微积分讲义Chap 1 Completeness axiom of R

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1、Chapter1Therealnumbersystem1.3.CompletenessaxiomofR1.16DefinitionLetand.(i).ThesetEissaidtobeboundedaboveifthereisans.tforall.(ii).AnumberMiscalledanupperboundofthesetEifforall.(iii).AnumberSiscalledasupremumofthesetEifSsatisfiesthefollowingconditions(1

2、)if,(2)ifMisanupperboundofEthen.RemarkThesupremumisalsocalledtheleastupperbound.1.17:ExampleIfE=[0,1],provethat1isasupremumofE.Proof.1..2.letMbeanupperboundthenforall.Wederivetheresult.1.18:RemarkIfasethasoneupperbound,ithasinfinitelymanyupperboundsProo

3、f:.LetEbeasubsetofR.Letforall.ThenMisanupperbound.LetthenM+bisalsoanupperbound.So,Ehasinfinitelymanyupperbounds.1.19.Theorem.LetEbeanonemptysubsetofR.ThentheleastupperboundofEisuniqueifitexists.Proof.SupposethataretheleastupperboundsofE.Thenareupperboun

4、dsofE..NotationThesupremumisalsocalledleastupperbound.WeusesupEtodenotethesupremumofnonemptysetE.1.20.Theorem[Approximationproperty]exists.Thens.t.Proof:.Supposetheconclusionisfalse.Thereisansuchthat.isanupperbound.s.t1.21.TheoremIfhasasupremum,thenProo

5、f.LetsupE=s.ByApproximationproperty,theres.t.Ifthenisobvious.If,thens.t.1..2..Itisacontradiction.●[CompleteaxiomofR]EverynonemptysubsetEofRthatisboundedabove,thenEhastheleastupperbound..1.22:[ArchimedeanPrinciple]s.tb

6、b,let..Eisboundedabove.ByCompletenessofR,supEexists.taken=supE+11.23:Example.LetandprovethatsupA=supB=1Proof.1..isanupperbound.LetMbeanotherbound.2.isanupperboundofB.LetMbeanupperboundofBToshow.SupposenotByArchimedeanprinciple,thereexistssuchthat,forsom

7、e.Misanupperbound.●[Well-OrderPrinciple]Ehasaleastelement(ie.s.t)1.24.Theorem(Densityofrational)Letsatisfya0,letByArchimedeanPrinciple.ByWell-OrderPrincipleEhasaleastelement,

8、says..Let.Wemustshowthata0.s.ta+k