资源描述:
《Bell _ Basic Set Theory》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、BasicSetTheoryJohnL.BellI.SetsandClasses.Wedistinguishbetweenobjectsandclasses.Anycollectionofobjectsisdeemedtoformaclasswhichisuniquelydeterminedbyitselements.Wewritea∈AtoindicatethattheobjectaisanelementormemberoftheclassA.Weassumethateverymemberofaclassisanobject.Lower-caselettersa,b,c,x,y,z,…wil
2、lalwaysdenoteobjects,andlater,sets.EqualitybetweenclassesisgovernedbytheAxiomofExtensionality:AxiomA=B⇔∀x[x∈A⇔x∈B].Oneclassissaidtobeasubclassofanotherifeveryelementofthefirstisanelementofthesecond.Thisrelationbetweenclassesisdenotedbythesymbol⊆.ThuswemaketheDefinitionA⊆B⇔∀x[x∈A⇒x∈B].AsubclassBofacl
3、assAsuchthatB≠AiscalledapropersubclassofA.Everypropertyofobjectsdeterminesaclass.Supposeϕ(x)isthegivenpropertyofobjectsx;theclassdeterminedbythispropertyisdenotedby{x:ϕ(x)},whichwereadastheclassofallxsuchthatϕ(x).Church’sschemeisanaxiomguaranteeingthattheclassnamedinthiswaybehavesinthemannerexpected
4、:Axiom∀y[y∈{x:ϕ(x)}⇔ϕ(y)].AmongclasseswesingleouttheuniversalclassVcomprisingallobjectsandtheemptyclass∅whichhasnomembers.ThuswemaketheDefinitionV={x:x=x}∅={x:x≠x}.2Weshallsometimeswrite0for∅.Asetisaclasswhichisalsoanobject.Thepurposeofatheoryofsetsistoformulateexistenceprincipleswhichensuretheprese
5、nceofsufficientlymanysetstoenablemathematicstobedone.Russell’sParadoxshowsthatnoteveryclasscanbeaset.Forconsidertheclass{x:x∉x}=R(herex∉xstandsfor“notx∈x”).Supposethisclasswereasetr.ThenitfollowsfromChurch’sschemethatr∈r⇔r∉r,acontradiction.ThereforeRisnotaset.Inthepresentformulationofthetheoryofsets
6、wequantifyonlyoverobjects,andnotoverclassesingeneral.Wedo,ontheotherhand,namemanyclassesandstateprincipleswhichapplytoallclasses.Definitions{a}=df{x:x=a}{a1,…,an}=df{x:x=a1∨…∨x=an}∀x∈Aϕ(x)⇔df∀x[x∈A⇒ϕ(x)]∃x∈Aϕ(x)⇔df∃x[x∈A∧ϕ(x)]{x∈A:ϕ(x)}=df{x:x∈A∧ϕ(x)}x,y,…,z∈A⇔dfx∈A∧y∈A∧…∧z∈Aa1∈a2∈…∈an⇔dfa1∈a2∧…∧an-
7、1∈anA∪B=df{x:x∈A∨x∈B}A∩B=df{x:x∈A∧x∈B}–A=df{x:x∉A}A–B=df{x:x∈A∨x∈B}Noticethat∪,∩and–satisfythefollowinglawsofBooleanalgebraA∪B=B∪A,A∩B=B∩A;A∪A=A,A∩A=A;(A∪B)∩B=B,(A∩B)∪B=BA∪(B∩C)=(A∪B)∩(A∪C),A∩(B∪C)=(A