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I

1 9 8 3 Y E A R THIS DEGREE C O I Y F E R E D / A W ~ D'WTENTIW DE CE GRADE

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wise reproduced without the auahor's mitten permission. w sutrement reproduits sans I'autaiscalim &rite de /*auteur. 1

David Hung Yiu Poon

Simon Fraser University, 1978 - - - -

THE REQUIRMENTS FOR THE DEGREE OF

MASTER OF SCIENCE

in the Departhnt -

8% of

@ David Hung Yiu Poon

SIMON FRASER UNIVERSITY =&

a December 1'983

- All rights reserved. This thesis may not be \

reproduced in whole or in part, by photocopy or other means, without permission of the author.

Name :

Degree :

Title of Thesis:

1%

\ f"

' L -

David Hung Yiu Poon .

Master of Science

Sperner's theorem i%nd extremal properties of - - /

f inite sets. ?, -

2- ,

Examining Committee: - - - -- -- ~ - ~

- - ~ --- -~

Chairman: Professor C. ille eggs

T.C. Brown Senior Supervisor

- -/,,

X:R. Freedman

.. e

K. Heinrich - - - - - -- - - --- -- External -- Examiner -- -

Associate Professor,'$4SERC . .- Department of Mathematics Simon Fraser Univkrsity

L C

-'.

I

1 hereby grant t o Simon Fraser Univers i ty the r i g h t t o lend z

rry thesis, p ro jec t o r extended essay. ( the t i t l e of which i s shcwn below)

to users o f the Simon Fraser Univers i ty t i brary, and t o make p a r t i.al o r

s i n g l e copies only f o r such- users o r i n rwponse to a request from the

t i brary of any other unive%sity, o r other educat ional Tn'sfi t u t i so , on -

2 L - - i t s own behalf o r f o r OM of i t s users: I f u r t he r agree t h a t permission

f o r m u l t i p l e copying o f t h i s work f o r scholar ly purposes may be granted

by me o r the Dean of Graduate Studies. It i s understood t h a t copying -

' uf thout my d t t e n permi ss ion.

T i t l e o f Thesis/Project/Extended Essay

Author: -C

-

This

I , - with sp&nef s

properties, we

is an expositpry paper surveying the main results connected

theorem and extremal properties of finite sets*. By extremal '

mean those involving maximum or m i n h m size'd families of - - - - - -- - -

, sets subject o eerthn restrie~ions, wGch are described in kerms of I' -

- -

r

containment, disjointness, union and intersection,- etc.

Chapter 1 deals vith Sperner's theorem, the starting point of - - - - - -- - - - pp - -

- - - pppp-p

the extremal theory of sets, and gives some generalizations of the theorem

in various directions.

In Chapter 2 we extend Sperner's idea to finite partially ordered

sets admitting B rank function. The LYM property is discussea and turns '

out to imply Sperner's theorem, giving many interesting results as well.

-- - -- --- - -- Chapter-3 descri-bes one of the most versatile of all theorems of

extremal set theory, the Kruskal-Kat-ona theorem. C

Firrally in the fast chapter, we give some applications of the

results in the previous chapters, to solve some problems in other areas

of mathematics. -

( iii) n

To my supervism~ Br. T. -3mwrr. f~ his- hefp and advice during the preparation

of this-thesis,

to my wife Lucy, for her. patient understanding during this work,

I express my sincere gratitude.

................................................................. Approval (ii)

Acknowledg~ents ........................................................ (iv) C- i

Table of Contents ........................................................ (v) -- - - --- -- --

- - - -- \% Chapter 1 . Sperner's Th'eorem.........................'.................. 1

................... Chapter 2 . The S p e h e r property and the L w r o p e r t y 15

f - ....................... Chapter 3 . he Kruskal-Katona Theorem............ 39

4

h I

Chapter 4 . pplicatioqs .................o........................r...'., 50

Bibliography ..........................

I ' 9-

Sperner 's Theorem

\

The problems s tudied i n t h i s paper\have evolved from the ideas \, $ r I

t -

developed by Sperner. . - I n 1927, he proved mkr& s i z e of a -

family of subsets of an .n element set, sub jec t t o the requirement that no

- s e t i n the family conta ins another, i s the same'as t h e maximum of t h e

- - - - - - - - - - - - - -- b'== --- icientF 7 R . -

lnomlal coeff ( ! , 0 5 k 5 n . The max- is a t t a i n e d when +

n k = L + where n denotes t h e integer p a r t of n/2. . There are,

many proofs o? t h i s theorem; t h e s imples t one is t h a t of Lube11 [.'27]. , .

I / Theorem 1.1. (Spern,er [ 3 2 ] ) . L e t S be a s e t with n

elements. Let F be a family of subsets o f ' S such that f o r a l l

Proof. (Lubell. 6271). Let F be a family of subsets of i

L S (IS1 = n) s a t i s f y i n g tbe above condit ion. We c a l l

C - (Ao,Al,, . . . , An} (where A. c S , 0 5 i c n). a maximal chain i f P

t * 1 -

Clear ly t h e t o t a l number of maximal chain6 i n S i s *n! . * --- , + -- ---p---- > 7-

L \

\

\ - I ' I

, . P b . .

r *

4

C . . - 2

* -- - -- - - - - - - - - - - Q - --

L

5

' I . C * -2.

- - -- - - P

A C F : ( f ~ f = k S'n) , .we count the nmher , \,

' a'* r' : '> .- , A e

'" - r

of tbaximal it,, ~ i i s t we have. k! poss>biLities . :* , ' * ' @

in hoosing AO,A1,. . ,A,---l , and then (*-kg posqihilities in choosing 4 - d

Ak+l , , . . - , A . Therefore* the number of maximal chains I 4 . n ' - a c .

contain A &s ' . - +Iz . P % 0'

, . - 'i . 5-

* . / A I ! (n - [A/S !' . - - - %. ' f* A. 2

-, i . C - * *

L e " t = Nov consider .. 'L 1, e. 'T 1

, e U .•’(A) , \ d

A€ F r'

C

A and )3 in F , A # B implies f(A) f ? f ( B ) ' =

. are n! maximal chains altogether, ve obtain

so that

Remark. The result above is best possible since by taking- I

- ~ - . -

*. . o

~perner's theorem can be generalized in many directions. We "I; '

" . + shalJ consider three of them. - . ,

- h

P I (i) ~rd~os'[8] improved Sperner's thebrem to show that the s i 2 . . - * .

of-a' family'of subsets o'f S ( I s I - n) such $hat no 2 . + 1 subsets

3 ... 3 Ai , *A. ,A. ,. -. .*As in the f m i l y f ~ m a chain A. .?. A. 1

* '1 .:2 ' R-kl 1 1 1 2 %+I* . . . - ,. above by'the sum of the k largest binomial coefficients

e

. of order n . . . ~- -- ~~ - - - -- -~ ~ - - ~ - ~ - - - - - - - - --- - -- - -- *- ~

(ii) Katona 1151 and'Kleitman [21] independently proved a sharp-

ening of Spyrner ' s theorem? = I )

f 4 -

Q >?-- Let ~ ( 1 ~ 1 = n)

. ' be into 'two disjoint sets and

t 1

let F be a family of subsets of S no w o of which are

ordered in one component of S and identical in the other. 'Then the

order n . - -

4

\ (iii) DeBrui jn, Tengberger, and Kruyswijk [4] gederaliz4d~eGnei.'s

' 9 - 1 Q theorem in the following manner:

I - v

' 4 Let f l. f 2 , . . . , f m be integer-vtlued funct ion3 defined on

S, 4 {x1,x2, ..., xn1 such that 0 5 fi(%) ak, wheres&; 'ak% arev - .i. given positive integers. If no two differentafuflctions satisfy , %

- - of funetions satisfying

We s h a i l g ive proofs f o r all! t h r e e of these r e s u l t s . Our , 4

^ proofs a r e essentia!Iy! t h e t dard ones.'found i n the l i t e r a t u r e , w i t l i v - 4 " ,

/ , .* *

. . v ' 3

minor*mdi • ’ i f a t ions f o r t h e s&e of c l a r i t y . ' C, % +'

P . . . fl.

We begin with t h e following d e f i n i t i o n . * ,

, t

D e f i n i 3 1.1. seque&e of subsets . Al ,A2,. . . 'of s ( 1 s 1 =- n)

is a s e t r i c chain i f .i;r

. *

Def in i t ion 1.2. A sequence o f subset; A1,A2,. . . ,I4,# of S i s a chain . '

i f i t s a t i s f i e s t h e f i r s t p a r t of d e f i n i t i o n 1.1. The length of a cbain

' i s the nTmBer of elements of t h e chain. .. /

w?+ + . The foldowing observations follow immediately from Def ini t ion

- d Observatiqn 1.2. A syuj&tric chain i s brained by omit t ing a maximal ?'

n cha in ' s 2 smal les t and E l a r g e s t elements f o r 0 5 8 5 . A

'. r ' . k.

f symmetric chain s o formed- w i l l have (n+l - 211) elements wi th . .

151 = 2 and

-/ . . - - - - - Observation 1.3. I f n i s -even, then k i s odd, and v i c e versa .

