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  • Lecture notes on Topology and Geometry

    Huy`nh Quang Vu

    F M I, U N S,V N U, 227 N V C, D 5, H C M C,V

    E-mail address: [email protected]: http://www.math.hcmuns.edu.vn/hqvu

  • A. This is a set of lecture notes prepared for a series of introductorycourses in Topology and Geometry for undergraduate students at the Universityof Natural Sciences, Vietnam National University in Ho Chi Minh City.

    This is indeed a lecture notes in the sense that it is written to be deliveredby a lecturer, namely myself, tailored to the need of my own students. I did notwrite it with self-study readers or with other lecturers in mind.

    Most statements are intended to be exercises. I provide proofs for someof the more difficult propositions, but even then there are still many details forstudents to fill in. More discussions will be carried out in class. I hope in thisway I will be able to keep this lecture note shorter and more readable.

    In general if you encounter an unfamiliar notion, look at the Index and theContents. A problem with a sign * is considered more difficult. A problem witha sign is an important one.

    This lecture notes will be continuously developed and I intend to keep itfreely available on my web page. Although at this moment it is only a draft I dohope that it is useful to the readers. Your comments are very welcomed.

    May 14, 2005 September 23, 2008.

  • Contents

    Part 1. General Topology 1

    Chapter 1. Theory of Infinite Sets 31.1. The Cardinality of a Set 31.2. The Axiom of Choice 8Guide for Further Reading in General Topology 10

    Chapter 2. Topological Spaces 112.1. Topological Spaces 112.2. Continuity 152.3. Subspaces 18

    Chapter 3. Connectivity 213.1. Connected Space 213.2. Path-connected Spaces 25Further Reading 28

    Chapter 4. Separation Axioms 294.1. Separation Axioms 29

    Chapter 5. Nets 315.1. Nets 31

    Chapter 6. Compact Spaces 356.1. Compact Spaces 35

    Chapter 7. Product Spaces 397.1. Product Spaces 397.2. Tikhonov Theorem 42Further Reading 44

    Chapter 8. Compactifications 458.1. Alexandroff Compactification 458.2. Stone-Cech Compactification 47

    Chapter 9. Urysohn Lemma and Tiestze Theorem 499.1. Urysohn Lemma and Tiestze Theorem 49Further Reading 53

    Chapter 10. Quotient Spaces 5510.1. Quotient Spaces 55

    Chapter 11. Topological Manifolds 6111.1. Topological Manifolds 61Further Reading 63

    iii

  • iv CONTENTS

    Part 2. Algebraic Topology 65

    Chapter 12. Homotopy and the fundamental groups 6712.1. Homotopy and the fundamental groups 67

    Chapter 13. Classification of Surfaces 6913.1. Introduction to Surfaces 6913.2. Classification Theorem 7013.3. Proof of the Classification Theorem 71

    Part 3. Differential Topology 75

    Chapter 14. Differentiable manifolds 7714.1. Smooth manifolds 7714.2. Tangent spaces Derivatives 80

    Chapter 15. Regular Values 8315.1. Regular Values 8315.2. Manifolds with Boundary 87

    Chapter 16. The Brouwer Fixed Point Theorem 9116.1. The Brouwer Fixed Point Theorem 91

    Chapter 17. Oriented Manifolds The Brouwer degree 9317.1. Orientation 9317.2. Brouwer Degree 9417.3. Vector Fields 98Further Reading 99

    Part 4. Differential Geometry 101Guide for Reading 102

    Chapter 18. Regular Surfaces 10318.1. Regular Surfaces 10318.2. The First Fundamental Form 10518.3. Orientable Surfaces 10618.4. The Gauss Map 107

    Bibliography 109

    Index 111

  • Part 1

    General Topology

  • CHAPTER 1

    Theory of Infinite Sets

    In general topology we often work in very general settings, in particular weoften deal with very large sets. Therefore we start with a deeper study of settheory.

    1.1. The Cardinality of a Set

    Sets. We will not define what a set is. That means we only work on the level ofthe so-called naive set theory.

    Even so here we should be aware of certain problems in naive set theory. Theseproblems are both educational and fascinating.

    Till the beginning of the 20th century, the set theory of German mathemati-cian George Cantor, in which set is not defined, was widely used and thought tobe a good basis for mathematics. Then certain critical problems were discoveredrelating to the liberal uses of the undefined notion of set.

    1.1.1. E (Russells Paradox). A famous version of it is the Barber para-dox. It is as follows: In a village there is a barber. His job is to do hair cut toall villagers who cannot cut his hair himself. The question is who would cut thebarbers hair? If he cut his hair, then he would have cut the hair of somebody whocould do that himself, thus violating the terms of his job. On the other hand, if hedoes not cut this hair he also violates those terms, because he did not cut the hair ofsomeone who could not do it himself. This means if we take the set of all villagerswho had his hair cut by the barber then we cant decide whether the barber himselfis a member of that set or not.

    1.1.2. E. Consider the set S = {x/x < x} (the set of all sets which are not amember of itself). Then whether S S or not is undecidable.

    These examples show that we are having too much liberty with the notion ofset. Deeper study of the notion of set is needed. There are two axiomatic sys-tems for the theory of sets, the Zermelo-Fraenkel system and the Von Neumann-Bernays-Godel one. In the Von Neumann-Bernays-Godel system a more generalnotion than set, called class, is used.

    For our purpose it is enough for us to be a bit careful when dealing with largesets, set of sets. In those occations we often replace the term set by the termsclass or collection. See [Dug66, p. 32].

    Indexed family. Suppose that A is a collection, I is a set and f : I A is asurjective map. The collection A together with the map f is called an indexedfamily. We often write fi = f (i), and denote the indexed family by { fi/ i I}. Notethat it can happen that fi = f j for some i , j.

    Thus a family is not a collection, an indexed set is not a set.On the other hand a collection can be associated with an indexed family by

    taking the index set to be the collection itself and the index map to be the identity.

    3

  • 4 1. THEORY OF INFINITE SETS

    Often it is convenient to write a collection of sets as an indexed family of sets,so we often say let {S i/ i I} be a collection of sets . . . .

    Cartesian product. Let {Ai}iI be a family of sets indexed by a set I. The Carte-sian product

    iI Ai of this family is defined to be the collection of all maps

    f : I iI Ai such that for each i I we have f (i) Ai. An element ofiI Ai is often denoted by (ai)iI , with ai Ai is the coordinate of index i, in

    analog to the finite product case.

    Equivalent sets. Two sets are said to be equivalent if there is a bijection fromone to the other. When A and B are equivalent we write A B.

    1.1.3. Two intervals [a, b] and [c, d] on the real number line are equivalent via alinear map. Two intervals (a, b) and (c, d) are equivalent.

    1.1.4. The interval (1, 1) is equivalent to R via the map

    x 7 11 |x| x.

    1.1.5. The ball B(0, 1) = {(x1, x2, . . . , xn) Rn/ x21 + x22 + + x2n < 1} is equivalentto Rn.

    1.1.6. Define the sphere S n to be the set

    {(x1, x2, . . . , xn+1) Rn+1/ x21 + x22 + + x2n+1 = 1}.

    Then S n \ {(0, 0, . . . , 0, 1)} is equivalent to Rn via the stereographic projection.

    Countable sets.

    1.1.7. D. A set is called countably infinite if it is equivalent to the set ofpositive integers. A set is called countable if it is either finite or countably infinite.

    A countably infinite set can be enumerated by the positive integers. Theelements of such a set can be indexed by the positive integers as a1, a2, a3, . . . .

    1.1.8. E. The set of intergers Z is countable. The set of even integers iscountable.

    1.1.9. A union between a countable set and a finite set is countable.

    1.1.10. A subset of a countable set is countable.

    1.1.11. T. A union of a countable collection of countable sets is a countableset

    P. The collection can be indexed as A1, A2, . . . , Ai, . . . (if the collection isfinite we can let Ai be the same set for all i starting from a certain number). Theelements of each set Ai can be indexed as ai,1, ai,2, . . . , ai, j, . . . .

    This means the union

    iI Ai can be mapped injectively to the index set Z+Z+by ai, j 7 (i, j).

    Thus it is sufficient for us to prove that Z+ Z+ is countable.We can index Z+ Z+ by the method shown in the following diagram:

  • 1.1. THE CARDINALITY OF A SET 5

    (1, 1) // (1, 2)

    {{wwwwwwww

    (1, 3) // (1, 4)

    {{wwwwwwww

    (2, 1)

    (2, 2)

    ;;wwwwwwww(2, 3)

    {{wwwwwwww

    (3, 1)

    ;;wwwwwwww(3, 2)

    {{wwwwwwww

    (4, 1)

    (4, 2)

    (5, 1)

    ;;wwwwwwww

    1.1.12. Theorem 1.1.11 can be reduced to the statement that Z+ Z+ is equivalentto Z+.

    Give another proof by checking that the map f : Z+Z+ Z+, (m, n) 7 2m3nis injective.

    1.1.13. T. The set Q of rational numbers is countable.

    P. Write Q =

    n=1

    {pq / p, q Z, q > 0, |p| + q = n

    }.

    Another way is to observe that the map pq 7 (p, q) from Q to ZZ is injective.

    1.1.14. Q Q Q.

    Inquiry minds would have noted that we did not define the set of integers or theset of real numbers. Such definitions are not easy. We contain ourself that thosesets have some familiar properties.

