A Concrete Approach to Classical Analysis (Instructor Solution Manual, Solutions)

83 Pages·2008·0.322 MB·English

by  Marian Muresan

Marian Mure(cid:24)san
A Concrete Approach to
Classical Analysis
Solutions to Exercises
February 29, 2012
Springer

Contents
1 Sets and Numbers ......................................... 5
2 Vector Spaces and Metric Spaces .......................... 25
3 Sequences and Series ...................................... 27
4 Limits and Continuity ..................................... 39
5 Differential Calculus on R ................................. 49
6 Integral Calculus on R .................................... 57
7 Differential Calculus on Rn ................................ 69
8 Double Integrals, Triple Integrals, and Line Integrals ...... 75

1
Sets and Numbers
1.1.(i)Supposethatthesetintheleft-sidecontainsanelement,say x. Then
x belongs to A, not to B, and to B. We get a contradiction, hence the set
in the left-side contains no element.
(ii) Obviously, A∩B ⊂A\(A\B). If there exists an x∈A\(A\B), then
x∈A and x∈/ A\B. It follows that x∈B, so that x∈A∩B.
(iii) Both sides of this identity contain elements belonging either to A or to
B.
(iv) Both sides of the identity contain elements in A and neither in B nor in
C.
(v) The three members of this double identity contain elements belonging to
A and C, but not to B.
1.2. Suppose that A \ B ⊂ C and pick up an x ∈ A. If x ∈/ B, then
x ∈ A\B and thus x ∈ C. If x ∈ B, then x ∈ B∪C. Conversely, suppose
that A ⊂ B∪C and pick up an x ∈ A\B. Then x ∈ A and x ∈/ B. Thus
x∈(B∪C)\B ⊂C.
1.3. (i) We have
x∈(A∪B)\C ⇐⇒(x∈A or x∈B) and x∈/ C
⇐⇒(x∈A and x∈/ C) or (x∈B and x∈/ C)
⇐⇒x∈(A\C)∪(B\C).
(ii) We have
x∈(A∩B)\C ⇐⇒(x∈A and x∈B) and x∈/ C
⇐⇒(x∈A and x∈/ C) and (x∈B and x∈/ C)
⇐⇒x∈(A\C)∩(B\C).
6 1 Sets and Numbers
(iii) By (i) we have
(A∪C)\B =(A\B)∪(C\B)⊂(A\B)∪C.
1.4. (i) One can easily show that Y =(B\C)\A and Y ⊂X. If C =∅ and
A=B ̸=∅, then ∅=Y ̸=A=X. We conclude that Y ⊂X.
1.5. (i)
x∈∪∞ B ⇐⇒∃n∈N, x∈B ⇐⇒∃n∈N, i∈{0,...,n}, x∈A
n=0 n n i
⇐⇒x∈∪∞ A .
n=0 n
(ii)
x∈∩∞ B ⇐⇒∀n∈N, x∈B ⇐⇒∀n∈N, i∈{0,...,n}, x∈A
n=0 n n i
⇐⇒x∈∩∞ A .
n=0 n
1.6. (i)
x∈∪∞ (∩∞ A ) =⇒ ∃m ∈N, x∈∩∞ A
m=0 n=0 m;n 0 n=0 m0;n
=⇒ ∃m ∈N,∀n∈N, x∈A
0 m0;n
=⇒ x∈∩∞ (∪∞ A )⊂∩∞ (∪∞ A ).
n=0 m=m0 m;n n=0 m=0 m;n
(ii) Not true.
1.7. (i) { ∅=X\∅=X and { X =X\X =∅.
X X
(ii) {(A\B)={(A∩{B))=({A)∩({({B))=({A)∪B.
(iii)
(A∪B)\(A∩B)=(A∪B)∩({(A∩B))=(A∪B)∩({A∪{B)
=(A∩({A∪{B))∪(B∩({A∪{B))
=((A∩{A)∪(A∩{B)∪((B∩{A)∪(B∩{B)=(A∩{B)∪((B∩{A).
(iv) x∈{B =⇒ x∈/ B =⇒ x∈/ A =⇒ x∈{A.
