(Series on analysis, applications and computation 12) wing sum cheung metric space topology examples, exercises and solutions (2023)

12: Metric Space Topology: Examples, Exercises and Solutionsby W-S CheungVol.. Title: Metric space topology : examples, exercises and solutions / Wing-Sum Cheung, the University of Hong

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ISSN: 1793-4702

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Vol 12: Metric Space Topology: Examples, Exercises and Solutions

by W-S Cheung

Vol 11: Hardy Operators on Euclidean Spaces and Related Topics

by S Lu, Z Fu, F Zhao & S Shi

Vol 10: Fractional Differential Equations and Inclusions:

Classical and Advanced Topics

by S Abbas, M Benchohra, J E Lazreg, J J Nieto & Y Zhou

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by P Popivanov & A Slavova

Vol 8: The Linearised Dam-Break Problem

by D J Needham, S McGovern & J A Leach

Vol 7: The “Golden” Non-Euclidean Geometry: Hilbert’s Fourth Problem,

“Golden” Dynamical Systems, and the Fine-Structure Constant

by A Stakhov & S Aranson, Assisted by S Olsen

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Names: Cheung, Wing-Sum (Mathematician), author

Title: Metric space topology : examples, exercises and solutions /

Wing-Sum Cheung, the University of Hong Kong, Hong Kong

Description: New Jersey : World Scientific, [2024] | Series: Series on analysis, applications and computation, 1793-4702 ; Vol 12 | Includes bibliographical references and index

Identifiers: LCCN 2023023176 | ISBN 9789811266973 (hardcover) |

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in heaven

This is a volume on the Topology of Metric Spaces, which is a ject the author has been teaching in the last couple of decades Itcan serve as a textbook or a reference on the subject concerned,mainly for advanced level undergraduates concentrating onMathematics, Physics, Economics and Finance, etc A commonserious drawback of most existing references/textbooks at this level

sub-is the lack of worked examples/exercsub-ises with detailed solutions Sothe readers are frequently frustrated for not being able to solve theexercises or not being able to tell whether their solutions are valid

To overcome this drawback, it is the purpose of the presentvolume to provide plenty of worked examples and exercises,including True-or-False type questions and open-ended questions,with detailed solutions True-or-False type questions and open-endedquestions are particularly effective in helping the readers to get abirds eye view of the subject and to master the materials Theyare also instrumental in nurturing the mathematical insight and themathematical maturity of the readers

The emphasis of most standard textbooks or reference books

on the subject is the abstract and theoretical aspect, but such anapproach may not be readily digestible by most readers Inorder to master a new concept, it is most effective to first keep

in mind a few simple, concrete and familiar examples, and thenwhen dealing with a problem in the general setting, consider firstthe problem in such simple concrete settings In many situations,the treatment of a problem in a simple setting could serve as a clue

to tackle the problem in the general setting On the other hand,pictorization or visualization of abstract problems into simplepictures is very often instrumental to the rigorous treatment of theproblems However, this is rarely provided in most standard

vii

references In view of this, in this volume, within each section ineach chapter, blended with the concise and rigorous treatments onthe materials and before tackling problems in the more abstractsettings, there will be a number of concrete examples/exercises,supplemented with simple pictorizations as appropriate, to help thereaders master the concepts.

The learning outcomes of the volume include:

• Demonstrate knowledge and understanding of the basicfeatures of mathematical analysis and point set topology (e.g.,able to identify objects that are topologically equivalent)

• Apply knowledge and skills acquired in mathematicalanalysis to analyze and handle novel situations in a critical way(e.g., able to determine whether a specific function is uniformlycontinuous)

• Think creatively and laterally to generate innovative examplesand solutions to non-standard problems (e.g., able to constructcounterexamples to inaccurate mathematical statements)

• Acquire sufficient background for further studies in FunctionalAnalysis, Real Analysis, Complex Analysis, Differential and

Physics, Engineering, Economics and Finance, etc

Throughout the 30+ years of teaching experience of the author, themost frequently asked question is “What is the use of the subject?”

