2jk1jk2jThisexamplegeneralizestoNNNNforany nitenumberoffactorsinthecartesianproductontheleftThisfollowsfromthenextexampleEx3LetABCDbesetsIfACandBDthenABCDProofByde ID: 450832
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Ex.2.NNN.Thismaylookcounterintutiveatrst:thereare\justasmany"orderedpairsofnaturalnumbersastherearenaturalnumbersthemselves!ThiscanbeproventhroughCantor's\diagonalcount":NN=f(1;1);(1;2);(2;1);(1;3);(2;2);(3;1);(1;4);(2;3);(3;2);(4;1);:::g:Thatis,weorderthepairsaccordingtothesumofthetwocoordinates,andwithineachsumlistthemintheorderoftherstcoordinate.Thiscorrespondstothebijectionf:NN!Ngivenby:f(j;k)=1 2(j+k�1)(j+k�2)+j:Thisexamplegeneralizesto:NN:::NN;forany(nite)numberoffactorsinthecartesianproductontheleft.Thisfollowsfromthenextexample.Ex.3.LetA;B;C;Dbesets.IfACandBD,thenABCD.Proof.Bydenitionofequipotent,weknowthereexistbijectionsf:A!Candg:B!D.Itisnaturaltodeneafunctionh:AB!CDbyh(a;b)=(f(b);g(c)).WeleaveitasanexercisetoshowthathisabijectionfromABtoCD.Thefollowingresultsareoftenusedtoestablishthatgivensetsarecount-ablyinnite.Proposition1.LetAN.AssumeAisneitheremptynornite.ThenAiscountablyinnite(thatis,AN).Proof.Deneafunctionf:N!Aasfollows.Letf(1)bethesmallestelementofA(intheusualorderingofN).ThisexistsbytheWell-OrderingPrinciple,sinceA6=;.Thenletf(2)bethesmallestelementinAnff(1)g.Notethatthissetisalsonon-empty(sinceA,beinginnite,cannotequalff(1)g),sotheWell-OrderingPrincipleappliesagain.Ingeneral,givenff(1);:::;f(n)g,weletf(n+1)bethesmallestelementinAnff(1);:::;f(n)g(whichisanon-emptysubsetofN).Thisdenesthefunctionfinductively;fisinjective,sincefromtheconstructionwehave:f(1)f(2)f(3):::f(n)f(n+1):::2 Proof.Denef:ZN!Qbyf(p;q)=p=q.Sinceeachrationalnumberisoftheformp=qforsomep2Zandsomeq2N(notnecessarilyuniqueones),itfollowsthatfissurjective.SinceZNN(usingexamples1,2,and3),itfollowsfrompart(2)ofProposition2andRemark1thatQiscountable(hencecountalyinnite.)Corollary3.LetA1;A2;A3;:::benon-emptycountablyinnitesets(notnecessarilydisjoint),Thentheirunioniscountablyinnite:[i1AiN:Proof.We'regiventhat,foreachi2N,thereexistsabijectionfi:N!Ai.Denef:NN![i1Ai;f(i;j)=fi(j):Toseethatfissurjective,letx2Si1Aibearbitrary.Thenx2Aiforsomei2N,sox=fi(j)forsomej2N;thusx=f(i;j).BeforedescribingtheproofofthefactthatRisuncountable,weneedtoreviewsomebasicfactsaboutdecimalexpansionsofrealnumbers.Thisexpansionisnotunique.Forexample:0:9999999:::=1:Thisequalityisnotapproximate,itisEXACT,afactthatsomestudentsndsurprising;bothsidesaredi erentdecimalrepresentations(inbase10)ofthesamerationalnumber.Onewaytoseethisis:0:99999:::=91Xj=110�j=9=10 1�1=10=1;usingthewell-knownformulaforthesumofaconvergentgeometricseries,P1j=0aq�j=a 