MC0063(A)-Unit5-fi

DiscreteMathematics Unit5

SikkimManipalUniversity PageNo:121

Unit5 PartiallyOrderedSetsandLatticesStructure

5.1 Introduction

Objectives

5.2 RelationMatrices

5.3 PartialOrderedSets

5.4 Lattices

5.5 Duality

5.6 ModularandDistributivelattices

SelfAssessmentQuestions

5.7 Summary

5.8 TerminalQuestions

5.9 Answers

5.1Introduction

Therearevarioustypesofrelationsdefinedonaset.Inthisunitourinterest

ispartiallyorderedrelationwhichisdefinedonaset,referredasapartially

ordered set. This would lead to the concepts of lattices and Boolean

algebras.Wediscussthedifferentpropertiesofpartialorderrelationsona

set,observedthepropertiesoflatticesusingitsaxioms.Someofthelattice

ordereddiagramswhichhaveapplicationstoseveralalgebraicsystems.

Objectives

Attheendoftheunitthestudentmustbeableto:

Writetherelationmatrixforthegivenrelation.

Diagramrepresentationofpartialorderedsets.

Knowthestructureofalatticeandmodularlattice.

Representationoflatticediagrams.



5.2RelationMatrices

5.2.1Definition

ArelationRbetweenthesetsA1,A2,,An isasubsetofA1 A2 An.

Thisrelation R iscalledannaryrelation(twoaryiscalledbinary,three

aryiscalledternary).Ingeneral,arelation meansbinaryrelationonaset

S(meansasubsetof S S).

Arelation R on S issaidtobe

i) reflexiveif(a,a) R forall a S

ii) symmetricif(a,b) R implies(b,a) R

iii) antisymmetricif(a,b) R and(b,a) R a=b

iv) transitiveif(a,b) R, (b,c) R implies(a,c) R

5.2.2Definition:

LetSbeafinitesetwithnumberofelementsn,and S={xi /1 i n}.

If R isarelationonS,thentherelationmatrixof R isthen nmatrix

(ji

s )nn wherej

i ji

1 , ) Rs

0

if(x x

otherwise

=

.

5.2.3 Example

Suppose S ={a,b,c,d}and R ={(a,a),(b,b),(c,d),(a,b),(d,a)}.

ThentherelationmatrixofR isgivenby

00011000

0010

0011

.

5.2.4Observation

i) Suppose A = (ji

s )nn with eachji

s is equal to 0 or 1. Write

S={xi /1 i n} anddefinearelation RonSas(xi,xj)isanelement



ofR ji

s =1.Then A istherelationmatrixof R.

ii) A relation R is reflexive each of the diagonal elements of the

relationmatrixisequalto1. A relationR issymmetric therelation

matrixissymmetric.

5.3PartiallyOrderedSets

5.3.1Definition

A partiallyorderedset(POset)isasetSwitharelation R on S whichis

reflexive,antisymmetricandtransitive.

Notethatif(a,b) R,thenwewrite a b.If a b and a b,then

wewrite a>b.

5.3.2Example

i) The relations and are the partial orderings on the set of real

numbers.

ii) LetXbethepowersetofthesetA. Thentherelationinclusionisa

partialorderingonX.

5.3.3Definition

AfinitePOsetcanbediagrammedontheplane.If S isaPOsetand a,b

arein S suchthat a>bandthereisno c in S suchthat a>c and

c>b,thenwesaythata covers b.

5.3.4Example

If a covers b, then represent thepoint corresponding to a, above the

pointfor b andjointhepoints(ThisfactisillustratedinthefollowingFig1).

NowconsidertheFig2.Inthis,wecanobservethat:

D covers E B covers C F covers C A covers F.

Also note that B joined to E by a sequence of line segments all going



downwards.

Sowehave B E.

SuppInthisexample, infig2, theelementEisin level0,elementDis in

level1,elementCisinlevel2,elementsFandBareinlevel3andelement

Aisinlevel4.

5.3.5Definition

Theelementsinlevel1arecalledatoms.

