Notesarxiv.codlab.net/book/note-wr-pma/note-wr-pma-0.3.1.pdf · Notes on Walter Rudin’s Principles of ... Codlab.net. Title: Notes on Walter Rudin’s Principles of Mathematical

Shengtian Yang

Notes

on Walter Rudin’sPrinciples of Mathematical Analysis

Version: 0.3.1Date: 2014-01-03

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Contents

Preface v

1 The Real and Complex Number Systems 11.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Basic Topology 132.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Numerical Sequences and Series 233.1 Notes on Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 23

Bibliography 39

iii

iv Contents

Preface

This document includes my solutions to the exercises of the book Principlesof Mathematical Analysis (Rudin, 1976), as well as my notes on some interestingfacts in the book. Because of copyright reasons, the original text of the exercisesis 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 itis very suitable for those (like me) who have already basic knowledge of calculusor analysis.

I am very glad for bug reports as well as any comments on all aspects of thisdocument.

Shengtian YangHangzhou, [email protected] 12, 2013

(Last revised on June 12, 2013)

v

mailto:[email protected]

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

(f(x))2 − 2 = (x2 − 2)

[1− 2x

g(x)+

x2 − 2

(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

17≈ 1.412, x2 =

140

99≈ 1.41414, . . . .

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−2a+ · · ·+ 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 isomor-phic.

b

Proof. We will show that any ordered field F with the least-upper-bound propertyis isomorphic to R. In other words, there is a ring-isomorphism from R to F thatis also monotone increasing.

Suppose that 0′ and 1′ are the additive identity and the multiplicative identityof F , respectively. We first define a map from Q, the subfield of R, to F . For anyx ∈ 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 wherem,n > 0, we have f(x) = (m1′)/(n1′) > 0, where the last inequality follows fromthe 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) := supy∈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 thatg(x) = f(x) for x ∈ Q. Furthermore, it is easy to show that g is a monotoneincreasing ring-monomorphism. Finally, let us show that it is in fact an isomor-phism. For any x′ ∈ F , it follows from Theorem 1.20 that there exists a sequencey′k = (mk1

′)/(nk1′) such that x′ − 1′/(k1′) < y′k < x′. Let x = supk(mk/nk). We

haveg(x) > g(mk/nk) = y′k > 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 ≥ supk(mk/nk) = x,

contrary to the monotone increasing property of g. Therefore g(x) = x′, so thatg 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) isrational, a contradiction.

2.

Proof. If there were a rational x such that x2 = 12, we could write x = m/nwhere 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 dividesm2. It then follows that n2 is divisible by 3, so that n is a multiple of 3, whichcontradicts our assumption.

3.

Proof. If xy = xz and x ̸= 0, then y = 1y = (1/x)xy = (1/x)xz = 1z = z. Thisproves (a). Take z = 1 in (a) to obtain (b). (a) with z = 1/x gives (c). (c) withidentity (1/x)x = 1 yields (d).

4.


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 forall −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 henceinf 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 followingargument:

[(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 for n ∈ Z ̸=0,

xmn = (xm)n for m,n ∈ Z,

x1/(mn) = (x1/m)1/n for m,n ∈ Z ̸=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)

= (bmqbnp)1/(nq)

= (bmq)1/(nq)(bnp)1/(nq) (Corollary of Theorem 1.21)

= brbs.


(c) We first show that br is strictly increasing in r ∈ Q. It is clear that br isstrictly 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 ≥ supB(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). Bydefinition, 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 ≤ bxby. 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 andbs ≤ by, so that

bxby ≥ brbs

= br+s (Part (b))

= bx+y − (bx+y − br+s)

> bx+y − (br+s+1/n − br+s) (Proof of part (c))

= bx+y − br+s(b1/n − 1)

> bx+y − br+s(b− 1)n−1.

This implies that bxby ≥ 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 = brbs,

wherebr > 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 thatbx+y > bxby − 2brbs(b− 1)n−1

and therefore bx+y ≥ bxby. The proof is complete.

7.


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) =bwb1/n < y.

(e) Since bw > y, or b−w < y−1, it follows from (d) that b−w+(1/n) < y−1 forsufficiently 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 isimpossible that bx > y, for otherwise it follows from (e) that x is not the leastupper 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 anordered 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 arecomparable.

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+ fi. Supposing z < w and w < x, wewill show that z < x.

The relation z < w shows that a < c or a = c but b < d. Likewise, w < xgives 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).


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 anupper 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 thisordered 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

z2 = a2 − b2 + 2abi = u+ |v|i = w for v ≥ 0

and(z)2 = z2 = u− |v|i = w for v ≤ 0.