    1.1.15. T. The set of real numbers is uncountable.

    P. The proof uses the Cantor diagonal argument.Suppose that the interval [0, 1] is countable and is enumerated as a sequence

    {ai/ i Z+}. Suppose thata1 = 0.a11a12a13 . . .

    a2 = 0.a21a22a23 . . .

    a3 = 0.a31a32a33 . . .

    ...

    There are rational numbers whose decimal presentations are not unique, suchas 1/2 = 0.5000 . . . = 0.4999 . . . . To cover this case we should choose bn , 0, 9.

    Choose a number b = 0.b1b2b3 . . . such that b1 , a11, b2 , a22, . . . , bn , ann,. . . . Then b , an for all n.

  • 6 1. THEORY OF INFINITE SETS

    Cardinality. A genuine definition of cardinality of sets requires an axiomatic treat-ment of set theory, therefore here we contain ourselves that for each set A thereexists an object called its cardinal |A|, and there is an order on the set of cardinalssuch that:

    (1) If a set is finite then its cardinal is its number of elements.(2) Two sets have the same cardinals if and only if they are equivalent:

    |A| = |B| (A B)(3) |A| |B| if and only if there is a injective map from A to B.

    The set Z+ has cardinal 0 (read aleph-0, aleph being the first character inthe Hebrew alphabet): |Z+| = 0.

    The set R has cardinal c (continuum): |R| = c.1.1.16. An infinite set contains a countably infinite subset.

    1.1.17. 0 is the smallest infinite cardinal, and 0 < c.Georg Cantor put forward the The Continuum Hypothesis: There is no cardinal

    between 0 and c. Godel (1939) and Cohen (1964) have shown that the ContinuumHypothesis is independent from other axioms of set theory.

    1.1.18. If A has n elements then |2A| = 2n.1.1.19. T. The cardinal of a set is strictly less than the cardinal of the set ofits subsets, i.e. |A| < |2A|.

    This implies that there is no maximal cardinal. There is no universal set, aset which contains everything.

    P. Let A , and denote by 2A the set of its subsets.(1) |A| |2A|: The map from A to 2A: a 7 {a} is injective.(2) |A| , |2A|: Let be any map from A to 2A. Let X = {a A/ a < (a)}.

    Then there is no x A such that (x) = X (assuming the contrary therewill be a contradiction), therefore is not surjective.

    1.1.20. 2N is equivalent to the set of sequences of binary digits.

    1.1.21. T (Cantor-Bernstein-Schroeder). If A is equivalent to a subsetof B and B is equivalent to a subset of A then A and B are equivalent.

    (|A| |B| |B| |A|) |A| = |B|P. Suppose that f : A 7 B and g : B 7 A are injective maps. Let

    A1 = g(B), we will show that A A1.Let A0 = A and B0 = B. Define Bn+1 = f (An) and An+1 = g(Bn). Then An+1

    An. Furthermore via the map g f we have An+2 An, and An \An+1 An+1 \An+2.Using the following identities

    A = (A \ A1) (A1 \ A2) (An \ An+1) . . . (

    n=1

    An),

    A1 = (A1 \ A2) (A2 \ A3) (An \ An+1) . . . (

    n=1

    An),

    we see that A A1.

  • 1.1. THE CARDINALITY OF A SET 7

    1.1.22. 20 = c.Hint: That |2N| |[0, 1]| follows from 1.1.20. For the reverse direction, observe that any real

    number can be written in binary form (or use the Continuum Hypothesis).

    1.1.23. 20 20 = 20 .Hint: An injective map from 2N2N to 2N can be constructed as follows. Two binary sequences

    a1a2 . . . and b1b2 . . . correspond to the sequence a1b1a2b2 . . ..

    1.1.24. R2 R, in other words c2 = c.Hint: Use the results of 1.1.22 and 1.1.23, or prove directly as in 1.1.23.

    Note: In fact for all infinite cardinal we have 2 , see [Dug66, p. 52], [Lan93, p. 888].

    More problems.

    1.1.25. Check that (

    iI Ai) ( jJ B j) = iI, jJ Ai B j.1.1.26. Which of the following formulas are correct?

    (a) (

    iI Ai) (iI Bi) = iI(Ai Bi).(b)

    iI(

    jJ Ai, j) =

    iI(

    jJ Ai, j).

    1.1.27. Let f be a function. Then:(a) f (

    i Ai) =

    i f (Ai).

    (b) f (

    i Ai) i f (Ai). If f is injective (one-one) then equality happens.(c) f 1(

    i Ai) =

    i f 1(Ai).

    (d) f 1(

    i Ai) =

    i f 1(Ai).

    1.1.28. Let f be a function. Then:(a) f ( f 1(A)) A. If f is surjective (onto) then equality happens.(b) f 1( f (A)) A. If f is injective then equality happens.

    1.1.29. A set which contains an uncountable subset is uncountable.

    1.1.30. If A is finite and B is inifinite then A B B.Hint: There is an infinitely countable subset of B.

    1.1.31. The set of points in Rn with rational coordinates is countable.1.1.32. Is the set of functions f : {0, 1} Z countable?1.1.33. The set of functions f : A {0, 1} is equivalent to 2A.1.1.34. A real number is called an algebraic number if it is a root of a polynomialwith integer coefficients. Show that the set of algebraic numbers is countable.

    1.1.35. A real number which is not algebraic is called transcendental. For exampleit is known that pi and e are transcendental (whereas it is still not known whetherpi + e is transcendental or not). Show that the set of transcendental numbers isuncountable.

    1.1.36. Show that [a, b] (a, b] (a, b).1.1.37. A countable union of continuum sets is a continuum set.

    Hint:

    n=1[n, n + 1] = [1,).

  • 8 1. THEORY OF INFINITE SETS

    1.2. The Axiom of Choice

    Ordered Sets. A relation on S is a non-empty subset of the set S S . A (partial)order on the set S is a relation R on S such that for all a, b, c S :

    (1) (a, a) R.(2) ((a, b) R (b, a) R) a = b.(3) ((a, b) R (b, c) R) (a, c) R.

    If any two elements of S are related by R then R is called a total order and (S ,R)is a totally ordered set. When (a, b) R we often write aRb and use the suggestivenotation for R.1.2.1. E. (1) (R,) is a totally ordered set.

    (2) Let S be a set. Then (2S ,) is a partially ordered set but is not totallyordered if S has more than one element.

    1.2.2. Let (S 1,1) and (S 2,2) be two ordered sets. Show that the following is anorder on S 1 S 2: (a1, b1) (a2, b2) if (a1 < a2) or ((a1 = a2) (b1 b2)). This iscalled the dictionary order.

    1.2.3. D. A totally order on a set S is a well-order if every non-emptysubset A of S has a smallest element, i.e. a A, b A, a b.

    For example (N,) is well-ordered while (R,) is not.The Axiom of Choice.

    1.2.4. T. The following statements are equivalent:

    (1) Axiom of Choice: Given an arbitrary set M, there exists a choice func-tion giving with any subset of M an element of M.

    (2) Given a family of disjoint non-empty sets Ai there exists a set A such thatA contains exactly one element from each Ai.

    (3) The Catersian product of a family of non-empty sets is non-empty.(4) The cardinalities of any two sets can be compared.(5) Zorn Lemma: If any totally ordered subset A of a partially ordered set X

    has an upper bound (i.e. b X, a A, b a) then X has a maximalelement (i.e. a X, (b X b a) a = b).

    (6) Hausdorff Maximality Principle: A totally ordered subset of a partiallyordered set belongs to a maximal totally ordered subset.

    The Axiom of Choice is needed for many important results in mathematics,such as the Tikhonov Theorem about products of compact sets, the Hahn-Banachand Banach-Alaoglu Theorems in Functional Analysis, the existence of bases in avector space, the existence of a Lebesgue unmeasureable set, . . . .

    Zorn Lemma is often a convenient form of the Axiom of Choice.

    1.2.5. Show that any vector space has a vector basis.Hint: Consider the collections B of all independent sets of vectors in a vector space V with

    the order of set inclusion. Suppose that {Bi} is a totally ordered collection of members of B. LetA =

    i Bi. Then A B and is an upper bound of {Bi}.

    The following is based on the Axiom of Choice.

    1.2.6. T (Zermelo, 1904). On any set there is a well-order.

  • 1.2. THE AXIOM OF CHOICE 9

    More problems.

    1.2.7. The set Z+ with the order m n m|n is a partially ordered set.1.2.8. Let (S 1,1) and (S 2,2) be two ordered sets. For a and b in S 1 S 2, definea b if (a 1 b) (a, b S 1), or (a 2 b (a, b S 2 \ S 1)). Show that this is anorder on S 1 S 2.1.2.9. Find a partial order on R2.

    1.2.10. A totally ordered finite set is well-ordered.

    1.2.11. A subset of a well-ordered set is well-ordered.

    1.2.12. The set of rational numbers on the interval [0, 1] is not well-ordered.

    1.2.13 (Transfinite Induction Principle). Let A be a well-ordered set. Let P(a)be a statement whose truth depends on a A. If

    (1) P(a) is true when a is the smallest element of A(2) if P(a) is true for all a < b then P(b) is true

    then P(a) is true for all a A.Hint: Assume the contrary.

  • 10 1. THEORY OF INFINITE SETS

    Guide for Further Reading in General Topology

    The book by Kelley [Kel55] has been a classics and a standard reference al-though it was published in 1955. Its presentation is rather abstract. The book hasno picture!