Observation 1.4. . I f a synrmetric chain contains -k elements, then 0 - - - -- --- - '--

#' C

Lemma 1.5. ~ h ' e c l a s s of a l l sudbets of a f i n i t 6 set S -ran L 4

be expressed a s t h e union of a c o l l e c t i o n of d i s j o i n t symmetric chains. ' . I - . .

Proof. We proceed by induction on t h e s i z e of S . * i When I S ! = 1 , .

{Q , S). i s a symmetric chain. - -

1 r (ii) Suppose t h e lemma i s t r u e f o r a l l sefs o @ c a r d i n a l i t y n .

Consider I s I = n+l . ' r ) ' .

Let x E S . Then 1,s - (x)-l = n and the re fo re . .. --, -- - - v -

- - - - - - - -- - - - - -- - - - --

- - ,

- -- - - - - - - - --

P(S - {GI) = U Ci f o r some v , where each C . i s i=l 1

. a symmetric chain of S - {XI , and t h e C i l s a r e d i s j o i n t . d

7 Let

a Ci = { A ~ i , A2 i , . . . , A k , } i , i = 1 ,2 , . . . ,v . 1

C o n ~ i d e ~ t h e chaihs C . ' and C." i n S , where 1 1

We s h a l l show t h a t =-=

m

(a) S ' c S - implies S f € C i l o r S' C C . " . - 1

(b) he chains C. ' and C i = 1 2 . , a r e d i s j o i n t . 1

(c) The chains C. ' and C i = 1 . v a r e symmetric 1

.' chains.

induction.

For p a r t ( a ) , i f x $! S' then S ' € C i l f o r some 1 . Z

ci 9

-- -.fa' and if x € S ' , t hen S' - {x) i s i n C . f o r some i . In t h i s - c a s e ,

1

S ' C Ci ' 6r S f € Ci" depending on whether o r n o t S' -- {x} i s t h e

la& element of C . . 1

P a r t (b) fo l lows from t h e f a c t tgat t h e C . ' s a r e d i s j o i n t . 1

i

Fof p a r t ( c ) , we observe t h a t Tr Ci ' we have

I + 1 = n + l ,

and f o r C-," we 1

have

v * Lemma 1.6. Let P(S)= U C . , where I S

1 i=l

a r e pa i rwi se d i s j o i n t symmetric chains, Then t h e

cha ins C . con ta in ing k o r more elements i s 1

1 = n and t h e C . ' s 1

number nf_ s y p u e t r i c -

Proof. The number of d i s t i n c t s u b s e t s of S con ta in ing

n+k n - elements i s . Each of t h e s e s u b s e t s l i e s i n o n e and - 6 ~ 3 , only one C. ' fur thermore Ci must con ta in k o r more elements ,

1

-

s i n c e i n a symmetric cha in of l e n g t h k , n+k i s odd and t h e l a r g e s t a

n+k-1 n+k elemen,t of t h e cha in has s i z e - =

2

We cons ide r two cases .

Case ( i ) . (n+k-1) i s even.

1 hen t h a number of symmetric cha ins conta in ing k o r more elements

-

Case (ii). (n+k-17 is odd.

Then the number of synrmetric chains containing exactly k elements I

is 0 . Therefore the number of ~ynunetric chains containing k or

more elements is the same as the number of symmetric chains containing

k+l or more elements, which equals, bif; case (i), .

1, v Lemma 1.7. Let P ( S ) = U C. as in Lemma 1.6; then for each

1 i- 1 r , 1 5 r 5 n+l , the number o_f symmetric chains C L containing =aGly - -

1

r elements is

theorem.

Proof. An immediate consequence\of Leuma 1.6.

We are now ready to prove ~rdss' theorem and Katona's t

Theorem 1.8. (~rdos [ 8 ] ) . Let F be a family of distinct

subsets of an n-element set S such that there exists no chain of length

r+l in I?. Then

IF^ 5 sum of the r largest binomial coefficients 2

of order n .

symmetric chains such that - - -

a 5 -

The existence of the C.'s is guaranteed by Lemna-1.5. 1

v 9L 5i= F = tf f%. f+ Fk a+-.% a r e d i s j o i n t ,

i=l 1 1

I

Also we a r e given t h a t t h e r e i s no cha in o f - l e n g t h more than r , so

Thus, -

v 5 2 m i n ( r , 1 ~ ~ 1 ) .

i=l

By Lemma 1.7, t h e number of C . ' s con ta in ing e x a c t l y j - -

-- I - - -

elements equa l s

and t h e l eng ths of t h e C . ' s run from 1 t o n+l . Therefore , 1

(The second i n e q u a l i t y above can be e a s i l y checked by w r i t i n g o u t t h e

sum and cance l l i n h appro~xiatc terms. ) - - - - - - -

- -- -- -~ - -- -

' Remark. It can e a s i l y be seen t h a t t h e upper bound bn F *

i s b e s t p o s s i b l e , and t h a t when r=l , t h e theorem reduces t o Sperner ' s

theorem. >

Tkemem 1.9. ftktona € 1 3 a n i K l e l t m ~ f 2 1 ] ) . L e t I s I = n b

and S = H U H z , where* H fl H = d . I f P i s a fami ly of s u b s e t s 1

\ 1 - 2 1

of S such t h a t no two members A, and B s a t i s f y L t *

( i ) A fl H1 = B f l H, and A fl H2 3 B fl H2

o r ( i i ) A n Hz,= B fl H2 and A fl H1 3 % fl H 7 . i *

then

3

Proof. Let F be a family of s u b s e t s of S s a t i s f y i n g t h e 4 - - - - - - - - -- - - -

h y p o t h e s i s of t h e theorem, and cons'ider a f i x e d subse t K of H2 . Let A1,A2, ..., A be s e t s i n F such t h a t

P

.c

Then by cond i t i on ( i i ) of t h e hypothes is ,

i s a Sperner system, i - e . , no s e t c o n t a i n s a n o t h e r .

NOW let K 2 K 2 ... 3 Kr 1 2 be, s u b s e t s of H . Let 2.

B(K. ) = {A fl H, : A fl H, = K . and A c F ) . J J

Then, as above, B(K.1 i s a Sperner system, 1 5 j 5 r . Also, i f 3

i j , B(K.1 fl B(K.1 = d . For i f n o t , t h e r e e x i s t s C such t h a t 1 J

C € B(K.1 fl B(K.1 ; then C U K . and C ' U K a r e i n F . But then 1 J 1 j =

and (C U K . ) n H. = K ~ , r

(C u K . ) n H~ = . K ~ , J

which is imposs ib le by c o n d i t i o n (i).

-

contain a chain of length &ore than - r , and U B ( K . ) C ' H ~ . f j =1

J

< I i

Therefore by Theorem 1 . 8 , s e t t i n g 1~~ 1 = nl , < V

B

Obvious ly ,

IFI =

H2 = U Ci for some v , i=l

uhbre the C . are pairwise d i s j o i n t symmetric chains i n 1 H2 '

Theref ore I

From Lemma 1 . 7 an$ -@) , s d t t ing 1 H2 1 = n2 ,

(ii) When H1 # d and H2 # d , Theorem 1.9 is

stronger than Sperner's theorem, for the result obtained is the same "

while the hypothesis is weaker.

Figure:l illustrates this.

P Sets such as A and B in*Figure 1 can be included in F

d in Theorem 1.9, bht not in Theorem 1.1.

(iii) Theorem 1.9 does not hold if we divide the

S into three or more parts. An example is the following. Let

--

where n is odd. " I

Let F = A U B U C U D , where .

n-3 -. A = {Ai : I A ~ " 1 = - 1 2 and I fl H~ 1 = 0 and

Y

n-3 B = { B ~ : n ~ ~ l = - 2 and IB. I. Cl ~~1 = 1 and

Then F does not have two elements E -and F , satisfying "1 - - - - -

w

But

De Bruijn [ 4 1 g e n e r a l i z e d Sperner 's theorem in number theore t i c

de Brui jn ' s

2

terms. W e need some d e f i n i t i o n s and a ' l e - before we can prove ~ ---- - - ~

~ ~ - ~ ~ - - = - = y . = ~ Y - - z - . . -=-=:=:=~== ===-= - -

theorem.

Def in i t ion 1.3. Let m be a n a t u r a l number and l e t m = l1 A2 k P i P2 ** 'Pk

be i t s canonical f a c t o r i z a t i o n i n t o powers of d i s t i n c t primes. Then t h e

Def in i t ion 1.4. A sequence dl,d2,- 9

f B of d i v i s o r s

symmetr$c chain i f

Lemma 1.10. The set of d i v i s o r s of m can be p a r t i t i o n e d

Proot . W e proceed by inductiga on the number v(m) of - d i f f e r e n t primes d ividing m.

X ( v(m) = 1 . Let m, = p where

2 h p, p , . . . ,p is a symmetric ciliain.

( i i ) Suppose t h e lemma is t r u e f o r v(m) = k, ~ e t

"2 %+l P P2 --*Pk+l be the canonical f a c t o r i z a t i o n of a n a t u r a l number n .

Then ( m ) = k , a n h b y hypothesis the set of a l l d i v i s o r s *

of m can be ~ a r t i t i o h e d i n t o ' p a i w i s e d i s j o i n t symmetric c h a h s . >

Let C - {dl ,d2,. . . ,411 be -a symmetric chain of d i v i s o r s of m.. - .