5.3.6HasseDiagram

Supposethereareni elementsof leveli for i=0,1,2, . Arrangen0

pointsforelementsoflevel0horizontally.Arrangen1pointsforelementsof

level1horizontallybutbelowthoseoflevel0.Afterarranging n(i1) points

for theelementsof level(i1),arrangeni pointsfor theelementsof leveli

horizontally,butbelowthoseoflevel(i1).Finally,jointheelementspiand

pj bya line segment if pi covers pj. Thediagramobtained is called the

Hassediagram.

5.3.7Definition

i) AnelementxofaPosetS issaidtobeaminimalelementifitsatisfies

thefollowingcondition:y S and x y y=x.

ii) Anelement aofS issaid tobeamaximalelement if itsatisfies the

followingcondition:b S and b a b=a.

a

b

acoversb

A

BF

C

D

E

Fig1Fig2



5.3.8Definition

APOset S issaidtobeatotallyordered(orordered)setiffor a,b in S

exactlyoneoftheconditions: a > b, a = b,or b > a holds.

5.3.9Problem

InafinitePOset S,showthatthereisalwaysatleastonemaximalelement

andoneminimalelement.

Solution:(Formaximalelement):Inacontraryway,supposeScontainsno

maximalelement.Let x1 S.Since x1 isnotmaximal,thereexists x2 in

Ssuchthatx2 > x1.Since x2 isnotmaximal,thereexists x3 in S such

that x3 > x2. If we continue this process, we get an infinite sequence of

distinct elements x1, x2, x3,, such that xi+1 > xi foreach i. This is a

contradictiontothefactthat S containsonlyafinitenumberofelements

(since S isafinitePOset).Henceweconcludethat Scontainsamaximal

element.

5.3.10Definition

Achain inaPOset isasequencea0,a1,,an ofelementsof the POset

suchthatai>ai+1.Thelengthofthischainisn.

5.3.11Definition

Let(P, )beaPOsetand A P.Anelement x P iscalledalower

boundfor A if a x,forall a A.Alowerbound xof A iscalleda

greatestlowerboundofA if x y foralllowerbounds y of A.

Anelement x P iscalledanupperboundforAif x a,forall a A.

Anupperbound x iscalledaleastupperboundofAif b a forallupper

boundsb of A.

5.3.12Note

LetR bethesetofall realnumbers, f A R. If A hasa lower

bound,thenitsgreatestlowerboundiscalledinfimumanditisdenotedby



inf A. If A has an upper bound, then its least upper bound is called its

supremum anditisdenotedbysupA.

Forany subset A of R (the set of all realnumbers), wehave that

infA = minA andsupA =maxA.

5.4Lattices

5.4.1Definition

ApartiallyorderedsetLissaidtobea lattice ifforanypairofelements x, y L,

glb{x,y}(thegreatestlowerboundofx andy isdenotedbyx y),andlub

{x, y}exist(theleastupperboundof x and y isdenotedby x y).

5.4.2Examples

i) LetZ+ bethesetofpositiveintegers.DefinearelationDon Z+ by

aDb a divides bforanya,b Z+ .Then(Z+,D)isalattice,in

which,a b =gcd{a,b}anda b =lcm{a, b}.

ii) The set L = {1,2,3,4,6,12}, the factors of 12 under the relation

divisibilityformsalattice.

iii) LetX beanonemptysetandconsider(P(X), ),thepowersetwith

the inclusionrelation. Thenforany A,B in P(X), wehavethat

A B = A B and A B = A B.

5.4.3Definition

Let(L, )bealattice.Ifeverynonemptysubsetof L hasgreatestlower

boundandleastupperbound,then L issaidtobeacompletelattice.

5.4.4Examples

i) LetP bethesetofallintegerswithusualordering.Clearlyitisalattice.

Thesetofallevenintegersisasubsetof P andithasnoupperbound

orlowerbound.Hence P isnotacompletelattice.

ii) If P={i/1 i n}and istheusualorderingofintegers,then P

isacompletelattice.



5.4.5Note

AtotallyorderedsetSisapartiallyorderedsetinwhicheither a b or

b a holdsforall a,b in S.

5.4.6Example

i) Thesetofallnaturalnumbersformsatotallyordered setunderusual.

ii) Thesetofallnaturalnumberswithadividesbisnotatotallyordered

set.