Given any w, the formula for z provides a way for computing one square rootof 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 thaty2 = w, then y2 = x2, or (y + x)(y − x) = 0, so that y = x or y = −x, acontradiction.

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 andw′ = 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|.


14.

Solution.

|1 + z|2 + |1− z|2 = 1 + 2Re(z) + |z|2 + 1− 2Re(z) + |z|2 = 4.

15.

Answer. The equality holds if and only if Baj = Cbj for some B,C ∈ R and allj = 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 equal-ity:

(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,{x1y1 + x2y2 + · · ·+ xkyk = 0,

x21 + x2

2 + · · ·+ x2k = s2.


Since y ̸= 0, with no loss of generality, we suppose yk ̸= 0. Now let x3, . . . ,xk−1 be zero. We get two reduced equations:{

x1y1 + x2y2 + xkyk = 0,

x21 + x2

2 + x2k = s2.

From the first equation we have xk = c1x1 + c2x2 where c1 = −y1/yk and c2 =−y2/yk. This combined with the second equation gives

(1 + c21)x21 + (2c1c2x2)x1 + (1 + c22)x

22 − s2 = 0.

Calculating the discriminant, we have

∆ = 4[(1 + c21)s2 − (1 + c21 + c22)x

22].

Then for any x2 such that

|x2| <

√(1 + c21)s

2

1 + c21 + c22,

we obtain two solutions for x1. This proves that there are infinitely many solutionsbecause x2 can take infinitely many distinct values.

(b) It suffices to show that for any y ̸= 0, there is exactly one x ∈ Rk suchthat x · y = 0 and |x| = 0, which is obvious.

(c) Obvious.If k = 2, then the equations in the proof of (a) become{

x1y1 + x2y2 = 0,

x21 + x2

2 = 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{x1y1 = 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.


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 parallelo-gram 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-upper-bound property.

Given two cuts α and β, we say α < β if α is a proper subset of β. Now ifα ̸= 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 itstransitive 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 provethat α ∈ R and α = supA. It is clear that α is nonempty. If p ∈ α, then p ∈ γfor 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 upperbound of A. Then γ ⊆ β for all γ ∈ A, so that α ⊆ β, that is, α ≤ β, andtherefore α = supA.

Second, we show that addition satisfies axioms (A1) to (A4).


Given two cuts α and β, we define the addition α+β to be the set of all sumsr + s where r ∈ α and s ∈ β. It is well defined because it is easy to show thatα+ 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 someβ 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.


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 followingequivalence:

A ⊆ B ⇔ A ∩Bc = ∅,

which establishes the relation between intersection problems and covering prob-lems.

According to the proof, the theorem also holds for a collection of closed subsetswith only one subset being compact. b

Note 2.2 (p. 40, Theorem 2.41) Conditions (b) and (c) are equivalent in any met-ric space (Ex. 26). The importance of the compactness of a space is that it enablesus to convert an infinite covering into a finite covering, so that many facts in a fi-nite setting can be easily generalized to a general compact space. Analogously, theseparable property (Ex. 22) enables us to reduce a problem involving uncountableset 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 andB 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 topologicalspace (or subspace). b


1.

Proof. This is trivially true because the empty set contains no element. In otherwords, we can say there is no element of the empty set such that it is not anelement of an arbitrary set.

2.

13

14 2. Basic Topology

Proof. Since every equation with integer coefficients yields only n algebraic num-bers, it suffices to show that the set of such equations is countable. We note thateach equation is identified with the n-tuple (a0, a1, . . . , an), so the whole set ofsuch equations is

∪∞N=3 EN , where

EN :=

{(a0, a1, . . . , an) ∈

N−1∪m=2

Zm : a0 ̸= 0, n+ |a0|+ · · ·+ |an| = N

}.

Since EN is finite for each N ∈ Z>0, it is clear that the number of equations iscountable (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 or2.43).

4.

Proof. No. If the set of all irrational real numbers were countable, the set of realnumbers 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). Thereforep 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 pbe a limit point of E. If p were not a limit point of E, then p would be a limitpoint 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.


Proof. (a) Observe that a point p is a limit point of Bn if and only if it is a limitpoint of Ai for some i (since Bn is a finite union of A1, . . . , An). Then,

Bn = Bn ∪B′n =

n∪i=1

Ai ∪n∪

i=1

A′i =

n∪i=1

(Ai ∪A′i) =

n∪i=1

(Ai).

(b) Use a similar argument of (a) with the fact that a limit point of Ai is alimit point of B.