J Associated with C is t h e following set f(C) ' of d i v i s o r s

It is obvious from t h e scheme below t h a t the set f(C) can be

1 par t i t ipr ied i n t o symmetric chains of d i v i s o r s of n .

chaips of d i v i s o r s of m , it fol lows t h a t t h e set of d i v i s o r s of n

, Theorem 1.11. *- (De Brui jn [ 4 I ) . Let ' m be a n a t k a l 'number

with degree n . Let F be a tset of d iv i sors of m such tha t fo r a i l

a,b € F , a b implies a k b . - ' ?

Then I

1 I? 5 the number of d iv i sors of m whose

n degree is I T ] .

- (The symbol a # b means a is not a d iv i sor of b) .

\ symmetric chains. This is possible by Lemma 1.10. Let U be the set-

n of a l l d ivisors whose degree is I T ] . 3

- Clearly, any element of U is contained i n j u s t one sym~letric ,

chain and each symmetric chain contains only one element of U . There-

fore the number of symmetric chains in the p a r t i t i o n equals Iu] . , -. =?

- - Ef F saliSf ies < ~ h t - 5 r e & r t + ~ e e & c k -- - - - -- ---

- ~ymnaetric~chain contains a t most one element of F . Therefore

Renqrk. When m is square f ree , de Bruijn's theorem -

reduces t o Spemer' s theorem. b

I

CHAPTER 2

The Sperner property and the LYM property 2 . - L

*

Tn the Jasb chapter, we dealt with two particular partially . '

ordered sets: the set of subsets of a given finite set and the set

of positive divisors of a,given positive integer. 1't is natural to , - i ask whether Spernerts theorem can be generalized to an arbitrary

ordered set, and,if not, for what.kind of partially ordered - -- --

- - ~ - ~ - - - ~ - - -- --- ~

-

a sets will a generalized version of Sperner,,'~ theorem be true? . ~

We first give some definitions.

- R Definition 2.1. A partially ordered set (P,c) is a set. P - together with a relation 5 on P suc,h that

\ \, x 5 y and y 5 x iyplies x = y. \ - y,

.. (iii) 5 is transitive, l.e., for all x,y,z E P , \

x 5 y and y c z implies x I z. ' .

\

definition 2.2. A subset C of a partially ordered set P is a. 7a

chain if it is totally ordered. That is, for all x,y € C , either 3 3

, x < y _ n r - - p - u --- - ---.- -3 - -- -

~efinition 2.3. A subset F of a partially ordered set P ie an \ - . a %

antichain if no two distinct elements are related. That is, 'if -7---

3 5

- -- ---

x,y 4 F , x #-y , then x d y aid y d x .

* I

A least element of P ' is an element x such 'that for

y. Similarly a greatest element of P is an element &

all y € P , x z

z such that for d

all y € P , y,5 z .

ib ? An element y cdvers g in P if Definition 2.5.

(i) x r y and

Cii) x 5 i c y implies z = x ' 1 Or

~efinition 2.6. A minimal element of P is an element x such thtt * -

there is no y (y#x) E P with y 5 x. - - - - - - - - - - - - - - - - - -

- - - - -

- - - -- - - - - - - - -- - -- -- - - - - - - - - - -- - -

Similarly's mqimal element of P is an element z such ,

that there is no y (y # z ) E P with z 5 y .

~efinitibn 2.7. A rank function r on P is an in6eger-valued - defined on P such that

- -- - -- - --- - -- - -

(ii) r(x) '= 0 for every minimal element of P . 1

(iii) For all x1,x2, € P , x ' covers x implies 1 . 2

Remark. If P -is finite, and if a rank function on P

then it is unique, exists,

Definition 1.8. If x 6 P , and . r is a rank function defined on P ,

For each k >

is denoted by Pk ' Thus <

* d - - I

I \ . . .

I- > -

-- - - -- - - - - - - - - - 17.

* . 3 ,

DefInTtTon ZT.- X F a X - k Z @ ,--tEe kth Whitneynnumber of P , denoted 0

, is the number of elements in P of rank k . . *Thus ' by Nk , -

--

-

Defini,tion 2.10. The rank of P is defined - q

, ~ C P ) = max r7(x) ,

L C xEP

.. if this maxi~ l r r r e exists. '4

In this paper we are concerned on!y with finite partially .-* r Y^ >

ordered sets P having a rank function. ~ence, from now on, P - - -- - - -- -- -- - - - - .- - - - - -- -- -- - - - --- -- *

always denotes the underlying set of a finite partially ordered set +

(P , 5 ) with a fixed rank function r . - . * . . w I .

-- From the definition of Pk, we can see that each P is

k - -=ex -

an antichain. Thus 'an analogue for Sperner's theorem in an arbitrary I B

finite partially ordered set is

4 % Definition 2.11. If the above equality holds in a given partially . 4

'd

ordered set P , we say that ,P has the Sperner property.

a Note that since Pk is always an antichain, then the

* 3 Sperner property for P is equivalent to i:

d

3 2 il f

max IF^ 5 max I P I . .F an ahtichain k k -3 $

4 $4 was conjectured by Rota [[29 1 that the Sperner properx~ -,- -

- - - -- --

holds in a Lertain .general class of partially ordered sets, finite t - - - - - - - 4

geometric lattices, but Dilworth and ,Greene [ 7 ] later gave *a counter- 1 + -

example to this conjecture. We are going to give some conditions on

3- partially ordered sets which are sufficient to imply the Sperner property. C 3

Definition 2.12. A finite partially ordered set. P is regular if -

for each k, 0 5 k 5 r(P) , there are %,bk such that each element a 2 x of Pk covers exactly bk elements of P and is d e r e d by.

exactly aku elements of P.

Theorem 2.1. (~aker [I]). If P is regulaf yith a least

d

element and a greatest element then P satisfies the Sperner property..

Proof. Let P be regular. '

rank k covers bk elements,and is covered by ak elements.

"For a .given element7 x of P , there are exactly - . .

maximal chains containing x , if 1 r x 5 1 . If r(x) = 0 3 4

or r(x> = Y~(P) (that is, if -.ither t h e greatest or the least 3

- - br (PI ... b b . . 2 1 fj 1 2

% \

Let t be the total nurpber of maximal chains in P. Since d 3

each maximal chain contains exactly ofie element of Pk , we have 1 I - i

,Z s (XI = t , O'E-k 5 r(P) . 4 9 A . x€Pk 4

contains at most one element of F. Therefore.

Since Nr (x) 5 max N = max IpkI , t h i s g ives

xE P r ( x )

O . =

Theref o r e

max ,P.

1'1 ' max l p k l 9

> '

k

4 as requi red .

Corol lary 2.2. F i n i t e Boolean l a t t i c e s , vec to r space l a t t i c e s

and f i n i t e p r o j e c t i v e l a t t i c e s a r e knovn t o be regu la r a n h t h u s s a t i s f y -- - -

-

*

t h e Sperner property. (For d e f i n t i o n s of t h e above terms, see

Birkhoff 121.)

Another s u f f i c i e n t condi t ion f o r the Sperner proper ty t o *

hold in P i s t h a t P--has t h e proper ty of u n i m d a l i t y and has complete

t matchings between success ive ranks. We f i r s t de f ine these two proper t ies .

Def in i t ion 2.13. A p a r t i a l l y ordered s e t P -is unimodal i f t h e r e e x i s t s

some in tqger m such t h a t - - - - - - -

Def in i t ion 2.14. Let X and Y be subse t s of P . We say t h a t t h e r e

-

i s a eamplete matching of X i n t o Y iff t h e r e ex l l s t s a 1-r map

cp : X + P such t h a t * - f o r a l l x E X , x zcp(x) o r , cp(x) 5 x .

We say t h a t t h e r e i s a complete matching between 2 and Y

i f t h e r e i s a complete matching of X i n t o Y o r t h e r e i s a complete *

matching of Y i n t o X . .'

*, Remark 1. By a famous theorem of Hal l [14], a comppte

matching of X into4 .Y e x i s t s i f f JR(A) I 2 I A ~ f o r every A c X , where - R(A) = {y € Y : t h e r e e x i s t s some x € A

wi th x 5 y o r .y 5 XI .

r Remark 2 . I f cp : P. -+ P I i+l is a complete matching, then

x 5 q(x) f o r each x E Pi .

Theorem 2 . 3 . I f P i s unimodal and t h e r e i s a complete

matching between Pk and Pk+l f o r a l l k , 0 5 k r (P) , teen P -

s a t i s f i e s the Sperner property.

Proof. Let F be an an t i cha in i n P . We proceed by

induct ion on

t ( F ) = max { r ( x ) - r(yY : x , y € F) . ( i ) I f t'(F) = 0 , then F C P f o r some k , t he re fo re

k

I F / 5 I P 1 5 max lpkl . k k

( i i ) Fix n >_ 0 , and suppose the above i s t r u e whenever - - - - - - -- - --

. t (F) 5 n , i . e . , every an t i cha in F wi th t (F) I n a

- - --

i n P s a t i s f i e s I F ~ 5 max lpkl . L k

I^_ - ? i s ,iifiimodal, t he re is an i n t e g e r m wi th

N o 5 N 1 5 ... 5 N t N & 1 2 ... r _ N m r (P> ' .

