ABookofSETTHEORY

CHARLESC.PINTEREmeritusProfessorBucknellUniversity

DOVERPUBLICATIONS,INC.Mineola,NewYork

Copyright

Copyright©1971,2014byCharlesC.PinterAllrightsreserved.

BibliographicalNote

ABookofSetTheory,firstpublishedbyDoverPublications,Inc., in2014,isarevisedandcorrectedrepublicationofSetTheory,originally published in 1971byAddison-WesleyPublishingCompany,Reading,Massachusetts.This bookhas been reprintedwith thecooperationofKyungMoonPublishers,SouthKorea.

LibraryofCongressCataloging-in-PublicationData

Pinter,CharlesC.,1932–author.Abookofsettheory/CharlesCPinter.

p.cm.“A revised and corrected republication of Set Theory, originally published in 1971 by Addison-Wesley Publishing Company,

Reading,Massachusetts.”Summary: “This accessible approach to set theory for upper-level undergraduates poses rigorous but simple arguments. Each

definitionisaccompaniedbycommentarythatmotivatesandexplainsnewconcepts.Ahistoricalintroductionisfollowedbydiscussionsofclassesandsets,functions,naturalandcardinalnumbers,thearithmeticofordinalnumbers,andrelatedtopics.1971editionwithnewmaterialbytheauthor”—Providedbypublisher.

Includesbibliographicalreferencesandindex.eISBN-13:978-0-486-79549-2

1.Settheory.I.Title.

QA248.P552013511.3’.22—dc23

2013024319

ManufacturedintheUnitedStatesbyCourierCorporation497089012014

www.doverpublications.com

Tomystudents,fromwhomIhavelearnedhowtoexplainandhowtoteach

Contents

Preface

Chapter0 HistoricalIntroduction1 Thebackgroundofsettheory2 Theparadoxes3 Theaxiomaticmethod4 Axiomaticsettheory5 ObjectionstotheaxiomaticapproachOtherproposals6 Concludingremarks

Chapter1 ClassesandSets1 Buildingsentences2 Buildingclasses3 Thealgebraofclasses4 OrderedpairsCartesianproducts5 Graphs6 Generalizedunionandintersection7 Sets

Chapter2 Functions1 Introduction2 Fundamentalconceptsanddefinitions3 Propertiesofcompositefunctionsandinversefunctions4 Directimagesandinverseimagesunderfunctions5 Productofafamilyofclasses6 Theaxiomofreplacement

Chapter3 Relations1 Introduction2 Fundamentalconceptsanddefinitions3 Equivalencerelationsandpartitions4 Pre-image,restrictionandquotientofequivalencerelations5 Equivalencerelationsandfunctions

Chapter4 PartiallyOrderedClasses1 Fundamentalconceptsanddefinitions2 Orderpreservingfunctionsandisomorphism3 DistinguishedelementsDuality

4 Lattices5 FullyorderedclassesWell-orderedclasses6 Isomorphismbetweenwell-orderedclasses

Chapter5 TheAxiomofChoiceandRelatedPrinciples1 Introduction2 Theaxiomofchoice3 Anapplicationoftheaxiomofchoice4 Maximalprinciples5 Thewell-orderingtheorem6 Conclusion

Chapter6 TheNaturalNumbers1 Introduction2 Elementarypropertiesofthenaturalnumbers3 Finiterecursion4 Arithmeticofnaturalnumbers5 Concludingremarks

Chapter7 FiniteandInfiniteSets1 Introduction2 Equipotenceofsets3 Propertiesofinfinitesets4 Propertiesofdenumerablesets

Chapter8 ArithmeticofCardinalNumbers1 Introduction2 Operationsoncardinalnumbers3 Orderingofthecardinalnumbers4 Specialpropertiesofinfinitecardinalnumbers5 Infinitesumsandproductsofcardinalnumbers

Chapter9 ArithmeticoftheOrdinalNumbers1 Introduction2 Operationsonordinalnumbers3 Orderingoftheordinalnumbers4 Thealephsandthecontinuumhypothesis5 Constructionoftheordinalsandcardinals

Chapter10 TransfiniteRecursionSelectedTopicsintheTheoryofOrdinalsandCardinals

1 Transfiniterecursion2 Propertiesofordinalexponentiation3 Normalform4 Epsilonnumbers5 Inaccessibleordinalsandcardinals

Chapter11 ConsistencyandIndependenceinSetTheory1 Whatisaset?2 Models3 Independenceresultsinsettheory4 Thequestionofmodelsofsettheory5 Propertiesoftheconstructibleuniverse6 TheGödelTheorems

Bibliography

Index

Preface

Over many years of selecting instructional materials for my courses, I came to understand thattextbooksthatarecongenialtostudentsobeyalawofreciprocitythatIwishtoproposeasanaxiom:

AxiomThemore effort an authorputs intowriting a text, the less effort is requiredof the reader tounderstandit.Iadoptedthisguidingprincipleasacategoricalimperativeduringthewritingofthisbook.Evenafter

allthemathematicswasinplace,Irereadandrewrotemanytimesandtestedexplanatorystrategieswithmystudents. Ihavededicated thisbook to thembecause theyhavebeenmymostpatientandhonestcritics.Thebookowes itspresent formlargely to themand to their (sometimesnaïve,oftenbrilliant)suggestions.Mathematicshasasuperblyefficientlanguagebymeansofwhichvastamountsofinformationcanbe

elegantly expressed in a few formal definitions and theorems. It is remarkable that the lifework ofconsecutivegenerationsofgreatthinkerscanoftenbesummedupinasetofequations.Theeconomyofthe languagemasks the richness and complexity of the thoughts that lie behind the symbols. Everymathematicsstudenthastomastertheconventionsforusingitslanguageeffectively.However,whatisfarmore important is that thestudentbe initiated into the inner lifeofmathematics—the images, theintuitions,themetaphorsthat,oncegrasped,makeussay,“Aha!NowIunderstand!NowIseeit!”Thisinnerseeingiswhatmakesmathematicsvitalandexciting.Whatismostuniqueaboutsettheoryisthatitistheperfectamalgamofthevisualandtheabstract.

Thenotionsofsettheory,andtheideasbehindmanyoftheproofs,presentthemselvestotheinnereyeinvividdetail.Thesepicturesarenotasovertlyvisualasthoseofgeometryorcalculus.Youdon’tseethem in the same way that you see a circle or a tangent line. But these images are the way intoabstraction.For thematuringstudent, the journeydeeper intoabstraction isa riteofpassage into theheartofmathematics.Settheoryisalsothemost“philosophical”ofalldisciplinesinmathematics.Questionsareboundto

comeupinanysettheorycoursethatcannotbeanswered“mathematically”,forexamplewithaformalproof.Thebigquestionscannotbedodged,andstudentswillnotbrookaflippantoreasyanswer.Isthecontinuumhypothesisafactoftheworld?Istheaxiomofchoiceatruth?Ifwecannotanswerwithadefiniteyesorno, inwhatmannerare they justified? (Thatone requiresa longanswer).What is themeaningoftheJacob’sladderofsuccessiveinfinitiessohighthattheverythoughtofitleadstoakindofintellectualvertigo?TowhatextentdoesmathematicsdwellinaPlatonicrealm—andifitdoes,theninthewordsofEugeneWigner,“howdoyouexplaintheunreasonableeffectivenessofmathematicsinthenaturalworld?”OrasDescarteswondered,wheredoyoufindthenexusbetweenthematerialworldandtheproductsofthought?In this book I have tried, insofar as possible, not to evade these questions nor to dwell on them

excessively. Students should perceive that mathematics opens doors to far-reaching and fascinatingquestions, but on the other hand theymust remain anchored inmathematics and not get lost in thenarcotichazeofspeculativethought.Ihavetriedtoprovidethenecessarybackgroundforunderstandingaxiomaticsandthepurposeandmeaningofeachoftheaxiomsneededtofoundsettheory.Ihavetriedto give a fair account of the philosophical problems that lie at the center of the formal treatment ofinfinitiesandotherabstractions.Aboveall,ofcourse,Ihaveendeavoredtopresentthestandardtopicsofsettheorywithuncompromisingrigorandprecision,andmadeitclearthattheformalismontheonehand,andtheintuitiveexplanationsontheotherhand,belongintwoseparatedomains,oneusefulforunderstanding,theotheressentialfordoingmathematics.

Thisbook is a revised and re-writtenversionof an earlier edition, published in1972byAddison-Wesley.Ihaveretainedmostoftheformaldefinitions,theoremsandproofs,withnothingmorethanafewcorrectionswhereneeded.Ihavealsoretainedtheinitialchapterthatnarratestheoriginsandearlyhistoryofsettheory,becausehistory(ingeneral)doesnotchange.Ihaveaddedcommentary,introducedsomenewdiscussions,andreorganizedafewproofsinordertomakethemcleanerandclearer.Finally,IhaveaddedanewchapteronmodelsofsettheoryandtheindependenceresultsofGödelandCohen.Ihave set the discussion of these topics at a level that is accessible to undergraduates while notconcealingthedifficultiesofthesubject.

0HistoricalIntroduction

1THEBACKGROUNDOFSETTHEORY

Although set theory is recognized to be the cornerstone of the “new”mathematics, there is nothingessentiallynewintheintuitiveideaofaset.Fromtheearliesttimes,mathematicianshavebeenledtoconsider setsofobjectsofonekindor another, and the elementarynotionsofmodern set theory areimplicitinagreatmanyclassicalarguments.However,itwasnotuntilthelatterpartofthenineteenthcentury,intheworkofGeorgCantor(1845–1918),thatsetscameintotheirownastheprincipalobjectofamathematicaltheory.Strangely,itwashisworkinthehighlytechnicalfieldoftrigonometricserieswhichfirstledCantor

to study thepropertiesof sets.At first,heconfinedhimself tocertainparticular setsof realnumberswhichoccurredinconnectionwiththeconvergenceofseries.ButCantorwasquicktounderstandthathisdiscoveriesappliedtosetsquitegenerally;inaseriesofremarkablepapers,publishedbetween1873and1897,hemovedprogressivelyfurtherfromtheconcreteproblemswhichhadinitiatedhisthinkingonsets,andtowardthepowerfulgeneralconceptswhichunderliesettheorytoday.TheboldeststepwhichCantorhadtaken—intheeyesofhiscontemporaries–washisuseofinfinite

sets,whichheconsideredasnolessnaturalthanusingfinitesets.Thequestionof“infinity”hadlongbeenoneof themost sensitiveproblemsofmathematics.The reader is undoubtedly acquaintedwithZeno’sfamous“paradox”, inwhichaunit linesegment isdividedintosubintervalsbythepoints1/2,1/4,1/8,1/16,etc.Eachsubinterval—nomatterhowsmall—hasadefinite,nonzero length,and thereare infinitely many subintervals; hence, the seemingly paradoxical conclusion that infinitely manynonzerolengthscanbeaddedtogethertoproduceafinitelength.Inordertoavoidsuchtraps,classicalmathematiciansmadeadistinctionbetweenthe“actual”infinite—inwhichinfinitelymanyobjectsareconceived of as existing simultaneously—and the “virtual” infinite,which is simply the potential toexceedanygivenfinitequantity.The“virtual”infinitewasregardedassafe,henceadmissible,whereasthe“actual”infinitewastaboo.Itisnotsurprisingthen,thatCantor’stheory—withitsuninhibiteduseofinfinitesets(thenotionof

infinitewas obviously understood here in the “actual” sense)—wasnot immediately accepted by hiscontemporaries.Itwasreceivedatfirstwithskepticism,sometimesevenwithopenhostility.However,by the 1890’s themore “palatable” parts ofCantor’s theorywerewidely used, for they provided anelegantframeworkforawidevarietyofmathematicaltheories.Andbeforetheturnofthecentury,eventhe most revolutionary aspects of set theory had been accepted by a great many mathematicians—chieflybecausetheyturnedouttobeinvaluabletools,particularlyinanalysis.Meanwhile, thework of several outstandingmathematicians, in particularDedekind,was taking a

turnwhichwouldcastsettheoryinitsmostpromisingrole—asthefundamental,“unifying”branchofmathematics.Fromtheearliesttimes,mathematicianshavegiventhoughttothepossibilityofunifyingtheentiredisciplineunderasmallnumberofbasicprinciples.Manyoftheancientschools,fromEuclidthroughtheMiddleAges,contendedthatthevariousbranchesofmathematicscouldbesubsumedundergeometry (numbersmight be conceived as geometric proportions); a far more successful attempt atunification came in the nineteenth century, when the work of Weierstrass, Dedekind, and otherssuggestedthatallofclassicalmathematicscouldbederivedfromthearithmeticofthenaturalnumbers

(positive integers). It was shown that every real number can be regarded as a sequence (called a“Cauchysequence”)ofrationalnumbers;hencethestudyoftherealnumbersisreducedtothatoftherational numbers.But the rational numbers can easilybe regarded aspairs of integers, so finally themathematics of the real numbers—which includes the calculus and (via analytic geometry) all ofgeometry—canbebasedonthenaturalnumbers.ItwasatthiscrucialpointintheevolutionofideasonthefoundationsofmathematicsthatDedekind,

in his little bookWas sind und was sollen die Zahlen (1888), revealed that the concept of naturalnumbers can be derived from the basic principles of set theory. Amodern way to show this is thefollowing:welet“0”betheemptyset(thatis,thesetwithnoelements,denotedbythesymbolØ);“1”isdefinedtobetheset{Ø},thatis,theset(ofsets)containingtheoneelementØ.Then,“2”isdefinedtobetheset{0,1},“3”isdefinedtobe{0,1,2},andsoon.Allthepropertiesofthenaturalnumberscanbeprovenusingthesedefinitionsandelementarysettheory.Bytheturnofthecentury,then,settheoryhadnotonlybeenacceptedasanindispensabletoolbya

largesegmentofthemathematicalcommunity,but,moreover,itwasaseriouscontenderforthepositionofprimacyamongthemathematicalsciences.Ironically,attheverytimewhenCantor’sideasseemedfinallytohavegainedacceptance,thefirstof

certain“paradoxes”wereannounced,whicheventuallycastseriousdoubtsastothebasicsoundnessofsettheoryinits“Cantorian”form.Theseparadoxeshadsuchwiderepercussionsthatitisworthlookingattheminsomedetail.

2THEPARADOXES

Between1895and1910anumberofcontradictionswerediscoveredinvariouspartsofsettheory.Atfirst,mathematicianspaidlittleattentiontothem;theyweretermed“paradoxes”andregardedaslittlemorethanmathematicalcurios.Theearliestoftheparadoxeswaspublishedin1897byBurali-Forti,butit had already been discovered, two years earlier, byCantor himself. Since theBurali-Forti paradoxappearedinarathertechnicalregionofset theory, itwashoped,atfirst, thataslightalterationofthebasicdefinitionswouldbesufficienttocorrectit.However,in1902BertrandRussellgaveaversionofthe paradox which involved the most elementary aspects of set theory, and therefore could not beignored.Intheensuingyearsothercontradictionswerediscovered,whichseemedtochallengemanyofthe“safest”notionsofmathematics.

The“paradoxes”ofsettheoryareoftwodifferentkinds,theonecalledlogicalparadoxes,theothercalledsemanticparadoxes.Thereasonforthenames“logical”and‘semantic”willbecomecleartouswhenwehaveseenafewexamplesoftheseparadoxes;essentially,the“logical”paradoxesarisefromfaultylogicwhereasthe“semantic”paradoxesarisefromthefaultyuseoflanguage.We will devote the remainder of this section to the presentation of two of the most celebrated

paradoxes,whichinvolveonlyelementaryconceptsofset theory.Thefirst isa“logical”paradox,thesecondisa“semantic”paradox;bothmaybeconsideredastypicaloftheirkind.ThesimplestofthelogicalparadoxesisRussell’sparadox,whichcanbedescribedasfollows:

IfAisaset,itselementsmaythemselvesbesets;thissituationoccursfrequentlyinmathematics—forexample,Amaybeasetof lines,whereeachline isregardedasasetofpoints.Nowthepossibility arises that A may be an element of itself; for example, the set of all sets has thisproperty.

LetSdenotethesetofallsetsthatarenotelementsofthemselves.IsSanelementofitself?Well,ifSisanelementofS,then—bytheverydefinitionofS—SisnotanelementofS.IfS is

notanelementofS,then(again,becauseofthewaySisdefined)SisanelementofS.Thus,wehaveproventhatSisanelementofSifandonlyifSisnotanelementS—acontradictionofthemostfundamentalsort.

Usually,inmathematics,whenwereachacontradictionofthiskind,weareforcedtoadmitthatoneof our assumptionswas in error. In this case,we are led to conclude either that it ismeaningless tospeakofasetasbeinganelementofitself,orthatthereisnosuchthingasa“setofallsetswhicharenot elements of themselves.”Wewill return to this question presently;meanwhile, let us say a fewwordsaboutthesemanticparadoxes.

TypicalofthesemanticparadoxesisBerry’sparadox:

Forthesakeofargument,letusadmitthatallthewordsoftheEnglishlanguagearelistedinsomestandarddictionary.LetTbethesetofallthenaturalnumbersthatcanbedescribedinfewerthantwentywordsoftheEnglishlanguage.SincethereareonlyafinitenumberofEnglishwords,thereareonlyfinitelymanycombinationsof fewer than twentysuchwords—that is,T isa finiteset.Quite obviously, then, there are natural numberswhich are greater than all the elements ofT;hencethereisaleastnaturalnumberwhichcannotbedescribedinfewerthantwentywordsoftheEnglish language. By definition, this number is not in T; yet we have described it in sixteenwords,henceitisinT.

Once again, we are faced with a glaring contradiction; since the above argument would beunimpeachableifweadmittedtheexistenceofthesetT,weareirrevocablyledtotheconclusionthatasetsuchasTsimplycannotexist.Beforetheparadoxes,thequestionoftheexistenceofsetshadneverbeenposed.Cantor“defined”a

settobe“acollectionofdefinitedistinguishableobjectsofourperceptionwhichcanbeconceivedasawhole.”Morespecifically,Cantorandhisearlyfollowersacceptedthe“common-sense”notionthatifwe can describe a property of objects, we can also speak of the set of all objects possessing thatproperty. The paradoxes had the singular merit of proving this native conception of sets to beunacceptable—ifonlybecausecertain“properties”leadtoparadoxicalsets.In the variousmovementswhich sprang up, during the early 1900’s,with the aim of revising the

foundations of set theory, the topic of central concern was the existence of sets. What propertieslegitimatelydefinedsets?Underwhatconditionsdopropertiesdefinesetsatall?Howcannewsetsbeformedfromexistingones?

3THEAXIOMATICMETHOD

Theappearanceof theparadoxesmarked thebeginningofacrisis in the foundationsofmathematicswhich has not been completely resolved to our day. It became abundantly clear that the intuitiveconceptionofaset,asembodiedinCantor’s“definition,”doesnotprovideasatisfactorybasisforsettheory—muchlessformathematicsasawhole.Minorattemptstoeliminatetheparadoxesbyexcludingspecific types of concepts and definitionswere doomed to failure; nothing less than an entirely newapproachwasneeded.Startingabout1905,severalwaysofdealingwiththeproblemwereproposedanddeveloped by their adherents; most of them can be classified into three major groups, called the“axiomatic,” the“logistic,”and the“intuitionist”schools.Theremainderof thischapter isdevoted to

presentingthesethreewaysofthought.First,however,weshallbrieflyreviewthedevelopmentoftheaxiomaticmethod.Theaxiomaticmethodinmathematicsemergedinahighlydevelopedform,about300B.C.,withthe

appearance of Euclid’s Elements. Although the method popularized by Euclid has become acharacteristic featureofeverybranchofmathematics today,only incomparativelyrecentyearshas itbeenappliedoutsideofgeometry.Forthisreason,ourmodernunderstandingofaxiomsystems,andofdeductivereasoninggenerally,has toagreatextentcomeoutofstudies in thefieldofgeometry.It isworth examining a few of themajor developments in geometrywhich influenced the growth of theaxiomaticmethod.To Euclid and his times, the axioms and postulates represent “truths” whose validity is beyond

question.Forexample,itwasthisbeliefintheabsolutetruthofgeometricpropositionswhichledtothemillenia-longcontroversyoverEuclid’s“parallelpostulate.”Thispostulateasserts that if twolines,AandB,intersectathirdlineC,andiftheinteriorangleswhichAandBmakewithC(onagivensideofC) add up to less than two right angles, thenA andB necessarily intersect. Because this statementappearedtobe“obviouslytrue”—yetitlackedthetersesimplicityoftheotheraxiomsandpostulates—geometers from Euclid to the 1700’s succeeded one another in vain attempts to prove it from theremainingassumptions.OnlyinthemidnineteenthcenturywasthequestionresolvedwhenBolyaiandLobachevski,eachreplacingtheparallelpostulatebyanassumptionwhichcontradicted it,developed“non-Euclidean”geometries.Thenon-Euclideangeometrieswereshowntobeno lessconsistent thanEuclidean geometry, since they could be given Euclidean interpretations (that is, by suitablyreinterpreting“point,”“line,”“angle,”andsoforth,thepostulatesofeitherBolyaiorLobachevskicanbemade to hold inEuclidean geometry). Thus, not only is the parallel postulate independent of theother axioms andpostulates ofEuclid’s system,but alternative, equally consistent geometries canbefoundedwhichdonotdescribe thespaceofoureverydayexperience.With thiscame the recognitionthataxiomsarenot“universal truths,”butarewhatever statementswewish touseaspremises inanargument.Perhaps the greatest defect in theElements is the number of tacit assumptionsmade by Euclid—

assumptionsnotgrantedbythepostulates.Forexample,inacertainproofitisassumedthattwocircles,eachpassingthroughthecenteroftheother,haveapairofpointsincommon—yetthepostulatesdonotprovide for the existence of these points. Elsewhere, Euclid speaks of a point as beingbetween twoothers,yethedoesnotdefine“betweenness’orpostulateanyofitsproperties.OtherargumentsintheElements involve the concept of rigidmotion—a concept which is not defined or mentioned in thepostulates.Thus,throughoutEuclid,theorderlychainoflogicalinferencesisfrequentlybrokenbytacitappealstovisualevidence.Withthediscoveryofthesegaps,mainlyinthenineteenthcentury,grewtheunderstandingthatamathematicalargumentmustbeabletoproceedwithoutthemediationofspatialorother intuition; that certain objects and relations (such as “point,” “line,” “betweenness”) must beregardedasundefinednotionsandtheirpropertiesfullyspecified;thatdeductionis,inaveryessentialmanner,independentofthemeaningofconcepts.In1882,M.Paschpublishedthefirstformulationofgeometryinwhichtheexclusionofanyappealtointuitionisclearlystatedasagoalandsystematicallycarriedout.Bytheendof thenineteenthcentury, then,amodernconceptionof theaxiomaticmethodbegan to

emerge.Initsbroadoutlines,itdidnotdifferfromtheideasheldbyEuclid:amathematicaltheoryis“axiomatic”ifcertainstatementsareselectedtobe“axioms,”andalltheremainingpropositionsofthetheoryarederivedfromtheaxiomsbylogicalinference.However,therewasanewunderstandingoftheformalnatureofmathematicalproof.Inasmuchaspossible,theaxiomsshouldbesufficientlydetailed,andtherulesoflogicaldeductionsufficientlyexplicit,thatneitherintuitionnorintelligenceisneededtogothroughthestepsofaproof.Ideally,itshouldbepossibleforacomputertoverifywhetherornota

proofiscorrect.Aslongasmathematicsisformulatedinordinarylanguages,suchasEnglish,humanunderstandingis

indispensable for interpreting statements and finding the structure of complex sentences. Thus, ifintuition is to be completely removed from mathematical proof, an essential prerequisite is thedevelopmentofaformalmathematicallanguage:the“rules”ofthislanguagemustbestrictlycodified,so that every statement is unambiguous and its structure clear. The creation of formal, symboliclanguageswasoneof themost importantdevelopmentsofmodernmathematics;here iswhat suchalanguagelookslike.Themostbasicmathematicalstatementslooklikethis:

“XisparalleltoY,”“yliesbetweenxandz,”“Xisanopenset,”etc.

Theyarestatementsaboutanobject,apairofobjects,ormoregenerally,aboutanorderedn-tupleofobjects.Thesestatementsarecalledelementarypredicates,and the lettersX,Y,x,y,zarecalled theirvariables.Itisconvenienttodenoteapredicatebyasingleletterfollowedbythelistofitsvariables.Thus,“Xis

paralleltoY”maybewrittenA(X,Y),“y liesbetweenxandz”maybewrittenB(x,y,z),andsoon.Now themature student is aware of the fact that the “meaning” of a predicate is immaterial in theprocessofmathematicalreasoning.Forexample,the“meaning”ofthewordparallelhasnobearingonthecourseofageometricalargument;all thatmatters is the relationshipbetween the statement“X isparallel toY ” andother statements suchas“X intersectsV ” and“Y is perpendicular toZ.”For thisreason,elementarypredicatesarealsocalledatomicformulas;theyareintegral,“indivisible,”nottobeanalyzedfurther;theyareonlytobedistinguishedfromeachother.Itisaremarkablefactthateveryknownbranchofmathematicsrequiresonlyafinitenumber(usually

a very small number) of distinct elementary predicates. For example, every statement of planeEuclideangeometrycanbeexpressedintermsofthefollowingbasicpredicates:

Set theory, aswe shall see,maybe formulated entirely in termsof theonepredicatex∈A (x is anelementofA).Predicatesalonearenotsufficienttoexpressall thestatementsofmathematics, justasnounsalone

wouldbeinadequatetowriteEnglishsentences.Forexample,wemaywishtosaythatif“xisparalleltoy”and“y isperpendicular toz,” then“x isperpendicular toz.”Suchstatementsconsistofpredicatesjoinedtogetherbymeansoflogicalconnectives.Thus,ifPandQarestatementsinourlanguage,thensoarethefollowing:

Finally,wemaywishtosay,forexample,thatif“xisapoint”and“yisapoint,”thenthereexistsapointzsuchthat“zisbetweenxandy.”Thisrequirestheuseofquantifiers.Thus,ifP(x)isastatementwithavariablex,thenthefollowingarealsostatements:

This completesour formalmathematical language.All ofknownmathematics canbe expressed intermsofelementarypredicates, logicalconnectives,andquantifiers.Toillustratehowthislanguageisused,letustakeasimpleexample.Thesentence

“Ifxandyaredistinctpoints,thenthereisapointzbetweenxandy”

canbesymbolizedas

wherethemeaningofthepredicatesisgivenin(1)above.Oneof themanybenefits tobederived from theuseof a formal language is that it is possible to

describedpreciselyandexplicitlytheprocessofdeductioninthislanguage.Afewclear,unambiguousrulesdecidewhenastatementTmaybeinferredfromastatementS.Afewsuchrulesarethefollowing:

These,andafewotherlaws,*arecalled“rulesofinference”inourlanguage.Aformalargument,fromgivenpremises,isasequenceofexpressionsoftheformallanguage,whereeachexpressioniseitherapremise, or is derived from a preceding expression (or expressions) by applying one of the rules ofinference.

ExampleConsideraformallanguagewithapredicateL(x,y);letusagreetowritex<yforL(x,y).Thefollowingisaverysimpleformalargumentinthislanguage.

Premisesi) a<b.ii) b<c.iii) [(a<b)∧(b<c)]→(a<c).

Theorem∃x∋[(a<x)∧(b<x)].

Proof.

Thereadershouldnotethattherulesofinferenceareappliedtoexpressionsinaperfectlymechanicalway.Toallintentsandpurposes,theexpressionscanberegardedasmeaninglessarraysofsymbols;thefactthattheyhaveameaningtousinirrelevanttothetaskofcarryingouttheproof.Thusintuitionistotallyabsentfromaformalmathematicalproof.Anaxiomatictheoryissaidtobeformalizedifitsaxiomsaretranscribedintheformallanguage(for

example,formula(2)onp.8isanaxiomofHilbert’splanegeometry),andallofitsproofsareformalproofs.While it is commonly accepted as the ideal, today, that every axiomatic theorybedevelopedformally,itwouldbefartootedious,inpractice,todoso.Symbolicstatementsaredifficulttodecipher,and formal proofs tend to be exceedingly long. Thusmathematicians are usually content to satisfythemselvesthatanaxiomatictheorycanbeformalized,andthenproceedtodevelopit inaninformalmanner.Thiswillbeourprocedureinthisbook.

4AXIOMATICSETTHEORY

Toagreatmanymathematiciansintheearly1900’s,theanswertotheproblemposedbytheparadoxeswastoprovideset theorywithanaxiomaticbasis.Theterm“set”andtherelation“isanelementof”wouldbe theundefinednotionsof such a theory, just as “point” and “line” areundefinednotions ingeometry; their “meaning”would be irrelevant, and their propertieswould be given formally by theaxioms. In particular, the axioms would be chosen in such a manner that all the useful results ofCantor’stheorycouldbeproven,whereastheparadoxescouldnot.Thefirstaxiomatizationofsettheorywasgivenin1908byZermelo.Zermelo’ssystem,withcertain

modifications due to Skolem and Fraenkel, iswidely used up to the present day. Zermelowrote hisworkbeforethetimewhenformalmethodsbecamewidelyunderstoodandaccepted;thushissettheoryisnotwritteninaformallanguage,butiscloserinstyletotheolderaxiomatictreatmentsofgeometry.InZermelo’ssystem,thereisoneprimitiverelation,denotedbythesymbol∈;theexpressionx∈Yis

toberead“xisanelementofY”.Thevariablesx,y,z,X,Y,etc.,whichweplacetotherightortotheleftofthesymbol∈,standforobjectswhichweagreetocall“sets.”Thereadermayfeelthereoughttobetwokindsofobjects,namelysetsandelements.Actually,this

distinctionisunnecessary:For,ontheonehand,therelationshipbetweenelementandsetisarelativeoneratherthananabsoluteone(infact,theelement-setrelationshipispreciselytherelation∈).Ontheotherhand,almosteverysetinmathematicsisasetofsets.Forexample,inplaneanalyticgeometry,alineisasetofpoints;apointisapairofrealnumbers(itscoordinates);arealnumberisregardedasasequence (that is, a set) of rational numbers; etc. Thus a useful simplification which is made inaxiomaticsettheoryistoregardtheelementsofeverysettobesetsthemselves;inotherwords,every

set is considered to be a setofsets. This simplification has no harmful effects, and has themerit ofreducingthenumberofprimitivenotionsandaxiomsofsettheory.Thissuggestsacommentonnotation.Althoughitiscustomarytousesmallandcapitallettersasinx

∈Y, it is innowaynecessary.Infact,wewillsometimeswritethingslikex∈YandX∈Y.All thevariablesintheseexpressionsdenotesets.Almosteverysetthatarisesinourthinkingisasetconsistingofalltheobjectsofaspecifiedkind—

that is, consisting of all the objectswhich satisfy a given condition.This is themost naturalway inwhichsetsoccur:weareabletodescribeaconditiononx—letussymbolizethisconditionbyS(x)—andweareledtospeakofthesetofallobjectsxwhichsatisfyS(x).

ExamplesThesetofallobjectsxwhichsatisfythecondition“x isanirrationalnumberand0 x 1” (looselyspeaking,thesetofallirrationalnumbersbetween0and1).

Thesetofallobjectsxwhichcanbedescribedbythesentence“xisaman”(looselyspeaking,thesetofallmen).

Sincethisisthemostnaturalwaythatsetsarise,itisclearlydesirabletohaveaprincipleinsettheorywhichmakesitpossible—givenanyconditionS(x)—toformthesetofallobjectsxwhichsatisfyS(x).However,aswenotedinSection2,ifsuchaprincipleisadoptedwithoutanyrestrictions,weareledtotheparadoxes(forexample,wecanformthesetofallsetswhicharenotelementsofthemselves).Thus,wemustdevisesuchrestrictionsonthisprincipleaswilleliminate theparadoxes.Zermeloconceivedthefollowingrestriction:LetS(x)beaconditiononx;wecannotformthesetofallxwhichsatisfyS(x);but,ifAisagivenset,wecanformthesetofallxinAwhichsatisfyS(x).Thus,roughlyspeaking,aproperty of objects cannot be used to form a “ new” set, but only to “ select,” from a setA whoseexistencehasalreadybeensecured,alltheelementswhichsatisfythegivenproperty.Zermelointroducedthisprincipleasanaxiominhissystem.Becauseitsroleistoselectelementsin

sets,hecalledittheaxiomofselectionandstateditasfollows:

LetAbeaset,and letS(x)beastatementaboutxwhich ismeaningful foreveryobjectx inA.ThereexistsasetwhichconsistsofexactlythoseelementsinxandAwhichsatisfyS(x).

Thesetwhoseexistenceisgivebytheaxiomofselectioniscustomarilydenotedby

[toberead:“thesetofallxinA such thatS(x)”].Thus thereadershouldnote thatZermelo’ssystemdoesnotallowustoform{x:S(x)},[the“setofallxwhichsatisfyS(x)”];but,foranysetA,wecanform{x∈A:S(x)}.Howdoes the axiomof selection avoid the paradoxes?First, let us seewhat happens toRussell’s

paradox: the crucial set in Russell’s argument is the “set of all sets which are not elements ofthemselves,”whichcanbesymbolizedas{x:x∉x}.Aswehavenoted,thissetcannotbeformedinZermelo’ssystem;thebestwecandoistoproduce{x∈A:x∉x},whereAisanysetwhichcanbeshown toexist. Ifwesubstitute{x∈A :x∉x} for {x :x∉x} inRussell’s argument, the outcomechangescompletely.Indeed,letusgothroughthestepsoftheargument,withSdenoting{x∈A:x∉x}:

S∈Sisimpossible,forS∈SimpliesS∉S,acontradiction!ThusS∉S.ItfollowsthatS∉A,forifSwereinA,then(becauseS∉S)wewouldhaveS∈S,whichwouldbeacontradiction.

HenceRussell’sargumentmerelyprovesthatifAisanyset,thentheset{x∈A:x∉x}cannotbeanelementofA.The other logical paradoxes disappear in similar fashion. The crucial sets in all of the logical

paradoxes have a common trait: they are overly comprehensive—that is, they are “too large,” theyincludetoomuch.InRussell’sparadoxitisthe“setofallsetswhicharenotelementsofthemselves”;inCantor’sparadox (which isclosely related to thatofRussell) it is the“setofall sets.”Theaxiomofselectioncannotcontributetotheformationofthese“excessivelylarge”sets,sinceitcanonlybeusedtoformsubsetsofexistingsets.

Theproblemofavoidingthesemanticparadoxesisamoredifficultone.ThecrucialsetsinparadoxessuchasBerry’sarenot“toolarge.”Thetroubleseems,rather,tobeinherentintheconditionS(x)whichdetermines the set; even the restriction imposedby the axiomof selection isnot an effectivebarrier.Thus, ifS(x) designates the sentence “x can be described in fewer than twentywords of theEnglishlanguage,”thentheoffendingsetinBerry’sparadoxis{x∈N:S(x)},whereNdenotesthesetofthenatural numbers. This set can be formed in Zermelo’s system, if we admit S(x) as an acceptableconditiononx.Thus,topreventthesemanticparadoxes,wemustplacerestrictionsonthetypeof“conditions”S(x)

whichareadmissiblefordeterminingsets.Zermeloattemptedtodothisbystipulating,intheaxiomofselection, that {x∈ A : S(x)} can be formed only if S(x) is meaningful for every element x in A.However,insodoing,heonlyraisednewquestions:Howarewetounderstand“meaningful”?HowdowedeterminewhetherS(x)ismeaningful?Weareforced,atlast,tofaceaquestionwhichthealertreadermayalreadyhaveaskedhimself:What

dowemeanbya“condition”S(x),bya“statementaboutanobjectx”?Wecannotbecontenttoregardtheconceptofa“statementaboutx”asintuitivelyknown,sinceourpurposenowistoaxiomatizesettheory,thatis,tofreeitofalldependenceonintuition.Zermelofailedtogiveasatisfactoryanswertothisquestion,becausehedidnotframehissysteminaformallanguage.However,in1922,SkolemandFraenkel, both working on formal axiomatizations of set theory, saw the natural way out of thedilemma:a“statementaboutx”issimplyastatementintheformallanguagewithone“free”variablex.(WesaythatxisfreeinS(x)ifxisnotgovernedbyaquantifier∃xor∀x;thus,in∃y∋(x<y),xisfreewhereasyisnot).In Zermelo’s system there is only one elementary predicate, denoted by the symbol ∈. Thus a

statementintheformallanguageisanexpressionwhichcanbewrittenusingonlypredicatesx∈Y,u∈V,etc.,logicalconnectives,andquantifiers.Ifwerestrict the“statements”S(x)whichcanbeused in theaxiomofselection to thosewhichare

expressibleintheformallanguage,weimmediatelyeliminateallthesemanticparadoxes.Forexample,thereisnowayofwritingthesentence“xcanbedescribedinfewerthantwentywordsoftheEnglishlanguage”intermsoftheformallanguage.Thissolution—thiswayofavoidingthesemanticparadoxes—is acceptable from the mathematical point of view, though it is hardly an ideal solutionphilosophically.Mathematically,wecanstillformallthesetsessentialformathematics;fromabroaderpointofview,though,wecannotformanythinglikethe“setofallmen,”the“setofallLatinverbs,”etc.Nobettersolutionhasbeendevisedtothisday.WehaveseenhowZermelo’s system,withmodificationsdue toSkolemandFraenkel,manages to

avoid the paradoxes. The remaining axioms of Zermelo’s system are similar to thosewhichwill bedeveloped in the following chapters. Essentially, they provide for the existence of the set of all the

natural numbers (from which we can construct the other number systems of mathematics), andguaranteetheexistenceofunions,intersections,andproductsofsets.Beforegoingon,wewillbrieflyreviewanotherwayofaxiomatizingsettheory,whichisofincreasinginterestinourday;theessentialideasareduetovonNeumann.VonNeumannnotedthattwofactscombinetoproducethelogicalparadoxes:inthefirstplace,aswe

haveseen, thecrucialsets(forexample, thesetS inRussell’sparadox)are“toolarge;”inthesecondplace,these“large”setsareallowedtobeelementsofsets(forexample,itisadmittedthatRussell’ssetSmaybeanelementofitself).Ofthesetwofacts,Zermelousedthefirst;heavoidedtheparadoxesbymakingitimpossibletoformthe‘large”sets.VonNeumannproposedtousethesecondofthesefacts:hewouldpermittheexcessivelylargesetstoexist,butwouldnotallowthemtobeelementsofsets.Briefly,vonNeumann’ssystemcanbedescribedasfollows.AsinZermelo’stheory,thereisonlyone

elementarypredicate,namelythepredicatex∈Y.Thevariablesx,y,X,Y,etc.standforobjectswhichwe agree to call classes; however,we distinguish between two kinds of classes, namely elements—whicharedefinedtobethoseclasseswhichareelementsofclasses—andproperclasses,whicharenotelements of any class. Zermelo’s axiom of selection is now replaced by a principle called the classaxiom,whichstatesthefollowing:

IfS(x)isanystatementaboutanobjectx,thereexistsaclasswhichconsistsofallthoseelementsxwhichsatisfyS(x).

Inotherwords,ifS(x)isanystatementaboutx,wecanformtheclass

To verify that Russel’s paradox does not “work” in this system, let us go through the steps ofRussell’sargument,withSdenoting{x:xisanelementandx∉x}.

S∈Sisimpossible,forS∈SimpliesS∉S,whichisacontradiction.ThusS∉S.ItfollowsthatSis not an element, for, if S were an element, then we would have S∈ S, which would be acontradiction.

ThusRussell’sargumentmerelyprovesthattheclassS,definedabove,isnotanelement.The semantic paradoxes are avoided, as in the revised Zermelo system, by admitting in the class

axiomonlythose“statements”whichcanbewrittenintheformallanguage.Variants of von Neumann’s system have been developed by Gödel and Bernays. They have an

advantageoverZermelo’ssysteminthattheclassaxiomisclosertothespiritofintuitivesettheorythantheaxiomofselection.Indeed,ifS(x)isanystatementaboutx,theclassaxiomguaranteestheexistenceofaclasscontainingalltheelementsxwhichsatisfyS(x).Inmathematics,systemsofthevonNeumanntypeprovideuswiththeconvenienceofbeingabletospeaktothe“classofallelements”andofbeingabletooperateonclasseswhicharenotelements.(Suchclassestendtooccuratvariouspointsinhighermathematics;theycanbeavoided,butonlyatprice.)Thechiefdisadvantageofthesesystemsisthatthedistinctionbetweenclasseswhichareelementsandclasseswhicharenotelements—ahighlyartificialone—must always be borne inmind; however, this disadvantage is undoubtedly outweighed by thegreater flexibility and naturalness of vonNeumann type systems. In this textwe shall use a slightlymodifiedformofvonNeumann’ssystemofaxioms.

5OBJECTIONSTOTHEAXIOMATICAPPROACH.OTHERPROPOSALS

Whatarethechiefgoalsofaxiomaticsettheory,andtowhatextenthavethesegoalsbeensuccessfullyattained? In order to answer that question we must remember the circumstances which ledmathematicians, in the early years of the twentieth century, to search for an axiomatic basis to settheory.TheideasofCantorhadalreadythoroughlypermeatedthefabricofmodernmathematics,andhadbecomeindispensable toolsof theworkingmathematician.Algebraandanalysiswereformulatedwithina frameworkof set theory,andsomeof themostelegant,powerfulnewresults in these fieldswere established by using the methods introduced by Cantor and his followers. Thus, when theparadoxeswere discovered and there arose doubts as to the basic validity of Cantor’s system,mostmathematicianswereunderstandablyreluctanttogiveitup;theytrustedthatsomewaywouldbefoundtocircumventthecontradictionsandpreserve,ifnotall,atleastmostofCantor’sresults.Hilbertoncewroteinthisconnection:“WewillnotbeexpelledfromtheparadiseintowhichCantorhasledus.”With thediscoveryofnewerparadoxes, and the failureof all the initial attempts to avoid them, it

became increasinglyclear that itwouldnotbepossible topreserve intuitive set theory in itsentirety.Something—possiblyquitealot—wouldhavetoberelinquished.Thebestthatonecouldhopeforwastoretainasmuchofintuitivesettheoryaswasneededtosavethenewresultsofmodernmathematicsandprovideanadequateframeworkforclassicalmathematics.Briefly, then, axiomatic set theorywas created to achieve a limited aim: it had to provide a firm

foundationforasystemofsettheorywhich—whileitdidnotneedtobeascomprehensiveasintuitivesettheory—mustincludeallofCantor’sbasicresultsaswellastheconstructions(suchasthenumbersystems,functions,andrelations)neededforclassicalmathematics.ThesystemsofbothZermeloandvonNeumannweresuccessfulinachievingthislimitedaim.But

the amount of intuitive set theory which they had to sacrifice was considerable. For example, inZermelo’ssystem,aswehavealreadyseen,theintuitivewayofmakingsets—bynamingapropertyofobjects and forming the set of all objectswhich have that property—does not take place at all. It isreplacedbytheaxiomofselection,inwhichpropertiesareallowedonlytodeterminesubsetsofgivensets.Furthermore,theonlyadmissible“properties”arethosewhichcanbeexpressedentirelyintermsofthesevensymbols∈,∨,∧,¬,⇒,∀,∃andvariablesx,y,z,…Asaresult,manyofthethingsthatwenormallythinkofassets—forexample,the“setofallapples,”the“setofallatomsintheuniverse”—arenotadmissibleas“sets”inaxiomaticsettheory.Infact,theonly“sets”whichtheaxiomsprovideforare,first, theemptysetØ,andthenconstructionssuchas{Ø},{Ø,{Ø}},etc.,whichcanbebuiltupfrom the empty set. It is remarkable fact that all ofmathematics can be based upon such ameagerconceptofset.Whilethevariousaxiomaticsystemsofsettheorysavedmathematicsfromitsimmediateperil,they

failed to satisfy a great many people. In particular, many who were sensitive to the elegance anduniversalityofmathematicswerequick topointout that the creationsofZermeloandvonNeumannmustberegardedasprovisionalsolutions—asexpedientstosolveatemporaryproblem;theywillhavetobereplaced,soonerorlater,byamathematicaltheoryofbroaderscope,whichtreatstheconceptof“set”initsfull,intuitivegenerality.Thisargumentagainstaxiomaticsettheory—thatitdealswithanamputatedversionofourintuitive

conceptionofaset—hasimportantphilosophicalramifications;it ispartofafarwiderdebate,onthenature ofmathematical “truth.”The debate centers around the following question:Aremathematicalconcepts creations (that is, inventions) of thehumanmind, or do they exist independentlyof us in a“platonic”realmofconcepts,merelytobediscoveredbythemathematician?Thelatteropinionisoftenreferredtoas“platonicrealism”andisthedominantviewpointofclassicalmathematics.Weillustratethese twoopposingpointsofviewbyshowinghowtheyapply toaparticularconcept—thenotionof

naturalnumbers.Fromtheviewpointofplatonicrealism,theconcepts“one,”“two,”“three,”andsoon,existinnatureandexistedbeforethefirstmanbegantocount.Ifintelligentbeingsexistelsewhereintheuniverse,then,nomatterhowdifferenttheyarefromus,theyhavenodoubtdiscoveredthenaturalnumbersandfoundthemtohavethesamepropertiestheyhaveforus.Ontheotherhand,accordingtothe opposing point of view,while three cows, three stones, or three trees exist in nature, the naturalnumber three is a creation of ourminds;we have invented a procedure for constructing the naturalnumbers(bystartingfromzeroandadding1eachtime,thusproducingsuccessively1,2,3,etc.)andhaveinthismannerfashionedaconceptualinstrumentofourownmaking.How does platonic realism affect the status of axiomatic set theory? From the point of view of

platonicrealism,mathematicalobjectsaregiventousready-made,withall theirfeaturesandall theirproperties.Itfollowsthattosayamathematicaltheoremistruemeansitexpressesacorrectstatementabouttherelevantmathematicalobjects.(Forexample,theproposition2+2=4isnotmerelyaformalstatement provable in arithmetic; it states an actual fact about numbers.) Now—if we admit thatmathematicalobjectsaregiventouswithalltheirproperties,itfollows,inparticular,thatthenotionofset isafixed,well-definedconceptwhichwearenotfree toalterforourownconvenience.Thus the“sets”createdbyZermeloandvonNeumanndonotexist,andtheoremswhichpurporttodescribethesenonexistent objects are false! In conclusion, if we were to accept a strict interpretation of platonicrealism,wewould be forced to reject the systems of Zermelo and vonNeumann asmathematicallyinvalid.Fortunately,thetrend,forsometimenow,hasbeenawayfromplatonismandtowardamoreflexible,

more “agnostic” attitude toward mathematical “truth.” For one thing, developments in mathematicshave been conforming less and less to the pattern dictated by platonic philosophy. For another, thecardinal requirement of platonism—that everymathematical object correspond to a definite, distinctobjectofourintuition(justas“point”and“line”refertowell-definedobjectsofourspatialintuition)—came to be an almost unbearable burden on the work of creativemathematicians by the nineteenthcentury.Theyweredealingwithahostofnewconcepts (suchas complexnumbers, abstract lawsofcomposition, and the general notion of function) which did not lend themselves to a simpleinterpretation inconcrete terms.Thecaseof thecomplexnumbers is agood illustrationofwhatwashappening.Classicalmathematicsneverfeltateasewiththecomplexnumbers,foritlackedasuitable“interpretation” of them, and as a result therewere nagging doubts as towhether such things really“existed.”Realnumbersmaybeinterpretedas lengthsorquantities,but thesquarerootofanegativereal number—this did not seem to correspond to anything in the real world or in our intuition ofnumber.Yetthesystemofthecomplexnumbersarisesinamostnaturalway—asthesmallestnumbersystemwhich contains the real numbers and includes the roots of every algebraic equationwith realcoefficients;whetherornot thecomplexnumbershaveaphysicalorpsychologicalcounterpartseemsirrelevant.Thecaseofthecomplexnumbersstrikesaparallelwiththeproblemofaxiomaticsettheory.Forthe

“sets”createdbyZermeloandvonNeumannarisequitenaturallyinamathematicalcontext.Theygiveusthesimplestnotionofsetwhichisadequateformathematicsandyieldsaconsistentaxiomatictheory.Whetherornotwecaninterpretthemintuitivelymayberelativelyunimportant.Bethatasitmay,manymathematiciansintheearly1900’swerereluctanttomakesosharpabreak

with traditionasaxiomaticset theoryseemed todemand.Furthermore, theyfelt,onestheticgrounds,that amathematical theory of sets should describe all the things—andonly those things—which ourintuitionrecognizestobesets.AmongthemwasBertrandRussell;inhiseffortstoreinstateintuitivesettheory,Russellwasledtotheideathatwemayconsidersetstobeorderedinahierarchyof“levels,”where,ifAand aresetsandAisanelementof ,then is“onelevelhigher”thanA.Forexample,inplanegeometry,acircle(regardedasasetofpoints)isonelevelbelowafamilyofcircles,which,in

turn, isonelevelbelowasetof familiesofcircles.Thisbasic ideawasbuiltbyRussell intoa theorycalledthetheoryoftypes,whichcanbedescribed,inessence,asfollows.Everysethasanaturalnumberassignedtoit,calleditslevel.Thesimplestsets,thoseoflevel0,are

called individuals—they do not have elements. A collection of individuals is a set of level 1; acollectionofsetsoflevel1isasetoflevel2;andsoon.Inthetheoryoftypestheexpressiona∈Bisonlymeaningfulif,forsomenumbern,aisasetoflevelnandBisasetofleveln+1.Itfollowsthatthestatementx∈xhasnomeaninginthetheoryoftypes,andasaresult,Russell’sparadoxvanishesforthesimplereasonthatitcannotevenbeformulated.Russell’s theoryof types isbuiltuponabeautifullysimple idea.Unfortunately, inorder tomake it

“work,”Russellwasforcedtoaddahostofnewassumptions,untilfinallytheresultingtheorybecametoocumbersometoworkwithandtoocomplicatedtobetrulypleasing.Foronething,correspondingtothehierarchyof“levels”ofsets, itwasnecessarytohaveahierarchyof“level”oflogicalpredicates.Then—asawayofavoidingthesemanticparadoxes—setsatthesamelevelwerefurtherdividedinto“orders.”Finally,Russellhadtoadmitaso-calledAxiomofReducibilitywhichwasjustasarbitrary,justas ungrounded in intuition, as any of theadhoc assumptionsmade byZermelo.A a result of theseshortcomings,thetheoryoftypeshasnotgainedwideacceptanceamongmathematicians,althoughitisstillaninteresting(andmaybepromising)areaofresearch.Afarmoreradicalapproachwastakenbyagroupofmathematicianscallingthemselvesintuitionists.

To the intuitionists,muchofmodernmathematics, including almost all ofCantor’s theoryof sets, isbasedontheuncriticaluseofrulesoflogicwhichtheyconsidertobeinvalid.Theintuitionistattitudetowardsettheorycanthereforebesummedupveryeasily:itisoneofalmosttotalrejection.Inordertoproperlyunderstandthephilosophyofintuitionism,wemustfirstgainanunderstandingof

itsattitudetowardlogic.Astheintuitionistseesit, therulesoflogicusedbymathematicianshaveanempiricalcharacter.Certainmethodsofproofcametobecommonlyusedbymathematicians,and,overthe years, were codified into a body of rules. These rules were observably correct in their originalcontext,but—aftertheywerecodified—theycametobeuseduncriticallyintotallydifferentcontextsinwhichtheynolongerapplied.Letusbemorespecific: inEuclideangeometry,whichis thesourceofmostmathematicsbeforethefifteenthcentury,everytheoreminvolvesonlyafinitenumberofobjects,and eachof theseobjects (geometric figures) is givenby an explicit construction.The rules of logicused by Euclid are perfectly valid in this context, say the intuitionists; it is only when they aretransposedtoproblemsinvolvinganinfinitedomainofobjects,orinwhichtheobjectsarenotgivenbyanexplicitconstruction,thattherulesareincorrect.Asanexample,letustakethelawoftheexcludedmiddle.ThisistherulewhichsaysthatifSisany

statement,theneitherS is trueorthedenialofS is true.Inparticular, letAbeasetandletP(x)beastatementwhich ismeaningful foreveryelementx inA.By the lawofexcludedmiddle,either thereexistsanxinAsuchthatP(x)istrue,orelse foreveryx inAP(x)isfalse.Nowthe intuitionistswillacceptthisruleifAisafinitesetandifeachofitselementscanbetestedtodeterminewhetherP(x)istrueorfalse.Infact,saytheintuitionist,thisiswheretheruleoriginated—ourexperiencetellsusthatifweexamineeveryelementxinAtodeterminewhetherornotP(x)istrue(todoso,Ahastobeafiniteset),therecanbeonlytwopossibleoutcomes:eitherwehavefoundanxforwhichP(x)istrue,orelse,foreveryxinAwehavefoundP(x)tobefalse.Ourexperience,then,confirmsthelawoftheexcludedmiddleinthecaseoffinitesets.But,saytheintuitionists,toassumeitistrueinthecaseofinfinitesets—inanareawherewehavenoexperienceandexperienceisimpossible—iswhollywithoutfoundation.On grounds such as these, the intuitionists deny the law of the excluded middle. Other rules oftraditionallogicaresimilarlyrejected,becausetheygobeyondtherealmofourexperience.The intuitionistpointsout thatmathematicsoriginatedasastudyofcertainmentalconstructions—

chiefly geometric figures and simple constructions involvingwhole numbers. The theorems of early

mathematicswereessentiallystatementstotheeffectthatifcertainconstructionsarecarriedout,certainresultswillbeachieved.Forexample,considerthefollowingtheoremofgeometry:Giventwotriangles,iftwosidesandtheincludedangleofonetriangleareequal,respectively,totwosidesandtheincludedangleoftheothertriangle,thenthetwotrianglesarecongruent.Whatisexpressedhereisthefactthatifweconstructtwotriangles,withtwosidesandanincludedangleequalasstatedabove,wewillbeableto verify (for example, by using a compass) that the remaining side of one triangle is equal to theremaining side of the other triangle. The proofs of Euclidean geometry, and of early mathematicsgenerally, have a constructive character. For example, the Pythagorean theorem is proven byconstructingafigureinwhichcorrespondingpartsarecongruent,hencehavethesamearea.Oncetheconstructioniscompleted,itremainsonlytopointoutthecongruentpartsandtherebyreachthedesiredconclusion. Thus, say the intuitionists, rules of logic were originally intended to describe situationswhich arose in the context of such constructions and determinations. For example, the rule of theexcludedmiddlewasintendedsimplytonotethefactthatifwearegivena(finite)setofobjectandamethod(forexample,usingarulerandcompass) to testeachobject forsomepropertyP, then,ifweperformthetestoneveryobject,eitheroneoftheobjectswillpasstherequirementsofthetest,orelseeveryobjectwillfailtherequirements.Tosumup,then,theintuitionistmaintainsthatamathematicaltheoremisnothingmorethanafactual

statementtotheeffectthatagivenmentalconstructionwillleadtoagivenresult.Everyproofmustbeconstructive. Ifweclaim that amathematicalobject exists,wemustprove itbygivingamethod foractually constructing the object. Ifwe assert that a relation holds among given pairs of objects, ourproof must include a method for testing every pair of objects in question. The “rules of logic” arenothingmore than simpleobservationson theprocess of performingmathematical constructions;wehavenogroundstobelievetheserulesapplyoutsidethecontextofconstructivemathematics—infact,itismeaninglesstoapplythemoutsidethiscontext.Logicisincidental,notessential,tomathematics.It is obvious that the intuitionist’s notion of set must be quite different from ours. Consider, for

example, Cantor’s principle that ifwe can name a property of objects, then there exists a set of allobjectswhichhavethatproperty.Nowthisprinciple—aswellasthelimitedversionsofitacceptedbyZermelo and von Neumann—is anathema to the intuitionists. An object exists only if it can beconstructed;hence,asetexistsonlyifweareabletodescribeaprocedureforbuildingit.A full discussion of intuitionist set theory is outside the scope of this book; however, it is worth

mentioning a particular kind of set which is important in intuitionist mathematics; this is called aspread.Aspreadisidentifiedwitharuleforproducingallofitselements.Thus,aspreadisnotregardedas an “already formed” totality, but rather as a “process of formation”; each of its elements can beformedifweapplytherulelongenough.Theparadoxesdonotoccurinintuitionistsettheorybecausethecrucialsetsintheparadoxescannot

be produced in intuitionist mathematics, and the essential arguments cannot be rendered usingintuitionistlogic.

