,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∪B x∈A∪C, x∈(A∪B)∩(A∪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∪)−C=(A−

(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(i=1

Ai)= ∁sAi.i=1∞ ∞x∈∁(

Ai), x∈S,

Ai, i,x∈A, x∈∁A,si=1

i=1

i si1∞∁sAi.

∞∁sAi,

i,x∈∁A,

x∈S,x∈A,

x∈S,

∞Ai,i=1 x∈∁(

i=1 Ai). ∁s(

si i∞∁sAi.

i=1si=15.

(1)(

i=1A)−B

i=1(A−B);

(2)(

A)−B

(A−B).Λ α

Λ

Λ α

α∈Λ α(1)

A−B=(

A)∩∁B

(A∩∁B)=

(A−B);Λ α

α

α sΛ

α∈Λ α(2) A−B=(

A)∩∁B=

(A∩∁B)=

(A−B).Λ

α

α

n−1

Λ α6. {A}

B=A,B

=A−(

Aν),n>1.

{B}n 1 1 n n n

ν=1 nν=1

ν=1

Bν,1≤n≤∞.i=j, i<j−1

Bi⊂Ai (1≤i≤n).i∩j⊂i∩(j−

n=1n

An)=Ai∩Aj∩∁sA1∩∁sA2∩···∩∁sAi∩···∩∁sAj−1=∅.nBi⊂Ai(1=i=n)n

i=1

i=1

Ai.n

x∈A, x∈B

Bi.

x∈A, in x∈A,i=1

1 1 i=1 1 inin−1 in−1 n n nx∈ Ai x∈A.

x∈

− A=B⊂

Bi.

= Bi.i=1

in

i=1

i

i=1 i=1 i=1A7.A2n−1

=0,

,

=(0,n),n=1,2,···,

{An}AlimAn→∞

n=(0,∞);x∈(0,∞), N, x<N,

0<x<n,

xx∈A,AlimAn→∞

n=(0,∞).

x An,

limx∈ x∈ A

2n→∞ nlimA⊂→∞ AlimAnn→∞n

n=∅;Ax∈Ax∈n→∞

=∅, N, n>N, x∈A. 2n−1>Nx∈A ,

0<x< 1.

n→∞ 0<x≤0,

=∅.2n−18. lim

n∞ ∞Am.

n→∞ nn→∞

n=1m=n

∞ ∞ ∞x∈limA,

N, n>N,x∈A, A

Am,n→∞ n∞ ∞ ∞

n m=n+1 ∞

n=1m=nlimA⊂

Am.

Am,

n, Am,

m≥n,n→∞n

n=1m=n n=1m=n

m=nx∈An, x

limA.limAn=

n→∞ n∞

Am.n→∞ n=1m=n29.(−∞,∞)10.

(−1,1) (−∞,+∞)ϕ:(−1,1)→(−∞,+∞).

x∈(−1,1),ϕ(x)=tanπ2x.ϕ

(−1,1)S:x+y+(z−)=(

(0,0,1)

xOy M2 2 122 122(x,y,z)∈S\(0,0,1),y,z)= xy

∈M.1−z1−zϕ S M11.△

G={△z|△zz

A A}, △z

rz,zG G12.

An

n n+1

n=1,

∞A=n=0

An.An0

n+1=a, §4

§44,A=a.13. A ( )AA :(x,y,r). (x,y)x,y r 0

rA=a.14.f (−∞,∞) E,(1) x∈(−∞,f(x=f(x+0)

limf(x+△x)=f