For a counter example, let Ai = {1/i} in R. We have

∞∪i=1

Ai =∞∪i=1

Ai ̸= {0} ∪∞∪i=1

Ai = B.

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 pointof 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, thenevery 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 Ec and hence its closure.

It remains to show that every point of (E◦)c is a point of the closure of Ec. Ifp /∈ E◦, then every neighborhood of p contains points of Ec. This implies that pis a point of Ec or a limit point of Ec (or both), and therefore p is a point of Ec.

(e) No. An interior point of E must be an interior point of E. However aninterior point of E is not necessarily an interior point of E. For example, considerthe subset E = (0, 1) ∪ (1, 2) of R. It is clear that the interior of the closure of Eis (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 haveE = E ∪ {0} while E◦ = ∅.

10.

Proof. First, we show that this is a metric. It suffices to show that it satisfies thetriangular inequality. For p, q, r ∈ X, if p ̸= q, then

d(p, q) = 1 ≤ d(p, r) + d(r, q),


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 asubset S is infinite, then we may consider the open cover (N1/2(p))p∈S , of whichany 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 alsosymmetric. 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 ametric. We now show that it satisfies the triangular inequality.

|x− y|1 + |x− y|

= 1− 1

1 + |x− y|

≤ 1− 1

1 + |x− z|+ |z − y|

≤ |x− z|1 + |x− z|

+|z − y|

1 + |z − y|.

So d5(x, y) is a metric.

12.

Proof. Suppose G is an open cover ofK. Then there is an open set A0 ∈ G covering0, and hence covering the numbers 1/n for n ≥ N where N is fixed integer. Nowfor each n < N , choosing an open set An ∈ G that covers 1/n, we thus obtain afinite open cover (An)

Nn=0 of K, and therefore K is compact.

13.


Solution. Let An := {1/2n} ∪ {1/2n + 1/k : k ∈ Z>0}. Let us consider the limitpoints 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

2min{x− 1/2m+1, 1/2m − x}.

Then the neighborhood Nr(x) contains at most finite points of An for n > m andcontains no points of An for n ≤ m. So it is not a limit point of A. Finally, it isalso 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 more-over, 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 and thatthe intersection of every finite subcollection of (An)

∞n=1 is nonempty. However∩∞

n=1 An = ∅.Let An = (0, 1/n) where n ∈ Z>0. It is clear that An is bounded and that

the intersection of every finite subcollection of (An)∞n=1 is nonempty. However∩∞

n=1 An = ∅.

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 itis not compact, since we may easily construct an infinite sequence of points in Ethat converges to

√2, which is not in E, so that the sequence has no limit points

in E.

17.


Solution. E is countable. Since we may establish a one-to-one correspondencebetween 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 thesegment (0.1, 0.2) contains 1.

For every point p ∈ [0, 1] \ E, its decimal expansion contains at least a digitnot equal to 4 or 7. We suppose the first such digit is at the ith position and thatthe decimal expansion of that point is in the form 0.d1d2 · · · di. With no loss ofgenerality, 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.d1d2 · · · di−1(di − 1)9, 0.d1d2 · · · di−1(di + 1)1)

are not in E; if di = 0, then any numbers in the segment

(0.d1d2 · · · (di−1 − 1)99, 0.d1d2 · · · di−111)

are not in E; if di = 9, then any numbers in the segment

(0.d1d2 · · · di−189, 0.d1d2 · · · (di−1 + 1)01)

are not in E. All the above facts show that Ec 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 expansionby 4 or 7. In this way, we can construct an infinite sequence of points (distinctfrom 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

Sn be the segment (rn − 2−n−1, rn + 2−n−1) and consider the set A = [0, 2] \∪∞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 containingno 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 Bc, so that A∩B ⊆ Bc ∩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 Theo-rem 2.19 also shows that B is open. Since A and B are disjoint, they are separated(part (b)).


(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 thatd(p, uδ) = δ, for otherwise the metric space would consist of two separated setsby (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 aconnected 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, incontradiction 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. SinceA ∩ 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. Forexample, 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 anyA ⊆ R1, if s is a limit point of A, then p(s) is a limit point of p(A). This showsthat 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 absurdbecause [0, 1] is connected.

(c) Let S be a convex subset of Rk. If it were a union of two separatednonempty 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, acontradiction.

22.


Proof. It is clear that the set Qk of points having only rational coordinates iscountable (Theorem 2.12). Furthermore, for every point of Rk, we can easilyfind a point with rational coordinates that is arbitrarily close to that point. Forexample, given p ∈ Rk, we may define the point

qn =

(⌈np1⌉n

,⌈np2⌉n

, . . . ,⌈npk⌉n

).