6CONCLUDINGREMARKS

During the early part of the twentieth century, as we have seen, various ways of building anoncontradictorytheoryofsetswereproposedanddevelopedbydifferent“schools”ofmathematicians.We have reviewed the basic principles of axiomatic set theory, Russell’s theory of types and theintuitionist (or “constructivist”) approach to sets. In addition to these, a greatmanyother ideaswereproposed,toovarioustodescribeinthisbriefintroduction.Ofallthewaysofdealingwithsets,theaxiomaticmethodseemedtobestsuittheneedsofmodern

mathematics. The notion of “set” embodied in the system of Zermelo and von Neumann is broad

enough for the purposes of mathematics, and therefore in a mathematical setting it is virtuallyindistinguishablefromtheCantoriannotionofset.Themethodsofproof,thesymbolism,therigor—allof thesecorrespondtocurrentmathematicalusage,Most importantofall,axiomaticset theoryseems“natural”tomostworkingmathematicians.Those who reject axiomatic set theory do so on the basis of some philosophical bias. Those

philosophicalpositions,however,whichrefusetoacceptaxiomaticsettheoryalsodenythevalidityofalarge part of modern mathematics. For example, the intuitionist school rejects the greater part ofcontemporaryanalysis,because it is foundedonnonconstructivistprinciples.While theargumentsofthesecriticpresentachallenge,andcertainlygiveusfoodforthought,theyarenotpowerfulenoughtodestroytheachievementsofthreebrilliantgenerationsofmathematicians.In the course of the past seventy years or so, set theory has come to bewidely recognized as the

fundamental,“unifying”branchofmathematics.Wehavealreadyseenhowthenaturalnumberscanbeconstructed,andtheirpropertiesderived,within theframeworkofset theory;fromthere, it iseasytodevelop the rational numbers, the real and complexnumbers, aswell as remarkable systems such asCantor’s “transfinite cardinals.” The notions of function, relation, operation, and so forth are easilydefinedintermsofsets,and,asaresult,everyknownbranchofmathematicscanbeformulatedwithinsettheory.Itisthereforelegitimate—and,infact,vital—toaskthequestion:“Howsecureafoundationdoessettheoryprovideforthewholeedificeofmathematics?”Inparticular,areweabsolutelycertainthataxiomaticsettheoryisconsistent,thatis,freeofcontradictions?Ifitis,theneverythingwedevelopwithin it—inotherwords, all ofmathematics—is consistent; and if it is not, thenwhateverwebuilduponitisworthless.Thefact is that there isnoknownproofof theconsistencyofaxiomaticset theory.This isnot too

surprising though, in view of some of the results ofmodern logic. For example, in 1931,K.Gödelproved that it is impossible to give a finitary proof of the consistency of ordinary arithmetic (of thenaturalnumbers).Thesituationinalmosteveryotherbranchofmathematicsismuchthesame.Thusthebestassurancethatwehave,atthistime,oftheconsistencyofaxiomaticsettheoryisthefactthatthefamiliarcontradictionscannotbeobtainedintheusualway.Wecannotdoanybetteratthistime.A result relating to relative consistency is of some interest. It has been proven recently that if

Zermelo’s axiomatization of set theory is consistent, then von Neumann’s axiomatization is alsoconsistent.Itisprobablethat,inthefinalanalysis,anyassuranceoftheconsistencyofmathematicswillhaveto

restonsomecombinationofbasicintuitionandempiricalevidence.

∗Thefourrulesgivenin(3)above,togetherwithsevenadditionalrules,makeupthesystemofnaturaldeductiondescribedinSlupeckiandBorkowski[8];theseelevenrulesaresufficientforeveryvalidlogicalargument.OthersystemsaredescribedinQuine[6]andSuppes[9].

1ClassesandSets

1BUILDINGSENTENCES

Beforeintroducingthebasicnotionsofsettheory,itwillbeusefultomakecertainobservationsontheuseoflanguage.Byasentencewewillmeanastatementwhich,inagivencontext, isunambiguouslyeither trueor

false.Thus

LondonisthecapitalofEngland.

Moneygrowsontrees.

Snowisblack.

areexamplesofsentences.WewilluselettersP,Q,R,S,etc.,todenotesentences;usedinthissense,P,forinstance,istobeunderstoodasassertingthat“Pistrue.”Sentencesmaybecombinedinvariouswaystoformmorecomplicatedsentences.Often,thetruthor

falsityofthecompoundsentenceiscompletelydeterminedbythetruthorfalsityofitscomponentparts.Thus, ifP is a sentence, one of the simplest sentences wemay form fromP is the negation ofP,denotedby¬P(toberead“notP”),whichisunderstoodtoassertthat“Pisfalse.”NowifPistrue,then, quite clearly, ¬P is false; and if P is false, then ¬P is true. It is convenient to display therelationshipbetween¬PandPinthefollowingtruthtable,

wheretandfdenotethe“truthvalues”,trueandfalse.

Anothersimpleoperationonsentencesisconjunction:ifPandQaresentences,theconjunctionofPandQ,denotedbyP∧Q(toberead“PandQ”),isunderstoodtoassertthat“PistrueandQistrue.”Itis intuitivelyclearthatP∧Q is trueifPandQarebothtrue,andfalseotherwise; thus,wehavethefollowingtruthtable.

ThedisjunctionofPandQ,denotedbyP∨Q(toberead“PorQ”),isthesentencewhichasserts

that“Pistrue,orQistrue,orPandQarebothtrue.”ItisclearthatP∨QisfalseonlyifPandQarebothfalse.

Anespeciallyimportantoperationonsentencesisimplication:ifPandQaresentences,thenP⇒Q(tobe read“P impliesQ”) asserts that “ifP is true, thenQ is true.”Aword of caution: in ordinaryusage,“ifPistrue,thenQistrue”isunderstoodtomeanthatthereisacausalrelationshipbetweenPandQ(asin“ifJohnpassesthecourse,thenJohncangraduate”).Inmathematics,however,implicationisalwaysunderstoodintheformalsense:P⇒QistrueexceptifPistrueandQisfalse.Inotherwords,P⇒Qisdefinedbythetruthtable.

The properties of formal implication differ somewhat from the propertieswewould expect “causal”implicationtohave.Forexample,

is true, even though there is no causal relationship between the two component sentences. To takeanotherexample,

istrue,eventhoughthetwocomponentsentencesarefalse.Thisshouldnotdisturbthereaderunduly,forformalimplicationstillhasthefundamentalpropertywhichwedemandofimplication—namely,ifP⇒Qistrue,then,necessarily,ifPistruethenQistrue.Certain compound sentences are true regardless of the truth or falsity of their component parts; a

typicalexampleisthesentenceP⇒P.RegardlessofwhetherPistrueorfalse,P⇒Pisalwaystrue;inotherwords,nomatterwhatsentencePis,P⇒Pistrue.Forfuturereferencewerecordafewsentenceswhichhavethisproperty.

1.5TheoremForallsentencesPandQ,thefollowingstatementsaretrue.

Proof

i) WewishtoprovethatifPandQareanysentences, thenP⇒P∨Q is true; inotherwords,wewishtoprovethatnomatterwhattruthvaluesareassumedbyPandQ,P⇒P∨Qisalwaystrue.Todothis,wederiveatruthtableforP⇒P∨Qasfollows.

Thebasicideaofthederivedtruthtableisthis:inline1,PandQbothtakethevaluet;thus,by1.3,P∨Qtakesthevaluet;now,PhasthevaluetandP∨Qhasthevaluet,so,by1.4,P⇒P∨Qtakesthevalue t.Wedo thesameforeach line,andwefind that inevery line(that is, foreverypossibleassignmentoftruthvaluestoPandQ)P⇒P∨Qhasthevaluet(true).Thisiswhatwehadsetouttoprove.

i)′ ThederivedtruthtableforQ⇒P∨QisanalogoustotheoneforP⇒P∨Q;theconclusionisthesame.

ii) InordertoprovethatP∧Q⇒PforallsentencesPandQ,wederiveatruthtableforP∧Q⇒P.

Ineveryline(thatis,foreverypossibleassignmentoftruthvaluestoPandQ),P∧Q⇒P takesthevaluet;thus,P∧Q⇒PistrueirrespectiveofthetruthorfalsityofitscomponentsentencesPandQ.

ii)′ ThetruthtableforP∧Q⇒QisanalogoustotheoneforP∧Q⇒P,andtheconclusionisthesame.

1.6TheoremForallsentencesP,QandR,thefollowingistrue:

Proof.Thereadershouldderivethetruthtablefor

andverifythatthissentencetakesthetruthvaluetineverylineofthetable.

1.7TheoremForallsentencesP,QandR,ifQ⇒Ristrue,then

i) P∨Q⇒P∨Ristrue,andii) P∧Q⇒P∧Ristrue.

Proof

i) WeassumethatQ⇒Ristrue,andderivethetruthtableforP∨Q⇒P∨R.

SinceweassumethatQ⇒Ristrue,wecannothave,simultaneously,QtrueandRfalse;thus,wemaydisregardthesixthlineofthetable.Inalloftheremaininglines,P∨Q⇒P∨R takes thevaluet.

ii) TheproofthatP∧Q⇒P∧Risanalogoustotheabove.WeagreethatP⇔Qistobeanabbreviationfor(P⇒Q)∧(Q⇒P).

1.8TheoremForallsentencesP,QandR,thefollowingaretrue:

Theproofofthistheoremisleftasanexerciseforthereader.Inthisandthesubsequentchapters,⇒willbeusedasanabbreviationforimplies,⇔willbeusedas

anabbreviationfor ifandonly if (wewill sometimeswrite “iff” insteadof⇔),∧willbeusedasanabbreviationforand,and∨willbeusedasanabbreviationforor.IfP,Q,R,…areanystatements,anexpressionoftheformP⇒Q⇒R⇒…shouldbeunderstoodtomeanthatP⇒Q,Q⇒R,andsoon;analogously,P⇔Q⇔R⇔…shouldbeunderstoodtomeanthatP⇔Q,Q⇔R,andsoon.Asiscustomary,∃istobereadthereexists,∀istobereadforall,and∋istobereadsuchthat.

EXERCISES1.1

1. ProveTheorem1.8.2. ProvethatthefollowingsentencesaretrueforallPandQ(DeMorgan’sLaws).

3. Provethatthefollowingsentencesaretrue,foreverysentenceP.

4. ProvethatthefollowingsentencesaretrueforallPandQ.

5. ProvethefollowingsentencesaretrueforallP,QandR.

6. Provethat,forallsentencesP,QandR,ifQ⇔Ristrue,thenthefollowingaretrue.

7. ProvethatforallsentencesP,Q,RandS,ifP⇒QandR⇒S,then

2BUILDINGCLASSES

Wewillnowbeginourdevelopmentofaxiomaticsettheory.Everyaxiomaticsystem,aswehaveseen,muststartwithacertainnumberofundefinednotions.For

example,ingeometry,thewords“point”and“line”aregenerallytakentobeundefined.Whilewearefreeinourownmindstoattacha“meaning,”intheformofamentalpicture,toeachofthesenotions,mathematicallywemustproceed“asif”wedidnotknowwhattheymeant.Nowan“undefined”notionhasnopropertiesexceptthosewhichareexplicitlyassignedtoit;therefore,wemuststateasaxiomsalltheelementarypropertieswhichweexpectourundefinednotionstohave.Oursystemofaxiomaticsettheoryisbasedonjusttwoundefinednotions:Thewordclassand the

membershiprelation∈.Alltheobjectsofourtheoryarecalledclasses.ItwasexplainedinChapter1that inorder toavoid logicalparadoxes,wehave todistinguishbetween twokindsofclasses—thosethatarecalledsetsandthosethatarecalledproperclasses.Weshallsaynomoreaboutthisdistinctionnow,butshallreturntoitlater.IfxandAareclasses, theexpressionx∈A isread“x isanelementofA”,or“xbelongs toA”,or

simply“xisinA”.Itisconvenienttowritex∉Afor“xisnotanelementofA“.

DefinitionLetxbeaclass.IfxisanelementofsomeclassAthenxiscalledanelement.

Weshallhavemoretosayaboutelementsandclassesinthefinalsectionofthischapter.Fromhereon,weshalluse thefollowingnotationalconvention: lower-case lettersa,b,c,x,y,…willbeusedonlytodesignateelements.Thus,acapitalletter,suchasA,maydenoteeitheranelementoraclasswhichisnotanelement,butalower-caseletter,suchasx,maydenoteonlyanelement.Intuitively,twoclassesshouldbecalledequaliftheyareelementsofthesameclasses.

1.9Definition LetA andB be classes.We defineA =B tomean that every class that hasA as anelementalsohasBasanelement,andvice-versa.Insymbols,

We have defined two classes to be equal if and only if they are members of the same classes.Informally,wemaythereforethinkofthemasinterchangeable.Equalclasseshaveanotherproperty.IfAandBareequal,weexpectthemtohavethesameelements.Thispropertyisstatedasourfirstaxiom:

A1.A=Biffx∈A⇒x∈Bandx∈B⇒x∈A.ThisaxiomissometimescalledtheAxiomofExtent.

1.10DefinitionLetAandBbeclasses;wedefineA⊆BtomeanthateveryelementofAisanelementofB.Insymbols,

IfA⊆B,thenwesaythatAisasubclassofB.

WedefineA⊂BtomeanthatA⊆BandA≠B;inthiscase,wesaythatAisastrictsubclassofB.

IfAisasubclassofB,andAisaset,wewillcallAasubsetofB.

Afewsimplepropertiesofequalityandinclusionaregiveninthenexttheorem.

1.11TheoremForallclassesA,BandC,thefollowinghold:i) A=A.ii) A=B⇒B=A.iii) A=BandB=C⇒A=C.iv) A⊆BandB⊆A⇒A=B.v) A⊆BandB⊆C⇒A⊆C.

Proof

i) Thestatementx∈A⇒x∈Aandx∈A⇒x∈Aisobviouslytrue;thus,byAxiomA1,A=A.ii) SupposeA=B;thenx∈A⇒x∈Bandx∈B⇒x∈A;henceby1.8(i)′x∈B⇒x∈Aandx∈A⇒x∈B;thus,byAxiomA1,B=A.

iii) SupposeA=BandB=C;thenwehavethefollowing:

From the firstand thirdof these statementsweconclude (by1.6) thatx∈A⇒x∈C. From thesecondandfourthofthesestatementsweconcludethatx∈C⇒x∈A.Thus,byAxiomA1,A=C.Weleavetheproofsof(iv)and(v)asanexerciseforthereader.

Wehaveseenthattheintuitivewayofmakingclassesistonameapropertyofobjectsandformtheclassofall theobjectswhichhave thatproperty.Oursecondaxiomallowsus tomakeclasses in thismanner.

A2. LetP(x)designateastatementaboutxwhichcanbeexpressedentirelyintermsofthesymbols∈,∨,∧,¬,⇒,∃,∀,brackets,andvariablesx,y,z,A,B,…ThenthereexistsaclassCwhichconsistsofalltheelementsxwhichsatisfyP(x).

AxiomA2iscalledtheaxiomofclassconstruction.

ThereadershouldnotethataxiomA2permitsustoformtheclassofalltheelementsxwhichsatisfyP(x),not the class of all the classesxwhich satisfyP(x);as discussed on page13, this distinction issufficienttoeliminatethelogicalparadoxes.ThesemanticparadoxeshavebeenavoidedbyadmittinginaxiomA2onlythosestatementsP(x)whichcanbewrittenentirelyintermsofthesymbols∈,∨,∧,¬,⇒,∃,∀,bracketsandvariables.TheclassCwhoseexistenceisassertedbyAxiomA2willbedesignatedbythesymbol

1.12Remark.Theuseof a smallx in theexpression{x :P(x)} is not accidental, but quite essential.Indeed,wehaveagreedthatlower-caselettersx,y,etc.,willbeusedonlytodesignateelements.Thus

assertsthatCistheclassofalltheelementsxwhichsatisfyP(x).

Wewillnowusetheaxiomofclassconstructiontobuildsomenewclassesfromgivenclasses.

1.13Definition LetA andB be classes; the union ofA andB is defined to be the class of all theelementswhichbelongeithertoA,orB,ortobothAandB.Insymbols,

Thus,x∈A∪Bifandonlyifx∈Aorx∈B.

1.14DefinitionLetAandBbeclasses;theintersectionofAandBisdefinedtobetheclassofalltheelementswhichbelongtobothAandB.Insymbols,

Thus,x∈A∩Bifandonlyifx∈Aandx∈B.

1.15Definition By the universal class we mean the class of all elements. The existence of theuniversal class is a consequence of the axiom of class construction, for if we take P(x) to be thestatementx=x, thenA2guarantees theexistenceofaclasswhichconsistsofall theelementswhichsatisfyx=x;by1.11(i),everyelementisinthisclass.

1.16DefinitionBytheemptyclasswemeantheclassØwhichhasnoelementsatall.Theexistenceoftheemptyclassisaconsequenceoftheaxiomofclassconstruction;indeed,A2guaranteestheexistenceofaclasswhichconsistsofalltheelementswhichsatisfyx≠x;byTheorem1.11(i),thisclasshasnoelements.

1.17TheoremForeveryclassA,thefollowinghold:

Proof

i) In order to prove that Ø⊆ A, we must show that x∈ Ø⇒ x∈ A. It suffices to prove thecontrapositiveofthisstatement,thatis,x∉A⇒x∉Ø.Well,supposex∉A;thencertainlyx∉Ø,forØhasnoelements;thusx∉A⇒x∉Ø.

ii) Ifx∈A,thenxisanelement;hence .

1.18DefinitionIftwoclasseshavenoelementsincommon,theyaresaidtobedisjoint.Insymbols,

1.19DefinitionThecomplementofaclassAistheclassofalltheelementswhichdonotbelongtoA.Insymbols,

Thus,x∈A′ifandonlyifx∉A.

Relationsamongclassescanbe representedgraphicallybymeansofausefuldeviceknownas theVenndiagram.Aclassisrepresentedbyasimpleplanearea(circularorovalinshape);ifitisdesiredtoshowthecomplementofaclass,thenthecircleorovalisdrawnwithinarectanglewhichrepresentstheuniversalclass.Thus,A∪BisrenderedbytheshadedareaofFig.1,A∩BbytheshadedareaofFig.2,andA′bytheshadedareaofFig.3.ThereaderwillfindthatVenndiagramsarehelpfulinguidinghisreasoning about classes, and that they givemoremeaning to set-theoretic formulas bymaking themmoreconcrete.Forexample,inSection3ofthischapterwewillprovetheformula

ThisformulaisillustratedinFig.4,wheretheshadedarearepresentsA∩ (B∪C);one immediatelynoticesthatthissameshadedarearepresents(A∩B)∪(A∩C).

Fig.1

Fig.2

Fig.3

Fig.4

EXERCISES1.2

1. SupposethatA⊆BandC⊆D;provethat

[Hint:UsetheresultofExercise7,ExerciseSet1.1.]2. SupposeA=BandC=D;provethat

[Hint:Usetheresultoftheprecedingexercise.]3. ProvethatifA⊆B,thenB′⊆A′.

[Hint:UsetheresultofExercise4(a),ExerciseSet1.1.]4. ProvethatifA=B,thenA′=B′.5. ProvethatifA=BandB⊆C,thenA⊆C.6. ProvethatifA⊂BandB⊂C,thenA⊂C.7. ProveTheorem1.11,parts(iv)and(v).8. LetS={x:x∉x};useRussell’sargumenttoprovethatSisnotanelement.9. DoesAxiomA2allowustoformthe“classofallclasses”?Explain.10. ExplainwhyRussell’sparadoxandBerry’sparadoxcannotbeproducedbyusingAxiomA2.

3THEALGEBRAOFCLASSES

One of the most interesting and useful facts about classes is that under the operations of union,intersection, and complementation they satisfy certain algebraic laws fromwhichwe candevelop analgebraofclasses.Weshallseelater(Chapter4)thatthealgebraofclassesismerelyoneexampleofa

structureknownasaBooleanalgebra; another example is the“algebraof logic,”where∨,∧, ¬ areregardedasoperationsonsentences.

1.20TheoremIfAandBareanyclasses,theni) A⊆A∪BandB⊆A∪B.ii) A∩B⊆AandB∩B⊆B.

Proof

i) ToprovethatA⊆A∪B,wemustshowthatx∈A⇒x∈A∪B:

Analogously,wecanshowthatB⊆A∪B.ii) ToprovethatA∩B⊆A,wemustshowthatx∈A∩B⇒x∈A.

1.21TheoremIfAandBareclasses,theni) A⊆BifandonlyifA∪B=B,ii) A⊆BifandonlyifA∩B=A.

Proof

i) LetusfirstassumethatA⊆B;thatis,x∈A⇒x∈B.Then

Thus,A∪B⊆B;butB⊆A∪Bby1.20(i);consequently,A∪B=B.Conversely,letusassumethatA∪B=B.By1.20(i),A⊆A∪B;thusA⊆B.

ii) Toproofisleftasanexerciseforthereader.

1.22Theorem(AbsorptionLaws).ForallclassesAandB,

Proof

i) By1.20(ii),A∩B⊆A;therefore,by1.21(i),A∪(A∩B)=A.ii) By1.20(i),A⊆A∪B;therefore,by1.21(ii),A∩(A∪B)=A.

1.23TheoremForeveryclassA,(A′)′=A.

Proof

1.24Theorem(DeMorgan’sLaws).ForallclassesAandB,

Proof

ii) Theproofisleftasanexerciseforthereader.

1.25TheoremForallclassesA,BandC,thefollowingaretrue.

CommutativeLaws: i)A∪B=B∪Aii)A∩B=B∩A

IdempotentLaws: iii)A∪A=Aiv)A∩A=A

AssociativeLaws: v)A∪(B∪C)=(A∪B)∪Cvi)A∩(B∩C)=(A∩B)∩C

DistributiveLaws: vii)A∩(B∪C)=(A∩B)∪(A∩C)viii)A∪(B∩C)=(A∪B)∩(A∪C)

Proof

Theproofsof(ii),(iii),(iv),(vi),and(viii)areexercisesforthereader.

Theemptyclassandtheuniversalclassareidentityelementsforunionandintersectionrespectively;theysatisfythefollowingsimplerules:

1.26TheoremForeveryclassA,

Proof

i) By1.17,Ø⊆A,andthereforeby1.21(i),A∪Ø=A.iii) By1.17(ii), ,andthereforeby1.21(i), .

Theproofsoftheremainingpartsofthistheoremareleftasanexerciseforthereader.

Byusingthelawsofclassalgebrawhichwehavedevelopedabove,wecanprovealltheelementarypropertiesofclasseswithoutreferringtothedefinitionsofthesymbols∪,∩,′,and⊆.Thefollowingisanexampleofhowsuchproofsarecarriedout.

ExampleProvethatA∩(A′∪B)=A∩B.

Proof

Thefollowingdefinitionisfrequentlyuseful:ThedifferenceoftwoclassesAandBistheclassofallelementswhichbelongtoA,butdonotbelongtoB.Insymbols,

ExampleProvethatA−B=B′−A′.

Proof

ItisusefultonotethatwiththeaidofTheorem1.21,relationsinvolvinginclusion(⊆),notmerelyequality,canbeprovedusingclassalgebra.

EXERCISES1.3

1. ProveTheorem1.21(ii).2. ProveTheorem1.24(ii).3. ProveTheorem1.25,parts(ii),(iii),(iv),(vi)and(viii).4. ProveTheorem1.26,parts(ii),(iv),(v)through(viii).5. Useclassalgebratoprovethefollowing.

6. Useclassalgebratoprovethefollowing.a)IfA∩C=Ø,thenA∩(B∪C)=A∩B.b)IfA∩B=Ø,thenA−B=A.c)IfA∩B=ØandA∪B=C,thenA=C−B.

7. Usingclassalgebra,proveeachofthefollowing.a)A∩(B−C)=(A∩B)−C.b)(A∪B)−C=(A−C)∪(B−C).c)A−(B∪C)=(A−B)∩(A−C).d)A−(B∩C)=(A−B)∪(A−C).

8. Wedefinetheoperation+onclassesasfollows:IfAandBareclasses,then

Proveeachofthefollowing.

9. Proveeachofthefollowing.a)A∪B=Ø⇒A=ØandB=Ø.b)A∩B′=ØifandonlyifA⊆B.c)A+B=ØifandonlyifA=B.

10. Proveeachofthefollowing.a)A∪C=B∪CifandonlyifA+B⊆C.b)(A∪C)+(B∪C)=(A+B)−C.

11. UseclassalgebratoprovethatA⊆BandC=B−A,thenA=B−C.

4ORDEREDPAIRSCARTESIANPRODUCTS

Ifaisanelement,wemayusetheaxiomofclassconstructiontoformtheclass

Itiseasytoseethat{a}containsonlyoneelement,namelytheelementa.Aclasscontainingasingleelementiscalledasingleton.Ifaandbareelements,wemayusetheaxiomofclassconstructiontoformtheclass

Clearly{a, b} contains two elements, namely the elementsa and b. A class containing exactly twoelementsiscalledanunorderedpair,or,moresimply,adoubleton.Inlikefashion,wecanformtheclasses{a,b,c},{a,b,c,d},andsoon.

1.27TheoremIf{x,y}={u,v},then

Proof

Suppose{x,y}={u,v};wewillconsidertwocases,accordingasx=yorx≠y.

Case1:x=y.Nowu∈{u,v}and{u,v}={x,y},sobyAxiomA1,u∈{x,y}.

Thus,u=xoru=y;ineithercase,u=x=y.Analogously,v=x=y,sowehaveu=v=x=y,andwearedone.

Case2:x≠y.Nowx∈{x,y}and{x,y}={u,v},sox∈{u,v};thusx=uorx=v.Wewillconsiderthesetwocasesseparately:i) x=u:Nowy∈{x,y},hencey∈{u,v},soy=uory=v;butx=u,soify=u,thenx=y,whichis

impossiblebecauseweassumex=y.Thusy=v.Inthiscase,wearedone.ii) x=v:Werepeat theargumentof (i),merelyswitching the rolesofuandv;wegety =u; hence,

again,wearedone.

Animportantnotioninmathematicsisthatofanorderedpairofelements.Intuitively,anorderedpairisaclassconsistingoftwoelementsinaspecifiedorder.Infact,theorderisnotreallyessential;whatisessentialisthatorderedpairshavethefollowingproperty.

1.28Let(a,b)and(c,d)beorderedpairs.If(a,b)=(c,d),thena=candb=d.

Wewouldliketodefineorderedpairsinsuchawayastoavoidintroducinganewundefinednotionof“order.”Itisaninterestingfactthatthiscan,indeed,beaccomplished;weproceedasfollows.

1.29DefinitionLetaandbbeelements;theorderedpair(a,b)isdefinedtobetheclass

Itisworthnotingthat

Hencethereisacleardistinctionbetweenthetwopossible“orders”(a,b)and(b,a):theyaredifferentclasses.Itremainstoprovethatorderedpairs,aswehavejustdefinedthem,haveProperty1.28.

1.30TheoremIf(a,b)=(c,d),thena=candb=d.

Proof.Supposethat(a,b)=(c,d);thatis,

ByTheorem1.27,either

or

wewillconsiderthesetwocasesseparately.

Case 1: {a}={c} and {a, b}={c, d}. From {a}={c}, it follows that a = c. From {a, b}={c, d} andTheorem1.27,itfollowsthateithera=candb=d,ora=dandb=c;inthefirstcase,wearedone;inthesecondcase,wehaveb=c=a=d,soagainwearedone.

Case2:{a}={c,d}and{a,b}={c}.Herec∈{c,d}and{c,d}={a},soc∈{a};thusc=a;analogously,d=a.Also,b∈{a,b}and{a,b}={c},sob∈{c};henceb=c.Thusa=b=c=d,andwearedone.

1.31DefinitionTheCartesianproductof twoclassesAandB is the class of all orderedpairs (x, y)wherex∈Aandy∈B.Insymbols,

ThefollowingareafewsimplepropertiesofCartesianproducts.

1.32TheoremForallclassesA,B,andC,i) A×(B∩C)=(A×B)∩(A×C).ii) A×(B∪C)=(A×B)∪(A×C).iii) (A×B)∩(C×D)=(A∩C)×(B∩D).

Proof

JustaswefounditinstructivetorepresentrelationsbetweenclassesbymeansofVenndiagrams,itisoftenconvenienttoillustraterelationsbetweenproductsofclassesbyusingagraphicdeviceknownasacoordinatediagram.AcoordinatediagramisanalogoustothefamiliarCartesiancoordinateplane;therearetwoaxes—averticaloneandahorizontalone—butweconsideronlyone“quadrant.”IfwewishtorepresentaclassA×B,thenasegmentofthehorizontalaxisismarkedofftorepresentAandasegmentof the vertical axis is marked off to represent B; A × B is the rectangle determined by these twosegments(Fig.5).Asanexampleoftheuseofcoordinatediagrams,Theorem1.32(iii)isillustratedinFig.6.

Fig.5

Fig.6

EXERCISES1.4

1. LetA={a,b,c,d},B={1,2,3},C={x,y,z}.FindA×B,B×A,C×(B×A),(A∪B)×C,(A×C)∪(B×C),(A∪B)×(B∪C).

2. ProveTheorem1.32(ii).3. ProvethatA×(B−D)=(A×B)−(A×D).4. Provethat(A×B)∩(C×D)=(A×D)∩(C×B).5. IfA,BandCareclasses,provethefollowing.

a)(A×A)∩(B×C)=(A∩B)×(A∩C).b)(A×B)−(C×C)=[(A−C)×B]∪[A×(B−C)].c)(A×A)−(B×C)=[(A−B)×A]∪[A×(A−C)].

6. Prove thatA andB aredisjoint if andonly if, for anynonempty classC,A ×C andB ×C aredisjoint.

7. IfAandCarenonemptyclasses,provethatA⊆BadC⊆DifandonlyifA×C⊆B×D.8. LetA,B,C,Dbenonemptyclasses.ProvethatA×B=C×DifandonlyifA=CandB=D.9. IfA,B,andCareanyclasses,prove

a)A×BandA′×Caredisjoint, b)B×AandC×A′aredisjoint.10. ProvethatA×B=ØifandonlyifA=ØorB=Ø.11. Proveeachofthefollowing.

a)Ifa={b},thenb∈a.b)x=yifandonlyif{x}={y}.c)x∈aifandonlyif{x}⊆a.d){a,b}={a}ifandonlyifa=b.

12. Wegivethefollowingalternativedefinitionoforderedpairs:(x,y)={{x,Ø},{y,{Ø}}}.Usingthisdefinition,provethat

5GRAPHS

Aclassoforderedpairsiscalledagraph.Inotherwords,agraphisanarbitrarysubclassof .TheimportanceofgraphswillbecomeapparenttothereaderinChapters2and3.Itmaybeshown,

for instance, that a function from A to B is a graph G⊆ A × B with certain special properties.Specifically,Gconsistsofallthepairs(x,y)suchthaty=f(x).Thisexamplemayhelptomotivatethefollowingdefinitions.

1.33DefinitionIfGisagraph,thenG−1isthegraphdefinedby

1.34DefinitionIfGandHaregraphs,then isthegraphdefinedasfollows:

Thefollowingareafewbasicpropertiesofgraphs.

1.35TheoremIfG,H,andJaregraphs,thenthefollowingstatementshold:

Proof

1.36DefinitionLetGbeagraph.BythedomainofGwemeantheclass

andbytherangeofGwemeantheclass

Inotherwords,thedomainofGistheclassofall“firstcomponents”ofelementsofG,andtherangeof

Gistheclassofall“secondcomponents”ofelementofG.

1.37TheoremIfGandHaregraphs,then

Proof

1.38CorollaryLetGandHbegraphs.IfranH⊆domGthendom =domH.

Theproofofthistheoremisleftasanexerciseforthereader.

EXERCISES1.5

1. Let

and

Find .2. ProveTheorem1.37,parts(ii)and(iv).3. ProveTheorem1.38.4. IfG,H,andJaregraphs,proveeachofthefollowing.

5. IfGandHaregraphs,proveeachofthefollowing.a)(G∩H)−1=G−1∩H−1,b)(G∪H)−1=G−1∪H−1.

6. IfG,H,J,andKaregraphs,provea)ifG⊆HandJ⊆K,thenG J⊆H K,b)G⊆HifandonlyifG−1⊆H−1.

7. IfA,B,andCareclasses,proveeachofthefollowing.

8. LetGandHbegraphs;proveeachofthefollowing.a)IfG⊆A×B,thenG−1⊆B×A.b)IfG⊆A×BandH⊆B×C,then ⊆A×C.

9. IfGandHaregraphs,proveeachofthefollowing.a)dom(G∪H)=(domG)∪(domH).b)ran(G∪H)=(ranG)∪(ranH).c)domG−domH⊆dom(G−H).d)ranG−ranH⊆ran(G−H).

10. LetGbeagraph,andletBbeasubclassofthedomainofG.BytherestrictionofGtoBwemeanthegraph

Proveeachofthefollowing.

11.LetGbeagraphandletBbeasubclassofthedomainofG.WeusethesymbolG(B)todesignatetheclass

Proveeachofthefollowing.

6GENERALIZEDUNIONANDINTERSECTION

Considertheclass{A1,A2,…,An};itselementsareindexedbythenumbers1,2,…,n.Suchaclassifoftencalledanindexedfamilyofclasses;thenumbers1,2,…,narecalledindicesandtheclass{1,2,…,n}iscalledtheindexclass.Moregenerally,wearefrequentlyledtothinkofaclassIwhoseelementsi,j,k,…serveasindicesto

designatetheelementsofaclass{Ai,Aj,Ak,…}.Theclass{Ai,Aj,Ak,…}iscalledanindexedfamilyofclasses,Iiscalleditsindexclass,andtheelementsofIarecalledindices.Acompactnotationwhichisoftenusedtodesignatetheclass{Ai,Aj,Ak,…}is

Thus,speakinginformally,{Ai}i∈IistheclassofalltheclassesAi,asirangesoverI.

Remark. The definition of an indexed family of classeswhichwe have just given is, admittedly, anintuitive one; it relies on the intuitive notion of indexing. This intuitive definition is adequate at the

presenttime;however,forfuturereference,wenowgiveaformaldefinitionofthesameconcept:Byanindexedfamilyofclasses,{Ai}i∈I,wemeanagraphGwhosedomainisI;foreachi∈IwedefineAiby

Forexample,consider{Ai}i∈IwhereI={1,2},A1={a,b},andA2={c,d}.Then,formally,{Ai}i∈Iisthegraph

If{Ai}i∈Iisanindexedfamilyofclassessuchthatforeachi∈I,Aiisanelement,thenwelet{Ai:i∈I}designatetheclasswhoseelementsarealltheAi,thatis,{Ai:i∈I}={x:x=Ai,forsomei∈I}.However,weshallfollowcurrentmathematicalusageandusethetwoexpressions,{Ai}i∈Iand{Ai:i∈I},interchangeably.

1.39DefinitionLet{Ai}i∈Ibeanindexedfamilyofclasses.TheunionoftheclassesAiconsistsofalltheelementswhichbelongtoatleastoneclassAiofthefamily.Insymbols,

The intersection of the classesAi consists of all the elementswhich belong to every classAi of thefamily.Insymbols,

Thefollowingaresomebasicpropertiesofindexedfamiliesofclasses.

1.40TheoremLet{Ai}i∈Ibeanindexedfamilyofclasses.

Proof

i) SupposethatAi⊆Bforeveryi∈I ;,thenx∈Ajforsomej∈I;butAj⊆B,sox∈B.

Thus .

Theproofof(ii)isleftasanexerciseforthereader.

1.41Theorem(GeneralizeddeMorgan’sLaws).Let{Ai}i∈Ibeanindexfamilyofclasses.Then,

Proof

Theproofof(ii)isleftasanexerciseforthereader.

1.42Theorem(GeneralizedDistributiveLaws).Let{Ai}i∈Iand{Bj}j∈Jbeindexedfamiliesofclasses.Then

Proof

Theproofof(ii)isleftasanexerciseforthereader.

Atheoremconcerningtheunionofgraphswillbeusefultousinthenextchapter.

1.43TheoremLet{Gi}i∈Ibeafamilyofgraphs.Then

Proof

Theproofof(ii)isleftasanexerciseforthereader.

Avariantnotationfortheunionandintersectionofafamilyofclassesissometimesuseful.If isaclass(itselementsarenecessarilyclasses),wedefinetheunionof ,orunionoftheelementsof , tobetheunionofalltheclasseswhichareelementsfor .Insymbols,

In other words, A if and only if there is a class A such that x ∈ A and A ∈ .

Analogously,wedefinetheintersectionof ,orintersectionoftheelementsof ,tobetheintersectionofalltheclasseswhichareelementsof .Insymbols,

1.46ExampleLet ={K,L,M},whereK={a,b,d},L={a,c,d},andM={d,e}.Then

1.47Remark.Itisfrequentpractice,intheliteratureofsettheory,towrite

and

Weshalloccasionallyfollowthatpracticeinthisbook.

EXERCISES1.6

1. ProveTheorem1.40(ii).2. ProveTheorem1.41(ii).3. ProveTheorem1.42(ii).4. ProveTheorem1.43(ii).

5.Let{Ai}i∈Iand{Bi}i∈IbetwofamiliesofclasseswiththesameindexclassI.Supposethat∀i∈I,Ai⊆Bi;provethat

6. Let{Ai}i∈Iand{Bj}j∈Jbeindexedfamiliesofclasses.Provethefollowing.

7. Let{Ai}i∈Iand{Bj}j∈Jbeindexedfamiliesofclasses.Supposethat∀i∈I,∃j∈JBj⊆Ai.Provethat

8. Let{Ai}i∈Iand{Bj}j∈Jbeindexedfamiliesofclasses.Provethat

9. Wesaythatanindexedfamily{Bi}i∈IisacoveringofAif Supposethat{Bi}i∈Iand

{Cj}j∈JaretwodistinctcoveringsofA.Provethatthefamily{(Bi∩Cj)}(i,j)∈I×JisacoveringofA.

10. Leta={u,v,w},b={w,x},c={w,y},r={a,b},s={b,c},andp={r,s}.Findtheclasses∪(∪p),∩(∩p),∪(∩p),∩(∪p).

11. Provethat .12. Proveeachofthefollowing.

7SETS

Undoubtedly, everythingpresented in this chapter is fairly familiar toyou.Even thoughweused theword class where you are more accustomed to hearing set,it is obvious that the “union” and“intersection”defined in this chapter are the sameas the familiarunionand intersectionof sets.TheCartesianproductofclassesisnodifferentfromtheCartesianproductofsets,andthesameistruefortheotherconceptsintroducedinthischapter.Thishasbeenveryconvenient,butit istimetofacethefactthatifwefailtodistinguishbetweenthetwokindsofclasses—namelysetsandproperclasses—weshallrapidlyfindourselvesinthemidstoflogicalcontradictions.Therearetwokindsofclasses:Setsandproperclasses.

1.48DefinitionIfX∈YforsomeclassY,thenXisaset.IfX∉YforanyclassY,thenXisaproperclassYouwillnoticeimmediatelythatasetisexactlyanelement.Thisistheawkwardaspect of our terminology: Itmay irk us, but axiomatic set theory requires us to accept thewords

“element” and “set” as synonyms, to be distinguished from proper classes, which are not sets, and

cannotbeelementsofanything.Proper classes are a little like embarrassing relatives thatwe are forced to acknowledge, butwe’d

rather keep at arm’s length. Those are the classes that may lead to paradox, and in order to domathematicssafelywewanttobesurethatwearedealingonlywithsets.Sothemostimportantaxiomsof set theory are designed to provide the assurance that whenwe carry out operations on sets—forexamplewhenweformageneralizedunionofsets—wearenotinadvertentlycreatingaproperclass.Tobeprecise,wewantaguaranteethatwhenperformingoperationsonsets,theresultsoftheoperationsaresetsandnotproperclasses.Since,intuitively,asetisanyclassthatisnot“toolarge”,wewouldcertainlyexpecteverysubclass

ofasettobeaset.SoifAisnot“toolarge”andB⊆A,thenBshouldnotbe“toolarge”.Westatethisasanaxiom.

A3.Everysubclassofasetisaset.

AxiomA3hasasimpleconsequence:FromTheorem1.20,A∩B⊆A.SobyAxiomA3,ifAisaset,thenA∩Bisaset.Thatis,theintersectionofanytwosetsisaset.Withallthistalkaboutsets,onewouldassumethatwecouldexhibitone—inotherwords,wecould

say,“Here,thisisaset!”Thefactisthatwecan’t:Ourdefinitionsandaxiomssofarhavenotgivenusanysets.Sowemustnowstateanaxiomwhoseonlypurposeistoassertthatsetsexist—well,atleastonesetexists.

A4.∅isaset.

Mightyoaksfromlittleacornsgrow.Weshallseelaterthatfromthehumbleemptyset,manymoresetscanbeshowntoexist.Asapreview,considertheset{∅}:Ithasoneelement,namelytheemptyset.Theset{∅, {∅}}has two elements.We’ll stophere for now.Actually, in a later chapterwepresentanotheraxiomthatprovidesfortheexistenceofsets,thistimeofinfinitesets.Afteraddingthataxiom,ourAxiomA4willberedundant.Meanwhileitwillserveuswell.Itisreasonabletoassumethatifaandbaresets,thenthedoubleton{a,b},withonlytwoelements,

isnottoolargetobeaset.

A5.Ifaandbaresets,then{a,b}isaset.

FromAxiomsA3andA5,itfollowsimmediatelythatifaisaset,thenthesingleton{a}isaset.Thenexttwoaxiomsareespeciallyimportant,becausetheyguarantee that ifyoucombinesets into largersets—forexamplebyformingthegeneralizedunionofafamilyofsets,oraCartesianproductofmanysets—theselargercollectionsarestillsets.

1.49DefinitionLetA be a set; by thepowerset ofAwemean the class of all the subsets ofA. Insymbols,thepowersetofAistheclass

NotethatbyAxiomA3, istheclassofallthesetsBwhichsatisfyB⊆A.

1.50ExampleIfA={a,b},then

Notethat ifandonlyifB⊆A.

It is easy to see that is a larger class thanA, for includes (among other things) all thesingletons{x}asxrangesoverA.Thuswemaylegitimatelyaskthefollowingquestion:ifAisaset,isit necessarily true that is a set?Or is it possible that may be “too large” to be a set? Ananalogousquestionmayberaisedinregardtotheunionofsets: if isasetofsets, isaset,ormightitbe“toolarge”tobeaset?Thesequestionsmaybeansweredintuitivelyasfollows:noneofthe“giant”collectionswhichcausecontradictionsinintuitivesettheorycanbeobtainedeitherasapowersetofasetorasaunionofasetofsets.Thuswearejustifiedinadoptingthefollowingasaxioms.

A6.If isasetofsets,then isaset.

A7.IfAisaset,thenpowersetofAisaset.

IfAandBaresets,thenbyAxiomA5,{A,B}isaset;itfollowsimmediatelyfromDefinition1.44that;thus,byAxiomA6,A∪Bisaset.Thisshowsthattheunionoftwosetsisaset.

Several other axioms for sets have been proposed, but are not essential in everydaymathematicalpractice.Oneaxiom,thatweshallencounteragaininthefinalchapterofthisbook,iscalledtheAxiomofFoundation.Itstatesthefollowing:

A8.IfAisanyset,thereisanelementa∈Asuchthata∩A=∅.ThisaxiomhasanequivalentformwhichhasapplicationsinChapter11:Anydescendingsequenceofsets…∈A4∈A3∈A2∈A1isfinite.Inotherwords,youcannothaveaninfinitedescendingsequenceofsets,eachanelementofthepreviousone.

Averyintriguingaxiom,whichhassurprisinglyfar-reachingconsequences,isthefollowing,whichwe do not include in the axiomatic system of this book: Every proper class is in one-to-onecorrespondencewiththeuniversalclass ,thatis,withtheclassofallsets.Thoughweshallnotadoptthisaxiomhere,ithelpsustoformanintuitiveimageofwhatproperclassesarelike:Theyareclasseswhosesizeisaslargeasthatoftheclasscontainingallsets.Threemoreaxioms,whosepurposewillbeexplainedlater,areintroducedinChapters5,6and7.

1.51TheoremIfAandBaresets,thenA×Bisaset.

Proof.LetAandBbesets.ByAxiomA6,A∪B isaset;byAxiomA7, isaset; finally,byAxiomA7again, is a set.Wewill prove thatA ×B⊆ , and itwill follow, byAxiomA3,thatA×Bisaset.Let(x,y)∈A×B.By1.29, (x,y)={{x},{x,y}}.Nowx∈A∪B,hence{x}⊆A∪B,so {x}∈

.Similarly,x∈A∪Bandy∈A∪B,so{x,y}⊆A∪B,hence{x,y}∈ .Wehave justshownthat{x}and{x,y}areelementsof ,hence

itfollowsthat

thusis,

Thus

ItfollowsfromTheorem1.51andAxiomA3thatifAandBaresets,thenanygraphG⊆A×Bisaset.ItiseasytoshowthatifGisaset,thendomGandranGaresets(seeExercise5,ExerciseSet1.7).

Usingthisfact,onecaneasilyshowthatifGandHaresets,thenG HandG−1aresets(seeExercise6,ExerciseSet1.7).

EXERCISES1.7

1. IfAandBaresets,provethatA−BandA+Baresets.(SeeExercise8,ExerciseSet1.3.)2. IfA isaproperclassandA⊆B,prove thatB isaproperclass.Conclude that theunionof two

properclassesisaproperclass.3. Prove that the“Russellclass”and theuniversalclassareproperclasses. [Hint.Use the resultof

Exercise8,ExerciseSet1.2.]4. Let{Ai}i∈Ibeanindexedfamilyofsets.Provethat isaset.5. LetGbeagraph.ProvethatifG isaset, thendomGandranGaresets.[Hint:Showthatboth

domGandranGaresubsetsof∪(∪G).]6. LetGandHbegraphs.ProvethatifGandHaresets,thenG−1and aresets.7. Letr={a,b},s={b,c},p={r,s}.Findthesets , ,and .8. LetAandBbesets;provethefollowing.

9. IfAand aresets,provethefollowing.

10. Exhibitthesets .

2Functions

1INTRODUCTION

The concept of a function is one of themost basicmathematical ideas and enters into almost everymathematical discussion. A function is generally defined as follows: IfA andB are classes, then afunctionfromAtoBisarulewhichtoeveryelementx∈Aassignsauniqueelementy∈B;toindicatethisconnectionbetweenxandyweusuallywritey=f(x).Forinstance,considerthefunctiony=sinx;ifwetakeAtobethesetofalltherealnumbersandBtobetheclosedinterval[−1,1],thenitiseasytoseethaty=sinxisarulewhich,toeverynumberx∈A,assignsauniquenumbery∈B.Thegraphofafunctionisdefinedasfollows:IffisafunctionfromAtoB,thenthegraphoffisthe

classofallorderedpairs(x,y)suchthaty=f(x).Forexample,letA={a,b,c}andB={d,e},andletfbethefunctiondefinedbythefollowingtable.

Thegraphoffis{(a,d),(b,e),(c,d)}.

Clearly,wemayusetheinformationcontainedinthetabletoconstructthegraphoff;wemayalsooperatetheotherway,thatis,wemayusetheinformationcontainedinthegraphtoconstructthetableoff.Thusafunctionfcompletelydeterminesitsgraph,andconversely,itsgraphcompletelydeterminesf.Hencethereisnoneedtodistinguishbetweenafunctionanditsgraph.Sinceafunctionanditsgraphareessentiallyoneandthesamething,wemay,ifwewish,definea

functiontobeagraph.Thereisanimportantadvantagetobegainedbydoingthis—namely,weavoidhaving to introduce the word rule as a new undefined concept of set theory. For this reason it iscustomary,inrigoroustreatmentsofmathematics,tointroducethenotionoffunctionviathatofgraph.Weshallfollowthatprocedurehere.

2FUNDAMENTALCONCEPTSANDDEFINITIONS

Webeginbygivingour“official”definitionofafunction.

2.1DefinitionAfunctionfromAtoBisatripleofobjects ,whereAandBareclassesandfisasubclassofA×Bwiththefollowingproperties.

F1.∀x∈A,∃y∈Bsuchthat(x,y)∈f.F2.If(x,y1)∈fand(x,y2)∈f,theny1=y2.

Itiscustomarytowritef:A→Binsteadof .

Inordinarymathematicalapplications,everyfunctionf:A→BisafunctionfromasetAtoasetB.However,theintuitiveconceptofafunctionfromAtoBismeaningfulforanytwocollectionsAandB,whetherAandBbesetsorproperclasses;henceitisnaturaltogivethedefinitionofafunctioninitsmostgeneralform,lettingAandBbeanyclasses.Onceagain,everysetisaclass,henceeverythingwehavetosayaboutfunctionsfromaclassAtoaclassBapplies,inparticular,tofunctionsfromasetAtoasetB.

Letf:A→Bbeafunction;if(x,y)∈f,wesaythatyistheimageofx(withrespecttof);wealsosaythatxisthepre-imageofy(withrespecttof);wealsosaythatfmapsxontoy,andsymbolizethisstatementby .(Thereadermay,ifhewishes,picturethesestatementsasinFig.1.)

Thus,F1statesthat

everyelementx∈Ahasanimagey∈B.

F2statesthatifx∈A,then

theimageofxisunique;

forif(x,y1)∈fand(x,y2)∈f,thatis,ify1andy2arebothimagesofx,thenF2dictatesthaty1=y2.ItfollowsthatF1andF2combinedstatethat

2.2Everyelementx∈Ahasauniquelydeterminedimagey∈B.

2.3TheoremLetAandBbeclassesandletfbeagraph.Thenf:A→Bisafunctionifandonlyifi) F2holds,ii) domf=A,andiii) ranf⊆B.

Proof.Supposef:A→Bisafunction;by2.1,F2holds.Furthermore,

By(a)and(b),domf=A;by(c),ranf⊆B.Thus,(i),(ii),and(iii)hold.

Fortheconverse,supposethat(i),(ii),and(iii)hold.

Thus,f⊆A×B.b) LetxbeanarbitraryelementofA.By(ii),x∈domf;hence∃y(x,y)∈f;byy∈ranf,soby

(iii),y∈B.ThisprovesthatF1holds.By(i),F2holds;thus,by2.1,f:A→Bisafunction.FromTheorem2.3weconclude,inparticular,thatiff:A→Bisafunction,thenAisthedomainoff

andBcontainstherangeoff.WecallBthecodomainoff:A→B.

2.4CorollaryLetf:A→Bbeafunction;ifC isanyclasssuchthatran f⊆C,thenf :A→C isafunction.

Proof.Iff:A→Bisafunction,thenby2.3,F2holdsanddomf=A;thus,ifranf⊆C,then,by2.3,f:A→Cisafunction.

Let f :A→Bbea functionand letx∈A; it is customary touse the symbol f(x) todesignate theimageofx.Thus,

y=f(x)hasthesamemeetingas(x,y)∈f.

Whenwewritey=f(x)insteadof(x,y)∈f,ConditionsF1andF2taketheform

F1.∀x∈A,∃y∈B,y=f(x).F2.Ify1=f(x)andy2=f(x),theny1=y2.

ItisoftenconvenienttowriteF2inaslightlydifferentway.F2statesthatif(x,y1)∈fand(x,y2)∈f,thatis,if

theny1=y2.Thisisthesameassayingthatif , andx1=x2,thatis,if

thenf(x1)=f(x2).Thus,F2maybewrittenintheform

.Ifx1=x2,thenf(x1)=f(x2).

2.5TheoremLetf:A→Bandg:A→Bbefunctions.Thenf=gifandonlyiff(x)=g(x),∀x∈A.

Proof.First,letusassumethatf=g.Then,forarbitraryx∈A,

thus,f(x)=g(x).Conversely,assumethatf(x)=g(x),∀x∈A.Then

thus,f=g.

Injective,SurjectiveandBijectiveFunctions

Thefollowingdefinitionsareofgreatimportanceinthestudyoffunctions.

2.6DefinitionAfunctionf:A→Bissaidtobeinjectiveifithasthefollowingproperty.

INJ.If(x1,y)∈fand(x2,y)∈f,thenx1=x2.

ThereadershouldnotethatINJstates,simply,thatifyisanyelementofB,then

yhasnomorethanonepre-image;

forif(x1,y)∈fand(x2,y)∈f,thatis,ifx1andx2arebothpre-imagesofy,thenINJdictatesthatx1=x2.

ItisoftenconvenienttowriteINJinaslightlydifferentway.INJstatesthatif(x1,y)∈fand(x2,y)∈f,thatis,if

thenx1=x2.Thisisthesameassayingthatifx1andx2areelementsofAandf(x1)=f(x2),thatis,if

thenx1=x2.Thus,INJmaybewrittenintheform

INJ .Iff(x1)=f(x2),thenx1=x2.

(ThefunctionofFig.2isinjective,whereasthefunctionofFig.3isnotinjective.)

2.7DefinitionAfunctionf:A→Bissaidtobesurjectiveifithasthefollowingproperty:

SURJ.∀y∈B,∃x∈Ay=f(x).

Clearly,conditionSURJstatesthateveryelementofBistheimageofsomeelementofA;thatis,B⊆ranf.Butranf⊆BbyTheorem2.3;hencef:A→Bissurjectiveifandonlyifranf=B. (ThefunctionofFig.3issurjective,whereasthefunctionofFig.2isnotsurjective.)

Fig.1

Fig.2

Fig.3

Fig.4

2.8DefinitionAfunctionf:A→Bissaidtobebijectiveifitisbothinjectiveandsurjective.

Tosaythatf:A→BisinjectiveistosaythateveryelementofBistheimageofnomorethanoneelementofA; tosay that f issurjective is tosay thateveryelementofB is the imageofat least oneelementofA;thus,tosaythatfisbijectiveistosaythateveryelementofBistheimageofexactlyoneelementofA(Fig.4).Inotherwords,iff:A→Bisabijectivefunction,everyelementofAhasexactlyoneimageinBandeveryelementofBhasexactlyonepre-imageinA;thusalltheelementsofAandalltheelementsofBareassociatedinpairs;forthisreason,iffisbijective,itissometimescalledaone-to-onecorrespondencebetweenAandB.

2.9DefinitionIfthereexistsabijectivefunctionf:A→B,thenwesaythatAandBareinone-to-onecorrespondence.

ExamplesofFunctions

2.10Identityfunction.LetAbeaclass;bytheidentityfunctiononAwemeanthefunctionIA:A→Agivenby

Inotherwords,

IAisclearlyinjective,forsupposeIA(x)=IA(y);nowIA(x)=xandIA(y)=y,sox=y;thus holds.IAissurjectivebecause,obviously,therangeofIAisA.ThusIAisbijective.

2.11Constantfunction.LetAandBbeclasses,andletbbeanelementofB.Bytheconstant functionKbwemeanthefunctionKb:A→Bgivenby

Inotherwords,Kb={(x,b):x∈A}.NotethatifAhasmorethanoneelement,Kbisnotinjective;ifBhasmorethanoneelement,Kbis

notsurjective.

2.12Inclusionfunction.LetAbeaclassandletBbeasubclassofA.BytheinclusionfunctionofBinAwemeanthefunctionEB:B→Agivenby

Note that ifB=A, the inclusion functioncoincideswith the identity function IA.By theargumentusedin2.10,EBisinjective;however,ifB≠A,thenEBisnotsurjective.

2.13Characteristicfunction.Let2designateaclassoftwoelements,saytheclass{0,1}.IfAisaclassandBisasubclassofA,thecharacteristicfunctionofBinAisthefunctionCB:A→2givenby

TheCBmapseveryelementofBonto0andeveryelementofA−Bonto1.

2.14Restrictionofafunction.Letf:A→BbeafunctionandletCbeasubclassofA.Bytherestriction

offtoCwemeanthefunctionf[C]:C→Bgivenby

Toputitanotherway,f[C]={(x,y):(x,y)∈fandx∈C}.Notethatf[C]⊆f.Restrictionsoffunctionshavethefollowingproperties,whichwillbeusefultouslater.

2.15TheoremIff:B∪C→Aisafunction,thenf=f[B]∪f[C].

Thesimpleproofofthistheoremisleftasanexerciseforthereader.

2.16TheoremLetf1:B→Aandf2:C→Abefunctions,whereB∩C=Ø.Iff=f1∪f2,thenthefollowinghold:i) f:B∪C→Aisafunction.ii) f1=f[B]andf2=f[C].

iii) Ifx∈Bthenf(x)=f1(x),andifx∈Cthenf(x)=f2(x).

Proof.Wewillbeginbyprovingthefollowingtworelations.

a) (x,y)∈fandx∈B⇔(x,y)∈f1.b) (x,y)∈fandx∈C⇔(x,y)∈f2.

If(x,y)∈f1,thenx∈Bbecausedomf1=B,and(x,y)∈fbecausef=f1∪f2.Conversely,suppose(x,y)∈fandx∈B:(x,y)∈fimpliesthat(x,y)∈f1or(x,y)∈f2;if(x,y)∈f2,thenx∈C(becausedomf2=C),whichisimpossiblebecausex∈BandB∩C=Ø;thus,(x,y)∈f1.Thisproves(a);theproofof(b)isanalogous.Next,wewillprovethat

c) domf=B∪Candranf⊆A.

Indeed,by1.43,

and

Ournextstepwillbetoprovethat

d) fsatisfiesConditionF2.

Suppose(x,y1)∈fand(x,y2)∈f;nowx∈domf,soby(c),x∈Borx∈C.Ifx∈B,then,by(a),(x,y1)∈f1and(x,y2)∈f1,soby2.1,y1=y2;ifx∈C,then,by(b),(x,y1)∈f2and(x,y2)∈f2,soby2.1,y1=y2;thisproves(d).From(c),(d),andTheorem2.3,weconcludethatf:B∪C→Aisafunction.By(a)and2.14,(x,y)∈f1⇔(x,y)∈f[B],thatis,f1=f[B];analogously,f2=f[C].

Finally,(a)statesthat

and(b)statesthat

thus(iii)holds.

EXERCISES2.2

1. Provethatthefunctionsintroducedin2.10through2.14qualifyasfunctionsunderDefinition2.1.2. Provethatiff:A→BisaninjectivefunctionandC⊆A,thenf[C]:C→Bisaninjectivefunction.3. LetAbeaclassandletf={(x,(x,x)):x∈A}.ShowthatfisabijectivefunctionfromAtoIA.4. Letf:A→Bandg:A→Bbefunctions.Provethatiff⊆gthenf=g.5. Letf:A→Bandg:C→Dbefunctions.Theproductoffandgisthefunctiondefinedasfollows:

Provethatf•gisafunctionfromA×CtoB×D.Provethatiffandgareinjective,thenf•g isinjective,andiffandgaresurjective,thenf•gissurjective.Provethatran[f•g]=(ranf)×(rang).

