实变函数与泛函分析基础(程其襄张奠宙着)高等教育出版社课后答案

1.A∪(B∩C)=(A∪B)∩(A∪C).x∈(A∪(B∪C)).x∈A,x∈A∪B,x∈A∪C,x∈(A∪B)∩(A∪C).x∈B∩C,x∈A∪Bx∈A∪C,x∈(A∪B)∩(A∪C),A∪(B∩C)⊂(A∪B)∩(A∪C).x∈(A∪B)∩(A∪C).x∈A,x∈A∪(B∩C).x∈A,x∈A∪Bx∈A∪C,x∈Bx∈C,x∈B∩C,x∈A∪(B∩C),(A∪B)∩(A∪C)⊂A∪(B∩C).A∪(B∩C)=(A∪B)∩(A∪C).2.(1)A−B=A−(A∩B)=(A∪B)−B;(2)A∩(B−C)=(A∩B)−(A∩C);(3)(A−B)−C=A−(B∪C);(4)A−(B−C)=(A−B)∪(A∩C);(5)(A−B)∩(C−D)=(A∩C)−(B∪D);(6)A−(A−B)=A∩B.(1)A−(A∩B)=A∩∁s(A∩B)=A∩(∁sA∪∁sB)=(A∩∁sA)∪(A∩∁sB)=A−B;(A∪B)−B=(A∪B)∩∁sB=(A∩∁sB)∪(B∩∁sB)=A−B;(2)(A∩B)−(A∩C)=(A∩B)∩∁s(A∩C)=(A∩B)∩(∁sA∪∁sC)=(A∩B∩∁sA)∪(A∩B∩∁sC)=A∩(B∩∁sC)=A∩(B−C);(3)(A−B)−C=(A∩∁sB)∩∁sC=A∩∁s(B∪C)=A−(B∪C);(4)A−(B−C)=A−(B∩∁sC)=A∩∁s(B∩∁sC)=A∩(∁sB∪C)=(A∩∁sB)∪(A∩C)=(A−B)∪(A∩C);(5)(A−B)∩(C−D)=(A∩∁sB)∩(C∩∁sD)=(A∩C)∩∁s(B∪D)=(A∩C)−(B∪D);(6)A−(A−B)=A∩∁s(A∩∁sB)=A∩(∁sA∪B)=A∩B.3.(A∪B)−C=(A−C)∪(B−C);A−(B∪C)=(A−B)∩(A−C).(A∪B)−C=(A∪B)∩∁sC=(A∩∁sC)∪(B∩∁sC)=(A−C)∪(B−C);(A−B)∩(A−C)=(A∩∁sB)∩(A∩∁sC)=A∩∁sB∩∁sC=A∩∁s(B∪C)=A−(B∪C).∞∞4.∁s(Ai)=∁sAi.i=1∞i=1∞x∈∁s(i=1Ai),x∈S,x∈i=1Ai,i,x∈Ai,x∈∁sAi,1limx∈A2n,limlimlimlimlimn.limlimlimlim∞∞∞x∈i=1∁sAi.∞x∈i=1∁sAi,∞∞i,x∈∁sAi,x∈S,x∈Ai,x∈S,x∈i=1Ai,x∈∁s(i=1Ai).∁s(i=1Ai)=i=1∁sAi.5.(1)(α∈ΛAα)−B=α∈Λ(Aα−B);(2)(α∈ΛAα)−B=α∈Λ(Aα−B).(1)α∈ΛAα−B=(α∈ΛAα)∩∁sB=α∈Λ(Aα∩∁sB)=α∈Λ(Aα−B);(2)α∈ΛAα−B=(α∈ΛAα)∩∁sB=α∈Λ(Aα∩∁sB)=α∈Λ(Aα−B).n−16.{An}nnB1=A1,Bn=An−(ν=1Aν),n1.{Bn}ν=1Aν=ν=1Bν,1≤n≤∞.i=j,ij.Bi⊂Ai(1≤i≤n).j−1Bi∩Bj⊂Ai∩(Aj−n=1nAn)=Ai∩Aj∩∁sA1∩∁sA2∩···∩∁sAi∩···∩∁sAj−1=∅.nBi⊂Ai(1=i=n)i=1Bi⊂i=1Ai.nnx∈i=1Ai,x∈A1,x∈B1⊂i=1Bi.x∈A1,inx∈Ain,in−1in−1nnnx∈i=1Aix∈Ain.x∈Ain−i=1Ai=Bin⊂i=1Bi.i=1Ai=i=1Bi.7.A2n−1=0,n1,A2n=(0,n),n=1,2,···,n→∞An=(0,∞);{An}Nx∈(0,∞),N,xN,xnNAn,0xn,x∈n→∞An,xn→∞An⊂(0,∞),n→∞An=(0,∞).n→∞An=∅;x∈n→∞An=∅,N,nN,x∈An.2n−1Nx∈A2n−1,0x1n→∞0x≤0,n→∞An=∅.