As n → +∞, qn can be arbitrarily close to p, so that every point of Rk is a limitpoint of Qk. Therefore Rk is separable.

23.

Proof. Let X be a separable metric space and S its countable dense subset. Forany open set G and any point x ∈ G, there is a neighborhood, say Nr(x), suchthat Nr(x) ⊆ G. Let r′ = ⌈kr⌉/3k and x′ ∈ S such that d(x, x′) < r/3. Then forsufficiently large k we have x ∈ Nr′(x

′) ⊆ Nr(x) ⊆ G. This shows that X has acountable base.

24.

Proof. According to the hint, this process must stop after finite steps, otherwisewe would obtain an infinite sequence of points that has no limit point, whichcontradicts to the condition that every infinite subset has a limit point. This alsoimplies that X is covered by finitely many neighborhoods of radius δ. Repeatingthis 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. SinceK is compact, it contains a finite subcover. Considering all the centers of suchneighborhoods for n = 1, 2, . . . . The set of all these centers is (at most) countableand 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 countablebase 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 amap exists but may not be unique.


Now we show that the countable subcollection (Gf(k))k∈K covers X. For everypoint p ∈ X, there is some α such that p ∈ Gα, so that p ⊆ Bk ⊆ Gα for somek ∈ 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. Ifno 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 pointfrom each Fn, then E is infinite. Consider a limit point of E, say p, and p ∈ Gk forsome 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 whichE ∩ 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 , acontradiction. Therefore P = W c and it is closed (since every limit point of Pis also a condensation point of E). Furthermore, since P c ∩ E = W ∩ E is atmost countable, every condensation point of E must contain uncountable pointsof P ∩ E. In other words, every point of P is a condensation point of P , so thatP is perfect.

Below is an alternative proof of a general result. Let X be a separable metricspace and E a uncountable subset of X. We will show that for any subset G ofX, if it contains no condensation points of E, it contains at most countably manypoints of E. For each point p of G, because it is not a condensation point, there isa neighborhood Vp of p which contains at most countably many points of E. ThenG is covered by (Vp) and hence is covered by a countable subcollection (V ′

q ) (seethe proof of Ex. 26), so that G contains at most countably many points of E. Thisresult thus shows that P c ∩E is at most countable and that any neighborhood ofa 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. SinceE 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 oth-erwise it would be a perfect set which is uncountable (Theorem 2.43).

29.


Proof. Let G be an open set in R1. Define the map f : G → 2R1

given byp 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 clearthat for every q ∈ f(p), f(q) = f(p), and hence for p, q ∈ G, if f(p) ̸= f(q), thenthey are disjoint, for otherwise, we may choose r ∈ f(p) ∩ f(q) and we wouldhave f(p) = f(r) = f(q). Therefore G is a union of disjoint segments. To provethat this collection of segments is at most countable, we may restrict the domainof f to the subset Q1 of rational numbers. Since Q1 is dense in R1, the inverseimage 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 two statements are equivalent. Let Fn = Gcn, if∩∞

n=1 Gn were empty, then∪∞

n=1 Fn = (∩∞

n=1 Gn)c = Rk, so that at least one

Fn has a nonempty interior, which implies that Gn is not dense, a contradiction.Similarly, let Gn = F c

n, we can also prove that the second statement implies thefirst.

Now let us prove the second statement. For every point p ∈ Rk and anyneighborhood N of p, since G1 is open, we may choose a neighborhood V1 suchthat V 1 ⊆ G1 ∩ N . Suppose Vn has been constructed, so that V n ⊆ Gn. SinceGn+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 isnonempty and must be contained in N and all Gn. Therefore

∩∞n=1 Gn is dense

in Rk.

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) mayserve as illustrations for the notion that the “boundary” conjecture is false, so itis interesting to carefully study these exercises.

Ex. 11.(b) provides an approach to construct a new divergent series from adivergent series. For example, given the obvious divergent series sn =

∑nk=1 1, it

follows from Ex. 11.(b) that the series s′n =∑

(1/n) is divergent. Furthermore, itis easy to see that the resulting series is always less than the base series, in thesense that

limn→∞

(sn − s′n) = +∞.

Similarly, Ex. 12.(b) gives an approach to construct from a convergent series anew convergent series with larger limit.

Therefore we cannot expect a clear boundary between convergent and diver-gent series, since the series at the boundary, if any, would be neither convergentnor divergent. b


1.

