n
: Suppose there exists a nelement b q 2( b C k \S n )nC k fo r so me k 2I n : Since 5 S n (J) (b q ) = b q if b q 2 S n \Q n (J) it follow s that b q k = 1: If b q j = 0 fo r every j 2 I n nfkg;
thenitfollowsby thecond itions ofT heo rem4 .2tha t b q2C k :Hence b q j >0fo rso mej 6=k;
giving a co ntra diction s ince b q k =1a nd b q2S n : Con sequently b C j \S n =C j ; 8j 2I n ; and b s2\ n j= 1 C j : Q.E.D.
In Theorem4.3 the Ichiis hilemma (see Ichiis hi (198 8)) isderivedfromTheorem 3.8. T he
Ichiishi lemma is illustra ted in Figure8 . In Figure 8 there is exactly one p oint fo r which
Figure 8 : Illustration of Ichiish iL emma,case n =3:
balan cedco llections ff1;2 g;f3 gga nd ff1 ;3 g;f2ggh avean on-empty intersection
consist-ing of the same p oint. No tice th at in Figure 8 wehave C f1g
=; a nd C f1;2;3g
= ;:D eno te
the collection of a ll no n-empty subs etsof I
n by T 3 n ; so T 3 n =T n nf;g:
Theorem 4.3 (Ichiishi Lemma)
Let fC T
j T 2 T 3
n
g be a collect ion of c losed sub sets of S n sat isfying [ T2T 3 n C T = S n :
Moreov er, for ev ery p 2 S n w ith I 0 (p ) 6= ; the re is a se t T 2 T 3 n satisfyin g p 2 C T and I 0
(p ) T: The n t he re is a bal anced collec tion fT 1 ;:::;T m g of set s in T 3 n such that \ m i= 1 C T i 6=;: Proof The ca sew here C In
6=; is trivial,hence consider the case C In =;:De ne b C ; = b C In =;:
De nefor every T 2T
n nf;;I n g b C T = [ J2T 0 n fq2Q n (J)j5 S n (J) (2q )2C T g[fq 2Q n jI n nT I 1 (q)g:
Itwillb esh ownth attheco llectionofsetsf b
C T
j T 2T
n
gsa tis esthecond itionsofTheorem
3.8. C learly,foreveryT 2T
n ; b C T isclosed,a nd[ T2Tn b C T =Q n :Mo reover,ifq2Q n nf0 n g with I 0 (q ) 6= ;; then q 2 Q n (I 0 (q )) an d 9T 2 T n such tha t I 0 (5 S n (I 0 (q )) (2q )) T and 5 S n (I 0 (q )) (2 q ) 2 C T : Hence q 2 b C T while I 0 (q ) I 0 (5 S n (I 0 (q )) (2q)) T: If q 2 Q n nf1 n g and I 1 (q )6= ;; then q 2 b C InnI 1 (q )
: C onsequently the co llection f b C T jT 2T n g sa tis es the
conditio ns of Th eo rem 3.8 a nd there exists a co nnected set b
S with the pro p erties sta ted
in Th eo rem 3.8 . As in the proof of Theorem 4.1 it follows tha t there exis ts b s 2 b S \fq 2 Q n j P n j=1 q j = 1
g:Hence thereis abalancedcollection fT 1 ;:::;T m go f sets inT n nf;;I n g
such tha t b s 2 \ m i= 1 b C T i : Since b s j 6= 1 ; 8j 2 I n ; it h olds tha t 8i 2 I m ; 9J 2 T 0 n such that b s2Q n (J)a nd 5 S n (J) (2b s)2C T i :Since b s2Q n (J)a nd P n j=1 b s j = 1 2 implies5 S n (J) (2 b s)= 2b s it holdstha t 2b s2\ m i=1 C T i : Q.E.D.
In T heo rem4.3 a coverof S n with s etsin T 3 n =T n
nf;g is considered, whichis th e usua l
formula tion . Clea rly, the sta tement o f Theorem 4.3 is still true if a cover with sets in T
n
is co nsidered, s ince in the case C ;
6= ; Theorem4 .3 is trivia lly tru e. It is clear tha t also
Theorem 3.9 can b e used to derive the Ichiis hi L emma. Similarly the KKMSL emmacan
easily be derivedfro m b oth Theorem3.8a nd Theorem3.9. In Th eo rem4.4 the derivation
usingTh eo rem3.9 willb eshown. Theorem4 .4isillustrated inFigure 9forthe casen =3 :
In the illu stration the un ique intersectio n point is given by the intersectio n o f th e sets in
Figu re 9 : Illustration of KKMS L emma,ca sen =3:
the balan cedcollectio nff1;2g;f1;3 g;f2 ;3gg:
Theorem 4.4 (KKM S Lemma) Let fC T j T 2 T 3 n
g be a collect ion of c losed sub sets of S n sat isfying [ T2T 3 n C T = S n :
Moreov er, for eve ry p 2 S n with I 0 (p ) 6= ; the re is a se t T 2 T 3 n satisfyin g p 2 C T and I 0 (p ) I n
nT: Then there is a balanced coll ection fT 1 ;:::;T m g of sets in T 3 n such that \ m i= 1 C T i 6=;:
The proof goes a lo ng the s ame lines as the proof of Theorem 4 .3 by using Theorem 3.9
insteado f Theorem3.8 a nd de ning fo r every T 2T
n nf;;I n g b C T = [ J2T 0 fq2Q n (J)j5 S n (J) (2q )2C T g[fq2Q n jT I 1 (q)g:
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