Shengtian Yang
Notes
on Walter Rudin’s
Principles of Mathematical Analysis
Version: 0.3.1
Date: 2014-01-03
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Title: Notes on Walter Rudin’s Principles of Mathematical Analysis.
Author: Shengtian Yang.
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c 2013 by Shengtian Yang.
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Contents
Preface
v
1 The Real and Complex Number Systems
1.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
3
2 Basic Topology
2.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
13
13
13
3 Numerical Sequences and Series
3.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . .
23
23
23
Bibliography
39
iii
iv
Contents
Preface
This document includes my solutions to the exercises of the book Principles
of Mathematical Analysis (Rudin, 1976), as well as my notes on some interesting
facts in the book. Because of copyright reasons, the original text of the exercises
is not included in the public release of this document.
Rudin’s book is very well known. As someone said, it can rightly be called
“the bible of classical analysis”. The book may be difficult for a beginner, but it
is very suitable for those (like me) who have already basic knowledge of calculus
or analysis.
I am very glad for bug reports as well as any comments on all aspects of this
document.
Shengtian Yang
Hangzhou, China
yangst@codlab.net
June 12, 2013
(Last revised on June 12, 2013)
v
vi
Contents
Chapter 1
The Real and Complex Number Systems
1.1
Notes on Text
√
Note 1.1 (p. 2, Example 1.1) The described trick is useful for estimating 2. It
essentially requires us to find a function f such that for any initial number x0 > 0,
the sequence
x0 , x1 = f (x0 ), x2 = f (x1 ), . . . ,
√
being ascending or descending, converges to 2. (An extra requirement is that
f (Q) ⊆ Q.)
A heuristic approach is to find a function having the following form:
f (x) := x −
x2 − 2
,
g(x)
√
where g(x) is positive
√ for x > 0. This ensures that f (x) >√x for x < 2 and
f (x) < x for x > 2. To make the sequence
converge
to 2, we
√
√
√ still need√to
choose appropriate g such that f (x) < 2 for x < 2 and f (x) > 2 for x > 2.
Since
[
]
2x
x2 − 2
2
2
(f (x)) − 2 = (x − 2) 1 −
+
,
g(x) (g(x))2
we have the following equivalent condition:
(g(x))2 − 2xg(x) + x2 − 2 > 0.
If we let g(x) = x + a, the above inequality further reduces to
a2 − 2 > 0.
Now we choose a = 32 , and then
f (x) =
3x + 4
.
2x + 3
Taking x0 = 0, we obtain the following sequence:
x0 = 0,
x1 =
4
,
3
x2 =
24
≈ 1.412,
17
x2 =
140
≈ 1.41414,
99
....
b
1
2
1. The Real and Complex Number Systems
Note 1.2 (p. 10, Theorem 1.21)
Fact 1.1. For b > a > 0 and n ∈ Z≥2 , nan−1 (b − a) < bn − an < nbn−1 (b − a).
b
Proof. Use bn − an = (b − a)(bn−1 + bn−2 a + · · · + ban−2 + an−1 ).
Note 1.3 (p. 21, Step 9)
Fact 1.2. Any two ordered fields with the least-upper-bound property are isomorphic.
b
Proof. We will show that any ordered field F with the least-upper-bound property
is isomorphic to R. In other words, there is a ring-isomorphism from R to F that
is also monotone increasing.
Suppose that 0′ and 1′ are the additive identity and the multiplicative identity
of F , respectively. We first define a map from Q, the subfield of R, to F . For any
x ∈ Q, write x = m/n where m and n are integers and n ̸= 0. We define
f (x) := (m1′ )/(n1′ ),
where n1′ denotes the sum of n 1′ s.
Clearly, f is well defined, and is in fact a ring-monomorphism, since
f (a/b + c/d) = f ((ad + bc)/bd)
= [(ad + bc)1′ ]/(bd1′ )
= (a1′ )/(b1′ ) + (c1′ )/(d1′ )
= f (a/b) + f (c/d),
f ((a/b)(c/d)) = f ((ac)/(bd))
= (ac1′ )/(bd1′ )
= [(a1′ )/(b1′ )][(c1′ )/(d1′ )]
= f (a/b)f (c/d),
f (1) = 1′ /1′ = 1′ .
The map f is monotone increasing since for any rational x = m/n > 0 where
m, n > 0, we have f (x) = (m1′ )/(n1′ ) > 0, where the last inequality follows from
the property of an ordered field (Definition 1.17 and Proposition 1.18).
Now we extend f from Q to R. We define the map g : R → F by
g(x) :=
sup
y∈Q:y<x
f (x).
1.2. Solutions to Exercises
3
It is well defined since F has the least-upper-bound property. It is also clear that
g(x) = f (x) for x ∈ Q. Furthermore, it is easy to show that g is a monotone
increasing ring-monomorphism. Finally, let us show that it is in fact an isomorphism. For any x′ ∈ F , it follows from Theorem 1.20 that there exists a sequence
yk′ = (mk 1′ )/(nk 1′ ) such that x′ − 1′ /(k1′ ) < yk′ < x′ . Let x = supk (mk /nk ). We
have
g(x) > g(mk /nk ) = yk′ > x′ − 1′ /(k1′ ),
so that g(x) ≥ x′ . If g(x) > x′ , then there would exist a rational y ′ = (m1′ )/(n1′ )
such that x′ < y ′ < g(x). It then follows that
y := m/n ≥ sup(mk /nk ) = x,
k
contrary to the monotone increasing property of g. Therefore g(x) = x′ , so that
g is surjective.
1.2
Solutions to Exercises
1.
Proof. Suppose that r + x (resp. rx) is rational. Since r is rational and nonzero,
both −r and 1/r are rational, so that x = −r + (r + x) (resp. x = (1/r)rx) is
rational, a contradiction.
2.
Proof. If there were a rational x such that x2 = 12, we could write x = m/n
where m and n are not both multiples of 3. Then x2 = 12 implies that
m2 = 12n2 .
This shows that 3 divides m2 , and hence, that 3 divides m, so that 9 divides
m2 . It then follows that n2 is divisible by 3, so that n is a multiple of 3, which
contradicts our assumption.
3.
Proof. If xy = xz and x ̸= 0, then y = 1y = (1/x)xy = (1/x)xz = 1z = z. This
proves (a). Take z = 1 in (a) to obtain (b). (a) with z = 1/x gives (c). (c) with
identity (1/x)x = 1 yields (d).
4.
4
1. The Real and Complex Number Systems
Proof. Choose x ∈ E and we have α ≤ x and x ≤ β, so that α ≤ β.
5.
Proof. Note first that there is a one-to-one correspondence between x ∈ A and
−x ∈ −A. By definition, x ≥ inf A for all x ∈ A, so that −x ≤ − inf A for
all −x ∈ −A, and hence sup(−A) ≤ − inf A, or inf A ≤ − sup(−A). Similarly,
y ≤ sup(−A) for all y ∈ −A, so that −y ≥ − sup(−A) for all −y ∈ A, and hence
inf A ≥ − sup(−A). Therefore inf A = − sup(−A).
6.
Proof. (a) It is clear that (xn )1/n = x = (x1/n )n for real x > 0 and integer n > 0
(Theorem 1.21). It is also clear that for positive integers m and n, xmn = (xm )n
and x1/(mn) = (x1/m )1/n , where the second identity can be proved by the following
argument:
[(x1/(mn) )n ]m = (x1/(mn) )mn = x,
which implies that x1/(mn) = (x1/m )1/n .
With the simple identities (1/x)n = 1/xn and (1/x)1/n = 1/x1/n where x > 0
and n ∈ Z>0 , it can be further shown that
(xn )1/n = x = (x1/n )n
mn
x
1/(mn)
x
m n
= (x )
= (x
1/m 1/n
)
for n ∈ Z̸=0 ,
for m, n ∈ Z,
for m, n ∈ Z̸=0 .
The convention here is x−m/n := 1/xm/n .
We are now ready to prove the identity. Since m/n = p/q, we have mq = np,
so that
(bm )1/n = {[(bm )q ]1/q }1/n
= (bmq )1/nq
= (bnp )1/nq
= {[(bp )n ]1/n }1/q
= (bp )1/q .
(b) Suppose r = m/n and s = p/q where n > 0 and q > 0. Then
br+s = b(mq+np)/(nq)
= (bmq+np )1/(nq)
= (bmq bnp )1/(nq)
= (bmq )1/(nq) (bnp )1/(nq)
r s
=b b .