    Munkres book [Mun00] is famous. Its treatment is somewhat more modernthan Kelley, with many examples, pictures and exercises. It also has a section onAlgebraic Topology.

    Hocking and Youngs book [HY61] contains many deep and difficult results.This book together with Kelley and Munkres contain many topics not discussed inour lectures, some are at more advanced levels than the level here.

    The more recent book by Roseman [Ros99] works mostly in Rn. Its strength isthat it is more down-to-earth and it contains many new topics such as space-fillingcurves, knots, and manifolds.

    Some other good books on General Topology are the books by Cain [Cai94],Duong Minh Duc [Duc01], Viro and colleagues [VINK08].

    If you want to have some ideas about the kinds of current research in GeneralTopology you can have a look at a collection of open problems in [MR90].

    All of the items in the References are available from me.

  • CHAPTER 2

    Topological Spaces

    On sets equipped with topological structures we can study continuity of func-tions.

    2.1. Topological Spaces

    2.1.1. D. Let X be a set. A topology on X is a family of subsets of Xsatisfying:

    (1) The sets and X are elements of .(2) A union of elements of is an element of .(3) A finite intersection of elements of is an element of .

    Elements of are called open sets of X in this topology.A complement of an open set is called a closed set.

    The set X together with the topology is called a topological space. We oftencall an element of a topological space a point.

    2.1.2. E. (1) On any set X there is the trivial topology {, X}.(2) On any set X there is the discrete topology whereas any point constitutes

    an open set. That means any subset of X is open, so the topology is 2X .

    2.1.3. E. Let X = {1, 2, 3}. The following are topologies on X:(1) 1 = {, {1}, {2, 3}, {1, 2, 3}}.(2) 2 = {, {1, 2}, {2, 3}, {2}, {1, 2, 3}}.

    2.1.4. E (The Euclidean topology). In Rn = {(x1, x2, . . . , xn)/xi R}, theEuclidean distance between two points x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn)is d(x, y) = [

    ni=1(xi yi)2]1/2. An open ball centered at x with radius r is the set

    B(a, r) = {y Rn/d(y, x) < r}. A subset of Rn is called open if it is either empty oris a union of open balls. This is the Euclidean topology of Rn.

    2.1.5. The finite complement topology on X consists of the empty set and all sub-sets of X whose complements are finite.

    2.1.6. A finite intersection of open sets is open is equivalent to an intersectionof two open sets is open.

    2.1.7. Show that in a topological space X:(a) and X are closed.(b) A finite union of closed sets is closed.(c) An intersection of closed sets is closed.

    A neighborhood of a point x X is a subset of X which contains an open setcontaining x. Note that a neighborhood doesnt need to be open.

    11

  • 12 2. TOPOLOGICAL SPACES

    Bases of a topology. A collection B of open sets is called a basis for thetopology of X if any non-empty member of is a union of members of B.

    2.1.8. E. In the Euclidean space Rn the set of balls with rational radii is abasis.

    2.1.9. In the Euclidean space Rn the set of balls with radii 12m , m 1 is a basis.2.1.10. A collection B of open sets is a basis if for each point x and each open setO containing x there is a U B containing x such that U is contained in O.

    A collection S of open sets is called a subbasis for the topology of X ifthe collection of finite intersections of members of S is a basis for .

    2.1.11. E. Let X = {1, 2, 3}. The topology 2 = {, {1, 2}, {2, 3}, {2}, {1, 2, 3}}has a basis {{1, 2}, {2, 3} {2}} and a subbasis {{1, 2}, {2, 3}}.2.1.12. The collection of open rays, that is sets of the forms (a,) and (, a) isa subbasis of R with the Euclidean topology.

    Generating topologies. Given a collection of subsets of a set, when is it a basisfor a topology?

    2.1.13. T. Let B be a collection of subsets of X. Then B {} is a basis fora topology on X if and only if the union of members of B is X and the intersectionof two members of B is a union of members of B.

    The collection consists of the empty set and all unions of members of B is atopology on X. It is called the topology generated by B.

    P. Verifying that is a topology is reduced to checking that the intersec-tion of two members of is a member of . In deed (

    i Bi)( j B j) = i, j(BiB j).

    But then Bi B j = i, j,k Bk. Next we want to know, given a collection of subsets of a set, is there a small-

    est topology which contains these subsets? When is this collection a subbasis fora topology?

    2.1.14. T. Let S be a family of subsets of X whose union is X. Then Sis a subbasis for a topology on X consisting of the empty set and unions of finiteintersections of members of S .

    This topology is called the topology generated by S . A basis for this topologyis the collection of finite intersections of members of S .

    By this theorem, just any collection of subsets covering the set generates atopology on that set.

    2.1.15. E. Let X = {1, 2, 3, 4}. The set {{1}, {2, 3}, {3, 4}} generates the topol-ogy {, {1}, {3}, {1, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}. A basis forthis topology is {{1}, {3}, {2, 3}, {3, 4}}.2.1.16 (Order topology). Let (X,) be a totally ordered set with at least twoelements. The subsets of the forms { X/ < } and { X/ > } generate atopology on X, called the order topology.

    2.1.17. E. The Euclidean topology on R is the ordering topology with re-spect to the usual order of real numbers.

  • 2.1. TOPOLOGICAL SPACES 13

    Metric spaces. Metric space is a motivation for generalization to topologicalspace. Metric spaces are also important examples of topological spaces.

    A ball is a set of the form B(x, r) = {y X/ d(y, x) < r}.2.1.18. In the theory of metric spaces, a set U X is said to be open if x U, > 0, B(x, ) U.

    Show that the family of balls is a basis for a topology on (X, d). When we speakabout topology on a metric space we mean this topology.

    Comparing topologies. Let 1 and 2 are two topologies on X. If 1 2 wesay that 2 is finer (or stronger) than 1 and 1 is coarser (or weaker) than 2.

    2.1.19. E. On a set the trivial topology is the coarsest topology and thediscrete topology is the finest one.

    2.1.20. The topology generated by a collection of subsets is the coarsest topologycontaining those subsets.

    More problems.

    2.1.21. Is the collection of all open half-planes (meaning not consisting the sepa-rating lines) a subbasis for the Euclidean topology of R2?

    2.1.22. Show that on R2 the basis consists of interiors of circles, the basis consistsof interiors of squares, and the basis consists of interiors of ellipses generate thesame topology.

    2.1.23. Show that two bases generate the same topology if and only if each mem-ber of one basis is a union of members of the other basis.

    2.1.24 (Rn has a countable basis). The set of balls each with rational radiuswhose center has rational coordinates forms a basis for the Euclidean topology ofRn.

    2.1.25. On Rn consider the following metrics, which are generated by norms. Findthe unit ball for each metric. Show that these metrics generate the same topology.

    (a) d1(x, y) =n

    i=1|xi yi|(b) d2(x, y) = max1in|xi yi|(c) d3(x, y) = (

    ni=1(xi yi)2)1/2

    2.1.26. Let d1 and d2 be two metrics on X. If there are , > 0 such that forall x, y X, d1(x, y) < d2(x, y) < d1(x, y) then the two metrics are said to beequivalent. This is indeed an equivalence relation. Two equivalent metrics generatethe same topology.

    Hint: Show that a ball in one metric contains a ball in the other metric with the same center.

    Note: It is shown in a course in Real Analysis or Functional Analysis that all norms on Rn

    generate equivalent metrics, so all norms on Rn generate the same topology, exactly the Euclidean

    topology.

    2.1.27. Is the Euclidean topology on R2 the same as the ordering topology on R2

    with respect to the dictionary order?

    2.1.28. An open set in R is a countable union of open intervals.

    2.1.29. The family of intervals of the form [a, b) generates a topology on R. Is itthe Euclidean topology?

  • 14 2. TOPOLOGICAL SPACES

    2.1.30. Describe all sets that have only one topology.

    2.1.31. * Describe all topologies that have only one basis.Hint: It must be that any open set is not a union of other open sets. But if this happens then

    the topology has no two open sets such that one is not contained in the order. This means that the

    topology is totally ordered with respect to the inclusion order. Thus this is a totally ordered topology

    such that any open set is not the union of its open strict subsets.

    2.1.32. On the set of integer numbers Z, consider all arithmetic progressions

    S a,b = a + bZ,

    where a Z and b Z+.(a) Show that these sets form a base for a topology on Z.(b) Show that with this topology each set S a,b is both open and closed.(c) Show that the set {1} is closed.(d) Show that if there are only finitely many prime numbers then the set {1}

    is open.(e) From (d) conclude that there are infinitely many prime numbers.

  • 2.2. CONTINUITY 15

    2.2. Continuity

    Continuity.

    2.2.1. D. Let X and Y be topological spaces. We say a map f : X Y iscontinuous at x X if for any neighborhood U of f (x) there is a neighborhood Vof x such that f (V) U.

    Equivalently, f is continuous at x if for any neighborhood U of f (x) the inverseimage f 1(U) is a neighborhood of x.

    We say that f is continuous on X if it is continuous everywhere on X.

    2.2.2. T. A map is continuous if and only if the inverse image of an open setis an open set.

    P. () Suppose that f : X Y is continuous. Let U be an open setin Y . Let x f 1(U), and let y = f (x). Since f is continuous at x and U is aneighborhood of y, the set f 1(U) is a neighborhood of x. Thus x is an interiorpoint of f 1(U), so f 1(U) is open.