6. Iff:B∪C→Aisafunction,provethatf=f[B]∪f[C].7. Letf1:A→Bandf2:C→Dbebijectivefunction,whereA∩C=ØandB∩D=Ø.Letf=f1∪

f2;provethatf:A∪C→B∪Disabijectivefunction.8. Letf:B→Aandg:C→Abefunctions,andsupposethatf[B∩C]=g[B∩C].Ifh=f∪g,provethat

h:B∪C→Aisafunction,f=h[B]andg=h[C].9. Letf:A→Bbeafunction;provethatfisinone-to-onecorrespondencewithA.

Bya functionalgraphwemean a graphwhich satisfiesConditionF2.ThusG is a functionalgraphifandonlyif

10. IfGisafunctionalgraph,showthateverysubclassofGisafunctionalgraph.11. LetGbeagraph.ProvethatGisafunctionalgraphifandonlyifforarbitrarygraphsHandJ.

12. LetGbeafunctionalgraph.ProvethatGisinjectiveifandonlyifforarbitrarygraphsJandH.

3PROPERTIESOFCOMPOSITEFUNCTIONSANDINVERSEFUNCTIONS

Thefollowingtheoremsexpressafewbasicpropertiesoffunctions.

2.17TheoremIff:A→Bandg:B→Carefunctions,then :A→Cisafunction.

Proofi) By1.38,dom =domf=A;by1.37(iv),ran ⊆rang⊆C.ii) Suppose(x,z1)∈ and(x,z2)∈ ;by1.34,∃y1(x,y1)∈fand(y1,z1)∈gand∃y2(x,y2)∈f

and(y2,z2)∈g.From(x,y1)∈fand(x,y2)∈fweconclude,by2.1,thaty1=y2;thus(y1,z1)∈gand(y1,z2)∈g.ItfollowsbyF2(appliedtog)thatz1=z2;thus, satisfiesF2.

From(i),(ii),and2.3,weconcludethat :A→Cisafunction.

By1.34(x,y)∈ ifandonlyifforsomeelementz,(x,z)∈fand(z,y)∈g.Thus, ifandonlyifforsomez, and .(Thereadermay,ifhewishes,picturethisstatementasinFig.5.)Thisisthesameassayingthaty=[ ](x)ifandonlyifforsomez,z=f(x)andy=g(z).Thus

Fig.5

2.19DefinitionAfunctionf:A→Bissaidtobeinvertibleiff−1:B→Aisafunction.

Letf:A→Bbeaninvertiblefunction;by1.33,(x,y)∈fifandonlyif(y,x)∈f−1.Thus ifandonlyif .(Thereadermay,ifhewishes,picturethisstatementasinFig.6.)Thus

2.20y=f(x)ifandonlyifx=f−1(y).

Fig.6

Thenexttwotheoremsgiveanecessaryandsufficientconditionforafunctiontobeinvertible.

2.21TheoremIff:A→Bisabijectivefunction,thenf−1:B→Aisabijectivefunction.

Proof.By2.3,domf=A,andby2.7,ranf=B;thus,by1.37,domf−1=Bandranf−1=A.Nowwewillprovethatf−1satisfiesF2:

Thus,byTheorem2.3,f−1:B→Aisafunction.Next,wewillprovethatf−1satisfiesINJ:

Finally,f−1satisfiesSURJbecause(seeabove)ranf−1=A.

2.22TheoremIff:A→Bisinvertible,thenf:A→Bisbijective.

Proof.Letf:A→Bbeinvertible;thatis,letf−1:B→Abeafunction.By2.3,domf−1=B, soby1.37(ii),ranf=B;thus,f:A→Bissurjective.Now

Thus,f:A→Bisinjective.

Theorem2.21and2.22maybesummarizedasfollows:

f:A→Bisinvertibleifandonlyifitisbijective;furthermore,iff:A→Bisinvertible,thenf−1:B→Aisbijective.

Thenexttwotheoremsgiveanotherusefulcharacterizationofinvertiblefunctions.

2.23TheoremLetf:A→Bbeaninvertiblefunction.Then

Proof

i) Letx∈Aandlety=f(x);thenby2.20,x=f−1(y).Thus

thisholdsforeveryx∈A,soby2.5,f−1 f=IA.ii) Theproofisanalogousto(i),andisleftasanexercise.

2.24TheoremLetf:A→Bandg:B→Abefunctions.If =IAand =IB,thenf:A→Bis

bijective(henceinvertible),andg=f−1.

Proofi) First,wewillprovethatf:A→Bisinjective.

ii) Next,wewillprovethatf:A→Bissurjective.Ify∈Btheny=IB(y)=[ ](y)=f(g(y));inotherwords,ifyisanyelementofB,theny=f(x),wherex=g(y)∈A.

iii) Finally,wewillprovethatg=f−1.Tobeginwith,

conversely,

Thus,∀y∈B,x=f−1(y)iffx=g(y);thatis,f−1(y)=g(y);itfollows(by2.5)thatf−1=g.

Theorem2.23and2.24maybesummarizedasfollows:

f:A→Bisinvertibleifandonlyifthereexistsafunctiong:B→Asuchthat =IAand =IB.Thefunctiong,ifitexists,istheinverseoff.

Ournexttheoremgivesanimportantcharacterizationofinjectivefunctions.

2.25TheoremLetf:A→Bbeafunction;f:A→Bisinjectiveifandonlyifthereexistsafunctiong:B→Asuchthat =IA.

Proofi) Supposethereexistsafunctiong:B→Asuchthat =IA.Toprovethatf:A→Bisinjective,

werepeatpart(i)oftheproofof2.24.ii) Conversely,supposethatf:A→Bisinjective;letC=ranf.By2.4,f:A→Cisafunction;f:A

→Cissurjective(becauseC=ranf),henceitisbijective;thusf−1:C→A isafunction.Ifa issomefixedelementofA,letKa:(B−C)→Abetheconstantfunction(see2.11)whichmapseveryelementofB−Contoa. Ifg= f−1∪Ka, then,by2.16(i),g :B→A isa function(seeFig.7).Finally,ifx∈A,lety=f(x);then

Thus∀x∈A,[ ](x)=IA(x);itfollowsby2.5that =IA.

Fig.7

InChapters5wewillproveacompaniontheoremto2.25,whichwillstatethefollowing:f:A→Bissurjectiveifandonlyifthereexistsafunctiong:B→Asuchthat[ ]= IB.Theorem2.25and itscompanionareoftenparaphrasedasfollows.

Let f :A→B be a function; f :A→B is injective if and only if it has a “left inverse” andsurjectiveifandonlyifithasa“rightinverse”.

2.26TheoremSupposef:A→B,g:B→C,and :A→Carefunctions.i) Iffandgareinjective,then isinjective.ii) Iffandgaresurjective,then issurjective.iii) Iffandgarebijective,then isbijective.

Proofi) Supposethatfandgbothsatisfy :then

thus satisfies .ii) SupposethatfandgbothsatisfySURJ:ifz∈C,then∃y∈Bz=g(y);sincey∈B,∃x∈Ay=f(x);

thusz=g(f(x))=[ ](x).Consequently, satisfiesSURJ.iii) Thisfollowsimmediatelyfrom(i)and(ii).

Itfollowsfrom2.26(iii)that

thecompositeoftwoinvertiblefunctionsisinvertible.

Furthermore,by1.35(iii),

EXERCISES2.3

1. Letf:A→Bbeafunction.Provethat and .2. Supposef:A→Bandg:B→Carefunctions.Provethatif isinjective,thenfisinjective;

prove that if is surjective, then g is surjective. Conclude that if is bijective, then f is

injectiveandgissurjective.3. GiveanexampletoshowthattheconverseofthelaststatementofExercise2doesnothold.4. Letf:A→Bandg:B→Abefunctions.Supposethaty=f(x)ifandonlyifx=g(y).Provethatf

isinvertibleandg=f−1.5. Letg:B→Candh:B→Cbefunctions.Supposethat = foreveryfunctionf:A→B.

Provethatg=h.6. Suppose g :A→B andh:A→B are functions. LetC be a setwithmore than one element;

supposethat = foreveryfunctionf:B→C.Provethatg=h.7. Letf:B→Cbeafunction.Provethatfisinjectiveifandonlyif,foreverypairoffunctionsg:A

→Bandh:A→B, = ⇒g=h.8. Letf:A→Bbeafunction.Provethatfissurjectiveifandonlyif,foreverypairoffunctionsg:B

→Candh:B→C, = ⇒g=h.9. Letf:A→Candg:A→Bbefunctions.Provethatthereexistsafunctionh:B→Csuchthatf=

ifandonlyif∀x,y∈A,

Provethathisunique.10. Letf:C→Aandg:B→Abefunctions,andsupposethatgisbijective.Provethatthereexistsh:

C→Bsuchthatf=g hifandonlyifranf⊆rang.Provethathisunique.11. Let f :A→B be a function, and letC⊆A.Prove that f[C] = EC,whereEC is the inclusion

functionofCinA(2.12).

4DIRECTIMAGESANDINVERSEIMAGESUNDERFUNCTIONS

2.27DefinitionLetf:A→Bbeafunction;ifCisanysubclassofA, thedirectimageofCunderf ,whichwewrite ,isthefollowingsubclassofB:

Thatis, istheclassofalltheimagesofelementsinC.

2.28DefinitionLetf:A→Bbeafunction;ifDisanysubclassofB,theinverseimageofDunderf,whichwewrite (D),isthefollowingsubclassofA:

Thatis, (D)istheclassofallthepre-imagesofelementsinD.If{a}and{b}aresingletons,wewillwrite (a)for ({a})and (b)for ({b}).

2.29TheoremLetf:A→Bbeafunction.i) ifC⊆AandD⊆A,thenC=D⇒ = (D).ii) ifC⊆BandD⊆B,thenC=D⇒ (C)= (D).

Proofi) SupposeC=D;then

ii) SupposeC=D;then

Caution. = (D)does not always imply that C = D; for a simple counterexample, see Fig. 8.Similarly, (C)= (D)doesnotalwaysimplyC=D;foracounterexample,thereadershouldlookatFig.9.(However,seeExercise3,ExerciseSet2.4.)

Fig.8

Fig.9

2.30TheoremLetAandBbesetsandletf:A→Bbeafunction;then

Proofi) By2.27,itiseasytoseethatdom = andran ⊆ .Theorem2.29(i)states that satisfies

ConditionF2;thusby2.3, : → isafunction.

ii) Analogously, : → isafunction.

2.31TheoremLetf:A→Bbeafunction,let{Ci}i∈IbeafamilyofsubclassesofA,andlet{Di}i∈IbeafamilyofsubclassesofB.Then

Proof

iii) Theproofisleftasanexerciseforthereader.

Caution.ItisimportanttonotethatthereisnocounterpartofTheorem2.31(iii)for ;moreprecisely,wehave

butwedonothaveinclusiontheotherway.Forasimplecounterexample,seeFig.8,where (C∩D)=(b)={2}≠{1,2}= ∩ (D).Forthisreason,avarietyoftheoremswhicharetrueforinverseimagesofsetsfailtoholdfordirectimagesofsets.

EXERCISES2.4

1. Supposethatf:A→Bisafunction,C⊆AandD⊆B.

2. Supposethatf:A→Bisafunction,C⊆AandD⊆B.a) Iffisinjective,provethatC= [ ].b) Iffissurjective,provethatD= [ (D)].

3. Letf:A→Bbeafunction.Provethefollowing.a) SupposeC⊆AandD⊆A;iffisinjective,then = (D)⇒C=D.b) SupposeC⊆BandD⊆B;iffissurjective,then (C)= (D)⇒C=D.[Hint:UsetheresultofExercise2.]

4. Letf:A→Bbeafunction.Provethefollowing:a) Iffisinjective,then isbijective.[Hint:UsetheresultofExercise2(a).]b) Iffissurjective,then isbijective.[Hint:UsetheresultofExercise2(b).]

5. Supposethatf:A→Bisafunction;letC⊆A.a) Provethat { [ ]}= .b) Usetheresultof(a)toprovethat = .

6. Letf:A→Bbeafunction.Provethefollowing:a) Iffisinjective,then isinjective.b) Iffissurjective,then issurjective.c) Iffisbijective,then isbijective.

7. Letf:A→Bbeafunction.Provethefollowing:a) Iffisinjective,then issurjective.b) Iffissurjective,then isinjective.c) Iffisbijective,then isbijective.

8. Letf:A→Bbeafunction.Provethat

foreverypairofsubclassesC⊆AandD⊆Aifandonlyiffisinjective.9. Supposethatf:A→Bisafunction,C⊆BandD⊆B.Provethat

10. Letf:A→Bbeafunction.Proveeachofthefollowing:a) IfC⊆AandD⊆A,then − (D)⊆ (C−D).b) − (D)⊆ (C −D) for every pair of subclassesC⊆A andD⊆A if and only if f is

injective.

5PRODUCTOFAFAMILYOFCLASSES

InthebeginningofChapters1wespokeoftheunionandintersectionoftwoclasses;later,weextendedthisnotionbydefiningtheunionandintersectionofanarbitraryfamilyofclasses. Inmuchthesamemanner,wewillnowextendthenotionoftheCartesianproductoftwoclassesbydefiningtheproduct

ofafamilyofclasses.TheproductoftwoclassesAandBhasbeendefinedtobetheclassA×Bofallorderedpairs(x,y),

wherex∈Aandy∈B.Thisdefinitionmaybeextended,inanaturalway,toafinitenumberofclassesA1,A2,…,An;wemay define the productA1 ×A2 ×…×An to be the class of all “orderedn-tuples”(a1,a2,…,an),whereai∈Aiforeachindexi=1,2,…,n.Now,wewishtoextendthisconcepttothecaseofanindexedfamilyofclasses,{Ai}i∈I,wheretheindexclassIisanyclasswhatsoever.Evidentlywe cannot speak of “I -tuples” of elements, because I may be an infinite class and may fail to beordered;therefore,wemustalterourapproachtotheproblem.LetustakeanotherlookattheproductA1×A2×…×An.Clearly{A1,A2,…,An}isafamilywhose

indexclassisI={1,2,…,n}.Now,anorderedn-tuplemayberegardedasafunction(whosedomainisI)whichmaps each element i∈ I ontoan element ai inAi. Indeed, if f is such a function, then f isdescribedbythefollowingtable.

Using the table,wemay construct the orderedn-tuple (a1,a2,…, an); conversely, ifwe are give theorderedn-tuple(a1,a2,…,an), thenwemayconstructthetable;(infact, theorderedn-tuple issimplythetablepresentedasahorizontalarray).Thus,thefunctionfandtheorderedn-tuple(a1,a2,…,an)are,essentially, one and the same thing. This simple observation leads to the following definition of theproductofafamilyofclasses.

2.32DefinitionLet{Ai}i∈Ibeanindexedfamilyofclasses;let

TheproductoftheclassesAiisdefinedtobetheclass

2.33ExampleLetI={1,2},A1={a,b},andA2={c,d}.By2.32, consistsofallthefunctionsf:{1,2}→{a,b,c,d}suchthatf(1)∈A1andf(2)∈A2.Therearefoursuchfunctions,givenbythefollowingtables.

We may identify these four functions with the four ordered pairs (a, c), (a, d), (b, c), and (b, d),respectively.Thus isexactlyA1xA2.

Weadoptthefollowingnotationalconvention:henceforth,wewilldesignateelementsofaproductbyboldfacelettersa,b,c,etc.

Ifaisanelementof andj∈I,weagreethatajwillhavethesamemeaningasa(j);wewillcallajthej-coordinateofa.Let{Ai}i∈Ibeanindexedfamily,and,foreachi∈I,letxi∈Ai.Wewillusethesymbol{xi}i∈Ito

designatetheelementin whosei-coordinate,foreachi∈I,isxi.LetA= ;correspondingtoeachindexi∈I,wedefineafunctionpifromAtoAiby

Thefunctionpiiscalledthei-projectionofAtoAi.

2.34DefinitionIfAandBarearbitraryclasses,thesymbolBAreferstotheclassofallfunctionsfromAtoB.Inparticular,if2denotesaclassoftwoelements,then2AdenotestheclassofallfunctionsfromAto

2.Thefollowingisanimportantresultwhichwillbeusedinalaterchapter.

2.35TheoremIfAisaset,then2Aand areinone-to-onecorrespondence.

Proof.Wewillshowthatthereexistsabijectivefunctionγ: →2A.IfB∈ ,letCBdenotethecharacteristicfunctionofBinA(see2.13);CBisanelementof2A.Wedefineγby

Bythewayγisdefined,itisclearthatγmapseveryB∈ ontoauniquelydeterminedelementof2A;henceγ: →2Aisafunction;itremainstoshowthatγisinjectiveandsurjective.i) LetB,D∈ ;ifγ(B)=γ(D),thenCB=CD;hence

thatB=D.Thusγsatisfies .

ii) Iff∈2A,andifweletB= (0),thenf=CB=γ(B).ThusγsatisfiesconditionSURJ.

ItiseasytoshowthatifAandBaresets,thenABisaset(seeExercise12,ExerciseSet2.5).Usingthisfact,itcaneasilybeshownthatif{Ai}i∈IisanindexedfamilyofsetssuchthattheindexclassIisaset,then isaset(seeExercise13,ExerciseSet2.5,andRemark2.38).

EXERCISES2.5

1. LetA={1,2,3},B={a,b}.FindAB,BA,2A,and .2. Supposethat{Bi}i∈IisafamilyofsubclassesofA.Provethat

3. Supposethat{Ai}i∈Iand{Bi}i∈IarefamiliesofclasseswiththesameindexclassI.ShowthatifAi⊆Bi,∀i∈I,then

Inthenextthreeexercises(Exercises4,5and6),assumethefollowing:

4. Suppose that {Ai}i∈I and{Bi}i∈I are families of nonempty classeswith the same index class I .Provethatif

thenAi⊆Biforeachindexi.5. Supposethat{Ai}i∈Iand{Bi}i∈IarefamiliesofclasseswiththesameindexclassI.Provethat

6. Let{Ai}i∈Iand{Bj}j∈Jbefamiliesofclasses.Provethefollowing:

7. Let{Ai}i∈Ibeafamilyofclasses,andforeachi∈I,letBibeasubclassofAi.Provethat

8. Let{Ai}i∈Ibeanindexedfamily,andlet

IfB⊆A,letBi=¯pi(B)foreachi∈I.ProvethatB⊆9. ProvethatAC∪BC⊆(A∪B)C.10. Provethat(A∩B)C=AC∩BC.

11. Provethat(A−B)C=AC−BC.

12. ProvethatifAandBaresets,thenABisaset.[Hint:EachelementofABisasubsetofB×A.Usetheaxiomsofthelasttwochaptersasneeded.]

13. Let{Ai}i∈Ibeanindexedfamily;supposethatIisaset,thateachAiisaset,andthat{Ai:i∈I}isaset.Provethatisaset.[Hint:UsetheresultsofExercises2and12.]

6THEAXIOMOFREPLACEMENT

AxiomsA3throughA7are“set”axioms,thatis,theyaredesignedforthepurposeofestablishingthepropertiesofsets.Wearenowinapositiontointroduceourlast“set”axiom.Thisaxiomismotivatedbythefollowingconsiderations.Wenotedearlierthatwearetothinkofasetasaclasswhichis“nottoolarge.”Now,ifAandBare

classesandf:A→B isasurjectivefunction, then, inanobvious intuitivesense,Bhas“asmany,orfewerelementsthanA“(seeFig.3).ThusifAis“nottoolarge”andf:A→Bisasurjectivefunction,itstandstoreasonthatBis“nottoolarge”.Theseremarksleadustostatethefollowingasanaxiom.

A9.IfAisasetandf:A→Bisasurjectivefunction,thanBisaset.

StatementA9istraditionallycalledtheaxiomofreplacement;ithasthefollowingconsequences.

2.36IfAisasetandAisinone-to-onecorrespondencewithB,thenBisaset.

Wenotedonpage47thatØisaset,hencebyAxiomA5{Ø,Ø}isaset,thatis,{Ø}isaset.Thus,by2.36,

2.37everysingletonisaset.

Sincetheunionoftwosetsisaset,itfollowsthateverydoubletonisaset;similarly,everyclassofthreeelementsisaset,everyclassoffourelementsisaset,andsoon,throughallthepositiveintegers.Thus,inanintuitivesense,everyfiniteclassisaset.

2.38Remark.Let{Ai}i∈Ibeanindexedfamilyofsets,wheretheindexclassIisaset.Itisclearthatthefunctionφdefinedbyφ(i)=AiisasurjectivefunctionfromIto{Ai:i∈I};thuswehave

If{Ai}i∈IisanindexedfamilyofsetsandIisaset,then{Ai:i∈I}isaset.

3Relations

1INTRODUCTION

Intuitively, a binary relation in a classA is a statementR(x,y)which is either true or false for eachorderedpair(x,y)ofelementsofA.Forinstance,therelation“xdividesy,”whichwemaywriteD(x,y),isarelationintheclass oftheintegers:D(x,y)istrueforeverypair(x,y)ofintegerssuchthatyisamultipleofx;itisfalseforeveryotherpairofintegers.TherepresentinggraphofarelationinAisagraphG⊆A×Awhichconsistsofallthepairs(x,y)

suchthatR(x,y) is true.Conversely, ifwearegivenanarbitrarygraphG⊆A×A, thenGdefinesarelationinA,namelytherelationRsuchthatR(x,y)istrueifandonlyif(x,y)∈G.Thus, aswe did in the case of functions,we are able to identify relationswith their representing

graphs.Inthiswaythestudyofrelationsispartofelementarysettheory.

2FUNDAMENTALCONCEPTSANDDEFINITIONS

3.1DefinitionLetAbeaclass;byarelationinAwemeananarbitrarysubclassofA×A.

3.2DefinitionLetGbearelationinA;then

Giscalledreflexiveif

Giscalledsymmetricif

Giscalledanti-symmetricif

Giscalledtransitiveif

3.3DefinitionThediagonalgraphIAisdefinedtobetheclass{(x,x):x∈A}.ItiseasytoseethatGisreflexiveifandonlyifIA⊆G.

Thereisavarietyofinterestingandusefulalternativewaysofdefiningtheabovenotions.Somearegiveninthenexttheorem.

3.4TheoremLetGbearelationinA.

i) GissymmetricifandonlyifG=G−1.

ii) GisantisymmetricifandonlyifG∩G−1⊆IA.iii) Gistransitiveifandonlyif ⊆G.

Proof

i) SupposeGissymmetric.Then

thusG=G−1.Conversely,supposeG=G−1.Then

ii) SupposeGisantisymmetric.Then

Conversely,supposethatG∩G−1⊆IA.Then

ii) SupposeGistransitive.Then(x,y)∈ ⇒∃z(x,z)∈Gand(z,y)∈G⇒(x,y)∈G.Thus .Conversely,suppose :Then(x,y)∈Gand(y,z)∈G⇒(x,z)∈ ⊆G.

3.5DefinitionArelationiscalledanequivalencerelationifitisreflexive,symmetric,andtransitive.

Arelationiscalledanorderrelationifitisreflexive,antisymmetric,andtransitive.

3.6DefinitionLetGbearelationinA.

Giscalledirreflexiveif

Giscalledasymmetricif

Giscalledintransitiveif

ExamplesLet designatethesetoftheintegers;theequalityrelationin isreflexive,symmetric,andtransitive; hence, it is an equivalence relation. The relation (“less than or equal to”) is reflexive,antisymmetric,andtransitive;henceitisanorderrelation.Therelation<(“strictlylessthan”)isnotanorder relation: it is irreflexive,asymmetric,and transitive;sucha relation iscalleda relationofstrictorder.

EXERCISES3.2

1.EachofthefollowingdescribesarelationinthesetZoftheintegers.State,foreachone,whetherithas any of the following properties: reflexive, symmetric, antisymmetric, transitive, irreflexive,asymmetric, intransitive. Determine whether it is an equivalence relation, an order relation, orneither.Proveyouranswerineachcase.a) G={(x,y):x+y<3}.b) G={(x,y):xdividesy}.c) G={(x,y):xandyarerelativelyprime}.d) G={(x,y):x+yisanevennumber}.e) G={(x,y):x=yorx=−y}.f) G={(x,y):x+yisevenandxisamultipleofy}.g) G={(x,y):y=x+1}.

2. LetGbearelationinA;proveeachofthefollowing:a)GisirreflexiveifandonlyifG∩I=Ø.b)GisasymmetricifandonlyifG∩G−1=Ø.c)Gisintransitiveifandonlyif( )∩G=Ø.

3. ShowthatifisanequivalencerelationinA,then =G.4. Let{Gi}i∈IbeanindexedfamilyofequivalencerelationsinA.Showthat isanequivalence

relationinA.5. Let{Gi}i∈IbeanindexedfamilyoforderrelationsinA.Showthat orderrelationinA.6. LetHbeareflexiverelationinA.ProvethatforanyrelationGinA,G⊆ andG⊆ .7. LetGandHberelationsinA;supposethatGisreflexiveandHisreflexiveandtransitive.Show

thatG⊆Hifandonlyif =H.(Inparticular,thisholdsifGandHareequivalencerelations.)8. ShowthattheinverseofanorderrelationinAisanorderrelationinA.9. LetGbearelationinA.ShowthatGisanorderrelationifandonlyifG∩G−1=IAand =G.

10. LetGandHbeequivalencerelationsinA.Showthat isanequivalencerelation inA ifandonlyif = .

11. LetGandHbeequivalencerelationsinA.ProvethatG∪HisanequivalencerelationinAifandonlyif ⊆G∪Hand ⊆G∪H.

12. LetGbeanequivalencerelationinA.ProvethatifHandJarereflexiverelationsinA,thenG⊆HandG⊆J⇒G⊆ .

3EQUIVALENCERELATIONSANDPARTITIONS

In the remainder of this chapter we will concern ourselves with equivalence relations in sets. Theconceptsweareabout to introducearisenaturally in termsof sets,but cannotbeextended toproperclasses; to understand why not, the reader should review our discussion in Section 7 of Chapter 1.Briefly,ifAisasetandP(X)isaproperty,thenby1.52itislegitimatetoformthesetofallthesubsetsX⊆AwhichsatisfyP(X).However,ifAwereanarbitraryclass,itwouldnotbepermissibletoformthe“classofallsubclassesofAwhichsatisfyP(X).”Thisrestrictioncompelsus toconfine thefollowingdiscussiontosets.Intuitively,thisshouldnotdisturbthereadertoomuch,forasetisalmostthesamethingasaclass:asetisanyclassexceptan“excessivelylarge”one.

3.7DefinitionLetAbeaset;byapartitionofAwemeanafamily{Ai}i∈IofnonemptysubsetsofAwiththefollowingproperties:

P1.∀i,j∈I,Ai∩Aj=ØorAi=Aj.

Intuitively, a partition is a family of subsets ofAwhich are disjoint fromone another, andwhoseunionisallofA(Fig.1).Thesubsetsarecalledthemembersofthepartition.Itiscustomarytoallowagivenmemberofthepartitiontobedesignatedbymorethanoneindex;thatis,wemayhaveAi=Aj ,wherei≠j.HencetheconditionthattwodistinctmembersbedisjointiscorrectlyexpressedbyP1.

Fig.1

PropertyP1statesthatanytwomembersAiandAjareeitherdisjointorequal;thatis,theyhaveeithernoelementsincommonoralltheirelementsincommon;inotherwords,iftheyhavesomuchasoneelementincommon,theyhavealltheirelementsincommon.Thus,P1mayalsobestatedasfollows:

. If∃x∈Ai∩Aj,thenAi=Aj.

P2 maybereplacedbythesimplercondition

P2′.

For, independently of Condition P2, we are given that eachAi is a subset ofA; hence, by 1,40(i),

.Consequently,itissufficienttostateP2′inordertohave .ItisconvenienttowriteP2′intheform.

. Ifx∈A,thenx∈Aiforsomei∈I.

Briefly,then,apartitionofAisafamily{Ai}i∈IofnonemptysubsetsofAsuchthat

. If∃x∈Ai∩AjthenAi=Ajand

. Ifx∈A,thenx∈Aiforsomei∈I.

Examplesofpartitionsaregivenintheexerciseswhichfollowthissection.TheresultswhichfollowstatetheconnectionbetweenequivalencerelationsinAandpartitionsofA.

Theyareofgreatimportanceinmanybranchesofmathematics.LetGbeanequivalencerelationinA;wewillsometimeswrite insteadof(x,y)∈G,andsay

that“xisequivalenttoymoduloG;”whenthereisnodangerofambiguity,wewillwritesimplyx∼yandsaythat“xisequivalenttoy.”NotethatsinceGisanequivalencerelationinA,wehavei) x∼x,∀x∈A.ii) x∼y⇒y∼x.iii) x∼yandy∼z⇒x∼z.

3.8DefinitionLetAbeasetandletGbeanequivalencerelationinA.Ifx∈A,thentheequivalenceclassofxmoduloGisthesetGxdefinedasfollows:

In otherwords,Gx is the set of all the element ofAwhich are equivalent to x. In themathematicalliterature,GxisalsodenotedbythesymbolsAx,[x],x/G.

3.9LemmaLetGbeanequivalencerelationinA.Then

Proofi) Supposex∼y;wehave

WehaveshownthatGx⊆Gy;analogously,Gy⊆Gx;henceGx=Gy.ii) SupposeGx=Gy;bythereflexiveproperty,x∼x,sox∈Gx;butGx=Gy;hencex∈Gy,thatis,x∼

y.

3.10TheoremLetAbeaset,letGbeanequivalencerelationinA,andlet{Gx}x∈AbethefamilyofalltheequivalenceclassesmoduloG.Then

Proof.Bydefinition,eachGxisasubsetofA;itisnonemptybecausex∼x,hencex∈Gx.Itremainstoprovethat and hold.

( ) z∈Gx∩Gy⇒z∈Gxandz∈Gy⇒z∼xandz∼y⇒x∼zandz∼y⇒x∼y⇒Gx=Gy(thelastimplicationfollowsby3.9).

( ) Ifx∈A,thenbythereflexivepropertyx∼x;hencex∈Gx.

IfGisanequivalencerelationinA,and{Gx}x∈AisthefamilyofalltheequivalenceclassesmoduloG,then{Gx}x∈AisreferredtoasthepartitioninducedbyG,orthepartitioncorrespondingtoG.Theorem3.10hasanimportantconverse,whichfollows.

3.11TheoremLetAbeaset, let{Ai}i∈IbeapartitionofA,andletGbe thesetofallpairs(x,y)ofelementsofAsuchthatxandyareinthesamememberofthepartition;thatis,

ThenG is an equivalence relation in A, and {Ai}i∈I is the partition induced byG.G is called theequivalencerelationcorrespondingto{Ai}i∈I.

Proof

Gisreflexive:x∈A⇒x∈Aiforsomei∈I⇒x∈Aiandx∈Ai⇒(x,x)∈G.Gissymmetric:(x,y)∈G⇒x∈Aiandy∈Ai⇒y∈Aiandx∈Ai⇒(y,x)∈G.Gistransitive:(x,y)∈Gand(y,z)∈G⇒x∈Aiandy∈Aiandy∈Ajandz∈Aj⇒Ai=Aj(becausey∈Ai∩Aj)⇒x∈Aiandz∈Ai⇒(x,z)∈G.Finally,eachAiisanequivalenceclassmoduloG;forsupposex∈Ai;theny∈Ai⇔(y,x)∈G⇔y∈Gx;thusAi=Gx.

The last twotheoremsmake itclear thateveryequivalencerelation inAcorrespondsuniquely toapartition of A, and conversely. Once again: if we are given a partition of A, the correspondingequivalencerelation is the relationwhichcallselementsxandy “equivalent” if theyare in the samememberofthepartition.Lookingattheothersideofthecoin,ifwearegivenanequivalencerelationinA, the corresponding partition is the one which puts elements x and y in the same member of thepartitionifftheyareequivalent.ThereadershouldnotethatGistheequivalencerelationcorrespondingto{Ai}i∈Iifandonlyif{Ai}i∈IisthepartitioncorrespondingtoG.

3.12ExampleLetA={a,b,c,d,e};letA1={a,b},A2={c,d}andA3={e}.LetG={(a,a),(b,b),(c,c),(d,d),(e,e),(a,b),(b,a),(c,d),(d,c)}.Itiseasytoseethat{A1,A2,A3}isapartitionofA(seeFig.2),andthatGisanequivalencerelationinA;Gistheequivalencerelationcorrespondingto{A1,A2,A3},

and{A1,A2,A3}isthepartitioncorrespondingtoG.ItshouldbenotedthatA1=Ga=Gb,A2=Gc=Gd,andA3=Ge.

Fig.2

IfG isanequivalencerelation insetA, then thesetofequivalenceclassesmoduloG iscalled thequotientsetofAbyG,andiscustomarilydenotedbyA/G.Thus,intheprecedingexample,A/Gistheset of three elements {Ga,Gc,Ge}. The concept of a quotient set plays a vital role inmany parts ofadvancedmathematics.

EXERCISES3.3

1. Let bethesetoftheintegers.Foreachintegern,letBn={m∈ :∃qm=n+5q}.Prove that{Bn}n∈ isapartitionof .

2. Let bethesetoftherealnumbers.Ineachofthefollowing,provethat{Br}r∈ isapartitionof× . Describe geometrically the members of this partition. Find the equivalence relationcorrespondingtoeachpartition.a)Br={(x,y):y=x+r}foreachr∈ ,

b)Br={(x,y):x2+y2=r}foreachr∈ .

[Hint:y=x+ristheequationofalineandx2+y2=ristheequationofacircle.]3. Let bethesetoftherealnumbers.Provethateachofthefollowingisanequivalencerelationin

× :a)G={[(a,b),(c,d)]:a2+b2=c2+d2}.b)H={[(a,b),(c,d)]:b−a=d−c}.c)J={[(a,b),(c,d)]:a+b=c+d}.Findthepartitioncorrespondingtoeachoftheseequivalencerelations,anddescribegeometricallythemembersofthispartition.[Hintfor(b):Ifb−a=d−c=k,notethat[(a,b),(c,d)]∈Hifandonlyif(a,b)and(c,d)bothsatisfytheequationy=x+k.Hintfor(c):Ifa+b=c+d=k,notethat[(a,b),(c,d)]∈Jifandonlyif(a,b)and(c,d)bothsatisfytheequationy=−x+k.]

4. IfHandJ are the equivalence relationsofExercise 3, describe the equivalence relationH∩J.DescribetheequivalenceclassesmoduloH∩J.

5. LetHandJbetheequivalencerelationsofExercise3.Provethat = ;concludethatisanequivalencerelation,anddescribetheequivalenceclassesmodulo .[Hint:SeeExercise10,ExerciseSet3.2.]

6. LetLbethesetofallthestraightlinesintheplane.LetGandHbethefollowingrelationsinL:G={(ℓ1,ℓ2):ℓ1isparalleltoℓ2},H={ℓ1,ℓ2):ℓ1isperpendiculartoℓ2}.Provethefollowing(argueinformally):

a)GisanequivalencerelationinL.b) =Hand =H.c)G∪Hisanequivalencerelation;describeitsequivalenceclasses.

7. LetA be an arbitrary set. Prove that IA andA ×A are equivalence relations inA. Describe thepartitionsinduced,respectively,byIAandA×A.

8. Let{Ai}i∈IbeapartitionofAandlet{Bj}j∈JbeapartitionofB.Provethat{Ai×Bj}(i,j)∈I×JisapartitionofA×B.

9. Supposef:A→Bisasurjectivefunction,and{Bi}i∈IisapartitionofB.Provethat{ (Bi)}i∈IisapartitionofA.

10. Supposef:A→Bisaninjectivefunction,and{Ai}i∈IisapartitionofA.Provethat{ (Ai)}i∈Iisapartitionof (A).

11. LetGandHbeequivalencerelationsinA.ProvethateachequivalenceclassmoduloG∩HistheintersectionofanequivalenceclassmoduloGwithanequivalenceclassmoduloH.Moreexactly,

12. LetGandHbeequivalencerelationsinA,andassumethatG∪HisanequivalencerelationinA.Prove thateachequivalenceclassmoduloG∪H is theunionofanequivalenceclassmoduloGwithanequivalenceclassmoduloH.Moreexactly,

4PRE-IMAGE,RESTRICTIONANDQUOTIENTOFEQUIVALENCERELATIONS

3.13DefinitionLetf:A→Bbeafunction,andletGbeanequivalencerelationinB.Thepre-imageofGunderfisarelationinAdefinedasfollows:

Itissimpletoshowthat (G)isanequivalencerelationinA.

3.14DefinitionLetG be an equivalence relation inA and letB⊆A.The restrictionofG toB is arelationinBdefinedasfollows:

ItissimpletoshowthatG[B]isanequivalencerelationinB.

3.15DefinitionLetGandHbeequivalencerelationsinA.WecallGarefinementofHifG⊆H;wealsosaythatGisfinerthanH,andthatHiscoarserthanG.

3.16TheoremLetGandHbeequivalencerelationsinA;supposeG⊆H.Thenz∈Hx⇒Gz⊆Hx.

Proof.Supposez∈Hx,thatis(z,x)∈H;thenwehave

ThusGz⊆Hx.

3.17CorollaryIfG⊆H,thenforeachx∈A,Gx⊆Hx.

Thisfollowsimmediatelyfrom3.16andthefactthatx∈Hx.

Itfollowsfrom3.16thatifGisarefinementofH ,theneachequivalenceclassmoduloH isanunionofequivalenceclassesmoduloG.Indeed,ifHx isanequivalenceclassmoduloHandz∈Hx,then,by3.16,HxcontainsthewholeclassGz;inotherwords,HxcontainsonlywholeclassesmoduloG.

3.18DefinitionLetGandHbeequivalencerelations inasetAand letGbea refinementofH.ThequotientofHbyG,whichisusuallydenotedbyH/G,isarelationinA/Gdefinedasfollows:

3.19TheoremH/GisanequivalencerelationinA/G:

Proof

H/Gisreflexive:ForeachequivalenceclassGx,(x,x)∈HbecauseHisreflexive;thus,by3.18,(Gx,Gx)∈H/G.

H/Gissymmetric:(Gx,Gy)∈H/G⇒(x,y)∈H⇒(y,x)∈H⇒(Gy,Gx)∈H/G.

H/G is transitive: (Gx,Gy)∈H/G and (Gy,Gz)∈H/G⇒ (x,y)∈H and (y,z)∈H⇒ (x, z)∈H⇒(Gx,Gz)∈H/G.

SinceH/GisanequivalencerelationinA/G,wemaywriteGx∼H/GGy insteadof(Gx,Gy)∈H/G.ThusDefinition3.18maybewritteninthemoresuggestiveform

Thereadermayeasilyverifythat ifandonlyifGxandGyaresubsetsofthesameequivalenceclassmoduloH.

3.21ExampleLetAandGbedefinedasinExample3.12;let

ItisobviousthatGisarefinementofH.ThepartitionofAinducedbyHis{Ha,Hc},whereHa={a,b}

andHc={c, d, e} (seeFig. 3). The readerwill note that each classmoduloH is a union of classesmoduloG.NowA/G={Ga,Gc,Ge};by3.18,H/GisthefollowingrelationinA/G:

Fig.3

Note,forinstance,that .ThepartitionofA/GinducedbyH/GisillustratedinFig.4.Inparticular,(A/G)/(H/G)istheset{α,β},whereα={Ga}andβ={Gc,Ge}.

Fig.4

EXERCISES3.4

1. LetA={a,b,c,d,e,f},andletGandHbethefollowingequivalencerelationsinA:

ClearlyHisarefinementofG.ExhibitthesetsA/G,A/H,G/H,(A/H)/(G/H).2. Let bethesetoftherealnumbers,andletGbethefollowingrelationin × :

Let f : → × be the function given by f(x) = (sin x, cos x). Describe (G);what are itsequivalenceclasses?

3. Let f :A→B be a function and letG be an equivalence relation inB. Prove that (G) is anequivalencerelationinA.

4. Letf:A→BbeafunctionandletGbeanequivalencerelationinB.Provethateachequivalenceclassmodulo (G)istheinverseimageofanequivalenceclassmoduloG.Moreprecisely,ifH=(G)andy=f(x),provethatHx= (Gy).

5. LetGbeanequivalencerelationinAandsupposethatB⊆A.ProvethatG[B] isanequivalencerelationinB.

6. LetGbeanequivalencerelationinAandsupposeB⊆A.Provethatforeachx∈B,(G[B])x=Gx∩B.

7. LetG,H,andJbeequivalencerelationsinA,andsupposethatG⊆HandH⊆J.ProvethatH/GisfinerthanJ/G.

8. LetG,H,andJbeequivalencerelationsinA,andsupposethatG⊆HandH⊆J.Proveeachofthefollowing.a)G⊆ .b)If isanequivalencerelationinA,then(H/G) (J/G)=(H J)/G.c)(H/G) (J/G)isanequivalencerelationinA/G.

9. SupposethatGandHareequivalencerelations inA,and thatG⊆H.Prove that ifandonlyifGxandGyaresubsetsofthesameequivalenceclassmoduloH.

10. SupposethatGisanequivalencerelationinA,andHisanequivalencerelationinB.TheproductofGandHisdefinedtobethefollowinginA×B:

ProvethatG•HisanequivalencerelationinA×B.11.ProvethateveryequivalencerelationinasetAisthepre-imageofanequivalencerelationinA×A.

[Hint:Letf:A→A×Abethefunctiongivenbyf(x)=(x,x);ifG isarelationinA,considertherelationG•G(seeExercise10)inA×A.]

12.Letf:A→BbeafunctionandletGbeanequivalencerelationinB.Provethat .

5EQUIVALENCERELATIONSANDFUNCTIONS

Iff:A→Bisafunction,wedefinearelationGinAasfollows:

ItiseasytoseethatGisanequivalencerelationinA,Giscalledtheequivalencerelationdeterminedbyf.

Conversely,ifGisanequivalencerelationinasetA,wedefineafunctionf:A→A/Gasfollows:

Itiseasytoseethatfisafunction;fiscalledthecanonicalfunctionfromAtoA/G.

3.22TheoremLetGbeanequivalencerelationinasetA.IffisthecanonicalfunctionfromAtoA/G,thenGistheequivalencerelationdeterminedbyf.

Proof.LetfbethecanonicalfunctionfromAtoA/G,andletHbetheequivalencerelationdeterminedbyf;wewillprovethatG=H:

LetAandBbesetsandletf:A→Bbeafunction;wewilldefine threefunctionsr,s, t,obtainedfrom f , which play an important role in many mathematical arguments. LetG be the equivalencerelationdeterminedbyf:

r:A→A/GisthecanonicalfunctionfromAtoA/G.s:A/G→ (A)isthefunctiongivenbys(Gx)=f(x),∀x∈A.t: (A)→Bisthefunctiongivenbyt(y)=y,∀y∈ (A).

Notethattistheinclusionfunctionof (A)inB(see2.12).

3.23Theorem LetA andB be sets, let f :A→B be a function, letG be the equivalence relationdeterminedbyf ,andletr,s,tbethefunctionsdefinedabove.Thenr issurjective,s isbijective, t isinjective,andf= .

Proofi) IfGx∈A/G,thenx∈Aandr(x)=Gx;thusrissurjective.ii) Iff(x)∈ (A),thenx∈A,Gx∈A/G,andf(x)=s(Gx);thussissurjective.iii) s(Gx)=s(Gy)⇒f(x)=f(y)⇒(x,y)∈G⇒Gx=Gy;thussisinjective.iv) t(y1)=t(y2)⇒y1=y2;thustisinjective.v) Letx∈A;t{s[r(x)]}=t[s(Gx)]=t(f(x))=f(x);thus

soby2.5, =f.

Wemaysumuptheforegoingresultsbysayingthatanyfunction f:A→Bcanbeexpressedasacompositeof threefunctionsr,s, twhichare, respectively, surjective,bijective, and injective.This isreferredtoasthecanonicaldecompositionoff,anditiscustomarilyexhibitedinadiagramsuchasthefollowing:

Oneof theresultsof3.23 isespeciallyuseful;namely, that if f :A→B isa functionandG is theequivalence relation determined by f , thenA/G and (A) are in one-to-one correspondence. This iscustomarilyexpressedbywritingA/G≈ (A).Inparticular,

3.24iffissurjective,thenA/G≈B.

LetAandBbesets,letf:A→Bbeafunction,andletHbetheequivalencerelationdeterminedbyf.LetGbeanyequivalencerelationinAwhichisfiner thanH.WedefineafunctionfromA/G toBasfollows:

Itiseasytoseethatf/GisafunctionfromA/GtoB;f/GiscalledthequotientoffbyG.

3.26TheoremLetf:A→Bbeafunction,letHbetheequivalencerelationdeterminedbyf,andletGbearefinementofH.ThenH/Gistheequivalencerelationdeterminedbyf/G.

Proof.LetJbetheequivalencerelationdeterminedbyf/G;wewillprovethatJ=H/G.Indeed,

AsanexampleoftheuseofTheorem3.26,considerthefollowingsituation:GandHareequivalencerelationsinA,G⊆H,fisthecanonicalfunctionfromAtoA/H(hence,by3.22,Histheequivalencerelationdeterminedby f ).Thus,by3.25,f/G isafunctionfromA/GtoA/H ,andby3.26,H/G is theequivalencerelationdeterminedby f/G. It iseasy tosee that f/G is surjective,because f issurjective.Therefore,by3.24,

EXERCISES3.5

1. Letf:A→Bbeasurjectivefunction,letGbetheequivalencerelationinducedbyf,andletHbeanequivalencerelationinAwhichiscoarserthanG.DefinetheimageofHasfollows:

Provethat (H)isanequivalencerelationinB.2. Letf:A→Bbeasurjectivefunction,andletGbetheequivalencerelationinducedbyf.LetJbe

anyequivalencerelationinB.Provethata) (J)iscoarserthanG.b)H= (J)ifandonlyifJ= (H).(SeeExercise1above.)Concludethatthereexistsaone-to-onecorrespondencebetweentheequivalencerelationsinBandtheequivalencerelationsinAwhicharecoarserthanG.

3. Letf:A→Bbeafunction,andletGbetheequivalencerelationdeterminedbyf.ProvethatG=f−1 f.

4. Letf:A→Bandg:B→Cbefunctions,andletGbetheequivalencerelationdeterminedbyg.Provethat (G)istheequivalencerelationdeterminedbyg f.

5. LetGandHbeequivalencerelationsinasetA,andsupposethatG⊆H.Letfbe thecanonicalfunctionfromAtoA/G.ProvethatH/G= (H).(SeeExercise1.)

6. LetGandHbeequivalencerelationsinA,andsupposethatG⊆H.LetfbethecanonicalfunctionfromAtoA/G,andletgbethecanonicalfunctionfromAtoA/H.Leth=g/G.Provethatg= .

7. LetGandHbeequivalencerelationsinA,andsupposethatG⊆H.IffisthecanonicalfunctionfromAtoA/G,provethatH= (H/G).

8. LetG,H,andJbeequivalencerelationsinAandsupposethatG⊆H⊆J.Letf:A→A/G,g:A→A/H ,andh:A→A/J be the canonical functions associated, respectively,withG,H , and J.Provethath/G=h/ /G.

9. LetGandHbeequivalencerelationsinA,andsupposethatG⊆H.LetfbethecanonicalfunctionfromAtoA/H.Provethatf/Gissurjective.

10. LetGandHbearbitraryequivalencerelationsinA.Provethata)A/( )≈(A/G)/( /G).b)A/G≈(A/G∩H)/(G/G∩H).

11. Letf:A→Abeafunction,andletGbetheequivalencerelationdeterminedbyf.Provethat =fifandonlyif

foreveryz,x∈A.

4PartiallyOrderedClasses

1FUNDAMENTALCONCEPTSANDDEFINITIONS

Byapartiallyorderedclasswemeanapairofobjects A,G ,whereA is aclassandG isanorderrelationinA.WesaythatA isorderedbyG,orthatGordersA.IfA isaset,wesay that A,G isapartiallyorderedset.In ordinary mathematical applications, every partially ordered class is a partially ordered set.

However,theintuitiveideaofan“orderedcollectionofelements”ismeaningfulforanycollectionA,whetherAbeasetoraproperclass;henceitisnaturaltogivethedefinitioninitsmostgeneralform,lettingAbeanyclass.Onceagain,sinceverysetisaclass,everythingwehavetosayaboutpartiallyorderedclassesapplies,inparticular,topartiallyorderedsets.Let A,G beapartiallyorderedclass; if it iswellunderstood, inagivendiscussion, thatG is the

orderrelationinA,thenwewillsaylooselythatAisapartiallyorderedclass.IfAisapartiallyorderedclass,orderedbyG,itiscustomarytowritex ytodenotethefactthat(x,y)∈G.Wefurtheragreethaty xhasthesamemeaningasx y,andthatx ymeansthat(x,y)∉G.Ifx∈Aandy∈Aandx y,thenwesaythat“xislessthanorequaltoy.”Weagreethatx<yisan

abbreviationfor“x yandx≠y.”Ifx<y,wesaythat“xisstrictlylessthany.”Notethat<isnotanorderrelation;itisirreflexive,asymmetric,andtransitive.IfAisapartiallyorderedclassandBisasubclassofA,wemayconsiderBtobeorderedbytheorder

relationinA.Specifically,ifx∈Bandy∈B,thenweletx yinBifandonlyifx yinA.

IfAandBarepartiallyorderedclasses,thereareseveralpossiblewaysoforderingtheclassA×B;thetwomostusefulwaysofdoingsoaregiveninthefollowingdefinitions.

4.1DefinitionLetAandBbepartiallyorderedclasses;bythelexicographicorderingofA×Bwemeanthe followingorder relation inA×B: If (a1,b1)∈A ×B and (a2,b2)∈A ×B, thenwe let (a1,b1)(a2,b2)ifandonlyif

i) a1<a2or

ii) a1=a2andb1 b2.Thelexicographicorderingissocalledbecauseitimitatesthewayweorderwordsinthedictionary

(forexample,beprecedesgobecausebprecedesg,andbeprecedesbybecauseeprecedesy).

4.2DefinitionLetAandBbepartiallyorderedclasses;bytheantilexicographicorderingofA×BwemeanthefollowingorderrelationinA×B:If(a1,b1)∈A×Band(a2,b2)∈A×B,thenwelet(a1,b1)(a2,b2)ifandonlyif

i) b1<b2or

ii) b1=b2anda1 a2.

4.3DefinitionLetAbeapartiallyorderedclass.TwoelementsxandyinAaresaidtobecomparableifeitherx yory x;otherwise,theyaresaidtobeincomparable.

4.4DefinitionLetAbeapartiallyorderedclass,andletBbeanarbitrarysubclassofA.IfeverytwoelementsofBarecomparable,thenwecallBafullyorderedsubclassofA,oralinearlyorderedsubclassofA,or,morecommonly,achainofA.IfeverytwoelementsofAarecomparable,thenA iscalledafullyordered,orlinearlyordered,class.

4.5DefinitionLetAbeapartiallyorderedclassandsupposea∈A.TheinitialsegmentofAdeterminedbyaistheclassSa,definedasfollows:

4.6Theorem LetA be a partially ordered class. IfP is an initial segment ofA, andQ is an initialsegmentofP,thenQisaninitialsegmentofA.

Proof.Byhypothesis,P={x∈A:x<a}forsomea∈A,Q={x∈P:x<b}forsomeb∈P.LetQ1={x∈A:x<b};Q1isobviouslyaninitialsegmentofA;wewillshowthatQ=Q1,andthetheoremwillthusbeproved.Clearly,Q⊆Q1;conversely,ifx∈Q1,thenx∈Aandx<b;butb<abecauseb∈P;hencex<a,anditfollowsthatx∈P;thusx∈Q.

Theorem4.6maybeparaphrasedasfollows:

AninitialsegmentofaninitialsegmentofAisaninitialsegmentofA.

4.7DefinitionIfAisapartiallyorderedclass,thenacutofAispair(L,U)ofnonemptysubclassesofAwiththefollowingproperties:i) L∩U=ØandL∪U=A.ii) Ifx∈Landy x,theny∈L.iii) Ifx∈Uandy x,theny∈U.

It is convenient to use a graphic device called a line diagram to illustrate simple properties ofpartially ordered classes.The elements of the class are represented by points on the diagram; if twopointsxandyareconnectedbyaline,andthelinerisesfromxtoy,thismeansthatx y.

4.8ExampleFig.1representsapartiallyorderedclasswithsixelements.Notethat{a,b,c}and{d,e,b,c}arechainsofA;Se={b,c,f}istheinitialsegmentdeterminedbye;ifL={a,b,c}andU={d,e,f},then{L,U}isacut.

Fig.1

4.9ExampleThemost importantorderrelation inmathematics is theclass inclusionrelation⊆; it isreflexive,forifAisanyclass,thenA⊆A;itisantisymmetric,becauseifA⊆BandB⊆A,thenA=B;itistransitive,forA⊆BandB⊆CimplythatA⊆C.If isaclass(ofclasses)andweconsiderAtobeorderedbytheinclusionrelation,wesaythat isorderedbyinclusion.Notethatif isachainof,thismeansthatforanytwoelements,A,B∈ ,eitherA⊆BorB⊆A.

EXERCISES4.1

1. LetA={a,b,c,d};if isorderedbyinclusion,drawitslinediagram.2. LetAbethepartiallyorderedclassdefinedbythefollowingdiagram.

ListallthechainsofA,alltheinitialsegmentsofA,andallthecutsofA.3. LetAbethepartiallyorderedclassofExcercise2.Drawthelinediagramforthefollowingclasses:

the class of all the chains ofA (ordered by inclusion), the class of all the initial segments ofA(orderedbyinclusion),theclassofallthecutsofA(orderedbyinclusiononthe“leftcomponent”L).

4. LetAandBbepartiallyorderedclasses,letCbeachainofA,andletDbeachainofB.IfA×Bisorderedlexicographically(4.1),provethatC×DisachainofA×B.

5. LetAandBbepartiallyorderedclasses,andletA×Bbeorderedantilexicographically(4.2).Provethatif(L,U)isacutofB,then(A×L,A×U)isacutofA×B.

6. LetAbeapartiallyorderedclass,andletGbeanequivalencerelationinA.Supposethefollowingconditionholds:If andx y z,then .DefinearelationHinA/GbyH={(Gx,Gy):∀w∈Gx,∃z∈Gy∋w z}.ProvethatHisanorderrelationinA/G.

2ORDERPRESERVINGFUNCTIONSANDISOMORPHISM

4.10DefinitionLetAandBbepartiallyorderedclasses;afunctionf:A→Bissaidtobeincreasing,ororder-preserving,ifitsatisfiesthefollowingcondition:Foreverytwoelementsx,y∈A,

Wesaythatf:A→Bisstrictlyincreasingifitsatisfiesthefollowingcondition:Foreverytwoelementsx∈Aandy∈A,

4.11DefinitionLetAandBbepartiallyorderedclasses;afunctionf:A→Biscalledanisomorphismifitisbijectiveandsatisfiesthefollowingcondition:Foreverytwoelementsx∈Aandy∈A,

Figures 2, 3, and 4 provide simple illustrations of the concepts we have just defined. Figure 5describesafunctionwhichisbijectiveandincreasing,butisnotanisomorphism[notethatf(b)<f(a)butaandbareincomparable].ThereadershouldcomparethisexamplewithDefinition4.11.

4.12TheoremIff:A→Bisanisomorphism,then

Fig.2

Fig.3

Fig.4

Fig.5

Proofi) Letusassumethatx<y;thenx y;hencef(x) f(y).Iff(x)=f(y),thenx=y,whichiscontraryto

ourassumption;thusf(x)<f(y).ii) Theconverseisprovedbythesameargument.

4.13TheoremLetAandBbepartiallyorderedclassesandletf:A→Bbeabijectivefunction.Thenf:A→Bisanisomorphismifandonlyiff:A→Bandf−1:B→Aareincreasingfunctions.

Proof.Note, first, that if f is bijective, then∀x∈A, f−1(f (x))=x.Now suppose that f and f −1 areincreasingfunctions:

fromthis,and thefact that f is increasing,wededuce that f isan isomorphism.Conversely, if f is anisomorphism,thencertainlyfisincreasing;furthermore,iff(x)andf(y)arearbitraryelementsofB,then

sof−1isincreasing.

4.14TheoremLetA,B,andCbepartiallyorderedclasses.i) TheidentityfunctionIA:A→Aisanisomorphism.

ii) Iff:A→Bisanisomorphism,thenf−1:B→Aisanisomorphism.iii) Iff:A→Bandg:B→Careisomorphisms,theng f:A→Cisanisomorphism.

Proofi) By2.10,IA:A→Aisbijective;nowIA(x)=xandIA(y)=y;hence

ii) Iff:A→Bisanisomorphism,thenitisbijective,hencef−1:B→Aisbijective;by4.13,f−1:B→Aisincreasing.Finally,

sof−1:B→Aisanisomorphism.

Theproofof(iii)isleftasanexerciseforthereader.

4.15DefinitionIfAandBarepartiallyorderedclassesandthereexistsanisomorphismfromAtoB,wesaythatAisisomorphicwithB.

Theorem4.14indicatesthattherelation“AisisomorphicwithB”isanequivalencerelationamongpartiallyorderedclasses.Indeed,by4.14(i),AisisomorphicwithA;by4.14(ii),ifAisisomorphicwithB,thenBisisomorphicwithA;by4.14(iii),ifAisisomorphicwithBandBisisomorphicwithC,thenAisisomorphicwithC.WewillwriteA BtodenotethefactthatAisisomorphicwithB.