(Corollary of Theorem 1.21)
1.2. Solutions to Exercises
5
(c) We first show that br is strictly increasing in r ∈ Q. It is clear that br is
strictly increasing in r ∈ Z (Proposition 1.18). Now suppose that r = m/n < s =
p/q where n > 0 and q > 0. It then follows that
br = (bmq )1/(nq) < (bnp )1/(nq) = bs .
This implies br ≥ sup B(r) for r ∈ Q, so that br = B(r) since br ∈ B(r).
It can be further shown that bx is strictly increasing in x ∈ R. For any real
x < y, choose rational r and s such that x < r < s < y (Theorem 1.20). By
definition, it follows that by ≥ bs > br ≥ bx .
(d) The following inequality will be needed in the proof:
b1/n − 1 < (b − 1)n−1
for n ∈ Z≥2 ,
which is an easy consequence of Fact 1.1.
We first show that bx+y ≤ bx by . Choose rational r ≤ x and s ≤ y such that
r > x − 1/(2n) and s > y − 1/(2n) where n ∈ Z≥2 . It follows that br ≤ bx and
bs ≤ by , so that
bx by ≥ br bs
= br+s
(Part (b))
=b
x+y
− (b
x+y
>b
x+y
− (b
r+s+1/n
−b
r+s
)
− br+s )
(Proof of part (c))
= bx+y − br+s (b1/n − 1)
> bx+y − br+s (b − 1)n−1 .
This implies that bx by ≥ bx+y .
On the other hand, choose rational r ≤ x and s ≤ y such that r > x − 1/n
and s > y − 1/n where n ∈ Z≥2 . It follows that
bx+y ≥ br+s = br bs ,
where
br > bx − (br+1/n − br ) > bx − br (b − 1)n−1 ,
bs > by − (bs+1/n − bs ) > by − bs (b − 1)n−1 .
This shows that
bx+y > bx by − 2br bs (b − 1)n−1
and therefore bx+y ≥ bx by . The proof is complete.
7.
6
1. The Real and Complex Number Systems
Proof. (a) bn − 1 = (b − 1)(bn−1 + bn−2 + · · · + 1) ≥ n(b − 1) (cf. Fact 1.1).
(b) b − 1 = (b1/n )n − 1 ≥ n(b1/n − 1).
(c) b1/n = 1 + b1/n − 1 ≤ 1 + (b − 1)/n < t.
(d) Taking n > (b−1)/(yb−w −1), and then it follows from (c) that bw+(1/n) =
w 1/n
b b
< y.
(e) Since bw > y, or b−w < y −1 , it follows from (d ) that b−w+(1/n) < y −1 for
sufficiently large n, that is, bw−(1/n) > y.
(f) From (d ) it follows that x ∈
/ A, so that bx ≥ y. On the other hand, it is
x
impossible that b > y, for otherwise it follows from (e) that x is not the least
upper bound of A. The proof is complete.
(g) Use the strict increasing property of bx (see the proof of Ex. 6.(c)).
8.
Proof. If there were an order defined in the complex field that turns it into an
ordered field, we would have −1 = i2 > 0 (Proposition 1.18), a contradiction.
9.
Proof. Enumerating all the possibilities shows that
a < c, a = c, or a > c
and
b < d, b = d, or b > d.
Every possibility gives z < w, z = w, or z > w. In other words, any z and w are
comparable.
To prove this is an ordered set, it remains to show that the order is transitive.
Suppose z = a + bi, w = c + di, and x = e + f i. Supposing z < w and w < x, we
will show that z < x.
The relation z < w shows that a < c or a = c but b < d. Likewise, w < x
gives c < e or c = e but d < f . Thus there are in total four possibilities:
1. a < c and c < e;
2. (a = c but b < d) and c < e;
3. a < c and (c = e but d < f );
4. (a = c but b < d) and (c = e but d < f ).
1.2. Solutions to Exercises
7
The first three cases yields a < e and the last case yields a = e but b < f .
Therefore z < x in all cases.
Let A consist of all complex numbers whose real part is 0. If z = a + bi is an
upper bound of A, then a must be positive, for otherwise we have z < (b+1)i ∈ A.
However, z = a + (b − 1)i is a smaller upper bound of A. This indicates that this
ordered set does not have the least-upper-bound property.
10.
Proof. At first, we note that
(
ab =
|w|2 − u2
4
)1/2
|v|
.
2
=
Then by conditions, we have
z 2 = a2 − b2 + 2abi = u + |v|i = w
for v ≥ 0
and
(z)2 = z 2 = u − |v|i = w
for v ≤ 0.
Given any w, the formula for z provides a way for computing one square root
of w, say x. It is clear that −x is another square root whenever x ̸= 0. In fact,
w has only two square roots, i.e., ±x. If there were another number y such that
y 2 = w, then y 2 = x2 , or (y + x)(y − x) = 0, so that y = x or y = −x, a
contradiction.
11.
Proof. Suppose z = a + bi. Let r = |z| and w = z/r. It is easy to show that r ≥ 0,
|w| = 1, and z = rw.
If z ̸= 0, it uniquely determines r and w. If there were another r′ and w′
such that r′ > 0, |w′ | = 1, and z = r′ w′ , then we would have r′ = |z| = r and
w′ = z/r′ = z/r = w.
12.
Proof. Use induction on n with Theorem 1.33.
13.
Proof. By Ex. 12, we have
|x| = |x − y + y| ≤ |x − y| + |y|,
so
that |x|
− |y| ≤ |x − y|, which also implies |y| − |x| ≤ |y − x| = |x − y|. Therefore
|x| − |y| ≤ |x − y|.
8
1. The Real and Complex Number Systems
14.
Solution.
|1 + z|2 + |1 − z|2 = 1 + 2 Re(z) + |z|2 + 1 − 2 Re(z) + |z|2 = 4.
15.
Answer. The equality holds if and only if Baj = Cbj for some B, C ∈ R and all
j = 1, 2, . . . , n (see the proof of Theorem 1.35).
16.
Proof. Consider the equation |z − x| = |z − y| = r, which is equivalent to
(z − x) · (z − x) = (z − y) · (z − y) = r2 .
After some manipulations, we get three equivalent equations from the first equality:
(2z − x − y) · (x − y) = 0,
2(z − x) · (y − x) = d2 ,
2(z − y) · (x − y) = d2 .
Furthermore, we obtain the following equation:
|2z − x − y|2 = |z − x|2 + 2(z − x) · (z − y) + |z − y|2
= 2r2 + 2|z − x|2 + 2(z − x) · (x − y)
= 4r2 − d2 .
It is then clear that |z − x| = |z − y| = r is equivalent to
{
[z − (x + y)/2] · (x − y) = 0,
√
|z − (x + y)/2| = r2 − d2 /4,
or more compactly,
{
x′ · y′ = 0,
√
|x′ | = r2 − d2 /4.
where y′ := x − y ̸= 0.
(a) It suffices to show that for any y ̸= 0 and s > 0, there are infinitely many
x ∈ Rk such that x · y = 0 and |x| = s, that is,
{
x1 y1 + x2 y2 + · · · + xk yk = 0,
x21 + x22 + · · · + x2k = s2 .
1.2. Solutions to Exercises
9
Since y ̸= 0, with no loss of generality, we suppose yk ̸= 0. Now let x3 , . . . ,
xk−1 be zero. We get two reduced equations:
{
x1 y1 + x2 y2 + xk yk = 0,
x21 + x22 + x2k = s2 .
From the first equation we have xk = c1 x1 + c2 x2 where c1 = −y1 /yk and c2 =
−y2 /yk . This combined with the second equation gives
(1 + c21 )x21 + (2c1 c2 x2 )x1 + (1 + c22 )x22 − s2 = 0.
Calculating the discriminant, we have
∆ = 4[(1 + c21 )s2 − (1 + c21 + c22 )x22 ].
Then for any x2 such that
√
|x2 | <
(1 + c21 )s2
,
1 + c21 + c22
we obtain two solutions for x1 . This proves that there are infinitely many solutions
because x2 can take infinitely many distinct values.
(b) It suffices to show that for any y ̸= 0, there is exactly one x ∈ Rk such
that x · y = 0 and |x| = 0, which is obvious.
(c) Obvious.
If k = 2, then the equations in the proof of (a) become
{
x1 y1 + x2 y2 = 0,
x21 + x22 = s2 .
The equations have only zero, one, or two solutions, corresponding to the cases
(c), (b), and (a), respectively.
If k = 1, then the equations in the proof of (a) become
{
x1 y1 = 0,
x21 = s2 .
The equations have only zero or one solutions. The former corresponds to (a) or
(c), while the latter corresponds to (b).
17.