    () Suppose that the inverse image of any open set is an open set. Let x X,and let y = f (x). Let U be a neighborhood of y, containing an open neighborhoodU of y. Then f 1(U) is an open set containing x, therefore the set f 1(U) will bea neighborhood of x.

    2.2.3. A map is continuous if and only if the inverse image of a closed set is aclosed set.

    2.2.4. Let (X, d1) and (Y, d2) be two metric spaces. In the theory of metric spaces,a map f : (X, d1) (Y, d2) is continuous at x X if and only if

    > 0, > 0, d1(y, x) < d2( f (y), f (x)) < .Show that this is a special case of the notion of continuity in topological spaces.

    2.2.5. R. Therefore from now on we can use any results in previous coursesinvolving continuity in metric spaces.

    Homeomorphism. A map from one topological space to another is said to bea homeomorphism if it is a bijection, is continuous and its inverse map is alsocontinuous.

    Two spaces X and Y are said to be homeomorphic, written X Y , if there is ahomeomorphism from one to the other.

    An open map is a map such that the image of an open set is an open set.A closed map is a map such that the image of a closed set is a closed set.

    2.2.6. A homeomorphism is both an open map and a closed map.

    2.2.7. P. If f : (X, X) (Y, Y ) is a homeomorphism then it induces abijection between X and Y .

    P. The map

    f : X YO 7 f (O)

    is a bijection.

  • 16 2. TOPOLOGICAL SPACES

    In the category of topological spaces and continuous maps, when two spacesare homeomorphic they are the same.

    An embedding (or imbedding) (phep nhung) from the topological space X tothe topological space Y is a map f : X Y such that its restriction f : X f (X)is a homeomorphism. This means f maps X homeomorphically onto its image. Wesay that X can be embedded in Y .

    2.2.8. E. The Euclidean line R can be embedded in the Euclidean plane R2.

    2.2.9. E. Suppose that f : R R2 is continuous under the Euclidean topol-ogy. Then R can be embedded onto the graph of f .

    Topologies generated by maps.

    2.2.10. Let (X, ) be a topological space and Y be a set. Let f : X Y be a map.Find the coarsest and finest topologies on Y such that f is continuous.

    2.2.11. Let X be a set and (Y, ) be a topological space. Let f : X Y be a map.Find the coarsest and finest topologies on X such that f is continuous.

    2.2.12. Let X be a set and (Y, ) be a topological space. Let fi : X Y, i I be acollection of maps. Find the coarsest topology on X such that all maps fi, i I arecontinuous.

    Note: This problem involves a useful construction. For example in Functional Analysis, there

    is the important notion of weak topology on a normed space. It is the coarsest topology such that

    all linear continuous (under the norm) functionals on the space are continuous. See for example

    [Con90].

    More problems.

    2.2.13. If f : X Y and g : Y Z are continuous then g f is continuous.2.2.14. Suppose that f : X Y and B is a basis for the topology of Y . Show thatf is continuous if and only if the inverse image of any element of B is an open setin X.

    2.2.15. A continuous bijection is a homeomorphism if and only if it is an openmap.

    2.2.16. Any two closed intervals in R are homeomorphic. Also (a, b) R.

    2.2.17. S n \ {(0, 0, . . . , 0, 1)} Rn via the stereographic projection.2.2.18. Any two balls in the Euclidean Rn are homeomorphic.

    2.2.19. The unit ball B(0, 1) in the Euclidean Rn is homeomorphic to Rn.Hint: Consider the map x 7 1

    1 ||x|| x.

    2.2.20. (a) A square and a circle are homeomorphic.(b) The region bounded by a square and the region bounded a the circle are

    homeomorphic.

    2.2.21. If f : X Y is a homeomorphism and Z X then X \ Z and Y \ f (Z) arehomeomorphic.

    2.2.22. Let X = A B where A and B are both open or are both closed in X.Suppose f : X Y , and f |A and f |B are both continuous. Then f is continuous.

  • 2.2. CONTINUITY 17

    2.2.23. On the Euclidean plane R2, show that:(a) R2 \ {0, 0} and R2 \ {1, 1} are homeomorphic.(b) R2 \ {{0, 0}, {1, 1}} and R2 \ {{1, 0}, {0, 1}} are homeomorphic.

  • 18 2. TOPOLOGICAL SPACES

    2.3. Subspaces

    Relative Topology. Let (X, ) be a topological space and A X. The relativetopology, or the subspace topology of A is {A O/O }. With this topology wesay that A is a subspace of X.

    Thus a subset of a subspace A X is open in A if and only if it is a restrictionof an open set in X to A.

    2.3.1. A subset of a subspace A X is closed in A if and only if it is a restrictionof a closed set in X to A.

    2.3.2. Suppose that X is a topological space and Z Y X. Then the relativetopology on Z with respect to Y is the same as the topology on Z with respect to X.

    2.3.3. R. Let Y be a subspace of a topological space X. It is not true that if aset is open or closed in Y then it is open or closed in X.

    When we mention that a set is open, we must know which topological space weare talking about.

    Interiors Closures Boundaries. Let X be a topological space and A X.The union

    A of all open sets of X contained in A is called the interior of A in X. It

    is the largest open set of X contained in A.A point is in the interior of A if and only if A is a neighborhood of this point in

    X. Such a point is called an interior point of A in X.

    2.3.4. A set is open if and only if it is equal to its interior. In other words, all of itspoints are interior.

    The intersection A of all closed set of X containing A is called the closure of Ain X. It is the smallest closed set of X containing A.

    A point in x X is said to be a contact point (or point of closure) of the subsetA of X if any neighborhood of x contains a point of A.

    2.3.5. A set is closed if and only if it contains all its contact points.

    A point in x X is said to be a limit point (or cluster point or acummulationpoint) of the subset A of X if any neighborhood of x contains a point of A otherthan x.

    2.3.6. A set is closed if and only if it contains all its limit points.

    2.3.7. The closure of a set is the union of the set and its set of limit points.

    If A X then define the boundary of A to be A = A X \ A. If x A we saythat it is a boundary point of A.

    2.3.8. A point is in the boundary of A if and only if every of its neighborhoods hasnon-empty intersections with A and the complement of A.

    More problems.

    2.3.9. Show that A is the disjoint union ofA and A.

    2.3.10. Show that X is the disjoint union ofA, A, and X \ A.

  • 2.3. SUBSPACES 19

    2.3.11. Verify the following properties.

    (a) X\ A= X \ A.(b) X \ A = X \ A.(c) If A B then A B.

    2.3.12. The set {x Q/ 2 x 2} is both closed and open in Q R.2.3.13. The map : [0, 1) S 1 R2 given by t 7 e2piit is a bijection but is nota homeomorphism.

    Hint: Compare the subinterval [1/2, 1) and its image via .

    2.3.14. Consider Q R. Then Q= and Q = R.2.3.15. Find the closures, interiors and the boundaries of the interval [0, 1) underthe Euclidean, discrete and trivial topologies of R.

    2.3.16. Find the closures, interiors and the boundaries of N R.2.3.17. In a metric space X, a point x X is a limit point of the subset A X ifand only if there is a sequence in A \ {x} converging to x.

    Note: This is not true in general topological spaces.

    2.3.18. In Rn with the Euclidean topology, the boundary of the ball B(x, r) is thesphere {y/ d(x, y) = r}. The closed ball B(x, r) = {y/ d(x, y) r} is the closure ofB(x, r).

    2.3.19. In a metric space, is the boundary of the ball B(x, r) the sphere {y/ d(x, y) =r}? Is the closed ball B(x, r) = {y/ d(x, y) r} the closure of B(x, r)?

    Hint: Consider a metric space consisting of two points.

    2.3.20. Suppose that A Y X. Then AY = AX Y . Furthermore if Y is closedin X then A

    Y= A

    X.

    2.3.21. Let On = {k Z+/k n}. Then {} {On/n Z+} is a topology on Z+.Find the closure of the set {5}. Find the closure of the set of even positive integers.2.3.22. Consider R with the finite complement topology. Find the closures, interi-ors and the boundaries of the subsets {1, 2} and N.2.3.23. Which ones of the following equalities are correct?

    (a)

    A B= A B.(b)

    A B= A B.

    (c) A B = A B.(d) A B = A B.

    2.3.24. Show that these spaces are not homeomorphic to each other: Z, Q, R withEuclidean topology, and R with the finite complement topology.

  • CHAPTER 3

    Connectivity

    3.1. Connected Space

    Connected space. A topological space is said to be connected if is not a unionof two non-empty disjoint open subsets.

    3.1.1. P. A topological space X is connected if and only if its only subsetswhich are both closed and open are the empty set and X.

    When we say that a subset of a topological space is connected we implicitlymean that the subset under the subspace topology is a connected space.

    3.1.2. In a topological space a set containning one point is connected.

    3.1.3. P. Let X be a topological space and A and B are two connectedsubsets. If A B , then A B is connected.

    Thus the union of two non-disjoint connected subsets is a connected set.

    P. Suppose that C is subset of AB that is both open and closed. Supposethat C , . Then either CA , or CB , . Without loss of generality, assumethat C A , .

    Note that CA is both open and closed in A. Since A is connected, CA = A.Then C B , , hence C B = B, so C = A B.

    The same proof gives a more general result:

    3.1.4. P. Let X be a topological space and let Ai, i I be connectedsubsets. If

    iI Ai , then i Ai is connected.

    3.1.5. T (Continuous image of connected set is connnected). If f :X Y is continuous and X is connected then f (X) is connected.3.1.6. If two spaces are homeomorphic and one space is connected then the otherspace is also connected.