5.4.7Note

Let(A, )beaPOset.(i) A hasatmostonegreatestandonesmallest

element. (ii) There may be none, one, or several maximal (or minimal)

elementsinaPOset.(iii)Everygreatestelementismaximal.(iv)Every

smallestelementisminimal.

5.4.8Example

ConsiderthePOset(A, )=(R, )whereRisthesetofrealnumbers

and" "istheusualorderonthesetofallrealnumbers.

i) Write B =theinterval[0,3).ThenitisclearthatinfB =0and

supB =3.

ii) Write C =theinterval(0,3].ThenitisclearthatinfC=0and

supC =3.

iii) From(i)and(ii),wecanunderstandthattheinfimum(orsupremum)of

aset X mayormaynotbeintheset X.

iv) Consider D =N,thesetofnaturalnumbers.ItisclearthatinfD =1,

butsupD doesnotexist.

5.4.9Note

i) Everychainislatticeordered.

ii) Let(L, )bealatticeorderedsetand x, y L.Thenwehavethe

following: x y sup{x,y}= y inf{x,y}= x.



5.4.10Definition

A lattice (L, , ) isaset L withtwobinaryoperations (calledas

meetor product)and (calledasjoinorsum)whichsatisfythefollowing

laws,forall x,y,z L:

x y = y x,and x y = y x (Commutativelaws).

x (y z)=(x y) z,andx (y z)=(x y) z (Associativelaws).

x (x y)= xand x (x y)= x(Absorptionlaws).

5.4.11Note

Let(L, , )beanalgebraiclatticeandx L.

i) x x = x

ii) x x = x

5.4.12Remark

Suppose(L, )beanalgebraiclattice.Nowweverifythat

x y = y x y = x forany x, y L.

i) Suppose x y = y. Then x y = x (x y) (by the

supposition)=x (byabsorptionlaw)

ii) Supposex y = x.Then x y=(x y) y (bysupposition)=

y (y x)(bycommutativelaw)= y (byabsorptionlaw).

By(i)and(ii),wehavethatx y = y x y = x.

5.4.13Problem

Letaandbbetwoelements ina lattice(L,).Showthata b=b ifand

onlyifa b=a.

Solution:Part(i):Supposea b=b.Now

a=a (a b) (byabsorptionlaw)

=a b (supposition).



Part(ii):

Supposea b=a.

Now

b=b (b a) (byabsorption)

=b (a b)(bycommutative)

=b a(supposition)

=a b(bycommutative)

5.4.14Theorem

Thefollowingtwoconditionsareequivalent.

i) (L,)isapartiallyorderedsetinwhicheverypairofelementsa,binL,

thelub{a,b}andglb{a,b}exist.

ii) (L, , )beanalgebraicsystemsatisfyingcommutative,associative,

absorptionandidempotentlawswithabifandonlyifa b=a.

5.5Duality

Duality:Theprincipleofdualityisanimportantconceptthatappearsinmany

differentplacesandcontexts.Webeginwithsomesimpleillustration.Inthe

United States, all passenger cars have the drivers seat at the left front.

Consequently,weobservetrafficregulationssuchasthefollowing.

Carsalwaystravelontherightsideofastreet

Carsmaymakerightturnsataredlight.

However,sinceallpassengercarsinIndiahavethedriversseatattheright

front,theyobservetrafficregulationssuchasthese:

Carsalwaystravelontheleftsideofastreet

Carsmaymakeleftturnsataredlight.

5.5.1DualityPrinciple

Any expression involving the operations and which is valid in any

lattice(L, , )remainsvalidifwereplace by ,and by everywherein

theexpression.



5.5.2Definition

IfalatticeL containsasmallest(greatest,respectively)elementwithrespect

to , thenthisuniquelydeterminedelement iscalledthezeroelement(unit

element, respectively). The zero element is denoted by 0, and the unit

elementisdenotedby1.Theelements0and1arecalleduniversallower

anduniversalupperboundsrespectively.Iftheelements0and1exist,then

wesaythatthelattice L isaboundedlattice.

5.5.3Note

(i)Ifalattice L isbounded(by0and1),thenevery x L satisfies

0 x 1,0 x =0,0 x = x,1 x = x,and1 x =1.

5.5.4Problem

Supposethat L isalattice.Showthat

i) Ifx1, x2,xn L,then x1 x2 xn Land x1 x2 xn L.

ii) If L isafinitelattice,then L isbounded.