Theconceptofisomorphismisofgreatimportanceinthestudyofpartiallyorderedclasses.SupposethatA andB are partially ordered classes and f :A→B is an isomorphism; let us agree towrite x′insteadoff(x);sincef:A→Bisbijective,everyelementxinAcorrespondswithauniqueelementx′inB:

Furthermore,by4.12,ifxandyareanytwoelementsinA,then

ThismeansthattheorderingofAisexactlythesameastheorderingofB;inparticular(ifitispracticaltodrawthelinediagramsofAandB),thelinediagramofAisthesameasthelinediagramofB.Thus,essentially,thereisnodifferencebetweenAandBexceptthelettersweusetodesignatetheirelements.

4.16ExampleIfB={l,m,n,o,p,q}isorderedasinthefollowingdiagram,

thenB is isomorphicwith the classA ofExample4.8. The isomorphism f :A→B is given by thefollowingtable.

Inconclusion,ifAisisomorphicwithB,andifweidentifycorrespondingelements,thenAandBareessentiallythesamepartiallyorderedclass.

EXERCISES4.2

1. Provethatiff:A→Bisaninjective,increasingfunction,thenitisstrictlyincreasing.2. Letf:A→Bbeanincreasingfunction.IfCisachainofA,provethat isachainofB.3. LetAandBbepartiallyorderedclasses.ProvethatifA×Bisorderedlexicographically,thenthe

projectionfunctionp1:A×B→Aisincreasing;ifA×Bisorderedantilexicographically,thentheprojectionfunctionp2:A×B→Bisincreasing.[Notethatp1(x,y)=x,p2(x,y)=y.]

4. AsubclassCofapartiallyorderedclassiscalledconvexifitsatisfiesthefollowingcondition:Ifa∈Candb∈Canda x b,thenx∈C.LetAandBbepartiallyorderedclasses,letf:A→Bbeanincreasingfunction,andletCbeaconvexsubclassofB.Provethat isaconvexsubclassofA.

5. LetAandBbepartiallyorderedclasses,letf:A→Bbeanincreasingfunction,andletHbetheequivalence relation determined by f. Prove that each equivalence class moduloH is a convexsubclassofA.

6. LetAandBbepartiallyorderedclasses,andletf:A→Bbeanincreasingfunction;assume =B.Provethatif(L,U)isacutofB,then isacutofA.

7. Provethatthecompositeoftwoincreasingfunctionsisincreasing.Usethisresulttoprove4.14(iii).8. LetAandBbepartiallyorderedclasses,andletf:A→Bbeanisomorphism.Proveeachofthe

following:a) IfCisaconvexsubclassofA,then isaconvexsubclassofB.

b) If(L,U)isacutofA,then isacutofB.c) If[a,b]isaclosedintervalofA,then isaclosedintervalofB.(Ifa,b∈A,thentheset

{x∈A:a x b}iscalledaclosedintervalofAandisdenotedbythesymbol[a,b].)9. LetE andF be partially ordered classes, and letg :E→F be an isomorphism. Prove that for

arbitraryx∈E, concludethatSx Sg(x).

10. LetAbeapartiallyorderedset.Foreacha∈A,letIa={x∈A:x a}.Let ={Ia}a∈Aandletbeorderedbyinclusion.Provethat isisomorphicwithA.

11. LetA,B,andCbemutuallydisjoint,partiallyorderedclasses.IfA B,provethat(A∪C) (B∪C).[Hint:Taketheunionoftwofunctions,asin2.16.]

12. LetAbeapartiallyorderedset.DefineLx={z∈A:z x}andUx={z∈A :z x}.Prove thefollowing:

a) Foreachx∈A,(Lx,Ux)isacutofA.b) Thefunctionϕdefinedbyϕ(x)=(Lx,Ux)isanisomorphismbetweenAandtheclassofallthe

cutsoftheabove-describedform.Thesetofcuts(L,U)isorderedbyinclusiononL.

3DISTINGUISHEDELEMENTS.DUALITY

Certaindistinguishedelementsplayanimportantpartinthestudyofpartiallyorderedclasses.Wenowdefinethem;ineachofthefollowingdefinitions,weassumethatAisapartiallyorderedclass.

4.18DefinitionAnelementm∈AiscalledamaximalelementofA ifnoneof theelementsofAarestrictlygreaterthanm;insymbols,thiscanbeexpressedasfollows:

Similarly,anelementn∈AiscalledaminimalelementofAifnoneoftheelementsofAarestrictlylessthann;insymbols,

4.19DefinitionAnelementa∈A is called thegreatestelement ofA ifa x for every x∈A. Anelementb∈AiscalledtheleastelementofAifb xforeveryx∈A.ItiseasytoseethatifAhasagreatestelement,thenthiselementisunique;forsupposeaanda′are

bothgreatestelementsofA.Thena a′anda′ a;hencea=a′.Analogously,theleastelementofAisunique.

4.20DefinitionLetBbeasubsetsofA.AnupperboundofBinAisanelementa∈Asuchthata xforeveryx∈B.AlowerboundofBinAisanelementb∈Asuchthatb xforeveryx∈B.Whenthereisnoriskofambiguity,wewillrefertoanupperboundofBinAsimplyasan“upperboundofB,”andtoalowerboundofBinAsimplyasa“lowerboundofB.”TheclassofalltheupperboundsofBwillbedenotedbyυ(B)andtheclassofallthelowerboundsofBwillbedenotedbyλ(B).

4.21DefinitionIftheclassoflowerboundsofBinAhasagreatestelement,thenthiselementiscalledthegreatestlowerboundofBinA.IftheclassofupperboundsofBinAhasaleastelement,thenthiselementiscalledthe leastupperboundofB inA.The leastupperboundofB inA isalsocalled thesupremumofB inA (abbreviated supAB), and thegreatest lowerboundofB inA is also called theinfinumofBinA(abbreviatedinfAB).Whenthereisnoriskofambiguity,wewillwritesupBforsupAB,andinfBforinfAB.Wehaveseenthatthegreatestelementandtheleastelementofanyclassareunique;hencethesup

andtheinf,iftheyexist,areunique.

Examples

4.22Figure6isthelinediagramofaclassthathasmaximalelementsbutnogreatestelement(aanddaremaximalelements).

4.23InFig.7,letA={a,b,c,d,e,f}andletB={b,c,e,f}.BhastwoupperboundsinA,namelyaandd,butnosup.

4.24InFig.7,letAandBbedefinedasabove,andletC={a,b,c,e,f}andD={d,b,c,e,f}.ThenBhasnosupinA,althoughsupCB=aandsupDB=d.

Fig.6

Fig.7

4.25TheclassNofallthepositiveintegershasaleastelementbutnogreatestelementandnomaximalelements.Theclass ofalltheintegershasneitheragreatestnoraleastelement.

4.26Let beaclass(ofclasses)whichisorderedbyinclusion;let ={Bi}i∈Ibeasubclassof ,and

letusassumethat and areelementsof .Thensup = :indead,eachBiis ,hence

isanupperboundof ;furthermore,ifCisanyotherupperboundof ,thismeansthatBi Cfor

everyi∈I;hence,by1.40(i), ;thisprovesthat istheleastupperboundof .Similarly,inf

= ,forclearly is eachBi,hence isalowerboundof ;furthermore,ifDisanyother

lowerboundof ,thismeansthatD Biforeveryi∈I,hence,by1.40(ii),D ;thisprovesthat

isthegreatestlowerboundof .

4.27Itisimportanttonotethatλ(Ø)=A.Indeed,ifx∈A,thenthestatement“x yforeveryy∈Ø”isnotfalse(fortodenyitwouldbetoassertthat“∃y∈Øx y,”whichisabsurd);henceitistrue.Thisholdsforeachelementx∈A, soweconclude thatλ(Ø)=A.Next,wenote that infØ is thegreatestelementofλ(Ø);thusifAhasagreatestelementa,thena=infØ.Analogously,ifbistheleastelementofA,thenb=supØ.

When we reason about partially ordered classes, we are led to the interesting notion of duality.Briefly,dualitycanbeexplainedasfollows.

IfGisanorderrelationinA,thenG−1isalsoanorderrelationinA(seeExercise8,ExerciseSet3.2).Let refertotheclassAorderedbyG,andlet refertoAorderedbyG−1.Thenx y in

ifandonlyifx yin ;itfollowsthataisamaximalelementof ifandonlyifaisaminimalelementof ;aisthegreatestelementof ifandonlyifa istheleastelementof

;ifB A,thenbisanupperboundofBin ifandonlyifbisalowerboundofBin;andb=supBin ifandonlyifb=infBin .Let beastatementaboutpartiallyorderedclasses.In ,supposethatwereplaceeachoccurrence

of by andviceversa;supposefurthermorethatwereplacethewords“maximal”by“minimal”andviceversa,“greatest”by“least”andviceversa,“upperbound”by“lowerbound”andviceversa,“sup”by“inf”andviceversa.Theresultingstatement ′iscalledthedualof .Inviewofwhatwehavesaidabove,itiseasytoseethatif isatruestatementinA,G−1,thenthe

dualof istrueinA,G.Inparticular,supposethat isatheoremforallpartiallyorderedclasses.IfAisanypartiallyorderedclassandGistheorderrelationinA,then istrueinA,G−1,hencethedualofistrueinA,G.Thusif isatheoremforallpartiallyorderedclasses,thedualof isalsoatheorem.The concept of duality permits a considerable economy in the presentation of theorems about

partiallyorderedclasses,foreverytimeweproveatheorem,weknowthatthedualofthetheoremisalsotrue.

Thefollowingareafewpropertiesofthedistinguishedelementsinapartiallyorderedclass.

4.28TheoremIfAhasagreatestelementa,andBhasagreatestelementb,andA⊆B,thena b.

Proof.Bydefinition,b xforeveryx∈B;buta∈A⊆B;henceb a.

DualIfAhasaleastelementa,andBhasaleastelementb,andA⊆B,thena b.

4.29TheoremLetBandCbesubclassesofA.IfB⊆C,thenv(C)⊆v(B).

Proof.x∈v(C)⇒x y,∀y∈C⇒x y,∀y∈B⇒x∈v(B).

DualIfB⊆C,thenλ(C)⊆λ(B).

4.30TheoremLetBandCbesubclassesofA,andsupposethatBandCeachhasasupinA.IfB⊆CthesupB supC.

Proof.By4.29,v(C)⊆v(B);henceby4.28(dual),supB supC.

DualIfBandCeachhasainfinA,andifB⊆C,theninfB infC.

4.31TheoremLetBbeasubclassofA.ThenB⊆v(λ(B)).

Proof.Supposex∈B;foreachy∈λ(B),y x,thatis,x y;hencex∈v(λ(B)).

DualB⊆λ(v(B)).

4.32LemmaLetBbeasubclassofAandsupposethatλ(B)hasasupinA.ThenBhasaninfinA,andinfB=supλ(B).

Proof.Leta=supλ(B).Supposeb∈B;foreveryc∈λ(B),c b;hencebisanupperboundofλ(B);thusa b.Thisistrueforeachb∈B,soweconcludethataisalowerboundofB.NowifdisanylowerboundofB,thend∈λ(B),soa dbecauseaisanupperboundofλ(B).WehaveprovedthataisthegreatestlowerboundofB.

DualIfv(B)hasaninfinA,thenBhasasupinAandsupB=infv(B).

4.33DefinitionLetA be a partially ordered class. If every nonempty subclass ofA that is boundedabovehasasup,thenAissaidtobeconditionallycomplete.

We have the following alternative definition of conditionally complete: A is called conditionallycomplete if every nonempty subclass of A that is bounded below has an inf. Our next theoremestablishestheequivalenceofthetwodefinitions.

4.34TheoremThefollowingtwoconditionsareequivalent:i) EverynonemptysubclassofAthatisboundedabovehasasup.ii) EverynonemptysubclassofAthatisboundedbelowhasaninf.

Proofa) Supposethat(i)holds;letBbeanonemptysubclassofAwhichisboundedbelow,thatis,λ(B)≠Ø.

EachelementofBisanupperboundofλ(B),henceλ(B)isboundedabove;thusλ(B)hasasup.But,by4.32,itfollowsthatBhasaninf.

b) Theconverseisthedualoftheresultwehavejustproven.

EXERCISES4.3

1. SupposeB⊆A;provethatifBhasagreatestelementb,thenb=supB.2. SupposeB⊆A;provethatv(B)=v(λ(v(B))).[Hint:Use4.29and4.31.]3. SupposeB⊆AandC⊆A;provethatλ(B∪C)=λ(B)∩λ(C).4. SupposeB⊆A;provethatifBhasasupb,thenλ(v(B))∩v(B)={b}.5. SupposeC⊆BandB⊆A;provethatsupAC supBC.6. LetBandCbesubclassesofapartiallyorderedclassA.ProvethatifsupB=supC,thenv(B)=

v(C).7. LetAandBbepartiallyorderedclassesandletf:A→Bbeastrictlyincreasingfunction.Prove

thatifbisamaximalelementofB,theneachelementof isamaximalelementofA.8. LetAandBbepartiallyorderedclasses,andletf:A→Bbeanincreasingfunction.Provethatifa

isthegreatestelementofA,thenf(a)isthegreatestelementof .9. LetAandBbepartiallyorderedclasses,andletf:A→Bbeanincreasingfunction;supposeC⊆

A.ProvethatifcisanupperboundofC,thenf(c)isanupperboundof .10. LetAandBbepartiallyorderedclasses,andletf:A→Bbeanisomorphism.Proveeachofthe

following:a) aisamaximalelementofAifff(a)isamaximalelementofB.b) aisthegreatestelementofAifff(a)isthegreatestelementofB.c) SupposeC⊆A;xisanupperboundofCifff(x)isanupperboundof .d) b=supCifff(b)=sup .

11. LetAbeapartiallyorderedclass.Provethefollowing:a) If every subclass ofA has a sup and an inf, inA, thenA has a least element and a greatest

element.[Hint:Use4.27.]b) Thefollowingtwostatementsareequivalent:EverysubclassofAhasasup;everysubclassofA

hasaninf.12. LetAandBbepartiallyorderedclasses.Provethefollowing:

a) SupposeA×Bisorderedlexicographically:if(a,b)isamaximalelementofA×B,thenaisamaximalelementofA.

b) SupposeA×Bisorderedantilexicographically.If(a,b)isamaximalelementofA×B,thenbisamaximalelementofB.

4LATTICES

4.35DefinitionLetAbeapartiallyorderedclass.Ifeverydoubleton{x,y}inAhasasupandaninf,thenAiscalledalattice.

Whendealingwithlatticesitiscustomarytodenotesup{x,y}byx∨yandinf{x,y}byx∧y.IfAisa lattice,x∨y isoftencalled the joinofxandy,andx∧y is often called themeet ofx andy; theexpressionx∨yisread“xjoiny“andtheexpressionx∧yisread“xmeety.”Note the following simple consequences of our definition. If a and b are arbitrary elements of a

latticeA,then

4.36a a∨bandb a∨bbecausea∨bisanupperboundofaandb.Furthermore,ifc∈A,then

4.37a candb c⇒a∨b c;inotherwords,ifcisanupperboundofaandb,thena∨b cbecausea∨bistheleastupperboundofaandb.Foranalogousreasons,wehave

4.38a∧b aanda∧b b

4.39c aandc b⇒c a∧b;

4.40TheoremLetAbealattice;thejoinandthemeethavethefollowingproperties:

L1. x∨x=xandx∧x=x.L2. x∨y=y∨xandx∧y=y∧x.L3. (x∨y)∨z=x∨(y∨z)and(x∧y)∧z=x∧(y∧z).L4. (x∨y)∧x=xand(x∧y)∨x=x.

Proof.L1andL2areimmediateconsequencesofthedefinitionsofsupandinf.

L3.First,wewillprovethat

by4.36,

hence,by4.37,x∨y x∨(y∨z)furthermore,by4.36,

soby4.37,(x∨y)∨z x∨(y∨z).Theinequalityx∨(y∨z) (x∨y)∨zisproveninthesameway,hence

Thedualofthisresultis(x∧y)∧z=x∧(y∧z).

L4.Toprovethat(x∨y)∧x=xistoprovethat

Nowxisalowerboundof{x∨y,x}becausex x∨yby4.36andobviouslyx x.Furthermore,ifzisanylowerboundof{x∨y,x},thenz x;thusxisthegreatestlowerboundof{x∨y,x}.Thisprovesthat(x∨y)∧x=x;thedualofthisresultis(x∧y)∨x=x.

A latticemay alternativelybedefined as an algebraic systemwith twooperations∨ and∧whichhave properties L1 throughL4. This fact,which is of great importance in the study of lattices, is aconsequenceofthefollowingtheorem.

4.41TheoremLetAbeaclassinwhichtwooperationsdenoted∨and∧aregivenandhavepropertiesL1throughL4.WedefinearelationinA,tobedenotedbythesymbol ,asfollows:

4.42x yifandonlyifx∨y=y.

Then isanorderrelationinA,andAisalattice.

Proof.First,letusprovethattherelation definedaboveisanorderrelation.

Reflexive.ByL1,x∨x=x;hence,by4.42,x x.

Antisymmetric.Supposethatx yandy x;by4.42,x∨y=yandy∨x=x;butbyL2,x∨y=y∨x,hence,x=y.

Transitive.Supposethatx yandy z;by4.42,x∨y=yandy∨z=z;thus

soby4.42,x z.Next,wewillprovethatx∨yistheleastupperboundofxandy:

soby4.42,x x∨y;analogously,y x∨y,hencex∨yisanupperboundofxandy.Nowifzisanyupperboundofxandy,thatis,x zandy z,thenx∨z=zandy∨z=z;thus

soby4.42,x∨y z.Thisprovesthatx∨y=sup{x,y}.Theproofthatx∧y=inf{x,y}isleftasanexerciseforthereader.WeconcludethatAisalattice.

Itfollowsfrom4.40and4.41thatalatticemaybedefinedintwodistinctways:asapartiallyorderedclassinwhicheverypairofelementshasasupandaninfor,alternatively,asanalgebraicsystemwithtwooperationssatisfyingrulesL1throughL4.

4.43DefinitionLetAbealattice,andletBbeasubclassofA.If

thenBiscalledasublatticeofA.

4.44DefinitionABooleanalgebraisdefinedtobealatticeAwiththefollowingadditionalproperties:

L5.Thereisanelement0∈Aandanelement1∈Asuchthatforeachx∈A,x∨0=xandx∧1=x.

L6.Foreachx∈Athereisanelementx′∈Asuchthat

L7.x∨(y∧z)=(x∨y)∧(x∨z)andx∧(y∨z)=(x∧y)∨(x∧z).

The algebra of classes is an example of a Boolean algebra; indeed, by 1.22, 1.25, and 1.26, theoperations ,∩and′satisfyL1throughL7.Wehavenotedindependently(4.26)thatA B=sup{A,B}andA∩B = inf{A,B}.Another example of aBoolean algebra is the algebra of sentences,with theoperations of conjunction, disjunction, and negation. In addition, Boolean algebra has a variety ofapplicationsinmanyareasofscienceandtechnology.

4.45DefinitionLetAbeapartiallyorderedclass;Aiscalledacompletelattice ifeverysubclassofAhasasup.Alternatively,AiscalledacompletelatticeifeverysubclassofAhasaninf.

Thepurposeofournext theoremis toshowthat these twoalternativedefinitionsareequivalent. ItwillfollowthatifAisacompletelattice(inthesenseofeitherdefinition),theneverysubclassofAhasa sup and an inf; in particular, every doubleton inA has a sup and an inf, hencewe are justified incallingAalattice.

4.46TheoremLetAbeapartiallyorderedclass;thefollowingtwoconditionsareequivalent:i) everysubclassofAhasasup;

ii) everysubclassofAhasaninf.

Proof.Letusassumethat(i)holds;itfollowsthatAhasasup,whichisnecessarilythegreatestelementofA,andØhasasup,whichistheleastelementofA(see4.27).LetMdesignatethegreatestelementofA,andletmdesignatetheleastelementofA.LetBbeanarbitrarysubclassofA.IfB=Ø,theninfB=M(see4.27);

ifB≠Ø, thenB isboundedbelowbym,hencebyTheorem4.34,B hasan inf.Thus, (ii)holds; theconverseisthedualofwhatwehavejustproven.

4.47ExampleLetAbeanarbitrarysetandlet bethesetofalltheequivalencerelationsinA,orderedbyinclusion. hasaleastelement,namelytherelationIA,andagreatestelement,namelytherelationA×A.Furthermore,if{Gi}i∈Iisanysubsetof ,then isanelementof (seeExercise4,Exercise

Set3.2);aswehaveseen(4.26), is thegreatest lowerboundof{Gi}i∈I.Thus, isacompletelatticeby4.46(ii).

EXERCISES4.4

1. LetAbeaclasswithtwooperations∨and∧whichsatisfyL1throughL4.Provethatx∨y=yifandonlyifx∧y=x.

2. InTheorem4.41,provethatx∧y=inf{x,y}.[Hint:UsetheresultofExercise1.]3. LetAbealattice.Provethatthefollowingstatementsaretrue.

a) Ifa b,then∀x∈A,a∨x b∨xanda∧x b∧x.b) Ifa bandc d,thena∨c b∨danda∧c b∧d.

4. LetAbealattice.If[a,b]and[c,d]areclosedintervalsofA,provethat

(SeeExercise8,ExerciseSet4.2,foradefinitionofclosedinterval.)5. LetAbealattice;provethateveryclosedinterval[a,b]ofAisasublatticeofA.6. LetAbealattice;ifa∈A,letIa={x∈A:x a}.ProvethatIaisasublatticeofA.7. ByadistributivelatticewemeanaclasswithtwooperationssatisfyingL1throughL4andL7.IfA

isadistributelattice,provethat

8. InanarbitrarylatticeA,provetheso-called“distributiveinequalities”

9. LetAbealatticeandletx,y,z∈A.Provethatifx z,then

10. DrawthelinediagramofalatticeAandasubclassB⊆AsuchthatBisnotasublatticeofA.11. Drawthelinediagramoftwolatticeswhoseintersectionisnotalattice.12. DrawthelinediagramofalatticewhichisnotaBooleanalgebra.13. LetAbeapartiallyorderedset;provethattheclassofallthecutsofA(orderedbyinclusiononthe

“leftcomponents”L)isacompletelattice.14. LetAbeapartiallyorderedset;provethat theclassofall theconvexsubsetsofA isacomplete

lattice.15. Draw the linediagramof a partially ordered classwhich is conditionally complete, but is not a

completelattice.

5FULLYORDEREDCLASSES.WELL-ORDEREDCLASSES

LetA be a partially ordered class. As previously stated,A is said to be fully ordered if every twoelementsofAarecomparable.Examplesoffullyorderedclassesare:theclass oftheintegers,theclass oftherationalnumbers,

andtheclass oftherealnumbers.Onecaneasilyseethateverysubclassofafullyorderedclassisfullyordered.Itisevident,too,thateveryfullyorderedclassisalattice.

4.48TheoremLetf:A→Bbeafunction,whereAisafullyorderedclassandBisapartiallyorderedclass.Iff:A→Bisbijectiveandincreasing,itisanisomorphism.

Proof.Supposef(x) f(y);sincexandyarecomparable,eitherx yory<x.Ify<x,thenf(y) f(x);butf(y)=f(x)wouldimplyy=x,sowemusthavef(y)<f(x).Thisiscontrarytoourassumption,hencex y.

4.49Definition LetA be a partially ordered class.A is said to bewell ordered if every nonemptysubclassofAhasaleastelement.

IfAiswellordered,thenAisfullyordered;forifx∈Aandy∈A,thenthedoubleton{x,y}hasaleastelement,whichiseitherxory;hencex yory x.IfAiswellordered,thenAisconditionallycomplete;forifBisasubclassofAandv(B)≠Ø,then

υ(B)hasaleastelementwhichisbydefinitionthesupofB.

4.50DefinitionLetAbeapartiallyorderedclass,andsupposea∈A.Anelementb∈Aiscalledtheimmediatesuccessorofaifa<bandthereisnoelementcinAsuchthata<c<b.

4.51Remark.IfA isawell-orderedclass, theneveryelementofA (withtheexceptionof thegreatestelementofA,ifitexists)hasanimmediatesuccessor.Indeed,ifx∈AandxisnotthegreatestelementofA,thentheclassT={y∈A:y>x}isnonempty,henceThasaleastelementwhichisobviouslytheimmediatesuccessorofx.LetAbeanonemptywell-orderedclass.By4.49,Ahasaleastelement,whichmaybedenotedbyx1;

ifx1 is not the greatest element (that is, the only element) ofA, then by 4.51, x1 has an immediatesuccessor,whichmaybedenotedbyx2;again,ifx2isnotthegreatestelementofA,thenby4.51,x2hasanimmediatesuccessor,whichmaybedenotedbyx3;andsoon.

Examples

4.52Theclassofnumbers{1,2,3,4,5},orderedintheusualway,isawell-orderedclass.

4.53TheclassN={1,2,3,4,…}ofallthepositiveintegers,orderedintheusualway,isawell-orderedclass.

4.54 , ordered in the usualway,isawell-orderedsubclassoftherealnumbers.

Wemaydefineytobeanimmediatepredecessorofxifandonlyifxisanimmediatesuccessorofy;notethat inExamples4.52and4.53,only the leastelementof theclass (namely1)doesnothaveanimmediatepredecessor;however,inExample4.54,therearethreeelements(namely0,1and2)whichdonothaveanimmediatepredecessor.Notethattheclass ofalltheintegers,theclass oftherationalnumbers,andtheclass of the

realnumbersarenotwellordered.Indeed,

isasubclassofeachoftheseclasses,andVdoesnothavealeastelement.

4.55DefinitionLetAbeapartiallyorderedclass;wedefineasectionofAtobeasubclassB⊆Awiththefollowingproperty:

4.56TheoremLetAbeawell-orderedclass;BisasectionofAifandonlyifB=AorB isaninitialsegmentofA.

Proof

i) IfB=AorBisaninitialsegmentofA,thenobviouslyBisasectionofA.ii) Conversely,supposeBisasectionofA;ifB=A,wearedone;thus,supposeB≠A,thatis,A−B≠

Ø.BecauseAisawell-orderedclass,A−Bhasaleastelementwhichwedenotebym;wewillshow

thatB=Sm.Well,

(becausemistheleastelementofA−B);conversely,supposex∈B : ifm x, thenm∈Bby4.55,andthiscontradictsourchoiceofm;thusx<m,sox∈Sm.

Oneof themost important featuresofwell-ordering is thefact that inductioncanbeused toprovetheoremsaboutalltheelementsofawell-orderedclass.Thisfactisgiveninthefollowingtheorem.

4.57Theorem (Principle of Transfinite Induction). LetA be awell-ordered class, and letP(x) be astatementwhichiseithertrueorfalseforeachelementx∈A;supposethefollowingconditionholds:

Ind.IfP(y)istrueforeveryy<x,thenP(x)istrue.

Inthatcase,P(x)istrueforeveryelementx∈A.

Proof.SupposethatP(x)isnottrueforeveryx∈A;thentheclass{y∈A:P(y)isfalse}isnonempty;hence,by4.49,hasaleastelementm.NowP(x)istrueforeveryx<m,sobyInd,P(m)istrue;butwechosemtobetheleastelementof{y∈A:P(y)isfalse},soP(m)isfalse.ThiscontradictionprovesthatP(x)mustbetrueforeveryx∈A.

EXERCISES4.5

1. LetAbeafullyorderedset.ProvethatthesetofallsectionsofA(orderedbyinclusion)isfullyordered.

2. LetAbeafullyorderedclass,letBbeapartiallyorderedclass,andletf:A→Bbeanincreasingfunction.Provethatfisinjectiveifandonlyiffisstrictlyincreasing.

3. LetA be fullyorderedclass, letB be apartiallyorderedclass, and let f :A→B be a bijectivefunction.Provethatiffisincreasing,thenfisanisomorphism.

4. LetAbeafullyorderedclassandlet{L,U}beapartitionofA.Provethat(L,U)isacutofAifandonlyif∀x∈Land∀y∈U,x y.

5. LetAbeafullyorderedclass.ProvethatifBandCareconvexsubclassesofAandB∩C≠Ø,thenB Cisconvex.

6. LetAbea fullyorderedclass.LetBandCbeconvexsubclassesofAandsupposeB∩C≠Ø.ProvethateveryupperboundofB∩CisanupperboundofBoranupperboundofC.Concludethat

7. LetAbeawell-orderedclass.Ifx∈A,provethattheimmediatesuccessorofxandtheimmediatepredecessorofx(ifitexists)areunique.

8. LetAbeapartiallyorderedclass;provethatBisasectionofAifandonlyif(B,A−B)isacutof

A.9. LetAbeawell-orderedclass;provethefollowing:

a)TheintersectionofanyfamilyofsectionsofAisasectionofA.b)TheunionofanyfamilyofsectionsofAisasectionofA.

10. LetAbeawell-orderedclassandletBandCbeinitialsegmentsofA.ProvethatifB⊂C,thenBisaninitialsegmentofC.

11. LetAbeafullyorderedclass.LetB⊆Aandm∈B;provethatBhasaleastelementifandonlyifSm∩Bhasaleastelement.(AssumeSm∩B≠Ø).

12. LetAbeafullyorderedclass.ProvethatAiswellorderedifandonlyifeveryinitialsegmentofAiswellordered.[Hint:UsetheresultofExercise11,above.]

13. LetAbeawell-orderedclass.Ifa∈A,leta′designate the immediatesuccessorofa,and leta″designatetheimmediatepredecessorofa(ifitexists).Provethefollowing:

14. LetAbeawell-orderedclass;ifa∈A,leta′designatetheimmediatesuccessorofa.AnelementqinAwillbecalleda limitelementofA ifq isnot the leastelementofAandqdoesnothaveanimmediatepredecessor.Provethefollowing:a) qisalimitelementofAiff[a<q⇒a′<q].b) qisalimitelementofAiffq=sup{x∈A:x<q}.

6ISOMORPHISMBETWEENWELL-ORDEREDCLASSES

Thepurposeofthissectionistoprovearemarkablepropertyofwell-orderedclasses:ifAandBareanytwowell-orderedclasses,eitherA is isomorphicwithB,orelseoneof the two is isomorphicwithaninitial segment of the other.What thismeans, roughly speaking, is thatwell-ordered classes do notdifferfromoneanotherexceptintheirsize.Thisfacthasmanyimportantapplicationsinmathematics,andwillbeessentialtoourlaterdiscussionofinfinitesetsandcardinalandordinalnumbers.Webeginbyprovingthreepreparatorylemmas.

4.58LemmaLetAbeawell-orderedclass,andletfbeanisomorphismfromAtoasubclassofA.Thenx f(x),∀x∈A.

Proof.Assume,onthecontrary,thattheclassP={x∈A:x>f(x)}isnonempty,andletabetheleastelementofP;hence,inparticular,f(a)<a.Wenowhave

sof(a)∈P ,which is impossiblebecausea is the leastelementofP.ThusP=Ø,and the lemmaisproved.

4.59LemmaLetAbeawell-orderedclass.ThereisnoisomorphismfromAtoasubclassofaninitialsegmentofA.

Proof.Assume,onthecontrary,thatfisanisomorphismfromAtoasubclassofaninitialsegmentSaofA.ByLemma4.58,a f(a), so f(a)∉Sa; this is impossible, for the range of f is assumed to be asubclassofSa.Henceafunctionfofthekindweassumedcannotexist.

4.60CorollaryNowell-orderedclassisisomorphicwithaninitialsegmentofitself.

4.61LemmaLetAandBbewell-orderedclasses.IfAisisomorphicwithaninitialsegmentofB,thenBisnotisomorphicwithanysubclassofA.

Proof.Let f :A→Sb be an isomorphism fromA to an initial segment ofB.Assume there exists anisomorphismg:B→CwhereC⊆A.Obviouslyg:B→Aisafunction;g:B→Aandf:A→Sbarebothinjectiveandincreasing,hencetheircompositefog:B→Sbisinjectiveandincreasing;thatis,by4.48,fogisanisomorphismfromBtoitsrangewhichisasubclassofSb.However,byLemma4.59,thisisimpossible;hencetheisomorphismgthatwasassumedcannotexist.

Thenexttheoremiswidelyusedinmathematics.Itwillplayasignificantroleindiscussionslaterinthisbook,andhasimportantapplications.

4.62TheoremLetAandBbewell-orderedclasses;exactlyoneofthefollowingthreecasesmusthold:i) AisisomorphicwithB.ii) AisisomorphicwithaninitialsegmentofB.iii) BisisomorphicwithaninitialsegmentofA.

Proof.Webeginbyprovingthatthefollowingholdsinanywell-orderedclassX.

I.LetSxandSybeinitialsegmentsofX;ifx<y,thenSxisaninitialsegmentofSy.

Indeed,ifx<y,thenclearlySx⊂Sy;furthermore,SxisasectionofSy,for

thusby4.56(notethatSx≠Sybecausex≠y)weconcludethatSxisaninitialsegmentofSy.NowletAandBbewell-orderedclasses,andletCbethefollowingsubclassofA:

Ifx∈C,thereisnomorethanoner∈BsuchthatSx Sr;forsupposeSx SrandSx St,wherer≠t,sayr<t.ByI,SrisaninitialsegmentofSt;butSr Sx St,andthisisimpossibleby4.60;thusforeach x∈C, the element r ∈ B such that Sx Sr is unique. Let us designate the unique r ∈ BcorrespondingtoxbyF(x);thusF:C→Bisafunction.Inparticular,ifD=ranF,thenF:C→Disafunction;wewillshownextthatF:C→Disanisomorphism.

Fisinjective.SupposeF(u)=F(v)=r,thatis,Su Sr Sv.Ifu≠v,sayu<v,thenbyI,SuisaninitialsegmentofSv,andthisisimpossibleby4.60.Weconcludethatu=v.

Fisincreasing.Supposeu v,whereF(u)=randF(v)=t;henceSu SrandSv St.Assumethatt<r,hencebyI,StisaninitialsegmentofSr;nowSu⊆Sv,soa) SvisisomorphicwithaninitialsegmentofSr,and

b) SrisisomorphicwithasubclassofSv.

Thisisimpossibleby4.61,henceweconcludethatr t,thatis,F(u) F(v).Itfollows,by4.48,thatF:C→Disanisomorphism.Next,wewillshowthatCisasectionofA;thatis,givenc∈Candx<c,wewillprovethatx∈C.If

F(c)=r,thenSc Sr,thatis,thereexistsanisomorphismg:Sc→Sr.Itisasimpleexercisetoprovethat

isanisomorphism;thedetailsareleftasanexerciseforthereader.ThusSx Sg(x),sox∈C.AnanalogousargumentshowsthatDisasectionofB.Thusby4.56,ourtheoremwillbeprovenif

wecanshowthatthefollowingisfalse:

CisaninitialsegmentofA,andDisaninitialsegmentofB.

Indeed, suppose theabove tobe true: sayC=SxandD=Sr;wehaveproven thatF :C→D is anisomorphism,thatis,C D,soSx Sr.Butthenx∈C,thatis,x∈Sx,whichisabsurd;thisprovesthatoneof theconditions(i), (ii)or (iii)necessarilyholds.Thefact thatno twoof theseconditionsholdssimultaneouslyfollowsfrom4.60and4.61.

4.63CorollaryLetA be awell-orderedclass; every subclassofA is isomorphicwithA or an initialsegmentofA.

Proof.IfBisasubclassofA,thenBiswellordered;henceby4.62,B A,orBisisomorphicwithaninitialsegmentofA,orAisisomorphicwithaninitialsegmentofB.Inordertoproveourresultwemustshowthatthelastcasecannothold;indeed,supposeitdoes:thenby4.61,BisnotisomorphicwithanysubclassofA.ButB BandBisasubclassofA,sowehaveacontradiction;thusthelastcasecannothold.

EXERCISES4.6

1. IntheproofofTheorem4.62,provethatg[Sx]:Sx→Sg(x)isanisomorphism.

2. IntheproofofTheorem4.62,provethatDisasectionofB.3. LetAbeawell-orderedclass.ProvethattheidentitymappingIAistheonlyisomorphismfromAto

A.4. LetAandBbewell-orderedclasses.Provethatiff:A→Bandg:B→Aareisomorphisms,theng

=f−1.

5. LetAandBbewell-orderedclasses.Provethatthereexistsatmostoneisomorphismf:A→B.6. LetAandBbewell-orderedclasses.Prove that ifA is isomorphicwithasubclassofB,andB is

isomorphicwithasubclassofA,thenAisisomorphicwithB.7. LetAandBbewell-orderedclasses.ProvethatifAisisomorphicwithaclasscontainingB,andBis

isomorphicwithaclasscontainingA,thenAisisomorphicwithB.8. LetAandB bewell-orderedclasses.Suppose thatA has no greatest element; suppose that every

elementofB (except the leastelement)hasan immediatepredecessor.Prove thatB is isomorphicwithasectionofA.

5TheAxiomofChoiceandRelatedPrinciples

1INTRODUCTION

Fromhereon,throughouttheremainderofthisbook,wewillbeconcernedmainlywithsets.

Inthischapterwewilldiscussaconceptwhichisoneofthemostimportant,andatthesametimeoneof the most controversial, principles of mathematics. In 1904, Zermelo brought attention to anassumptionwhichisusedimplicitlyinavarietyofmathematicalarguments.Thisassumptiondoesnotfollowfromanypreviouslyknownpostulatesofmathematicsorlogic,henceitmustbetakenasanewaxiom;Zermelo called it theAxiomofChoice. TheAxiom ofChoice has significant implications inmanybranchesofmathematics,andconsequencessopowerfulas,sometimes, todefycredibility.Thecontroversyoverthisprinciplecontinuesinourday;wewillpresentsomeofitsaspectsinthischapter.InordertoillustratewheretheAxiomofChoiceintrudesincommonmathematicalarguments,letus

examinethefollowingstatement.

5.1LetAbeanonempty,partiallyorderedset,andsuppose that therearenomaximalelements inA;thenthereexistsanonterminating,increasingsequencex1<x2<x3< ofelementsofA.

Proof.A isnonemptybyhypothesis,hencewemayselectanarbitraryelementofAandcall itx1.Byinduction,supposethatwearegivenx1<x2< <xn;wedefineAntobethesetofalltheelementsx∈Asuchthatx>xn.Anisnonempty,forifitwereempty,thenxnwouldbemaximal,contradictingoneofourassumptions.WeselectanarbitraryelementofAnandcallitxn+1;thuswehavex1< <xn<xn+1.ThisinductiveprocessdefinesanincreasingsequenceSn={x1,x2,…,xn}foreachnaturalnumbern;

thatis,itgiveus

and so on. Now if we let (where N is the set of all the positive integers), then S is thenonterminatingsequencex1<x2<x3< thatweareseeking.

Acarefulexaminationoftheaboveargumentwillrevealthatwehaveusedanassumptionwhichisbynomeansself-evidentorundisputablyplausible.Whatwehave,infact,doneistoassumethatwecanmakeaninfinitesuccessionofarbitrarychoices.Itiscommonenough,inmathematics,tomakeonearbitrarychoice(wedothiseverytimewecansay“letxbeanarbitraryelementofA”),andexperienceconfirmsthatwecanmakeafinitesuccessionofchoices;buttomakeaninfinitesuccessionofchoicesistocarryanargumentthroughaninfinitenumberofsteps—andnothinginourexperienceorinthelogicwehabituallyusejustifiessuchaprocess.In the proof of 5.1, it was necessary to choose the elements x1,x2,x3, etc., in succession, for each

choicedependedontheprecedingones.Thefactthatthechoicesaresuccessivemayappeartobethemostdisturbing element in thewholeproof, for this involves a time factor (an infinite succession of

acts,eachonrequiringacertainamountoftime,wouldtakeinfinitelylong).However,theargumentcanbealteredinsuchawaythatallthechoicesaremadesimultaneouslyandindependentlyofoneanother;weproceedasfollows.LetusadmitthatfromeachnonemptysubsetB⊆AitispossibletochooseanarbitraryelementrB,

tobecalledthe“representative”ofB.Notethat in thiscaseeachchoiceis independentof theothers;hence,inamannerofspeaking,allthechoicescanbemadesimultaneously.Returningtotheproofof5.1,ifxnandAnaregiven,wemaydefinexn+1tobetherepresentativeofAn;inotherwords,insteadofchoosingrepresentativesforA1,A2,A3,etc., insuccession,wehavechosenrepresentativesforall thenonemptysubsetsofAinadvance.(Ofcourse,thisrequiresthatwemakemanymorechoicesthanareneededforouroriginalargument,butthisisthepricewemustpaytosubstitutesimultaneouschoicesforsuccessiveones.)Theprecedingparagraphmakesitclearthatthesuccessivenatureofthechoicesisnotthecruxofthe

problem;theproblemis:Canwemakeinfinitelymanychoices—betheysuccessiveorsimultaneous?Itisworthnotingthatincertainparticularcasestheanswertothisquestionisanobvious“yes.”For

example,ifAisawell-orderedset,wemaydefinethe“representative”ofeachnonemptysubsetB⊆AtobetheleastelementofB;becauseAiswellordered,wehavealawatourdisposalwhichprovidesuswith a representative for each nonempty subset of A. The situation in 5.1, however, is completelydifferent,forwedonothaveanyready-maderulewhichisabletofurnishuswithrepresentatives.Itisonlythelattercase—asin5.1—whichisofinteresttoushere.Intheproofof5.1wespeakof“selecting”anelementofAn;clearly,wedonotwishtointroducethe

notionof “selecting” as anewundefined conceptof set theory, soweavoid theuseof thiswordbylettinganappropriatefunction“select”representatives.

5.2DefinitionLetAbeaset; letusagree towrite .Byachoice function forAwemeanafunctionr: suchthat

WewillsometimeswriterBforr(B)andcallrBtherepresentativeofB.

5.3ExampleLetA={a, b, c}; an exampleof a choice function forA is the functionr given in thefollowingtable.

InthelightofDefinition5.2,thequestionwehavebeenaskingcanbeexpressedasfollows:ifAisaset,doesthereexistachoicefunctionforA?Acrucialcommentneedstobemadeatthispoint:the

proof of 5.1 does not require that we construct a choice function, it requires merely that a choicefunctionexist!Indeed,ifrisachoicefunctionforA,then—inthecontroversialstepoftheproof—weletxn+1=r(An);ifweareassuredthatrexists,thereisnofurtherdifficulty.TheAxiomofChoiceassertsthateverysethasachoicefunction;itsintentistostatetheexistenceof

achoice function forevery set, evenwhere, admittedly,nonecanbeactuallyconstructed. Itmustbeemphasizedherethattostatetheexistenceofachoicefunctionisquiteadifferentthingfromproducingone,orevenclaiming thatonecanbeproduced.Forwearemerelyasserting thatamongallpossiblefunctionsfrom toA,thereisoneatleastwhichmapseveryBontoanelementx∈B.Theessenceof theAxiomofChoice is that it isanexistential statement rather thanaconstructive

one.Onceagain,itstatesthatamongallpossiblefunctionsfrom toA,thereisatleastonewhichsatisfiestheconditionofDefinition5.2;thisdoesnotseemgrosslyunreasonable.TheAxiomofChoicemakesnoclaimthatachoicefunctioncanbeconstructed;hence,itdoesnotassertthatthesequenceofchoicesdescribedintheproofof5.1canbeeffectivelycarriedout—andindeedthisisnotnecessaryinorderfortheprooftowork.Before the reader decideswhether or not theAxiomofChoice seems plausible to him, he should

examine some of the equivalent propositions which are developed in the next few sections of thischapter. Some of them are very powerful indeed, and a rejection of any one of themwould entail arejectionofallofthem(including,ofcourse,theAxiomofChoice).Itisimportanttonotethatwhatalloftheseprincipleshaveincommonisthefactthattheyarenonconstructive:theyasserttheexistenceofmathematicalobjectswhichcannotbeexplicitlyproduced. It isprecisely this aspectof theAxiomofChoiceandrelatedprincipleswhichmakesthemunacceptabletotheintuitionists(seepage17,Section5,Chapter0),whoclaimthatmathematicalexistenceandconstructibilityareone.Without reentering into the arguments for and against admitting nonconstructive propositions into

mathematics,wecansaythis:intuitively,theAxiomofChoicecannotberejectedoutright,norcanwefeel trulycertainof itsvalidity.Amore importantconsideration,however, is the fact that ithasbeenproven that the Axiom of Choice does not contradict the other axioms of set theory, nor is it aconsequence of them.Thus it has the same status as another famous axiom inmathematics, namelyEuclid’s “FifthPostulate.”Wecanhave a “standard” set theory inwhichwepostulate theAxiomofChoice,and“nonstandard”settheoriesinwhichwepostulatealternativestotheAxiomofChoice.Inconclusion,sincetheAxiomofChoiceisneitheraconsequenceoftheotheraxiomsofsettheory

nor in conflictwith them, it is impossible tomake a decisionpro orcon on purely logical grounds.SincetheAxiomofChoiceinvolvesanareaofmathematics(namely,infinitesets)whichisoutsidetherealmofourexperience,itwillneverbepossibletoconfirmitordenyitby“observation.”Inthefinalanalysis,thedecisionmustbeapurelypersonaloneforeachindividualmathematiciantomake;itisamatterofpersonaltaste.

2THEAXIOMOFCHOICE

Intheprecedingsection,wehavegiventhebackgroundforournextaxiom:

A10(AxiomofChoice).Everysethasachoicefunction.

IntheliteraturethereareseveralotherwaysofstatingtheAxiomofChoicewhichareequivalenttoour Axiom A10. We shall present two of these statements here, and several more in the exercisesfollowingthissection.

Ch1Let beasetwhoseelementsaremutuallydisjoint,nonemptysets.ThereexistsasetCwhichconsistsofexactlyoneelementfromeachA∈ .

If isafamilyofdisjoint,nonemptysets,thenthesetCdescribedinCh1iscalledachoicesetfor.Thus,Ch1assertsthateverysetofdisjoint,nonemptysetshasachoiceset.

Ch2Let{Ai}i∈Ibeasetofsets.IfIisnonemptyandeachAiisnonempty,then isnonempty.Let us show, first, that A10⇒ Ch 1. Suppose is a set whose elements are mutually disjoint,

nonemptysets,andlet

Clearly, ; by A10, there is a function r : such that r(B) ∈ B for each,itfollowsimmediatelythatCisthesetrequiredinCh.1.

Next,Ch1⇒A10.IfAisasetandB⊆A,letQB={(B,x):x∈B}.IfBandDaredistinct,thenQBandQDaredisjoint,forQBconsistsofpairs(B,x),whereasQDconsistsofpairs(D,x).Thusthefamily

isasetofdisjoint,nonemptysets*;itfollowsbyCh1thatthereexistsasetCwhichconsistsofexactlyoneelement(B,x)fromeachQB;itiseasilyverifiedthatCischoicefunctionforA.ThefactthatA10⇒Ch2followseasilyfromthedefinitionofaproductofsets.Indeed,let{Ai}i∈Ibe

asetofnonemptysets,andlet

byA10,thereexistsafunctionr: suchthatr(B)∈BforeachB∈ ;hence,inparticular,r(Ai)∈Aiforeachi∈I.Ifwedefineabya(i)=r(Ai),thenaisafunctionfromItoAsuchthata(i)∈Aiforeachi∈I;thatis,a∈ .Thus isnonempty.ThefactthatCh2⇒A10canbeprovenbyanargumentsimilartotheabove;thedetailsareleftto

thereader.InChapter2wepromisedtogiveacharacterizationofsurjectivefunctionswhoseproofdependson

theAxiomofChoice.

5.4TheoremLetAbeasetandletf:A→Bbeafunction;f:A→Bissurjectiveifandonlyifthereexistsafunctiong:B→A∋f g=IB.

Proofi) Supposethereexistsafunctiong:B→Asuchthatf g=IB;theproofthatf:A→Bissurjective

isgiveninpart(ii)oftheproofof2.24.ii) Conversely,supposethatf:A→Bissurjective;foreachy∈B, isanonemptysubsetofA.Ifr

isachoicefunctionforA,wedefineg:B→Abyg(y)=r[ ],∀y∈B.Insimpleterms,foreachy∈Bweletg(y)beanarbitraryelementof .Itisobviousthatifx=g(y),thenx∈ ,hencef(x)=y;thus

ForthesakeofsimplicitywehaveprovenTheorem5.4inthecasewhereAisaset;usingaslightlystrongerformoftheAxiomofChoicewecanprove5.4inthemoregeneralcasewhereAisanyclass;weomitthedetails.

EXERCISES5.2

1. LetAbeasetandletf:A→Bbeasurjectivefunction.ProvethatthereexistsasubsetC⊆AsuchthatCisinone-to-onecorrespondencewithB.[Hint:Use5.4.]

2. LetAbeaset,letf:B→Candg:A→Cbefunctions,andsupposethatranf⊆rang.Provethatthereexistsafunctionh:B→Asuchthatg h=f.[Hint:UsetheAxiomofChoice.]

3. Let{Ai}i∈Ibeanindexedfamilyofclasses,whereIisaset.ProvethatthereexistsJ⊆Isuchthat

and,in{Aj}j∈J,eachAjisindexedonlyonce(thatis,Ai=Aj⇒i=j).[Hint:UseRemark2.38andtheAxiomofChoice.]

4. ProvethatthestatementofTheorem5.4impliestheAxiomofChoice.

Ineachofthefollowingproblemsapropositionisstated.ProvethatthispropositionisequivalenttotheAxiomofChoice.5. Let beasetofdisjoint,nonemptysets.Thereexistsafunctionf,whosedomainin ,suchthatfor

allA∈ ,f(A)∈A.6. LetEbeasetandsupposeG⊆E×E.LetA=domGandB=ranG;thenthereexistsafunctionf:

A→Bsuchthatf⊆G.7. Let be a setwhose elements are nonempty sets, and let . Then, corresponding to every

functiong: ,thereexistsafunctiong*: suchthatg*(B)∈g(B).8. LetBbeasetandletf:A→Bbeafunction;thenthereexistsubsetsC⊆Aandg⊆fsuchthatg:

C→Bisaninjectivefunctionandrang=ranf.

3ANAPPLICATIONOFTHEAXIOMOFCHOICE

The purpose of this section is to develop a consequence of theAxiomofChoice.The resultwe areabout to prove is valuable as a stepping stonewhichwill enable us to prove the importantmaximalprinciplesthatfollowinthenextsection.LetAbeapartiallyorderedsetsuchthateverychainofAhasasupinA;assumethatAhasaleast

elementp.Weintendtoshowthatthereexistsanelementa∈Asuchthatahasnoimmediatesuccessor.Inorder to show this,wewill suppose that everyelementx∈A has an immediate successor; this

assumptionwillleadtoacontradiction.IfeveryelementofAhasanimmediatesuccessor,thenwecandefineafunctionf:A→Asuchthat

for eachx∈A, f (x) is an immediate successor of x. Indeed, letTx be the set of all the immediatesuccessors ofx; by theAxiomofChoice, there exists a choice functiong such thatg(Tx)∈Tx.We

definefbylettingf(x)=g(Tx);clearly,f(x)isanimmediatesuccessorofx.

5.5DefinitionAsubsetB⊆Aiscalledap-sequenceifthefollowingconditionsaresatisfied.α) p∈B,β) ifx∈B,thenf(x)∈B,γ) ifCisachainofB,thensupC∈B.

Therearep-sequences;forexample,Aisap-sequence.

5.6LemmaAnyintersectionofp-sequencesisap-sequence.Theproofisleftasanexerciseforthereader.

LetPbetheintersectionofallthep-sequences.(NotethatP≠Øbecausep∈P).By5.6,Pisap-sequence.

5.7DefinitionAnelementx∈Piscalledselectifitiscomparablewitheveryelementy∈P.

5.8LemmaSupposexisselect,y∈P,andy<x.Thenf(y) x.

Proof.y∈P,Pisap-sequence,henceby(β),f(y)∈P.Now,xisselect,soeitherf(y) xorx<f(y).Byhypothesisy< x;so ifx< f(y),wehavey< x< f(y), which contradicts the assertion that f(y) is theimmediatesuccessorofy.Hencef(y) x.

5.9LemmaSupposexisselect.Let

ThenBxisap-sequence.

Proof.WewillshowthatBxsatisfiesthethreeconditionswhichdefineap-sequence.

α) SincepistheleastelementofA,p x,hencep∈Bx.β) Supposey∈Bx;theny xory≥f(x).Considerthreecases:

1) y<x.Thenf(y) xby5.8,hencef(y)∈Bx.2) y=x.Thenf(y)=f(x),thusf(y)≥f(x);hencef(y)∈Bx.3) y f(x).Butf(y)>y,sof(y)>f(x);hencef(y)∈Bx.Ineachcaseweconcludethatf(y)∈Bx.

γ) IfCisachainofBx,letm=supC.Foreachy∈Bx,y xory≥f(x).If∃y∈Cy≥f(x),then(sincem≥y)m≥f(x),som∈Bx.Otherwise,∀y∈C,y x; thusx isanupperboundofC,som x.Thusagainm∈Bx.

5.10CorollaryIfxisselect,then∀y∈P,y xory≥f(x).

Proof.Bx is a p-sequence;P is the intersection of all p-sequences; henceP⊆Bx. ButBx⊆ P bydefinition,henceP=Bx.So∀y∈P,y xory≥f(x).

5.11LemmaThesetofallselectelementsisap-sequence.

Proofα) pisselectbecauseitislessthan(hencecomparableto)eachy∈P.β) Supposexisselect,by5.10.∀y∈P,eithery x(inwhichcasey f(x)becausex<f(x))ory≥f(x).

Thusf(x)isselect.γ) LetCbeachainofselectelementsandletm=supC;lety∈P.If∃x∈Cy x,theny m(because

x m).Otherwise,∀x∈C,x y,henceyisanupperboundofC,som y.Thusmisselect.

5.12CorollaryPisfullyordered.

Proof.ThesetSofalltheselectelementsisap-sequence;Pistheintersectionofallthep-sequences;henceP⊆S.ButS⊆P(bydefinitionaselectelementisinP),soP=S.ThuseachelementofP isselect,thatis,iscomparabletoeachelementofP.

Corollary5.12producesacontradiction.Indeed,letm=supP;bycondition(γ),m∈PbecausePisachainofP.Butbycondition(β),f(m)∈P,hencef(m) m;thiscontradictstheassertionthatf(m)isanimmediatesuccessorofm.Weconclude:

5.13TheoremLetAbeapartiallyorderedsetsuchthat(1)Ahasaleastelementpand(2)everychainofAhasasupinA.Thenthereisanelementx∈Awhichhasnoimmediatesuccessor.

4MAXIMALPRINCIPLES

The propositions we are about to develop are widely used in mathematics to prove theorems by“nonconstructive” methods. They assert the existence of mathematical objects which cannot beconstructed.Forexample,inlinearalgebraamaximalprinciplecanbeusedtoprovethateveryvectorspacehasabasis,although,ingeneral,itisimpossibletoexhibitsuchabasis.ThemaximalprinciplesareconsequencesoftheAxiomofChoice.Furthermore,asweshallverifyin

Section6,theyareequivalenttotheAxiomofChoice.

5.14Theorem(Hausdorff’sMaximalPrinciple).Everypartiallyorderedsethasamaximalchain.

Proof.LetAbeapartiallyorderedset,andlet bethesetofallthechainsofA,orderedbyinclusion.hasaleastelement,namelytheemptyset.Nowlet beachainof andlet

wewillshowthatK∈ .Indeed,ifx,y∈K,thenx∈Dandy∈EforsomeelementsD∈ andE∈

;but isachainof ,henceE⊆DorD⊆E,sayE⊆D;thusx,y∈D.ButD isachainofA(rememberthat isthesetofallthechainsofA),soxandyarecomparable;thisprovesthatKisachainofA, that is,K∈ .By 4.26K = sup ; it follows that the conditions ofTheorem5.13 aresatisfiedby .Thus,by5.13,thereisanelementC∈ whichhasnoimmediatesuccessor; thatis,thereexistsnox∈A−CsuchthatC∪{x}isachainofA.Thus,clearly,Cisamaximalchain.

5.15DefinitionApartiallyorderedsetAissaidtobeinductiveifeverychainofAhasanupperboundinA.

5.16Theorem(Zorn’sLemma).Everyinductivesethasatleastonemaximalelement.

Proof.LetAbeaninductiveset;by5.14,AhasamaximalchainC;by5.15,Chasanupperboundm.Nowsupposethereexistsanelementx∈Ax>m;thenx∉C,butxiscomparablewith(tobeexact,xis greater than) every element ofC.Thus,C∪{x} is a chain, contradicting the assertion thatC is amaximalchain;hencethereexistsnoelementx∈Asuchthatx>m,somisamaximalelementofA.

Theorems5.14and5.16canbestatedinasomewhatstrongerformasfollows:

5.17TheoremEverypartiallyorderedsethasamaximalwell-orderedsubset.

5.18Theorem(LetuscallapartiallyorderedsetAweaklyinductiveifeverywell-orderedsubsetofAhasanupperboundinA.)Everyweaklyinductivesethasatleastonemaximalelement.

TheproofsofthelasttwotheoremsaresimilartothoseofTheorems5.14and5.16;theyareleftasanexerciseforthereader.

EXERCISES5.4

1. LetAbeapartiallyorderedsetandlet bethesetofallthewell-orderedsubsetsofA.ForC∈andD∈ ,define ifandonlyif isasectionofD.a) Provethat isapartialorderrelationin .b) Provethat ,orderedby ,isinductive.c) Usingpart(b)andZorn’sLemma,proveTheorem5.17.