10
1. The Real and Complex Number Systems
Proof.
|x + y|2 + |x − y|2 = 2|x|2 + 2|y|2 + x · y + y · x − x · y − y · x
= 2|x|2 + 2|y|2 .
Geometrically, this shows that the sum of the squares of the sides of a parallelogram equals the sum of the squares of its diagonals.
18.
Proof. With no loss of generality, we suppose xk ̸= 0. Let
y = (xk , 0, . . . , 0, −x1 ),
which is nonzero. It is easy to see that x · y = 0.
It is clearly false for k = 1 if x ̸= 0 (by the property of field).
19.
Solution. The trick is to find an equivalent form of the equation that is like |x −
c| = r. Taking squares in both sides of the equation, we obtain
3|x|2 − 2x · (4b − a) + 4|b|2 − |a|2 = 0
|x − (4b − a)/3|2 = |2(b − a)/3|2 .
This shows that c = (4b − a)/3 and r = 2|b − a|/3.
20.
Proof. First, we show that the resulting set is an ordered set with the least-upperbound property.
Given two cuts α and β, we say α < β if α is a proper subset of β. Now if
α ̸= β, then without loss of generality, we assume that there is a p ∈ β but p ∈
/ α.
This implies that p is an upper bound of α, so that α is a proper subset of β,
and therefore α < β. This shows that the resulting set is an ordered set since its
transitive property is an easy consequence of the set-inclusion
relation.
∪
Let A be a nonempty subset of R. We define α := γ∈A γ. We shall prove
that α ∈ R and α = sup A. It is clear that α is nonempty. If p ∈ α, then p ∈ γ
for some γ ∈ A, so that q ∈ γ ⊆ α for all q ∈ Q with q < p. Therefore α ∈ R.
By definition, α is an upper bound of A. Assume that β ∈ R is an upper
bound of A. Then γ ⊆ β for all γ ∈ A, so that α ⊆ β, that is, α ≤ β, and
therefore α = sup A.
Second, we show that addition satisfies axioms (A1) to (A4).
1.2. Solutions to Exercises
11
Given two cuts α and β, we define the addition α + β to be the set of all sums
r + s where r ∈ α and s ∈ β. It is well defined because it is easy to show that
α + β is a cut. This also asserts that the set of cuts is closed under addition, i.e.,
(A1).
(A2) is obviously true, because α + β = {r + s : r ∈ α, s ∈ β} = {s + r : r ∈
α, s ∈ β} = β + α. The same trick also applies to (A3).
Let 0∗ = {t ∈ Q : t ≤ 0}. It is easy to show that α + 0∗ = α for any cut α.
This proves (A4).
Finally, we show that (A5) fails. Let α = {r ∈ Q : r < 0}. If there were some
β such that α + β = 0∗ , then there is an r ∈ α and s ∈ β such that r + s = 0.
Since r < 0, we can choose a p ∈ α between r and 0 so that p + s > 0 and hence,
the cut 0∗ contains positive rationals, which is absurd.
12
1. The Real and Complex Number Systems
Chapter 2
Basic Topology
2.1
Notes on Text
Note 2.1 (p. 38, Theorem 2.36) The crucial idea of the proof is the following
equivalence:
A ⊆ B ⇔ A ∩ B c = ∅,
which establishes the relation between intersection problems and covering problems.
According to the proof, the theorem also holds for a collection of closed subsets
with only one subset being compact.
b
Note 2.2 (p. 40, Theorem 2.41) Conditions (b) and (c) are equivalent in any metric space (Ex. 26). The importance of the compactness of a space is that it enables
us to convert an infinite covering into a finite covering, so that many facts in a finite setting can be easily generalized to a general compact space. Analogously, the
separable property (Ex. 22) enables us to reduce a problem involving uncountable
set operations to a problem with only countable set operations.
b
Note 2.3 (p. 42, Definition 2.45) Let Y = A∪B. The condition A∩B = A∩B =
∅ implies that A ∩ Y = A and B ∩ Y = B, which further implies that both A and
B are open relative to Y (Theorem 2.30). In some books (e.g., Munkres, 2000,
Sec. 23), this (equivalent) condition is used to define a separation of a topological
space (or subspace).
b
2.2
Solutions to Exercises
1.
Proof. This is trivially true because the empty set contains no element. In other
words, we can say there is no element of the empty set such that it is not an
element of an arbitrary set.
2.
13
14
2. Basic Topology
Proof. Since every equation with integer coefficients yields only n algebraic numbers, it suffices to show that the set of such equations is countable. We note that
each equation is identified
with the n-tuple (a0 , a1 , . . . , an ), so the whole set of
∪∞
such equations is N =3 EN , where
{
EN :=
(a0 , a1 , . . . , an ) ∈
N∪
−1
}
m
Z
: a0 ̸= 0, n + |a0 | + · · · + |an | = N
.
m=2
Since EN is finite for each N ∈ Z>0 , it is clear that the number of equations is
countable (Theorem 2.12).
3.
Proof. If all real numbers were algebraic, then R would be countable (Ex. 2),
which is in contradiction to the fact that R is uncountable (Theorem 2.14 or
2.43).
4.
Proof. No. If the set of all irrational real numbers were countable, the set of real
numbers would also be countable, in contradiction to Theorem 2.14.
5.
Solution. Define A(x) := {x+1/n : n ∈ Z>0 }. It is clear that A(−1)∪A(0)∪A(1)
is bounded and has three limit points −1, 0, and 1.
6.
Proof. For any p ∈
/ E ′ , there is a neighborhood Nr (p) that is disjoint from E \{p}.
This implies that Nr (p) contains no limit points of E (Theorem 2.20). Therefore
p is an interior point of E ′c , and hence E ′c is open, so that E ′ is closed.
It is enough to show that a limit point of E is also a limit point of E. Let p
be a limit point of E. If p were not a limit point of E, then p would be a limit
point of E ′ , so that p ∈ E ′ , and therefore p is a limit point of E, a contradiction.
No, they may have different limit points. For example, let E = {1/n : n ∈
Z>0 } in R, so that E ′ = {0}, which has no limit points.
7.
2.2. Solutions to Exercises
15
Proof. (a) Observe that a point p is a limit point of Bn if and only if it is a limit
point of Ai for some i (since Bn is a finite union of A1 , . . . , An ). Then,
Bn = Bn ∪ Bn′ =
n
∪
Ai ∪
i=1
n
∪
A′i =
i=1
n
∪
(Ai ∪ A′i ) =
i=1
n
∪
(Ai ).
i=1
(b) Use a similar argument of (a) with the fact that a limit point of Ai is a
limit point of B.
For a counter example, let Ai = {1/i} in R. We have
∞
∪
Ai =
i=1
∞
∪
i=1
Ai ̸= {0} ∪
∞
∪
Ai = B.
i=1
8.
Solution. It is true for open set but false for closed sets.
9.
Proof. (a) If p ∈ E ◦ , then there is a neighborhood Nr (p) of p such that Nr (p) ⊆ E.
Since Nr (p) is open, its every point is its interior point, and hence an interior point
of E, so that p is an interior point of E ◦ . This shows that E ◦ is open.
(b) By (a), E ◦ = E implies that E is open. Conversely, if E is open, then
every point of E is an interior point of E, so that E ◦ = E.
(c) Observe that every (interior) point of G is an interior point of E.
(d) The complement of E ◦ is a closed set containing E c and hence its closure.
It remains to show that every point of (E ◦ )c is a point of the closure of E c . If
p∈
/ E ◦ , then every neighborhood of p contains points of E c . This implies that p
is a point of E c or a limit point of E c (or both), and therefore p is a point of E c .
(e) No. An interior point of E must be an interior point of E. However an
interior point of E is not necessarily an interior point of E. For example, consider
the subset E = (0, 1) ∪ (1, 2) of R. It is clear that the interior of the closure of E
is (0, 2), which is however not equal to E ◦ = E.
(f) No. Consider the subset E = {1/n : n ∈ Z>0 } of R. Then we have
E = E ∪ {0} while E ◦ = ∅.
10.
Proof. First, we show that this is a metric. It suffices to show that it satisfies the
triangular inequality. For p, q, r ∈ X, if p ̸= q, then
d(p, q) = 1 ≤ d(p, r) + d(r, q),
16
2. Basic Topology
since r may equal p or q but not both; otherwise, we have p = q, so that
d(p, q) = 0 ≤ d(p, r) + d(r, q).
Since every point is open with respect to this metric, all subsets of X are open,
and hence also closed. The compact subsets of X are its finite subsets. For if a
subset S is infinite, then we may consider the open cover (N1/2 (p))p∈S , of which
any proper subset cannot cover S.