Also, t (F) > 0 implies t h a t the re is an x C F with r ( x ) # m . A s s u m e f o r t h e time being t h a t the re is x C F with r (x ) < m

and l e t i = min { r ( x ) : x C F}. L

Let F = F U Fly where 0 F~ = F n p i , F, = F \ F ~ .

Since i < m , then I pi I I 1 , - hence by assumption, the re

i s a complete matching ip : Pi + Pi+l. Define

h e n FO n 'P (F1) = 41 , s i n c e i f x C F and x C q(F1) , then 0

t h e r e e x i s t s y C F1 with q(y) = x . Thus x,y C F ' and y 5 x , -

con t rad ic t ing the f a c t t h a t F i s an q t i c h a i n .

Thus I F ' / = IF1 . Moreover, F' is an an t i cha in , s ince i f n o t , t h e r e is

- x C Fo , y C q (F1) and y 5 x . (This follows from r (x) > i and

r (y) = i+l. ) Now t h e r e e x i s t s z C F1 with ip (z) = y , s o jz I y . By t r a n s i t i v i t y , z 5 x , again con t rad ic t ing t h e f a c t t h a t F is an

an t i cha in . . 4

Thus F' is an an t i cha in and t ( F ' ) 5 t (F ) - 1 = n , t he re fo re

- - - - - -- - -

Ln t h e case t h a t min { r (x ) : x € F) 2 m y we l e t - - -

j = max { r ( x ) : x C F) , and proceed a s before , using a complete matching

Corollary 2.4. Boolean l a t t i c e s and vector , space lat t i c&- - - -

are kncm t o s a t i s f y t h e condi t ions of t h e above theorem and thus have

the spe rne r property.

The last condi t ion w e a r k going t o s tudy is t h e LYM proper ty ,

which, s t a t e s t h a t f o r every an t i cha in F of P ,

This i n e q u a l i t y is named a f t e r Lube11 [27 1, Yamamoto [ 3 3 ] , and

Mes chalkin [ 28 I .

Theorem 2.5. I f P has t h e LYM property, then P has t h e

Sperner property.

Proof. Assunre t h a t P has t h e LYM proper ty and l e t . F

b e an an t i cha in i n P . Then 6

I::

Since I P ~ ( ~ ) ~ 5 max , we have

i k

which g ives

T

* - A n a t u r a l ques t ion t o ask is what k ind of p a r t i a l l y ordered 3

> - ---- - - - 9 7

sets w i l l have t h e LYM property. From the proof of theorem 2.1, i t is $ 4

-- L - -

c l e a r t h a t every regu la r p a r t i a l l y , ordered s e t has the LYM property. - L -

We a r e going t o g ive two equivalent condi t ions , namely t h e nobnalized - matching condi t ion and t h e r egu la r covering cond i t ion . '

D e f i n i t i o n 2.15. A p a r t i a l l y ordered s e t P i s s a i d t o have a r e g u l a r

cover ing i f t h e r e e x i s t s a fami ly C of maximal cha ins of P (not

n e c e s s a r i l y d i s t i n c t ) s u c h t h a t , f o r each k , every element of rank k

occurs i n t h e same number of cha ins i n C .

~ e f i n i t i o n 2.16. A p a r t i a l l y ordered s e t P s a t i s f i e s t h e normalized

matching cond i t i on i f f o r eve ry k = O,l, ..., r,(P)-1 ,

I A I \ - < \ - f o r a l l A c P lPkl IPk+ll k '

$(A) = {Y Pk+l : t h e r e i s some x E A . wi th x 5 y ) . q&,

4P

Theorem 2.6. F o r Z a - p a r t i a l l y ordered s e t P , t h e fo l lowing

cond i t i ons a r e e q u i v a l e n t : . a L

3 (1) P has t h e LYM proper ty . 1

: (2) P s a t i s f i e s the normalized matching cond i t i on . -

-T 7 3

(3) P has a r e g u l a r covering. - 3

Proof. F i r s t we prove t h a t (1) Tmplies . (2) . Suppose P has 4 +

t h e LYM p rope r ty and l e t A c Pk f o r some k = O , l , ..., r ( P ) - 1 . 4 3 f

C l e a r l y F = A U (Pk+l - %(A)) i s an a n t i c h a i n . Therefore , P

* 3

so t h a t

\ 'i Hence P has t h e normalized matching p rope r ty .

Now we prove t h a t (2) imp l i e s ( 3 ) . Suppose P s a t i s f i e s i

\ t h e normalized matching proper ty . Le t

r (PI N = I[ ] p i ] .

'A- 1-0

- -- --

Def i n e a p a r t i a l l y ordered s e t P , 5 ' a s fo l lows . \

For every a € P , t h e r e a r e = &-, corresponding elements i n 4 rY

\/ .+ P ' , namely al ,a2, ..., a and ,

f (a) ' . \

-

Note t h a t t h e elements ai i n P' cor ;espondingto t h e e l e m e n t a t

i n P have rank r ( a ) i n P ' . C l e a r l y P ' i s a p a r t i a l l y rdered s e t and f o r each

k , 0 5 k 5 r ( P 1 ) , t h e number of o t j r a n k k i n P I - i s N . - Now we want t o show t h a t t h e r e e x i s t s a comp1e:e matching

between any two succes s ive ranks (k and k+l) i n P ' , where e k = O , l , . . . , r (P' 1-1 . By Remark 1 i n D e f i n i t i o n 2.14, we only need

t o show t h a t

x € A' with x 5' y) . Let A' c P; and let

a

A = {a € P . there is a; least one i with k .

Then

. . and

By the normalized matching cbndition in P , it follows that . -

-C

Now putting these complete matchings together for all

consecutive ranks in the natural way, we have a family of N maximal 3 1

chains in P' . By rephcing every a ~ 7 - PI- b y a T T ~ r r e 0 6 t a i K B ~ 1

\ family of U f k a i n s h + - a h t- ilemmtafnmlr k

- - in P occurs in ex+ctly - of these chains. Thus P has a regular

lPkl -3 covering. 3

$ 4 3 9 . ". 3

-* 2 2 - i.T,

Let F be an antichain in P . hen no two elements of F

can occur in a maximal chain. Therefore g

regular covering C. Then every element of rank k occur

Ict - maximal chains in C . - I 'k 1,

Therefore P %has the LYM property.

In par.tially ordered sets satisfying the LYM property, many I

results in extremal set theory can be deduced. We begin with the .

following theorem.

Theorem 2.7. Let' P be a partially ordered set with* - - - - - - -- - - - --- - -

regular covering C of P by chains. Let X be a real-valued

, function defined on P. Then for any subset F 2 P ,

C - n max . Z A . XEF Nr(x) CEC yccm Y . .

Proof. Consider9+the sum

C C X . ' CEC x€CflF

X

_ - - -- - -- Since C isaregularcoverlng, an element of ,rank k in P

s o we have

I c I max .Z A . CEC xECnF X

Upon cance l l ing t h e *

c.

we get t h e theorem.

- 1 -+ " 7.

- - term ~ c I roqrboth s i d e s of t h e inequa l i ty ,

r egu la r covering C of P by chains. For any subset F c P , '\ -

Proof. By p u t t i n g = . - - x Nr(x)- i n Theorem 2.7, w e have

rnax C 1 ' 1 ccc X E C ~ F r ( x )

-

Theorem 2.7 has many powerful intpl icat ions f o r p a r t i a l l y

ordered s e t s s a t i s f y i n g the LYM proper ty , because we a r e f r e e t o cho,ose

bo th X and t h e r e s t r i c t i o n on F . We give severa l r e s u l t s where P ? '

X

Theorem 2.7. \ - \

,

-

subsets of an n-element set S such that there exists do

Let F be a family of

distinct

chais of length r+l . Then

IF1 5 sum of the r largest binomial

coefficients of order n.

,

Proof. Let C be a regular covering of the subsere of

by chains. From Corollary 2.8,

max Z I F / ' ccc xccns Nr(x, .

- ---- - - - - --- -- -- - - -- - - -

In this case, =r any x E S , . 9

~lso, since C is a chain,

, I c ~ F J 5 r.

/ - - -

Thus the ~rd& result follows.

F be a family of subsets

of an

Then

n-element set such that no two members A , B in F satisfy

A c B with ,/A - B I 1 r. -

sum of the r largest binomial

of order n . coefficients

again I C fl F I 5 r , where C is any chain.

* ?j

\ 5 \ 3

I -

- -

- - - - - - - -- - 4

29 . - *

C o r o T ~ a ~ 2 ~ ~ F be a famil; of -.. - subse t s of an n-element s e t S such ' tha t no two members . A , B i n . -

F s a t i s f y .

Then

A q B with I A - B J < r .