2. UsetheresultofExercise1,above,toproveTheorem5.18.3. DeriveHausdorff’sMaximalPrinciplefromZorn’sLemma.4. ProvethatZorn’sLemmaisequivalenttothefollowing:LetAbeaninductivesetandleta∈A;

thenAhasatleastonemaximalelementbsuchthat .5. ProvethatHausdorff’sMaximalPrincipleisequivalenttothefollowing:IfAisapartiallyordered

setandBisachainofA,thenAhasamaximalchainCsuchthatB⊆C.6. LetAbeanysetwithmorethanoneelement.Provethatthereexistsabijectivefunctionf:A→A

suchthatf(x)=x,∀x∈A.7. Asetofsets issaidtobedisjointedif∀C,D∈ ,C∩D=Ø.Let beasetofsets;provethat

hasamaximaldisjointedsubset.8. LetAbeasetandlet beasetofsubsetsofA;let havethefollowingproperty:B∈ iffevery

finite subset ofB belongs to ; then is said to be of finite character. Let be ordered byinclusionandsuppose isoffinitecharacter.a) Provethat isaninductiveset.b) Provethat hasamaximalelement.

9. Prove that every vector space V has a basis. [Hint: Consider the set of all the linearlyindependent subsets of V. Use Zorn’s Lemma: it is easily verified that any maximal linearlyindependentsubsetofVisabasisofV.]

10. LetGbeagroupandletAbeanarbitrarysubsetofGsuchthatAincludestheidentityelementofG.ProvethatamongthesubgroupsofGwhicharesubsetsofA,thereisamaximalone.

5THEWELL-ORDERINGTHEOREM

The well-ordering theorem, which will be presented in this section, is one of the most importantconsequencesoftheAxiomofChoiceandisanoutstandingexampleofanonconstructiveproposition.Itassertsthatanysetcanbewellordered;thatis,ifAisanyset,thereexistsanorderrelationG suchthatA,orderedbyG,isawell-orderedset.Theproofofthewell-orderingtheoremgivesnoindicationhow such awell-ordering of the elements ofA is to be accomplished; it assertsmerely that awell-orderingexists.AfinitesetAcanobviouslybewellordered;forexample,ifA={a,b,c},thena<b<c,b<a<c,

c<b<aaredifferentwell-orderingsofA.However,nomethodhasyetbeendiscoveredtowell-ordersets such as the set of the real numbers; in fact, in the opinion of most mathematicians, it isimpossible to construct a well-ordering of . (This would mean giving a rule for rearranging theelementsof insuchawaythat wouldtherebybecomeawell-orderedset.)Onceagain,thewell-orderingtheoremdoesnotassertthateverysetcanbeeffectivelywellordered;it

merely states that, amongallpossiblegraphsG⊆A×A, there exists one at leastwhich is an orderrelationwhichwell-ordersA.Let us now prove the well-ordering theorem; note that the proof relies heavily on the Axiom of

Choice.LetAbeanarbitraryset.Wewillconsiderpairs(B,G),whereBisasubsetofA,andGisanorder

relationinBwhichwell-ordersB.Let bethefamilyofallsuchpairs(B,G).Weintroducethesymbol anddefine(B,G) (B′,G′)if

andonlyif

(Notethatthelastconditionasserts,roughly,thatalltheelementsofBprecedealltheelementsofB′−B.)Itiseasytoverifythat isanorderrelation ;thedetailsareleftasanexerciseforthereader.

5.20LemmaLet

beachainof ;let

Then(B,G)∈ .

Proof.By1.40(i),B⊆A;thus,ourresultwillbeestablishedifwecanshowthatGwell-ordersB.FirstweverifythatGisanorderrelationinB.

Reflexive.x∈B⇒x∈Biforsomei∈I⇒(x,x)∈Gi⊆G;thusGisreflexive.

Antisymmetric.(x,y)∈Gand(y,x)∈G⇒(x,y)∈Giand(y,x)∈Gjforsomei∈Iandj∈I;but isachainof ,soGi⊆GjorGj⊆Gi,sayGi⊆Gj.Thus(x,y)∈Gjand(y,x)∈Gj;butGjisanorderrelation,sox=y.ThisprovesthatGisantisymmetric.

Transitive.(x,y)∈Gand(y,z)∈G⇒(x,y)∈Giand(y,z)∈Gjforsomei∈Iandj∈I;but isachain,soGi⊆GjorGj⊆Gi,sayGi⊆Gj.Then(x,y)∈Gjand(y,z)∈Gj,so(x,z)∈Gj⊆G.ThusGistransitive.

NowwemustshowthatBiswell-orderedbyG.SupposethatD≠ØandD⊆B;thenD∩Bi≠Øforsomei∈I.NowD∩Bi⊆Bi,henceD∩Bihasaleastelementbin(Bi,Gi);thatis∀y∈D∩Bi,(b,y)∈Gi.WewillproceedtoshowthatbistheleastelementofDin(B,G);thatis,∀x∈D,(b,x)∈G.Indeed,letx∈D:ifx∈Bi,then(b,x)∈Gi⊆G.Nowsupposex∉Bi;inthiscase,x∈Bjforsomej

∈I;Bj Bibecausex∈Bjandx∉Bi,hence(Bj,Gj) (Bi,Gi);itfollowsthat(Bi,Gi) (Bj,Gj).Nowwehaveb∈Bi,x∈(Bj−Bi),and(Bi,Gi) (Bj,Gj);thus,by5.19(c),(b,x)∈Gj⊆G.ThisprovesthatbistheleastelementofDin(B,G).

5.21LemmaIf ,B,andGaredefinedasabove,(B,G)isanupperboundof .

Proof.Let (Bi,Gi)∈ ; clearlyBi⊆B andGi⊆G.Now suppose that x∈Bi, y∈B, and y∉Bi;certainlyy∈Bjforsome j∈ I.NowBj Bibecausey∈Bjandy∉Bi,so (Bj,Gj ) (Bi,Gi),hence(Bi,Gi) (Bj,Gj).Nowx∈Biandy∈(Bj−Bi),soby5.19(c),(x,y)∈Gj⊆G.Thus(Bi,Gi) (B,G).

5.22Theorem(Well-orderingTheorem).AnysetAcanbewellordered.

Proof.Bylemma5.20and5.21,wecanapplyZorn’sLemmato ;thus hasamaximalelement(B,G).WewillshowthatB=A;henceAcanbewell-ordered.Otherwise,∃x∈(A−B);bydefiningxtobegreaterthaneachelementofB,wegetanextensionG∗ofGthatwell-ordersB∪{x}.(Moreexplicitly,G∗=G∪{(a,x):a∈B}.)Thisisacontradiction,since(B,G)wasassumedtobemaximal.

6CONCLUSION

ItisclearthattheAxiomofChoicecanbederivedfromthewell-orderingtheorem.Indeed,letAbeanyset;bythewell-orderingtheorem,Acanbewellordered;ifBisanonemptysubsetofA,letf(B)betheleastelementofB.ThenfisachoicefunctiononA.

Wehavenowproventhefollowingimplications:

AxiomofChoice⇒Hausdorff’sMaximalPrinciple

⇒Zorn’sLemma⇒well-orderingtheorem⇒AxiomofChoice.

Thuswehaveestablishedthecompleteequivalenceoftheabovefourpropositions.

IntheremainderofthisbookwewillaccepttheAxiomofChoiceasoneoftheaxiomsofsettheory.ThuswewillfeelfreetousetheAxiomofChoiceinallofourarguments,exceptifwemakeanexplicitstatementtothecontrary.

*Notethatforeach thus

hencebyA3,andA7, isaset.

6TheNaturalNumbers

1INTRODUCTION

Probablythemostfundamental—andthemostprimitive—ofallmathematicalconceptsisthatofnaturalnumber.Thenatural,or“counting,”numbersare thefirstmathematicalabstractionwhichwelearnaschildren;everyhumansociety—eventhemostbackwardandremote—possessesasystemofsomekindforcountingobjects.Weallhaveaclearintuitiveunderstandingofwhatthenaturalnumbersare:theyare0,1,2,3,andso

forth.Buthisintuitiveperception—nomatterhowclearandimmediateitmaybe—isnotsufficientforthepurposesofmathematics;inordertodomathematicswithnumbers,wemustarticulatethisvagueperceptionandtransformitintoaprecisedefinition.Thedefinitionmust,ofcourse,faithfullyreflecttheintuitivenotionfromwhichitsprang.It isouraimin thissectiontodefine thenaturalnumbers; tobeexplicit,wewillconstructasetof

objectstobecalled“naturalnumbers,”andthese“naturalnumbers”willbeendowedwithallofthepropertieswhichareassociatedwiththenaturalnumbersinourmind.Weshouldcarefullynotetwoimportantrequirementsofourdefinition.Inthefirstplace,wewould

liketopresentthenaturalnumberswithoutintroducinganynewundefinednotions.Wewerefacedwithasimilarproblemwhenwewereabouttointroducethenotionoffunction;wesolveditbydefiningafunctiontobeacertainkindofclass(specifically,aclassoforderedpairs).Followingthisexample,itisclear thatweought todefinenaturalnumbers tobeclasses (morespecifically,sets)of somekind; infact,foreachnwewilldefine“n”tobeasetwhich(intuitively)hasnelements.Secondly,eachnaturalnumbermustbeuniquelydefined;thatis,foreachntheremustbejustone,uniqueobjectwhichcanberecognizedasthe“naturalnumbern.”Thuswemustdeviseameansoffixingexactlyoneset,amongallthesetswithnelements,andcallingthisset“n.”Thenumbers“0,”“1,”“2,”etc.,whichweareabout todefinewill serveasstandards inmuch the

samewaythatthestandardyardinWashington,D.C.,servesasanormformeasuringlength.Itmatterslittlewhether thestandardyard ismadeofplatinumorstainlesssteel,orwhether it isdecoratedwithfiguresofdancingmermaids; itsonlyuse is as a standardof reference, soanythinghaving the samelengthis,bydefinition,oneyardlong.Analogously,itdoesnotmattertoomuchwhichspecificsetswedefine“0,”“1.”“2,”etc.,tobe;theywillbeusedasstandardsofreference,soasetwillbesaidto“havenelements”ifitisinone-to-onecorrespondencewiththenaturalnumber“n.”

Wewillproceedasfollows:wedefine

Inordertodefine“1,”wemustfixasetwithexactlyoneelement;thus

Continuinginthisfashion,wedefine

Thereadershouldnotethat0=Ø,1={Ø},2={Ø,{Ø}},3={Ø,{Ø},{Ø,{Ø}}},etc.Ournaturalnumbersareconstructionsbeginningwiththeemptyset.Inviewofwhathasjustbeenindicated,thefollowingalsoaretrue:

Thisshouldexplainthefollowingdefinition:Theprecedingdefinitionscanberestated,alittlemoreprecisely,asfollows.IfAisaset,wedefine

thesuccessorofAtobethesetA+,givenby

Thus,A+isobtainedbyadjoiningtoAexactlyonenewelement,namelytheelementA.Nowwedefine

Itisclearthatthesedefinitionscoincidewiththosegivenintheprecedingparagraph.Fromthesedefinitions,itisworthnotingthat0⊂1⊂2⊂3⊂...,andlikewise0∈1∈2∈3∈....Wehavejustoutlinedamethodforproducingsetswhichwecall“naturalnumber;”beginningwith

theemptyset,wehavegivendirections forconstructing,successively,0=Ø,1=0+,2=1+, and soforth.An important questionnow is the following: Is there such a thing as the set of all thenaturalnumbers(oreventheclassofallthenaturalnumbers)?Thatis,isthereaset(oraclass)whichcontainsØ,andwhichcontainsX+wheneveritcontainsX?Certainly,ourmethoddoesnotenableustoconstructit—wearemerelygiveninstructionsforproducingnumbers0,1,2,...,nuptoanyn.Thuswecannotyetspeakoftheset(orclass)ofallthenaturalnumbers.AsetAiscalledasuccessorsetifithasthefollowingproperties:

i) Ø∈A.ii) ifX∈A,thenX+∈A.

Itisclearthatanysuccessorsetnecessarilyincludesallthenaturalnumbers.Motivatedbythisobservation,weintroducethefollowingimportantaxiom.

A11(AxiomofInfinity).Thereexistsasuccessorset.

Aswehavenoted,everysuccessorsetincludesallthenaturalnumbers;thusitwouldmakesenseto

definethe“setofthenaturalnumbers”tobethesmallestsuccessorset.Nowitiseasytoverifythatanyintersectionofsuccessorsetsisasuccessorset;inparticular,theintersectionofallthesuccessorsetsisasuccessorset(itisobviouslythesmallestsuccessorset).Thus,wearelednaturallytothefollowingdefinition.

6.1DefinitionBythesetofthenaturalnumberswemeantheintersectionofallthesuccessorsets.Theset of the natural numbers is designated by the symbolω; every element ofω is called a naturalnumber.

2ELEMENTARYPROPERTIESOFTHENATURALNUMBERS

In this sectionwewill show that thenatural numbers, aswehave just defined them, satisfy the fiveconditionscommonlyknownasthePeanoaxioms.Thissetofconditions—itiswellknown—isanotherwayofdefiningandcharacterizingthenaturalnumbers.

6.2TheoremForeachn∈ω,n+≠0.

Proof.Bydefinition,n+=n∪{n};thusn∈n+foreachnaturalnumbern;but0istheemptyset,hence0cannotben+foranyn.

6.3Theorem(MathematicalInduction).LetXbeasubsetofω;supposeXhasthefollowingproperties:i) 0∈X.ii) Ifn∈X,thenn+∈X.

ThenX=ω.

Proof.Conditions(i)and(ii)implythatXisasuccessorset.By6.1,ωisasubsetofeverysuccessorset;thusω⊆X.ButX⊆ω;soX=ω.

6.4LemmaLetmandnbenaturalnumbers;ifm∈n+,thenm∈norm=n.

Proof.Bydefinition,n+=n∪{n};thus,ifm∈n+,thenm∈norm∈{n};but{n}isasingleton,som∈{n}iffm=n.

6.5DefinitionAsetAiscalledtransitiveif,foreachx∈A,x⊆A.

Forexample,thenumber3isatransitiveset;indeed,itselementsare0,1,2,thatis,Ø,{Ø},{Ø,{Ø}}.Itisclearthateachoftheseelementsisasubsetof3.Thesameistrueofeverynaturalnumber,asweshallprovenext.

6.6LemmaEverynaturalnumberisatransitiveset.

Proof. Let X be the set of all the elements of ω which are transitive sets; we will prove, usingmathematical induction (Theorem 6.3), that X = ω; it will follow that every natural number is atransitiveset.

i) 0∈X,forif0werenotatransitiveset,thiswouldmeanthat∃y∈0suchthatyisnotasubsetof0;butthisisabsurd,since0=Ø.

ii) Nowsupposethatn∈X;wewillshowthatn+∈X;thatis,assumingthatnisatransitiveset,wewillshowthatn+isatransitiveset.Letm∈n+;by6.4,m∈norm=n.Ifm∈n,then(becausenistransitive)m⊆n;butn⊆n+,som⊆n+.Ifm=n,then(becausen⊆n+)m⊆n+; thus ineithercase,m⊆n+,son+∈X.Itfollowsby6.3thatX=ω.

6.7TheoremLetnandmbenaturalnumbers.Ifn+=m+,thenn=m.Proof.Supposen+=m+;nown∈n+,hencen∈m+;thus,by6.4,n∈morn=m.Bytheverysameargument,m∈norm=n.Ifn=m,thetheoremisproved.Nowsupposen≠m;thenn∈mandm∈n.Thus,by6.5and6.6,n⊆mandm⊆n,hencen=m.

ThePeanoaxiomsforthenaturalnumbersare:P1 0∈ω.P2 Ifn∈ω,thenn+∈ω.P3 Foreachn∈ω,n+≠0.P4 IfXisasubsetofωsuchthat

i) 0∈X,andii)ifn∈X,thenn+∈X,thenX=ω.

P5Ifn,m∈ωandn+=m+,thenn=m.

P1andP2followimmediatelyfromourdefinitionofω.P3 isgivenbyTheorem6.2,P4 isgivenbyTheorem6.3,andP5isgivenbyTheorem6.7.ThusoursetωsatisfiesthePeanoaxioms.

EXERCISES6.2

1. ProvethatAisatransitivesetifandonlyifthefollowingholds:IfB∈CandC∈A,thenB∈A.2. ProvethatifAandBaretransitivesets,thenA∪BandA∩Baretransitivesets.3. LetAandBbesets.ProvethatifA=B,thenA+=B+.4. Use6.3,toprovethatforeverynaturalnumbern,n∉n.5. Provethefollowing,wherem,n,p∈ω.

6. a)Provebyinduction:IfA∈nandn∈ω,thenA∈ω.Concludethatωisatransitiveset.

b)ProvethatifA+∈ω,thenA∈ω.

7. Provethatnonaturalnumberisasuccessorset.8. Provethatnonaturalnumberisasubsetofanyofitselements.9. Provebyinduction:Ifn∈ω,theneithern=0orn=m+forsomem∈ω.10. Letn∈ω.Provethefollowing.

a) ∪n+=n.b) ∪ω=ω(seeRemark1.47).

11.LetAbeanonemptysubsetofω.Provethatif∪A=AthenA=ω.

3FINITERECURSION

Induction is commonly used not only as amethod of proof but also as amethod of definition. Forexample,afamiliarwayofintroducingexponentsinarithmeticisbymeansofthe“inductivedefinition”I. a0=1,

II. an+1=ana,∀n∈ω.

ThepairofConditionsIandIIismeanttobeinterpretedasarulewhichspecifiesthemeaningofanforeachnaturalnumbern.Thus,byI,a0=1;byIandII,

byIIagain,

continuinginthisfashion—usingConditionIIrepeatedly—thenumbersa0,a1,a2,…,anaredefinedinsuccessionuptoanychosenn.Inductive definitions such as the onewe have just seen abound inmathematics. The situation, in

almosteverycase,isthefollowing:WehaveasetA,afunctionf:A→A,andafixedelementc∈A.Wedefineafunctionγ:ω→AbymeansofthetwoConditions.I. γ(0)=c,

II. γ(n+)=f(γ(n)),∀n∈ω

The reader should recognize that exactly this situation prevails in our preceding example. In thatexample,Aisthesetoftherealnumbers,cis1,γ(n)isdenotedbyan,andfisthefunctiondefinedbyf(x)=xa.Foranotherexample,let bethesetoftherealnumbersandletf: → bethefunctiondefinedby

f(x)=x2.Wedefineafunctionγ:ω→ byI. γ(0)=2,

II. γ(n+1)=f(γ(n))=[γ(n)]2.

Thereaderwillrecognizethisasaninductivedefinitionofthefunctionγ(n)=22n.Ineachoftheforegoingexamples,itisreasonabletobelievethatifγexists,thenConditionsIandII

determinethevaluesγ(n)foreveryn∈ω.However,ConditionsIandIIareinsufficientinthemselvestoguaranteethatafunctionsuchasγexists.Ifwearetoacceptdefinitionbyinductionasalegitimateway of constructingmathematical objects, thenwemust first establish the fact that γ—the functionwhich we purport to be defining—actually exists and is uniquely determined. The purpose of thefollowingtheoremistoperformthisimportanttask.

6.8RecursionTheorem LetAbeaset,cafixedelementofA,andfafunctionfromA toA.Thenthereexistsauniquefunctionγ:ω→AsuchthatI. γ(0)=c,and

II. γ(n+)=f(γ(n)),∀n∈ω.

Proof.First,wewillestablishtheexistenceofγ.Itshouldbecarefullynotedthatγ isasetoforderedpairswhichisafunctionandsatisfiesConditionsIandII.Morespecifically,γisasubsetofω×Awiththefollowingfourproperties:

Properties(1)and(2)expressthefactthatγisafunctionfromωtoA,whileproperties(3)and(4)areclearlyequivalenttoIandII.Wewillnowconstructagraphγwiththesefourproperties.Let

isnonempty,becauseω×A∈ .Itiseasytoseethatanyintersectionofelementsof isanelementof ;inparticular,

isanelementof .Weproceedtoshowthatγisthefunctionwerequire.Byconstruction,γsatisfies(3)and(4),soitremainsonlytoshowthat(1)and(2)hold.

1) Itwillbeshownbyinductionthatdomγ=ω,whichclearlyimplies(1).By(3),(0,c)∈γ,so0∈domγ;nowsupposen∈domγ.Then∃x∈A(n,x)∈γ;by(4),then,(n+,f(x))∈γ,son+∈domγ.Thus,byTheorem6.3,domγ=ω.

2) Let

ItwillbeshownbyinductionthatN=ω.Toprovethat0∈N,wefirstassumethecontrary;thatis,weassumethat(0,c)∈γand(0,d)∈γwherec≠d.Letγ∗=γ−{(0,d)};certainlyγ∗satisfies(3);toshowthatγ∗satisfies(4),supposethat(n,x)∈γ∗.Then(n,x)∈γ,so(n+, f (x))∈γ;butn+≠0(Theorem6.2),so(n+,f(x))≠(0,d),andconsequently(n+,f(x))∈γ∗.Weconcludethatγ∗satisfies(4),soγ∗∈ ;butγistheintersectionofalltheelementsof ,soγ⊆γ∗.Thisisimpossible,hence0∈N.Next,weassumethatn∈Nandprovethatn+∈N.Todoso,wefirstassumethatcontrary—thatis,wesupposethat(n,x)∈γ,(n+,f(x))∈γ,and(n+,u)∈γwhereu≠f(x).Letγ =γ−{(n+,u)};γsatisfies(3)because(n+,u)≠(0,c)(indeed,n+≠0byTheorem6.2).Toshowthatγ satisfies (4),

suppose(m,ν)∈γ ;then(m,ν)∈γ,so(m+,f(ν))∈γ.Nowweconsidertwocases,accordingas(a)m+≠n+or(b)m+=n+.

a) m+≠n+.Then(m+,f(ν))=(n+,u),so(m+,f(ν))∈γ .

b) m+=n+.Thenm=nby6.7,so(m,ν)=(n,ν);butn∈N,so(n,x)∈γfornomorethanonex∈A;itfollowsthatν=x,andso

Thus,ineithercase(a)or(b),(m+,f(ν))∈γ ;thusγ satisfiesCondition(4),soγ ∈ .Butγistheintersectionofalltheelementsof ,soγ⊆γ ;thisisimpossible,soweconcludethatn+∈N.ThusN=ω.Finally,wewillprovethatγisunique.Letγandγ′befunctions,fromωtoAwhichsatisfyIandII.

Wewillprovebyinductionthatγ=γ′.Let

Nowγ(0)=c=γ(0),so0∈M;next,supposethatn∈M.Then

hencen+∈M.

6.9CorollaryLetf,c,andγbeasinTheorem6.8.Iffisinjectiveandc∉ranf,thenγisinjective.

Proof.Wewishtoshowthatifγ(m)=γ(n),thenm=n;theproofisbyinductiononm.

i) m=0.Ifn=0wearedone;ifn≠0,thenn=k+forsomek∈ω,so

whichisimpossiblebecausecisnotintherangeoff.Thusn=0=m.

ii) Supposethecorollaryistrueform;letγ(m+)=γ(n).Ifn=0thenwehaveγ(0)=γ(m+),which,aswehavejustshown,isimpossible;thusn≠0,son=k+forsomek∈ω.Thusγ(m+)=γ(k+),thatis,f(γ(m))=f(γ(k));butfisinjective,soγ(m)=γ(k).Bythehypothesisofinduction,itfollowsthatm=k;hencem+=k+=n.

EXERCISES6.3

1. Leta∈ where isthesetoftherealnumbers;defineanbythefollowingtwoconditions.

Provethatforeachn∈ω,anisauniquelydefinedrealnumber.(Inotherwords,provethatγ(n)=anisauniquelydeterminedfunctionω→ ;useTheorem6.8.)

2. LetAbeasetandletf:A→Abeafunction.Definefnby

Provethatforeachn∈ω,fnisauniquelydeterminedelementofAA.3. LetAbeasetandlet f:A→Bbeaninjectivefunction,whereB⊂A.ProvethatAhasasubset

whichisinone-to-onecorrespondencewithω.[Hint:Use6.9toprovethatthereisaninjectiveγ:ω→DwhereD⊆A.]

4. IfA ispartiallyorderedset,byastrictlyincreasingsequenceinAwemeana functionγ :ω→Asuchthatγ(0)<γ(1)<γ(2)<···LetAbeapartiallyorderedsetwhichhasnomaximalelements;provethatthereisastrictlyincreasingsequenceinA.

4ARITHMETICOFNATURALNUMBERS

One of themost important applications of the recursion theorem is its use in defining addition andmultiplicationofnaturalnumbers.

Ifmisanaturalnumber,therecursiontheoremguaranteestheexistenceofauniquefunctionγm:ω→ωdefinedbythetwoConditionsI. γm(0)=m,

II. γm(n+)=[γm(n)]+,∀n∈ω.

Additionofnaturalnumbersisnowdefinedasfollows:

forallm,n∈ω.ConditionsIandIIimmediatelyabovecanberewrittenthus:

6.10

Weproceedtoderiveafewsimplepropertiesofaddition.

6.11Lemman+=1+n,where1isdefiedtobe0+.

Proof.Thiscanbeprovenbyinductiononn.Ifn=0,thenwehave

(this lastequalityfollowsfrom6.10),hencethelemmaholdsforn=0.Now,assuming the lemmais

trueforn,letusshowthatitholdsforn+:

6.12Lemma0+n=n.

Proof.LetX={n∈ω:0+n=n};itwillbeshownbyinductionthatX=ω.Indeed,0+0=0by6.10,hence0∈X.Nowsupposethatn∈X,thatis,0+n=n.Then

ItfollowsbyTheorem6.3thatX=ω.

6.13Theorem(m+n)+k=m+(n+k).

Proof.Theproofisbyinduction.Forarbitraryelementsm,n∈ω,let

itwillbeshownthatLmn=ω.First,

hence0∈Lmn.Nowsupposek∈Lmn,thatis,

Then

sok+∈Lmn.

6.14Theoremm+n=n+m.

Proof.Foranarbitrarynaturalnumberm,let

ItwillbeprovenbyinductionthatLm=ω.Nowm+0=m=0+mby6.10and6.12,hence0∈Lm.

Nextsupposen∈Lm,thatis,m+n=n+m.Then

Ifmisanaturalnumber,therecursiontheoremguaranteestheexistenceofauniquefunctionβm:ω→ωdefinedbythetwoConditionsI. βm(0)=0,

II. βm(n+)=βm(n)+m,∀n∈ω.

Multiplicationofnaturalnumbersisnowdefinedasfollows:

forallm,n∈ω.ConditionsIandIIimmediatelyabovecanberewrittenthus:

6.15

Thefollowingareafewsimplepropertiesofmultiplication.

6.16Lemma0n=0.

Proof.LetN={n∈ω:0n=0};itwillbeshownbyinductionthatN=ω.Indeed,00=0by6.15,hence0∈N.Nowsupposethatn∈N,thatis,0n=0.Thenby6.15,6.10,andthehypothesisofinduction,

hencen+∈N.ItfollowsbyinductionthatN=ω.

6.17Lemma1n=n.Theproof(byinduction)isleftasanexerciseforthereader.

6.18Theorem(DistributiveLaw).i) m(n+k)=mn+mk.ii) (n+k)m=nm+km.

Proofi) Ifmandnarenaturalnumbers,let

ItwillbeshownbyinductionthatLmn=ω.Now

hence0∈Lmn.Next,supposethatk∈Lmn,thatis,m(n+k)=mn+mk;then,by6.10,6.13and6.15,

hencek+∈Lmn.ii) Theproofisleftasanexerciseforthereader.

6.19Theorem(AssociativeLawforMultiplication).(mn)k=m(nk).Proof.LetLmn={k∈ω:(mn)k=m(nk)};itwillbeprovedbyinductionthatLmn=ω.First,by6.15,

hence0∈Lmn.Next,assumethatk∈Lmn,thatis(mn)k=m(nk).Then,by6.15and6.18,

hencek+∈Lmn.

6.20Theorem(CommutativeLawforMultiplication).mn=nm.Theproofisleftasanexerciseforthereader.

Oneofthemostimportantaspectsofthenaturalnumbersistheirordering.Beforeproceedingwithaformaldefinitionoftheorderrelationinω,thereadershouldreviewthedefinitionofω.Specifically,itshouldbenotedthatforeachn,thenaturalnumbernisthesetofallthenaturalnumbersprecedingn:

Itisclear,now,howwearetodefineorderinω:nistoprecedemifandonlyifnisanelementofm.Motivatedbythisobservation,wemakethefollowingformaldefinition.

6.21DefinitionArelation isdefinedinωasfollows:

First,itisrequiredtoprovethatthisisindeedanorderrelationinω.

6.22TheoremLetm ndenotethefactthatm∈norm=n.Thentherelation isanorderrelationinω.Proofi) Foreachm∈ω,m=m,hencem m(reflexivelaw).ii) Supposem nandn m;thismeansthateitherm=n,orm∈nandn∈m.Inthelattercase,m⊆

nandn⊆mby6.6,henceagainm=n(antisymmetriclaw).iii) Supposem nandn p;wehavefourpossiblecases.

1. m∈nandn∈p:thusm∈nandn⊆p,som∈p.2. m∈nandn=p:thusm∈p.3. m=nandn∈p:thusm∈p.4. m=nandn=p:thusm=p.

Ineachcase,m p(transitivelaw).

Wewillshownextthatωiswellordered;thiswillrequirethefollowingtwolemmas.

6.23LemmaIfmisanaturalnumber,0 m.

Proof.LetL={m∈ω:0 m};bythereflexivelaw6.22(i),0 0,so0∈L.Nowsupposem∈L,thatis,0 m.Fromm∈m+itfollowsthatm m+;thusbythetransitivelaw6.22(iii),0 m+,som+∈L.Soby6.3,L=ω.

6.24LemmaIfn<mthenn+ m.

Proof.Ifnisanaturalnumber,let

Wewilluseinduction(6.3)toprovethatLn=ω.Notethatm∉Lniffn<mandn+ m,thatis,n∈mandn+ m. In particular, 0∉ Ln iff n∈ 0 and n+ 0, which is impossible (specifically, n∈ 0 isimpossible);thus0∈Ln.Nowassumethatm∈Ln,thatis,nm⇒n+ m,andletusshowthatm+∈Ln,thatis,

Ifn<m+,thatis,n∈m+,thenby6.4,n∈morn=m.Ifn=m,thenn+=m+,andwearedone.Ifn∈m,thatis,n<m,thenbythehypothesisofinduction,n+ m<m+,sowearedoneagain.

6.25Theoremωiswellordered.

Proof.Suppose,onthecontrary,thatAisanonemptysubsetofωwithoutaleastelement.Let

By6.23,0∈L.Nowsupposethatn∈L,thatis,n mforeverym∈A.Ifn=pforsomep∈A,thenpistheleastelementofA,contrarytoourhypothesis;thusnmforeverym∈A.Itfollowsby6.24thatn+mforeverym∈A,son+∈L.Thusby6.3,L=ω.ButL∩A=ØbecauseAhasnoleastelement.ThusA=Ø.

EXERCISES6.4

1. ProveTheorem6.18(ii).2. ProveTheorem6.20.3. Proveeachofthefollowing.

a) m=n⇒m+k=n+k(seeExercise3,ExerciseSet6.2),b) m=n⇒mk=nk.

4. Proveeachofthefollowing.a) m<1⇒m=0.b) Thereisnonaturalnumberksuchthatm<k<m+.

5. Proveeachofthefollowing.a) n<k⇒m+n<m+k.b) m+n=m+k⇒n=k.

6. Proveeachofthefollowing.a) Ifm<nandk≠0,thenmk<nk.b) Ifmk=nkandk≠0,thenm=n.

7. Provethatifm n,thenthereexistsauniquep∈ωsuchthatm+p=n.8. Proveeachofthefollowing.

a) m+k<n+k⇒m<n.b) mk<nk⇒m<n.

9. Givean inductivedefinitionofexponentiationofnaturalnumbers; that is,definemn in amannersimilar to6.10and6.15, justifyingyourdefinitionin termsof therecursiontheorem.Thenproveeachofthefollowing.

a) mn+k=mnmk,b) (mn)k=mknk,c) (mn)k=mnk.

5CONCLUDINGREMARKS

ByAxiomA11,thereexistsasuccessorsetX;by6.1,ω⊆X,hencebyAxiomA3,ωisaset.Itfollows,byAxiomA3again,thateverynaturalnumberisaset.In the next chapterwewill define a class to be finite if it is in one-to-one correspondencewith a

natural number; it follows by 2.36 that every finite class is a set. In view of this remark, we willhenceforthspeakoffinitesetsratherthanfiniteclasses.

Wehavejustseenthatωisaset;thus,by1.53,ω×ωisaset.Now,ifweidentifyeachfractionn/m(wheren/m isassumedtobe in“lowest terms”)with theorderedpair (n,m), then theclassofall thepositiverationalnumbersisasubclassofω×ω,hencebyA3,itisaset.Analogously,theclassofallthenegativerationalnumbersisaset,hencebyA6,theclass ofalltherationalnumbersisaset.Itiswell known that every real number can be regarded as a sequence (called a Cauchy sequence) ofrationalnumbers;hence,roughlyspeaking,*asanelementof .Inotherwords,theclass oftherealnumbersisasubclassof ;hence,byA7andA3,itisaset.Insimilarfashion,theclass ofthecomplexnumbersisaset.Wehaveseeninprecedingsectionsthattheunionofanysetofsetsisaset;ifAisaset,then isa

set;if{Ai}i∈Iisafamilyofsets,whereIisaset,then isaset;andifAandBaresets,thenABisaset.Thusitisclearthateveryobjectwecanproducebytheclassicalconstructionprocessesisaset.Alltheinfinite(thatis,notfinite)classeswhichoccurintraditionalmathematicsaresets;thus,inthe

nextchapter,wewillconfineourattentiontofiniteandinfinitesets.

*Tobemoreprecise,everyCauchysequenceisanelementofQω.

7FiniteandInfiniteSets

1INTRODUCTION

Oneof themost fundamentaldistinctions inmathematics is thatbetween finiteand infinite sets.Thedistinctionissointuitivelycompellingthat,evenintheabsenceofaprecisedefinition,therecannotbeanydoubtastowhetheragivensetisfiniteorinfinite.Insimpleterms,afinitesetisonewhich“hasnelements,”wherenisanaturalnumber,andaninfinitesetisonewhichisnotfinite.Although the dichotomy between finite and infinite has always fascinatedmathematicians—it has

beenthesourceofthemostcelebratedriddles,paradoxes,andclassicalerrorsofmathematics—asoundtheoryofinfinitesetsdidnotappearuntilveryrecenttimes.Ithadtoawaitthearrivaloftherigorousconcepts of set mapping, and one-to-one correspondence. Once the use of these new tools becamefamiliartomathematicians,towardtheendofthenineteenthcentury,themoderntheoryofinfinitesetsdevelopedrapidly;itwaslargelytheworkofGeorgCantorandhissuccessors.Usingfamiliarconcepts,andargumentswhichareremarkablefortheirsimplicity,Cantorwasableto

draw conclusions which surprisedmathematicians and laymen alike. Cantor’s ideas are well knowntoday;theyhavebeenpopularizedininnumerableexpositorybooksandarticles,andhaveenteredtheloreofmodernmathematics.Weproceed,intheremainderofthissection,togivethebareoutlinesofCantor’s theory. In this discussion the words “finite” and “infinite” will be used informally; as weremarkedearlier,afinitesetcanbedescribedasonewhich“hasnelements”(n isanaturalnumber),andaninfinitesetis,simply,onewhichisnotfinite.TwofinitesetsAandBhave the“samenumberofelements” ifandonly if theyare inone-to-one

correspondence. Even though we cannot speak of two infinite sets as having the “same number ofelements,”wehavethefeeling,nonetheless, that ifAandBareinfinitesetsandthereisaone-to-onecorrespondencebetweenthem,then,inacertainsense,theyareofthe“samesize.”ThisintuitivenotionisformalizedbydefiningtwosetsAandBtobeequipotent,ortohavethesamepower,ifthereisaone-to-onecorrespondencefromAtoB.WesaythatAisofalesserpowerthanBifthereexistsaone-to-onecorrespondencebetweenAandapropersubsetofB,butnonebetweenAandB.Here,again,wehavetakenourcuefromthefinitecase:forifAandBarefinitesetsandAhasfewerelementsthanB, thencertainlythereexistsaone-to-onecorrespondencebetweenAandapartofB,butnonebetweenAandallofB.IfAandBhavethesamepower,wewriteA≈B;ifthepowerofAislessthanthatofB,wewriteA B; finally, if the power ofA is less than or equal to that ofB (that is,A is in one-to-onecorrespondencewithasubsetofB),thenwewriteA B.It isacuriousfact, firstprovenbyCantor, thatmanysetswhichappear tobesmaller—or larger—

thanωactuallyhavethesamepowerasω.Forexample,ifEisthesetoftheevennaturalnumbers,itiseasytoseethatthefunctionf(n)=2nisabijectivefunctionfromωtoE.Thus,althoughEisapropersubsetofω(infact,Eappearstohaveonly“halfasmany”elementsasω),actuallyEisequipotentwithω.Amoresurprisingexampleinvolvesωandtheset oftherationalnumbers;ourintuitionsuggests,in themostcompellingway, that isa“larger”set thanω; for notonly includesω, but is, in anobviousmanner,“infinitelydense”withrespect toω.Yet itcaneasilybeshownthatωand are inone-to-onecorrespondence;theproofconsistsin“enumerating”therationalnumbers—thatis,makingalistr1,r2,r3,…ofrationalnumberswhichincludesthemall;thecorrespondencei↔riisthenaone-to-

onecorrespondencebetweenωand .Weproceedasfollows.First,wegroupall thepositive rationalnumbers intoclassesA1,A2,… ,whereAi contains all the

fractionsn/msuchthatn+m=i.WithineachclassAi,weorderthenumbersn/minincreasingorderofthenumeratorn.Thusthefirstfewfractionsinthisorderingwouldbeasfollows:

Nowwedeleteallfractionswhicharenotin“lowestterms;”thisleaves

This is clearly an enumeration r1, r2, … of the positive rational numbers. If t1, t2, … is a similarenumeration of the negative rational numbers, then 0, r1, t1, r2, t2,… is an enumeration of all therationalnumbers.A set which is in one-to-one correspondence withω is said to be denumerable. Faced with the

unexpecteddiscoverythat isdenumerable,wearenaturallyledtowonderwhethereveryinfinitesetisdenumerable.ThisquestionwasansweredbyCantor—inthenegative:thesetofalltherealnumbers,forexample,isnotdenumerable.Toprovethis,weusetheso-calleddiagonalmethod.First,wenotethatthefunctiony=tan(πx−π/2)isaone-to-onecorrespondencebetweenthesetofall

therealnumbersandtheopeninterval(0,1);henceitwillbesufficientforourpurposestoprovethatthesetofalltherealnumbersrsuchthat0<r<1cannotbeenumerated.Wearguebycontradiction.Supposethatr1,r2,…isanenumerationofalltherealnumbersbetween0and1.Leteachrealnumberbeexpressedasanonterminatingdecimal;thus ,whereeachrijisadigit0,1,…,9.

Nowwedefinestobeanumber ,wheres1≠r11,s2≠r22,s3≠r33,s4≠r44,andsoforth(forexample,wemightdictatethatsi=1ifrii≠1andsi=2ifrii≠1).Now,since0<s<1,itfollowsthatsisoneofthenumbersintheenumeration,says=rk.Butthisisimpossible,becausethekthdigitofsisskandthekthdigitofrkisrkk,andsk≠rkk.Becauseofthiscontradiction,itisclearthatthereisnowayofenumeratingtherealnumbers.Yetωhasthesamepowerasasubsetoftherealnumbers—namelythoserealnumberswhichhappentobepositiveintegers.Thusω .WehavejustrevealedoneofthemostsignificantfactsinCantor’stheoryoftheinfinite:whileωandareboth infinitesets,oneof themisstrictly larger than theother; inotherwords infinitesets, like

finitesets,comeindifferent“sizes.”Ournextstep,naturally,istofindasetwhichisstrictlylargerthan.Inordertosettlethisquestion,however,weintroducearesultoffargreatergenerality,whichwill

provideusatoncewithastrictlyincreasingsequence ofinfinitesets.Ourresultissimplythis:ifAisanyset,thenthepowersetofA, ,isstrictlylargerthanA.Toprovethis,weuseavariantofthediagonalmethodwhichservedusintheprecedingparagraph.Webeginbyassumingthatthereisaone-to-onecorrespondenceϕ:A→ .WedefineasetBas

follows:

BisasubsetofA,soB=ϕ(y)forsomey∈A.Nowify∈ϕ(y),theny∉B,thatis,y∉ϕ(y);yetify∉ϕ(y),theny∈B,thatis,y∈ϕ(y).Quiteobviouslythisisimpossible;hencethereexistsnoone-to-onecorrespondencebetweenAand .However,Ahasthesamepowerasasubsetof ,namelythesetofallthesingletons{x}.WeconcludethatA .Theargumentoftheprecedingparagraphisaproofforthefollowingtheorem.

7.1TheoremIfAisaset,thereexistsnosurjectivefunctionA→ .

7.2CorollaryNosubsetofAcanbeequipotentwith .

7.3CorollaryAcannotbeequipotentwithanysetcontaining .Itfollowsfrom7.2and7.3that

7.4 ifB⊆A,thenB ;

if ⊆D,thenA D.

We have proved earlier (Theorem 2.35) that if A is a set, and 2A are in one-to-onecorrespondence.Thusalloftheabovestatementsholdtruewhenwereplace by2A.Ifweletω=K1,P(K1)=K2,P(K2)=K3,andsoforth,thenwehavethestrictlyincreasingsequence

ofinfinitesets

Nowconsider foreachi,Ki+1⊆L1,thatis, (Ki)=L1.Itfollowsby7.4thatforeachi,KiL1.NowweletL2= (L1),L3= (L2),andsoforth;hencewehavethestrictlyincreasingsequenceofinfinitesets

Thus,speakinginformally,therearemanymore“sizes”ofinfinitesetsthantherearedifferent“sizes”offinitesets.Itisworthnothingthattheset oftherealnumbersisequipotentwith2ω.Indeed,wehavealready

notedthat isequipotentwiththeopeninterval(0,1)of ,hencewiththeclosedinterval[0,1].(ThislastfactfollowstriviallyfromExercise2inExercises7.3.)Now,eachelementr in theinterval[0,1]canbewritteninbinarynotation

whereeachri,iseither0or1.Thisexpressionforrcanbeidentifiedwiththefunctionϕr:ω→{0,1}given by ϕr(i) = ri, ∀ i ∈ ω. It is easy to see that the correspondence r ↔ ϕr is a one-to-one

correspondencebetweentheinterval[0,1]and2ω.Wehaveseenthattheset oftherationalnumbersisequipotentwithω;itiseasytoshowthatthe

set of thecomplexnumbers isequipotentwith ; thus,bytheprecedingparagraph,allofclassicalmathematics deals with only two sizes of infinite sets, namely, sets equipotent with ω and setsequipotentwith2ω; the powerof 2ω is often called thepower of the continuum. Now an interestingquestionwhicharisesisthefollowing:isthereapowerbetweenthatofωandthatof2ω?Thatis,doesthereexistanysetAsuchthatω A 2ω?Sincenosuchsetoccursanywhereinclassicalmathematics,and there appears to be no way of constructing one, it was conjectured by Cantor and hiscontemporariesthattheanswertothatquestionmustbe“no;”thisconjectureisknownasthecontinuumhypothesis.Acloselyrelatedconjecture is thegeneralizedcontinuumhypothesis,whichproposes thatforeverysetB,thereisnosetAsuchthatB A 2B.Thesehypotheseshaveneverbeeneitherprovenordisproven;weshallhavemoretosayabouttheminChapter11.Theaimofthischapteristoexploitthevariousideaswhichhavebeenmotivatedinourintroduction.

Wewilldefinerigorouslytheconceptsoffiniteandinfiniteset,“power,”andcardinality,andgivetheclassicalresultsofCantor’stheory.

2EQUIPOTENCEOFSETS

Intheprecedingsection,wehavedefinedthesymbols≈, ,and asfollows:

A≈BiffAisinone-to-onecorrespondencewithB.

A BiffAisinone-to-onecorrespondencewithasubsetofB.

A B iff A is in one-to-one correspondence with a subset of B and A is not in one-to-onecorrespondencewithB.

Itisimmediatefromthesecondofthesestatementsthat

7.5 A BiffthereexistsaninjectivefunctionA→B.

Furthermore,wehavethefollowing.

7.6LemmaThereexistsaninjectivefunctionf:A→Bifandonlyifthereexistsasurjectivefunctiong:B→A.

Proof

i) Supposef:A→Bisinjective;by2.25,thereexistsafunctiong:B→Asuchthatg f=IA;thus,by5.4,gissurjective.

ii) Supposeg:B→Aissurjective;by5.4,thereexistsafunctionf:A→Bsuchthatg f=IA;thus,by2.25,fisinjective.By7.5and7.6wehave

7.7 A BiffthereexistsasurjectivefunctionB→A.

7.8TheoremLetA,B,C,andDbesetswhereA∩C=ØandB∩D=Ø.Iff:A→Bandg:C→Darebijectivefunctions,thenf∪gisabijectivefunctionA∪C→B∪D.

Proof.Iff:A→Bandg:C→Darefunctions,thenclearlyf:A→B∪Dandg:C→B∪Darefunctions,henceby2.16,

isafunction.Nowf:A→Bandg:C→Darebijective,henceby2.21,

arefunctions;thus,asabove,

isafunction.Butclearlyf−1∪g−1=(f∪g)−1,hence

isafunction;thus,by2.22,

isbijective.

7.9CorollarySupposeA∩C=ØandB∩D=Ø;ifA≈BandC≈D,thenA∪C≈B∪D.

7.10TheoremIfA≈BandC≈D,thenA×C≈B×D.

Proof.Letf:A→Bandg :C→Dbebijectivefunctions,and letusdefineh :A×C→B×Dasfollows:

h(x,y)=(f(x),g(y)),∀(x,y)∈A×C.

Itcaneasilybeshownthath:A×C→B×Disbijective;thedetailsarelefttothereader.

7.11TheoremIfA≈BandC≈D,thenAC≈BD.

Proof. Let f :A→B and g :D→C be bijective functions, and let us define h :AC→BD in thefollowingway.Foreachα∈AC,thatis,foreachfunctionα:C→A,leth(α)=f α g;clearlyf α gisafunctionD→B, thatis, f α g∈BD. Itcanbeshownroutinely thath :AC→BD isabijectivefunction;thedetailsarelefttothereader.

7.12CorollaryIfA≈B,then .

Thisfollowsimmediatelyfrom7.11and2.35.

EXERCISES7.2

1. CompletetheproofofTheorem7.10.2.CompletetheproofofTheorem7.11.3.Provethatif(A−B)≈(B−A),thenA≈B.4.SupposeA≈B,a∈A,andb∈B.Provethat(A−{a})≈(B−{b}).5.SupposethatA≈B,C≈D,C⊂AandD⊂B.Provethat(A−C)≈(B−D).6.Let{Bi}i∈Iand{Ci}i∈Ieachbeafamilyofmutuallydisjointsets.IfBi≈Ciforeachi∈I,provethat

7.Let{Bi}i∈Iand{Ci}i∈Ibefamiliesofsets.IfBi≈Ciforeachi∈I,provethat

3PROPERTIESOFINFINITESETS

AsetAissaidtobefiniteifAisinone-to-onecorrespondencewithanaturalnumbern;otherwise,Aissaidtobeinfinite.Severalotherdefinitionsof“finite”and“infinite”aretobefoundinthemathematicalliterature;foremostamongthemarethefollowing:

i) AisinfiniteifandonlyifAhasadenumerablesubset.ii) AisinfiniteifandonlyifAisequipotentwithapropersubsetofitself.

Ineachoftheabovetwocases,asetiscalled“finite”ifitisnotinfinite.Itwillbeshownnextthat(i)and(ii)areeachequivalenttoourdefinitionof“infinite,”givenabove.

7.13LemmaIfAisadenumerablesetandx∈A,thenA−{x}isadenumerableset.

Proof.IfAisdenumerable,thenthereexistsabijectivefunctionf:ω→A.Correspondingtox,thereisann∈ωsuchthatf(n)=x;defineg:ω→Aasfollows:

ItiseasytoseethatgisabijectivefunctionfromωtoA−{x};thedetailsarelefttothereader.

7.14TheoremAisaninfinitesetifandonlyifAhasadenumerablesubset.

Proof

i) Well-orderA;byTheorem4.62,exactlyoneofthefollowingcasesholds:(α)ωisisomorphicwithA;(β)ωisisomorphicwithaninitialsegmentofA;(γ)Aisisomorphicwithaninitialsegmentofω.IfAdoesnothaveadenumerablesubset,then(α)and(β)cannothold,hence(γ)holds;thereforeAisequipotentwithaninitialsegmentn=Snofω,soAisfinite.*WehavejustprovedthatifAdoesnothaveadenumerablesubset,thenAisfinite.

ii) Toprovetheconverse,wewillfirstuseinductiontoshowthatanaturalnumberncannothaveadenumerablesubset.Thisassertionisclearlytrueforn=0;letitbetrueforn,andsupposen+hasadenumerablesubsetS.Ifn∉S,thenSisadenumerablesubsetofn(recallthat“n“wasdefinedtobetheset{0,1,…,n−1});bythehypothesisofinduction,thiscannothappen.Ifn∈S,thenS−{n}⊆n;butS−{n}isdenumerable(Lemma7.13),sobythehypothesisofinductionthiscannothappen. We conclude that n+ cannot have a denumerable subset. Now suppose that A has adenumerablesubsetBandA is finite; that is,A≈n,B⊆A,andB≈ω.Thenwehave injectivefunctionsasfollows:ω→B→A→n; theircomposite isan injectivefunctionω→n,andwehavejustproventhistobeimpossible.ThusifAhasadenumerablesubset,thenAisinfinite.

7.15CorollaryEverysetwhichhasaninfinitesubsetisinfinite.

7.16CorollaryEverysubsetofafinitesetisfinite.

7.17CorollaryIfAisaninfinitesetandBisnonempty,thenA×BandB×Aareinfinitesets.

Proof.IfyisafixedelementofB,thefunctionf:A→A×Bgivenbyf(x)=(x,y)isclearlyinjective.Thusifg:ω→Aisabijectivefunction,thenf g:ω→A×Bisinjective.ItfollowsthatA×Bhasadenumerablesubset,henceA×Bisinfinite.

7.18TheoremAisaninfinitesetifandonlyifAisequipotentwithapropersubsetofitself.

Proof

i) SupposethatAisinfinite;byTheorem7.14,AhasadenumerablesubsetB={a0,a1,a2 ,…}.Letthefunctionf:A→Abedefinedby

Clearly,fisaone-to-onecorrespondencebetweenAandA−{a0}.

ii) Supposethereexistsabijectivefunction f :A→B,whereB isapropersubsetofA.Letcbeanarbitrary element ofA −B; by the recursion theorem, there exists a function γ :ω→A whichsatisfiestheconditions

(α)γ(0)=cand(β)γ(n+)=f(γ(n)).

Nowranf=Bandc∈A−B,soc∉ranf;thusby6.9,γisinjective.TherangeofγisobviouslyadenumerablesubsetofA,soby7.14,Aisinfinite.

EXERCISES7.3

1. LetAandBbeapairofdisjointfinitesets.UseinductiontoprovethatifA≈mandB≈n,thenA∪B≈m+n.Concludethattheunionoftwofinitesetsisfinite.

2. UsingtheresultofExercise1,provethatifAisaninfinitesetandBisafinitesubsetofA,thenA−Bisinfinite.ProveA−B≈A.

3. Provethatanaturalnumberisnotequipotentwithapropersubsetofitself.ConcludethatifA≈mandn>m,thenA n.

4. ProvethatAisaninfinitesetifandonlyif∀n∈ω,AhasasubsetBsuchthatB≈n.5. LetA andB be finite sets.Use induction to prove that ifA ≈m andB ≈ n, thenA ×B ≈mn.

ConcludethattheCartesianproductoftwofinitesetsisfinite.6. AssumingthatAisaninfinitesetandBisdenumerable,provethatA≈(A∪B).7. Supposex∈A;provethatAisaninfinitesetifandonlyifA≈(A−{x}).8. UseinductiontoprovethatifA≈n,then ≈2n.ConcludethatifAisafiniteset,then isa

finiteset.9. ProveCorollary7.15.10. ProveCorollary7.16.

4PROPERTIESOFDENUMERABLESETS

Onceagain,asetiscalleddenumerableifitisinone-to-onecorrespondencewithω.Thefundamentalpropertiesofdenumerablesetsarepresentedinthissection.

7.19TheoremEverysubsetofadenumerablesetisfiniteordenumerable.

Proof.Firstwenotethateverysubsetofωisfiniteordenumerable.Indeed,letE⊆ω;by4.63,eitherEω orE Sn =n for somen∈ω.Now letA be a denumerable set and letB⊆A; there exists abijectivefunctionf:A→ω.Nowf(B)⊆ω,sof(B)isfiniteordenumerable;butfisbijective,soB≈f(B);henceBisfiniteordenumerable.

7.20Theoremω×ω≈ω.

Proof.Wewillusetherecursiontheoremtoestablishtheexistenceofabijectivefunctionfromωtoω×ω.LetA=ω×ω;wedefineafunctionf:A→Aasfollows:

Wenotethatfisinjective,forsuppose

Ifr=0,then(becauseofthewayfisdefined)m=0andp=0;hence

Sok=n.Ifr≠0,then

sok=nandm=p.Thusfisinjective.Nowwemakeuseoftherecursiontheorem:wedefineafunctionγ:ω→Abythetwoconditionsi) γ(0)=(0,0),and

ii) γ(n+)=f(γ(n)).

Wenote(again,becauseofthewayfisdefined)that(0,0)cannotbeintherangeoff;itfollowsby6.9thatγisinjective.Finally,weshowthatγissurjective;indeed,wewillshowthatifk,m∈ω,then(k,m)=γ(n)forsomen∈ω.Theproofisbyinductiononk+m:

I. Ifk+m=0,then(k,m)=(0,0)=γ(0).

II. Supposek+m=n+;ifk=0,then(k,m)=f(m−1,0);bythehypothesisofinduction,(m−1,0)=γ(q)forsomeq∈ω,so

Ifk≠0,then(k,m)=f(k−1,m+1);bythehypothesisofinduction,(k+m−1,0)=γ(p)forsomep∈ω;thus

7.21CorollaryIfAandBaredenumerablesets,thenA×Bisadenumerableset.

Proof.IfA≈ωandB≈ω,thenA×B≈ω×ω(7.10).Butω×ω≈ω;thusA×B≈ω.

7.22TheoremLet{An}n∈ωbeadenumerablefamilyofdenumerablesets,andlet

thenAisdenumerable.[Adenumerableunionofdenumerablesetsisdenumerable.]

Proof.Tosay thateachAn isdenumerablemeans that thereexistsa family{fn}n∈ωof functionssuchthat,∀n∈ω,fn:ω→Anisbijective.Wedefineσ:ω×ω→Aby:σ(k,m)=fk(m).Itiseasytoseethatσissurjective:forifx∈A,thenx∈Anforsomen∈ω,andifx∈An,thenx=fn(m)=σ(n,m) forsomem∈ω.ByTheorem7.20,thereexistsabijectivefunctionϕ:ω→ω×ω;henceσ ϕ:ω→Aissurjective.

It follows (7.7) thatA ≈E for some subsetE⊆ω; nowE is either finite or denumerable (Theorem7.19);henceAiseitherfiniteordenumerable.ByCorollary7.15,Aisnotfinite,soAisdenumerable.

7.23CorollaryTheunionoftwodenumerablesetsisdenumerable.

EXERCISES7.4

1. Provethattheunionoftwodenumerablesetsisdenumerable.(Corollary7.23.)2. Let A be a denumerable set. Prove that A has a denumerable subset B such that A − B is

denumerable.3. Prove thatωn ≈ω. [Hint: Use the definitionsω1 =ω andωn+ =ω n ×ω; use 7.10, 7.20, and

induction.]ConcludethatifAisadenumerableset,thenAnisadenumerableset.4. Provethatω∪ω2∪ω3∪…isadenumerableset.5. Prove that the set of all finite subsets ofω is denumerable.Thenprove that the set of all finite

subsetsofadenumerablesetisdenumerable.6. LetAbeaninfiniteset.ProvethatAisdenumerableifandonlyifA≈BforeveryinfinitesubsetB⊆A.

7. ProvethatifAisanonemptyfinitesetandBisdenumerable,thenA×Bisdenumerable.8. Let bethesetofallpolynomialsa0+a1x+…+anxnwithintegercoefficients.Provethat is

denumerable.[Hint:Thismaybeprovedbyusinganargumentofthekindusedintheintroductiontoprovethat isdenumerable.]

9. Analgebraicnumberisanyrealrootofanequationa0+a1x+…+anxn=0,wherethecoefficientsaiareintegers.Provethatthesetofallalgebraicnumbersisdenumerable.

10. Arealnumberiscalledtranscendentalifitisnotalgebraic.Provethatthesetofalltranscendentalnumbersisnondenumerable.

11. Use the results ofExercises 1 and 5, above, to prove that the set of all infinite subsets ofω isequipotentwith2ω.

*Notethat,bydefinition,thenaturalnumbernistheset{0,1,2,…,n−1},thatis,nisexactlytheinitialsegmentSnofω.