1.8. Successively we have
{(X∪A)∪(X∪{A)=B ⇐⇒(({X∩{A)∪{A)∪X =B
⇐⇒(({X∪{A)∩({A∪{A))∪X =B ⇐⇒(({X∪{A)∩{A)∪X =B
⇐⇒{A∪X =B ⇐⇒A∩({A∪X)=A∩B ⇐⇒A∩X =A∩B.
Thus X is any subset of U satisfying A∩X =A∩B.
1.1 Solutions 7
1.9. (i)
X ∈P(A)∪P(B) =⇒ X ∈P(A) or X ∈P(B)
=⇒ X ⊂A or X ⊂B =⇒ X ⊂A∪B =⇒ X ∈P(A∪B).
(ii) Pick a nonempty set X ∈ P(A∪B). Then X ⊂ A∩B. If X ⊂ A or
X ⊂ B, the equality is true. Suppose that X = X ∪X , with X ̸= ∅,
1 2 1
X ⊂A, X ̸=∅, X ⊂B, and X ̸=X . Then X ∈/ P(A)∪P(B).
1 2 2 1 2
(iii) and (iv) are straightforward.
1.10. (i) One-to-one.
(ii) Onto.
(iii) One-to-one and onto.
1.11. (a) (i) =⇒ (ii) Consider f∗(M1) = f∗(M2). This is equivalent to
f(M ) = f(M ). We show that M = M . Pick an arbitrary x ∈ M . Then
1 2 1 2 1
f(x)∈f(M )=f(M ). Itmeansthatthereexists y ∈M with f(x)=f(y).
1 2 2
Since the function f is one-to-one, it follows that x = y and x ∈ M . Thus
2
M ⊂M . Inasimilarwayweshowthat M ⊂M , andtherefore M =M .
1 2 2 1 1 2
Thus the function f∗ is one-to-one.
(ii) =⇒ (iii) Let M ∈ P(A) be arbitrary. We show that there exists N ∈
P(B) with f∗(N) = M, that is, for M ⊂ A there exists N ⊂ B with
f−1(N) = M. Define N = f(M). From the injectivity of f we indeed have
that M =f−1(N)=f∗(N).
(iii) =⇒ (iv) The inclusion f(M ∩N) ⊂ f(M)∩f(N) is obvious. We now
show its reverse. Let N ,N ∈P(B) be with f−1(N )=M and f−1(N )=
1 2 1 1 2
M be arbitrary. We have f−1(N ∩ N ) = f−1(N ) ∩ f−1(N ). For y ∈
2 1 2 1 2
f(M )∩f(M ) there exists x ∈ M ,x ∈ M with f(x ) = f(x ) = y.
1 2 1 1 2 2 1 2
Since f(x ) ∈ N , f(x ) ∈ N , it follows that y ∈ N ∩N . Thus f−1(y) ∈
1 1 2 2 1 2
f−1(N ∩N )=f−1(N )∩f−1(N )=M ∩M , andthereexists x∈M ∩M
1 2 1 2 1 2 1 2
with f(x)=y. So f(M)∩f(N)⊂f(M ∩N).
(iv) =⇒ (v) Taking M =M and M ={M, we have that f(M)∩f({M)=
1 2
f(M ∩{M)=f(∅)=∅ and thus f({M)⊂{f(M).
(v) =⇒ (i) Consider x,y ∈ A with f(x) = f(y) and define M = A\{x}.
Then {M = {x} and f(x) = f(y) ∈ f({M) ⊂ {f(M). Thus f(y) ∈/ f(M)
and y ∈/ M. Therefore y =x and the function f is one-to-one.
(b) The proof is similar to (a).
(c) Follows from (a) and (b).
1.12. (a) (i) =⇒ (ii) For every ξ ∈ C we have that (f ◦g)(ξ) = (f ◦h)(ξ),
i.e., f(g(ξ))=f(h(ξ)). Becauseof f isone-to-one,itfollowsthat g(ξ)=h(ξ)
and since ξ was taken arbitrarily in C, we have that g =h.