Proof. Again we apply'Teorem 2.8 t o t h e p a r t i a l l y ordered \

s e t 'P of a l l subsets- - - - - - of - an - n - e l ~ n ~ s e t . , u s i n g - - a r e g ~ l a r r ~ v e r i n g - - - ----

C of P by chains. A s before, '\

and

The r e s u l t fol lows; \

Theorem 2 .7 is derived by making use of Lubel l ' s method and a

3

it turned out t o give many i n t e r e s t i n g r e s u l t s . It is poss ib le t o ,

extend t h e idea of the LYM proper ty t o formulate a moregeneral theorem,

- - - - -- - -- - - - --- - -

which w i l l give many r e s u l t s i n extremal set theory. We f i r s t de f ine

sets a = (A A A ) whereeach Ai is a subset of S. 1' 2' P

~efidition 2 .18 . s is the collection of all ordered set systems

t 8. which can be obtained from by the elements in S .

Let n be the permutation of S defined by n(1) = 3 , ~ ( 2 ) = 2 , A

LJ -F - 3 = 1 Then the a lcation of .rr to S transforms a into the

-f

set system B = ( {3} , {'2,31, 1 1 ~ 3 1 , {31 1 . -t -

- -

Theorem 2.12. ~ e t a = A , . A be an ordered *set -

" P

system on s (IS[ = n) , qith I A ~ I = a . f z i s p . Let F bea \ a 1 ' '\

family ef'subsets of S sucp that F has at most k members in

'\

cokon with each 3 € s(;) . h ~ t f denote the number of sets in j

F of size j . Then '\ \ \ fa '\

\ i Z - < - - k. - -

i=1 ( " ) a:

Proof. For each

Then

: the ith component of

3 is an element of F 1 . -r

f-

.which gives

as required. , /

+ Remark. If a represents a maximal chain of sets in the

-+ lattice of subsets of 1 2 , . . n ; i .e., a = (AI ,A2,. . . ,An+l) where

- -

A1 = 0 C A2 c . . . c An+l and F is an antichain, then Theorem 2.12

gives<he LYM property and thus implies Sperner's theorem.

Now we Can deduce some results in extremal set theory from

b --

Corollary 2.13. Let S be a set with n elements. Let F --

be a family of subsets of S such that

A n B # 0 forall A , B E F .

Then

-b Proof. For each i , 0 5 i 5 n , let a . = ( A ~ . s-A~) ,

1

where A. is any fixed i-element set. Then from Theorem 2.12, we have 1

-P since at most one element in a. can appear in F. .So

1

n fi + f 5 , 0 5 i 2 s n . n-i

- - - - Adding these' inequalities for i = 0 , 1, ..., n , gives

subsets' of

(Of course, a simple proof can be obtained by putting all the

into pairs The family can contain

at most one set from each pair. The proof given here is included for

the sake of "practice" in applying Theorem 2.12.) - 1 - -

Corollary 2.14. (~rd6s-KO-Rado 191). Let S be a set of n

elements. Let F be a family of subsets of S such that

(i) for all A E F , I A J = k 5 "/2 Hnd

Then

Proof. Consider a cyclic ordering of the elements of

S , (s1,s2,. 1. ,sn, sl)., S. # s for i # j . Let = (A1,A2,... ,An) , 1 j,

where Ai consists of the k consecutive elements, according to this -/

ordering, starting with s. . 1

. We shall show that F has at most k elements in common

-h 3 with a . By symmetry we can assume A1 T F . Then-the sets in a

-- --

But since k 5 n/2,

having non-empty' intersection - Therefore, of the (2k-2) A.'s 1

with A only k-1 of them can be in 1' F . So F has at 'most k

-+ elements in common with a .

-- - --- +- - -

By symmetry, no sequence f3 E s(;) has more than k members

in common with F . Therefore by heo or& 2.12, with a = C% = 1 2

...=a = k , n

we have

We now prove an analog of Theorem 2.7 for ordered set systems.

4 .. Theorem 2.15. Let F - b e a family of subsets of an n-element

- - - - - -

set S. Let hl ,T2,. . . , X beYG1-&bers (one can think of Xi as a n

-f ''weightu-Ct-t-ecs of size i ) and let a= (A1,A 2y...,A

P

be an ordered set system on S with ] A i 1 = ai for i = 1,2,. . . ,p . Let fa be the number of sets in F of size a. . $hen.

i 1

p C l i < - max

-f i=1 (n) BC s (g) L r m J

a,

Proof. Consider the sum

As in the proof of Theorem 2.12, let ,

f(i) = {d € s(;) : the ith component of

is an element of F} . Then for each i , 1 5 i 5 p , we have a contribution of

f a.! (n-a.)! A a. 1 1 a. 1 1

to the above sum. (Thus we are just interchanging the order of summation.)

Thus the above sum equals.

C f a . ! (n-a,)! A . a. 1 i=l 1 I. a .

1

Since

and

we have

which gives

f a . ! (n-a.)! A n ! maxi a. 1 1 1 i BES(a) ( A E F ~ ~

I-

and the proof is complete.

We shall use the above theorem to deduce some resuIts of

Kleitman [ 111 and Bollobas [ 31.

Corollary 2.46. (Greene, Katona, Kleitman [ll]).

Let S be a set of n elements. Let F be a family of

subsets of S such that

(ii) for all A,B € F , A # B implies A @ B ,

(iii) for all A,B E F , A n B # 0.

Then

where f. denotes the number of sets in F of size i. 1

Proof. Consider a cyclic ordering of the elements of S ,

where Let

where the A.,'s consist of all possible segments, of all lengths, 1

of consecutive elements of the cyclic ordering. --- 1 Applying Theorem 2.15 with A . = , we get

1 1

L P a. C 1 5 max -+ i=1 ai(an.)

1

+ + ~ r & the definition of a , there are n A: '.s in a of

n size i for each i = 1,2,. . . . Also, by condition (i) of the

-+-- corollary, f = O if a . >n/2. a. 1 - 1

Therefore, the leit side of the inequality is

Now cons ide r t h e r i g h t s i d e of t h e i n e q u a l i v , and suppose t h a t

t h e s m a l l e i t A occur r ing i n F n 3 has t h e s i z e k . Then j u s t i s i n

t h e proof of t h e ~rdos-KO-Rado theorem (Corol la ry 2.14 above) , i t fo l lows

t h a t IJ? fl $ 1 5 k .

Therefore

f o r each 3 t s(;) , so r .

-. as r e q u i r e d .

\

Remark. For t h e proof of t h e next r e s u l t , i t i s canvenient

t o r e s t a t e t h e l a s t conclus ion a s

Coro l l a ry 2.17. (Bollobas [ 3 ] ) . Let S be an n-element, s e t .

Let F be a fami ly of s u b s e t s of S such t h a t

, ( i ) F i s an a n t i c h a i n ,

( i i ) f u r a l l A 6 F , S-A f F . Then

. . S w(A) 5 2 ,

A f F

where *

-

' , Q

37. . -

Proof. Let F be a family of subsets satisfying the above

n . n conditions. Let A,B € F with I A ~ s 2 , I B I 5 - . Then since 2 ..

S-A € F , B $ S-A . heref fore A f l B # (3. Applying Corollary 2.16, . C

Z t

1 AEF 5 1 . n-L I A I 5 - 2 - (/A[-1 1

n Clearly the elements of F having sizes greater than 7 .

must be only the complements of all the sets in F of size less than

n - 2'

therefore, applying Corollary 2.16 again (to the collection of - n -

complements of the elementsof F of sizes greater than 7 ) , we obrain

Adding the two inequalities gives the desired result, since ?

n if 1 5 a r n , then a > - if and only if ( n- 1

2 n-a-1 > < P-5 a- 1

I

corollary 2.18. (~chgnheim [ 3 1 ) ) . Let S be a set of n

elements. Let F be a family of subsets of S such that

(i) F is an antichain,

(ii) for all A,B € F , A fl B # 0 and A U B # S . i%

Then

Proof. Let F be a family of subsets satisfying the above

conditions. Let

F' = { B C S : S - B C F } . - Consider A € F and B € F' .

Cleadly A f B and B $ A , for if A c B , then - - (s-B) fl A = 0 and (S-B) € F , which is impossible.

Also if B c A , then (S - B) U A = S , which is impossible. - Therefore F U F' satisfies the conditions in Corollary 2.17,

and we get

But

for all sets A . Therefore

A

But I F U F'I = 2 IF^ , so ,

and the result follows. '

. -

CHAPTER 3

The Kruskal-Katona Theorem

C

Ln this chapter, we describe some results which were first

found by Kruskal [25], and later independently by Katona [16]. We

begin by stating the question:

Let F be a family of R different subsets (each of

cardinality k) of a finite set S , where I S I = n . Let

A F = {B : I B I = k-1 and B c A for s,ome A € F).

What is min I A F ~ if n,k,R are fixed?

. This problem has a very simple answer that, when looked at

in the right'uay, 'implies the ~rdgs-KO- ado theorem, and many other

refinements . The first b w e r bound far / AF/ , -

' ,

was given by Sperner.

The above can be deduced by counting the number ofv"containment

'3 trhere B C AF , A ( F and B c A.

.?%

*i

-Fdr eaqh-element A of F , we can discard any one of the) k 1 .

elements of A to give a set in AF , therefore there are exactly k\Fl

containment pairs'. * * . - - - - - --

For.each element B of AF, we can add any one of the

n-(k-1) elements of S - B to B to give a set in F , therefore there - ,

are at most (n-k+l) ~ A F I containment pairs.