11.
Proof. The function d1 (x, y) is not a metric, because d1 (1, 3) = 4 > 2 = d1 (1, 2) +
d1 (2, 3).
The function d2 (x, y) is nonnegative and is zero only if x = y. It is also
symmetric. Moreover, it satisfies the triangular inequality because
√
√
√
√
|x − y| ≤ |x − z| + |z − y| ≤ |x − z| + |z − y|.
Therefore it is a metric.
The function d3 (x, y) is not a metric because d2 (1, −1) = 0.
The function d4 (x, y) is not a metric because d2 (2, 1) = 0.
It is easy to verify that d5 (x, y) satisfies the first and second condition of a
metric. We now show that it satisfies the triangular inequality.
|x − y|
1
=1−
1 + |x − y|
1 + |x − y|
1
≤1−
1 + |x − z| + |z − y|
|x − z|
|z − y|
≤
+
.
1 + |x − z| 1 + |z − y|
So d5 (x, y) is a metric.
12.
Proof. Suppose G is an open cover of K. Then there is an open set A0 ∈ G covering
0, and hence covering the numbers 1/n for n ≥ N where N is fixed integer. Now
for each n < N , choosing an open set An ∈ G that covers 1/n, we thus obtain a
finite open cover (An )N
n=0 of K, and therefore K is compact.
13.
2.2. Solutions to Exercises
17
n
n
Solution. Let An := {1/2
∪∞ } ∪ {1/2 + 1/k : k ∈ Z>0 }. Let us consider the limit
points of the set A := n=1 An . It is clear that 0 and 1/2n (n ∈ Z>0 ) are its limit
points. We shall show there are no other limit points.
At first, it is clear that any negative number is not a limit point of A. Next,
for any positive x such that 1/2m+1 < x < 1/2m where m ∈ Z>0 , let
r=
1
min{x − 1/2m+1 , 1/2m − x}.
2
Then the neighborhood Nr (x) contains at most finite points of An for n > m and
contains no points of An for n ≤ m. So it is not a limit point of A. Finally, it is
also clear that all points greater than 1/2 are not limit points of A.
Therefore A is closed and hence compact (because it is bounded), and moreover, its limit points form a countable set (see also Ex. 6).
14.
Solution. Let An = (2/4n+1 , 3/4n ) for n ∈ Z≥0 . It is clear that the segment (0, 1)
is covered by the collection (An )∞
n=0 which however contains no finite subcover.
15.
Proof. Let An = [n, +∞) where n ∈ Z>0 . It is clear that An is closed
the intersection of every finite subcollection of (An )∞
n=1 is nonempty.
∩
∞
A
=
∅.
n
n=1
Let An = (0, 1/n) where n ∈ Z>0 . It is clear that An is bounded
∞
the
∩∞ intersection of every finite subcollection of (An )n=1 is nonempty.
n=1 An = ∅.
and that
However
and that
However
16.
Proof. Note that Q is a subset of R with the same distance d(p, q). Since
√
√
√ √
E = Q ∩ ((− 3, − 2) ∪ ( 2, 3))
and
√
√ √
√
Q \ E = Q ∩ ((−∞, − 3) ∪ (− 2, 2) ∪ ( 3, +∞)),
E is both open and closed in Q (Theorem 2.30), and it is bounded. However it
is not compact, since
√ we may easily construct an infinite sequence of points in E
that converges to 2, which is not in E, so that the sequence has no limit points
in E.
17.
18
2. Basic Topology
Solution. E is countable. Since we may establish a one-to-one correspondence
between the points of E and binary sequences by mapping 4 to 0 and 7 to 1,
respectively.
E is not dense since, for example, the decimal expansion of any point in the
segment (0.1, 0.2) contains 1.
For every point p ∈ [0, 1] \ E, its decimal expansion contains at least a digit
not equal to 4 or 7. We suppose the first such digit is at the ith position and that
the decimal expansion of that point is in the form 0.d1 d2 · · · di . With no loss of
generality, we assume that i ≥ 2 (since the case of i = 1 is trivial). If di ̸= 0 or 9,
then any numbers in the segment
(0.d1 d2 · · · di−1 (di − 1)9, 0.d1 d2 · · · di−1 (di + 1)1)
are not in E; if di = 0, then any numbers in the segment
(0.d1 d2 · · · (di−1 − 1)99, 0.d1 d2 · · · di−1 11)
are not in E; if di = 9, then any numbers in the segment
(0.d1 d2 · · · di−1 89, 0.d1 d2 · · · (di−1 + 1)01)
are not in E. All the above facts show that E c is open and hence E is closed.
Because E is bounded, it is compact (Theorem 2.41).
For any point p of E, we may replace the nth digit of its decimal expansion
by 4 or 7. In this way, we can construct an infinite sequence of points (distinct
from p) in E such that it converges to p. Therefore E is perfect.
18.
Solution. Yes, there is. Let (rn )∞
n=1 be an enumeration of rational numbers. Let
−n−1
S
be
the
segment
(r
−
2
, rn + 2−n−1 ) and consider the set A = [0, 2] \
n
∪n∞
n=1 Sn . It is clear that A is a closed and uncountable set containing no rational
numbers. By Ex. 27, the set of condensation points of A is a perfect set containing
no rational numbers.
19.
Proof. (a) It is clear that A ∩ B = A ∩ B = A ∩ B = ∅, so A and B are separated.
(b) Suppose that A and B are disjoint open sets. This shows that A is a subset
of the closed set B c , so that A ∩ B ⊆ B c ∩ B = ∅. Likewise, we have A ∩ B = ∅.
Therefore A and B are separated.
(c) It is clear that A is open (Theorem 2.19). A similar argument to Theorem 2.19 also shows that B is open. Since A and B are disjoint, they are separated
(part (b)).
2.2. Solutions to Exercises
19
(d) Suppose that p and q are two distinct points of the connected metric space.
Let r = d(p, q). Then for every δ ∈ (0, r), there is at least one point uδ such that
d(p, uδ ) = δ, for otherwise the metric space would consist of two separated sets
by (c). Since the segment (0, r) is uncountable and uδ is a one-to-one mapping of
(0, r) into the metric space, the metric space is uncountable.
20.
Solution. The closure of a connected set is also connected. Suppose that A is a
connected set. If A were a union of two nonempty separated sets, say B and C,
then A would be a union of two nonempty separated sets A ∩ B and A ∩ C, in
contradiction to our hypothesis.
To see that A ∩ B and A ∩ C is separated, let us show that A ∩ B ∩ (A ∩
C) = (A ∩ B) ∩ A ∩ C = ∅. It is clear that A ∩ B is a subset of A ∩ B. Since
A ∩ B ∩ (A ∩ C) = A ∩ (B ∩ C) = ∅, we have A ∩ B ∩ (A ∩ C) = ∅. Likewise,
(A ∩ B) ∩ A ∩ C = ∅.
However, the interior of a connected set is not necessarily connected. For
example, let
A = {x ∈ R2 : |x| ≤ 1} ∪ {x ∈ R2 : |x − (2, 0)| ≤ 1}.
It is clear that A◦ is not connected.
21.
Proof. (a) Since p(t) − p(t′ ) = (t − t′ )(b − a), it is easy to show that for any
A ⊆ R1 , if s is a limit point of A, then p(s) is a limit point of p(A). This shows
that p(A) ⊆ p(A) for any A ⊆ R1 , and hence we have
p(A0 ) ⊆ p(A0 ) ⊆ A
and similarly p(B0 ) ⊆ B. Therefore A0 and B0 must be separated.
(b) Let A1 = A0 ∩ [0, 1] and B1 = B0 ∩ [0, 1]. It is clear that 0 ∈ A1 and
1 ∈ B1 . If there were no t0 ∈ (0, 1) such that p(t0 ) ∈
/ A ∪ B, the interval [0, 1]
would be the union of two separated nonempty sets A1 and B1 , which is absurd
because [0, 1] is connected.
(c) Let S be a convex subset of Rk . If it were a union of two separated
nonempty subsets A and B, then by (b), there would exist a point p(t0 ) ∈
/ A∪
B, which is however a point of S according to the definition of a convex set, a
contradiction.
22.
20
2. Basic Topology
Proof. It is clear that the set Qk of points having only rational coordinates is
countable (Theorem 2.12). Furthermore, for every point of Rk , we can easily
find a point with rational coordinates that is arbitrarily close to that point. For
example, given p ∈ Rk , we may define the point
)
(
⌈npk ⌉
⌈np1 ⌉ ⌈np2 ⌉
.
qn =
,
,...,
n
n
n
As n → +∞, qn can be arbitrarily close to p, so that every point of Rk is a limit
point of Qk . Therefore Rk is separable.