(ii) =⇒ (i)Supposethat f isnotone-to-one.Thenthereexist x,y ∈A, with
x̸=y and f(x)=f(y). Consider C ={u,v} and the functions g,h:C →A
8 1 Sets and Numbers
satisfying g(u)=x, g(v)=y, and h(u)=h(v)=y. Thenwehave f◦g =f◦g
and g ̸=h.
(b)(i) =⇒ (ii)Let y ∈B bearbitrary.Since f isonto,thereexists x∈A so
that f(x)=y. We have the sequence of implications g(f(x))=h(f(x)) =⇒
g(y)=h(y) =⇒ g =h.
(ii) =⇒ (i) Suppose f is not onto. Then there exists y ∈B so that for every
x∈A, f(x)̸=y. Consider C ={0,1} and g(z)=0, for all z ∈B, h(z)=0,
for all z ∈B\{y} and h(y)=1. Then g◦f =h◦f and g ̸=h.
1.13. (i) There exists an injection N∗ ∋n→2n ∈N∗.
(ii) The set is at least countable since the coordinates a given vertice runs
on the countable set Q×Q. The same set is a subset of the countable set
(Q×Q)×(Q×Q)×(Q×Q), Theorem 2.17.
(iii) The set coincides with Q×Q which is countable.
(iv) It follows from Theorem 2.17.
1.14. (i) Denote N ={a ,a ,...,a } and M ={b ,b ,...,b } Then to a
1 2 n 1 2 m 1
onecanassignanarbitraryelementfrom M, i.e.,thereare m possibilities,to
a onecanassignanarbitraryelementfrom M, i.e.,thereare m possibilities,
2
and so on.
(ii) Follows from (iii).
(iii) To a one can assign an arbitrary element from M, let it be {b }, i.e.,
1 i
there are m possibilities, to a one can assign an arbitrary element from
2
M \{b }, let it be {b }, i.e., there are m−1 possibilities, to a one can
i j 3
assign an arbitrary element from M \{b ,b }, let it be {b }, i.e., there are
i j k
m−2 possibilities, and so on till to each element from M is assigned one
from N.
(iv) Denote the set of onto functions from M into N by Sn, m ≥ n. For
m
each i = 1,...,n, let A be the set of functions from M into N such that
i
b does not belongs to the image of any such function. Then the set Sn
i m
of onto functions from M into N coincides with the set of all functions
from M into N except the functions belonging to some of the sets A . Then
i
|Sn| = nm−|A ∪A ∪···∪A |. By the first equality in Exercise 1.18, we
m 1 2 n
expand |A ∪A ∪···∪A |, and remark that
1 2 n
|A |=(n−1)m, |A ∩A |=(n−2)m,...,∩n A =∅.
i i j i=1 i
( ) ( )
Wealsonotethatthereare n sets A , n sets A ∩A , ....Nowtheresult
1 i 2 i j
follows.
1.15. First approach. One proves them by induction.
Second approach. One proves them by combinatorial argues. For the first
identity a combinatorial argue is the next one. Choose a set A of n ∈ N
elementsandaset B withtwoelements 0 and 1. Weidentifyasubset M ⊂A
with a mapping f : M → B defined as f(x) = 1 if x ∈ M and f(x) = 0
1.1 Solutions 9
otherwise. Thus we obtain a bijection between the family of subsets of A
and the set of functions from M into B. The cardinal of the later set is 2n,
according to (i) in Exercise 1.14. For the next two identities consider C and
D a partition of family of subsets of A so that C contains subsets of odd
cardinal whereas D contains subsets of even cardinals. We already saw that
|C∪D|=2n. Pick up an element in A, denote it x. We define a bijection f
from C onto D such that f(X) = X ∪{x} if x ∈/ X and f(X) = X \{x}
otherwise.Itfollowsthatthesets C and D areofthesamecardinalandthus
the last two identities are proved.
One can try using the next Mathematicar commands.