Proof. If (sn) converges, say to s, then for any ϵ > 0, there exists an integer Nsuch 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.

limn→∞

(√

n2 + n− n) = limn→∞

n√n2 + n+ n

= limn→∞

1√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

2−(√

n2 + n− n)=

1/4

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 supposethat 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

pn+1 =1

2(pn + 1), qn+1 =

1

2

(qn +

1

2

).

Simple algebraic manipulations further show that

pn+1 − 1 =1

2(pn − 1) =

(1

2

)n

(p1 − 1) = −(1

2

)n


and

qn+1 −1

2=

1

2

(qn − 1

2

)=

(1

2

)n(q1 −

1

2

)= −

(1

2

)n+1

.

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 supn→∞

(an + c) = lim supn→∞

an + c

for c ∈ R. The case of lim sup an = ±∞ is trivially true, so we assume thatlim sup an = a ∈ R. Then for every ϵ > 0 there exists an integer N such thatn ≥ N implies that an < a+ ϵ (Theorem 3.17). It follows that

lim supn→∞

(an + c) ≤ lim supn→∞

(a+ ϵ+ c) = a+ c+ ϵ. (Theorem 3.19)

Since ϵ is arbitrary, we have

lim supn→∞

(an + c) ≤ a+ c = lim supn→∞

an + c.

On the other hand, we have

lim supn→∞

an = lim supn→∞

(an + c− c) ≤ lim supn→∞

(an + c)− c.

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 supn→∞

(an + bn) ≤ lim supn→∞

(an + b+ ϵ) = lim supn→∞

an + b+ ϵ. (Theorem 3.19)

Since ϵ is arbitrary, we have

lim supn→∞


an + b = lim supn→∞

an + lim supn→∞

bn.

Finally, if lim sup bn = −∞ and lim sup an < +∞, then there exist a realnumber M and an integer N such that an < M for all n ≥ N , so that

lim supn→∞


(M + bn) = −∞.


6.

Solution. (a) Calculate the partial sum

sn =

n∑k=1

an =√n+ 1− 1,

which clearly diverges.(b) The term an can be rewritten as

an =1

n(√n+ 1 +

√n)

<1

n3/2,

which implies that∑

an converges (Theorems 3.25 and 3.28).(c) Resorting to the root test (Thoerem 3.33), we have

lim supn→∞

n√

|an| = lim supn→∞

n√n− 1 = 0, (Theorem 3.20)

so∑

an converges.(d) If |z| ≤ 1, then

limn→∞

|an| = limn→∞

1

|1 + zn|≥ lim

n→∞

1

1 + |z|n> 0,


an diverges (Theorem 3.23).As |z| > 1, we have

lim supn→∞

|an+1||an|

= lim supn→∞

|1 + zn||1 + zn+1|

≤ lim supn→∞

|z|n + 1

|z|n+1 − 1=

1

|z|.

Therefore∑

an converges (Theorem 3.34).In summary,

∑an converges for |z| > 1 and diverges otherwise.

7.

Proof.m∑

n=1

√ann

≤m∑

n=1

nan + n−1

2n=

1

2

m∑n=1

an +1

2

∞∑n=1

1

n2.

Since∑

an and∑

(1/n2) converge (Theorem 3.28),∑

(√an/n) is bounded and

hence converges (Theorem 3.24).

8.


Proof. Since (bn) is monotonic and bounded, it converges, say to b. Without lossof generality we assume that bn ≤ b. Then

m∑n=1

anbn =m∑

n=1

an[b− (b− bn)] = bm∑

n=1

an −m∑

n=1

an(b− bn).

Since∑

[an(b− bn)] is convergent (Theorem 3.42),∑

(anbn) converges.

9.

Proof. (a) Since α = lim supn→∞n√n3 = 1, the radius of convergence is 1/α = 1

(Theorem 3.39).(b) Because

lim supn→∞

|(2z)n+1/(n+ 1)!||(2z)n/n!|

= lim supn→∞

2|z|n+ 1

= 0,

the radius of convergence is +∞ (Theorem 3.34).(c) Since α = lim supn→∞

n√2n/n2 = 2, the radius of convergence is 1/α = 1

2(Theorem 3.39).

(d) Since α = lim supn→∞n√n3/3n = 1

3 , 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 supn→∞

n√

|an| ≥ 1

and hence the radius of convergence is at most 1 (Theorem 3.39).

11.

Proof. (a) If (an) is unbounded, then

lim supn→∞

an1 + an

= 1,

and therefore∑

[an/(1 + an)] diverges (Theorem 3.23). Otherwise, there exists apositive real number M such that an < M for all n, so that

an1 + an

>an

1 +M,

and hence∑

[an/(1 + an)] diverges (Theorem 3.25).