∞∞8.limAn=n→∞n=1m=nAm.∞∞∞x∈n→∞An,N,nN,x∈An,x∈m=n+1Am⊂n=1m=nAm,∞∞∞∞∞n→∞An⊂n=1m=nAm.x∈n=1m=nAm,n,x∈m=nAm,m≥n,x∈An,x∈n→∞An.limAn=∞∞Am.n→∞n=1m=n2S:x2+y2+(z−12)2=(12)2→→0f(x1−0)f(x1+0)≤f(x2−0)f(x2+0),(f(x−0),f(x+0)),9.(−1,1)(−∞,+∞)(−∞,∞)ϕ:(−1,1)→(−∞,+∞).x∈(−1,1),ϕ(x)=tanπ2x.ϕ(−1,1)10.(0,0,1)(x,y,z)∈S\(0,0,1),xOyMϕ(x,y,z)=xy,1−z1−z∈M.ϕSM11.AA△zrzGG={△z|△zz},G△zrz,12.∞Annn=1,2,···,A=n=0An.Ann+1nn+106,An=a,13.A§44,A=a.()§4AA:(x,y,r).(x,y)rx,yr0A=a.14.f(−∞,∞)E,(1)(2)x∈Ex∈(−∞,∞),xlim0+f(x+△x)=f(x+0)f(x+0)f(x−0).xlim−f(x+△x)=f(x−0)(3)x∈E,15.x1,x2∈E,(0,1)x1x2,11[0,1]3(3)Exϕ(0)=r1,ϕ(rn)=rn+2,n=1,2,···ϕ(x)=x,x∈((0,1)\R),√xi∈Ai,Ai=c,i=1,2,···.xi∈Ai,(a1,a2,···),(a1,a2,a3,···)∈E∞,ai∈R,i=1,2,···,′′′′ϕ16.(0,1)[0,1](0,1)AR={r1,r2,···},ϕ(1)=r2,AAn.A={x1,x2,···},AAn2nA=∞A.An={x1,x2,···,xn},AnAn,AAn=1A17.[0,1]c.[0,1]B=A,[0,1]√2√2,23,···,√2n,···{r1,r2,···},⊂Aϕ(ϕ(√2)=2n22n+1√2,n+1)=rn,n=1,2,···n=1,2,···ϕ(x)=x,x∈B.ϕ18.xiA[0,1]Ac[0,1]Ac,c.Ac.A={ax1x2x3···},E∞AiRϕi.ax1x2x3···∈A.ϕ(ax1x2x3···)=(ϕ1(x1),ϕ2(x2),ϕ3(x3),···).ϕAϕϕ(ax1x2x3···)=ϕ(ax′1x′2x′3···),i,ϕi(xi)=ϕi(x′i).ϕixi=xi,ax1x2x3···=ax1x2x3···.∞ϕϕi(xi)=ai.ax1x2x3···∈A,ϕ(ax1x2···)=(ϕ1(x1),ϕ2(x2),···)=AE∞c.ϕi19.Anc,n0,An0c.n=1∞E∞=c,n=1An=E∞.Anc,n=1,2,···.PiE∞Rx=(x1,x2,···,xn,···)∈E∞,Pi(x)=xi.A∗i=Pi(Ai),i=1,2,···,4ξ∈∈Ai,iiϕ:{ξ1,ξ2,···}→{ξ2,ξ3,···},A≤E∞=c,x=0.ξ1ξ2···,T≥(0,1]=c.A∗A∞c,i=1,2,···.An.n=1ξ∈∞∞n=1An,i,ξi∈R\A∗i,i,ξ=(ξi,ξ2,···,ξn,···)∈E∞.ξi=Pi(ξ)∈Pi(Ai)=A∗i,ξ∈R\A∗iξ∈n=1An=E∞,ξ∈E∞i0,Ai0=c.20.01T,Tc.T={{ξ1,ξ2,···}|ξi=0or1,i=1,2,···}.Tx∈(0,1](0,1]TE∞ϕf((0,1])(0,1]2ξi01,TE∞ϕ(T)f(x)={ξ1,ξ2,···},A=c.f5P0∈E),P0∈E′,P0∈E.sin1,′o(−∞,∞)EoE′E¯1.P0P0P0∈E′P1U(P,δ)(E(P0P0P1)U(P,δ)(oU(P,δ)⊂E.P0)P0P0P1∈E∩U(P0)⊂E∩U(P,δ)E.U(P,δ),