Solution:

i) Let x1, x2,xn L.Weprovethat x1 x2 xn L,byusing

themathematicalinduction.If n=2,thensince L isalattice,weget

that x1 x2 L.Nowassumetheinductionhypothesisthat x1, x2,

xn1 L

x1 x2 xn1 L.Supposethat x1, x2,xn L

x1 x2 xn1 L and xn L (byinductionhypothesis)

x1 x2 xn1 xn L (bythedefinitionoflattice).

Bymathematicalinduction,weconcludethat x1 x2 xn L for

anyinteger n and x1, x2,xn L.

Inasimilarway,wecanprovethat x1 x2 xn L.

ii) Supposethat L isafinitelatticewith m elements.

Thenwecan take L = {x1, x2, xm}. By (i), x1 x2 xm ,



x1 x2 xm L.Itisclearthat x1 x2 xm xi and xi

x1 x2 xm for1 i m.

Therefore x1 x2 xmisanupperboundforLandx1 x2 xmisalowerboundfor L.Thisshowsthat L isbounded.

5.5.5Problem

LetLbealattice,andx,y,z L.Then Lsatisfiesthefollowingdistributive

inequalities:

i) x (y z) (x y) (x z)

ii) x (y z) (x y) (x z)

Solution:

i) Weknowthat x y x,and x y y y z.

Sox y isalowerboundfor x and y z

x y x (y z).

Now x z x and x z z y z

x z isalowerboundfor x and y z

x z x (y z).

Thereforex (y z)isanupperboundfor x yandx z.

Thus(x y) (x z) x (y z).Thiscompletestheprooffor(i).

ii) Weknowthat x x y and x x z

x isalowerboundfor x y and x z

x (x y) (x z).

Also y z y x y andy z y x z

y z isalowerboundfor x y and x z

y z (x y) (x z).

Therefore(x y) (x z)isanupperboundfor x and y z.

Thus x (y z) (x y) (x z).



5.6ModularandDistributiveLattices

5.6.1Definition

A lattice (L, , ) is called amodular lattice if it satisfies the following

condition: x z x (y z)=(x y) z forall x, y, z L.

Thisconditioniscalledas modularidentity.

5.6.2Example

ConsiderthelatticeL1={0,a,b,c,1}whoseHassediagramisgiven.This

latticeL1 isamodularlattice.Thislatticecalledasdiamondlattice.

5.6.3Example

Consider the lattice L2 = {0,a,b,c,1} whoseHassediagram isgiven.

ThislatticeL2 isnotamodularlattice.Since b c,bymodularlaw,

wehavethat b (a c)=(b c) c b 0=1 c b = c,a

contradiction.HenceL2 isnotamodularlattice.

1

0

a cb

0

1

b

ac



5.6.4Problem

Alattice(L, , )isModularlattice

x {y (x z)}=(x y) (x z)forall x,y,z L.

Solution:SupposeL isamodularlatticeand x,y,z L.Since x x

z,bymodularlaw,wehavethat x {y (x z)}=(x y) (x z).

Converse:Suppose x {y (x z)}=(x y) (x z)forall x, y,

z L.

Let x,y, z L and x z.Then x z = z.

Now x (y z)= x {y (x z)}(since z = x z)

=(x y) (x z)(bytheconversehypothesis)

=(x y) z.

Thisshowsthat L isamodularlattice.

5.6.5Definition

Alattice L issaidtobeadistributivelatticeifitsatisfiesthefollowinglaws:

(i)a (b c)=(a b) (a c),and(ii)a (b c)=(a b) (a c),for

alla,b,c L.Thesetwolawsarecalledthedistributivelaws.

5.6.6Example

i) ForanysetX,thelattice(P(X), , )isadistributivelattice.

ii) Everychainisadistributivelattice.

1

a

b c

0



iii) Considerthelatticegivenbythediagram

Nowb(cd)=b a=b,and(bc) (bd)=e e=e.Thereforethisisnot

adistributivelattice.

5.6.7Problem

Provethatthefollowingpropertiesofalattice L areequivalent:

i) a (b c)=(a b) (a c)forall a, b, c L

ii) (a b) c = (a c) (b c)forall a, b, c L

iii) (a b) (b c) (c a)=(a b) (b c)(ca)forall a, b, c L.