1�qifjqj1.Thesamewouldhappenforanyrationalnumberwithaterminatingdecimalexpansion,forexample:0:1234567=0:12345669999999:::Thusifwewanttoassigntoeachrealnumberauniquedecimalexpansion,weneedtomakeachoice;forthepurposeoftheargumentthatfollows,we'llchoosethe\non-terminatingexpansion",thatis,theonethatiseventuallyaninnitesequenceof9's.4 Notethatdn6=0foralln,sothisnonterminatingdecimalexpansionosoftheallowedkind,anddenesarealnumberin(0;1].Weclaimthatforalln2Nf(n)6=x,contradictingthefactthatfisonto.Toseethis,observethatthen-th.digitsinthedecimalexpansionofxisdn,andintheexpansionoff(n)isdnn;thesearedi erent(fromtheconstructionabove).Thisconcludestheproof.Numericalexample.Wehavenocontroloverthe\listing"fthatisassumedtoexistatthestartoftheargument,butsuppose(forexample)therstveentrieswere(highlightingthediagonalentriesofthearray):f(1)=0:12034506:::f(2)=0:13579017:::f(3)=0:24608046:::f(4)=0:31415926:::f(5)=0:21784143::::::Thentherstvedigitsofxwouldbe:x=0:24725:::,anditisclearthatxcan'tbeanyoftherstveelementsonthelist(andindeedcan'tbeanyelementonthelist).Remark3.ItfollowsfromthetheoremandCorollary1thatRisun-countable.Remark3.Asimilarproofgivesthefollowingresult:thesetofinnitesequencesof0sand1sisuncountable.Bycontradiction,supposewehadsuchalisting(asabove),andmodifythediagonalentries(intheonlywaypossible)tondasequencewhichcan'tpossiblybeincludedonthelist.Thisresultaboutinnitesequencesof0sand1scanbestatedasaresultaboutthepowersetofN,P(N).Aninnitesequenceisafunctionf:N!f0;1g;thesetofsuchsequencesisusuallydenotedf0;1gN.AsubsetANdenesafunctionf:N!f0;1gvia:f(n)=1ifn2A;f(n)=0ifn62A:Andconversely,anyfunctionf:N!f0;1gdeterminesthesubsetA=f�1(f1g)N.Thuswehaveabijectivecorrespondence:P(N) !f0;1gN:Combiningtheseobservations,weobtainthefollowingresult:Theorem.(Cantor1891)P(N)isuncountable.6 PARTTWO(Outline)1.Cantor-Bernstein-Schroedertheorem:LetA;Bbesets.Ifthereexistinjectivefunctionsf:A!Bandg:B!A,thenAB.Ex.1.P(N)R.Ex.2.RnQR.Ex.3.(0;1]R.2.Cardinalitytrumpsdimensionality.Theunitsquare(0;1)(0;1)R2isequipotentwiththeunitinterval(0;1)R.Ideaofproof.Constructaninjection(0;1)(0;1)!(0;1)intermsofdecimalexpansions:(0:a1a2a3:::;0:b2b2b3:::)7!0:a1b1a2b2a3b3:::Ex.4.R2R;infactRnRforeachn.Ex.5.(i)ThesetRRofallfunctionsf:R!RisequipotentwithP(R).Hint:notethatRRisasubsetofP(RR),andP(RR)P(R)(sinceR2R).Fortheotherinjection,considercharacteristicfunctions.(ii)Thesetofallfunctionsf:Q!RisequipotentwithR.(iii)Thesetofallcontinuousfunctionsf:R!RisequipotentwithR.(Hint:iffandgarecontinuousfunctionsandtakethesamevalueateachrationalnumber,thenf=g.)Ex.6.Thesetofallalgebraicrealnumbers(rootsofpolynomialswithintegercoecients)iscountable.Hencetranscendentalnumbersexist.3.Theorem.(G.Cantor)LetAbeanynonemptyset.ThenP(A)andAarenotequipotent.NotethatthereisanobviousinjectionfromAtoP(A):x7!fxg.8