23.
Proof. Let X be a separable metric space and S its countable dense subset. For
any open set G and any point x ∈ G, there is a neighborhood, say Nr (x), such
that Nr (x) ⊆ G. Let r′ = ⌈kr⌉/3k and x′ ∈ S such that d(x, x′ ) < r/3. Then for
sufficiently large k we have x ∈ Nr′ (x′ ) ⊆ Nr (x) ⊆ G. This shows that X has a
countable base.
24.
Proof. According to the hint, this process must stop after finite steps, otherwise
we would obtain an infinite sequence of points that has no limit point, which
contradicts to the condition that every infinite subset has a limit point. This also
implies that X is covered by finitely many neighborhoods of radius δ. Repeating
this process with δ = 1/n thus yields a countable (or precisely, at most countable)
dense subset of X.
25.
Proof. For every n ∈ Z>0 , consider the open cover (N1/n (x))x∈K of K. Since
K is compact, it contains a finite subcover. Considering all the centers of such
neighborhoods for n = 1, 2, . . . . The set of all these centers is (at most) countable
and is dense in K.
An alternative proof is using Theorem 2.37 and Ex. 24.
26.
Proof. By Exs. 23 and 24, X has a countable base. Let (Bn )n∈Z>0 be a countable
base of X. Suppose that X has an open cover (Gα )α∈I . Let
K := {k ∈ Z>0 : Bk ⊆ Gα for some α},
and we define a map f from K to I given by k 7→ α such that Gα ⊇ Bk . Such a
map exists but may not be unique.
2.2. Solutions to Exercises
21
Now we show that the countable subcollection (Gf (k) )k∈K covers X. For every
point p ∈ X, there is some α such that p ∈ Gα , so that p ⊆ Bk ⊆ Gα for some
k ∈ Z>0 . It is clear that k ∈ K and therefore p ∈ Bk ⊆ Gf (k) .
Suppose that X is covered by a countable collection (Gn )n∈Z>0 of open sets. If
no finite subcollection of (Gn∩) covers X, then the complement Fn of G1 ∪ · · · ∪ Gn
∞
is nonempty for each n, but n=1 Fn is empty. If E is a set which contains a point
from each Fn , then E is infinite. Consider a limit point of E, say p, and p ∈ Gk for
some k. It follows that Gk contains infinitely many points of E, which is absurd,
because it contains at most k − 1 points of E according to our hypothesis.
27.
Proof. Let (Vn ) be a countable base of Rk , and W the union of those Vn for which
E ∩ Vn is at most countable. It is clear that P ⊆ W c . Moreover, if p ∈ W c \ P ,
then p is not a condensation point and hence there exists some k such that p ∈ Vk
and Vk contains at most countably many points of E, so that p ∈ Vk ⊆ W , a
contradiction. Therefore P = W c and it is closed (since every limit point of P
is also a condensation point of E). Furthermore, since P c ∩ E = W ∩ E is at
most countable, every condensation point of E must contain uncountable points
of P ∩ E. In other words, every point of P is a condensation point of P , so that
P is perfect.
Below is an alternative proof of a general result. Let X be a separable metric
space and E a uncountable subset of X. We will show that for any subset G of
X, if it contains no condensation points of E, it contains at most countably many
points of E. For each point p of G, because it is not a condensation point, there is
a neighborhood Vp of p which contains at most countably many points of E. Then
G is covered by (Vp ) and hence is covered by a countable subcollection (Vq′ ) (see
the proof of Ex. 26), so that G contains at most countably many points of E. This
result thus shows that P c ∩ E is at most countable and that any neighborhood of
a condensation point of E contains uncountably many points of P ∩ E.
28.
Proof. Let E be a closed set in a separable metric space. If it is at most countable,
then we are done. Otherwise, let P be the set of condensation points of E. Since
E is closed, P is a subset of E which is perfect and E \ P is at most countable
(see the second proof of Ex. 27).
As a corollary, every countable closed set must have isolated points, for otherwise it would be a perfect set which is uncountable (Theorem 2.43).
29.
22
2. Basic Topology
1
Proof. Let G be an open set in R1 . Define the map f : G → 2R given by
p 7→ (pa , pb ) where
pa := sup{x ̸∈ G : x < p},
pb := inf{x ̸∈ G : x > p}.
It is clear that (pa , pb ) ⊆ G and pa , pb ∈
/ G (because G is open). It is also clear
that for every q ∈ f (p), f (q) = f (p), and hence for p, q ∈ G, if f (p) ̸= f (q), then
they are disjoint, for otherwise, we may choose r ∈ f (p) ∩ f (q) and we would
have f (p) = f (r) = f (q). Therefore G is a union of disjoint segments. To prove
that this collection of segments is at most countable, we may restrict the domain
of f to the subset Q1 of rational numbers. Since Q1 is dense in R1 , the inverse
image of each segment S ∈ f (G), i.e, S, contains a rational number, so that
|f (G)| = |f (G ∩ Q1 )| ≤ |Q|.
30.
Proof. We first show that the
Let Fn = Gcn , if
∩
∪∞two statements
∩∞ are cequivalent.
∞
k
n=1 Gn were empty, then
n=1 Fn = ( n=1 Gn ) = R , so that at least one
Fn has a nonempty interior, which implies that Gn is not dense, a contradiction.
Similarly, let Gn = Fnc , we can also prove that the second statement implies the
first.
Now let us prove the second statement. For every point p ∈ Rk and any
neighborhood N of p, since G1 is open, we may choose a neighborhood V1 such
that V 1 ⊆ G1 ∩ N . Suppose Vn has been constructed, so that V n ⊆ Gn . Since
Gn+1 is open and dense, there is a neighborhood Vn+1 such∩that V n+1 ⊆ Vn
∞
and V n+1 ⊆ Gn+1 . Proceed this construction and consider ∩
n=1 V n , which is
∞
nonempty and must be contained in N and all Gn . Therefore n=1 Gn is dense
k
in R .
Chapter 3
Numerical Sequences and Series
3.1
Notes on Text
Note 3.1 (p. 63, before Definition 3.30) It says that Exs. 11.(b) and 12.(b) may
serve as illustrations for the notion that the “boundary” conjecture is false, so it
is interesting to carefully study these exercises.
Ex. 11.(b) provides an approach to construct a new divergent series
∑nfrom a
divergent series. For example, given the obvious
divergent
series
s
=
n
k=1 1, it
∑
follows from Ex. 11.(b) that the series s′n = (1/n) is divergent. Furthermore, it
is easy to see that the resulting series is always less than the base series, in the
sense that
lim (sn − s′n ) = +∞.
n→∞
Similarly, Ex. 12.(b) gives an approach to construct from a convergent series a
new convergent series with larger limit.
Therefore we cannot expect a clear boundary between convergent and divergent series, since the series at the boundary, if any, would be neither convergent
nor divergent.
b
3.2
Solutions to Exercises
1.
Proof. If (sn ) converges, say to s, then for any ϵ > 0, there exists an integer N
such that n ≥ N implies that |sn − s| < ϵ, so that
|sn | − |s| ≤ |sn − s| < ϵ
for all n ≥ N , and therefore (|sn |) converges to |s|.
The converse is false. For example, put sn = (−1)n (1 + 1/n). It is clear that
(|sn |) converges to 1 but (sn ) diverges.
2.
23
24
3. Numerical Sequences and Series
Solution.
√
n
lim ( n2 + n − n) = lim √
2
n→∞
n→∞
n +n+n
1
= lim √
n→∞
1 + 1/n + 1
1
= .
2
√
Here we utilize the continuity of x, which is still unknown in this chapter.
However, we may prove this result by the following way:
)
1/4
1 (√ 2
√
−
n +n−n =
,
2
n + 1/2 + n2 + n
which converges to 0 as n → ∞.
3.
Proof. At first, we show that sn < 2 for all n. It is clear that s1 < 2. Now suppose
that sn < 2, and then we have
√
√
√
√
√
sn+1 = 2 + sn < 2 + 2 < 4 = 2,
so that (sn ) is bounded above by 2.
Next, we show that (sn ) is monotonically increasing. It is clear that s1 < s2 .
Suppose sn < sn+1 , it is easy to see that
√
√
√
√
sn+2 = 2 + sn+1 > 2 + sn = sn+1 .
Thus (sn ) is bounded and monotonic, and therefore converges (Theorem 3.14).
4.
Solution. Let pn = s2n−1 and qn = s2n . It is clear that
(
)
1
1
1
pn+1 = (pn + 1), qn+1 =
qn +
.