To be absolutely frank, many branches of pure mathematics do notfind any direct application in the real world whatsoever In fact,except for a small portion of the students who would go on forfurther studies and take mathematics research as their careers,most theorems or results they have learnt in their mathematicscourses will not be seen or used for the rest of their lives So it isperfectly fine for them to forget all the mathematics contents theyhave learnt at school or in college But one thing they shouldnever forget is the study process of and the way of thinking in

mathematics, that is, the mathematics training they haveundergone It is not the mathematical contents they have learnt thatare important but the mathematics training they received Propermathematics training nurtures logical arguments, analytical power,critical thinking, and rational thinking With these attributes, onecould excel in any career one may go into So the target of thisvolume is, through numerous worked examples and exercises, toguide the readers argue logically, think critically, analyze variouspossible scenarios, and to nurture a mathematical and logical mindthat is able to tackle novel and ill-posed problems they may encounter

in the future With that, the so-called life long self-learning would

no longer be a burden but instead, would turn into a basic instinct.The volume is perfect as a textbook or a reference for a one-semester course on Introduction to Metric Space Topology It is theintention of the author to make it concise instead of stuffing it withunessential materials, as the latter may distract the readers from thenatural flow of the development of the subject Less is more Foradvanced undergraduate level students, mastering the central theme

of a subject would be much more beneficial than dabbling in a widerrange of materials shallowly and prematurely

Finally, the author would like to express his gratitude to themany students who have taken his course on the subject Theirenthusiastic comments and feedback on the teaching materials haveprovided strong incentive for his determination to write up this book.The professional assistance rendered by the staff of World Scientific,especially Lai Fun Kwong, is also gratefully acknowledged

A Note on the Convention

exclusively adopting the following convention:

xi

About the Author

Wing-Sum Cheung was a full Professor of the Department ofMathematics of the University of Hong Kong before he retired in

2022 He is currently the Director of Undergraduate Admissions ofthe Faculty of Science and an Honorary Professor of the Department

of Mathematics of the University of Hong Kong He holds a BSc(with 1st class honors) from the Chinese University of Hong Kong, an

MA and a PhD from Harvard University, USA He has served as Head

of Department of Mathematics and Associate Dean of the Faculty ofScience of the University of Hong Kong, Vice-President of the HongKong Mathematical Society, Council Member of the Southeast AsianMathematical Society, Council Member of the Hong Kong Institute ofScience, Leader of the Hong Kong International Olympiad Team, andHonorary Consultant of the Ministry of Education, Youth and Sports

of the Government of Cambodia He has published over 200 journalarticles, conference proceedings and book chapters, in which over

140 appeared in ISI journals He has been named a top 1% highlycited researcher in the world by Clarivate Analytics’ Essential ScienceIndicator for 6 times in the last decade He is on the editorial board of

a number of international mathematical journals including Abstract

xiii

and Applied Analysis, Asian European Journal of Mathematics, FarEast Journal of Mathematical Sciences, Journal of Inequalities andApplications, etc His current research interests include Differen-tial Geometry, Exterior Differential Systems, Calculus of Variations,Analytic Inequalities, and Differential Equations.

A Note on the Convention xi

1.1 Definitions and Examples 1

Exercise 1.1: Part A 10

Exercise 1.1: Part B 14

1.2 Topology of Metric Spaces 36

Exercise 1.2: Part A 50

Exercise 1.2: Part B 64

1.3 Compactness 85

Exercise 1.3: Part A 90

Exercise 1.3: Part B 94

1.4 Compactness in the Euclidean Space R n 108

Exercise 1.4: Part A 115

Exercise 1.4: Part B 118

2 Limits and Continuity 129 2.1 Convergence in a Metric Space 129

Exercise 2.1: Part A 134

Exercise 2.1: Part B 138

2.2 Complete Metric Spaces 145

Exercise 2.2: Part A 150

Exercise 2.2: Part B 155

2.3 Continuity and Homeomorphism 172

Exercise 2.3: Part A 193

Exercise 2.3: Part B 204

3 Connectedness 233 3.1 Connectedness 233

Exercise 3.1: Part A 245

Exercise 3.1: Part B 249

xv

3.2 Path-connectedness 266

Exercise 3.2: Part A 278

Exercise 3.2: Part B 281

4 Uniform Continuity 295 4.1 Uniform Continuity 296

Exercise 4.1: Part A 301

Exercise 4.1: Part B 309

4.2 Contraction and Banach’s Fixed Point Theorem 322

Exercise 4.2: Part A 330

Exercise 4.2: Part B 332

5 Uniform Convergence 349 5.1 Sequence of Functions 349

Exercise 5.1: Part A 368

Exercise 5.1: Part B 377

5.2 Series of Functions 389

Exercise 5.2: Part A 395

Exercise 5.2: Part B 401

Metric Spaces

In this chapter, the basic concept of metric spaces will be introduced.Naively, they are simply nonempty sets equipped with a structurecalled metric For the less matured students, at the beginning, thisstructure may appear to be a bit abstract and difficult to master.But in practice, this seemingly new concept is nothing more than

needs to do is that whenever one needs to work on a problem in an

would be able to see the clue of how to proceed in the general case

In fact, in general, the most effective way to master a new concept

in any branch of mathematics is to keep in mind a couple of typicalconcrete examples and think of these examples all the time It is justthat easy