Solution:(i) (ii):Supposea (b c)=(a b) (a c)foralla,b,c L

(ac)(bc)=[(ac)b] [(ac)c] (by(i))

=[(ac)b]c (bycommutativeandabsorptionlaws)

=[(a b) (c b)] c (by(i))

=(a b) [(c b) c] (byassociativelaw)

=(a b) c (byabsorptionlaw).

Thisproves(ii).

(ii) (iii):Suppose(ii).

(a b) (b c) (c a)=(a b) [(b c) (c a)]

={a[(bc)(ca)]}{b[(bc)(ca)]}(by(ii))

={a (b c)} {b (c a)}(bycommutative,associativeandabsorption)

={(a b) (a c)} {(b c) (b a)}(by(ii))

=(a b) (b c) (c a)(byidempotentlaw)

a

e

b dc



(iii) (i): Supposethat a c.Then ab c b (ab)(cb)=

(cb)..(A)Also a c= c.Now(a c) (b c)

=(a c) [(a b) (c b)] (by(A))

=(a b) (b c) (c a)

=(a b) (b c) (c a)(by(iii))

=(a b) (b c) c (since a c)

=(a b) c (byabsorptionlaw).

Nowweprovedthat(a b) c=(a c) (b c).

Thisshowsthat(i)istrue.Thiscompletestheproof.

5.6.8Problem

IfLisadistributivelattice,thenitisamodularlattice.

Solution:Assumethat L isadistributivelattice.Let x, y, z L and x z.

Wehavethat(x y) (y z) (z x)=(x y) (y z) (z x).

Since x z,wehavethat x z = x and x z = z,andso

(x y) (y z) x =(x y) (y z) z .

Thisimplies x (y z)=(x y) z (byabsorptionlaws).

Thisshowsthat L isamodularlattice.

Theconverseof theaboveproblemisnottrue.That is, thereexistmodular

lattices which are not distributive. The following example is a modular

lattice,butnotdistributive.1

0

a cb



5.6.9Problem

Foragivenlattice L,thefollowingtwoconditionsareequivalent:

(a) x(yz)=(x y)(xz),and

(b) x (y z)=(x y)(x z)forallx,y,z L.

Solution:Supposethat x (y z)=(x y) (x z)(i).Now

(x y) (x z)=[(x y) x] [(x y) z](by(i))

=x [(x y) z] (by commutative and absorptionlaws)

=x [z (x y)] (bycommutativelaw)

=x [(z x) (z y)] (by(i))

=[x (z x)] [z y] (byassociativelaw)

=x (z y) (by commutative andabsorption law)

Supposethat x (y z)=(x y) (x z)(ii).

Now(x y) (x z)= [(x y) x] [(x y) z](by (ii))

= [(x y) z](bycommutativeandabsorptionlaws)

= x [z (x y)] (bycommutativelaw)

= x [(z x) (z y)] (by(ii))

=[x (z x)] [z y] (byassociativelaw)

= x (z y) (bycommutativeandabsorptionlaw).

Therefore x (z y)=(x y) (x z).

Thiscompletestheproof.

5.6.10Definition:

AlatticeLwith0and1iscalledcomplementedifforeachx L thereexists

atleastoneelementysuchthatx y=0and x y=1.Eachsuch y is

calledacomplementof x.Wedenotethecomplementof x by x1.

5.6.11Example

i) Inaboundedlattice,1isacomplementof0,and0isacomplementof1

ii) Everychainwithmorethantwoelementsisnotacomplementedlattice.



iii) The complement need not be unique. For example, in the diamond

lattice, both the two elements b and c, are complements for the

element a.

5.6.12Problem

Inadistributivelattice,ifanelementhasacomplement,thenitisunique.

Solution:

Supposethatanelementahastwocomplements,saybandc.Thatis,

a b=1,a b=0,a c=1,a c=0.

Wehaveb=b 1 (since1istheuniversalupperbound)

=b (a c) (sincea c=1)

=(b a) (b c) (bythedistributivelaw)

=(a b) (b c) (since iscommutative)

=0 (b c) (sincea b=0)

=(a c) (b c) (sincea c=0)

=(a b) c (bydistributivelaw)

=1 c (sincea b=1)

=c (since1istheuniversalupperbound).