    Connected Component.

    3.1.7. T. Let X be a topological space. Define a relation on X whereastwo points are related if both belong to a connected subset of X. Then this is anequivalence relation.

    3.1.8. P. An equivalence class under the above equivalence relation isconnected.

    P. Consider the equivalence class [p] represented by a point p. By thedefinition, q [p] if and only if there is a connected set containing both p and q.Thus [p] =

    q[p] Oq where Oq is a connected set containing both p and q. By

    3.1.4, [p] is connected.

    21

  • 22 3. CONNECTIVITY

    3.1.9. D. Under the above equivalence relation, the equivalence classesare called the connected components of the space.

    Thus a space is a disjoint union of its connected components.

    3.1.10. A connected component is a maximal connected subset under the set in-clusion.

    3.1.11. P. The closure of a connected set is a connected set.

    P. Suppose that A is connected. Let B A be both open and closed in Aand is non-empty.

    If B does not contain a limit point of A then B A, therefore B = A.If B contains a limit point of A then B A , . Then B A = A, so B A,

    therefore B A.

    3.1.12. The proof above actually showed that if A is connected and A B Athen B is connected.

    3.1.13. A connected component must be closed.

    Connected sets in the Euclidean real number line.

    3.1.14. P. A connected set in R under the Euclidean topology must bean interval.

    P. Suppose that A X is connected. Suppose that x, y A and x < y. Ifx < z < y we must have z A, otherwise the set {a A/a < z} = {a A/a z}will be both closed and open in A.

    3.1.15. T. The interval (0, 1) as a subset of the Euclidean real number lineis connected.

    A proof of this fact must deal with some fundamental properties of real num-bers.

    P. First note that a set is open in (0, 1) if and only if it is open in R.Let C (0, 1) be both open and closed in (0, 1). Suppose that C , , and

    C , (0, 1). Then there is an x (0, 1) \C. We can assume that there is a number inC which is smaller than x, the other case is similar.

    The set D = C \ (x, 1) is the same as the set C \ [x, 1), which means D is bothopen and closed in (0, 1).

    If D , we can let s = sup D (0, 1). Then s x < 1. Since D is closed,s D. Since D is open s must belong to an open interval contained in D. But thenthere are points in D which are bigger than s.

    If D = we let E = C \ (0, x) and consider t = inf E. 3.1.16. T. An interval in the Euclidean real number line is connected.

    Therefore a subset of the Euclidean real number line is connected if and onlyif it is an interval.

    P. By homeomorphisms we just need to considers the intervals (0, 1),(0, 1], and [0, 1]. Note that [0, 1] is the closure of (0, 1), and (0, 1] = (0, 3/4) [1/2, 1].

    Or we can modify the proof of 3.1.15.

  • 3.1. CONNECTED SPACE 23

    More problems.

    3.1.17. S 1 is connected in the Euclidean topology of the plane.Hint: The circle is a continuous image of an interval.

    3.1.18 (Intermediate Value Theorem). If f : R R is continuous under theEuclidean topology then the image of an interval is an interval.

    3.1.19. If f : R R is continuous then its graph is connected in the Euclideanplane.

    3.1.20. Let X be a topological space and A Y X. If A is connected in Y then Ais connected in X, and vice versa.

    3.1.21. Let X be a topological space and let Ai, i I be connected subsets. IfAi A j , for all i , j then i Ai is connected.3.1.22. Let X be a topological space and let Ai, i Z+ be connected subsets. IfAi Ai+1 , for all i 1 then i=1 Ai is connected.

    Hint: Note that the conclusion is stronger than thatn

    i=1 Ai is connected for all n Z+.

    3.1.23. Is an intersection of connected sets connected?

    3.1.24. Let X be a topological space. A map f : X Y is called a discrete map ifY has the discrete topology and f is continuous.

    Show that X is connected if and only if all discrete maps on X are constant.Use this criterion to prove some of the results in this section.

    3.1.25. What are the connected components of N R under the Euclidean topol-ogy?

    3.1.26. What are the connected components of Q R under the Euclidean topol-ogy?

    Hint: Suppose that C is a connected component of Q. If C contains two different points a and

    b, then there is an irrational number c between a and b, and (, c) C is both open and closed inC.

    3.1.27. * What are the connected components of Q2 R2 under the Euclideantopology?

    3.1.28. If a space has finitely many components then each component is both openand closed.

    3.1.29. Let X = {0} {1n/n Z+} R where R has the Euclidean topology. Find

    the open connected components of X.

    3.1.30. If X and Y are homeomorphic then they have the same number of con-nected components, i.e. there is a bijection between the sets of connected compo-nents of the two sets.

    3.1.31. The Euclidean space Rn is connected.3.1.32. No interval on the Euclidean real number line is homeomorphic to S 1.

    3.1.33. The Euclidean spaces R and R2 are not homeomorphic.Hint: Delete a point from R. Use 2.2.21 and 3.1.30.

  • 24 3. CONNECTIVITY

    That R2 and R3 are not homeomorphic is not easy. It is a consequence of thefollowing difficult theorem:

    3.1.34. T (Invariance of Domain). If two subsets of the Euclidean Rn arehomeomorphic and one set is open then the other is also open.

    As a consequence, Rm and Rn are not homeomorphic if m , n.

    This result allows us to talk about topological dimension.

    3.1.35. Show that R with the finite complement topology and R2 with the finitecomplement topology are homeomorphic.

  • 3.2. PATH-CONNECTED SPACES 25

    3.2. Path-connected Spaces

    Let X be a topological space and a, b X. A path in X from a to b is acontinuous map f : [0, 1] X such that f (0) = a and f (1) = b.

    The space X is said to be path-connected if for any two different points a andb in X there is a path in X from a to b.

    3.2.1. The Euclidean Rn is path-connected.

    3.2.2. A ball in the Euclidean Rn is path-connected.

    Let f be a path from a to b. Then the inverse path of f is defined to be the pathf 1(t) = f (1 t) from b to a.

    Let f be a path from a to b, and g be a path from b to c, then the compositionof f with g is the path

    h(t) =

    f (2t), 0 t 12

    g(2t 1), 12 t 1.

    The path h is often denoted as f g.3.2.3. Show that f g is continuous.

    Hint: If f (t) = f (2t) and g(t) = g(2t 1) then h1(U) = f 1(U) g1(U).

    3.2.4. Let X be a topological space. Define a relation on X whereas a point xis related to a point y there is a path in from x to y. Then this is an equivalencerelation.

    3.2.5. An equivalence class under the above equivalence relation is path-connected.

    Each equivalence class is called a path-connected component.

    3.2.6. Let X be a topological space and A and B are two path-connected subsets.If A B , then A B is path-connected.3.2.7. A path-connected component is a maximal path-connected subset under theset inclusions.

    3.2.8. T. A path-connected space is connected.

    P. The proof is based on the fact that the image of a path is a connectedset.

    Let X be path-connected. Let a, b X. There is a path from a to b. The imageof this path is a connected subset of X. That means any two points of X are in thesame connected component of X. Therefore X has only one connected component.

    Generally, the reverse statement of 3.2.8 is not correct. However we have:

    3.2.9. T. An open, connected subset of the Euclidean space Rn is path-connected.

    P. Let A be open, connected in Rn. Let B be a path-connected componentof A. Then B must be both open and closed in A. Indeed, if x B then there is aball B(x, r) A. Then B B(x, r) is still path-connected, so B B(x, r), thus B isopen. Similarly we can show that A \ B is open. 3.2.10. If f : X Y is continuous and X is path-connected then f (X) is path-connected.

  • 26 3. CONNECTIVITY

    Locally connected and locally path-connected spaces. A topological spaceis said to be locally connected if every neighborhood of a point contains a con-nected neighborhood of that point.

    A topological space is said to be locally path-connected if every neighborhoodof a point contains a path-connected neighborhood of that point.

    3.2.11. E. All open sets in the Euclidean space Rn are locally connected andlocally path-connected.

    Generalize 3.2.9 we get:

    3.2.12. T. A connected, locally path-connected space is path-connected.

    P. Suppose that X is connected and locally path-connected. Let C be apath-connected component of X. If x X has a path-connected neighborhood Usuch that U C , , then U C is path-connected, and so U C. This means thatC is both open and closed in X, hence C = X.

    Topologists Sine Curve. The closure of the graph of the curve y = sin 1x , x > 0is often called the Topologists Sine Curve.

    -1

    -0.8

    -0.6

    -0.4

    -0.2

    0

    0.2

    0.4

    0.6

    0.8

    1

    0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

    sin(1/x)

    F 3.1. Topologists Sine Curve.

    Denote A = {(x, sin 1x )/0 < x 1} and B = {0} [1, 1]. Then A B = andthe Topologists Sine Curve is X = A B.3.2.13. P. The Topologists Sine Curve is connected.

    P. By 3.1.19 the set A is connected. And A X A. 3.2.14. P. The Topologists Sine Curve is not path-connected.

    P. Suppose that there is a path (t) = (x(t), y(t)), t [0, 1] from the origin(0, 0) on B to the point (1, sin 1) on A, we show that there is a contradiction.