1.1 Definitions and Examples

Definition 1.1.1 Let X be a nonempty set A metric on X is areal-valued function

distance between x and y with respect to d The pair (X, d) is called

a metric space and elements in X are referred to as points in X Forthe sake of convenience, in case there is a clearly defined metric d on

X, we shall simply call X a metric space

1

δxy :=

= y

1 if x = y Then d is a well-defined metric called the discrete metric and(X, d) is called a discrete metric space Notice that in a discretemetric space, all distinct points have the same distance 1, nomatter how “far” or “close” they are from each other

(iii) Let X :=Rn and de(x, y) :=x − y for any x, y ∈ Rn, where

z :=



n

i=1

for any x = (x1, , xn), y = (y1, , yn) ∈ Rn Then de is

(Rn, de) is called a Euclidean metric space In general, when

metric space (Rn, de) In the sequel, unless it is specifically

Rn for (Rn, de) in case no confusion may arise

(iv) Let X :=C and dC(x, y) :=|x − y| for any x, y ∈ C, where, as

Euclidean space (R2, de) In fact, it is easy to see that under the

points in R2, and under this identification, dC(x, y) = de(x, y)

100(x1− y1)2+ (x2− y2)2 forany x = (x1, x2) and y = (y1, y2)∈ R2 Then ˜d is a well-defined

R, ˜d= de

(vi) Let X := S1⊂ R2and d(x, y) := the arc length of the smaller arc

is a typical example of “intrinsic metric” that can be defined on ametric space It is called intrinsic as the distance between twopoints can be measured without referring to the ambient space

Then for you, the “distance” between two points A, B on therubber band is d(A, B) instead of de(A, B), because the latter isthe Euclidean distance between the two points A, B which isattained by the straight line segment joining the two points, butfor you, the ant, unless you can fly, you cannot travel from A to Balong the straight line segment joining the two points

there is a unique “great circle” on X passing through x, y

Define d(x, y) := the arc length of the smaller arc on the

circles In this case we define d(x, y) := π = half of the arclength of any such great circles Then, similar to Example (vi),

d is a well-defined intrinsic metric on X

(viii) Let X :=Rn and dS : X× X → R be defined by

dS(x, y) :=

n

i=1

|xi− yi|k

1/k

for any x = (x1, , xn), y = (y1, , yn)∈ Rn Then dk is awell-defined metric on Rn Note that dk = de, dS, nor d∞.(xi) Let X := C[a, b ] Clearly, X is a real vector space under point-

d1(f, g) :=

b

where, as usual,

|f − g|(x) := |f(x) − g(x)| for any x ∈ [a, b ]

space [See Exercise 1.1, Part B, Problem #4.]

(xii) Similar to Example (xi) above, let X := C[a, b ] For any f ,

Then (X, d∞) is a metric space Clearly, d∞is different from dp,

example, Exercise 1.2, Part B, Problem #19 for details),(X, d∞), (X, d1), and (X, d2) are very different metric spaces

with sup{|xn| : n ∈ N} < ∞ For any elements x = {xn}n ∈N

d(x, y) := sup{|xn− yn| : n ∈ N} Then it is elementary to verify that (∞, d) is a metric space

Remark From Example 1.1.2, we see that on the same nonemptyset, different metrics could be defined and they could in turn make thesame underlying set into metric spaces with very differentproperties

be a nonempty subset The distance from x to A is defined as

d(x, A) := inf

d(x, a) : a∈ A

and the diameter of A is defined as

d(a1, a2) : a1, a2∈ A

(see Exercise 1.1, Part B, Problem #1.) A is said to be bounded if

bounded if its image is bounded

d(2, 0, 0), S2

= 1,d(S2) = 2

Note that both values are attained in S2 That is, there exist a point

p∈ S2(namely, p = (1, 0, 0)) such that d

(2, 0, 0), p

= d(2, 0, 0), S2

S2will do) such that d(r, s) = d(S2) = 2

d(2, A) = 1,d(A) = 1

Note that neither of these values is attained in A

Definition 1.1.6 Let (X, d) be a metric space, and A, B be nonemptysubsets of X The distance between A and B is

d(A, B) := inf{d(a, b) : a ∈ A, b ∈ B}

general not true

Example 1.1.7 In (R, d), let A := (0, 1), B := [3, 4], and C :=

neither of the infima is attained

“inner-product spaces” In general, an inner-“inner-product space is an orderedpair (V,

called the inner-product in V satisfying

(I1)