2
2
2
Simple algebraic manipulations further show that
( )n
( )n
1
1
1
pn+1 − 1 = (pn − 1) =
(p1 − 1) = −
2
2
2
3.2. Solutions to Exercises
and
25
(
) ( )n (
)
( )n+1
1
1
1
1
qn −
=
q1 −
=−
.
2
2
2
2
1
1
qn+1 − =
2
2
Thus limn→∞ pn = 1 and limn→∞ qn = 12 . In other words, the upper and lower
limits of (sn ) are 1 and 12 , respectively.
5.
Proof. We first show that
lim sup(an + c) = lim sup an + c
n→∞
n→∞
for c ∈ R. The case of lim sup an = ±∞ is trivially true, so we assume that
lim sup an = a ∈ R. Then for every ϵ > 0 there exists an integer N such that
n ≥ N implies that an < a + ϵ (Theorem 3.17). It follows that
lim sup(an + c) ≤ lim sup(a + ϵ + c) = a + c + ϵ.
n→∞
(Theorem 3.19)
n→∞
Since ϵ is arbitrary, we have
lim sup(an + c) ≤ a + c = lim sup an + c.
n→∞
n→∞
On the other hand, we have
lim sup an = lim sup(an + c − c) ≤ lim sup(an + c) − c.
n→∞
n→∞
n→∞
Therefore, lim sup(an + c) = lim sup an + c.
Now, with no loss of generality, we assume that lim sup an ≥ lim sup bn .
If lim sup bn = +∞, the statement is clearly true.
If lim sup bn = b, then for every ϵ > 0 there is an integer N such that bn < b + ϵ
for all n ≥ N . It follows that
lim sup(an + bn ) ≤ lim sup(an + b + ϵ) = lim sup an + b + ϵ.
n→∞
n→∞
(Theorem 3.19)
n→∞
Since ϵ is arbitrary, we have
lim sup(an + bn ) ≤ lim sup an + b = lim sup an + lim sup bn .
n→∞
n→∞
n→∞
n→∞
Finally, if lim sup bn = −∞ and lim sup an < +∞, then there exist a real
number M and an integer N such that an < M for all n ≥ N , so that
lim sup(an + bn ) ≤ lim sup(M + bn ) = −∞.
n→∞
n→∞
26
3. Numerical Sequences and Series
6.
Solution. (a) Calculate the partial sum
sn =
n
∑
an =
√
n + 1 − 1,
k=1
which clearly diverges.
(b) The term an can be rewritten as
an =
1
1
√
√ < 3/2 ,
n
n( n + 1 + n)
∑
which implies that
an converges (Theorems 3.25 and 3.28).
(c) Resorting to the root test (Thoerem 3.33), we have
√
√
(Theorem 3.20)
lim sup n |an | = lim sup n n − 1 = 0,
n→∞
n→∞
so
∑
an converges.
(d) If |z| ≤ 1, then
lim |an | = lim
n→∞
n→∞
1
1
≥ lim
> 0,
|1 + z n | n→∞ 1 + |z|n
∑
which implies that
an diverges (Theorem 3.23).
As |z| > 1, we have
lim sup
n→∞
|an+1 |
1
|1 + z n |
|z|n + 1
= lim sup
≤ lim sup n+1
=
.
n+1
|an |
|
−1
|z|
n→∞ |1 + z
n→∞ |z|
∑
Therefore
an converges
(Theorem 3.34).
∑
In summary,
an converges for |z| > 1 and diverges otherwise.
7.
Proof.
m √
m
∞
m
∑
∑
an
1∑
1∑ 1
nan + n−1
≤
=
an +
.
n
2n
2 n=1
2 n=1 n2
n=1
n=1
∑
∑
∑√
Since
an and (1/n2 ) converge (Theorem 3.28), ( an /n) is bounded and
hence converges (Theorem 3.24).
8.
3.2. Solutions to Exercises
27
Proof. Since (bn ) is monotonic and bounded, it converges, say to b. Without loss
of generality we assume that bn ≤ b. Then
m
∑
an bn =
n=1
Since
m
∑
an [b − (b − bn )] = b
n=1
m
∑
an −
n=1
m
∑
an (b − bn ).
n=1
∑
∑
[an (b − bn )] is convergent (Theorem 3.42), (an bn ) converges.
9.
Proof. (a) Since α = lim supn→∞
(Theorem 3.39).
(b) Because
lim sup
n→∞
√
n
n3 = 1, the radius of convergence is 1/α = 1
|(2z)n+1 /(n + 1)!|
2|z|
= lim sup
= 0,
|(2z)n /n!|
n→∞ n + 1
the radius of convergence is +∞
√ (Theorem 3.34).
(c) Since α = lim supn→∞ n 2n /n2 = 2, the radius of convergence is 1/α = 12
(Theorem 3.39).
√
(d) Since α = lim supn→∞ n n3 /3n = 13 , the radius of convergence is 1/α = 3
(Theorem 3.39).
10.
Proof. Because infinitely many of (an ) are nonzero integers, we have
√
α = lim sup n |an | ≥ 1
n→∞
and hence the radius of convergence is at most 1 (Theorem 3.39).
11.
Proof. (a) If (an ) is unbounded, then
lim sup
n→∞
an
= 1,
1 + an
∑
and therefore [an /(1 + an )] diverges (Theorem 3.23). Otherwise, there exists a
positive real number M such that an < M for all n, so that
and hence
∑
an
an
>
,
1 + an
1+M
[an /(1 + an )] diverges (Theorem 3.25).
28
3. Numerical Sequences and Series
(b)
aN +1
aN +k
aN +1
aN +k
+ ··· +
≥
+ ··· +
sN +1
sN +k
sN +k
sN +k
sN
=1−
.
sN +k
Because limn→∞ sn = +∞, for any ϵ ∈ (0, 1) and N ∈ Z∑
>0 there exists an
integer k such that 1 − sN /sN +k > ϵ. This implies that
(an /sn ) diverges
(Theorem 3.22).
(c) For n ≥ 2,
an
an
1
1
≤
=
− ,
s2n
sn−1 sn
sn−1
sn
so that
)
n
n (
∑
∑
1
1
1
2
ak
1
2
≤
+
−
−
< ,
=
2
s
s1
sk−1
sk
s1
sn
s1
k=1 k
k=2
∑
and hence [an /s2n ] converges (Theorem 3.24).
(d) First, if lim inf(nan ) > 0, then there exist a positive real number α and an
integer N such that n ≥ N implies that nan > α, so that
an
an
1
>
=
1 + nan
(1 + 1/α)nan
n(1 + 1/α)
∑
for n ≥ N . Hence [an /(1 + nan )] diverges (Theorems 3.25 and 3.28).
Second, if lim sup(nan ) < +∞, then there exist a positive real number M such
that nan < M for all n, so that
an
an
>
.
1 + nan
1+M
∑
Hence [an /(1 + nan )] ∑
diverges (Theorems 3.25).
In general, however, [an /(1 + nan )] may converge or not. Let

if n = 2k with k ∈ Z≥0 ,
1
an =
1

otherwise.
n2
∑
It is clear that
an diverges (Theorem 3.23), but
n
∑
k=1
ak
<
1 + kak
⌊log2 n⌋
∑
j=0
⌊log2 n⌋
<
∑
j=0
< 3,
∑ 1/k 2
1
+
j
1+2
1 + 1/k
n
k=1
1
+
2j
n (
∑
k=1
1
1
−
k k+1
)
,
3.2. Solutions to Exercises
29
∑
which implies that [an /(1
∑ + nan )] converges (Theorem 3.24).
In contrast, the series [an /(1 + n2 an )] surely converges because
an
an
1
< 2
= 2.
1 + n2 a n
n an
n
(Theorem 3.25)
12.
Proof. (a)
am
an
am
an
+ ··· +
>
+ ··· +
rm
rn
rm
rm
rm − rn+1
=
rm
rn
>1−
.
rm
∑
Because
an converges, rn can be arbitrarily small, so that for any ϵ ∈ (0, 1) and
N ∈∑
Z>0 there exists an integer n ≥ N such that |1 − rn /rN | > ϵ. This implies
that (an /rn ) diverges (Theorem 3.22).
(b)
an
2an
√ <√
√
rn
rn + rn+1
√
√
2an ( rn − rn+1 )
<
rn − rn+1
√
√
= 2( rn − rn+1 ).
Therefore
and hence
∑
n
∑
√
√
ak
√
√ < 2( r1 − rn+1 ) < 2 r1
rk
k=1
√
(an / rn ) converges (Theorem 3.24).