Spernerts lower bound for I A F ~ depends on n , which does

not seem to be necessary. Kruskal and Katona give the exact value of

min I A F I , which does not depend on n . We now state the Kruskal-

Katona theorem.

Theorem 3.1. (Kruskal [25], Katona [161.) Let F be a

family of k-element sets. If

where

then \

where + AF = {B : I B I = k-1 and B c A for some A E F) .

Before we proceed to prove the theorem, we need the following

interesting lemma.

where

Lemma 3.2. Given any two natural numbers R and k , there

exist unique natural numbers ak,akal, ..., a such that t i

7

1 a

\

a - >

k %-l) q ... + ( tt) , I Q = ( , ) +(k-l \ t

\ *.

\ \ i , -

- - \- -

t r l \, 3- a k > %-1 > > a t '

- - - -

' 3 -

\ % t

a. i for i = t,t+l, ..., k . 1 3

' A - --

Proof. (aT We prove the existence part by induction over k.

(i) For k = 1 , put. 4 = 2 .

(ii) Assume the statement is true for k=m-1 , m 2 2.

Let a - be the maximal integer satisfying the inequality \ -

a 1f' ( m, = , then we are done.

m \ \ ,' * , . Otherwise, 2 - ( am ) is a natural number and we can apply , m 1.

the anduction hypothesis to get >, .

where \ t r 1 , -

a > a > ... > a and m-1 m-2 t

a 2 i for i = t,t+l,...,m-1 . . i -- - -

From the above,

* .

We need only verify a > a and a I rn . m m-1 m

a a Suppose a 5 a . Now since L 2 ( ) + ( m-l m-l ) , then m m-1

contradicting the choice of a . Therefore +> -ap-%-- m - -- - --

Since s-T?~l a r r d - b - -ine$u?lit m m - F --

then follows.

(b) The uniqueness part is also proved by induction over k . '\

\

# he' case k ='l ' is trivial. Now asswe that the statement' \ is true for k=m-1 , m 2 2.

Suppose the expression is not unique for k = m . Then

and

with

We consider two cases.

Case (i)

a m Z

Then R - ( has two different expressions, contradicting m

the induction hypothesis.

Case (ii) am # bm .

Since b > a b s a + 1 , therefore m m' m m

C

which is a contradiction.

Therefore the expression is u&ique for k=m . Thus the 1

-A '\

proof of the lemma is complete. $

1

We a r e now ready t o prove Theorem 3.1.

Proof of Theorem 3.1. ( a ) F i r s t we show t h a t t h e r e indeed e x i s t s a

family F where

-

wi th

Without, l o s s of g e n e r a l i t y , let t h e s e t s i n F be s u b s e t s of

1 . n . We in t roduce an o rde r ing f o r k-subsets of {1,2, . . . ,n} a s f oliows :

Define

1

Then

F& X , Y being k-subsets ,

X < Y i f max { a : a E (X-Y) U (Y-X)} E Y .

F = Ek U FkAl U ... U Ft , w h e r e

Y . = U { I + a . } f o r i = t , t + l , . . . , k . 1

i < j ~k 3

a . Obviously, IF^ I = ( I) f o r i = t ,t+l, . . . ,k , and t h e F. 's. a r e

1

pai rwise d i s j o i n t . Thus,

Note t h a t with respect t o our l i n e a r order ing , the l a s t s e t - *

i n F i s

L .

By the d e f i n i t i o n of F , every s e t of c a r d i n a l i t y k l e s s than = j --L

[ l - t + a t , a t ] U Yt i s i n F . This means t h a t F c o n s i s t s of tfrs

f i r s t I F / s e t s of c a r d i n a l i t y k . ' - P. -

Next w e must show t h a t

Since

we have

Thus

Again no te t h a t t h e last s e t i n QF is obta ined by d e l e t i n g

t h e f i r s t e l e m e n t from t h e l a s t s e t i n F , hence i t is --

- Also i f d a s e t of s i z e k-1 i s l e s s than the l a s t s e t i n A F , i t i s

a l s o i n AF . Therefore AF c o n s i s t s of t h e f i r s t I A F I s e t s of

c a r d i n a l i t y k-1 . - -

(b) Let F be t h e f m i l y o f t h e f i r s t P s e t s of c a r d i n a l i t y

k and l e t F be any o t h e r family of k s e t s of c a r d i n a l i t y k . To -

prove t h e theorem, we need t o prove l h ~ l 5 I A F I . . Let m = m i n {~w-xI : I w I = k , W Q F , X E F and W < X I .

S ince F # F , t h e r e f o r e m 1 and t h e r e e x i s t s Wo , Xo such t h a t

m = 1 w0 .- x0

Now l e t M =

\ Next put

wi th = k , w, Xo E F and W < X O . 0

Wo - Xo and N = X - Wo . Obviously, y

J M ~ = I N ! = m and +

B = { X : X f F , M ? X = O , N Z X , ( X - N ) U M P F ! ,

C = { ( x - N) M : X f B L and

D = C Q ( F - B ) .

of M and 'N ,

'Moreover,

3 Thus D # F . - /

- The way t h a t we.obtain D from F i s by rep lac ing N a s a --

_, subset of an element of F whenever poss ib le by M . Note t h a t every

time when a s e t X i s replaced by (X - N) U M y then (X - N) U M < X .

I n p a r t i c u l a r Xo/ is replaced by Wo which is < X s - 0

Since D # F , so D i s nea re r t o F than F , i n t h e sense i r

t h a t by repeatin+the above process , F w i l l be transformed t o F ,

- d >

eventual ly . Hence t h e theorem w i l i fol low i f we can show t h a t / A D [ 5 1 A F ~ . Let Z E (AD) - (AF) ,. we s h a l l show t h a t

( i ) Z n N = 0 and A

( i i ) M c Z . +* L

b \ i

To prove ( i ) , observe t h a t i f . Z € - (AD) - (AF) , then t h e r e

e x i s t s a s e t Y E D-F = c such t h a t Z c Y . By t h e def in i t - ion of C ,

t h e r e e x i s t s a s e t X k B such t h a t Y = (X - N) U M . Obviously Y ' -$

Y Q N = @ . Since Z c Y , then Z fl N = O . 3, * k

. To prove ( i i ) , suppose M q! Z . Then from the f a c t t h a t ,

/ z / = / Y / - 1 , Z c Y and ?I C Y , we have I M f l Z I = m - 1 . Now l e t

P = M f l Z , and

Q = N - { B : 6 i s t h e smal les t in t ege r i n N) . 'b it:

Then I Q I = m-1 . Also s i n c e M fl N = 0 , t he re fo re P fl Q = 0. -

Now e i t h e r P = Q = 0 o r P < Q . Since X € B y the re fo re M ; f l X = $

and N = X , so P O X = 0 and Q c X . Let W 1 = ( X - Q) U P , then -

Iwl - X( = I P / = m-1 and

By t h e d e f i n i t i o n of , W1 E F . . But from Z R N = 0 and Q c N , we have Q fl Z = 0. Also M f l Z = P ,

therefore- Z c W g i v i n g t h e c o n t r a d i c t i o n Z E AF. Thus M c Z . 1"

To prove ]AD1 5 I A F I , we s h a l l f i n d an i n j e c t i o n

cp : (AD) - (AF) i n t o (AF) - (AD) . If such a cp e x i s t s , t hen we a r e

done. - -

r---' For every Z E (AD) - (AF) , d e f i n e

q(Z) = (Z - M) U N .

S ince M c Z c Y = (X - N) U M , t h e r e f o r e - ( z - A) U N c X and I(Z - M) U N J = k-1

l . e . , cp(Z> c X and IQ(Z) 1 = k-1 , - - -

l . e . , . cp(Z> E AF . From t h e two p o i n t s we j u s t shown , namely, Z f l N = Ql and M c Z ;

z q i s an i n j e c t i o n of (AD) - (AF) i n t o AF . To cbmplete ou r prpof ,

we. c laim t h a t q(Z) # AD . Suppose t h e c o n t r a r y , i . e . , cp(Z) € A D . Then t h e r e e x i s t s

a s e t V such t h a t cp(Z) c V E D . Therefore , N c V . So from t h e

d e f i n i t i o n of D (D = C U (F - B)) , V (! C b u t V E F - - B .

Since V # B , u s ing t h e d e f i n i t i o n of B , we have k i ther - - -

i M fl V = 0 and (V - N) U M E F .

C ~ n s i d e r c a s e ( i i ) , s i n c e (Z - M) U N c V , t h e r e f o r e y8. Z c (V - N) U M , i - e . , Z € AF, which i s imposs ib le by t h e d e f i n i t i o n

of z .

M n ~ ( 2 ) = @ , s o M fl V must be {a} i n o r d e r t o s a t i s f y M V # Ql . A s b e f o r e , l e t

Q = N - { B : B is t h e smallest i n t e g e r i n N ) . Then

Q c N c V and P f l V = @ . Next l e t W2 = (V-Q) U P . C lea r ly

I w 2 / = / v / = k and

I w - V I = / P I = n i - 1 and F 2

W* 5 V c F ' . .