8ArithmeticofCardinalNumbers

1INTRODUCTION

In theprecedingchapterwedefinedwhat ismeantbya finite set; it follows fromourdefinition thateveryfinitesetisequipotentwithexactlyonenaturalnumbern.Thisfacthasanimportantconsequence,namely,thatthenaturalnumbersmaybeusedasasetofstandards—ascale,asitwere—tomeasurethesize of finite sets. IfA is any finite set, thenAmay be “measured” by comparisonwith the naturalnumbers,andwillbefoundtocorrespond—thatis,tobeequipotent—withexactlyoneofthem.Whenthe natural numbers serve in this capacity—as standards to measure the size of sets—they arecommonlycalledcardinalnumbers.Anaturalandfascinatingquestionarisesnow:Canwefindawayofextendingoursystemofcardinal

numberssoastocreateasetofstandardsformeasuringthesizeofallsets?Toputitanotherway:Canwedefine“infinitecardinalnumbers,”andcanweconstructasufficientsupplyofthemsothateverysethasacardinalnumber(ifAhasnelements,wesaythatAhascardinalnumbern)?Theansweris“yes:”We can generalize the concept of cardinal numberwith such remarkable ease that almost all of thepropertiesofthefinitecardinals—theirordering,theirarithmetic,andsoforth—applyasnaturallytotheinfinitecardinalsastheydidtothefiniteones.Totakeoneexample:Whatismeantbythesumm+noftwocardinalnumbers?Theidea,clearly,isthatifAhasmelementsandBhasnelements—andifAandBaredisjoint—thenm+nisthecardinalnumberofA∪B.Whatcouldbemorenaturalthantoextendthisnotionofcardinalsumtoallsets(orrather,toall“setsizes”)?Before giving a general definition of cardinal numbers, let us take one more look at the natural

numbers and seewhy they can be used as standards tomeasure the size of finite sets.Aswe statedearlier,everyfinitesetisequipotentwithexactlyonenaturalnumbern;thatis,thenaturalnumbersarewelldefinedsets,andthereisauniquenaturalnumberforeachandeveryfinite“setsize.”Itisclear,now,whatweexpectofourdefinitionofcardinalnumbers:Thecardinalnumbersaretobewell-definedsets,andeverysetistobeequipotentwithexactlyonecardinalnumber.Itisimmaterialwhatsetsthecardinalnumbersare;theonlyrequirementisthattherebeexactlyonecardinalnumberofeach“size.”A simple way of constructing the cardinal numbers would be the following.We observe that the

relation“AisequipotentwithB”(A≈B)isanequivalencerelationamongsets.Thuswemightpartitionthe class of all sets into equipotence classes, and select one representative of each class: therepresentatives would be our cardinal numbers. This process seems quite natural—andwill, indeed,serveastheintuitivebasisofourdefinition.However,itcannotbeappliedliterally.Note,forexample,thatifAisaset,theequipotenceclass{B:B≈A}maybeaproperclass;henceitisnotlegitimatetospeakofthe“classofalltheequipotenceclasses.”Furthermore,evenifwecouldspeakofthe“classofall the equipotence classes,” it would not be legitimate to use the Axiom of Choice to pick arepresentativeofeachclass;indeed,theAxiomofChoice(seestatementCh1onpage115)allowsustopickrepresentativesfromasetofsets,notfromanarbitraryclassofclasses.Since we cannot literally “select” our cardinals by using the Axiom of Choice, how are we to

proceed? A simple way, and one which is sanctioned by mathematical tradition, is to posit theirexistence(thatis,toposittheexistenceofarepresentativesetfromeach“equipotenceclass”)bymeans

ofanewaxiom.

A12AxiomofCardinalityThereisaclassCDofsets,calledcardinalnumbers,withthefollowingproperties:

K1IfAisanyset,thereexistsacardinalnumberasuchthatA≈a.K2IfAisasetanda,barecardinalnumbers,thenA≈aandA≈b⇒a=b.

WewilladdtheAxiomofCardinalitytoourlistofaxiomsforsettheory—butonlyonaprovisionalbasis,forinthenextchapterwewilldescribeamethodforconstructingsetswithpropertiesK1andK2—that is,wewill produce actual sets (inmuch the sameway aswe produced the natural numbers)whichwillserveascardinalnumbers.Wewilluselower-caseRomanletters,suchasa,b,c,d,etc.,todenotecardinalnumbers.Itisworthnothing,incidentally,thattheclassCDofallthecardinalnumbersisaproperclass.For

supposeCDisaset:sinceeachcardinalnumberisaset,itfollowsbyAxiomA6that

thatis,theunionofallthecardinalnumbers,isaset.ThusbyAxiomA7, (V)isaset;butthen,byconditionK1ofAxiomA12,thereisacardinalnumberesuchthate≈ (V).Nowe∈CD,hencee⊆V ,whichis impossiblebyCorollary7.2.ThiscontradictionprovesthatCD isnotaset,butaproperclass.

2OPERATIONSONCARDINALNUMBERS

IfAisaset,aisacardinalnumberandA≈a,thenwesaythataisthecardinalnumberofA.Wedenotethisbywriting

NowconditionsK1andK2canbeconvenientlyrestatedasfollows:

K1IfAisanyset,thereexistsacardinalnumberasuchthata=#A.K2IfAisasetanda,barecardinalnumbers,thena=#Aandb=#A⇒a=b.

8.1LemmaIfaandbarecardinalnumbersanda≈b,thena=b.

TheproofisanimmediateconsequenceofK2.

8.2LemmaIfA≈B,then#A=#B.

Proof.ByK1,therearecardinalsa,bsuchthata=#Aandb=#B.Nowa≈Aandb≈B;thus,ifA≈B,itfollowsthata≈Aandb≈A,sobyK2,a=b.

We now proceed to define the addition and multiplication of cardinal numbers. Our definitionsrequire no comment; they correspond in themost natural way to our intuitive understanding of theprocessofaddingandmultiplyingwholenumbers.

Letaandbbetwocardinals.LetAandBbedisjointsetssuchthata=#Aandb=#B.Thena+bisthecardinalnumberdefinedby

Note. In theprecedingdefinition ithasbeenassumed thatwecanalways finddisjoint setsA andBsuchthata=#Aandb=#B.Thisisobviouslytrue.Forexample,takeA=a×{0}andB=b×{1};thenAconsistsofpairs(x,0),whereasBconsistsofpairs(x,1).

Letaandbbetwocardinals.LetAandBbesetssuchthata=#Aandb=#B.Theabisthecardinalnumberdefinedby

Note.Sinceaandbaresets,wecanwriteab=#(a×b).When introducing a newoperation, it is necessary to show that the operation iswell-defined. For

cardinaladditionandmultiplication,thismeansthefollowing.Foraddition:IfA1≈AandB1≈BthenA1∪B1≈A∪B.Formultiplication:IfA1≈AandB1≈BthenA1×B1≈A×B.

Thisguarantees that the sumandproductdonotdependon the specific setschosen, so longas theircardinality is the same. The proof of these two assertions are given as exercises at the end of theSection.The usual algebraic laws for addition andmultiplication follow from the elementary properties of

sets.

8.3TheoremIfa,b,carecardinalnumbers,thefollowinglawshold:i) a+b=b+a,ii) ab=ba,iii) a+(b+c)=(a+b)+c,iv) a(bc)=(ab)c,v) a(b+c)=ab+ac.

Proof. Properties (i), (iii), and (v) are immediate consequences of Theorems 1.25(i), 1.25(v), and1.32(ii) respectively. To prove (ii), it must be shown that there exists a one-to-one correspondencebetweenA×BandB×A; the functionφ(x,y)=(y,x)isobviously sucha correspondence.Finally, toprove(iv),wemustshowthatthereexistsaone-to-onecorrespondencebetweenA×(B×C)and(A×B)×C;thefunction

isclearlysuchacorrespondence.

LetAandBbefinitesets.InChapter2wedefinedABtobethesetofallfunctionsfromBtoA.Nowsuppose thatA hasm elements and B has n elements, and consider the process of constructing anarbitraryfunctionfromBtoA.SinceBhasnelements,andeachelementcanbeassignedanimageinm

possibleways,thismeansthereareexactlymndistinctfunctionsfromBtoA.Thissimpleobservationsuggeststhefollowingdefinitionofcardinalexponentiation:

Letaandbbetwocardinals.LetAandBbesetssuchthata=#Aandb=#B.Thenabisthecardinalnumberdefinedby

Fornotationalconvenience,weagreethat0a=0anda0=1.

8.4TheoremForanycardinalsa,b,cthefollowingruleshold:

i) ab+c=abac,

ii) (ab)c=acbc,

iii) (ab)c=abc.

Proof.LetA,B,Cbesetssuchthat

(Forpart(i)oftheproof,assumeB∩C=Ø.)

i) Wemustshowthatthereexistsabijectivefunctionσ:AB∪C→AB×AC.Wedefineσasfollows:Iff∈AB∪C,thenσ(f)=(f[B],f[C]),wheref[B] istherestrictionofftoBandf[C]istherestrictionofftoC.ItisimmediatethatσisafunctionfromAB∪CtoAB×AC.

σisinjective.Supposeσ(f)=σ(g),wheref,g∈AB∪C;then

thatis,

Thus,byTheorem2.15,

σissurjective.Forif(f1,f2)∈AB×ACthen,byTheorem2.16,

ii) Wewillshowthatthereexistsabijectivefunction

Wedefineσasfollows:If(f1,f2)∈AC×BC,thenσ(f1,f2)isthefunctionfdefinedby

certainlyf∈(A×B)C.NowitisimmediatethatσisafunctionfromAC×BC→(A×B)C.σisinjective.Foriff=σ(f1,f2)=σ(f1,f2)=f,then

sof1(c)=f′1(c)andf2(c)=f′2(c).Itfollowsthatf1=f′1andf′2=f′2,hence

σissurjective.Foriff∈(A×B)C,wemaydefine

and

Itiseasilyshownthatf1∈AC,f2∈BC,andf=σ(f1,f2);thedetailsarelefttothereader.

iii) Wewillshowthatthereexistsabijectivefunctionσ:(AB)C→AB×C.Notethatiff∈(AB)Candc∈C,thenf(c)∈AB;thus,ifb∈B,then[f(c)](b)∈A.Nowdefineσ(f)tobethefunction givenby

Certainly,σ(f)= ∈AB×C.Nowitisimmediatethatσisafunctionfrom(AB)CtoAB×C.σisinjective.Forif =σ(f)=σ(f′)= ',then

Thus∀c∈C,f(c)=f′(c),sofinally,f=f′.σissurjective.Forifg∈AB×Candc∈C,letfcbedefinedby

Clearlyfc∈AB,Now,iff isgivenby f(c)=fc, it iseasilyverifiedthat f isafunctionfromC toAB;clearlyg=σ(f).

Thefinitecardinalnumbersaredesignated,asusual,bythesymbols0,1,2,andsoforth.Itistobeespeciallynoted that 0 is the cardinal numberof the empty set, and1 is the cardinal numberof anysingleton. The cardinal number of ω—that is, the cardinal number of any denumerable set—iscustomarilydesignatedbythesymbol 0(“aleph-null”).

Itisusefultodistinguishbetweenfinitecardinalnumbers—thatis,cardinalnumbersoffinitesets—and infinite,or transfinite,cardinalnumbers,which are the cardinal numbers of infinite sets.Wewillsee,shortly,thatinfinitecardinalshaveseveralpropertieswhichdonotholdforfinitecardinals.

EXERCISES8.2

1. Proveeachofthefollowing,whereaisanycardinalnumber.

2. Proveeachofthefollowing,whereaisanycardinalnumber.

3. Ifa,barearbitrarycardinalnumbers,provethatab=0ifandonlyifa=0orb=0.4. Ifa,barearbitrarycardinalnumbers,provethatab=1ifandonlyifa=1andb=1.5. Giveacounterexampletotherule:a+b=a+c⇒b=c.6. Giveacounterexampletotherule:ab=ac⇒b=c.7. Ifnisafinitecardinalnumber,useinductiontoprovethatna=a+a+…+a,wheretheright-hand

sideoftheequalityhasnterms.

8. Ifnisafinitecardinalnumber,useinductiontoprovethatan=aa…a,wheretheright-handsideoftheequalityhasnfactors.

9. Leta,bbecardinals,andletA,Bbesetssuchthata=#Aandb=#B.Provethata+b=#(A∪B)+#(A∩B).

10. Provethatifaisaninfinitecardinalnumberandnisafinitecardinalnumber,thena+n=a.11. Provethatifa+1=a,thenaisaninfinitecardinalnumber.12. Ifbisaninfinitecardinalnumber,provethat 0+b=b.

13. Prove:IfA1≈AandB1≈BthenA1∪B1≈A∪B.14. Prove:IfA1≈AandB1≈BthenA1×B1≈A×B.

3ORDERINGOFTHECARDINALNUMBERS

Sincecardinalnumbersmeasurethesizeofsets,wenaturallyexpectthecardinalnumberofasmallersettobe“lessthan”thecardinalnumberofalargerset.Thissuggestsanaturalorderingofthecardinalnumbers:

Leta andb be cardinals, and letA andB be sets such thata = #A andb = #B. The relation isdefinedby

Note. Clearly,a b if andonly ifa b. Inparticular,a b if andonly if there exists an injectivefunctionf:a→b.Ourgoalinthissectionistoshowthattherelation definedaboveisanorderrelationamongthe

cardinal numbers, and, in particular, that the class of all the cardinal numbers is well ordered with

respecttothisrelation.

8.5Theorem(Schröder-Bernstein).Letaandbbecardinalnumbers;ifa bandb a,thena=b.

Proof.Supposea bandb a;ifAandBaresetssuchthata=#Aandb=#B,thenA BandB A,thatis,thereexistinjectivefunctionsf:A→Bandg:B→A.IfC⊆A,let(c)=A− [B− (C)];itiseasytoseethatifCandDaresubsetsofA,then

Now,letS={B:B⊆AandB⊆Δ(B)},andletA1= B.WewillprovethatA1=Δ(A1).

i) Ifa∈A1,thena∈BforsomeB∈S;butB⊆A1,soby(1),(B)⊆(A1).Thuswehave

ii) Wehave justshownthatA1⊆ (A1),henceby(1),Δ(A1)⊆Δ[Δ(A1)],soΔ(A1)∈S.ButA1 is theunionofalltheelementsofS,soΔ(A1)⊆A1.Thus,wehaveprovedthatA1=(A1),whichisthesameas

Byelementaryclassalgebra(seeExercise11,ExerciseSet1.3)thisgives

Now,fandgareinjectivefunctions,henceA1≈ (A1)and,by(2),

But (A1)≈A1;thus,by7.9,A≈B.

It is immediate that therelation betweencardinalnumbers isreflexiveandtransitive;by8.5 it isantisymmetric,henceitisanorderrelation.Infact,thecardinalnumbersarelinearlyorderedby .Thatis,anytwocardinalsarecomparable.

8.6TheoremIfaandbarecardinalnumbersthena borb a.

Proof.Sinceaandbaresets,itfollowsfrom5.22thataandbcanbewell-ordered.Thus,from4.62,thereexistsaninjectiona→borb→a.

8.7TheoremEveryclassofcardinalnumbershasaleastelement.

Proof.Let beanarbitraryclassofcardinalnumbers,andleta∈ ;ifaistheleastelementof ,wearedone;otherwise,let ={b∈ :b<a}.Usingthewell-orderingtheorem,letuswellordera; foreachb∈ ,letφ(b)betheleastelementx∈asuchthatb≈Sx.Nowtheset{φ(b):b∈ }hasaleastelementφ(d)becauseitisasubsetofa;wewillshowthatdistheleastelementof .Indeed,letbbeanarbitraryelementof ;φ(d) φ(b),henceSφ(d)⊆Sφ(b).Thuswehaveinjectivefunctions

(λistheinclusionfunction),henced b.ThusdistheleastelementofLet ,hencetheleastelementof .

Weareabletoconclude:

8.8Theclassofallthecardinalnumbers,orderedby ,iswellordered.

Thefamiliar“rulesofinequality”applytothecardinalnumbers,asweshallseenext.

8.9TheoremLeta,bbecardinalnumbers.Thena bifandonlyifthereexistscsuchthatb=a+c.

Proofi) Supposeb=a+c;letA,B,Cbesets(assumeA∩C=Ø)suchthat

Thenthereexistsabijectivefunctionf:A∪C→B.Clearlyf[A]isaninjectivefunctionfromAtoB,soA B.

ii) Supposea b;letA,Bbedisjointsetssuchthat

Thereexistsaninjectivefunctionf:A→B;sincefisinjective,A≈ (A),soa=#f(A).IfC=B−f(A)andc=#C,clearlyb=a+c.

8.10TheoremLeta,b,c,dbecardinalnumbers.Ifa candb d,thenwehavethefollowing:

i) a+b c+d,ii) ab cd,iii) ab cd.

Proof.ByTheorem8.8,thereexistsr,ssuchthatc=a+randd=b+s.i) c+d=a+r+b+s=(A+b)+(r+s),soby8.8,a+b c+d.ii) cd=(a+r)(b+s)=ab+as+rb+rs=ab+(as+rb+rs),sobyTheorem8.8,ab cd.

iii) Firstwemustshowthatab (a+r)b;thatis,ifA,R,Baresetssuchthata=#A,b=#Bandr=#R,wemustshowthatthereexistsaninjectivefunctionσ:AB→(A∪R)B.Wedefineσby

andnot(see2.4)thatafunctionf:B→Aisalsoafunctionf:B→A∪R.Itisimmediatethatσisinjective,henceab (a+r)b,thatisab cb.Finally,usingpart(ii),wehave

Remark.Itisimportanttonotethat8.8givesusvaluablenewinformationontherelationA Bamongsets.Indeed,thefollowingaretwoimmediateconsequencesof8.8:1) IfAandBarearbitrarysets,thenA BorB A.2) IfA BandB A,thenA≈B.

Item(2)isespeciallyusefulwhenweneedtoprovethattwosetsareinone-to-onecorrespondence,foritisnowsufficienttoshowthatthereisaninjectivefunctionfromAtoBandaninjectivefunctionfromBtoA(alternatively,asurjectivefunctionfromAtoBandasurjectivefunctionfromBtoA).

EXERCISES8.3

1. ProvethatifAandBarearbitrarysets,thenA BorB A.2. ProvethatisA BandB A,thenA≈B.3. Provethefollowing,wherea,b,c,darecardinalnumbers.

a) Ifac<bc,thena<b.

b) Ifac<ad,thenc<d.4. Provethatifa+b=c,then(r+s)c≥rasb.5. Provethatthereexistsastrictlyincreasingsequencea1<a2<…ofcardinalnumbers,eachwith

thepropertyai 0=ai.[Hint:Takea1= 00;thentakea2=2a1 0,a3=2a2 0,etc.]

6. LetAbeadenumerableset.Proveeachofthefollowing:

a) AA⊆ (A×A).ConcludethatAA (A×A),henceAA (A).

b) Verifythatthefunctionφgivenby:φ(f)=rangef,∀f∈AA,isasurjectivefunctionAA→ (A)−φ.Concludethat (A) AA.

c) AA≈( (A);thatis,AA≈2A.Concludethat 00=2 0.

7. UsetheargumentoutlinedintheprecedingproblemtoprovethatthesetofallinjectivefunctionsA→Aisequipotentwith2A.

4SPECIALPROPERTIESOFINFINITECARDINALNUMBERS

Afewremarkablearithmeticrulesholdexclusivelyforinfinitecardinals.Asaresultoftheserules,thearithmeticofinfinitecardinalnumbersisaverysimplematter.

8.11TheoremIfaisaninfinitecardinalnumber,thenaa=a.

Proof.LetAbeasetsuchthata=#A.SinceAisinfinite,AhasadenumerablesubsetD.ByCorollary7.21,D≈D×D;thatis,thereexistsabijectivefunctionφ:D→D×D.NowletAbethesetofallpairs(B,f)whichsatisfythefollowingconditions:i) BisasubsetofAandfisabijectivefunctionfromBtoB×B.ii) D⊆B.

iii) φ⊆f.

Weorder bytherelation(B1,f1) (B2,f2)iffB1⊆B2andf1⊆f2. isnonempty,for(D,φ)∈A.Nowitiseasytoverifythat satisfiesthehypothesesofZorn’sLemma(thedetailsareleftasanexerciseforthereader).Thus hasamaximalelement(C,g);itremainsonlytoshowthat#C=a.Wewillprovethisbycontradiction—assumingthat#C<aandprovingthistobeimpossible.Letb=#Candassumethatb<a.SinceC×C≈C,itfollowsthatbb=b;furthermore,

and

henceb=b+b.Nowletd=#(A−C);CandA−Caredisjoint,so

Wenotethatb<d,ford bimpliesthat

whichwouldcontradictourassumptionthatb<a.Fromb<ditfollowsthatA−ChasasubsetEsuchthat#E=b.

Now

whereC×C,C×E,E×C,E×Earemutuallydisjointsets,eachofwhichhasthecardinalbb=b.Thus

hencethereexistsabijectivefunction

Itfollowsby7.8thatg∪hisabijectivefunctionfromC∪Eto

hence(C∪E,g∪h)>(C,g),whichisimpossiblebecause(C,g)isamaximalelementof .Theassumptionthatb<ahasledtoacontradiction;thusb=a,soaa=a.

8.12CorollaryLetaandbbecardinals,whereaisinfiniteandb≠0.Ifb a,thenab=a.

Proof.Sinceb 1,thusa=a1 ab;butab aa=a,henceab=a.

8.13CorollaryIfaisaninfinitecardinal,a+a=a.

Proof.Wehavea=1a 2a aa=a;but2a=(1+1)a=a+a,soa+a=a.

8.14CorollaryLetaandbbecardinals,whereaisinfinite.Ifb a,thena+b=a.

Proof.Wehavea=a+0 a+b;buta+b a+a=a,soa+b=a.

8.15CorollaryLetaandbbeinfinitecardinalnumbers.Then

8.16TheoremLeta>1beacardinalnumberandletbbeaninfinitecardinalnumber.Ifa b,thenab

=2b.

Proof.By7.4,a<2a,soab (2a)b=2ab.ButbyCorollary8.12ab=b,soab 2b.Ontheotherhand,2a,so2b ab.Consequentlyab=2b.

Remark.Theorem8.11 and its corollaries can be interpreted very profitably in terms of sets and therelationA Bamongsets.Forexample,Theorem8.11tellsusthatifAisaninfiniteset,thenA≈A×A.Thishasaninterestingconsequence:A×Ahasapartition{Bx}x∈AwhereBx={(x,y):y∈A}.HencethebijectivefunctionfromA×AtoAinducesacorrespondingpartition{Cx}x∈AofA,whereAistheindexsetandeachmemberofthepartitionisequipotentwithA.

EXERCISES8.4

1. Ifaisaninfinitecardinalnumberanda bc,provethata bora c.2. Letabeacardinalnumber>1,andletbbeaninfinitecardinalnumber.Provethatifa=ab,thenb

<a.3. Aninfinitecardinalnumberaissaidtobedominantifitsatisfiesthefollowingcondition:ifband

carecardinalnumberssuchthatb<aandc<a,thenbc<a.Provethataisadominantcardinalnumberifandonlyifd<a⇒2d<a.

4. Ifa,c,anddarearbitrarycardinalnumbersandbisaninfinitecardinalnumber,provethat

5. Leta,b,c,dbecardinalnumbers.Provethatifa<bandc<d, thenac<bdanda+c<b+d.[Hint:Forthecasewherebanddarebothfinite,thisresulthasbeenproveninExercises5and6,ExerciseSet 6.4. Ignore this case, and assume thatb is infinite ord is infinite (this assumptionincludesthreecases).]

InExercises6through8,a,b,andcarearbitrarycardinals.Foreachoftheseproblems,thecasewherea,b, andc are all finite has been considered inChapter 6. Ignore this case, and treat the remainingcases.6. Provethatifa+a=a+b,thena b.7. Provethatifa+b<a+c,thenb<c.8. Provethatifab<ac,thenb<c.

5INFINITESUMSANDPRODUCTSOFCARDINALNUMBERS

Earlyinthisbookwespokeoftheunionoftwoclasses;laterweextendedthisnotionbydefiningtheunionofanarbitrary familyofclasses.Similarly,we introduced theCartesianproductof twoclassesand later generalized this to theproduct of a familyof classes. Inboth cases, extendingouroriginaldefinitionseemedlikeaperfectlynaturalthingtodo,fortheintuitiveconceptsofunionandproductcanbeappliedaseasilytoafamilyofclassesastoapairofclasses.Thesameholdstruefortheprocessofaddingandmultiplyingcardinalnumbers;theylendthemselvestothefollowingobviousgeneralization.

Let{ai}i∈Ibeafamilyofcardinalnumbers;let{Ai}i∈Ibeafamilyofdisjointsetssuchthatai=#Aiforeachi∈I.Then isthecardinalnumberdefinedby

Let{ai}i∈Ibeafamilyofcardinalnumbers,andlet{Ai}i∈Ibeafamilyofsetssuchthatai=#Ai foreachi∈I.Then isthecardinalnumberdefinedby

Inelementaryarithmeticwelearnthatabistheresultof“addingatoitselfbtimes,”andthatabistheresult of “multiplying a by itself b times.” It is useful to know that this holds true for all cardinalnumbersaandb.

8.17TheoremLetaandbbecardinalnumbers,andletIbeasetsuchthatb=#I.Ifa=ai,∀i∈ I ,then

Proofi) Let{Ai}i∈Ibeafamilyofdisjointsetssuchthata=ai=#Aiforeachi∈I,andletAbeasetsuch

thata=#A.SinceAi≈Aforeachi∈I,thereexistsafamily{fi:A→Ai}i∈Iofbijectivefunctions.Wedefine

by

itiselementarytoverifythatfisbijective.Thus

thatis,

ii) WewishtoshowthatAI≈ ,whereAi=Aforeachi∈I.ButaglanceatthedefinitionsofAI

and Ai(whereAi=A,∀i∈I)willrevealthattheybothrefertothesameset—thesetofallfunctionsfromItoA.

Theorem8.9hasthefollowinganalogueforinfinitesumsandproducts.

8.18TheoremLet{ai}i∈Iand{bi}i∈Ibefamiliesofcardinalnumbers.Ifai biforeachi∈I,then

Proof

i) Let{Ai}i∈Iand{Bi}i∈Ibeafamiliesofdisjointsetssuchthatai=#Aiandbi=#Biforeachi∈ I.Sinceai biforeveryi∈I,thereexistsafamily{fi:Ai→Bi}i∈Iofinjectivefunctions.Itiseasyto verify that is an injective function from (The details are left as anexerciseforthereader.)

ii) Giventhefamily{fi:Ai→Bi}i∈Iintroducedabove,wedefineafunction:

asfollows:if ,then

Weverifythatfisinjective:Iff(u)=f(v),then

Buteachfiisinjective,soui=viforeveryi∈I;henceu=v.

Theorem8.17and8.18havethefollowingusefulcorollary.

8.19CorollaryLet{ai:i∈I}beasetofcardinalnumbers,andletbandcbecardinalnumbers.Ifaibforeachi∈Iandif#I=c,then

Theproof,whichfollowsimmediatelyfrom8.17and8.18,isleftasanexerciseforthereader.

EXERCISES8.5

1. Provethat1.Provethat =0ifandonlyifai=0forsomei∈I.

2. Supposeai a,∀i∈I,and#I a,whereaissomefixedcardinal.Provethat a,[Hint:Use

Theorems8.17and8.18.]3. Supposeai a,∀i∈I,and#I a.Provethat 2a.4. Let{ai}i∈I be a set of a cardinal numbers, and suppose there is no greatest element in this set.

Provethat∀j∈I,aj<

5. Provethata•

6. UseTheorem8.18tojustifyeachofthefollowing.(Eachsumisunderstoodtohave 0terms.)

a) 1+2+3+…= 0,b) n+n+…= 0,c) 0+ 0+···= 0.

7. Let f :A→B be a surjective function,whereB is an infinite set. If,∀y∈B, f−1(y)is finite ordenumerable,provethatA≈B.

8. LetAbeaninfiniteset,andletF(A)designatethefamilyofallfinitesubsetsofA.ProvethatF(A)≈A.[Hint:Foreachn∈ω,letFndesignatethefamilyofn-elementsubsetsofA.ThereexistsanobvioussurjectivefunctionfromAntoFn;thereare 0setFn.]

9. If{Ci}i∈Iisafamilyofsets,provethat#

10. Let{ai}i∈Iand{ai}i∈Jbefamiliesofcardinalnumbers.Provethat

9ArithmeticoftheOrdinalNumbers

1INTRODUCTION

In elementary school we learn that there are cardinal numbers and ordinal numbers. The cardinalnumbers, we are told, are the “counting” numbers: 1, 2, 3, and so on; the ordinal numbers are the“ranking”numbers: first, second, third,etc.Thedistinctionmayappear tobesomewhatpedantic, forthe natural numbers serve in both capacities, as ordinals and as cardinals, and there is no need inelementaryarithmetictodifferentiatebetweenthetwo.However,oneoftheunexpecteddiscoveriesofmodern set theory is that, just as the infinitecardinalsbehavedifferently from the finiteones, so theinfinite ordinals exhibit a strikingly different behaviour from the cardinals. It is the purpose of thischapter to introducetheordinalnumbersandexplore theirproperties—especially thoseof the infiniteordinals.The dichotomy between cardinal and ordinal, from the scholastic point of view, arises from two

differentwaysofusingthenaturalnumbers.Intheirroleascardinals,thenaturalnumbersmeasurethe“size,”orpower,of sets; in their roleasordinals, theyserve todesignate the rank,orposition,ofanobject ina linearlyorderedarray.Wewilluse this insight—althoughit issomewhatoutdated—asthestartingpointofourdiscussion.Whenwespeakofrankingelementsinsomeorder,whatkindoforderdowehaveinmind?There

mustbeafirstelement,asecondelement,andsoon—inotherwords, theorder is thatof thenaturalnumbers,whichisawell-ordering.Nowthereadershouldnotethatthegeneralnotionofwell-orderingisanextensionoftheorderofthenaturalnumbers.Everyinfinitewell-orderedsethasafirstelement,asecondelement,and—foreachnaturalnumbern—annthelement;butitmayalsohaveelementswhichare“beyondthereach”ofthefiniteordinals.Thustheset

hasafirstelementx1,asecondelementx2,andsoforth;buty1,forexample,thoughithasaperfectlywell-defined“position”intheset,cannotbedescribedasthe“nthelement”foranyfiniten.Situationsofthiskindarisefrequentlyinalmosteverybranchofmathematics.Forexample,onpage

141wedefinedasequenceofsetsbytheseconditions:ω=K1, ,andsoon; ,etc.Continuinginthismanner,we

getthewell-orderedfamilyofsets

NowK1isthefirstelementofthisfamily,and,ingeneral,Knisthenthelement;butwhatof(say)L1?Its position in the family is unambiguous: L1 immediately follows all of the setsKi ; yet classicalmathematicshasnotprovideduswithanyordinalnumbertodescribethepositionofL1.Thus,as inourstudyofcardinalnumbers,weare led toaskan intriguingquestion:Canwefinda

wayofextendingoursystemofordinalnumberssoastocreateasetofstandardsfordesignatingthe

positionofanyelementinanywell-orderedset?Theanswer,onceagain,is“yes;”wecangeneralizetheconceptofordinalnumberwithsuchremarkableeasethatnobarrier,eitherlogicalorintuitive,seemstoseparatethefiniteordinalsfromtheinfiniteones.Wewillapproachtheordinalsinmuchthesamewaythatweapproachedthecardinals.Wewillbegin

bydefiningarelationof“havingthesameordinalnumber,”andlaterdefinetheordinals,essentially,toberepresentativeofthedistinctclassesinducedbythisrelation.IfAandBarewell-orderedsets,andifx∈Aandy∈B,thentosaythat“xhasthesamerankasy”

(forexample,xisthethirdelementofAandyisthethirdelementofB)isthesameassayingthattheinitialsegmentSxisisomorphicwiththeinitialsegmentSy.Tosaythat“xhasalowerrankthany”isthesameassayingthatSxisisomorphicwithaninitialsegmentofSy.Thereadershouldstophereuntilhehasthoroughlyunderstoodthisfact,foritisthepointofdepartureforachievinganunderstandingofthemodern approach to the ordinal numbers. Isomorphism plays the same role in the study of ordinalnumbersthatone-to-onecorrespondenceplaysinthestudyofcardinalnumbers.An important warning needs to be given here. The alert reader may question the necessity of

introducingtheconceptofisomorphism.Afterall,hemayask,whynotsaythatxandyhavethesamerankifandonlyifSxisequipotentwithSy?Surelyiftenelementsprecedexandtenelementsprecedey,thenxandyarebotheleventhintheirclass.Thisistruewhenwearedealingwithfiniterank,butuntrueinthecaseofinfiniterank;asimpleexamplewillconvincethereader.Intheset

bothy1andy2areprecededbyadenumerablenumberofelements—thatis,Sy1isequipotentwithSy2—yety2clearlyfollowsy1.Whenwesay thatx has the same rankasy,orx has a lower rank thany,we are onlyapparently

speakingofxandy;actually,wearecomparingtheinitialsegmentsSxandSy.Hencewelosenothingifwe confine our attention to the study of initial segments ofwell-ordered sets.Butwe can go a stepfurther:An initial segment (of awell-ordered set) is awell-ordered set, and conversely, everywell-ordered setA is an initial segment (adjoin a last elementx toA—thenA isSx ).Hence the study ofordinalityis,essentially,thestudyofwell-orderedsets.Motivatedbytheforegoingremarks,weintroducethefollowingdefinitions:

LetAandBbewell-orderedsets.WesaythatAandBaresimilar(orhavethesameordinality)ifAisisomorphicwithB;wewriteA B.IfAisisomorphicwithaninitialsegmentofB,wesaythatBisacontinuationofA,orAhasalowerordinalitythanB,andwewriteA B.ItfollowsfromTheorem4.62thatifAandBareanytwowell-orderedsets,thenA B,orA B,

orB A.Inconclusion,tosaythatxhasthesamerankasyisthesameassayingthatSx Sy,andtosaythatx

hasalowerrankthany is thesameassayingthatSx Sy.Thuswehavecompletelycaptured—andformalized—the intuitive concept of “rank,” and have extended it beyond the unnatural confines offiniteordinality.As for the ordinal numbers, we simply imitate the procedure we followed for the cardinals by

introducingthe

A13AxiomofOrdinalityThereisaclassORofwell-orderedsets,calledordinalnumbers,with thefollowingproperties:

O1IfAisanywell-orderedset,thereexistsanordinalnumberαsuchthatA α.O2IfAisawell-orderedsetandα,βareordinalnumbers,thenA αand .

WewilladdtheAxiomofOrdinalitytoourlistofaxiomsforsettheory—butonlyonaprovisionalbasis, for in the last section of this chapter we will describe a method for constructing sets withproperties01and02;thosesetswillthenservethepurposeofordinalnumbers.Itisworthnoting,incidentally,thattheclassofalltheordinalnumbersisaproperclass.Indeed,let

ORbetheclassofalltheordinalnumbers,andsupposeORisaset;fromthisassumptionwewillderivea contradiction. Let , where since each ordinal number α is a set and OR(underourassumption)isaset,itfollowsbyAxiomsA6andA7thatB,andthereforeA,aresets.Letuswell-orderA;byO1,thereisanordinalnumberαsuchthatα A.Butα∈OR,henceα B,soby7.2,αcannotbeequipotentwith .ThiscontradictionprovesthatORisaproperclass.

2OPERATIONSONORDINALNUMBERS

Following our “naive” introduction to ordinal numbers in the preceding section,we now proceed tostudy the ordinals from a formal point of view.We will henceforth consider the ordinals to be theobjectsdefinedbyConditionsO1andO2.Thereadershouldadjusthisthinkingaccordingly;heshouldceasethinkingofordinalsas“symbolsfordesignatingrank,”andbegintothinkofthemascertainwell-orderedsets.

9.1DefinitionIfAisawell-orderedset,αisanordinalnumber,andA α, thenwesaythatαistheordinalnumberofA.Wedenotethisbywriting

NowConditionsO1andO2canbeconvenientlyrestatedasfollows:

O1IfAisawell-orderedset,thereexistsanordinalnumberαsuchthatα= A.

O2IfAisawell-orderedsetandα,βareordinalnumbers,thenα= Aandβ= A⇒α=β.

9.2LemmaIfαandβareordinalnumbersandα β,thenα=β.

Theproofisanimmediateconsequenceof02.

9.3LemmaIfA B,then A= B.

TheproofisanalogoustothatofLemma8.2.

Beforedefining theadditionandmultiplicationofordinalnumbers,weneed to introduce twonewoperationsonwell-orderedsets.

9.4DefinitionLetAandBbedisjoint,well-orderedsets.A⊕B,calledtheordinalsumofAandB,isthesetA Borderedasfollows.Ifx,y A B,thenx yifandonlyifi) x∈Aandy∈Aandx yinA,or

ii) x∈Bandy∈Bandx yinB,oriii) x∈Aandy∈B.

Thus,inA⊕B,theelementsofAareorderedasbefore,theelementsofBareorderedasbefore,andeveryelementofBisgreaterthaneveryelementofA.

Havingdefinedtheordinalsumoftwowell-orderedsets,itisnaturaltodefinetheordinalsumofanarbitraryfamilyofwell-orderedsets.

9.5DefinitionLetIbeaset,let{Ai}i∈Ibeafamilyofdisjointwell-orderedsetsandlettheindexsetIbewell-ordered.

,calledtheordinalsumofthefamily{Ai}i∈I,istheset orderedinthefollowingway:if

,thenx yifandonlyif

i) forsomei∈I,x∈Aiandy∈Aiandx yinAi,or

ii) x∈Aiandy∈Ajandi<j.Thus,in Ai,eachsetAiisorderedasbefore,and,fori<j,everyelementofAiislessthaneveryelementofAj.Aneasystepleadsus,now,tothenotionofordinalproduct.Toputitsimply,theproductA Bisthe

resultof“addingAtoitselfBtimes.”Moreprecisely,if{Ai}i∈Bisafamilyofdisjoint,well-orderedsets,indexedbyB,whereeachAiissimilartoA,thenA Bistheset .Theonlyremainingdifficultyistoproducethefamily{Ai}i∈B.Todosoiseasyenough:foreachi∈B,wedefineAitobetheset{(x,i):x∈A}—thatis,Ai=A×{i}.Butahappythoughtstrikesusnow,aswerealizethattheset isnoneotherthantheCartesianproductA×Borderedbytheantilexicographicordering(Definition4.2).Weexploitthisfortunatecoincidencetogivethefollowingelegantdefinitionofordinalproduct:

9.6DefinitionLetAandBbewell-orderedsets.ThenA B,calledtheordinalproductofAandB,isthesetA×Borderedbytheantilexicographicordering.

9.7ExampleLetA={a, b, c}bewell ordered as follows:a<b< c.LetB = {1<2<3}bewellorderedasfollows:1<2<3.ThenA BisthesetA×Bwellorderedasfollows(seeFig.9.1):

Fig.9.1

Now,backtotheordinalnumbers.

9.8DefinitionLetαandβbeordinalnumbers,andletAandBbedisjoint,well-orderedsetssuchthatα= Aandβ= B.Wedefinethesumα+βtobetheordinalnumbergivenby

Letαandβbeordinalnumbers,andletAandBbewell-orderedsetssuchthatα= Aandβ= B.Wedefinetheproductαβtobetheordinalnumbergivenby

Inordertotiedownthedefinitionofordinaladditionandmultiplication,itmustbeshownthattheseoperationsarewell-defined.Thatis,ifα1=α2andβ1=β2,thenα1+β1=α2+β2andα1β1=α2β2.ThisiseasytoproveandleftasanexerciseattheendofthisSection.Theelementarypropertiesofordinaladditionandmultiplicationaregiveninthefollowingtheorem.

9.9TheoremLetα,βandγbeordinalnumbers.Theni) α+(β+γ)=(α+β)+γ,ii) α(βγ)=(αβ)γ,iii) α(β+γ)=αβ+αγ.

Proof.LetA,B,andCbedisjoint,well-orderedsetssuchthatα= A,β= B,andγ= C.

i) Wemustshowthat

ButitfollowsimmediatelyfromourdefinitionofordinalsumsthatbothA⊕(B⊕C)and(A⊕B)⊕CdesignatethesetA B Cwiththefollowingorder:IfxandyarebothinA,bothinB,orbothinC,theyareorderedaccordingtotheirorderinA,B,orCrespectively;furthermore,everyelementofCisgreaterthaneveryelementofB,andeveryelementofBisgreaterthaneveryelementofA.

ii) Wemustshowthat

Wehaveseenearlierthatthefunction

isaone-to-onecorrespondencebetweenA×(B×C)and(A×B)×C.Inordertoestablishthatfis

anisomorphism,weneedsimplyshowthat

ifandonlyif

Thedetails,whichfollowimmediatelyfromthedefinitionofordinalproduct,areleftasanexerciseforthereader.

iii) Wemustshowthat

BothA (B⊕C)and(A B)⊕(A C)designatethesameset,

withcertainorderings;itiseasytoshowthatthetwoorderingsarethesame.Thedetailsareleftasanexerciseforthereader.

Asusual,0istheordinalnumberoftheemptyset,and1istheordinalnumberofanysingleton.Anordinalnumberμissaidtobefiniteifμissimilartoanaturalnumbern;ifμisnotafiniteordinal,thenμ iscalledan infinite,or transfinite,ordinal. It iscustomary todesignate theordinalnumberofωbymeansofthesymbolω.Itismostimportanttonotethatadditionandmultiplicationofordinalnumbersarenotcommutative.

Twosimpleexampleswillsufficetoestablishthisfact.First,letustakeaddition.Ifwecompareω+1with1+ω,weobservethat1+ωissimilartoω,whereasω+1isnot(ithasalastelement!);thusω+1≠1+ω.Next,letustakemultiplication.Weobservethat

andthat

thesesetsareobviouslynotisomorphic,henceω2≠2ω.

Wenotealsothatthe“rightdistributivelaw“(α+β)γ=αγ+βγdoesnothold.Forexample,

whereas

andwenotedintheprecedingparagraphthat2ω≠ω2;thus

EXERCISES9.2

1. LetA1,A2,B1,B2bewell-orderedsets.ProvethatifA1 A2andB1 B2,thena)A1⊕B1 A2⊕B2,andb)A1 B1 A2 B2.

2. IfAandBarewell-orderedsets,provethatA⊕BandA Barewell-orderedsets.3. Ifαisanordinalnumber,provethat1+α=αiffαisaninfiniteordinal.4. Letαandβbenonzeroordinalnumbers.Provethatifα+β=ω,thenαisafiniteordinalnumber

(thatis,similartoanaturalnumber)andβ=ω.Nowassumeβ≠1.Provethatifαβ=ω,thenαisfiniteandβ=ω.[Hint:Considerthewell-orderedsetsA⊕BandA B,whereα= Aandβ=B.]

5. Proveeachofthefollowing,whereμdesignatesafiniteordinal.a)μ+ω=ω,b)μω=ω,c)Ifαisaninfiniteordinal,thenμ+α=α.

6. Provethat(ω+ω)ω=ωω.7. Giveacounterexampletothe(false)rule

8. Provethefollowing,foreveryordinalnumberα.

9. Provethatαβ=0ifandonlyifα=0orβ=0.10. Proveeachofthefollowing,whereμ,ν,πarefiniteordinals.

11. Giveacompleteproofoftheisomorphism[Theorem9.9(ii)]

12. Giveacompleteproofoftheisomorphism[Theorem9.9(iii)]

3ORDERINGOFTHEORDINALNUMBERS

Intheintroductiontothischapter,wespokeofcomparingtheordinalityofwell-orderedsets.IfAandBare well-ordered sets, we say that A has a lower ordinality than B, (in symbols A B) if A isisomorphicwithaninitialsegmentofB.Itisconvenientnowtoadd:theordinalityofAislessthanorequaltotheordinalityofBifandonlyifAisisomorphicwithBoraninitialsegmentofB;inthiscase,wewriteA B.This is the sameas saying that thereexists an injective,order-preserving functionfromAtoB,whoserangeisasectionofB.(See4.48and4.56.)

9.10LemmaLetAandBbewell-ordered sets.A B if there exists an injective, orderpreservingfunctionf:A→B.

Proof.IfA Bthenclearlythereexistsaninjective,orderpreservingfunctionfromAtoB.

Conversely,supposethereexistsaninjective,orderpreservingfunctionf:A→B.If ,thenf:A→Cisanisomorphism.By4.63,thereexistsanisomorphismg:C→D,whereDisBoraninitialsegmentofB.Nowg f:A→Disanisomorphism,henceA B.

9.11CorollaryIfA BandB CthenA C.

Proof. Clearly the composite of two injective, order preserving functions is injective and orderpreserving.

IfAandBarewell-orderedsetsandAhasa lowerordinality thanB,wequitenaturallyexpect theordinalnumberofA tobe“less than” theordinalnumberofB.Accordingly,wedefine the“natural”orderingoftheordinalnumbersasfollows.

9.12DefinitionLetαandβbeordinals,andletAandBbewell-orderedsetssuchthatα= Aandβ=B.Therelation isdefinedby

Wenotethatα Bifandonlyifα β.

Therelation whichwehavejustdefinedisobviouslyreflexive;itisantisymmetricbyLemma4.61;itistransitivebyLemma9.10;henceitisanorderrelationamongtheordinalnumbers.Next, we are able to show that any two ordinal numbers are comparable. That is, ifα and β are

ordinals,thenα βorβ α.Thisfactfollowsimmediatelyfrom4.62.Moreover:

9.13TheoremEverynonemptyclassofordinalnumbershasaleastelement.

Proof.Let beanonemptyclassofordinalnumbers,andletαbeanarbitraryelementof .Ifαistheleastelementof ,wearedone;otherwise,let Itfollowsfromourdefinitionoftherelation<thatevery issimilartoaninitialsegmentofα.Foreach ,letϕ(β)betheleastelementx∈αsuchthatβ Sx.Nowtheset{ϕ(β): }hasaleastelementϕ(δ)becauseitisasubsetofα.Wewillshowthatδistheleastelementof .Indeed, let ; thenϕ(δ) ϕ(β),hence , soby4.63, .Thuswe

have

so,by9.10,δ β.

Thus,theclassofalltheordinalnumbersiswellordered.

9.14Theorem

i) Ifβ>0,thenα<α+β.ii) β α+β.

Proof

i) LetAandBbewell-orderedsetssuchthatα= Aandβ= B.IfbistheleastelementofB,thenclearlyAistheinitialsegmentSbofA⊕B.ThusA A⊕B,soα<α+β.

ii) B A⊕B,henceby4.63,BisisomorphicwithA⊕BoraninitialsegmentofA⊕B.ThusBA⊕B,soβ α+β.

9.15TheoremLetαandβbeordinalssuchthatα<β.Thenthereexistsauniqueordinalγ>0suchthatα+γ=β.

Proof.IfAandBarewell-orderedsetssuchthatα= Aandβ= B,thenA B;thatis,A Sxforsomex∈B.LetC=B−Sx;CiswellorderedandC≠Ø,soifγ= C,thenγ>0.NowB=Sx⊕C,α= Sx(becauseSx A),soβ=α+γ.Foruniqueness,supposeβ=α+γ=α+γ′,where,say,γ<γ′;thatis,γ′=γ+δ(δ>0).Then

whichisincontradictionwiththeresultofTheorem9.14(i).Thusγ=γ′.

9.16TheoremForanyordinalnumbersα,β,γ,thefollowingruleshold:

Proof.In(i),(iii),(v),and(vii)weassumethatα<β,henceweassumethatthereexistsδ>0suchthatβ=α+δ.[Note:In(iii)and(vii),thecaseα=βiseasilydisposedof;indeed,ifα=β,thenα+γ=β+γandαγ=βγ(seeExercise1,ExerciseSet9.2).]

i) γ+β=γ+(α+δ)=(γ+α)+δ>γ+α[thislastrelationisaconsequenceof9.14(i)],soγ+β>γ+α.

ii) Supposeγ+α<γ+β.Ifα=β,thenγ+α=γ+β(Exercise1,ExerciseSet9.2).Ifβ<α,thenγ+β<γ+αby(i).Henceα<β.

iii) Suppose,onthecontrary,thatβ+γ<α+γ;thatis,α+(δ+γ)<α+γ.Thenδ+γ<γby(ii),andthisisimpossibleby9.14(ii);thusα+γ β+γ.

iv) Supposeα+γ<β+γ.Ifα=β,thenα+γ=β+γ.Ifβ<α,thenβ+γ α+γby(iii).Thusα<β.v) γβ=γ(α+δ)=γα+γδ>γα[thislastrelationholdsby9.14(i)].

vi) Supposeγα<γβ.Ifα=β,thenγα=γβ.Ifβ<α,thenγβ<γαby(v).Thusα<β.vii) Wemustshowthatαγ (α+δ)γ.Letα= A,γ= C,δ= D;A⊆A⊕D,so

Itfollowsby4.63that

orαγ (α+δ)γviii) Supposeαγ<βγ.Ifα=β,thenαγ=βγ.Ifβ<α,thenβγ αγby(vii).Henceα<β.

9.17Theoremi) Ifγ+α=γ+β,thenα=β.ii) Assumeγ>0.Ifγα=γβ,thenα=β.

Theproof,animmediateconsequenceof9.16(i)and(v),isleftasanexerciseforthereader.

9.18LemmaIfγ<βα,thenthereexistordinalsδandεsuchthatγ=βδ+ε,δ<α,andε<β.

Proof.LetA,B,andCbewell-orderedsetssuchthatα= A,β= Bandγ= C.OurassumptionisthatC B A,thatis,C S(b,a)forsome(b,a)∈B A.

LetE={(x,a):x<b},thatis,E=Sb {a};clearlyE Sb.Wewillshowthat

(thisrelationisillustratedinFig.9.2).

Fig.9.2

Letx∈B,y∈A.Then

Nowit iseasytoverifythattheorderingof(B Sa)⊕E is thesameas theorderingofS(b,a); thedetailsareleftasanexerciseforthereader.Weconcludethat

Letε= E= Sbandδ= Sa.Bythedefinitionoftherelation<,ε<βandδ<α.Nowγ= C= S(b,a);thusγ=βδ+ε.

Itisveryusefultonotethatthe“divisionalgorithm”forthenaturalnumberscanbegeneralizedtoalltheordinalnumbers.

9.19TheoremIfαandβ>0areordinals,thenthereexistuniqueordinalsξandρsuchthatα=βξ+ρandρ<β.

Proof

Existence.Sinceβ 1,wehaveβα α.Ifβα=α,wearedone;otherwise,α<βα,sobyLemma9.18thereexistordinalsδ<αandε<βsuchthatα=βδ+ε,henceagainwearedone.

Uniqueness.Assumeα=βξ+ρ=βξ′+ρ′,where(say)ξ′<ξ;thatis,ξ=ξ′+μ(μ>0).Then

soρ′=βμ+ρ>βμ β,whichcontradictsourassumptionthatρ′<β.Thuswecannothaveξ′<ξ;bysymmetry,wecannothaveξ<ξ′;thusξ=ξ′.Itfollowsby9.17(i)thatρ=ρ′.

Ifαisanordinalnumber,itiseasytoseethatα+1istheimmediatesuccessorofα.Nowletβbeanon-zeroordinalnumber;ifβhasnoimmediatepredecessor—thatis,ifβisnotequaltoα+1foranyordinalα—thenβiscalledalimitordinal.Otherwise—thatis,ifβhasanimmediatepredecessor—thenβiscalledanonlimitordinal.Limitordinalshavethefollowingusefulproperties.

9.20Theoremi) Ifαisalimitordinal,thereexistsauniqueordinalξsuchthatα=ωξ.ii) Ifαisanonlimitordinal,thereexistsauniqueordinalξandauniquefiniteordinaln≠0suchthatα

=ωξ+n.

Proof.Wewillprovetheexistenceassertions;theuniquenessassertions'areleftasanexerciseforthe

reader.

i) ByTheorem9.19,thereexistuniqueordinalsξandρsuchthatα=ωξ+ρandρ<ω.Butifρ<ω,thenρmustbefinite.Butthenρmustbe0;forotherwiseρ=m+1forsomefinitem,henceα=ωξ+m+1wouldnotbealimitordinal.

ii) Asabove,α=ωξ+ρ,whereρisfinite.Nowρ≠0;forifρ=0,thenα=ωξ,whichisimpossiblebecauseωξisalimitordinal(seeExercise6,ExerciseSet9.3).

EXERCISES9.3

1. Provethat1+α=αifanonlyifα ω.2. Anordinalnumberρ>0iscalledirreducibleifthereexistsnopairofordinalsα,βsuchthatα<ρ,

β<ρ,andα+β=ρ.Provethefollowing:a) Anordinalρisirreducibleifandonlyifπ+ρ=ρforeveryordinalπ<ρ.b) Supposeρ>1andε>0;ερirreducible⇒ρirreducible.c) Ifρisirreducibleand0<μ<ρ,thenthereexistsanirreducibleordinalξsuchthatρ=μξ.d) Supposeα>0;thesetofallirreducibleordinals αhasagreatestelement.[Hint:Considerthe

setofallβsuchthatα=ρ+βforsomeirreducibleρ.]3.Showthatanordinalαisalimitordinalifandonlyif

4. Letγbeanonlimitordinal.Provethefollowing.

5. Letβ≠0.Provethatα+βisalimitordinalifandonlyifβisalimitordinal.6. Letα,β≠0.Provethatαβisalimitordinalifandonlyifαisalimitordinalorβisalimitordinal.

[Hint:UseExercise3andLemma9.18.]7. Provethatnω=ω,∀n∈ω.[Hint:UseTheorem9.19to“divide”ωbyn.]8. Letα≠0.Provethatαisalimitordinalifandonlyifnα=α,foreveryfiniten.[Hint:UseExercise

7andTheorem9.20(i).Fortheconverse,useExercise4(b).]9. a) Useinductiontoprovethatifγisaninfiniteordinal,then(γ+1)n=γn+1forallfiniten.[Use

Exercise1.]b) Provethat∀γ>0,(γ+1)ω=γω.[Hint:Ifγisinfinite,assumeγω<(γ+1)ωanduseTheorem

9.18toarriveatacontradiction.Ifγisfinite,useExercise7.]c) Concludethatifβisanylimitordinal,then

10. Letαbeanonlimitordinal.Provethat∀γ>0,(γ+1)α>γα.[Hint:UseExercise4.]11. Ifαisaninfiniteordinalandβ≠0isanonlimitordinal,provethat

[Hint:UseTheorem9.20(ii)andassumeαβ+1<(α+1)βtoarriveatacontradiction.UseExercise9.]

12. a) Ifαisalimitordinal,provethatα=sup{β:β<α}.[UseExercise3.]b) Ifαisalimitordinalandβisanyordinal,provethat

[Hint:Ifγisanupperboundof{β+μ:μ<α},thenγ>β,thatis,γ=β+δ;δprovestobeanupperboundof{μ:μ<α}.]

13. Ifαisalimitordinalandβisanyordinal,provethat

[Hint:Ifγisanupperboundof{βμ:μ<α},thenγ=αδ+ρ(ρ<α).Notethatμ<α⇒γ α(μ+1)andconcludethatδisanupperboundof{μ:μ<α}.]

14. ProveTheorem9.17.15. ProvetheuniquenessassertionsofTheorem9.20.

4THEALEPHSANDTHECONTINUUMHYPOTHESIS

In Chapter 8 it was proven that the relation among cardinals is a well-ordering; hence there is asmallest infinite cardinal, a next greater infinite cardinal, and so on; every infinite cardinal has auniquelydeterminedimmediatesuccessor.Itfollowsthattheinfinitecardinalnumberscanberankedin“first, second, third,…” order. This opportunity of ranking the cardinals—and using the ordinals todesignatetheranks—hasvaluableapplicationsinmathematics.Wewillproceedtoshowthatthereisanisomorphismbetweentheclassofalltheinfinitecardinals

andtheclassofalltheordinals.

9.21TheoremLetICbetheclassofalltheinfinitecardinalsandletORbetheclassofalltheordinals.ThereexistsanisomorphismfromICtoOR.

ProofWebeginbynotingthateveryinitialsegmentofICisaset.Ifaisaninfinitecardinal,then{b∈IC:b<a}isaninitialsegmentofICdenotedherebyIa.Wewell-ordera,andforeachx∈a,Sx={u∈a :u<x} isan initial segmentofa.Letϕ :a→ Iabedefinedbyϕ(x)=#Sx for eachx∈a.ϕ issurjective,becauseifb∈Ia, (henceb<a) thenb is isomorphic (henceequipotent)withsome initialsegmentofa.(Thealternative,namelyaisomorphicwithsomeinitialsegmentofb,contradictsb<a).Thusϕissurjective.Sinceaisaset,IaisasetbyAxiomA9.Likewise,everyinitialsegmentofORisaset.

Now IC andOR are bothwell-ordered classes, hence by 4.62, exactly one of the following threecasesmusthold:(a)IC OR,(b)ICissimilartoaninitialsegmentofOR,(c)ORissimilartoaninitialsegmentofIC.Supposeforamomentthat(c)holds.Fromthepreviousparagraph,everyinitialsegmentofICisaset,henceby2.36,ORisaset.ButwehaveprovedthatORisaproperclass,hence(c)cannothold.Analogously,(b)cannothold,whichprovesthat(a)holds.

