9 B k I S k k I B S k Proof Let x B S k I S k Then x B and x S k I S k This

9 b k i s k k i b s k proof let x b s k i s k then x

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9.B[kISk=[kI(BSk).Proof.LetxBSkISk.ThenxBandxSkISk.This meanskI,xSk. ChoosemIwithxSm.We now havexBandxSm; hencexBSm.Therefore,kI,xBSk. Thus,xSkI(BSk).Therefore,BSkISkSkI(BSk).Conversely, letxSkI(BSk).This meanskI,xBSk. ChoosenIwithxBSn.Now,xBSnmeansxBandxSn. In particular,xB.Also, sincexSn, we havekI,xSk, which meansxSkISk.We now havexBandxSkISk. Thus,xBSkISk.Therefore,SkI(BSk)BSkISk, and soBSkISk=SkI(BSk).11.B\[kISk=\kI(B\Sk).Proof.LetxB\SkISkThenxBandx<SkISk.x<SkISkmeanskI,x<Sk.LetkI.Thenx<Sk, and sincexB, we havexB\Sk.Therefore,kI,xB\Sk, which meansxTkI(B\Sk).Therefore,B\SkISkTkI(B\Sk).Conversely, letxTkI(B\Sk).SinceI,, choosemI.ThenxB\Sm, which meansxBandx<Sm. In particular,xB.SupposexSkISk. This meanskI,xSk. ChoosenIwithxSn.SincenIandxTkI(B\Sk), we must havexB\Sn.This gives usxBandx<Sn, which is a contradiction, sincexSn.Therefore,x<SkISk.We now havexBandx<SkISk. This meansxB\SkISk.Therefore,TkI(B\Sk)B\SkISk. Thus,B\SkISk=TkI(B\Sk).
110CHAPTER 2. SETS13.[kISkc=\kI(Sk)c.Proof.Letx(SkISk)c.Thenx<SkISk.That is,kI,x<Sk.Therefore,kI,xSck.This meansxTkI(Sk)c.Therefore, (SkISk)cTkI(Sk)c.Conversely, letxTkI(Sk)c.This meanskI,xSck.In other words,kI,x<Sk.This meansx<SkISk.Hence,x(SkISk)c.Therefore,TkI(Sk)c(SkISk)c.Therefore, (SkISk)c=TkI(Sk)c.15. If[kISk=, thenmI,Sm=.Proof.SupposeSkISk=andmI,Sm,.ChoosemIwithSm,, and sinceSm,, choose an elementxSm.Now, sincexSmandmI, we haveki,xSk. Thus,xSkISk.This gives usx∈ ∅, which is a contradiction.Therefore, ifSkISk=, thenmI,Sm=.17.J∈ P(I),[kJSk[kISk.Proof.LetJ∈ P(I). This meansJI.LetxSkJSk.Then,kJ,xSk. ChoosekJwithxAk.