1 By t h e d e f i n i t i o n of m , then W &st be i n F . S ince a C V , we

2 (

have M c W Therefore Z c W (from W = V - Q ) U P and 2 2

Z c (V - N) u M) , and s o Z C AF , which is impbss ib le .

Thus $ (Z) P! AD , and (D i s an i n j e c t i o n of (AD) - ' ( A F )

i n t o A - A D ) . So \ A D ] 5 I A F ~ , and t h e proof is complete.

Many i n t e r e s t i n g r e s u l t s of ex t remal s e t t heo ry fo l l ow from b

Theorem Ld; w e a r e going t o l i s t two of them. The f i r s t one is t h e

~rdos-KO- ado theorem.

Coro l l a ry 3.3. Le t S be a set of n element$ Le t F be

a fami ly of k-element s u b s e t s of S such t h a t

> ( i i ) f o r a l l A , B E F , A n B # 0 .

- 1 . L e t Proof . Suppose t h e c o n t r a r y , i. e . , I F I >, (k-i

G = { S - A ' : A E F) and

H = { B : I B I = k and B i s a s u b s e t o f s o m e s e t i n G I .

Then G c o n s i s t s of s e t s of s i z e n-k and

n-1 (n-k) '

t imes , By Theorem 3.1, app l i ed n-2k

n-1 (n-k - (n-2k)

Theref o r e t

So F Yl H # 0 , and t h e r e is a s e t A in F and in H . But s i n c e t h e I h

t h e r e is a s e t B i n G con ta in ing A , so by t h e s e t A is , i n H ,

d e f i n i t i o n of G , S - B is in F , which impl ies

( s - B ) n A = 0 .

proves t h e theorem. This c o n t r a d i c t i o n

The fo l lowing c o r o l l a r y s o l v e s a con jec tu re of Milner .

Coro l l a ry 3.4. Le t F be a family of k-element s u b s e t s of

an n-element s e t S and l e t G - b e a family of R-element s u b s e t s of

S wi th k+R 5 n . If no member of F is d i s j o i n t from a member of

G and 1 F J 2 (:I:) , then

Proof . The argument i s the same a s t h a t f o r Coro l l a ry 3.2.

?'

Remark. P u t t i n g Q=k i n Corol la ry 3.4, w e o b t a i n t h e

~rdos-Ko-Rado theorem (Corol la ry 3.3) .

A p p l i c a t i o n s "

I n t h i s c h a p t e r we s h a l l g i v e some a p p l i c a t i o n s o f S p e r n e r ' s

I

theorem and t h e , r e l a t e d r e s u l t s . The f i r s t problem i s t h a t of L i t t l ewood

and Offord . I n t h e i r s t u d y o f t h e z e r o s o f random po lynomia l s i n 1943,

t h e y r a i s e d t h e q u e s t i o n : 4

Let a l , a 2 , ..., a be complex numbers w i t h ' a ' 2 1 , n i

1 5 i r n . What i s t h e maximum number of l i n e a r combina t ions o f t h e . form

n A

- . a. 1 1 i= 1

where . = 1 , which can l i e i n s i d e t h e u n i t c i r c l e ? 1

The answer t h a t L i t t l m o d and Offord [ 2 6 ] gave i s i - -

C 2 " ( l o g n ) n 2 , and ~ r d o s 181 in h i s paper w a s a b l e t o g i v e a s i m p l e

s o l u t i o n t o t h e one-dimension c a s e , which i s t h e f o l l o w i n g .

Theorem 4 .1 . Let a l , a 2 , . . . , a be n r e a l . numbers w i t h n

n -- - l a . ' 2 1 , 1 s i n n . Then t h e number o f sums - E. a i l which 1 1 1

i=l

f a l l i n t h e i n t e r i o r of an a r b i t r a r y i n t e r v a l I o f l e n g t h 2 does -

n o t exceed G I ) -

P r o o f . W e can as-sume t h a t a l l t h e a 's a r e p o s i t i v e , s i n c e i

.changing t h e s i g n of an a . does n o t change t h e set o f l i n e a r combina- 1

n t i o n s . To e v e r y sum I E a . we a s s o c i a t e a s u b s e t A o f t h e s e t

i=l i 1

1 , . . . n a s f o l l o w s :

i E A i f f E . = + l . 1

n On - , Consider two d i f f e r e n t sums C E a and Z E . a . which

i i i=l . 1 1 i=l

a r e b o t h i n I . Let A and A ' be t h e c o r r e s p o n d i n g s u b s e t s o f

1 , n r e s p e c t i v e l y . We c l a i m t h a t A f A ' and d A . For i f

A z A ' , t h e n t h e r e e x i s t r l , r2 , ..., r f o r some s 3 1 such t h a t S f

- - - . . . - - i - t- = - - - = -1 , and r r r 1 2 S

-

f o r

T h e r e f o r e

which is i m p o s s i b l e . S i m i l a r l y A ' $ A . -

From t h e abdve argument , t h e c o r r e s p o n d i n g s u b s e t s o f a l l t h e

l i n e a r combina t ions which f a l l i n t h e i n t e r i o r o f I mus t - fo rm a n a n t i - - u

c h a i n . By S p e r n e r ' s theorem, t h i s number must n o t exceed *

t h e theorem f o l l o w s .

Remark. The upper bound i n Theorem 4 . 1 i s b e s t p o s ~ i b l e +

--dr s i n c e i f we choose

( i ) n t o be even , i . e . , n = 2k f o r some k ,

( i i ) ai = 1 for i = l Y 2 , . . . ,* ( i i i ) I = (-1, + l ) ,

n t h e n t h e sum C E . a . will be i n I

1 1 i=l a r e +1 , and t h e r e a r e ( ) such sums

i f and o n l y i f h a l f o f t h e ti's

From Theorem 4.1, ~ r d o s con jec tu red t h a t t h e same upper bound

w i l l hold f o r t h e c a s e s of two and t h r e e dimensions. Katona 1151 and

Kleitman [21] independent ly proved the ca se of dimension two, which i s v

t h e Litt lewood-Offord problem and l a t e r Kleitman [ 2 3 ] gave a proof f o r

a r b i t r a r y dimensions. The proof of t he Litt lewood-Offord problem makes

u s e o f theorems i n t h e f i r s t c h a p t e r .

Theorem 4 .2 . L e t a l , a 2 , ..., a n

1 a . l 1 1 . and E . = 2 1 , 1 5 i 5 n . Then 1 1

rl

form z . a . which l i e i n t h e i n t e r i o r 1 1 i= 1

r a d i u s 1 i s a t most

be complex numbers w i t h

t h e number of sums of t h e

of an a r b i t r a r y c i r c l e of

Proof . We may assume R ' ( a . ) 1 0 f o r i = , 2 n , s i n c e e 1

changing t h e s i g n of a does n o t change t h e s e t of sums. Let i

n T o e v e r y s u m C & . a a s s o c i a t e a s u b s e t A o f t h e s e t

1 i ' i=l

, 2 . n a s fo l l ows :

i € A i f f E . = +l . 1

t Consider two d i f f e r e n t sums C E . a . and C E . a which a r e both

1 1 i=l i=l 1 i *

i n a f i x e d u n i t c i r c l e . C l e a r l y ,

Le t A and A ' be t h e subseCs corresponding t o the two sums. We want

t o show t h a t i f A fl H 3 A ' f l H2 then A f l H1 # A ' fl H1 . 2

4 * \ 'i

- 1 . " \ 53. \

To prove this, suppose that A fl H 2 A' f l H2 and 2. & \

A ? H = A' n H then 1 1'

\

Since A 1 HZ -.

for any i € H' 2

are two complex

A ' fl H2, we have c. - E:) = 0 or ( E ~ - €1) = 2 1 1

. Also from the theory of complex numbers, if u and v i \

number such that Re(u) 2 0 , Re(v) 1 0 , Im(u) -= 0 , / . I (v) < 0 , then

m

f t i + v t 3 min {tuf , ivij .

Therefore

I ( E ~ - ~:,)a.\ 2 min {21ai/.) 2 - 2 , I i C H 2 iCH2 --

contradicting the assumption that the two sums are in the interior of a

unit circle.

A similar argument will show that A fl HI 2 A' fl H1 and. '1,

A Q HZ = A ' f l H, cannot both be true. Thus the corresponding subsets n

of the sums : E. a. satisfy the conditions in heo or em 1.9 of Chapter 1 - 1 1 i=l

- and therefore the number of sums cannot exceed

It would be interesting to know if the length of the interval ',

in Theorem 4.1 is different from 2, what will happen to the upper bound?

~rdos [8] solved the above problem, which is

Theorem 4.3. Let r be any integer and let al,a2, ..., a n

be n real numbers with lai/ t 1. Then the number of sums n T.

L E . a. (E. = + 1) which fall into the interior of any interval of 1 1 1 i=l

56,

length 2r is not greater than the sum of the r largest binomial

4

cAefficients of order n . B

1 Proof. Using the same argument as in Theorem 4.l,,we can

Tt t

show that for any two different sums 2 E. a. and C E. a. lying 1 1 i=l 1 1 i=l -

in the fixed interval of length 2r, their corresponding subsets A

/ and A ' cannot satisfy bogh A 3 A ' and I A - A ' [ 2 r. The result -

-

follows immediately from Corollary 2.10.