TraditionalForm[Sum[Binomial[n, k], {k, 0, n},Assumptions
\[RightArrow] n\[Element]Integers &&n\[GreaterSlantEqual] 0]]
TraditionalForm[Sum[Binomial[n,2*k],{k,0,n/2},Assumptions
\[RightArrow] n\[Element]Integers &&n\[GreaterSlantEqual] 0]]
TraditionalForm[Sum[Binomial[n,2*k+1],{k,0,n/2},Assumptions
\[RightArrow] n\[Element]Integers &&n\[GreaterSlantEqual] 0]]
1.16. Let A be the family of partitions of S. Then
S ={1} =⇒ A={S},
S ={1,2} =⇒ A={S, {{1},{2}}},
S ={1,2,3} =⇒ A={S, {{1},{2},{3}}, {{1,2},{3}}, {{1,3},{2}},
{{2,3},{1}} },
S ={1,2,3,4} =⇒ A={S, {{1},{2},{3},{4}}, {{1,2,3},{4}},
{{1,2,4},{3}}, {{1,3,4},{2}}, {{2,3,4},{1}}, {{1,2},{3,4}},
{{1,2},{3},{4}},{{1,3},{2,4}}, {{1,3},{2},{4}}, {{1,4},{2,3}},
{{1,4},{2},{3}}, {{2,3},{1},{4}}, {{2,4},{1,3}}, {{3,4},{1},{2}}}.
Thus the equalities in (1.11) of the statement hold. Some results can be ob-
tained by Mathematicar, Figure 1.1.
One can also try using the next Mathematicar command.
Table[BellB[n], {n, 0, 4}]
Now suppose S = {a1,a2,...,an,an+1}. Pick up(a)n arbitrary element in S,
say a . Select k elements in S \{a } (in n ways) and denote by A
n+1 n+1 k
thesetofremaining n−k elements.By A onehas Bn−k partitions.Thuswe
get (1.12).
1.17. Every x∈A belongs to at least a set A . The set of indices of the sets
i
A containing x is a nonempty subset to {1,2,...,n}. So, there are 2k −1
i
possibilities for x belonging to at least a set A . Since each x∈A belongs to
i
at least a set A , the conclusion follows.
i
10 1 Sets and Numbers
H* Set partitions*L
SetPartitions@4D
8881,2,3,4<<,881<,82,3,4<<,881,2<,83,4<<,881,3,4<,82<<,
881,2,3<,84<<,881,4<,82,3<<,881,2,4<,83<<,881,3<,82,4<<,
881<,82<,83,4<<,881<,82,3<,84<<,881<,82,4<,83<<,881,2<,83<,84<<,
881,3<,82<,84<<,881,4<,82<,83<<,881<,82<,83<,84<<<
H* Bell numbers*L
BellB@4D
15
Fig. 1.1. Set partitions and Bell numbers for n=4
1.18. By induction.
1.19. Both sides increase with ⌈y⌉, if x is substituted by x+1. Therefore
we can consider that 0 ≤ x < 1. Both sides vanish for x = 0 and both sides
increase by 1 if x increases over 1−k/y, for 0≤k <y.
⌊ ⌋
n+2k
1.20. There exists k ∈ N∗ such that n < 2k0. Then = 0 for all
0 2k+1
k ≥ k . Thus the left-hand side of (1.13) in the statement contains precisely
0
a finite sum.
For any real x we have
⌊ ⌋
1
x+ =⌊2x⌋−⌊x⌋.
2
Applying the above relation to each nonzero term in the left-side of (1.13),
we get the result.
1.21.Suppose (a−r)n+an =(a+t)n. Denote t=a/r. Then (t−1)n+tn =
(t+1)n, that is
(( ) ( ) )
n n
tn−2 tn−1+ tn−3+···+1 =0.
1 3
Since the coefficient of the leading term is 1 and all the other coefficients are
integers, the rational roots are integers. If t is odd, we get a contradiction.
If t is even, the leading term is divisible by 4, while the product is divisible
only by 2. We get a contradiction again. Thus the polynomial equation has
no integer root. So it has no rational root. ⊓⊔
1.22. Suppose that |A|=n. Then P(A) is equivalent with the set of n-term
sequences of 0 and 1.