    Let t0 = sup{t [0, 1]/x(t) = 0}. Then x(t0) = 0, t0 < 1, and for all t > t0we have x(t) > 0. Thus t0 is the moment when the path departs from B. Wecan see that the path jumps immediately when it departs from B. Thus we willshow that (t) cannot be continuous at t0, by showing that for any > 0 there are

  • 3.2. PATH-CONNECTED SPACES 27

    t1, t2 (t0, t0 +) such that y(t1) = 1 and y(t2) = 1, therefore y(t) is not continuousat t0.

    To find t1, note that the set {x(t)/t0 t t0 + } is an interval [0, x0] wherex0 > 0. There exists an x1 (0, x0) such that sin 1x1 = 1: we just need to takex1 = 1pi

    2 +k2piwith sufficiently large k. There is t1 (t0, t0 + ) such that x(t1) = x1.

    Then y(t1) = sin 1x(t1) = 1.

    More problems.

    3.2.15. R2 \ {one point} is path-connected under the Euclidean topology.3.2.16. R2 \ A where A is a finite set is path-connected under the Euclidean topol-ogy.

    Hint: Bound A in a rectangle. For any point in R2 \A there is a straight line through it that doesnot intersect A, by an argument using cardinalities of sets.

    3.2.17. * R2 \ Q2 is path-connected under the Euclidean topology. Indeed if A iscountable then R2 \ A is path-connected.

    Hint: Let a R2 \ A. Let `a be a line passing through a and does not intersect A. If b R2 \ Alet `b be a line passing through b and does not intersect A, but does intersect `a.

    3.2.18. The sphere S n, n 1 is connected under the Euclidean topology.Hint: Either show that S n is path-connected, or show that it is a union of connected parts which

    are not mutually disjoint.

    3.2.19. The Topologists Sine Curve is not locally path-connected.

    3.2.20. The Topologists Sine Curve is not locally connected.

    3.2.21. * (a) Classify the alphabetical characters up to homeormophisms, that is,which of the following characters are homeomorphic to each other as subsets ofthe Euclidean plane?

    A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

    Hint: Use 2.2.22 to modify a letter part by part.

    (b) Note that the result depends on the font you use! Indeed, what if instead ofa sans-serif font as above you use a serif one?

    A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

  • 28 3. CONNECTIVITY

    Further Reading

    Jordan Curve Theorem. The following is an important and deep result of planetopology.

    3.2.22. T (Jordan Curve Theorem). A simple, continuous, closed curveseparates the plane into two disconnected regions.

    More concisely, if C is a subset of the Euclidean plane homeomorphic to thecircle then R2 \C has two connected components.Space Filling Curves. A rather curious and surprising result is:

    3.2.23. T. There is a continuous curve filling a rectangle on the plane.More concisely, there is a continuous map from the interval [0, 1] onto the

    square [0, 1]2.

    Note that this map cannot be injective, in other words the curve cannot besimple.

    Such a curve is called a Peano curve. It could be constructed as a limit of aniteration of piecewise linear curves.

  • CHAPTER 4

    Separation Axioms

    4.1. Separation Axioms

    4.1.1. D. The following are called separation axioms.

    T0: A topological space is called a T0-space if for any two different pointsthere is an open set containing one of them but not the other.

    T1: A topological space is called a T1-space if for any two points x , ythere is an open set containing x but not y and an open set containing ybut not x.

    T2: A topological space is called a T2-space or Hausdorff if for any twopoints x , y there are disjoint open sets U and V such that u U andv V.

    T3: A T1-space is called a T3-space or regular (chnh tac) if for any pointx and a closed set F not containing x there are disjoint open sets U andV such that x U and F V.

    T4: A T1-space is called a T4-space or normal if for any two disjoint closedsets F and G there are disjoint open sets U and V such that F U andG V.

    These are called separation axioms because they involve separating certainsets from one another by open sets.

    4.1.2. P. A space is T1 if and only if a set containing exactly one pointis a closed set.

    4.1.3 (T4 T3 T2 T1 T0). If a space is Ti then it is Ti1, for 1 i 4.4.1.4. Any space with the discrete topology is normal.

    4.1.5. Give an example of a space which is T0 but not T1.

    4.1.6. The real number line under the finite complement topology is T1 but is notT2.

    4.1.7. A metric space is regular.

    4.1.8. P. A metric space is normal.

    P. We introduce the notion of distance between two sets in a metric spaceX. If A and B are two subsets of X then we define the distance between A andB as d(A, B) = inf{d(x, y)/x A, y B}. In particular if x X then d(x, A) =inf{d(x, y)/y A}. Using the triangle inequality we can check that d(x, A) is acontinuous function with respect to x.

    Now suppose that A and B are disjoint closed sets. Let U = {x/d(x, A) d(x, B)}. Then A U, B V , U V = , and bothU and V are open.

    4.1.9. A subspace of a Hausdorff space is Hausdorff.

    29

  • 30 4. SEPARATION AXIOMS

    4.1.10. P. A T1-space X is regular if and only if given a point x and anopen set U containing x there is an open set V such that x V V U.

    P. Suppose that X is regular. Since X \ U is closed and disjoint from Cthere is an open set V containing x and an open set W containing X \ U such thatV and W are disjoint. Then V (X \W), so V (X \W) U.

    Now suppose that X is T1 and the condition is satisfied. Given a point x and aclosed set C disjoint from x. Let U = X \C. Then there is an open set V containingx such that V V U. Then V and X \ V separate x and C. 4.1.11. A T1-space X is normal if and only if given a closed set C and an open setU containing C there is an open set V such that C V V U.More problems.

    4.1.12. If a finite set is a T1-space then the topology is the discrete topology.

    4.1.13. * There are examples of a T2-space which is not T3, and a T3-space whichis not T4, but they are rather difficult, see 9.1.13 and [Mun00, p. 197].

    4.1.14. * Let X be normal, f : X Y is surjective, continuous, and closed. Provethat Y is normal.

  • CHAPTER 5

    Nets

    5.1. Nets

    In metric spaces we can study continuity of functions via convergence of se-quences. In general topological spaces, we need to use a notion more general thansequences, called nets. Roughly speaking in general topological spaces, sequences(countable indices) might not be enough to describe the neigborhood systems at apoint, we need things of arbitrary infinite indices.

    A directed set is a (partially) ordered set I such that

    i, j I,k I, k i k j.A net on a topological space X is a map n : I X. Writing xi = n(i), we often

    denote n as (xi)iI .

    5.1.1. E. Nets on I = N with the usual order are exactly sequences.

    5.1.2. E. Let X be a topological space and x X. Let I be the family ofopen neighborhoods of x. Define an order on I by U V U V . Then Ibecomes a directed set.

    It will be clear in a moment that this is the most important example of directedsets concerning nets.

    Convergence. A net (xi) is said to be convergent to x X if neighborhood U of x,i I, j i x j U.

    The point x is called a limit of the net (xi), and we often write xi x.5.1.3. E. Convergence of nets on I = N is exactly convergence of sequences.

    5.1.4. E. Let X = {x1, x2, x3}with topology {, X, {x1, x3}, {x2, x3}, {x3}}. Thenet (x3) converges to x1, x2, and x3. The net (x1, x2) converges to x2.

    5.1.5. P. A point x X is a limit point for a set A X if and only if thereis a net in A \ {x} convergent to x.

    This proposition allows us to describe topologies in terms of convergences.With it many statements about convergence in metric spaces could be carried totopological spaces by simply replacing sequences by nets.

    P. () Suppose that x is a limit point of A. Consider the directed setI consisting of all the open neighborhoods of x with the partial order U V ifU V .

    For any open neighborhood U of x there is an element xU U A, xU , x.Consider the net (xU)UI . It is a net in A \ {x} convergent to x. Indeed, given anopen neighborhood U of x, for all V U, xV V U.

    () Suppose that there is a net (xi)iI in A \ {x} convergent to x. Let U be anopen neighborhood of x. There is an i I such that for j i we have x j U, inparticular xi U (A \ {x}).

    31

  • 32 5. NETS

    5.1.6. R (When can nets be replaced by sequences?). By examiningthe above proof we can see that the term net can be replaced by the term sequenceif there is a countable collection F of neighborhood of x such that any neighbor-hood of x contains a member of F. In other words, the point x has a countableneighborhood basis. A space having this property at every point is said to be a firstcountable space. A metric space is such a space.

    Similar to the case of metric spaces we have:

    5.1.7. T. Let X and Y be topological spaces. Then f : X Y is continuousif and only if whenever a net n in X is convergent to x then the net f n is convergentto f (x).

    In more familiar notations, f is continuous if and only if for all nets (xi) andall points x in X, xi x f (xi) f (x).

    The proof is simply a repeat of the proof for metric spaces.

    P. () Suppose that f is continuous. Let U is a neighborhood of f (x).Then f 1(U) is a neighborhood of x in X. Since (xi) is convergent to x, there is ani I such that for all j i we have x j f 1(U), which implies f (x j) U.

    () We will show that if U is open in Y then f 1(U) is open in X. Indeed,suppose that let x f 1(U) but is not an interior point, so it is a limit point ofX \ f 1(U). There is a net (xi) in X \ f 1(U) convergent to x. Since f is continuous,f (xi) Y \ U is convergent to f (x) U. That contradicts the assumption that U isopen.

    5.1.8. A map f is continuous at x if and only if for all nets (xi), xi x f (xi)f (x).

    5.1.9. P. If a space is Hausdorff then a net has at most one limit.

    P. Suppose that a net (xi) is convergent to two different points x and y.Since the space is Hausdorff, there are disjoint open neighborhoods U and V of xand y. There is i I such that for i we have x U, and there is j I suchthat for j we have x U. Since there is a I such that i and j, thepoint x will be in U V , a contradiction. 5.1.10. The converse statement of 5.1.9 is also true. A space is Hausdorff if andonly if a net has at most one limit.