P1=P,P0P0U(P0)⊂U(P,δ),P1P0P1P0E,P0′P1E,P0U(P0)P0∈Eo,U(P0)⊂E.P0∈U(P,δ)⊂E,U(P0)⊂U(P,δ)⊂E,P0∈Eo.2.E1[0,1]E1R1E1′,E1o,E¯1.E1′=[0,1],E1o=∅,E¯1=[0,1].3.E2={(x,y)|x2+y21}.E2R2E2′,E2o,E¯2.E2′={(x,y)|x2+y2≤1},E1o={(x,y)|x2+y21},E¯1={(x,y)|x2+y2≤1}.4.E3R2y=0,xE3x=0,x=0E3E3.E3′=E3∪{(0,y)|−1≤y≤1},E3o=∅.5.R22E1′,E1o,E¯1E1′={(x,0)|0≤x≤1,},E1o=∅,E¯1=E1′.6.FF¯=F.FFF′⊂F,F¯=F∪F′=F.F¯=F,F′⊂F∪F′=F¯=F,7.GF−G=F∩∁GF∁G∁FG−F=G∩∁F8.f(x)E={x|f(x)≥a}1a,E={x|f(x)a}limGn=x|d(x,F),Gny∈F,d(x0,y)≥d(x0,y0)=δ(n.n,x0∈Gn,d(x0,F)n,n,y∈Fn,y∈Fn.x0∈E,(−∞,∞),|x−x0|δf(x0)a.f(x)a,f(x)x∈U(x0,δ)x∈E,δ0,U(x0,δ)⊂E,Ex∈xn∈E,xn→x0(n→∞).f(xn)≥a,f(x)f(x0)=n→∞f(xn)≥a,x0∈E,9.Ey0∈F,F11n11d(x0,F)=infd(x0,y)≥1d(x0,F)1n).ǫ=1n−δ0,x∈U(x0,ǫ),d(x0,x)ǫ.d(x,y0)≤d(x0,x)+d(x0,y0)ǫ+δ=ǫ+1n−ǫ=1n.d(x,F)=infd(x,y)≤d(x,y0)1x∈Gn.U(x0,ǫ)⊂Gn,Gn∞x∈n=1Gn,n,x∈Gn,d(x,F)1n→∞,d(x,F)=0.∞Fx∈F(x∈F,∞yn∈F,d(x,yn)→0,∞x∈F′⊂F,),n=1Gn⊂F.Gn⊃F,n=1,2,···,n=1Gn⊃F,n=1Gn=F,F∞G∁GGn,∁G=Gn,n=1∞∞∁GnGG=∁(∁G)=∁(n=1Gn)=n=1∁Gn,10.[0,1][0,1][0,1]77(0.7,0.8).7······(0.07,0.08)(0.17,0.18)···(0.97,0.98).[0,1]n7(0.a1a2···an−17,0.a1a2···an−18),ai(i=1,2,···,n−1)n−1097{a1,a2,···,an−1}∞Ann=12x0∈E(xn∈E={x|f(x)≥c},Pt0∈E.t∈[0,1]t0t≤1,Pt0∈E,Ik,k=1,2,···,2n−1Ik[0,1]7∞∁n=1An∪(−∞,0)∪(1,∞).An,(−∞,0),(1,∞)[0,1]711.f(x)E1={x|f(x)≤c}[a,b]c,E={x|f(x)≥c}f(x)[a,b]8EE1EE1x0∈[a,b].f(x)f(xn)≤f(x0)−ǫ0,f(x0)f(x0)+ǫ0=c),x0Eǫ00,xn→x0,f(xn)≥f(x0)+ǫ0c=f(x0)+ǫ,f(x)[a,b]12.§25:E=∅,E=Rn,E(∂E=∅).P0=(x1,x2,···,xn)∈E,P1=(y1,···,yn)∈E.(1−t)x2,···,tyn+(1−t)xn),0≤t≤1.t0=sup{t|Pt∈E}.Pt=(ty1+(1−t)x1,ty2+Pt0∈∂E.t0,tn→t0Ptn∈E,Ptn→Pt0,Pt0∈∂E.t0=1.Pt∈E.tn,1tnPt0∈∂E.∂E=∅.t0=0,tn,0tnt0,tn→t0,Ptn→Pt0,Ptn∈E,P13.c.P1,P12,3312,9978,99······=(0.1,0.2),=(0.01,0.02),=(0.21,0.22),(P),n2n−1(n)(n)=