13.
Proof. Suppose that
∑
an and
∑
bn converge absolutely, so that
m ∑
n
∑
|ak ||bn−k |
n=0 k=0
converges (Theorem 3.50). On the other hand,
n
m ∑
m ∑
n
∑
∑
ak bn−k ≤
|ak ||bn−k |,
n=0 k=0
n=0 k=0
30
3. Numerical Sequences and Series
so the Cauchy product of the series
rem 3.24).
∑
an and
∑
bn converges absolutely (Theo-
14.
Proof. (a) Since lim sn = s, for every ϵ > 0 there exists an integer N such that
n ≥ N implies that |sn − s| < ϵ. So for n ≥ N ,
∑
N −1 (s − s) + ∑n (s − s) i=1 i
i=N i
|σn − s| = n+1
∑N −1
∑n
|si − s| + i=N |si − s|
≤ i=1
n+1
∑N −1
(N − 1)|s| + i=1 |si | + ϵ(n − N + 1)
≤
n+1
∑N −1
(N − 1)|s| + i=1 |si | − ϵN
=ϵ+
,
n+1
which is bounded by 2ϵ for sufficiently large n. Therefore lim σn = s.
(b) Let sn = k + 1 for n = 2k , otherwise put sn = 1/n2 . It is clear that sn
does not converge but lim σn = 0.
(c) See the solution of (b).
(d)
nsn − (s0 + s1 + · · · + sn−1 )
n+1
(s1 − s0 ) + 2(s2 − s1 ) + · · · + n(sn − sn−1 )
=
n+1
n
∑
1
=
kak .
n+1
sn − σn =
k=1
If lim(nan ) = 0, it follows from (a) that lim(sn − σn ) = 0, so that lim sn = lim σn .
(e) We follow the outline provided by the exercise. If m < n then
m+1
n+1
m+1
(σn − σm ) + sn −
σn +
σm
n−m
n−m
n−m


n
m
∑
∑
m+1
1 
=
(n − m)sn −
si +
sj 
(σn − σm ) +
n−m
n−m
i=0
j=0
sn − σn =
=
n
∑
m+1
1
(sn − si ).
(σn − σm ) +
n−m
n − m i=m+1
3.2. Solutions to Exercises
31
For i = m + 1, . . . , n,
n
∑
|sn − si | ≤
≤
|sj − sj−1 |
j=i+1
n
∑
M
j
j=i+1
(n − i)M
i+1
(n − m − 1)M
≤
.
m+2
≤
Fix ϵ > 0 and associate with each n the integer m that satisfies
m≤
n−ϵ
< m + 1.
1+ϵ
Then (m + 1)/(n − m) ≤ 1/ϵ and |sn − si | < M ϵ. Hence
|sn − σn | ≤
|σn − σm |
+ M ϵ,
ϵ
so that lim supn→∞ |sn − σ| ≤ M ϵ. Since ϵ was arbitrary, lim sn = σ.
15.
Fact 3.1 (Theorem 3.22).
an integer N such that
∑
an converges if and only if for every ϵ > 0 there is
m
∑ ak ≤ ϵ
k=n
if m ≥ n ≥ N .
∑
Proof. By Cauchy criterion,
an converges if and only if for every ϵ > 0 there is
an integer N such that
m
n
∑
∑
ak −
ak ≤ ϵ
for all m ≥ n ≥ N ,
k=1
k=1
which is clearly equivalent to the condition of the theorem.
Fact 3.2 (Theorem 3.23). If
∑
an converges, then limn→∞ an = 0.
32
3. Numerical Sequences and Series
Proof. Taking m = n in Fact 3.1, we have
|an | ≤ ϵ
for sufficiently large n.
In other words, each component of an converges to zero.
Fact 3.3 (Theorem
3.25(a)). If |an |∑≤ cn for n ≥ N0 , where N0 is some fixed
∑
integer, and if
cn converges, then
an converges.
Proof. Given ϵ > 0, there exists N ≥ N0 such that m ≥ n ≥ N implies
m
∑
ck ≤ ϵ.
(Theorem 3.22)
k=n
Hence
m
m
m
∑
∑
∑
|ak | ≤
ck ≤ ϵ,
ak ≤
k=n
k=n
k=n
which concludes the theorem.
Fact 3.4 (Theorem 3.33). Given
∑
an , put
√
α = lim sup n |an |.
n→∞
Then
∑
(a) if α < 1, ∑ an converges;
(b) if α > 1,
an diverges;
(c) if α = 1, the test gives no information.
Proof. (a) Use Fact 3.3 with ck = β k where β ∈ (α, 1).
(b) Use Fact 3.2.
(c) Use Theorem 3.33(c).
∑
Fact 3.5 (Theorem 3.34). The series
an
|an+1 |
(a) converges if lim sup
< 1,
|an |
n→∞
|an+1 |
(b) diverges if
≥ 1 for all n ≥ n0 , where n0 is some fixed integer.
|an |
Proof. If condition (a) holds, we can find β < 1 and an integer N such that
|an+1 |
<β
|an |
for n ≥ N .
3.2. Solutions to Exercises
33
Then we have
|an | ≤ |aN |β n−N
for n ≥ N ,
so that (a) follows from Fact 3.3.
(b) Use Fact 3.2.
Fact 3.6 (Theorem 3.42). Suppose
∑
(a) the partial sums An of
an form a bounded sequence;
(b) b0 ≥ b1 ≥ b2 ≥ · · · ;
(c)∑
limn→∞ bn = 0.
Then
an bn converges.
∑
Proof. Use Theorem 3.42 for each component of
an b n .
∑
∑
Fact 3.7 (Theorem 3.45). If
an converges absolutely, then
an converges.
∑m
∑m
Proof. Use the inequality k=n ak ≤ k=n |ak |.
∑
∑
∑
Fact 3.8 (Theorem
3.47). If
an = A, and
bn = B, then (an + bn ) =
∑
A + B, and
can = cA, for any fixed c.
Proof. Use Theorems 3.4 and 3.47 for each component of an and bn .
∑
Fact 3.9 (Theorem 3.55). If
an is a∑
series of vectors of Rk which converges
absolutely, then every rearrangement of
an converges, and they all converge to
the same sum.
∑
Proof. Use Theorems 3.25 and 3.55 for each component of
an , and then apply
Theorem 3.4.
16.
Proof. (a) Observe that
√
√ )2
√
1 (
1
α=
xn − α < (xn − α),
2xn
2
√
where the first equality ensures
√ that xn > α for all n ≥ 1, so that xn decreases
monotonically and lim xn = α.
(b) The first identity
has been√shown in the proof of (a), and we immediately
√
have ϵn+1 < ϵ2n /2 α since xn > α. This inequality can rewritten as
xn+1 −
ϵn+1
<
β
n
so that ϵn+1 < β(ϵ1 /β)2 .
(
ϵn
β
)2
34
3. Numerical Sequences and Series
(c) It suffices to show that ϵ1 /β < 1/10 and β < 4.
√
√
ϵ1
2− 3
2 3−3
2 × 1.8 − 3
0.6
1
√
=
=
<
=
=
.
β
6
6
6
10
2 3
√
√
β = 2 3 < 2 4 = 4.
17.
Proof. At first, note that
√
√
√
α + xn √
( α − 1)( α − xn )
xn+1 − α =
− α=
.
1 + xn
1 + xn
Furthremore,
xn+2 −
√
√
√
√
√
( α − 1)2 (xn − α)
( α − 1)2 (xn − α)
=
,
α=
(1 + xn )(1 + xn+1 )
1 + α + 2xn
√
√
√
√
(α − 2 α + 1)(xn − α)
< xn − α.
α<
1+α
Parts (a), (b), and (c) are easy consequences of this fact. It is also clear that the
formula described in Ex. 16 converges more rapidly.
so that
xn+2 −
18.
Proof. At first,
(p − 1)xpn − pα1/p xp−1
+α
n
p−1
pxn
p−1
(p − 1)xn (xn − α1/p ) − α1/p (xnp−1 − α(p−1)/p )
=
pxp−1
n
∑p−2
(xn − α1/p )[(p − 1)xnp−1 − α1/p k=0 α(p−2−k)/p xkn ]
=
pxnp−1
∑
p−2
(xn − α1/p ) k=0 (xnp−1 − α(p−1−k)/p xkn )
=
pxnp−1
∑p−2
∑p−2−k
(xn − α1/p )2 k=0 xkn m=0 α(p−2−k−m)/p xm
n
=
pxp−1
n
∑p−2 ∑
(xn − α1/p )2 l=0 k+m=l α(p−2−k−m)/p xk+m
n
=
pxnp−1
∑p−2
(xn − α1/p )2 l=0 (l + 1)α(p−2−l)/p xln
.