    Hint: Suppose that there are two points x and y that could not be separated by open sets.

    Consider the directed set whose elements are pairs (Ux,Vy) of open neighborhoods of x and y, under

    set inclusion. Take a net n such that n(Ux,Vy) is a point in Ux Vy. Then this net converges to bothx and y.

    More problems.

    5.1.11. Consider the Euclidean line R. Let I = {(0, a) R} with the order (0, a) (0, b) if a b. Check that I is a directed set. Define a net n : I R withn((0, a)) = a/2. Is n convergent?

    5.1.12. Suppose that 1 and 2 are two topologies on X. Furthermore suppose thatfor all nets xi and all points x, xi

    1 x xi 2 x. Show that 1 2.In other words, if convergence in 1 implies convergence in 2 then 1 is finer

    than 2.

  • 5.1. NETS 33

    5.1.13. Let X be a topological space, R have the Euclidean topology and f : X R be continuous. Suppose that A X and f (x) = 0 on A. Show that f (x) = 0 onA, by:

    (a) using nets.(b) not using nets.

  • CHAPTER 6

    Compact Spaces

    6.1. Compact Spaces

    A cover of a set X is a collection of subsets of X whose union is X. A coveris said to be an open cover if each member of the cover is an open subset of X. Asubset of a cover which is itself still a cover is called a subcover.

    A space is compact if every open cover has a finite subcover.

    6.1.1. E. Any finite topological space is compact.

    Let A be a subset of a topological space X. We say that A is a compact set if Ais a compact space under the topology restricted from X.

    6.1.2. R. Let I be an open cover of A. Each O I is an open set of A, so itis the restriction of an open set UO of X. Thus we have a collection of open sets{UO/O I} whose union contains A. On the other hand if we have a collection Iof open sets of X whose union contains A then the collection {U O/O I} is anopen cover of A.

    6.1.3. Any subset of R with the finite complement topology is compact.Hint: Any open set covers a subset of R except finitely many points.

    6.1.4. T. In a Hausdorff space compact sets are closed.

    P. Let A be a compact set in a Hausdorff space X. We show that X \ A isopen.

    Let x X\A. For each a A there are disjoint open sets Ua containing x and Vacontaining a. The family {Va/a A} covers A, so there is a finite subcover {Vai/1 i n}. Let U = ni=1 Uai and V = ni=1 Vai . Then U is an open neighborhood of xdisjoint from V , a neighborhood of A.

    6.1.5. T (Continuous image of compact set is compact). If X is com-pact and f : X Y is continuous then f (X) is compact.6.1.6. P. If X is compact and A X is closed then A is compact.

    P. Add X \ A to an open cover of A. 6.1.7. R. When A is a compact subset of X and X is a subspace of Y thenA is also a compact subset of Y , because the topology of X restricted to A and thetopology of Y restricted to A are the same.

    Compactness is an absolute property, not depending on outer spaces.

    The following proposition is quite useful later:

    6.1.8. P. If X is compact, Y is Hausdorff, f : X Y is bijective andcontinuous, then f is a homeomorphism.

    6.1.9. P. In a compact space an infinite set has a limit point.

    35

  • 36 6. COMPACT SPACES

    P. Let A be an infinite set in a compact space X. Suppose that A has nolimit point. Let x X, then there is an open neighborhood Ux of x that contains atmost one point of A. The family of such Ux cover A, so there is a finite subcover.But that implies that A is finite.

    Characterization of compact sets in terms of closed sets. A collection A ofsubsets of a set is said to have the finite intersection property if the intersection ofevery finite subcollection of A is non-empty.

    6.1.10. T. A space is compact if and only if any family of closed subsetswith the finite intersection property has non-empty intersection.

    Compact sets in metric spaces. Most of the following results have been stud-ied in a course in Real Analysis or Functional Analysis. It is very useful to have alook back at notes for those courses.

    6.1.11. In a metric space, if a set is compact then it is closed and bounded.

    6.1.12. If X is a compact space and f : X (R,Euclidean) is continuous then fhas a maximum value and a minimum value.

    6.1.13. A metric space is compact if and only if every sequence has a convergentsubsequence.

    6.1.14. In the Euclidean space Rn a subset is compact if and only if it is closed andbounded.

    More problems.

    6.1.15. Find a cover of the interval (0, 1) with no finite subcover.

    6.1.16. A discrete compact topological space is finite.

    6.1.17 (An extension of Cantor Lemma in Calculus). Let X be compact andX A1 A2 An . . . be a ascending sequence of closed, non-empty sets.Then

    n=1 An , .

    6.1.18. A compact Hausdorff space is normal.Hint: Use 6.1.20.

    The proof of 6.1.4 contains the following:

    6.1.19. In a Hausdorff space a point and a compact set disjoint from it can beseparated by open sets.

    More generally:

    6.1.20. In a Hausdorff space two disjoint compact sets can be separated by opensets.

    Hint: We already have that for each x in the first set there are an open set Ux containing x and

    an open set Vx containing the second set and disjoint from Ux. Since the first set is compact, it is

    covered byn

    i=1 Uxi . Considern

    i=1 Vxi .

    6.1.21. In a regular space a compact set and a disjoint closed set can be separatedby open sets.

    6.1.22. In a topological space a finite unions of compact subsets is compact.

  • 6.1. COMPACT SPACES 37

    6.1.23. In a Hausdorff space an intersection of compact subsets is compact.

    6.1.24. The set of nn-matrix with real coefficients, denoted by M(n;R), could benaturally considered as a subset of the Euclidean space Rn

    2by considering entries

    of a matrix as coordinates, via the map

    (ai, j) 7 (a1,1, a2,1, . . . , an,1, a1,2, a2,2, . . . , an,2, a1,3, . . . , an1,n, an,n).The Orthogonal Group O(n) is defined to be the group of matrices representing

    orthogonal linear maps ofRn, that are linear maps that preserve inner product. Thus

    O(n) = {A M(n;R)/ A AT = In}.The Special Orthogonal Group SO(n) is the subgroup of O(n) consisting of all

    orthogonal matrices with determinant 1.(a) Show that any element of SO(2) is of the formcos() sin()sin() cos()

    .This is a rotation in the plane around the origin with an angle . Thus SO(2) is thegroup of rotations on the plane around the origin.

    (b) Show that SO(2) is path-connected.(c) How many connected components does O(2) have?(d) Show that SO(n) is compact.(e) Show that any unit vector in Rn could be rotated to the unit vector e1 =

    (1, 0, 0, . . . , 0) using an orthogonal transformation.(f) Show that any matrix R SO(n) is path-connected in SO(n) to a matrix of

    the form 1 R1 ,

    where R1 SO(n 1).(g) Show that S O(n) is connected.The General Linear Group GL(n;R) is the group of all invertible nn-matrices

    with real coefficients.(h) Show that GL(n;R) is not compact.(k) Find the number of connected components of GL(n;R).

  • CHAPTER 7

    Product Spaces

    7.1. Product Spaces

    Finite products of spaces. Let X and Y be two topological space, and considerthe Cartesian product X Y . The product topology on X Y is the topology gener-ated by the collection F of sets of the form U V where U is an open set in X andV is an open set in Y .

    The collection F is a subbasis for the product topology. Actually, since theintersection of two members of F is also a member of F, the collection F is a basisfor the product topology.

    Thus every open set in the product topology is a union of products of open setsof X with open sets of Y .

    7.1.1. Note that, as sets:(a) (A B) (C D) = (A C) (B D).(b) (AB) (CD) (AC) (BD) = (AB) (AD) (CB) (CD).

    7.1.2. If 1 is a basis for X and 2 is a basis for Y then 1 2 is a basis for theproduct topology on X Y .

    Similarly the product topology onn

    i=1(Xi, i) is defined to be the topologygenerated by

    ni=1 i.

    7.1.3. If each i is a basis for Xi thenn

    i=1 i is a basis for the product topology onni=1(Xi, i).

    7.1.4. E (Euclidean topology). Recall that Rn = R R R n times

    . The

    Euclidean topology on R is generated by open intervals. An open set in the producttopology of Rn is a union of products of open intervals.

    Since a product of open intervals is an open rectangle, and an open rectangle isa union of open balls and vice versa, the product topology is exactly the Euclideantopology.

    Infinite products. Let {(Xi, i), i I} be a family of topological spaces. Theproduct topology on

    iI Xi is the topology generated by the collection F consist-

    ing of sets of the form

    iI Ui, where Ui i and Ui = Xi for all except finitelymany i I.7.1.5. Check that F is really a basis for a topology.

    For j I the projection p j : iI Xi X j is defined by p j((xi)) = x j. It is theprojection to the j coordinate.

    7.1.6. P (Product topology is the topology such that projectionsare continuous). (a) If

    iI Xi has the product topology then pi is continuous for

    all i I.39

  • 40 7. PRODUCT SPACES

    (b) The product topology is the coarsest topology on

    iI Xi such that all themaps pi are continuous. This is the topology generated by the all the maps pi, i I,as in 2.2.12.

    P. (a) Note that if O j X j then p1j (O j) =

    iI Ui with Ui = Xi for all iexcept j, and U j = O j.

    (b) The topology generated by all the maps pi is the topology generated by setsof the form p1i (Oi) with Oi i. A finite intersections of these sets is exactly amember of the basis of the product topology as in the definition.