(b)

aN+1

sN+1+ · · ·+ aN+k

sN+k≥ aN+1

sN+k+ · · ·+ aN+k

sN+k

= 1− sNsN+k

.

Because limn→∞ sn = +∞, for any ϵ ∈ (0, 1) and N ∈ Z>0 there exists aninteger k such that 1 − sN/sN+k > ϵ. This implies that

∑(an/sn) diverges

(Theorem 3.22).(c) For n ≥ 2,

ans2n

≤ ansn−1sn

=1

sn−1− 1

sn,

so thatn∑

k=1

aks2k

≤ 1

s1+

n∑k=2

(1

sk−1− 1

sk

)=

2

s1− 1

sn<

2

s1,

and hence∑

[an/s2n] converges (Theorem 3.24).

(d) First, if lim inf(nan) > 0, then there exist a positive real number α and aninteger N such that n ≥ N implies that nan > α, so that

an1 + nan

>an

(1 + 1/α)nan=

1

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

an1 + nan

>an

1 +M.

Hence∑

[an/(1 + nan)] diverges (Theorems 3.25).In general, however,

∑[an/(1 + nan)] may converge or not. Let

an =

1 if n = 2k with k ∈ Z≥0,

1

n2otherwise.

It is clear that∑

an diverges (Theorem 3.23), but

n∑k=1

ak1 + kak

<

⌊log2 n⌋∑j=0

1

1 + 2j+

n∑k=1

1/k2

1 + 1/k

<

⌊log2 n⌋∑j=0

1

2j+

n∑k=1

(1

k− 1

k + 1

),

< 3,



[an/(1 + nan)] converges (Theorem 3.24).In contrast, the series

∑[an/(1 + n2an)] surely converges because

an1 + n2an

<an

n2an=

1

n2. (Theorem 3.25)

12.

Proof. (a)

amrm

+ · · ·+ anrn

>amrm

+ · · ·+ anrm

=rm − rn+1

rm

> 1− rnrm

.

Because∑

an converges, rn can be arbitrarily small, so that for any ϵ ∈ (0, 1) andN ∈ Z>0 there exists an integer n ≥ N such that |1 − rn/rN | > ϵ. This impliesthat

∑(an/rn) diverges (Theorem 3.22).

(b)

an√rn

<2an√

rn +√rn+1

<2an(

√rn −√

rn+1)

rn − rn+1

= 2(√rn −√

rn+1).

Thereforen∑

k=1

ak√rk

< 2(√r1 −

√rn+1) < 2

√r1

and hence∑

(an/√rn) converges (Theorem 3.24).

13.

Proof. Suppose that∑

an and∑

bn converge absolutely, so that

m∑n=0

n∑k=0

|ak||bn−k|

converges (Theorem 3.50). On the other hand,

m∑n=0

∣∣∣∣∣n∑

k=0

akbn−k

∣∣∣∣∣ ≤m∑

n=0

n∑k=0

|ak||bn−k|,


so the Cauchy product of the series∑

an and∑

bn converges absolutely (Theo-rem 3.24).

14.

Proof. (a) Since lim sn = s, for every ϵ > 0 there exists an integer N such thatn ≥ N implies that |sn − s| < ϵ. So for n ≥ N ,

|σn − s| =

∣∣∣∣∣∑N−1

i=1 (si − s) +∑n

i=N (si − s)

n+ 1

∣∣∣∣∣≤∑N−1

i=1 |si − s|+∑n

i=N |si − s|n+ 1

≤(N − 1)|s|+

∑N−1i=1 |si|+ ϵ(n−N + 1)

n+ 1

= ϵ+(N − 1)|s|+

∑N−1i=1 |si| − ϵN

n+ 1,

which is bounded by 2ϵ for sufficiently large n. Therefore limσn = s.(b) Let sn = k + 1 for n = 2k, otherwise put sn = 1/n2. It is clear that sn

does not converge but limσn = 0.(c) See the solution of (b).(d)

sn − σn =nsn − (s0 + s1 + · · ·+ sn−1)

n+ 1

=(s1 − s0) + 2(s2 − s1) + · · ·+ n(sn − sn−1)

n+ 1

=1

n+ 1

n∑k=1

kak.

If lim(nan) = 0, it follows from (a) that lim(sn−σn) = 0, so that lim sn = limσn.(e) We follow the outline provided by the exercise. If m < n then

sn − σn =m+ 1

n−m(σn − σm) + sn − n+ 1

n−mσn +

m+ 1

n−mσm

=m+ 1

n−m(σn − σm) +

1

n−m

(n−m)sn −n∑

i=0

si +m∑j=0

sj

=

m+ 1

n−m(σn − σm) +

1

n−m

n∑i=m+1

(sn − si).