=
pxnp−1
xn+1 − α1/p =
3.2. Solutions to Exercises
35
This implies that xn > α1/p for every integer n ≥ 2, so that
∑p−2 p−1
∑p−2 p−1
− α(p−1−k)/p xkn )
− α(p−1)/p )
p−1
k=0 (xn
k=0 (xn
<
<
,
p−1
p−1
p
pxn
pxn
and therefore
xn+1 − α1/p <
(p − 1)(xn − α1/p )
p
so that (xn ) decreases monotonically (for n ≥ 2) and converges to α1/p . Moreover,
this process has the same speed of convergence as the one in Ex. 16.
19.
Proof. At first, it is clear that x(a) is well defined (Theorem 3.33).
Next, we need a formal definition of the Cantor set. Define the piecewise
function
{
3x
x ∈ [0, 23 ),
f (x) :=
3x − 2 x ∈ [ 32 , 1].
∩∞
Let En := f −n ([0, 1]) (where f −(n+1) := f −1 ◦ f −n ). Then P := n=1 En is just
the Cantor set.
∑
Now we show that every point x of P can be expressed as the form (αn /3n ).
Define the step function
{
0 x ∈ [0, 32 ),
g(x) :=
2 x ∈ [ 23 , 1].
Let b1 = x, αn = g(bn ), and bn+1 = 3bn − αn = f (bn ). By the definition of P , it
is easy to see that bn ∈ [0, 1] for all n. By induction on n, it is also easy to show
that
(
)
n
∑
α
k
bn+1 = 3n x −
for n ≥ 1.
3k
k=1
∑
This thus implies that x = (αn /3n ).
Finally, we show that every x(a) is a point of P , which is equivalent to show
that x(a) ∈ En for all n. In other words, we have to show that f n (x(a)) ∈ [0, 1]
for all n (where f n+1 := f ◦ f n ). Since
∑ αk
1
0≤
≤ n,
3k
3
k≥n+1
it is easy to show that
f n (x(a)) =
∑
k≥n+1
αk
∈ [0, 1]
3k−n
for all n.
36
3. Numerical Sequences and Series
20.
Proof. Fix ϵ > 0. By the definition of Cauchy sequence, there is an integer
N1 such that d(pm , pn ) < ϵ for all m, n ≥ N1 . On the other hand, since (pni )
converges to p, there is an integer I2 such that i ≥ I2 implies that d(pni , p) < ϵ.
Let N = max{N1 , N2 } where N2 = nI2 . Then for all n ≥ N and some i such that
ni ≥ N , we have
d(pn , p) ≤ d(pn , pni ) + d(pni , p) < 2ϵ,
so that (pn ) converges to p.
21.
Proof. For every n, choose one point xn in En . It is easy to see that (xn ) is a
Cauchy sequence. Since X is complete, (xn ) converges to a point x ∈ X. Thus x
is a limit point of the range of (xk )k≥n . Note that the range of (xk )k≥n
∩∞is a subset
of En and En is closed, so that x ∈ En for all n and hence x ∈ E = 1 En . If E
contains another point y, then diam E > 0. This contradicts the assumption that
diam En → 0.
22.
Proof. Since G1 is open, there is a neighborhood E1 such that diam E1 < 1 and
E ⊆ G . Suppose that En has been constructed, so that diam En < 1/n, En ⊆
∩1n−1 1
∩n
Ek , and En ⊆ 1 Gk . Since Gn+1 is dense, En ∩Gn+1 is not empty. Since En
1
and Gn+1 are both open, there is a neighborhood En+1 such∩that diam En+1 <
n
1/(n + 1) and En+1 ⊆ En ∩ Gn+1 , so that En+1 ⊆ En = 1 En and En+1 ⊆
∩n+1
Gk .
1
∩∞
Let E = 1 En . It is clear that ∩
limn→∞ diam
follows
∩∞ En = 0, and hence it ∩
∞
∞
from Ex. 21 that there is a point x ∈ 1 En ⊆ 1 Gn . This proves that 1 Gn
is not empty. In fact, it is dense in X, which can be proved by a similar argument
starting from a neighborhood near any point y ∈ X.
23.
Proof. For any m and n,
|d(pm , qm ) − d(pn , qn )| ≤ |d(pm , pn ) + d(pn , qn ) + d(qn , qm ) − d(pn , qn )|
= d(pm , pn ) + d(qm , qn ),
which can be arbitrarily small as long as m and n are large enough. This thus
shows that (d(pn , qn )) is a Cauchy sequence in R and hence converges.
24.
3.2. Solutions to Exercises
37
Proof. (a) It is easy to prove that the relation is reflexive and symmetric. We
only prove that the relation is transitive. We denote the relation by ∼.
Suppose that the Cauchy sequences (pn ), (qn ), and (rn ) satisfy (pn ) ∼ (qn )
and (qn ) ∼ (rn ). Then lim d(pn , qn ) = 0 and lim d(qn , rn ) = 0. Since d(pn , rn ) ≤
d(pn , qn ) + d(qn , rn ), we have lim d(pn , rn ) = 0, so that (pn ) ∼ (rn ).
(b) First, we show that ∆ is well defined, that is, it is independent of the
choice of representatives of the equivalence classes. Let (rn ) ∈ P and (sn ) ∈ Q.
Then we have
lim d(rn , sn ) ≤ lim (d(rn , pn ) + d(pn , qn ) + d(qn , sn ))
n→∞
n→∞
= lim d(pn , qn ),
n→∞
where the last equality follows from (a). Then by symmetry, we must have
lim d(rn , sn ) = lim d(pn , qn ).
n→∞
n→∞
Second, we show that ∆ is a metric in X ∗ . By definition ∆ is nonnegative.
(a) shows that ∆(P, P ) = 0 and that ∆(P, Q) > 0 if P ̸= Q. The symmetry and
triangular inequality follow easily from the symmetry and triangular inequality of
d.
(c) Let (Pn ) be a Cauchy sequence in X ∗ . For every n, choose a sequence
(pn,k ) ∈ Pn . Fix ϵ > 0. For every n, there is an integer Kn such that d(pn,k , pn,l ) <
ϵ for all k, l ≥ Kn . Define qk := pk,Kk . Let us show that Pn converges to the
equivalence class of (qk ).
Since Pn is a Cauchy sequence, there is an integer N such that ∆(Pm , Pn ) < ϵ
for all m, n ≥ N . Then
lim d(pm,k , qn,k ) < ϵ
k→∞
for m, n ≥ N ,
′
′
so that there is an integer Km,n
such that k ≥ Km,n
implies that d(pm,k , qn,k ) < 2ϵ
when m, n ≥ N .
For k, l ≥ N ,
d(qk , ql ) ≤ d(qk , pk,K ′′ ) + d(pk,K ′′ , pl,K ′′ ) + d(pl,K ′′ , ql )
< d(pk,Kk , pk,K ′′ ) + 2ϵ + d(pl,K ′′ , pl,Kl )
< 4ϵ,
′
where K ′′ := max{Kk , Kl , Kk,l
}. This proves that (qk ) is a Cauchy sequence.
Let Q be the equivalence class of (qk ). For n ≥ N and k ≥ K ′′′ with K ′′′ :=
′
max{Kn , Kk , Kn,k
},
d(pn,k , qk ) ≤ d(pn,k , pn,K ′′′ ) + d(pn,K ′′′ , pk,K ′′′ ) + d(pk,K ′′′ , qk )
< 3ϵ + d(pk,K ′′′ , pk,Kk )
< 4ϵ,
38
3. Numerical Sequences and Series
which implies that ∆(Pn , Q) < 4ϵ for all n ≥ N , so that (Pn ) converges to Q.
(d) By definition.
(e) For every P ∈ X ∗ , choose (pn ) ∈ P , so that (φ(pn )) converges to P , and
therefore φ(X) is dense in X ∗ . Furthermore, if X is complete, then (pn ) converges
to p ∈ X, so that P = φ(p) ∈ φ(X). In other words, X ∗ = φ(X).
25.
Solution. The completion of X, according to Ex. 24, is X ∗ , of which each element
is identified with a real number x, or a sequence (xn ) of rational numbers, where
xn is obtained by truncating x to n decimal digits (right of the decimal point).
In other words, the completion of X is R.
Bibliography
J. R. Munkres. Topology. Prentice Hall, Inc., Upper Saddle River, NJ, second
edition, 2000. 13
W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, Inc., New York,
third edition, 1976. v
39