,實(shí)變函數(shù)與泛函分析基礎(chǔ)(程其襄張奠宙著)高等教育出版社課后答案

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) (?∞,+∞)?:(?1,1)→(?∞,+∞).

x∈(?1,1),?(x)=tanπ2x.?

(?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?z? 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