Ifaisaninfinitecardinalnumber,theordinalΩ(a)iscalledtheordinalrankofa.NotethatΩisanisomorphism; thus, if a is the least infinite cardinal, then Ω(a)= 0; if a is the next greater infinitecardinal,thenΩ(a)=1,andsoon.Theinfinitecardinalsareoftencalledalephs.Ifaisaninfinitecardinalandα=Ω(a),wefrequently

write

Thusthefirstfewinfinitecardinalsare 0, 1, 2,…Theorem9.21hasthefollowingsimpleconsequences:

9.22i) α= β⇒α=β,ii) α=β⇒ α= β,

iii) α< β⇒α<β,iv) α<β⇒ α< β

Wehaveseenthateveryinfinitecardinalnumberahasanimmediatesuccessor—butwhatexactlyisthe immediate successor ofa?We know that 2a is greater than a, but is there any cardinal numberbetweenaand2a?Letusaskamorespecificquestion:Wehaveseenthat 0isthecardinalnumberofdenumerablesets(forthecardinalnumberofωistheleastinfinitecardinal),andthat2 0isthecardinalnumberoftherealnumbers;isthereacardinalnumberbetweenthesetwo?Theearlyset theoristsproposedthehypothesis that there isnocardinalbetween 0and2 0 ,and

nameditthecontinuumhypothesis—foritisequivalenttosayingthateverysetofrealnumberswhichisnotdenumerablehasthepoweroftherealnumbers,calledthe“powerofthecontinuum.”Anobviousextensionofthisconjectureisthestatement:Foreveryinfinitecardinalnumbera,thereisnocardinalbetweenaand2a;thisiscalledthegeneralizedcontinuumhypothesis.

ContinuumHypothesis.Theredoesnotexistanycardinalcsuchthat 0<c<2 0

GeneralizedContinuumHypothesis.Ifaisanyinfinitecardinal,theredoesnotexistanycardinalcsuchthata<c<2a.

It has been proven in recent years that the continuum hypothesis and the generalized continuumhypothesiscannotbeproven from theotheraxiomsof set theory,anddonotcontradict these.Hencetheir status is analogous to thatofEuclid’s “FifthPostulate” ingeometry.Wemaypostulate themordenythem,ineachcasegettingaconsistenttheoryofcardinalnumbers.

EXERCISES9.4

1. Provethatthegeneralizedcontinuumhypothesisisequivalentto

2. Assumingthegeneralizedcontinuumhypothesis,provethefollowing:

3. Assumingthegeneralizedcontinuumhypothesis,provethefollowing:

FurtherproblemsonthealephsaregiveninExerciseSet9.5.

5CONSTRUCTIONOFTHEORDINALSANDCARDINALS

Wesaid,intheintroductiontothischapter,thatitispossibletoconstructsetswhichsatisfyConditions01and02of theAxiomofOrdinality.Thechiefpurposeof thisconstruction is toprove thatwecandispensewiththeAxiomofOrdinalitybyactuallyproducingthesetswhoseexistencetheaxiomasserts.Our process of construction is basedupon the same idea—outlinedonpage125—thatweused to

constructthenaturalnumbers.Webeginbydefining

IfAisaset,wedefinethesuccessorofAtobethesetA+,givenby

Thus0=Ø,1=0+,2=1+,andsoon.Thistime,however,wewillgofurtherthanwedidinChapter7.Startingwithω,wedefine

Then,startingwithω+ω=ω2,weget

andsoon.This is thebasic idea ofour constructionprocess, butwewill notproceedexactly in this fashion.

Instead of starting with 0 and constructing successive sets one by one, we will define all the setssimultaneously.Thiscanbeaccomplishedinthefollowingway.The“elementhood”relation∈isnot,generallyspeaking,anorderrelation;forexample,ifx∈Aand

A∈B,itdoesnotnecessarilyfollowthatx∈B.However,therearespecialcaseswhere∈doesbehaveasifitwereanorderrelation;oneofthesecasesconcernsushere.

9.23DefinitionAclassAissaidtobe∈-orderedifitisorderedbytherelation∈.Itisunderstoodherethattherelation∈isastrictorderrelation<.Consequently,tosaythatAis∈-orderedistosayithasthefollowingproperties∀a,b,c∈A:

a)a∈a.b)a∈b⇒b∈a.c)(a∈b)∧(b∈c)⇒a∈c

9.24Remark.Itisimmediatethatthetransitivelaw(c)holdsiffb∈c⇒b⊆c.

ByDefinition6.5, a setA is said to be transitive if: (x∈A)∧ (y∈ x)⇒ y∈A. Note that thispropertyisnotthesameassayingthat(c)aboveholdsinA.Property(c)tellsusthateveryelementofAistransitive,butnotthatAitselfistransitive.ItisimmediatethatAistransitiveiffx∈A⇒x⊆A.Hereisour“abstract”definitionofordinals:

9.25DefinitionA setA is called anordinal ifA is transitive and ordered by∈. Thismeans that∈satisfiesthethreeconditionsofDefinition9.23onA,andAisatransitivesetbyDefinition6.5.

9.26LemmaLetαbeanordinal.i) Everyelementofanordinalisanordinal.ii) Ifx∈α,thenx=Sx,whereSxistheinitialsegment{y∈α:y∈x}.

Proof

i) Supposex∈α;showthatxisanordinal.Becauseαistransitive,x⊆α.Soifa,b,careelementsofx,theyarealsoelementsofα,hencesatisfyconditions(a),(b),(c)of9.23.Likewise,ifu,v,wareelementsofx,theyareelementsofα,henceu∈v∈x⇒u∈x.

ii) Ify∈xthen(becauseAisatransitiveset)y∈α.Thusy∈Sx.Conversely,ify∈Sx thenbythedefinitionofSx,y∈x.Thus,x=Sx.

FromLemma9.26wedrawtwoimportantconclusions:(1)Everyordinalisasetofordinals,and(2)Everyordinalαisthesetofallordinalsβ<α.

WeletthesymbolORstandfortheclassofalltheordinals.Sinceordinalsaresets,∈isarelationbetweenthem,andourfirstobjectiveistoshowthatORsatisfies9.23(a),(b)and(c).Weaimalsotoshowthatanytwoordinalsαandβarecomparablebytheorderrelation∈.Infact,let’sbeginwiththat:

9.27TheoremLetαandβbeordinals.Ifα≠β,theneitherα∈βorβ∈α.

Proof.Weshallcalltwosetsxandyincomparableifx≠y,x∉yandy∉x.LetAbe thefollowingsubsetofOR:A={x∈OR:∃y∈ORsuchthatxandyare incomparable}.WewillshowthatAisempty.WereasonbycontradictionandassumeAisnonempty.AtthispointweshallmakeourfirstuseoftheAxiomofFoundation(A8),whichstatesthateverynonemptysetAcontainsanelementdisjointfromA.Thatis:∃a∈Asuchthata A= .Inthatcase,thesetB={y∈OR:yisincomparablewitha}isnotempty.Asabove,by(A8)thereisanelementb∈Bsuchthatb B= .Itwillnowbeshownthata⊆b.

Ifz∈athenzisanordinal,andz∉Abecausea A= .Foreveryy∈OR,yandzarecomparable.Inparticular,bandzarecomparable.Nowifitwerethecasethatz=bthatwouldimplyb∈ahenceb∉B,whichiscontradiction.Similarly,ifitwerethecasethatb∈z,thenb∈abecauseaistransitive,henceb∉B.Againacontradiction.Thus,z∈b,andsincezwasanarbitraryelementofa,thisproves

that a⊆b. In the same fashion you can show b⊆A, hence a = b. This contradiction proves thetheorem.

9.28LemmaIfαandβareordinalsthenα∈β⇔α⊂β.(α⊂βmeansα⊆βbutα=β.)

Proof.HalfofthisclaimistruefromRemark9.24with9.23(a).Nowsupposeα⊂β:Itmustbeshownthatα∈β.ByTheorem9.27,eitherα∈βorβ∈α.Thelatterisimpossible,becausefromα⊆βwewouldgetβ∈β,contraryto9.23(a).Thusα∈β.

TheclassofalltheordinalsisdenotedbyOR.Therelation∈isawell-definedrelationonOR,andwewishtoshowthat∈isanorderrelationonOR.Inordertodothis,wemustshowthat∈satisfiesProperties9.23(a),(b)and(c)onOR.WealsoshowthatORisatransitiveclass.

9.29TheoremTheclassORofalltheordinalsisorderedby∈.

Proof.Letα,β,γ∈OR:

Ifα∈αthenαisanelementofanordinal,hencesatisfies9.23(a).Thus,α∉α. Weshowthetransitivelawfirst:Supposeα∈βandβ∈γ:WemakethreeusesofLemma9.28:α⊂β

andβ⊂γhenceα⊂γsoα∈γ. Supposeα∈β.Ifβ∈αthen(becauseeveryordinalistransitive)α∈α,whichisfalsefrom(a).

9.30LemmaIfαisanordinalthenα+=α {α}isanordinal.Theproofofthislemmaisasimpleverification,andisleftasanexercise.

9.31Theorem

a) Ifαisanordinalthenα=Sα.

b) ORisatransitiveclass.

Proof

a) Itmustbeshownthatifαisanordinalthenα=Sα.From9.26(ii),allweneedistoshowthatαisanelementofanordinal.Itis,becauseα∈α+.

b) Provethatx∈OR⇒x⊂OR.Inotherwords,ifxisanordinal,thenxisasetofordinals.ThisfollowsfromPart(a).

9.32TheoremEverynon-emptyclassAofordinalshasaleastelement.

ProofItwillbeshownthatifAisanyclassofordinals,then Aisanordinal,isanelementofA,andisthereforetheleastordinalinA.Ifx∈ Athenforanyy∈A,x∈y.Sonow,ifz∈xthenz∈ybecauseordinalsaretransitive.Sincethisistrueforeachy∈A,itfollowsthatz∈ A.Thisprovesthat A istransitive.It isclear that A isorderedby∈,becauseeachordinal inAsatisfies9.23(a), (b)and(c),hencesodoestheirintersection.Thus, Aisanordinal.

Finally,either A=A(whichimpliesAisasingleton{α}soα∈AistheleastelementofA)or A⊂A.Thenfrom9.28, A∈A.Itisimmediatethatifγ∈Athen A⊂γ,so A∈γ.Thus A is theleastelementofA.

Thistheoremisveryimportant,becauseitshowsthattheclassORofallordinalsiswell-orderedbytherelation∈.Moreover,sincefrom9.26(i)everyordinalα isasubsetofOR, thisshows thateveryordinaliswell-orderedby∈.OneessentialfactaboutORisthis:

9.33TheoremORisaproperclass.

Proof.WehaveshownthatORhasallthepropertiesofanordinal.ThusifORwereaset,itwouldbeanordinal,andthenwewouldhaveOR∈OR,whichisimpossibleby9.23(c).

Finally,wecometothemostimportantpropertyofOR—thereasonforinventingordinalsinthefirstplace:

9.34TheoremEverywell-orderedsetAisisomorphicwithauniqueordinal.

Proof Since A and OR are well-ordered classes, it follows from Theorem 4.62 that either A isisomorphicwithaninitialsegmentofOR,orORisisomorphicwithaninitialsegmentofA,orORisisomorphicwithA.Thelasttwoareimpossible,becausefromAxiomA9,thiswouldmakeORaset,whereasORisaproperclass.Thus,AisisomorphictotheinitialsegmentSα=αofOR.Theproofofuniquenessisleftasanexercise.

9.35Remark.Ordinalnumbershavemanyapplicationsinmathematics.Hereisabriefsummaryofthehighpointsthatitisimportanttoretain:Everyordinalαisan∈-well-orderedsetequaltothesetofallordinalsprecedingit:α={β∈OR:β<α}.Moreover,α∈β⇔α⊂β.Thus,whenwethinkofordinalsassets(whichtheyare),eachordinalisachainofsets:αisthechainofallthesetsβ⊂α.Asaset,α={β∈OR:β⊂α}.Infact,ifAisanyclassofordinals,thenAmaycorrectlybeviewedasachainof

sets.Consequently,anyboundedclassAofordinalshasaleastupperboundsup(A)= A.

9.36Remark.LetC⊆ORbeaclassofordinalnumbers,andsupposethefollowingthreeconditionshold:(a)0∈C.(b)Ifα∈Cthenα+∈C.(c)Letαbealimitordinal:Ifβ∈Cforallβ<αthenα∈C.ThenC=OR.Thisfactiseasytoprove:Indeed,ifC≠OR,letDbetheclassofallordinalsγsuchthatγ ∉ C. By assumption, D ≠ , so from 9.32, D has a least element μ. It is easy now to get acontradiction.

IfAisanarbitraryset,considertheclassofalltheordinalsequipotentwithA;thisclasshasauniqueleastelement,whichwecall the initialordinalequipotentwithA. It is trivial,now, toverify that theclassofalltheinitialordinalssatisfiesConditionsK1andK2oftheAxiomofCardinality.Thuswearejustifiedinmakingthefollowingdefinition.

9.37DefinitionByacardinalnumberwemeananinitialordinal.

Thus,theclassCDofthecardinalnumbersistheclassofalltheinitialordinals.Wehavethusfulfilledourpromiseofactuallyproducingsetstoserveasthecardinalnumbersandthe

ordinalnumbers.We noted earlier that every natural number is a transitive,∈-well-ordered set, that is, an ordinal

number. It is immediate, too, that every natural number is an initial ordinal number. Thus, in ourconstruction,thenaturalnumberscoincidewiththefiniteordinals,aswellaswiththefinitecardinals.

ThereadershouldnotethateverythingwehaveprovedinChapters8and9abouttheclassCDofthecardinalnumbershasdependedsolelyuponConditionsK1andK2oftheAxiomofCardinality.Thuseverything we have already said about the cardinals holds, without any alteration, for the class CDdefinedby9.37.Inparticular,9.22stillholds,thatis,

isanisomorphismbetweenORandtheclassofalltheinfinitecardinals.It is important to note that by 9.37, α is both a cardinal and an ordinal (specifically, an initial

ordinal).Inordertoavoidanyconfusion,itiscommonpracticeinmathematicstowrite

foreveryinfinitecardinal α ,andtotreat αasacardinalnumberandωαasanordinalnumber.Inotherwords,thenumberinquestionisdenotedby αwhenitisusedasacardinal,andbyωαwhenitisusedasanordinal.Forexample, α+ βdesignatesthecardinalsumofthetwonumbers,whereasωα+ωρdesignatestheirordinalsum.

EXERCISES9.5

1. Foreachordinalnumberα,provethatα ωα;concludethat#α α.[Hint:Use4.58.]2. Provethatifβisalimitordinal,then β=sup{ γ:γ<β}.[Hint:Use9.22.]

3. Provethat .

4. Ifμisalimitordinal,provethat .

10TransfiniteRecursion.SelectedTopicsintheTheoryofOrdinalsand

Cardinals

1TRANSFINITERECURSION

InChapter6wediscussedthenotionsoffiniteinductionandfiniterecursion;finiteinductionismethodof proof by induction and is familiar to most students of elementary algebra. Finite recursion is amethodofdefinitionbyinduction.Proofbyinductionworksasfollows:Atheoremisfirstshowntobetruefor0.Thenitisshownthatifthetheoremistrueforn,itmustlikewisebetrueforn+1.Wearethen able to conclude that the theorem is true for all natural numbers. Induction rests on the sameprinciple,butisawayofdefiningafunction.First,wedefinethevalueofthefunctionat0.Thenweuseitsvalueatntodefineitsvalueatn+1.Inthisway,itsvalueisdefinedforeverynaturalnumber.Ourpurposehereistogobeyondthenaturalnumbersandusethesameprincipleontheclassofall

the ordinals. Transfinite induction has been described briefly in Section 4.5. The chief differencebetweenconventionalinductionandtransfiniteinductionisthattherearetwokindsofinductionstep:Oneforthecasewhereαisasuccessorordinal,andoneforthecasewhereαisalimitordinal.Thetwocasesmustobviouslybetreateddifferently,becauseifαisasuccessorordinal,sayα=β+,

thenthetaskistoshowthatifapropertyistrueforβthenitistrueforβ+.Ontheotherhand,ifαisalimitordinal,thetaskistoshowthatifthepropertyistrueforallβ<α,thenitistruealsoforα.Thedifferencebetweeninductionandrecursionisclear:Ininductiontheobjectiveistoprovethata

propertyP(α) is trueforeveryordinalα. In recursion theobjective is toassignavalue toa functionF(α)foreveryordinalα.Theideaisthesame,buttheaddedcomplicationisthatthevalueyouassignisinasetAwhichistherangeofF.Supposeα isasuccessorordinal,α=β+.Theidea,inrecursionasininduction,isthatthevalueof

F(β+) is obtaineddirectly from thevalueofF(β)—inotherwords, by performing someoperation onF(β).IfGisthesymbolassignedtothatoperation,then

Ontheotherhand,ifαisalimitordinal,whatyoudotoallthevaluesF(β),β<αinordertogetF(α)istoassignthe“nextvalue”toF(α),whichistheirleastupperbound:

Aswehavejustseen,GisanoperationontherangeofF.Sothenextquestionis:WhatistherangeofF?Let’susethesymbolAfortherangeofF.DifferentchoicesofAgiveyoudifferentversionsoftherecursiontheorem.Youwanttobeasnoncommittalaspossiblesoastoallowtherecursiontheoremtobeusedinawiderangeofapplications.So,let’ssaythatAisanysubclassoftheclassVofallsets.Sinceanythinginmathematicscanbeconstruedasaset,thisisverysatisfactory.Givenanysetofsets,theirleastupperboundisusuallytheirunion.Thus,ourinductionstepforlimitordinalsmaybewrittenthus:

Itisgoodtorecallhere,from9.35,thatORisachainofsetsorderedby⊂.Thus,anysubclassC⊆ORislikewiseachainofsets,soitsunionisitsleastupperbound.Thatis,supC= C.ItisalsogoodtorecallthateveryordinalαisequaltotheinitialsegmentSαofOR,andsinceSα isachainofsets,

.Soherewehaveit:

TransfiniteRecursionTheoremLetA⊆Vbeanyclassofsets,letG:A→Abeafunction,andleta∈A.ThenthereexistsauniquefunctionF:OR→Asuchthat

Proof.Theproofwillflowsmoothlyifwethinkofallourfunctionsassetsoforderedpairs—whichisactuallywhattheyare.Ourtasknowistoshowthatforeveryordinalαthereisexactlyoneelementxαsuchthattheorderedpair(α,xα)isinF.Inthemind’seye,weshouldthinkofFasbeingbuiltupstagebystage—addinganeworderedpairateachstage—untilFis“full”.Condition(i)ofthetheoremtellsusthatthepair(0,a)isinF.Thus,Fisnotempty.Ifdom(F)=OR,

wearedone.Ifnot,letαbetheleastordinalsuchthatnopair(α,xα)isinFforanyelementxα. (UseTheorem9.32.)Considertwocases:

a) αisasuccessorordinal,α=β+.Byassumption,Fcontainsanorderedpair(β,xβ)forsomeelementxβ.ThenfromCondition(ii), theorderedpair(β+,G(xβ)) is inF:Thiscontradicts theassumptionthatnopair(α,xα)isinF.

b) αisalimitordinal.Thenbyassumption,forallβ<α,thereareorderedpairs(β,xβ) inF.Thatis,{(β,xβ):β<α}⊆F.Thenbytheruleofinduction(iii),thereis suchthat(α,xα)∈F.Thisshowsthatforeveryα,somepair(α,xα)isinF.

Weuse inductiononα toprove that foreachordinalα, there isonlyonexα such that (α,xα)∈F.Supposethatforallβ<α,thereisjustoneorderedpair(β,xβ)∈F.Thatis,if(β,xβ)∈Fand(β, )∈Fthenxβ= .Letαbeasuccessorordinal,α=β+,andsupposetherearedifferentelementsxαandsuchthat(α,xα)∈Fand(α, )∈F.FromCondition(ii), .Nowlet α be a limit ordinal, and suppose (α, xα) ∈ F and (α, ) ∈ F. From (iii),

.ThisprovesthatthesetFoforderedpairsisafunction.From9.36,thedomainofFisallofOR.IthasthusbeenshownthatFisafunctionwhosedomainis

ORandrangeisinA,andthatConditions(i)–(iii)determineFuniquely.

Remark.ThereasoninginParagraph3ofthisproofmaybeinterpretedasfollows:Foreveryordinalα,ifF SαistheonlyfunctionwithdomainSαthatsatisfiesConditions(i)–(iii),thenF Sα+=(F Sα)∪(α,xα)istheonlyfunctionwithdomainSα+thatsatisfies(i)–(iii).But ,soFistheuniquefunctionwithdomainORthatsatisfies(i)–(iii).

Just as finite recursionwas used inChapter 6 to define the addition andmultiplication of naturalnumbers, transfinite recursion may be used to give alternative definitions for the addition andmultiplicationofordinalnumbers.

10.2DefinitionForanyarbitraryα∈OR,wedefineafunctionσα:OR→ORasfollows:

a) σα(0)=α,

b) σα(β+)=[σα(β)]+,c) σα(β)=sup{σα(γ):γ<β}ifβisalimitordinal.

Theorem10.1guaranteestheexistenceofauniquefunctionσαsatisfying(a),(b),(c).

10.3TheoremForarbitraryordinalsα,β,σα(β)=α+β.

Proof.Ifβ=0,thenσα(0)=α=α+0.Byinduction,letussupposenowthatthetheoremholdsforallγ<β.Ifβisalimitordinal,then

Ifβisanonlimitordinal,β=δ+,then

Theorem10.3tellsusthatifwedefinetheadditionofordinalnumbersby

then10.4isequivalenttoDefinition9.8.10.4mayalsobewritteninthefollowingform:

Since10.5isequivalenttoDefinition9.8,wewillhenceforthconsider10.5todefinetheadditionofordinalnumbers.

10.6DefinitionForanarbitraryα∈OR,wedefineafunctionπα:OR→ORasfollows:

a) πα(0)=0,

b) πα(β+)=πα(β)+α,c) πα(β)=sup{πα(γ):γ<β}ifβisalimitordinal.

Theorem10.1guaranteestheexistenceofafunctionπαsatisfyingtheconditions(a),(b),and(c).

10.7TheoremForarbitraryordinalsα,β,πα(β)=αβ.

Proof.Ifβ=0,thenπα(0)=0=α0.Byinduction,letussupposethatthetheoremholdsforeveryγ<β;thatis,πα(γ)=αγ,∀γ<β.Ifβisalimitordinalthen

Ifβisanonlimitordinal,β=δ+,then

Theorem10.7tellsusthatifwedefinethemultiplicationofordinalnumbersby

then10.8isequivalenttoDefinition9.8.10.8mayalsobewritteninthefollowingform:

Since 10.9 is equivalent to 9.8, we will henceforth consider 10.9 to be the definition of ordinalmultiplication.

2PROPERTIESOFORDINALEXPONENTIATION

10.5and10.9provideuswithanalternativewayofdefiningtheadditionandmultiplicationofordinalnumbers, using transfinite recursion. We will use this new (and in many ways, more convenient)approachtodefineordinalexponentiation.

10.10DefinitionIfα≠0isanyordinal,wedefineafunctionηα:OR→ORasfollows:

a) ηα(0)=1.

b) ηα(β+)=ηα(β)α,c) ηα(β)=sup{ηα(γ):γ<β}ifβisalimitordinal.

Theorem10.1guaranteestheexistenceofafunctionηα:OR→ORwhichsatisfiesconditions(a),(b),and(c)above.

10.11DefinitionWedefineordinalexponentiationasfollows:Ifαandβarearbitraryordinals,welet

and0β=0.Inviewof10.10,10.11maythereforebewrittenasfollows:

10.12a) α0=1,

b) αβ+=(αβ)α,

c) αβ=sup{αγ:γ<β},ifβisalimitordinal,

d) 0β=0.

Wewillnowdevelopthefundamentalpropertiesofordinalexponentiation.

10.13LemmaIfγisalimitordinalandε<αγ,then .

Proof.Suppose,onthecontrary,that∀π<γ,απ ε;thismeansthatεisanupperboundoftheset{απ:π<γ};butαγisthesupofthissameset,soαγ ε,whichcontradictsourassumptionthatε<αγ.Thusforsomeπ<γ,ε<απ.

10.14TheoremForanyordinalnumbersα>1,β,γ,

i)β<γ⇒αβ<αγ,ii)αγ<αβ⇒γ<β.

Proofi) Theproofisbyinductionofγ.Ifγ=0,theconditionissatisfiedvacuously.Nowsupposethat(i)

holds∀δ<γ,thatis,

Supposefirstthatγ isalimitordinal.Ifβ<γ,thenβ+1<γ,andbythehypothesisof inductionwehave

butαγ=sup{αβ:β<γ},soαβ+1 αγ;thusαβ<αγ.Nowsupposethatγisanonlimitordinal,γ=δ+1.Ifβ<γ,thenβ=δorβ<δ.Ifβ=δ,thenwehave

Ifβ<δ,thenbythehypothesisofinduction,αβ<αδ,sowehave

ii) αγ αβ⇒γ βisthecontrapositiveof(i).Nowsupposeαγ<αβ;ifγ=β,thenαγ=αβ,henceγ<β.

10.15TheoremForanyordinalnumbersα,β,andγ,

i) α β⇒αγ βγ,

ii) βγ<αγ⇒β<α.

Proofi) Theproof isby inductiononγ. Ifγ=0, the condition is satisfied trivially.Nowsuppose that (i)

holds∀δ<γ,thatis,

Suppose first thatγ is anonlimit ordinal, γ =δ+1.Weassumeα β and, by the hypothesis ofinduction,αδ βδ.Thus,by10.12(b)and9.16(v)and(vii),wehave

Next,supposethatγisalimitordinal;thenαγ=sup{αδ:δ<γ}.Ifδ<γ,thenbythehypothesisofinductionαδ βδ;butβδ βγbecauseβγ=sup{βδ:δ<γ},henceαδ βγforeveryδ<γ.Itfollowsthatβγisanupperboundof{αδ:δ<γ},henceαγ βγ.

ii) Thisissimplythecontrapositiveof(i).

10.16Theoremαβαγ=αβ+γforanyordinalnumbersα,β,andγ.

Proof.Theproofisbyinductiononγ;thetheoremholdstriviallyifγ=0,henceweassumethatαβαδ=αβ+δforeveryordinalδ<γ.

i) Letussupposefirstthatγisanonlimitordinal,γ=δ+1;then

ii) Nowletussupposethatγisalimitordinal;weshallprovethetwoinequalities(a)αβ+γ αβαγand(b)αβαγ αβ+γ.a)Ifγisalimitordinal,thenclearlyβ+γisalimitordinal,henceby10.12(c),

Nowifδ<β+γ,theneitherδ β,orifδ>β,thenby9.15,δ=β+ρforsomeρ<γ[ρ<γbecauseδ=β+ρ<β+γ⇒ρ< γ by9.16(ii)]. In the first case, namelyδ β, it follows by 10.14(i) and9.16(v)thatαδ αβ αβαγ.(Weassumethatα≠0,hence1 αγ;ifα=0,then10.16holdstrivially.)Inthesecondcase,namelyδ=β+ρwhereρ<γ,itfollowsby10.14(i)thatαρ<αγ,hencebythehypothesisofinductionand9.16(v),

Thus,ineitherofthetwocases,αδ αβαγforeveryδ<β+γ,soαβαγisanupperboundof{αδ:δ<β+γ},henceαβ+γ αβαγ.

b) Weareassumingthatγisalimitordinal;itfollowsveryeasily(seeExercise1(a)attheendofthissection)thatαγisalimitordinal,henceby10.9(c),

Nowifε<αγ,thenby10.13,∃π<γε<απ;henceby9.16(v),10.14(i),andthehypothesisofinduction,

Thusαβ+γisanupperboundof{αβε:ε<αγ},henceαβαγ αβ+γ.

10.17Theorem(αβ)γ=αβγforanyordinalsα,β,andγ.

Proof. The proof is by induction on γ. If γ = 0, the theorem follows trivially from 10.12(a). Let usassume,then,that(αβ)δ=αβδforeveryδ<γ.i) Supposefirstthatγisanonlimitordinal,γ=δ+1;thenby10.12(b), thehypothesisofinduction,

and10.16,wehave

ii) Next,weshallsupposethatγisalimitordinalandwewillprovethetwoinequalities(a)(αβ)γ αβγ

and(b)αβγ (αβ)γ.

Butifδ<γ,thenβδ<βγ,soby10.14(i),αβδ<αβγ;itfollowsthatαβγisanupperboundof{αβδ:δ<γ},so(αβ)γ αβγ.

b) If γ is a limit ordinal, then (see Exercise 6, Exercise Set 9.3) βγ is a limit ordinal; thus by10.12(c),

Nowifδ<βγ,thenby9.18,δ=βξ+ε,whereξ<γandε<β;thus

Nowξ<γandγisalimitordinal,soξ+1<γ;thus,bythehypothesisofinductionof10.14(i),

Thus,using10.14(i)onceagain,αδ<αβ(ξ+1)<(αβ)γ.Itfollowsthat(αβ)γisanupperboundof{αδ:δ<βγ},soαβγ (αβ)γ.

Notethatby10.12(c),2ω=sup{2n:n<ω}=ω.Wenotedintheprecedingchapterthatthe“usual”arithmeticlawsdonotallapplyto transfiniteordinalnumbers; inparticular, thecommutativelawformultiplicationdoesnothold,nordoestherightdistributivelaw.Asweshallnowsee,thelaw(αβ)γ=αγ

βγdoesnotapplygenerallytoordinalnumbers.

10.18Example

i) (2·2)ω=4ω=ωbecause4ω=sup{4n:n<ω}=ω.

ii) 2ω2ω=ωω=ω2.Sinceω<1,itfollowsby9.16(v)thatωω>ω1=ω;thus(2·2)ω≠2ω2ω.

EXERCISES10.2

1. Provethefollowing:

a) Ifγisalimitordinalandα>1,thenαγisalimitordinal.

b) Ifαisalimitordinalandγ≠0,thenαγisalimitordinal.2. Provethatforanyordinalsα>1,β,andγ,αβ=αγ⇒β=γ.3. Use10.5,10.9,and10.12toprovethatforanyfiniteordinaln.

a) n+ω=ω,b)nω=ω,c)nω=ω.4. Useinductiontoprovethatforeveryordinalnumberβ,2β β.Considerthatforeveryα>0andβ,

αβ β.5. Provethatifα>1andβ≠0,thenαβ αβ.

6. Prove that ifα is a limit ordinal and p andq are finite ordinals, then (αp)q = αqp. [Hint: UseExercise8,ExerciseSet9.3.]

7. Letalimitordinalγbecalledsimpleifitcannotbewrittenγ=δ+ωforanyordinalδ.Provethatγissimpleifandonlyif∀ε<γ,thereexistsalimitordinalλε<λ<γ.

8. Ifα is a denumerable ordinal (that is, if #α = ), use 9.8 to prove that αω is denumerable.Concludethattheordinalsω,ω2,…,ωn,…(nfinite)arealldenumerable.

9. a)Let{γi:i∈I}beasetofordinals;provethat

[Use9.25and9.26(ii).]b) Prove that if {γn :n∈α} is a set of denumerable ordinals,whereα is a denumerable, then

sup{γn:n∈α}isadenumerableordinal.Note:ItfollowsimmediatelyfromExercises2and9,ExerciseSet8.5that

c) UsetheresultofExercise8abovetoprovethatωωisadenumerableordinal.(Notethatωω=sup{ωn:n∈ω}.)

d) Concludesimilarlythat , ,etc.,aredenumerableordinals.

10. Provethatifαandβaredenumerableordinals,thenαβisadenumerableordinal.[Use9(ii)above.]

3NORMALFORM

It is a well-known fact of elementary number theory that every natural number n has a uniquelydetermineddecimalrepresentation.Thatis,givenn,thereexistuniquenaturalnumbersk,m0,m1,…,mk(eachmi<10)suchthat

Thedecimalrepresentationofnisalsocalleditsrepresentationwithbase10;itcaneasilybeshownthateverynaturalnumbernalsohasauniquerepresentationwithbaseb,foranyb>1.The idea of giving every number a representation with base b can easily be extended to ordinal

numbersgenerally.Inthesequelwewillonlyconsiderbaseω,butitshouldbecleartothereaderthatanyotherbasewilldo.

10.19DefinitionLetγbeanordinalnumber;supposetherearenonzeronaturalnumbersa1,…,anandordinalα1>α2>…>αnsuchthat

Then10.20iscalledanormalformrepresentationofγ.

10.21TheoremEveryordinalγ≠0hasanormalformrepresentation.

Proof.Theproofwillbebyinductiononγ.Ifγ=1,thenγ=ω01isanormalformrepresentationofγ.Nowassumethetheoremistrue∀ρ<γ.LetA={μ:ωμ>γ};Aisnonempty,asmayeasilybeseenbyusingExercise4,ExerciseSet10.2.ThusAhasaleastelementν;ων>γ.Supposeνisalimitordinal,ων=sup{ωδ:δ<ν}.Foreachδ<ν,ωδ γbecauseνistheleastelementofA;thusγisanupperboundof{ωδ:δ<ν},soων γ.Thisiscontrarytoourchoiceofν,henceν=α1+1forsomeordinalα1.

By9.19, ,where ;clearlyξ<ω,forifξ ωthen

which is impossible by 9.14(i). It follows that ξ is a natural number a1,so γ = , where.Bythehypothesisofinduction,thereexistnonzeronaturalnumbersa2,…,anandordinals

α2>…>αnsuchthat

clearlyα1>α2,forα1 α2,then

whichisfalsebecause .Thus,

wherea1,…,anarenonzeronaturalnumbersandα1>…>αn.

Wewillprovenextthatthenormalformrepresentationofγisunique.

10.22LemmaIf isanormalformrepresentationof ,

Proof.By10.19,αi<α1fori=2,…,n;thus fori=2,…,n.Thus,

10.23TheoremThenormalformrepresentationofanyordinalγisunique.

Proof.Suppose ,wherea1,…,an,b1,…,bmarenonzeronaturalnumbers,α1>α2>…>αnandβ1>…>βm.Letuswrite

thus,

Supposeα1<β1;thenα1+1 β1,so

Butby10.22, ,sowehaveacontradiction.Analogously,wecannothaveβ1<α1,henceα1=β1.Thus,

Nowsupposea1<b1,hencea1+1 b1;then

thatis,

Thus

This gives us , hence . But this is impossible,forα2<α1,hence ,soby10.22, .Consequently,wecannothavea1<b1;analogously,wecannothaveb1<a1,soa1=b1.Thus, ,soρa=ρb.Byinduction,wemaynowassumethatthenormalform

representationofρa=ρbisunique,henceα2=β2,…,αn=βn,a2=b2,…,an=bn.

Thetheoremwhichfollowsmakesiteasytoaddandmultiplyordinalnumberswhentheyarewritteninnormalform.

10.24Theorem

i) Ifα<β,thenωαa+ωβb=ωβb.ii) If isthenormalformrepresentationofγandifβ≠0,theniii) If is the normal form of γ and if b is finite, then

.

Proof

i) Ifα<β,thenby9.15,β=α+δforsomeδ>0.Thus,

Sinceaisfinite,itisclearthata+ωδb=ωδb(see,forexample,Exercise3,ExerciseSet10.2).Thus

ii) Itcanbeprovenveryeasilythatnω=ωforeveryn∈ω(see,forexample,Exercise3,ExerciseSet10.2).Itfollowsthatnωβ=ωβforeveryβ>0andn∈ω.Indeed,forβ=1thishasjustbeengiven:ifβ>1,thenby9.14(i),β=1+δforsomeδ>0,sowehave

Thus

Ontheotherhand, ,hence .iii) Theproofisbyfiniteinductiononb.Ifb=1,thereisnothingtoprove.Nowsuppose(iii)holdsfor

b,andletusproveitforb+1.Wehave

by the hypothesis of induction. The reader should note that in the above sum, the termsallprecedetheterm ,andthatα1>α2,…,α1>αn; thus,by10.24(i), they

disappearfromthesum.Thus,

Whenaddingormultiplyingordinalnumbersinnormalform,itisalsousefultorememberthatifβisaninfiniteordinalandnisfinite,thenn+β=β(see,forexample,Exercise3,ExerciseSet10.2).

10.25ExampleLet

Weshallformthesumsα+βandβ+αandtheproductsαβandβα.

Notethattheterms precede ,soby10.24(i)theydisappearfromthesum.

EXERCISES10.3

1. Ineachofthefollowing,computeα+β,β+α,αβandβα.

a) α=ωω32+ωω4+ω105;β=ωω+17+ω29+14.

b) α=ωωω9+ωω7+ω2;β=ωω58+ω72+1.c) α=ωω322+ωω184+71;β=ωω2+312+100.2. Letγbeanordinalnumber;prove thatγ is irreducible (SeeExercise2,ExerciseSet9.3.) ifand

onlyifγ=ωβforsomeordinalβ.Ineachofthefollowingexercise,wewillassumethat

isthenormalformrepresentationofγ3. Provethatγisalimitordinalifandonlyifαn≠0.4. Letusdefinethemagnitudeofγtobetheordinalα1.Provethatα+β=βifandonlyifmagnitude

α<magnitudeβ.5. Provethatωγ=γifandonlyifα1,…,αnareallinfiniteordinals.Concludethatωγ=γifandonly

ifγ=ωωβforsomeordinalβ.6. Letγbealimitordinalandletbbeafiniteordinal.Usefiniteinductiononbtoprovethat

7. UsetheresultofExercise6abovetocomputeα6andβ15,whereαandβaregiveninExercise1(a).

4

Cantor investigated the properties of an interesting class of limit ordinals which he called epsilonnumbers.Thesenumbersshedsomelightonthestructureofthewell-orderedclassORandhaveusefulapplicationsinanalysisandelsewhere.Weshallgiveabriefreviewoftheirpropertiesinthissection.

10.26DefinitionLetαbeanordinalnumber;αiscalledanepsilonnumberifα=ωα.

Itisimmediatethateveryepsilonnumberisnecessarilyalimitordinal.Now,thefirstquestionweareledtoaskis:Arethereanyepsilonnumbers?Toanswerthisquestion,wefirstneedalemma.

10.27LemmaLet{βi:i∈I}beasetofordinals,andletβ=sup{βi:i∈I};thenαβ=sup{αβi:i∈I};

Proof.i) Supposethatβ=βiforsomei∈I.Thus, Itfollowsthat

ii) Nowsupposethat∀i∈I,β≠βi;βmustbealimitordinal,forifβ=δ+1,then∃i∈Iβi>δ,soβi=δ+1=β,whichiscontrarytoourassumption.Nowαβ=sup{αγ:γ<β}; foreach i∈ I,βi<β,henceαβi<αβ;thus .Ontheotherhand,ifγ<β,thenβi>γforsomei∈I,hence

.Thus

Consequently,

Wenow return to thequestion:Are thereanyepsilonnumbers?Theanswer is “yes,” and this caneasilybeshownasfollows:

Wedefineafunctionf0:ω→ORasfollows:

Theexistenceoff0isguaranteedbythefiniterecursiontheorem,6.8.Clearly,wehave

Now,letε0=sup{f(n):n∈ω};weclaimthatε0isanepsilonnumber.Indeed,by10.27,

Thusthereisatleastoneepsilonnumber,namelyε0;wecaneasilyshow,infact,thatε0istheleastepsilonnumber.

10.29ε0istheleastepsilonnumber.

Proof. Ifα is anepsilonnumber, then f0(0)=1 α (for clearly 0 is not an epsilon number).Nowsupposethatf(n) α;thenf(n+1)=ωf(n) ωα=α. It followsbyfinite inductionthat f (n) α foreveryn∈ω;thus,ε0 α.

Isε0theonlyepsilonnumberorarethereothers?Itiseasytoanswerthisquestion,forthemethodweusedtoconstructε0maybeusedtoconstructinfinitelymanyotherepsilonnumbers.Indeed,wehavethefollowing:

10.30Ifαisanyordinalnumber,letfαbethefunctiondefinedinductivelybythepairofconditions

andletS(α)=sup{fα(n):n∈ω}.ThenS(α)isanepsilonnumber;furthermore,S(α)istheleastepsilonnumbergreaterthanα.

Proof.By10.27,

Clearlyα>S(α).Nowletγbeanepsilonnumbersuchthatα>γ;thenfα(0)=α+1 γ.Furthermore,assumingthatfα(n) γ,wehave

It follows,by finite induction, that fα(n) γ for everyn∈ω, henceS(α) γ. Thus S(α) is the leastepsilonnumbergreaterthanα.

Itiseasytoseethatε0isS(0);thus,thefirstfewepsilonnumbersareS(0),S(1),S(2),S(3),etc.Thenextquestionweareledtoaskis:ArethereanyepsilonnumbersgreaterthanalltheS(n),(n∈ω)?Thequestioniseasilyansweredinthefollowingtheorem.

10.31Let{αi:i∈I}beasetofepsilonnumbers;ifβ=sup{αi:i∈I},thenβisanepsilonnumber.

Proof.By10.27,

Letε:OR→ORbethefunctiondefinedrecursivelyasfollows[weagreetowriteεiforε(i)]:

10.32

10.33εisanisomorphismfromORtotheclassofalltheepsilonnumbers.

Proof.Toprovethatεβisanepsilonnumberforeveryβ∈OR,wearguebyinduction:i) Wehavealreadyseenthatε0isanepsilonnumber.

ii) Nowassumethatεαisanepsilonnumberforeveryα<β.Ifβisanonlimitordinal,β=δ+1,then

isanepsilonnumberby10.30.Ifβisalimitordinal,then

isanepsilonnumberby10.31.Thus,∀β∈OR,εβisanepsilonnumber.Itisimmediatethatεisastrictlyincreasingfunction,henceitisinjective.ToshowthattherangeofεistheclassEoftheepsilonnumbers,letδ∈Eandletεγbetheleastεζsuchthatδ<εζ.Nowγcannotbealimitordinal,forifitis,thenby10.32,

hence(becauseδ<εγ)∃π<γδ<επ,whichcontradictsourchoiceofγ.Thus,γisanonlimitordinal,γ=π+1;bythechoiceofγ,επ δ>επ+1,soby10.30,δ=επ.Thefactthatεisanisomorphismfollowsnowby4.48.

It turnsout, then, that there are “asmany”epsilonnumbers as there areordinals.Theclassof theepsilonnumbersis

asαrangesoveralltheordinals.Afeweasyarithmeticrulessimplifycomputationswhichinvolveepsilonnumbers.Theyaregivenin

thenexttheorem.

10.34Letεdesignateanarbitraryepsilonnumber.Theni) Ifα>ε,thenα+ε=ε.ii) Ifα>ε,thenαε=ε.

iii) Ifα>ε,thenαε=ε.

Proof.Letα>ε,andwriteαinnormalform:

Thus . Since every epsilon number is a limit ordinal, so by10.34(ii),(α1+1)ε=ε.Thus,finally,

Butwealwayshaveαε ε(seeExercise4,ExerciseSet10.2.),soαε=ε.

Itisworthnotingthatαisanepsilonnumberifandonlyifαsatisfiesthefollowingcondition:

10.35Ifβ>αandγ>α,thenβγ>α.

Theproofofthisstatementisleftasanexerciseforthereader(Exercise4below).

EXERCISES10.4

1. Provethatifαisanepsilonnumber,thenαisalimitordinal.2. Ifαisanepsilonnumber,provethatβω

α=βωαforeveryordinalnumberβ.

3. Provethatifαisaninfiniteordinalandαβ=β,thenβisanepsilonnumber.4. a) Provethatifα(α>ω)satisfies10.35,thenαisalimitordinal.

b) Provethatα(α>ω)isanepsilonnumberifandonlyifαsatisfies10.35.5. Provethatα(α>ω)isanepsilonnumberifandonlyif2α=α.

5INACCESSIBLEORDINALSANDCARDINALS

Inaccessible ordinals and cardinals play an important role in current investigations on the axiomaticfoundations of set theory. They also have applications in functional analysis, topology, algebra,mathematicallogicandotherareasofadvancedmathematics.Inthissectionwewillintroducethemandgiveafewoftheirbasicproperties.Ifαisanyordinalnumbers,letZ(α)designatetheclassofalltheordinalswhichareequipotentwith

α; the leastelementofZ(α) is called the initialordinalbelonging toα, and isdenotedby Io(α). It isimmediatethat

10.36∀α∈OR,Io(α) α,

and

10.37Io(α)isthelargestinitialordinallessthanorequaltoα.

(Toprove10.37,notethatifIo(α)>γ α,thenIo(α)⊆γ⊆α,henceIo(α)≈α≈γ.)WehaveseenthattheclassofalltheinitialordinalssatisfiesconditionsK1andK2oftheaxiomofcardinality,hencewearejustifiedinmakingthefollowingdefinition(see9.33):

Byacardinalnumberwemeananinitialordinal;thus,theclassCDofallthecardinalnumbersistheclassofalltheinitialordinals.

Wehavenotedthatifαisanyordinalnumber,itiscommonpracticetowrite α=ωα,treating αasacardinalandωαasanordinal.Ifαisanyordinalnumber,itiseasytoseethat

10.38

Itfollowseasilyfrom10.36,10.37,10.38thatifAisanywell-orderedset,

10.39

(Toprove10.39, set γ = hence Io(γ)=#γ = #A; the simpledetails are left as an exercise for thereader.)Wewillnowbegintheprocessofdefininginaccessibleordinalsandcardinals.

10.40DefinitionLetAbeawell-orderedclassandletB⊆A;wesaythatBisacofinalsubclassofAif

10.41Lemma IfC is a cofinal subclass ofB andB is a cofinal subclass ofA, thenC is a cofinalsubclassofA.

Proof.Letx∈A;then∃y∈B∈y>x;hence∃z∈C∋z>y,soz>x.

10.42LemmaLetαbeanordinalandletB⊆α;BisacofinalsubsetofαifandonlyifsupB=α.

Proof. The reader should note that by 9.26(ii) and9.24, every element of an ordinal number is anordinalnumber,andγ>αiffγ∈α.i) LetBbeacofinalsubsetofα;∀x∈B,x∈α,thatis,x<α,soαisanupperboundonB.Nowifγ<α,thenγ∈α,soby10.40,∃β∈B∋β>γ,soγisnotanupperboundonB;thisprovesthatαistheleastupperboundofB.

ii) SupposesupB=α;ifγ∈α,thatis,γ<α,then(becauseγisnotanupperboundofB)∃β∈B∋β>γ.ThusBisacofinalsubsetofα.

10.43DefinitionLetαbealimitordinal;bythecofinalityofαwemeantheleastordinalβsuchthatαhasacofinalsubsetsimilartoβ.Ifβisthecofinalityofα,wewriteβ=cf(α).

10.44Ifβ=cf(α)forsomeα∈OR,thenβ=cf(β).

Proof.Letγ=cf(β);nowαhasacofinalsubsetsimilartoβandβhasacofinalsubsetsimilar toγ,henceby10.41,αhasacofinalsubsetsimilartoγ.Butβistheleastordinalγsuchthatαhasacofinalsubsetsimilartoγ,soβ γ;nowγ& βbecauseγissimilartoasubsetofβ,soβ=γ.

10.45Letβbealimitordinal;ifβ=cf(β),thenβisaninitialordinal.

Proof.Letωα=Io(β),andletfbeabijectivefunctionf:ωα→β;letA={γ∈ωα:∀δ<γ,f(δ)<f(γ)}.Wewillshowfirstthat isacofinalsubsetofβ.Indeed,ifν∈β,then(becauseβisalimitordinalanda limitordinalhasnogreatestelement), thereareelementsξ∈ωα f(ξ)>ν. (Remember that f isbijective!)Letπbetheleastsuchelement;then∀ξ<π,f(ξ) ν<f(π),henceπ∈A;thisprovesthat∀ν∈β,∃π∈A∋f(π)>ν;thus isacofinalsubsetofβ.BecauseofthewaywehavedefinedA,itiseasytoseethatf[A]:A→ isanisomorphism;thus,

ifδ= ,thenδissimilartoacofinalsubsetofβ,socf(β) δ.ButAisasubsetofωα,soδ ωα;thus,β=cf(β) ωα.Butωαistheleastordinalequipotentwithβ,soωα β.Thusωα=β,soβisaninitialordinal.

10.46Alimitordinalβiscalledregularifcf(β)=β.

Itfollows,byLemma10.44,thattheclassoftheregularordinalsistherangeofthefunctioncf:OR→OR.ByTheorem10.45,every regular ordinal is an initial ordinalωα. Thus, in particular, every

regularordinalisacardinal.

10.47Ifωαisaregularordinal,then αiscalledaregularcardinal.

The significance of regular cardinals in set theory will become apparent once we have given analternativedefinitionforthem.Webeginwiththefollowingtwolemmas:

10.48Let{γi:i∈I}beasetofordinals,andlet#γi= δiforeachi∈I.Ifωα=sup{γi:i∈I},thenα=sup{δi:i∈I}.

Proof.Ifωα=sup{γi:i∈I},then,foreachi∈I,γi ωα,henceby10.39, δi=#γi α,soby9.22,δiα.Nowsupposethatforeveryi∈I,δi<π;then∀i∈I,#γi= δi< π,henceby10.39γi>ωπ.Butωα=sup{γi:i∈I},soωα ωπ,henceby9.22,α π.Itfollowsthatαistheleastupperboundof{δi:i∈I}.

10.49Let{δi:i∈I}beasetofordinals,andletαbeanyordinalsuchthat#I< α.Then = α

ifandonlyifα=sup{δi:i∈I}.

Proof.IfIisfinite,theresultfollowstrivially;thus,wemayassumeIisinfinite.i) Suppose

Thenforeachj∈I,

henceδj α.Thusαisanupperboundof{δi:i∈I}.Beforegoingon,wenotethatthereexistsani∈Isuchthat#I< δi;forif δi #Iforeveryi∈I,thenby8.19, (#I)(#I)=#I< α,which

iscontrarytoourassumption.Nowsupposethatδi<πforeveryi∈I.Thenby8.19,

Butwehavejustseenthatforsomei∈I,#I< δi,hence#I π;thus(#I) π= π.Itfollowsthat

π,thatis, α π,soby9.22,α π.Wehaveprovedthatαistheleastupperboundof

{δi:i∈I}.ii) Supposeα=sup{δi:i∈I}.Foreachi∈I,δi α,soby9.22, .Thusby8.19,

Nowlet .Thenforeachj∈I,

henceby9.22,δj δ.Butα=supδi,soα δ,hence

Thus,finally,

10.50Theorem isaregularcardinalifandonlyifitsatisfiesthefollowingcondition:

10.51If{ai:i∈I}isanysetofcardinalssuchthatai< foreachi∈Iand#I< ,then .

Proofi) Suppose isaregularcardinaland{ai:i∈I}isasetofcardinalssuchthatai< foreachi∈I

and#I< .Wemayassumetheaiareallinfinitecardinals,foranyfiniteαiamongthemwouldnotaffectourresult;thus,wemaysetai= foreachi∈I.Wenowhave < foreachi∈Iand#I< ,soby8.19, .Nowif ,thenby10.49,α=sup{δi:i∈I},henceby9.22,

soby10.42,{ωδi:i∈I}isacofinalsubsetofωα.Butthisisimpossible,forthefollowingreason:#I> ,hence

soby10.39, whichisincontradictionwiththefactthatωαisaregularordinal.

ii) Conversely,suppose satisfies10.51,andlet{γi:i∈I}beacofinalsubsetofωα,thatis,

[Wewillassumethattheγiarealldistinct,hence#{γi:i∈I}=#I.]Set =#γiforeachi∈I.Then,by10.48,α=sup{δi:i∈I},henceby10.49,

Thus,by10.51,wemustnecessarilyhave#I ,soby10.39,

Thisprovesthatanycofinalsubsetofωαhasordinality ωα,soωαisaregularordinal.

10.52CorollaryIfαisanonlimitordinal,then isaregularcardinal.

Proof.Ifα=δ+1,thensetai= foreachi∈I,where#I= .Thenby8.17,

Thus,by10.50, isaregularcardinal.

10.53DefinitionLeta be a cardinal number;a is called an inaccessible cardinal (more precisely, astronglyinaccessiblecardinal)ifi) aisaregularcardinal,and

ii) b<aandc<a⇒bc<a.

ωαiscalledaninaccessibleordinalif isaninaccessiblecardinal.

InviewofTheorem10.50,aninfinitecardinalnumberbisinaccessibleifandonlyifitsatisfiesthefollowingpairofconditions:i) If{ai:i∈I}isasetofcardinalssuchthatai<bforeachi∈Iand#I<b,then

ii) Ifa<bandc<b,thenac<b.

Inotherwords,aninfinitecardinalnumberbisinaccessibleifandonlyifitcannotbeobtainedeitherasa sumof fewer thanb cardinals smaller thanb, or by raisinga cardinal smaller thanb to a powersmallerthanb.Thisexplainstheuseoftheword“inaccessible.”Itcanbeshown,furthermore,thatifbisaninaccessiblecardinal,thenbcannotbeobtainedasaproductoffewerthanbcardinalssmallerthanb(seeExercise4,below).

WemaycallanonemptysetAinaccessibleifitcannotbeconstructedfromsmallersetsbyusingtheset-theoreticaloperationsofunion,intersection,product,orpowerset.Tobetechnical,thismeansthati) Aisnotequalto foranysetofsets{Bi:i∈I},where{Bi:i∈I} AandBi A

foreachi∈I;ii) AisnotthepowersetofanysetBsuchthatB A.

Forexample,ifwebeginwiththeemptysetandstarttoconstructsetssuchas

and if we proceed to construct larger and larger sets from these by using the operations of union(includinginfiniteunion),product(includinginfiniteproduct)andpowerset,wewillneverendupwithan inaccessible set. Now, in view of what we have said in the preceding paragraph, it is clear thatinaccessiblecardinalsarethecardinalsofinaccessiblesets.Aratherobviousquestionwhicharisesinaxiomaticsettheoryis,“Arethereanyinaccessiblesets?”

Inotherwords,doinaccessiblecardinalsexist?Theexistenceofinaccessiblecardinalscannotbeprovedbymeansoftheaxiomswehavealreadyintroduced;however,anewaxiommaybeaddedtosettheory,calledtheaxiomforinaccessiblecardinals,assertingtheirexistence.Ithasbeenproveninrecentyearsthatthisaxiomisnotaconsequenceoftheotheraxiomsofsettheory;whetheritisconsistentwiththeotheraxiomsofsettheoryisstillanopenquestion.Thefollowingdefinitionisuseful:

10.54Definition iscalledaweaklyinaccessiblecardinalifi) isaregularcardinal,ii) αisalimitordinal.

ByCorollary10.52,ifαisanonlimitordinal,then isnecessarilyregular;thus,itisnaturaltoaskwhether there are any regular cardinals where α is a limit ordinal. Again, we cannot prove theexistenceofsuchcardinalsfromtheusualaxiomsofsettheory,buttheirexistencefollowsimmediatelyfromtheaxiomforinaccessiblecardinals;indeed,wehave:

10.55TheoremIfaisstronglyinaccessible,thenaisweaklyinaccessible.

Proof.Suppose isstronglyinaccessible,andassumeαisanonlimitordinal,α=δ+1.By8.16,

whichiscontrarytoourhypothesisthat isstronglyinaccessible;thusαisalimitordinal.

Thereisanotherinterestingconnectionbetweenstrongandweakinaccessibility:

10.56TheoremAssumingthegeneralizedcontinuumhypothesistobetrue,ifaisweaklyinaccessiblethenaisstronglyinaccessible.

Proof.Assumethegeneralizedcontinuumhypothesis,let beweaklyinaccessible,andsuppose .By9.22,β>α,andsinceαisalimitordinal,β+1<α,so Butbythegeneralizedcontinuumhypothesis, ,hence Itfollows(seeExercise3,ExerciseSet10.5.)that isstronglyinaccessible.

Thus,ifweassumethegeneralizedcontinuumhypothesis, thenthenotionsofstronglyinaccessibleandweaklyinaccessibleareequivalent.

EXERCISES10.5

1. Prove10.37and10.38.2. Prove10.39.3. Provethataisstronglyinaccessibleifandonlyif

a) aisregular,and

b) ∀b<a,2b<a.[SeeExercise3,ExerciseSet8.4.]4. Provethatifbisaninaccessiblecardinal,thanbsatisfiesthefollowingcondition:

If{ai:i∈I}isasetofcardinalssuchthata<bforeachi∈Iand#I<b,then

5. Prove that ifan infinitecardinalb satisfies thecondition inExercise4, thenb isan inaccessiblecardinal.

11ConsistencyandIndependenceinSetTheory

1WHATISASET?