Remark. Again the upper bound in Theorem 4 : 3 is best possible

since if we choose a. = 1 for i = 2 , n, the maximum is attained. k: 1

- d o n a also considered a different type of problem involving n

sums C E . a. . The 1 1 i=l

Theorem 4.4.

positive integers such n 1 E . a. ( E ~ = 0 or

1 1 i=1

following theorem states his result.

(Katana 1191). Let al,a2,. . . ,a n' a,b , be

that 1 5 a. 5 a. Then the number of sums 1 .

1) which.are congruent modulo ab is at most

E n (2) .

iE (mod b)

\ Proof. Consider all the sums, C E. a. c (mod ab) for 1 1 i=1 ,

some c. Let A Y E be two such sums and let

Then if F 3 G and JF - G I < b ,

A - B = E a - Z ai = E',ai5 IF-GI a c a b . iEF

i iEG i€ F-G;,,

\ \ But - B ] 2 1 , contradicting that A is\ ongruent to. B (mod ab) . 8

So there is no F , G satisf;ing F i G and 4 - G I i b . ~ ~ p l ~ i n g

Corollary 2.11, the number of such sums is at mos a, C n n

i-17] (mod b) (i' . -

We conclude the Littlewood-Offord problem by stating some

results (without proof) with different lengths of the radius, i.e., n

the maximum number of linear combinations of the form C E. a.(~. = f l 1 1 1

i=l

and the a. 's are complex numbers with la. 1 1 1) that can lie within 1 1 -

a circle of radiusa r .

(i) For r = 1, the upper bound is 6

([;J , as can be seen from

Theorem 4.2.

(ii) For r = , the upper bound is the-sum of the largest

two binomial coefficients of order n.

(iii) For r = 6 and n 1 5 , the upper bound is the sum of

the largest three binomial coefficients of order n.

(iv) For r = k , the upper bound is the sum of the largest

" 2[kfi ] binomial coefficients of order n .

for fixed n , Jhe number oa linear combinations inside a circle can

grow at most linearly with the radius, and does not grow linearly with ,

.the area of the circle.

5 6. -

/ - -

We now turn our attention to number-theoketic-problems. We \

\ have seen from Chapter 1 that de &uijn made use of Sperner 's theorem

to prove that the number of divisors of a natural number m with

degree n satisfying the condition that no divisor divides another

n must not exceed the number of divisors of m whose degree equals I T ] .

De Bruijn, in his paper [ 4 1 discovered another result.

First we need some definitions. - -.

Definition 4..1. Let rn be a natural number. A set S of divisors

of m is convex if

a € S , c E S and alblc imply b € S.

Definition 4.2. The ~gbius function, whose domain is the set of natural -

numbers, is defined as follows:

(i) ~ ( 1 ) = 1 , -

fii) ~ ( m ) = 0 if m has a square factor,

k (iii) u(plp 2...pk) = (-1) if all the primes p1,p2, ...,

Pk

are different.

Now-we can see that the following theorem is an immediate

consequence of Lemma 1.10.

Theorem 4.5. Let u be the number of different primes

dividing a natural number m . Let S be a convex set of divisors 8

of rn . Then

- - P r o o f . S i n c e p ( a ) = 0 i f a i s n o t s q u a r e - f r e e , we may

r e s t r i c t o u r s e l v e s . t o t h e c a s e t h a t m i s s q u a r e - f r e e . T h e r e f o r e t h e

d e g r e e o f m i s v . By Lemma 1.10, t h e s e t o f d i v i s o r s o f m can

be d i v i d e d i n t o mut&ally d i s j o i n t symmetric c h a i n s

v where n = .

L e t where e a c h f o r

i = 1 , 2 , ..., n . -

S i n c e 'S i s convex, each S . i s e i t h e r empty o r c o n s i s t s 1

of a number o f c o n s e c u t f v e e l e m e n t s of C . . Now u ( a ) assumes t h e 1

v a l u e s +1 and -1 a l t e r n a t e l y a s "a" runs th rough Ci . Hence - P

1 2 ( a ) = 0 o r 1 o r -1 . -aC Si

/

a s r e q u i r e d .

Schgnheirn [ 301 extended de B r u i j n ' s i d e a t o g e t r e s u l t s i n

n ~ r n b e r ~ t h e o - r y which a r e s i m i l a r t o t h e g e n e r a l i z a t i o n s of S p e r n e r ' s

theorem i n Chap te r 1. Before we s t a t e h i s r e s u l t , we d e f i n e some te rms .

D e f i n i t i o n G . 3 . Let rn be a n a t u r a l number w i t h d e g r e e n . The

k number o f d i v i s o r s o f rn wi? degree k will be deno ted by Sn(m) .

~ e f i & t i o n 4.5. A sequence of n a t u r a l numbers a 1 < a2 < - ' : < ak

i s c a l l e d a cha in of l eng th k i f l

a . I ai+l f o r i = 1 ,2 , ..., k-I 1 -

Reca l l t h e d e f i n i t i o n of symmetric cha ins and Lemma 1.10 i n

Chapter 1. Let U be a s e t of mutual ly d i s j o i n t symmetric cha ins

~ a r t i t i o n i n g t h e d i v i s o r s of a n a t u r a l number q wi th degree n .

I t i s obvious t h a t t he fo l lowing i s t r u e .

-----

obse rva t ion 4 . 6 . Every member of U con ta ins a d i v i s o r of degree

n Observat ion A . 7 . Any d i v i s o r of degree of m i s placed i n t he

*

middle o f a c e r t a i n member of U.

Observat ion n .8. js:(m) = s:-' (m) . ' P -

n 1 2 I$

Observat ion L . 9 . The sequence S (m),Sn(rn), ..., S ( m ) i s non-decreasing. J n n

Lemma A.10. I f a member C . of U c o n t a i n s a d i v i s o r of 1

n degree s 5 - then Ci con ta ins a t l e a s t n-2s d i v i s o r s of m of

2

degree higher than s .

Proof. Let C . = { a a L 2 , . . . , % l . Since d e g ( a l ) + d e g ( % ) = n ,

w e have >

The number of d i v i s o r s of m i n C having degree h igher than s is exactly i

B u t deg(a ) I s , t h e r e f o r e in Ci 1

there are a t least n-2s

d i v i s o r s of hegree h.igher than s .

Ttteorem 4 - 3 1 . L m be a natural number with degree n . Let A = ial,a7, ...,\ ! be a set of divisors of m such that no subset - of A is a chain of length c+l . hen

i k 5 sum of I largest S (rn) .

El

io+Z i0+f.-1 ~rooi. ~ e t sA0(rn) , sn (rn] , ... , S (m) be the 2

n i

latgest values of S (m) . Then by Observation 4.9, we can choose n

1 i0 5 (n-~+2) . Suppose there exists a: ; A

1

L e t

such that deg(a.) i i 1 0 -

= j and

L' = :C ,C,,...,Ct ] be a partition of all divisors of m into 1 ,

symmetric chains, Each element of A is cuntained.in exactly one j

member pf U . Suppose a C A is contained in v CV, then by Lemma 4.10,

- . j . .

C contains at least n - 2j divisors of m of degree higher than j . v

Since j + C cn-j , then n-2j 1 3 . A

I" '

Moreover C is a symmetric chain, so there is an element in v

C with degree at least j + k . Denote this element by x . Obviously v v - the portion of C from a to x is a chain of length k + l .

v v v

Therefore there exists 'an element v

of Cv (between a and x v v

inclusive) which is not in A . Choose Yv

to have smallest degree.

Let 1

A = the set of all such yV1s obtained by j

considering each element of A . j

I

Consider (A - A . ) U A. = B . ' - -

J 3

Clea r ly [ B I = I A ~ , and B s a t i s f i e s t h e requirement of

r Theorem 4.11 wi th e l e m e ~ t s of degree h igher than j . Also s i n c e

( j+ l ) 5 (iO-1) + i , no element of d r e e h igher than i +!?,-I w i l l e9 0

be in t roduced .

Repeating t h e same procedure w i l l even tua l ly g ive a s e t wi th

elements having degree a t l e a s t i . 0

S i m i l a r l y we can r e p l a c e elements of A having degree g r e a t e r

than i +<-I by elements w i th degree l e s s than i +Q without i n t r o - 0 0

ducing elements of degree l e s s t han i 0 '

So, f o r every s e t A s a t i s f y i n g t h e c o n d i t i o n of Theorem 4.11,

t h e r e i s a s e t B of d i v i s o r s of m s a t i s f y i n g t h j same cond i t i on and

B ' = / A ; , but con ta in ing only elements of degree between i and 0

i 1 . This completes t h e proo!. 0

Remark 1.' The r e s u l t i n Theorem 4.11 i s b e s t p o s h b l e s i n c e

i f we choose a l l d i v i s o r s of m wi th degrees between i0 and io+P-1

i n c l u s i v e , t he s e t does not con ta in a cha in of l eng th 2+1 and t h e s i z e

i of t he s e t i s t h e sum of t h e 2 l a r g e s t numbers Sn(m) :

Remark 2 . When m i s square- f ree , Theorem 4.11 reduces t o

Theorem 1.8 i n Chapter 1.

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