(I2)

(I3)

α1x1+ α2x2, y 1 x1, y 2 x2, yfor any x1, x2, y∈ V and any α1, α2∈ R

Note that by (I2) and (I3),

is bilinear If (V,

  : V → Rby

V Hence in particular, every inner-product space is automatically anormed vector space

Now if (V, ) is a normed vector space, define

by

Then it is easily checked that d satisfies (M1) – (M3) and thus (V, d)

is a metric space In general, we have the relations

{inner product spaces}  {normed vector spaces}

 {metric spaces}

gives rise to the usual metric spaceRn Similarly, C[a, b ] is an innerproduct space with

a

and it gives rise to the metric space in Example 1.1.2(xii)

d|Y is a well-defined metric on Y and hence (Y, d|Y) is also a metricspace which is known as a metric subspace of X For the sake of sim-plicity, unless ambiguity may arise, we normally drop the restriction

induced by d on Y

Example 1.1.10

d = de In particular, for any x, y ∈ Y , dY(x, y) = de(x, y) =

|x − y|

(ii) Let Y := N ⊂ R Then (Y, d) is a metric subspace of R with

d = de In particular, for any m, n∈ Y, dY(m, n) = de(m, n) =

|m − n|

Example 1.1.2 (vi), (X, d) is a well-defined metric space On

while de(n, s) = 2

For the rest of this chapter, unless otherwise specified, X will ways stand for a metric space, and S, T , etc., will stand forarbitrary subsets of X

al-Exercise 1.1

Part A: True or False Questions

For each of the following statements, determine if it is true or false

If it is true, prove it If it is false, give a counterexample or provideproper justification

Then dsatisfies (i) and (ii) but it fails to satisfy (M2).

Proof (ii) and (iii) are exactly (M2) and (M3) For (M1), for anyx,

y ∈ X, by (i), (iii) and (ii), we have

Proof (M1) and (M2) are clearly satisfied For any x, y, z ∈ R,

d((x1, y1), (x2, y2)) :=

|x1− x2| + |y1− y2|2,for any (x1, y1), (x2, y2)∈ R2, is a metric on R2

|xi− yi| : i = 1, , nfor any x = (x1, , xn) and y = (y1, , yn) ∈ Rn is a well-

Part B: Problems

Solution: Recall that we definedd(S) := inf{d(x, y) : x, y ∈ S}for anyφ= S ⊂ X So if we want to the extend the domain of definition

of diameter so as to include the empty set, in order to make things defined in a “consistent” or “continuous” manner, we should define

whatinf φshould be Now observe that for anyA,B ⊂ R, ifA⊂ B,

we must have inf A ≤ sup B As φ ⊂ B for all B ⊂ R, we must haveinf φ≤ sup Bfor allB⊂ R Hence this leaves us no choice but

to define inf φ :=−∞and sod(φ)must also be defined as−∞.

R by

˜d(x, y) := min(M, d(x, y))for any x, y∈ X Is (X, ˜d) a metric space?

Answer: Yes.

Proof We need to verify the conditions (M1)-(M3).

(M1): Since d(x, y) ≥ 0 and M > 0, we haved(x, y)˜ ≥ 0 for any

x, y ∈ X Furthermore, for any x, y ∈ X, if x = y, then

≤ M ≤ ˜d(x, z)≤ ˜d(x, z) + ˜d(y, z)

Case 2:d(y, z)˜ ≥ M Similar to Case 1, we have

˜d(x, y) = min(M, d(x, y))

Combining, we see that (M3) also holds.

(a) d(x, y) := [|x − y| ]

[x] := the largest integer that is less than or equal to x,

(b) d(x, y) := (x− y)2

1+ |x|.(d) d(x, y) :=

(c) Yes.

Proof It is clear thatd(x, y) ≥ 0 andd(x, x) = 0for allx,

Taking absolute values on both sides and simplify, we have|x| =

|y| Putting it back to (*), we have x = y and so (M1) is satisfied (M2) is obvious (M3) follows immediately from the usual triangle inequality for absolute value:

≤ |f(x) − f(z)| + |f(z) − f(y)|

= d(x, z) + d(z, y)

(d) Yes.

from the usual triangle inequality for absolute value.