Mosttextbooksofsettheorybeginbyaskingthisquestion.However,itisonlynow—atthisstageofthe course—that it is possible to give ameaningful answer.WhenGeorgCantor first beganwritingabout sets, he defined a set to be any collectionof definite objects.That’s thewaymost people stillthinkof sets, and they’renotwrong. In thematerialworld, sets are finite collectionsof objects, andwhatiscallednaïvesettheoryisperfectlyadequateandcorrectforfinitesets.Itisonlyinthecaseofinfinitesetsthatdifficultiesmaketheirappearance.Peoplehavenoexperiencewithinfinitecollectionsofthings,andthatiswhyourintuitionsfallshort.Cantor constructed his theory as a theory of infinite sets, and fully understood that they are very

differentfromfinitesets.Hesawnoreasontodistinguishsetsinmathematicsfromarbitrarycollectionsofobjects,becausethediscoveryofthefamous“paradoxes”,orcontradictions,wasstilldecadesaway.Moreover,thetheoryheconstructedwassobeautifulinitsclarityandsimplicitythatwhenwerecollectit today, we refer to it as “Cantor’s Paradise”. In fact, when the earliest of the paradoxes wereannounced,nobodytookthemseriously,becauseitwassocomfortableinCantor’sParadisethatnobodywantedtoleave.Aswehaveseeninpreviouschapters,insettheorywecanconstructtheintegers,andfromthemwe

maydefinerationalnumbersasorderedpairsofintegers.Everyrealnumbermaybeidentifiedwithaconvergentsequenceofrationalnumbers.Thereisnoneedtogoon:Allofmathematics,piecebypiece,canbeconstructedoutofsets.Althoughsetsarevieweddifferentlytoday,itstillremainstruethatallofmathematicscanbebuiltwithintheframeworkofsettheory.Axiomaticset theory isnot for themasses: It isaspecializedbranchofmathematicsbuiltonfirm,

solidfoundations,designedtobeusedasaframeworkforallothermathematicaltheories.Thevariousaxiomatic systems of set theory used today are lean and mean: They assume the bare minimumnecessaryfortheconstructionofnumbersystems,analysis,andmathematicsgenerally.Tounderstandthe motivation behind them, you must keep in mind that they are designed to be minimalist andpragmatic incharacter:Theypresuppose theabsolute least that suffices toget theirenterpriseoff theground,andaredirectedsolelytowardthegoalofconstructingmathematicalmodels.The class OR of all the ordinal numbers is indispensable in mathematics, hence we begin by

establishing the existence ofOR.The class of the ordinals is defined recursively beginningwith theemptysetasfollows:

AsnotedinChapter9,theclassORoftheordinalsisaproperclass.InspiredbythesimplicityoftheclassOR,Zermelodefinedahierarchyofsetswhich—heproposed

—wasadequateforcarryingoutallofmathematics.ItisknownastheCumulativeHierarchy,because

itsmembersarearrangedinsuccessivelevelsindexedbytheordinalnumbers.

Finally,Visdefinedtobetheunionofallthelevels:This construction begins with just one set, which is the empty set. Then, by using the power set

axiomandtheaxiomofunions,webuildprogressivelylargersets.Notethatthesearetheonlysetsthatareguaranteedtoexistinaxiomaticsettheory.Theessenceoftheconstructionprocessisthatwebeginatastage0withthesimplestpossibleset,whichistheemptyset,andworkupwardstagebystage.Ateverystage,letussaystageα+1,theelementsofVα+1areallthesetsinVα(thatis,thesubsetsofVα).Andifαisalimitordinal,thenalltheelementsofVαaresetsthatexistatthepreviouslevelsVβforβ<α.Sotheelementsateachlevelarethesetsthatwerecreatedatthepreviouslevels.The value of this hierarchy is that it classifies sets according to their complexity, that is, their

“distance”upfromtheemptyset.Thedivisionofallthesetsintosuccessivelevelsmeansthatyoucanprovethingsbytransfiniteinduction,anddefineobjectsusingtransfiniterecursion.The“distanceofasetfromtheemptyset”iscalleditsrankanddefinedasfollows:

11.1DefinitionTherankofasetAistheleastordinalαsuchthatA∈Vα.

ZermeloproposedthatforthepurposesofmathematicseverypossiblesetisamemberofV,andhecalledVtheuniverseofsets.Inorderforthisclaimtobeplausible,itmustbepossibletoshowthatthemembersofVsatisfyalltheaxiomsofsettheory.Todothis,wemustfirstshowthatthemembersofVhavetwoessentialpropertieswhichtheyshare(almost)withtheordinals.Thefirstpropertyisgiveninthetwodefinitionsthatfollow:

11.2DefinitionIfwetreatthemembershiprelation∈asarelationonasetA,thenanelementx∈Aissaidtobe∈-minimalifthereisnoelementy∈Asuchthaty∈x.

Itisveryimportant,here,todistinguishaminimalelementofAfromaleastelementofA:TheleastelementiscomparablewitheveryelementofA.Butaminimalelementisnot.

11.3DefinitionAsetA is said to bewell-founded if everynon-empty subset ofA has an∈-minimalelement.

11.4Remark.Notethataleastelementofasetisalwaysminimal,buttheconverseisnottrue.Thus,ifasetAis∈-well-ordered(seeChapter9)thenitiswell-founded,butawell-foundedsetisnotnecessarily∈-well-ordered.IfasetAiswell-founded,thisfacthasseveralfar-reachingconsequencesthatareveryeasytoprove:

11.5TheoremLetAbeawell-foundedset.ThenthefollowingaretrueinA:a) Foreveryx∈A,x∉x.

b) Foranyx,y∈A,ifx∉ytheny∉x.c) InA,thereisnoinfinitedescendingsetofelements…∈x3∈x2∈x1∈x0.d) Aiswell-foundediff∃x∈Asuchthatx∩A=∅.

Proof.For(c),theset{xi:i∈ω},ifitexisted,wouldhavenominimalelement.(a) Iftherewereanx∈Asuchthatx∈x,we’dhaveadescendingsequence{…x∈x∈x}.(b) Herewewouldhaveadescendingsequence{…y∈x∈y∈x}.Finally,theproofof(d)isleftas

anexercise.

Itwasmentionedabovethatinaxiomaticsettheory,asetiswhatevercanbebuiltupfromtheemptysetbyiteratingthepowersetandunionsteps.That’sthepointoftheconceptofwell-founded.Thesetmembershiprelationcangodownonlyafinitenumberoftimes,thusprovingthateverypossiblesetisatsomestageinahierarchyofsetmembership.

11.6DefinitionAsetAissaidtobetransitiveifeveryelementofAisasubsetofA.Equivalently:(x∈A)∧(y∈x)⇒(y∈A).Wehaveseen(Chapter9)thateveryordinalnumberisatransitiveset.

FromParts(a)and(b)ofthelasttheorem,ifasetAistransitiveandwell-founded,thismeansthat∈isanorderrelationonA.(Specifically,itisastrictorderrelation<.)However,∈isnotalinearorderonA,becausetwoarbitraryelementsx,y∈Aarenotnecessarilycomparable. Infact, if∈werea linearorderonA, then awell-founded setAwouldbe an∈-well-ordered set, and thereforeAwouldbeanordinal.

11.7Lemma

a) Everysubsetofatransitivesetistransitive.b) Everyunionoftransitivesetsistransitive.c) Thepowersetofatransitivesetistransitive.

(Thesimpleproofsareleftasexercises.)

11.8TheoremForanytwoordinalnumbersξandα,

a) Vαisatransitiveset,andb) ξ<α⇒Vξ⊂Vα.

Proof.Weprove(a)and(b)jointlybyinductiononα.Assume(a)and(b)aretrueforallβ<α.Theclaimsaretrivialforα=0.Ifαisalimitordinal,then(a)and(b)followimmediatelyfromLemma11.7(b).Ifαisasuccessorordinal,α=β+1,then .SinceVβ is transitive, it follows fromLemma11.7(c) thatVα is transitive.Finally,Vβ⊂Vα from thedefinitionofthecumulativehierarchy.

ThesetsVαhaveapropertythatis,moreorless,aconverseoftransitivity:

11.9LemmaIfx⊂Vαthenx∈Vα+1.

Proof.Fromthedefinitionofthecumulativehierarchy,theelementsofVα+1arethesubsetsofVα.

Toreachourgoal,wemustprovenowthateveryVαiswell-founded:

11.10TheoremForeveryα∈OR,Vαiswell-founded.

Proof.Webeginbyshowingthefollowing:(x∈Vα)∧(y∈x)⇒y∈Vβforsomeβ<α.Theproofisbyinductiononα.Ifα isalimitordinalthen .Thenx∈Vα⇒x∈Vβ forsomeβ<α.By thehypothesisofinduction,y∈Vβ⊂Vα.Next,ifα=β+1,then .Soifx∈Vαthenx⊂Vβ,soy∈x⇒y∈Vβ.Nowforourmainresult:LetYbeanon-emptysubsetofVα.LetβbetheleastordinalsuchthatY∩Vβ≠∅.Obviouslyβ<α,

sinceY∩Vα=Y.LetxbeanyelementofY∩Vβ.Bythepreviousparagraph,y∈x⇒y∈Vδforsomeδ<β.ButβwastheleastordinalforwhichY∩Vβ≠∅,hencey∉Y,whichisacontradiction.Thus,xis∈-minimal.

WehaveshownthateveryVα istransitiveandwell-founded.CanthesamebesaidforsetsthataremembersofVα?Let’stry:

11.11TheoremIfA∈Vα,thenAiswell-founded.

Proof.IfXisanynon-emptysubsetofA,wewishtoprovethatXhasan∈-minimalelement.Well,letαbetheminimalrankamongallelementsinX,andlety∈Xbeanelementwhoserankisα,hencey∈Vα. Now suppose there is an element x∈X such that x∈ y: Then x∈ y⇒ x∈Vα becauseVα istransitive.Butthatisimpossiblebecausex∈yimpliesthattherankofxmustbelessthantherankofywhichisminimalinX.Soxcannotexist,henceyisminimalinX.

Next, itwouldbenice ifwewereable to show that for everyordinalα, everymemberofVα is atransitive set. If that were the case, we could characterizeV as the collection of all transitive well-foundedsets.Unfortunately,wecannotdothisbecauseitturnsoutthatnoteverysetinVistransitive.Whenmathematiciansfindaroadblockof thiskind, theyusuallysearchforadetour that leads to thesameresult.Theroadblockwehavejustencounteredmotivatesthefollowingdefinition:

11.12DefinitionThetransitiveclosureofasetAisthesmallesttransitivesetthatcontainsA.

The transitive closure of a set A is easy to construct by using the Axiom of Union. Begin withandforeveryn, .ThenthetransitiveclosureofAisdefinedto

betheset

Intuitively,tc(A)isthesetofallobjectswhichareelementsofelementsof…ofelementsofA(iteratedafinitenumberoftimes).Itisasimpleexercisetoverifythattc(A)isatransitivesetthatcontainsA,andmoreover,thatifBisatransitivesetsuchthatA⊆B,thentc(A)⊆B.

11.13LemmaForanyordinalα,A∈Vαifftc(A)∈Vα.

Proof.From thepreviousparagraph: tc(A) is a transitive set that containsA, andmoreover, ifB is atransitivesetsuchthatA⊆B,thentc(A)⊆B.BecauseVαistransitive,ifA∈VαthenA⊆Vα.Thus,(withVαintheroleofBabove),tc(A)⊆Vα.Conversely,iftc(A)∈Vα,thentc(A)⊆VαsoA⊆tc(A)⊆Vα.SofromLemma11.9,A∈Vα+1.

Inparticular, ifA is amemberof someVα, then tc(A)∈Vα andbyTheorem11.11, tc(A) iswell-founded.Thus everymember ofV has a transitivewell-founded closure. In our last theoremof thissectionweshallprove theconverseof theabove,namely:Everysetwhose transitiveclosure iswell-foundedisamemberofV.ItwillfollowfromthisthatwehavefullycharacterizedV:Itistheclassofallthesetswhosetransitiveclosureiswell-founded.Compare thiswith a corresponding result about ordinal numbers inChapter9,Section5:OR (the

class of all ordinal numbers) is the class of all transitive∈-well-ordered sets. Since every∈-well-orderedsetiswell-founded—andsinceeverytransitivesetisitsowntransitiveclosure—itfollowsthateveryordinalnumberisamemberofV.

11.14TheoremEverytransitivewell-foundedsetisamemberofV.

Proof.LetA be a transitive,well-founded set. It suffices to prove thatA⊆V, for byLemma 11.9 itfollowsthatA∈V.IfA V,letybeaminimalelementofA−V,andletz∈y.Thenz∈AbecauseAistransitive.Wehavejustshownthaty⊆A.ByLemma11.9y∈V.Butthiscontradictsy∈A−V,andfromthiscontradictionweconcludethatA⊆V.

ThecentralfactaboutV—thereasonforitsimportanceinmathematics—isthatalltheaxiomsofsettheory holdwhen theword “set” is replaced by thewords “transitivewell-founded set”. That is thereasonwhyV is called theuniverseofsets.The classV contains all the ordinal numbers (hence thecardinalsaswell),andalltheconstructionsneededinmathematicscanbecarriedoutonthemembersofV.InV,setsarebuiltfromthebottomup.ThemembersofVareclear,clean,welldefinedobjectswhich

satisfysimpleaxioms,sosettheoryonVisabranchofmathematicsasrigorousasotherareassuchasalgebraandanalysis.NoticethatwehavestatedthatalltheaxiomsofsettheoryapplytoV.Theproofsare straightforward verifications, and several of them are given in the exercises at the end of thissection. It is therefore perfectly reasonable to assert that all the sets that exist in mathematics aremembersofV. If youwish to consider themembersofV tobe all the sets there are, thenyoumustlegislatethisfactasanaxiom:

AxiomofFoundationVistheclassofallsets,thatis,V= .

Fromthispointonward,wetaketheAxiomofFoundationtobeoneofouraxioms.

EXERCISES11.1

1. SupposethatinasetAthereisnodescendingsetofelements…∈x2∈x1∈x0∈A.ProvethatAiswell-founded.

2. ProvethatasetAiswell-foundediff∃x∈Asuchthatx∩A=∅.

3. Provethateverysubsetofawell-foundedsetiswell-founded,andeveryelementofawell-foundedsetiswell-founded.

4. Provethatifxandyarewell-founded,soare .5. Provethateverysubsetofatransitivesetistransitive.6. Provethatanyunionoftransitivesetsistransitive.7. Provethatthepowersetofanytransitivesetistransitive.8. ProvethatasetAistransitiveiff .9. ProvethatasetAistransitiveiff .10. Supposethatαisalimitordinal.Prove(a)Ifξ<αthenVξ⊆Vα.

(b)IfVξistransitiveforeveryξ<αthenVαistransitive.

11. IfAisasetandtc(A)isitstransitiveclosure,prove:(a)tc(A)istransitive.(b)A⊆tc(A).

12. Prove:IfAisaset,Bisatransitiveset,andA⊆B,thentc(A)⊆B.13. Prove(a)IfAisatransitivesetthenA=tc(A).(b)Ifx∈Athentc(x)⊆tc(A).14. IfA∈VandB⊆AthenB∈V.(YoumayuseLemma11.9).15. Prove:ifx,y∈V,then{x},{x,y},x×y,andx∩yareinV.16. Prove:Foralln∈ω,|Vn|isfinite.(Provebyinductiononn.)17. Everywell-foundedsetisinV.ConcludethatVistheclassofwell-foundedsets.18. Theaxiomofpairsstatesthatifxandyaresets,then{x,y}isaset.Toprovetheaxiomofpairsfor

V is to show that if x, y∈ V then {x, y}∈ V. Prove this, using the rule for constructing thecumulativehierarchy.

19. TheaxiomofunionassertsthatifAisaset,then isaset.ToprovetheaxiomofuniononVistoshowthatifA∈Vthen .Provethis.

20. ThepowersetaxiomstatesthatifAisaset,sois .ProvethatthepowersetaxiomholdsinV.Thatis,ifA∈V,then ∈V.

2MODELS

Whenevermathematicsisdonerigorously,onebeginswithaformallanguage whichhassymbolstodenote the relations and operations pertaining to the mathematical theory under discussion. Forexample,whenstudyingorderedsets,oneusesalanguage withtwobinaryrelationsymbols<and=,togetherwiththelogicalconnectives∧,∨,¬andthequantifiers∃and∀.Forthetheoryofringsweusea language with a binary relation symbol =, operation symbols ·, + and −, and the logicalconnectivesandquantifiers.If isaformallanguage,oneobtainsatheoryTinthelanguage bylayingdownasetofaxioms

forthetheory.Forexample,ifTisthetheoryofpartiallyorderedsets,itsaxiomsare:

Finally,amodelofatheoryTisasetAtogetherwitharelationonAcorrespondingtoeachrelationsymbol,andanoperationonAcorrespondingtoeachoperationsymbol.Forexample,amodelforthetheoryoforderedsetswouldconsistofasetA togetherwitharelationonA(that is,asetoforderedpairsofA)satisfyingthethreeaxiomsabove.Inplainlanguage,amodelforthetheoryoforderedsetsisanorderedset.Likewise,amodelforthetheoryofringsisaring.Amathematical theory issaid tobeconsistent if itdoesnotharboranycontradictions, that is,you

cannotusetheaxiomsof thetheorytoproveastatementFaswellas itsnegation¬F.ProvingthatatheoryT isconsistent isnotoriouslydifficult,because in theworstcase,youwouldhavetoseeeverytheoremofTtoknowwhetherornottherearecontradictions.However,oneofthefirstresultsofmodeltheoryiscalledtheCompletenessTheoremandstatesthatatheoryTisconsistentifandonlyifithasamodel. In practice, this is the easiest—and sometimes the only—way to prove that a theory isconsistent.Inplainlanguage,itisonlyifyouproduceamodelofatheorythatyoucanbecertainitisconsistent.One of the most productive branches of mathematics during the 20th Century (it occupied

mathematical geniuses such asGödel and vonNeumann duringmuch of their careers) has been thestudyofmodelsofsettheory.Notethatsettheorycannotbetakenseriouslyasarigorousdisciplineinmathematicsuntilitisshownthattherearemodelsofsettheory,becauseitisonlyonceweseeamodeloftheaxiomsofsettheorythatweknowitisconsistent.However,amodelofsettheoryisnotlikeanyothermodel:Forexample,amodelofgrouptheoryis

a setwithanoperation that satisfies theusualgroupaxioms. In thesameway,amodelof set theorywouldhavetobeasetwhosemembers(sets)satisfytheaxiomsofsettheory.Achievingsuchathingisfraughtwithdifficulties.Inthefirstplace,youwantasettoserveasamodelofsettheory,butitisonlyinsideset theory thatyoucometoknowwhataset is.Secondly, if it isamodelofset theoryweareconstructing,then—inthatmodel—weareabletoconstructeveryordinalnumber,andweknowthattheclass of all the ordinal numbers is a proper class:So the set, in ourmodel,wouldhave to contain aproperclass,whichisimpossible.Attheverytimewhenthesequestionswerebeingasked,aremarkableresultaboutmodels—called

theLöwenheim-SkolemTheorem—was discovered. Itwas proved thatevery consistent theory has acountablemodel.Inotherwords,ifatheoryhasamodelofanysize,italsohasacountablemodel.Soifsettheoryisconsistent,notonlymustithaveamodel,butitmusthaveacountablemodel.Imagine:acountablemodelforallofsettheory!Inthe1920s,ingeniousmodelswereconstructedforsubsetsoftheaxioms,butneverforthefullsetofaxiomsofsettheory.Thisquestwasabandonedatthebeginningofthe1930swhenKurtGödeldroppedabombshellonthemathematicalworld.

3INDEPENDENCERESULTSINSETTHEORY

In1931,aspartofhisdoctoraldissertation,KurtGödelprovedthatifaconsistentsystemofaxiomsissufficienttoprovethetheoremsofarithmetic,thentheconsistencyofsuchasystemcannotbeprovedbythestandardmethodsofmathematics.Aswehaveseeninthisbook,it ispossibletoconstructthenatural numbers and carry out arithmeticwithin axiomatic set theory. Thus, byGödel’s Theorem, ifaxiomaticsettheoryisconsistent,thenitsconsistencycannotbeproved.Inparticular,wemustabandonourquesttoconstructamodelofsettheory.Where does this leave us? We appear to have just two choices: Either we choose to accept the

consistency of axiomatic set theory on faith, or alternatively, we abandon the ideal of foundingmathematicsonaxiomsandrigorousproof,andputourfaithinintuitioninstead.Mostmathematiciansaccept theconsistencyof thevariousaxiomaticsystemsofset theorybecause theaxiomsstrikeusasabsolutelyplausibleandalmostundeniablytrue.Primarily,theyarewillingtoaccepttheaxiomsofone

system of set theory (axioms similar to the ones we have been using) called the Zermelo-FraenkelaxiomsanddiscussedinthenextSection.Thissystem,knownatZFsettheory,doesnotrequireproperclassesbecauseitplacesotherlimitationsonwhatisallowedtobeaset.I have said that the axioms of set theory seem perfectly plausible. That is true with two notable

exceptions:TheAxiomofChoiceandtheContinuumHypothesis.Welikethembecausetheysimplifyandbeautifymathematics,but theydonotstrikeusasunquestionably true.Thus, formanyyears theHolyGrail afterwhich themost gifted knights ofmathematics questedwas a proof of the axiomofchoice(AC)andthecontinuumhypothesis(CH)fromtheotheraxiomsofsettheory.Onceagain,itwasKurtGödelwhosettledtheissueinamostprovocativeway:In1936hepublishedaproofthatbothACandCHareconsistentwiththeotheraxiomsofsettheory.Specifically,heshowedthat—assumingtheotheraxiomsofsettheoryareconsistent—then¬ACand¬CHcannotbededucedfromthem.Thisdiscoverysetoffanewrace toprovethat thenegationsofACandCH(denotedby¬ACand

¬CH)arelikewiseconsistentwiththeotheraxioms.Itwasnotuntilthirtyfouryearslaterthatthisresultwasachieved,usingverynewideas,byPaulCohen.Takentogether,whattheseresultsshowisthatACandCHareindependentofotheraxiomsofsettheory.IfyouwantACinsettheory,youmuststateitasanindependentaxiom,andifyouwantCHthenyoumuststateCHasanindependentaxiom.Andforthatmatter,youmayadd¬ACasanewaxiomwithoutanydangerofcontradiction,andlikewiseyoumayadd¬CH.TheproofsoftheseresultsarecalledtheIndependenceProofsofsettheory.Therewas a timewhen people asked: “After all, is theContinuumHypothesis true or not? Is the

AxiomofChoicetrueinrealityorisitnot?”Wenolongeraskthesequestionstodaybecausewehavebecome a littlemore cynical about absolute truth. Few people today believe in the Platonic heavenwhere all truths are enthroned forever.Mostmathematicians don’t ask such questions, but generallyaccept AC and CH because it makes their work easier. For the Philosophical Few, it is clear thatmathematics is a great fountain of abstract ideas, perfect if viewed fromwithin, somewhat flawed ifviewed fromwithout. In anaxiomatic systemyou laydowna setof axiomsandwork todeduce theconsequencesofyouraxioms.Youdonotaskifyouraxiomsare“true”.Aperfectexemplarforthisis theParallelPostulateinEuclideangeometry.Fortwothousandyears

brilliant thinkers burned out their synapses in vain attempts to prove the Parallel Postulate from theotheraxiomsofgeometry.OnefinedaypeoplerealizedthatthereareotherkindsofgeometryinwhichtheParallelPostulatedoesnotholdinitsEuclideanform.Itturnsout,then,thattheParallelPostulateisneither true nor false. That is the very situation we have today with the Axiom of Choice and theContinuumHypothesis. It seemsvery likely that therewill beuseful applications inmathematics for¬ACand¬CH.

4THEQUESTIONOFMODELSOFSETTHEORY

Wehaveseen,above,thataxiomaticsettheoryisconsistentifandonlyifithasamodel.Theforemostproblemwithbuildingamodelofsettheoryisthatitwouldbetoobig.Forexample,Visaclassofsetsthatsatisfiesalltheaxiomsofsettheory,butitistoolargetobeamodel:Visaproperclass,whereasamodel must be a set. The only way out of this trap is to ingeniously construct a modelM whoseelementsmimicall thepropertiesofsets,butsuch thatMdoesnot trulycontaineveryset. In fact, inviewoftheLöwenheim-Skolemtheorem,themodelMweconstructmaybecountable.The idea for achieving this is deceptively simple: We mimic the construction of the cumulative

hierarchy{Vα}α∈ONwhileplacingrestrictionsonthesetsthatareadmittedintoeachVα.Thisisdoneinsuchawayastokeeponlythosesetsthatareindispensabletoourarguments,whileexcludingallthe“inconvenient”sets.TheresultinguniverseofsetsM isa set—andwerecall that“sets” includesuch

thingsasfunctions,relations,thenaturalnumbers,andsoon.Asamodelofsettheory,Mmustincludeanuncountablesety(infactmanyuncountablesets).Aset

yisuncountableifthereisno1-1correspondencebetweenyandω.RememberthatMdoesnotcontainallsets—many“true”setsareabsentfromM—andamongthesetsabsentfromMareallthebijectivefunctionsbetweenωandy.Asaresult,inthemodelMthereisno1-1correspondencebetweenωandy,and that makes y uncountable by definition. This is true even though from outside the model, y iscountable—perhapsevenfinite.Theidea,then,isthattherearetwoparalleluniverses:TheuniverseofsetsasobservedfrominsidethemodelM,andtheuniverseof“true”setsasobservedfromoutsidethemodel.Forexample,inMletxbethesetthatplaystheroleof andythesetthatplaystheroleof .

Thenydoesnotreallycontaineverysubsetof ,itcontainsonlythosesubsetsofxthatareinM.SoifM is a countablemodel then ymust be countable, and so clearly y cannot be equal to for theoutsideobserver.Thesetythatwecallthepowersetof inMisnotthesameasthe“real”powersetofbecausemanysubsetsof aremissinginM.WhatwehavejustrelatedistheunderlyingideabehindGödel’sproof—butthedevilisinthedetails.

AfullaccountofGödel’sresultsandthoseofPaulCohenisbeyondthescopeofthisbook.ButIshalloutlinetheprincipalstagesoftheargument.ThefirstproblemistheconstructionofamodelMofsettheoryhavingthepropertiesdescribedinthepreviousparagraphs.Thekeyideaisthatofconstructiblesets.Inatrueuniverseofsets,ifAisanysetthenbytheaxiomsofsettheory,theuniversemustcontain

suchthingsas andallthecombinationsofAwithothersets in theuniverse.Sothesupplyofsetsrapidlyoutstripstheboundariesofacountableuniverse,oranyuniversewhichisaset.Whatmustbedoneattheverystart,then,istorestrictwhatcountsasaset.Theideaisthattheonlykindsofsetthatcanplayanyroleinset-theoreticproofsaresetsthatcanbedescribedinformulasofthelanguageofsettheory.Itstandstoreasonthatindescribablesetscanplaynopartinanyreasoningaboutsets.Recallthatanexpressionintheformallanguageofsettheoryisanyformulathatcanbewrittenasa

valid combinationof the logical connectives∨,∧, ¬, the quantifiers∀ and∃, and the relation∈.Aformulamayalsocontainparameters,thatis,constantsymbolsreferringtoobjectsalreadyconstructedatanearlierstageofaproof.Anarbitraryformulawithfreevariablesx1,x2,…,xnmaybewrittenasϕ(x1,x2,…,xn)withoutdisplayingtheparameters.This isa timelymoment tosayawordabout theZermelo-Fraenkel (ZF)axiomsofset theory,and

howtheydifferfromtheaxiomswehavebeenusingthroughoutthisbook.ThedistinctionboilsdowntoanimportantdifferencebetweentheAxiomofClassConstruction(AxiomA2)usedinthisbook,andthe corresponding axiom in the ZF system, called the Axiom of Selection. The Axiom of ClassConstruction asserts the existence of a classC consisting of all setsx that satisfy a formulaϕ(x). Inmany cases, the classC formed in thismanner is a proper class. TheAxiomof Selection in theZFsystem is more conservative: It asserts that if A is any set and ϕ(x) is a formula, there is a set Sconsistingofallxsuchthatx∈Aandϕ(x).Itassertstheexistenceofsetsthataresubsetsofexistingsets,henceitcannotgiverisetoproperclasses.InZFsettheory,ifAandBaresets,thenAissaidtobeconstructibleoverBifthereexistsaformula

ϕ(x)intheformallanguageofsettheorysuchthatAconsistsofalltheelementsofBwhichsatisfyϕ(x).Theformulaϕmay containparameters (constant symbols)which refer to elementsofB.That is,wemayform:

Wenow revise the cumulative hierarchyby restricting sets to constructible sets.Theconstructible

universeisbuiltasahierarchyofsetsindexedbytheordinals,justasthecumulativehierarchywas.Ineachsuccessorstep,insteadofaddingallsubsetsofthecurrentset,onlythedefinableonesareadded.Thatis,wereplacethepowersetoperationbytheconstructiblepowersetoperationdefinedasfollows:

Thissmalldifferencemakesabigdifference:Thereasonisthatinacountableformallanguagethereareonlycountablymanyformulas,hencethereareonlycountablymanyconstructiblesets.Thus,evenifAisaninfiniteset, hasnomorethancountablymanymembers.Exceptforthisonedifference,theconstructiblehierarchyisdefinedlikethecumulativehierarchy.

Finally,Lisdefinedtobetheunionofallthelevels:Liscalledtheconstructibleuniverse.Nowlet’sslowdownamoment,becauseareadersittinginthe

backoftheroomhasaquestion:—“Author:YoutellusontheonehandthatbyGödel’sincompletenesstheorem,itisimpossibleto

prove the consistency of the ZF axioms. You also tell us that consistencymeans that there exists amodel.And finally, you speak tous about constructingmodelsof set theory.Surely if you constructsuchamodel,youhaveprovedtheconsistencyofZFwhichyousayisimpossible.Whatgives?”—Thank you, that’s an excellent question!Themodelwewant to construct is builtwithinZF set

theory.SoifZFhappenstobeinconsistenttobeginwith,thenthemodelwehavebuiltisflawed,sinceitisbuiltwithintheconstraintsofZF.SoourmodelisonlyamodelofZFonthepriorassumptionthatZFisconsistent.Consequently,anyresultweareabletoderivefromthismodel isconditionalontheconsistencyoftheZFaxioms.

EXERCISES11.4

1. a) Writethedefinitionofx=yasaformulaϕinthelanguageofsettheory.(SeeDefinition1.9fortheformaldefinitionoftheequalityofsets.)b) WriteaformulaψwhichstatesAxiomA1inthelanguageofsettheory.c) Theset{a,b}istheset{x:ϕ(x)}.Writethecorrectformulaforϕ(x).d) Theset Writethecorrectformulaforψ(x).e) Theset .Writetheformulaforξ(x).f)Theorderedpair(a,b)={x:χ(x)}.Writetheformulaforχ(x).(Harderthan(a)–(e).)

2. IfAisaclassandϕisaformulainthelanguageofsettheorythenϕAistheformulaobtainedbyreplacingeveryquantifier∃xinϕby(∃x∈A)andevery∀xby(∀x∈A).a) ProvethatifB⊆A,thenϕistrueinBiffϕBistrueinA.b) Give an informal example (e.g. using finite sets) showing that there are classesB⊆A and

formulasϕsuchthatϕBistrueinAbutϕisnottrueinA,andwhereϕistrueinAbutϕBisnot

trueinA.3. Supposeaandbaresetssuchthata,b∈L.

a) Provethat{a,b}∈L,andexplainfromthiswhytheaxiomofpairsistrueinL.b) Provethattheorderedpair(a,b)∈L.c) Provethat .ExplainwhytheaxiomofunionistrueinL.

4. Eachofthefollowingisaset{x:ϕ(x)}whereϕmaycontainparameters.Foreachofthefollowingsets,givetheformulaϕinthelanguageofsettheory,andindicatewhicharetheparametersinϕ.Explainwhyeachis—orisnot—aconstructibleset.a) a∪b.b) a×b.c) (a,b),where(a,b)isanorderedpair.d) .

5PROPERTIESOFTHECONSTRUCTIBLEUNIVERSE

Thehierarchyofconstructiblesetshasmanyofthesamepropertiesasthecumulativehierarchy,butthestyleof reasoningaboutconstructible sets ismore subtle than thewaywe reasonaboutconventionalsets.Lookcarefullyattheproofofthenexttheorem.

11.15TheoremForeachordinalα,thefollowingaretrue:

a) Lαistransitive.b) Forβ<α,Lβ⊆Lα.c) Forβ<α,Lβ∈Lα.

Proof.Thefirsttwostatementsareprovedjointlybyinductiononα.Thus,suppose(a)and(b)aretrueforβ.Inparticular,Lβistransitive.Ifα=β+1,then .Ifx∈Lβ,letA={a∈Lβ:a∈x}.NotethatAisthesetofallelementsa∈Lβthatsatisfytheformulaa∈x,wherexisaparameterthatrefers to an element of Lβ. (You are free to replace the parameter x by an n-tuple x1,x2, …xn ofparameters.) Thus,A is a constructible set. Moreover,A consists of all the elements of Lβ that areelementsofx:InotherwordsA=x.Thus,x∈Lβ+1=Lα,whichshowsthatLβ⊆Lα.Moreover, if x∈ Lα then x⊆ Lβ, hence x⊆ Lα from the previous line. Thus, Lα is transitive.

Consequently,(a)and(b)aretrueforαifαisasuccessorordinal.Ifαhappenstobealimitordinal,theresultisverysimple.Now for Part (c): Note that the formula x = x is trivially a formula (without parameters) in the

languageofsettheory,hence{a∈Lβ:a=a}=LβisanelementofLβ+1.ThisprovesPart(c)ifαisasuccessorordinal,andtheproofisimmediateifαisalimitordinal.

11.16CorollaryListransitive.

Proof.Indeed,ifx∈Lthenx∈Lαforsomeordinalα.SinceLαistransitive,itfollowsthatx⊆Lα⊆L.

TwomoresimplefactsaboutLwillproveuseful,andarestatedasalemma:

11.17LemmaForeveryordinalα,

a) α∈Lα.b) Lα⊆Vα.

Proof.(a)isaneasyinductiononα:SupposeourclaimistrueforLα,andproveitforLα+1.Thenα⊆LαbecauseLα is transitive.Thus,α+1=α∪{α}⊆Lαhenceα+1∈Lα+1.Thecasewhereα isa limitordinalissimilar,butsimpler.Asfor(b),wenotethat ,henceLα+1⊆Vα+1.

FromLemma11.17(a) it follows thatLcontainsall theordinalnumbers.L is transitiveas shownabove,andasweareabouttosee,LsatisfiesalltheZFaxioms.AnytransitiveclassMthatcontainsalltheordinalsandsatisfiestheZFaxiomsiscalledaninnermodelofsettheory.Asmentionedearlier,thereisastrongintuitivebasisforconsideringLtobetheclassofallsets.By

definition,Lcontainsall thesets thataredescribablebyaformula in the languageofset theory.Andthereisnopracticalreasontoadmitsetswhichlackanydescription,forwewouldnevermakeuseofsuchsets.Theywouldmerelysit thereandmuddythewaters.Thus, fromthispointonwardweshallassumethefollowingimportantaxiom:

AxiomofConstructibilityEvery set is constructible, that is, every set is inL.This axiom is usuallydenotedbythesymbolV=L.

Asyouhavejustseen,inordertoprovestatementsaboutaconstructiblesetA,itisnecessarytokeepinmindthatAconsistsofalltheelementsx(insomeset)thatsatisfyaformulaϕ(x).Acrucialsubtletyisthatanassertionϕ(x)mayholdinasetAbutmayfailtoholdinasubsetB⊂A,orvice-versa.Forexample,let ,andletB={{a},{a,b},{a,b,c}}.ThenBisstrictlyorderedby⊂(henceBsatisfiesaformulawhichstatesthatitisstrictlyordered)butAdoesnotbecauseAisnotstrictlyorderedby⊂. To be precise, B satisfies the formula but A does not satisfy thatformula.However,Asatisfies .Thislatterformula,denotedbyϕB,iscalledtheformulaϕrelativizedtoBinA.ItisobviousthatϕdoesnotholdinA,butϕBdoes.YoumaythinkofϕBasavoiceinAthatmakesastatementaboutB.Thenotionofrelativizedformulasisclearlyessentialforreasoningaboutconstructiblesetsandtheir

subsets.Wenowgiveaprecisedefinitionforit:

11.18DefinitionIfA isanyclassandϕ isa formula in the languageofset theory thenϕA, called therelativizationofϕtoA,istheformulaobtainedbyreplacingeveryquantifier∃xinϕby(∃x∈A),andlikewisereplacingeveryquantifier∀xby(∀x∈A).

ItisclearthatifAandBareclassessuchthatB⊆A,thenϕistrueinBiffϕBistrueinA.However,ifϕBistrueinAthisdoesnotentailthatϕistrueinA.Thisfactisillustratedinthepreviousexample,andmotivatesthedefinitionthatfollows.

11.19DefinitionLetA⊆V.WesaythatϕisabsoluteforAinVifforallx1,…,xn∈A,

Informally,ϕandϕAareequivalentinVifallthefreevariablesofϕtakevaluesinA.WesaythatϕisabsoluteifitisabsoluteforeveryA.

If all the important formulaswereabsolute, itwouldmake life simpler for set theorists.Mainly, itwouldbepossibletoform“small”modelsofsettheory.Unfortunately,thatisnotthecase.Asthenextexampleshows,evenasimpleformulasuchasx⊆yisnotabsolute.

ExampleLetA⊆Vbethesetwhoseonlyelementsare∅and{{∅}}.IfϕAistheformula(∀x∈A)(x∈{{∅}}⇒ x∈∅)—equivalently {{∅}}⊆∅—then ϕ is true in V. [Note that the only x∈ {{∅}} is{∅},but{∅}isnotinA].However,ϕisnottrueinV.

11.20DefinitionAformulaϕissaidtobeabsoluteovertransitivedomainsif,foreverytransitiveclassA,ϕisabsoluteforAinV.

Thegoodnews is thatagreatmanyformulasareabsoluteover transitivedomains.All themodelsthatweshallbeworkingwithfromthispointonwardwillhavetransitivedomains.Tobegin,itwillbeshownthatAxiomA1isabsoluteovertransitivedomains.Thatis,AxiomA1istrueineverytransitiveclass.Pleasenotecarefullyhowtheproofworks.

11.21LemmaIfA is a transitive class, thenA satisfiesAxiomA1. (In otherwords, the formula forAxiomA1isabsolutefortransitiveclasses.)

Proof.AxiomA1isthefollowingformulaϕ:(∀x,y)[(x=y)⇔((∀u)(u∈x⇔u∈y)].ThenϕAis:(∀x,y∈A)[(x=y)⇔((∀u∈A)(u∈x⇔u∈y)].ItisobviousthatifϕistrueinV,thena

fortioriϕAistrue.Fortheconverse,supposeϕAholdsinV,andshowthatϕholdsinVforelementsx,y∈A.Ourproofisbycontradiction:Weshallassumethereisausuchthatu∈xandu∉y.SinceAistransitive, u∈A. Since we assume that ϕA holds, it follows from u∈ x that u∈ y, and this is acontradiction.Thus,ϕistrueinA.

At theheartof theproof is thefact that—becauseAis transitive—everyelementofx is inA,andeveryelementofyisinA.NotethattheargumentwouldnotworkifAwerenottransitive.Manyotherpropertiessimilarlyillustratetheimportanceoftransitivity.Forinstance,lookatthenotionoffunction,and suppose f is a function inV:Then f is a set of ordered pairs (x,y) = {{x}, {x, y}}. IfA is nottransitive,wecannotprovethatxisinAoryisinA,hencewecannotprovethatfisafunctioninA.(ButifAistransitive,wecan.)

Many other properties and relations are absolute for transitive models. These include: being anorderedpair,afunction,a1-1function,arelation,thedomainorrangeofarelation,thesetx×y,beingan ordinal number, the setω, andmany others. In fact, the next theorem yields a treasure-trove offormulasthatareabsoluteontransitivedomains.

11.22DefinitionA formulaϕ is called aΔ0 formula if all of its quantifiers are bounded. That is, allquantifiersinϕareoftheform∃x∈yor∀x∈yforasety.

11.23TheoremIfAistransitiveandϕisaΔ0formula,thenϕisabsoluteforA.

Proof.It is clear thatx =y andx∈y are absolute for anyA, because theydonot changewhenyourelativize them.Infact,allquantifier-freeformulasϕareunchangedwhenrelativized, that is,ϕ=ϕA.The proof of the theorem is by induction on the length ofϕ, which is defined to be the number ofoccurrencesoflogicaloperators(∧,∨,¬,∃,∀)inϕ.Wehavealreadygiventheproofforformulasoflength0(namelyx∈yandx=y).Now,ifϕandψareabsoluteforA,thenclearly,soare¬ϕ,ϕ∧ψandϕ∨ψ.Thedelicatestepofthe

proofbeginshere:Assumethatϕisoftheform(∃x∈y)ψandsupposethatϕAistrueinV,whereyandotherpossiblefreevariablesinψrepresentelementsinA.SinceAistransitive,x∈A,andtherefore(∃x∈A)[x∈y∧ψA]holdsinV.Bythehypothesisofinduction,ψholdsiffψAdoes,hence(∃x∈A)(x∈y∧ψ)=(∃x∈y)(ψ)=ϕholdsinV.Conversely,supposeϕ is true inV, that is, (∃x∈y)ψ is true inV,wherey andother possible free

variablesinψrepresentelementsinA.SinceAistransitive,y∈A.Now,ϕA=(∃x∈y)ψA,andtheproofisessentiallythesameasinthepreviousparagraph.

Other broad categories of formulas can also be proved to be absolute over transitive domains.Actually,whatisevenmoreimportantinmathematicsisthefactthatmanysimplepropertiesfailtobeabsolutefortransitivemodels.ThereasonthisisdesirableisthatwewishtoconstructmodelsMhavingproperties that hold in M but are not consequences of ZF. For example, we would like to find atransitivemodelMinwhichthenegationoftheaxiomofchoice(¬AC)holds.ThiswouldimplythatZF(withoutAC)doesnotimplyAC.Theverypurposeoftheendeavorofcreatingmodelsofsettheoryistofindmodelshavingpropertiesthatarenotabsolute.ThebasisofGödel’sindependenceresultsisthattheconceptofawell-ordering,aswellastheconceptofcardinality,arenotabsolute.

OurmissionnowistoprovethatifLisamodelofZF,thenanumberofadditionalpropertiesalsoholdinL.AndbecausethesepropertiesholdinL,theymustbeconsistentwithZF.Tobeexplicit,ifapropertyϕistrueinL(whichisamodelofZF)thenitisnotpossibletoprove¬ϕinZF,becauseweareassumingthatZFisconsistentonaccountofitshavingamodel.

11.24Remark. InordertoprovethataformulaϕholdsinthemodelL,whatweneed toshowis that.Thereasonisthatweareassuming(thisismerelyanassumptionhencerequiresnoproof)that

L is amodel of ZF. So if thenϕL is true in everymodel of ZF, hence inL. Thus,ϕL, andthereforeϕitself,aretrueinthemodelL.

FromthepreviousRemark,inordertoshowthattheaxiomsinZFholdinL,wemustshowthatforeachaxiomϕ,theformuaϕListrue.Itisthistaskthatweundertakenext.WearesubstantiallyaidedbythefactthatLisatransitiveclass.

11.25TheoremForeveryaxiomϕofZF,ϕListrueinL.

Proof.Tocarryoutthisproof,itisessentialfirst,tobeclearabouttherelationshipbetweenanaxiomϕofZFandthecorrespondingformulaϕL.WehavealreadyseenthatinV,theunorderedpair{a,b}istheset{x:x=a∨x=b}.Notethatthesetisdefinedbytheformulaϕ=(x=a∨x=b).ButinL,{a,b}istheset{x∈L :x=a∨x=b}.Youwillnote thatϕ is the sameasϕL, so theoperationof formingorderedpairsisabsolute.

Forunions,supposethata∈L:Then isthesetofallxthatsatisfytheformulaϕ=(∃y∈a)(x∈y).So relativized toL, = {x∈L : (∃y∈a)(x∈y)}.Note the importance of assuming thatL istransitive:Ifa∈Landy∈athenyouconcludethaty∈L.Asyoucansee,inthiscaseϕListhesameformulaasϕ,henceϕisabsoluteoverL.SincewehaveshownthatunionsarethesameinLasinV,theAxiomofUnionsholdsinL.Likewise,forsubsets,z⊆xisanabbreviationforϕ=(∀v)(v∈z⇒v∈x).Bycontrast,ϕL=(∀v∈L)

(v∈z⇒v∈x)wherezandxareparametersrepresentingelementsinL.ItisobviousthatifϕistrueinV,thenϕLislikewisetrue.Conversely,supposeϕListrueinV,whichisthesameassayingthatϕistrueinL.SinceL is transitive,z⊆x isequivalent toz∈x,hence(v∈z⇒v∈x).Soϕ is true inV forparametersx,z∈L.Thisshowsthat⊆isabsoluteonanytransitivedomain.ThatmeansthatsubsetsarethesameinLas

inV.Consequently,thepowersetaxiomistrueinL.Ithasalreadybeenshown(Lemma11.21)thatifϕistheformulaforAxiomA1,thenϕLisprovable

fromZF,becauseLisatransitiveclass.Next,theAxiomofFoundationistrueinL:Indeed,supposea∈L.RecallthatL⊆V.Now,sincethe

AxiomofFoundationistrueinV,thereisanelementb∈Vsuchthat(b∈a)∧(b∩a=∅).SinceListransitive,itfollowsfromb∈aanda∈Lthatb∈L.Thus,thefollowingholdsinL:(b∈a)∧(b∩a=∅).FromTheorem11.5(d),Liswell-founded.SeveraloftheZFaxiomshavenowbeenshowntoholdinL;othersareverytechnicalandrequire

moreelaboratemachineryfortheirproof.Theyareomittedhere.ItturnsouttobeanimportantfactthattheAxiomofConstructibility(V=L)holdsinL.Youmay

recall that the Axiom of Constructibility has been added to ZF, and it might appear that, as aconsequence,itistrueinL,sinceLisamodelofZF.However,asmentionedabove,whatweneedtoestablishisthat(V=L)LholdsinL—andthatturnsouttobemoredifficultthanitappearstobe.TheAxiomofConstructibility is the claim that every set is constructible, in otherwords it is the

formula(∀x)(∃α)[On(α)∧x∈Lα]whereOn(α)standsfor“α isanordinal”.Asexplainedpreviously,whatwereallyneedtoshowisthatitistheformulafor(V=L)LthatholdsinL.Now(V=L)Listheformula

ItcanbeshownthattheformulasOn(α)aswellas(x∈Lα)LareabsoluteinL.Theproofsarefairlytechnical,andomittedhere.Usingthesefacts,itfollowsthat(V=L)Lisabsolute,andthereforeistrueinthemodelL.Ourconclusionis:

11.26TheoremLisamodelofZF+(V=L).

Consequently,theAxiomofConstructibilityV=LcannotberefutedinZF,soitmaybeaddedasanewaxiomwithoutdangerofcontradiction.Fromthispointonward,weassumethatV=Lisoneofouraxioms.Wearenowreadytoreapthewhirlwind!

EXERCISES11.5

1. ForanyclassA,proveeachofthefollowing:

a) IfAistransitive,then .b) IfX⊆AandXisfinite,thenX∈ .IfAisaninfiniteset,then .

2. Provethatforeveryordinalα,Lα∩On=α,whereOnistheclassoftheordinals.3. Forallfiniten,provethatLn=Vn.4. Forallinfinitecardinalsα,provethat|Lα|=|α|.5. Ofall theaxiomsofset theory(includingAF,V=L,andAC),whicharetrueintheemptyset?

WhicharetrueinthefinitemodelAwhoseelementsare:a,b,{a},{b},{a,b}?6. Referring to the previous exercise, let B be the subset of A whose elements are a, b.Write a

formulaϕsuchthatϕistrueinAbutϕBisnottrueinA.7. Explainwhyeachofthefollowingformulasisabsoluteovertransitivedomains:

a) a⊆b.b) c=(a,b)where(a,b)denotestheorderedpair.c) c=a∪b.e) c=a∩b.f) c=a∪{a}.g) aisatransitiveset.

8. ItwasmentionedinChapter1thatthereisanaxiomproposedbyvonNeumann(butnotusedhere)thatstatesthefollowing:AisaproperclassiffAisin1-1correspondencewithV.Equivalently:AisasetiffAisnotin1-1correspondencewithV.CallthisaxiomVN:a) ProvethatVNimpliestheAxiomofChoice.b) ExplainwhyVNimpliesastrengthenedversionoftheAxiomofChoicethatappliesnotonlyto

setsbuttoallclasses.Whatdoesthisdotothewell-orderingtheorem?c) ProvethatVNimpliestheAxiomofReplacement(ourAxiomA9).d) ProvethatAxiomA9togetherwiththestrengthenedAxiomofChoiceimplyVN.

6THEGÖDELTHEOREMS

UsingZF+ (V=L),Gödelwasable, first, to show that theAxiomofChoice is true inL.Hedidnotshowthisdirectlybyprovingthattherearechoicefunctions.Rather,heprovedthatitwaspossibletoconstruct awell-ordering ofL, fromwhich it is obvious that every set can bewell-ordered becauseeveryset isasubsetofL.Andasweknow, thewell-ordering theoremisequivalent to theAxiomofChoice.

11.27TheoremIfV=L, then theAxiom ofChoice holds. This implies that if ZF is consistent, then.

Proof.Theproof consists of defining a constructible relationwhich is anorder relation onL, andshowingthat well-ordersL.LetxandybetwoarbitraryelementsofL,andletusdefinetheconditionwhichmakesx ytrue.Therearethreecases,treateddifferently:1. Letα<βbe twocardinalssuch thatx firstappears inLα,whereasy firstappears inLβ.We then

decreethatx yintheorderingofLwearedefining.

2. NowsupposethatxandyhavetheirfirstoccurrenceinthesameLγ,callitLα+1.Forthiscase,let{ϕi : i < ω} be an enumeration of all the formulas of our countable language. We reason byinduction, sowe assume that the ordering has already been defined onLα andwell-orders it.Supposexisdefinedbyaformulaϕxandyisdefinedbyaformulaϕy,wherebothformulashave(forsimplicity)justoneparameter—saypinϕxandrinϕy:

Wedecreethatx yifϕxprecedesϕyintheenumerationofformulas.Or,incaseϕx=ϕy,thenx yifpprecedesrintheorderrelationonLα.Recallthattheclassoftheordinalnumberiswell-ordered,andbyinductiononα,Lβiswell-ordered

by foreveryβ<α.Usingthesefacts,theorderrelation wehavejustdefinedwell-ordersL.

11.28TheoremLsatisfiesCH(theContinuumHypothesis),thatis, .ItfollowsthatifZFisconsistentthen .

The technicalmachinery required to show thatCH is true inLgoeswellbeyond thescopeof thisbook.Weshallthereforeconfineourselvesheretooutliningtheintuitivebasisfortheproof.WehavealreadydiscussedtheimplicationsofthefactthatLcontainsonlyconstructiblesets:Forα<ω1,(whereω1 is the firstuncountableordinal)Lα is countable:Every set inLα isdefinedbya formulaϕ in thecountable language of set theory. Moreover, the language includes no more than countably manyparameters,becausetheparametersdenoteelementsinthecountablesetLα.Sothereareonlycountablymanyformulas.Sincetherearenomorethancountablymanyformulasavailabletoconstructsets,therearenomore

thancountablymanysetsinLαforeachα<ω1.Sofinally,theunionofastrictlyincreasingfamilyofω1= manycountable setshascardinalityno less than norgreater than .A formalversionof thisproofcanbegeneralizedtohighercardinals,andyieldsthefollowingtheorem:

11.29TheoremLsatisfiesGCH(theGeneralizedContinuumHypothesis), .Consequently,.

For a brief recapitulation of what has been done, we have shown that if you assume ZF to beconsistent, then so isZF+AF+V=L+AC+GCH.AfterGödeldemonstrated that theAxiomofChoice and the Continuum Hypothesis are relatively consistent with the ZF axioms—that is, if weassume theZFsystem is consistent, then soareACandGCH—thequestionwas immediately raisedwhetherthenegationsofACandGCHarelikewiseconsistentwithZF.Iftheyare,thatmeansthatACandGCHare independentof the ZF axioms—in otherwords, theymay be added to the ZF axiomswithout contradiction, and likewise, their negations may be added to ZF without producingcontradictions.Itwasnotuntil1964thatthisquestionwasansweredaffirmatively:Inagroundbreakingpaper,Paul

Cohen invented a method called forcing to show that indeed (if ZF is consistent) it is possible toconstructmodelsofZFthatsatisfy¬ACand¬GCH.Perhapstheseresultsdonotadvancethepracticeofmathematics toanyappreciabledegree,but theirphilosophical impact is tremendous.Theyshowthattheconceptofset—thoughfoundedonveryconcrete intuitionsandmentalpictures—isnotnearlyas

elementaryaswearepreparedtobelieve.Thereisnooneuniquetruthconcerningthepropertiesofsets:Rather,therealityofwhatasetisbifurcatesintoseveralalternativerealities,allequallyplausibleandallequallytrue.Lastly,ithasbeenshownthatwewillneverhaveabsolutecertaintythatsettheory—ormathematics

generally—is free of contradictions. It is not merely a question of the state of current knowledge:Rather,what has been shown is that it is fundamentally impossible ever to prove the consistency ofmathematics.For set theory, is that really surprising? Think of it! It is already a remarkable fact that animals

(includinghomosapiens)areabletoabstractoutofexperiencethefact that thereissuchathingasacluster,abatch,abundleofsimilarobjects—andthatsuchabundleisaseparateunitofreality.Thenwebuildonthatandthinkofcollectionswe’veneverexperienced,suchasallthetreesinaforest,orallthenaturalnumbers.Thenweabstractfurtherandthinkofasetasathinginitself,irrespectiveofwhatitsmembersare.Thenotionofset is theabstractionofanabstractionofanabstraction.Thatkindofiteratedabstractingseems tobe theessenceof thehuman intellectualenterprise.More than that, it isperhapsthelong-termecologicalfunctionofbrains.

Bibliography

1. Bourbaki,N.,ThéoriedesEnsembles,Paris,Hermann,1963.2. Fraenkel,A.A.,SetTheoryandLogic,Reading,Mass.,Addison-Wesley,1966.3. Halmos,P.,NaiveSetTheory,Princeton,VanNostrand,1960.4. Kamke,E.,TheoryofSets,NewYork,Dover,1950.5. Monk,J.D.,IntroductiontoSetTheory,NewYork,McGraw-Hill,1969.6. Quine,W.V.,MathematicalLogic,Cambridge,Mass.,HarvardUniversityPress,1951.7. Rubin,J.E.,SetTheoryfortheMathematician,SanFrancisco,Holden-Day,1966.8. Slupecki,J.andL.Borkowski,ElementsofMathematicalLogicandSetTheory,Oxford,Pergamon

Press,1967.9. Suppes,P.,IntroductiontoLogic,Princeton,VanNostrand,1957.10. Suppes,P.,AxiomaticSetTheory,Princeton,VanNostrand,1960.

Index

alephs,180anti-lexicographicordering,87axiom(s),5class,12ofcardinality,151ofchoice,110,112ofclassconstruction,26ofconstructibility,227ofextent,25ofinfinity,125ofordinality,169ofreplacement,70ofselection,10

axiomaticmethod,4

bijectivefunction,54,58Booleanalgebra,29,101

Cantor,Georg,1,4cardinalnumbers,150,186Cartesianproduct,35chain,87choicefunction,112class(es),12,24algebraof,29empty,27equivalence,76universal,27

codomain,51cofinality,207comparableelements,87complement,28conditionallycomplete,97conjunction,20constructiblesets,224constructivemathematics,18continuumhypothesis,141,180CumulativeHierarchy,214

denumerablesets,139,146diagonalmethod,139,140disjoint,28disjunction,21domain,38duality,93,95

elements,4,25,26epsilonnumbers,202∈-orderedclass,182∈-well-orderedclass,185equipotent,138,142finitesets,138,144formallanguage,6,7fullyorderedclass(set),87,103function(s),49canonical,83characteristic,55constant,54identity,54inclusion,54increasing,90invertible,58,59orderpreserving,89restrictionof,55strictlyincreasing,89

Gödeltheorems,232graph(s),37,49diagonal,72

greatestelement,94greatestlowerbound,94

image,50direct,62inverse,62

immediatepredecessor,104immediatesuccessor,103implication,21inaccessiblecardinals,206,210inaccessibleordinals,206,210indexedfamilyofclasses,66induction,126mathematical,126transfinite,105,188

inductiveset,118infinitesets,1,138,140initialsegment,87,107injectivefunction,56,60intersection,27,40,43intuitionism,16isomorphism,89,106

lattice,98,99complete,101

sub,101lawoftheexcludedmiddle,17leastelement,94leastupperbound,94lexicographicordering,86limitordinals,178,189linearlyorderedclass(set),87lowerbound,94,95greatest,94,95

mathematicalinduction,126maximalelement,94maximalprinciples,118Hausdorff’s,118

minimalelement,94

naturalnumber(s),123negation,20normalformofordinalnumbers,197

one-to-onecorrespondence,54orderedpairs,33,34ordinalnumbers,166,168,186

paradoxes,2,3,10–12,13Berry’s,4,11Russell’s,3,10–12

partiallyorderedclasses,86partition,74Platonicphilosophy,15powerset,46pre-image,50predicate,6productofclasses,66,67properclasses,25,151,168

quotientset,78

rangeofgraph,38recursion,128theorem,128transfinite,188

regularcardinal,208relations,71anti-symmetric,71asymmetric,73equivalence,72,73intransitive,73irreflexive,73

order,72,86reflexive,71symmetric,71transitive,71

Russell,B.,3,16

section,104sentence,20set(s),9,13,15,25,44similar,168singleton,34subset,26successorset,125surjective,53,60

theoryoftypes,16transfinite,155cardinals,155induction,105,188recursion,188,189recursiontheorem,189

transitiveclosure,217transitiveset,126,215truthtable,20

union,27,41,43upperbound,94least,94

Venndiagram,28vonNeumann,12,14

well-foundedset,215well-orderedclass(set),103well-orderingtheorem,120

Zermelo,9,14Zorn’sLemma,118

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