For i = m+ 1, . . . , n,

|sn − si| ≤n∑

j=i+1

|sj − sj−1|

≤n∑

j=i+1

M

j

≤ (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

1 + ϵ< m+ 1.

Then (m+ 1)/(n−m) ≤ 1/ϵ and |sn − si| < Mϵ. Hence

|sn − σn| ≤|σn − σm|

ϵ+Mϵ,

so that lim supn→∞ |sn − σ| ≤ Mϵ. Since ϵ was arbitrary, lim sn = σ.

15.

Fact 3.1 (Theorem 3.22).∑

an converges if and only if for every ϵ > 0 there isan integer N such that ∣∣∣∣∣

m∑k=n

ak

∣∣∣∣∣ ≤ ϵ

if m ≥ n ≥ N .

Proof. By Cauchy criterion,∑

an converges if and only if for every ϵ > 0 there isan integer N such that∣∣∣∣∣

m∑k=1

ak −n∑

k=1

ak

∣∣∣∣∣ ≤ ϵ for all m ≥ n ≥ N,

which is clearly equivalent to the condition of the theorem.

Fact 3.2 (Theorem 3.23). If∑

an converges, then limn→∞ an = 0.


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 fixedinteger, and if

∑cn converges, then

∑an converges.

Proof. Given ϵ > 0, there exists N ≥ N0 such that m ≥ n ≥ N implies

m∑k=n

ck ≤ ϵ. (Theorem 3.22)

Hence ∣∣∣∣∣m∑

k=n

ak

∣∣∣∣∣ ≤m∑

k=n

|ak| ≤m∑

k=n

ck ≤ ϵ,

which concludes the theorem.

Fact 3.4 (Theorem 3.33). Given∑

an, put

α = lim supn→∞

n√|an|.

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 = βk where β ∈ (α, 1).(b) Use Fact 3.2.(c) Use Theorem 3.33(c).

Fact 3.5 (Theorem 3.34). The series∑

an

(a) converges if lim supn→∞

|an+1||an|

< 1,

(b) diverges if|an+1||an|

≥ 1 for all n ≥ n0, where n0 is some fixed integer.

Proof. If condition (a) holds, we can find β < 1 and an integer N such that

|an+1||an|

< β for n ≥ N.


Then we have|an| ≤ |aN |βn−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∑

anbn converges.

Proof. Use Theorem 3.42 for each component of∑

anbn.


an converges absolutely, then∑

an converges.

Proof. Use the inequality∣∣∑m

k=n ak∣∣ ≤∑m

k=n |ak|.


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.


an is a series of vectors of Rk which convergesabsolutely, 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 applyTheorem 3.4.

16.

Proof. (a) Observe that

xn+1 −√α =

1

2xn

(xn −

√α)2

<1

2(xn −

√α),

where the first equality ensures that xn >√α for all n ≥ 1, so that xn decreases

monotonically and limxn =√α.

(b) The first identity has been shown in the proof of (a), and we immediatelyhave ϵn+1 < ϵ2n/2

√α since xn >

√α. This inequality can rewritten as

ϵn+1

β<

(ϵnβ

)2

so that ϵn+1 < β(ϵ1/β)2n .


(c) It suffices to show that ϵ1/β < 1/10 and β < 4.

ϵ1β

=2−

√3

2√3

=2√3− 3

6<

2× 1.8− 3

6=

0.6

6=

1

10.

β = 2√3 < 2

√4 = 4.

17.

Proof. At first, note that

xn+1 −√α =

α+ xn

1 + xn−√α =

(√α− 1)(

√α− xn)

1 + xn.

Furthremore,

xn+2 −√α =

(√α− 1)2(xn −

√α)

(1 + xn)(1 + xn+1)=

(√α− 1)2(xn −

√α)

1 + α+ 2xn,

so that

xn+2 −√α <

(α− 2√α+ 1)(xn −

√α)

1 + α< xn −

√α.

Parts (a), (b), and (c) are easy consequences of this fact. It is also clear that theformula described in Ex. 16 converges more rapidly.

18.