4 Prove Example 1.1.2 (xi) and (xii), that is, on C[a, b ],

Solution: For p = 1, it is clear that d1(f, g) ≥ 0for all f, g ∈C[a, b ] andd1(f, g) = 0 forf = g Furthermore, if d1(f, g) = 0,

Since |f − g| ≥ 0and is continuous on[a, b ], this forcesf − g = 0

on [a, b ] and sof = g Hence (M1) is satisfied Next, it is obvious

that (M2) also holds Finally, for (M3), letf,g,h∈ X, we have

Hence (M3) also holds and sod1is a metric.

Next, for p = 2, it is clear thatd2(f, g) ≥ 0for allf, g ∈ C[a, b ]

andd2(f, g) = 0forf = g Furthermore, ifd2(f, g) = 0, then

b a(f − g)2(x)dx1

= 0

By the continuity of f − g, this forces f − g = 0 on[a, b ] and so

f = g Hence (M1) is satisfied Next, it is obvious that (M2) also holds Finally, for (M3), let f,g,h∈ X By the elementary Cauchy– Schwarz inequality, we have

+ 2

0(f (x)− h(x))2dx

0(h(x)− g(x))2dx

= (d(f, h) + d(h, g))2

Hence (M3) also holds Sod2 is a metric.

However, d(f, g) :=

b a(f− g)2(x)dx is not a metric, as triangle inequality (M3) is violated For instance, take f :≡ 1, g :≡ 2, and

5 Let{(Xn, dn)}n ∈Nbe a sequence of metric spaces and {cn}n ∈N

(a) We need to show (M1)-(M3).

(M1): For any x, y ∈ PN and every n = 1, , N, since

dn is a metric on Xn , all dn(xn, yn) are non-negative real numbers Hence being a finite sum of non-negative real numbers,

dN(x, y)≥ 0 Furthermore, for any x,y∈ PN ,

dN(x, y) = 0⇐⇒

N

n=1

cndn(xn, zn)

1 + dn(xn, zn) + dn(zn, yn)+

N

n=1

cndn(xn, zn)

1 + dn(xn, zn) +

N

n=1

cndn(zn, yn)

1 + dn(zn, yn)

= dN(x, z) + dN(z, y)

(c) (M1): Again, it is easily seen thatdN(x, y)≥ 0 for anyx, y∈

PN Furthermore, for anyx,y∈ PN ,

6 Let dk(k = 1, 2) and d∞ be metrics onRn defined by

dk(x, y) :=

n

i=1

|xi− yi|2≥

n

i=1

|xi− yi|

2

Taking the square root on both sides, we have

n

i=1

|xi− yi|2

1/2

≤ maxi=1, ,n|xi− yi| = d∞(x, y)

The third inequalityd∞≤ d2 is obvious Finally, it is easy to see that

max

i=1, ,n|xi− yi|2≤

n

i=1

|xi− yi|2≤

n

i=1

|xi− yi|

2,

which is exactly the last inequality.

7 Let X = C[0, 1] be the set of continuous real-valued functions

2−kmin{|f(rk)− g(rk)|, 1}

Prove that (X, d) is a metric space

Proof.

(a) (X, d∞) is a metric space:

Clearly, d∞(f, g) ≥ 0 and d∞(f, f ) = 0 for all f, g ∈ X Furthermore, ifd∞(f, g) = 0, then for allt∈ [0, 1],

Next, for(X, d2), (M1) and (M2) are clear For (M3), letf,g,

h∈ X By Cauchy–Schwarz inequality,

d22(f, g) =

0(f (x)− g(x))2dx

=

0(f (x)− h(x) + h(x) − g(x))2dx

=

0(f (x)− h(x))2dx +

0(h(x)− g(x))2dx+ 2

0(h(x)− g(x))2dx

+ 2

0(f (x)− h(x))2dx

1

×

0(h(x)− g(x))2dx

1

=[d(f, h) + d(h, g)]2

Hence (M3) also holds.

(b) (M1): It is obvious that d(f, g) ≥ 0 for all f, g ∈ X and

there exists  ∈ N such that f (r) = g(r), then d(f, g) ≥

2−|f(r)− g(r)| > 0 Hence we must havef (rk) = g(rk)

for allk∈ N That is,f = g on{rk : k ∈ N}which is dense

in[0, 1] Since bothf andg are continuous, this forcesf = g

on[0, 1].

(M2): Trivial.

(M3): Observe first the following elementary inequality

for alla,b,c≥ 0witha≤ b + c Hence for anyf,g,h ∈ X,