Thereforethecomplementisunique.

5.6.13Definition

Let L bealatticewithzero.Anelement a L issaidtobeanatomif

a 0andifitsatisfiesthefollowingcondition:bL,0



2. Determine which of the following are equivalence relations and / or

partialorderingrelationsforthegivensets

i) S={linesintheplane}xRy xisparalleltoy

ii) N={setofnaturalnumbers}xRy |xy| 5.

3. Determinewhichofthefollowingarepartialorder?

i) R1={(a,b) ZxZ/|ab| 1}onZ

ii) R2={(a,b) ZxZ/|a| |b|}onZ

iii) R3={(a,b) ZxZ/adividesbinZ}onZ

iv) R4={(a,b) ZxZ/ab 0}

4. DefinearelationRonZ,thesetofallintegersas:aRb a+biseven

foralla,b Z. IsRapartialorderrelationonZ?

5. Let A= {1, 2, 3, 4, 5, 6}. The relation | (divides) isa partial order

relationonA.DrawtheHassediagramof(A,|).

6. Let S = {a, b, c}. Define on P(S), the power set of S as set

inclusion.DrawtheHassediagramforthepartiallyorderedset(P(S), ).

7. Findtheatomsinthefollowinglattice.

5.7Summary

Thestructuresofpartial orderedsetsand latticesareuseful in sortingand

searchprocedures,andconstructionsoflogicalrepresentationsforcomputer

circuits. The diagrammatic forms of lattices are useful in search, path

a

g

b

e

c

d

f



procedures.Weobservedtheinterrelationsbetweenthealgebraicstructures

POsets and Lattices and obtained some of their important equivalences.

TheseconceptsarebasefortheBooleanalgebraandlogicalcircuits.

5.8TerminalQuestions

1. Consider thepartialorderedsetS = {1, 2, 3,4,5,6,7, 8}under the

relationwhoseHassediagramsshownbelow. Consider thesubsets

S1={1,2},S2={3,4,5}ofA.Find(i).Allthelowerandupperbounds

of S1andS2

(ii).glbS1,lubS1,glbS2,lubS2.

2. Verifywhetherornotthefollowingaremodularlattices.

3 21o o

4

6

oo

o o

o

o

8

7

5

u

c ba

0

o

o

o oo

(i)

a

u

o

o

o

(ii)0

o

o

o

(ii)

a

o

c

b

o

0



3. ConsiderthelatticeA={0,a1,a2,a3,a4,a5,1}givenbelow.

i) IsAisadistributivelattice

ii) Whatarethecomplementsofa1anda2

3. Writethecomplementsa,bandc fromthegivenlattice.

5.9Answers

SelfAssessmentQuestions

1. (i) No(ii).No

2. (i) It is anequivalance relation, but not partial orderingasR isnot

antisymmetric.

(ii) Nottransitiveandsoitisneither.

3. (i) No,(ii).No,(iii).No,(iv).Yes.

4. Notapartialorderrelation. (Reason:1R3(since1+3 iseven),3R1

(since3+1iseven),but13).

0

b

a

c

1

o

o

oo

o

a1

a4

a3oo

o o

o

o

o

1

a5

a2

0



5.

6.

7. eandfareatoms

TerminalQuestions

1. UpperboundsofS1are3,4,5,6,7and8

LowerboundsofS1arenone.

glb(S1):none

lub(S1):none

UpperboundsofS2are6,7and8

LowerboundsofS2are1,2and3

glb(S2)=3

lub(S2)=none.

2. (i) Modularlattice

4

5

o

o

6

32

1

o

o

o o

{a,b}

{a} {b}{c}

{a,c}{b,c}

{a,b,c}

f

o

o

oo

o

o

o

o



(ii) Notamodularlattice,sinceforb c,bymodularlaw

b (a c)=(b c) c b 0=c c b=c,acontradiction.

3. (i) Aisnotadistributivelattice.

(ii) Thecomplementofa1anda2area5anda4 respectively.

4. Complementof a = cComplement of b = cComplement of c are a

andb.

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MC0063(A)-Unit5-fi