Proof. At first,

xn+1 − α1/p =(p− 1)xp

n − pα1/pxp−1n + α

pxp−1n

=(p− 1)xp−1

n (xn − α1/p)− α1/p(xp−1n − α(p−1)/p)

pxp−1n

=(xn − α1/p)[(p− 1)xp−1

n − α1/p∑p−2

k=0 α(p−2−k)/pxk

n]

pxp−1n

=(xn − α1/p)

∑p−2k=0(x

p−1n − α(p−1−k)/pxk

n)

pxp−1n

=(xn − α1/p)2

∑p−2k=0 x

kn

∑p−2−km=0 α(p−2−k−m)/pxm

n

pxp−1n

=(xn − α1/p)2

∑p−2l=0

∑k+m=l α

(p−2−k−m)/pxk+mn

pxp−1n

=(xn − α1/p)2

∑p−2l=0 (l + 1)α(p−2−l)/pxl

n

pxp−1n

.


This implies that xn > α1/p for every integer n ≥ 2, so that∑p−2k=0(x

p−1n − α(p−1−k)/pxk

n)

pxp−1n

<

∑p−2k=0(x

p−1n − α(p−1)/p)

pxp−1n

<p− 1

p,

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

f(x) :=

{3x x ∈ [0, 2

3 ),

3x− 2 x ∈ [ 23 , 1].

Let En := f−n([0, 1]) (where f−(n+1) := f−1 ◦ f−n). Then P :=∩∞

n=1 En is justthe Cantor set.

Now we show that every point x of P can be expressed as the form∑

(αn/3n).

Define the step function

g(x) :=

{0 x ∈ [0, 2

3 ),

2 x ∈ [ 23 , 1].

Let b1 = x, αn = g(bn), and bn+1 = 3bn − αn = f(bn). By the definition of P , itis easy to see that bn ∈ [0, 1] for all n. By induction on n, it is also easy to showthat

bn+1 = 3n

(x−

n∑k=1

αk

3k

)for n ≥ 1.

This thus implies that x =∑

(αn/3n).

Finally, we show that every x(a) is a point of P , which is equivalent to showthat x(a) ∈ En for all n. In other words, we have to show that fn(x(a)) ∈ [0, 1]for all n (where fn+1 := f ◦ fn). Since

0 ≤∑

k≥n+1

αk

3k≤ 1

3n,

it is easy to show that

fn(x(a)) =∑

k≥n+1

αk

3k−n∈ [0, 1] for all n.


20.

Proof. Fix ϵ > 0. By the definition of Cauchy sequence, there is an integerN1 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 thatni ≥ 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 aCauchy sequence. Since X is complete, (xn) converges to a point x ∈ X. Thus xis a limit point of the range of (xk)k≥n. Note that the range of (xk)k≥n is a subsetof 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 diamE > 0. This contradicts the assumption thatdiamEn → 0.

22.

Proof. Since G1 is open, there is a neighborhood E1 such that diamE1 < 1 andE1 ⊆ G1. Suppose that En has been constructed, so that diamEn < 1/n, En ⊆∩n−1

1 Ek, and En ⊆∩n

1 Gk. SinceGn+1 is dense, En∩Gn+1 is not empty. Since En

and Gn+1 are both open, there is a neighborhood En+1 such that diamEn+1 <1/(n + 1) and En+1 ⊆ En ∩ Gn+1, so that En+1 ⊆ En =

∩n1 En and En+1 ⊆∩n+1

1 Gk.Let E =

∩∞1 En. It is clear that limn→∞ diamEn = 0, and hence it follows

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 argumentstarting 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 thusshows that (d(pn, qn)) is a Cauchy sequence in R and hence converges.

24.


Proof. (a) It is easy to prove that the relation is reflexive and symmetric. Weonly 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 thechoice of representatives of the equivalence classes. Let (rn) ∈ P and (sn) ∈ Q.Then we have

limn→∞

d(rn, sn) ≤ limn→∞

(d(rn, pn) + d(pn, qn) + d(qn, sn))

= limn→∞

d(pn, qn),

where the last equality follows from (a). Then by symmetry, we must have

limn→∞

d(rn, sn) = limn→∞

d(pn, qn).

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 andtriangular inequality follow easily from the symmetry and triangular inequality ofd.

(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 integerKn such that d(pn,k, pn,l) <ϵ for all k, l ≥ Kn. Define qk := pk,Kk

. Let us show that Pn converges to theequivalence class of (qk).

Since Pn is a Cauchy sequence, there is an integer N such that ∆(Pm, Pn) < ϵfor all m,n ≥ N . Then

limk→∞

d(pm,k, qn,k) < ϵ for m,n ≥ N ,

so that there is an integerK ′m,n such that k ≥ K ′

m,n implies that d(pm,k, qn,k) < 2ϵwhen 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,K′k,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,K

′n,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,


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) convergesto 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 elementis identified with a real number x, or a sequence (xn) of rational numbers, wherexn 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, secondedition, 2000. 13

W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, Inc., New York,third edition, 1976. v

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