    7.1.7. Show that each projection map pi is a an open map, mapping an open set toan open set.

    Hint: Only need to show that the projection of an element of the basis is open.

    7.1.8 (Map to product space is continuous if and only if each componentmap is continuous). Show that f : Y iI Xi is continuous if and only iffi = pi f is continuous for every i I.7.1.9. P (Convergence in product topology is coordinate-wise con-vergence). A net n : J iI Xi is convergent if and only if all of its projectionspi n are convergent.

    P. () Suppose that each pi n is convergent to ai, we show that n isconvergent to a = (ai)iI .

    A neighborhood of a contains an open set of the form U =

    iI Oi with Oi areopen sets of Xi and Oi = Xi except for i K, where K is a finite subset of I.

    For each i K, pi n is convergent to ai, therefore there exists an index ji Jsuch that for j ji we have pi(n( j)) Oi. Take an index j0 such that j0 ji forall i K. Then for j j0 we have n( j) U. More problems.

    7.1.10. Check that in topological sense R3 = R R R = R2 R.Hint: Look at their bases.

    7.1.11. If X is homeomorphic to X1 and Y is homeomorphic to Y1 then X Y ishomeomorphic to X1 Y1.7.1.12. Give an example to show that in general the projection pi is not a closedmap.

    7.1.13. Fix a point x = (xi) iI Xi. Define the inclusion map f : Xi iI Xiby

    y 7 ( f (y)) j =x j if j , iy if j = i .

    The inclusion f is a homeomorphism to its image (an embedding of Xi).The image of the inclusion map is Xi =

    jI A j where A j = {x j} if j , i and

    Ai = Xi. Thus Xi is a parallel copy of Xi in

    iI Xi.The spaces Xi have x as common point.Hint: Use 7.1.8 to prove that the inclusion map is continuous.

    7.1.14. (a) If each Xi, i I is Hausdorff then iI Xi is Hausdorff.(b) If

    iI Xi is Hausdorff then each Xi is Hausdorff.

    Hint: (b) Use 7.1.13.

  • 7.1. PRODUCT SPACES 41

    7.1.15. (a) If

    iI Xi is path-connected then each Xi is path-connected.(b) If Xi, i I is path-connected then iI Xi is path-connected.Hint: (b) Let (xi) and (yi) be in

    iI Xi. Let i(t) be a continuous path from xi to yi. Let

    (t) = (i(t)). Use 7.1.8.

    7.1.16. (a) If

    iI Xi is connected then each Xi is connected.(b) If X and Y are connected then X Y is connected.(c)* If each Xi, i I is connected then iI Xi is connected.Hint: (b) Use 7.1.13. (c) Fix a point x iI Xi. Use (b) to show that if a point differ from x

    atmost finitely many coordinates then that point and x belong to a certain connected subset of

    iI Xi,

    such that the union of these connected subsets is

    iI Xi.

    7.1.17. (a) If

    iI Xi is compact then each Xi is compact.(b)* Prove that if X and Y are compact then X Y is compact, without using

    the Axiom of Choice.Hint: Use 7.1.13. It is enough to prove for the case an open cover of X Y by open sets of

    the form a product of an open set in X with an open set in Y . For each slice {x} Y there is finitesubcover {(Ux)i (Vx)i/ 1 i nx}. Take Ux = nxi=1(Ux)i. The family {Ux/ x X} covers X sothere is a subcover {Ux j/ 1 j n}. The family {(Ux j )i (Vx j )i/ 1 i nx j , 1 j n} is a finitesubcover of X Y .

    7.1.18. (a) If Oi is an open set in Xi for all i I then is iI Oi open?(b) If Fi is a closed set in Xi for all i I then is iI Fi closed?

    7.1.19 (Zariski Topology). Let F = R or F = C.A polynomial in n variables on F is a function from Fn to F which is a finite

    sum of terms of the form axm11 xm22 xmnn , where a, xi F and mi N. Let P be the

    set of all polynomials in n variables on F.If S P then define Z(S ) to be the set of common zeros of all polynomials in

    S , thus Z(S ) = {x Fn/ p S , p(x) = 0}. Such a set is called an algebraic set.(a) Show that if we define that a set in Fn is closed if it is algebraic, then this

    gives a topology on Fn, called the Zariski topology.(b) Show that the Zariski topology on F is exactly the finite complement topol-

    ogy.(c) Show that if both F and Fn have the Zariski topology then all polynomials

    on Fn are continuous.(d) Is the Zariski topology on Fn the product topology?Note: The Zariski topology is the fundamental topology in Algebraic Geometry.

  • 42 7. PRODUCT SPACES

    7.2. Tikhonov Theorem

    7.2.1. T (Tikhonov theorem). A product of compact spaces is compact.

    Applications of this theorem include the Banach-Alaoglu Theorem in Func-tional Analysis and the Stone-Cech compactification.

    Tikhonov theorem is equivalent to the Axiom of Choice, and needs a difficultproof. However in the case of finite product it can be proved more easily (7.1.17).Different techniques can be used in special cases of this theorem (7.2.3 and 7.2.4).

    A proof of Tikhonov theorem.

    P. Let Xi be compact for all i I. We will show that X = iI Xi is com-pact by showing that if a collection of closed subsets of X has the finite intersectionproperty then it has non-empty intersection.

    Let F be a collection of closed subsets of X that has the finite intersectionproperty. We will show that

    AF A , .

    Note: Have a look at the following argument, which suggests that proving the Tikhonov theoremmight not be easy. If we take the closures of the projections of the collection F to the i-coordinatethen we get a collection {pi(A), A F} of closed subsets of Xi having the finite intersection property.Since Xi is compact, this collection has non-empty intersection.

    From this it is tempting to conclude that F must have non-empty intersection itself. But that isnot true, see the figure.

    In what follows we will overcome this difficulty by enlarging the collection F to the extent thatthe intersections of the closures of the projections of the collection will give common elements ofthe collection.

    (1) We show that there is a maximal collection F of subsets of X that containsF and still has the finite intersection property. We will use Zorn Lemmafor this purpose. (This is a routine step; it is easier for the reader to carryit out instead of reading.)

    Let K be the collection of collections G of subsets of X such that Gcontains F and has the finite intersection property. On K we define anorder by the usual set inclusion.

    Now suppose that L is a totally ordered subcollection of K. Let M =AL A. We will show that M K, therefore M is an upper bound of L.

    First M contains F. We also need to show that M has the finite in-tersection property. Suppose that Mi M, 1 i n. Then Mi Ai forsome Ai L. There is an i0 such that Ai0 contains all Ai. Then Mi Ai0for all 1 i n, and since Ai0 has the finite intersection property, wehave

    ni=1 Mi , .

    (2) Since F is maximal, it is closed under finite intersections.

  • 7.2. TIKHONOV THEOREM 43

    (3) We will show that

    ,iI

    [AF

    pi(A)] AF

    A.

    To show that

    iI[

    AF pi(A)] is non-empty means to show thatAF pi(A) is non-empty for each i I. Since F has the finite intersection

    property, the collection {pi(A), A F} also has this property, and so is thecollection {pi(A), A F}. Furthermore {pi(A), A F} is a collection ofclosed subsets of the compact set Xi. So

    AF pi(A) is non-empty.

    (4) Let x iI[AF pi(A)], we will show that x A,A F. Since Ais closed we need to show that any neighborhood of x has non-emptyintersection with A. It is suffice to prove this for neighborhoods of theform

    iD p1i (Ui) where Ui is open in Xi, and D is a finite subset of I.

    Let A F. We have pi(x) = xi pi(A). For each i D, since Ui is aneighborhood of xi we have Ui pi(A) , . Therefore p1i (Ui) A , .Then {p1i (Ui)}F has the finite intersection property. Since F is maximalwe must have p1i (Ui) F. Therefore [

    iD p1i (Ui)] A , .

    More problems.

    7.2.2. Let [0, 1] have the Euclidean topology. If I is an infinite set then [0, 1]I iscalled the Hilbert cube. The Hilber cube is compact.

    7.2.3. Using the characterization of compact subsets of Euclidean spaces, provethe Tikhonov theorem for finite products of compact subsets of Euclidean spaces.

    7.2.4. Let (X, dX) and (Y, dY ) be metric spaces.(a) Prove that the product topology on X Y is given by the metric

    d((x1, y1), (x2, y2)) = max{dX(x1, x2), dY (x2, y2)}.(b) Using the characterization of compact metric spaces in terms of sequences,

    prove the Tikhonov theorem for finite products of compact metric spaces.

  • 44 7. PRODUCT SPACES

    Further Reading

    Strategy for a proof of Tikhonov theorem based on net. The proof that wewill outline here is based on further developments of the theory of nets and a char-acterization of compactness in terms of nets.

    7.2.5. D (Subnet). Suppose that I and I are directed sets, and h : I Iis a map such that

    k I,k I, (i k h(i) k).If n : I X is a net then n h is called a subnet of n.

    The notion of subnet is an extension of the notion of subsequence. If we takeni Z+ such that ni < ni+1 then (xni) is a subsequence of (xn). In this case the maph : Z+ Z+ is given by h(i) = ni. Thus a subsequence of a sequence is a subnetof that sequence.

    On the other hand there are subnets of sequences which are not subsequences.Suppose that x1, x2, x3, . . . be a sequence. Then x1, x1, x2, x2, x3, x3, . . . is a subnetof that sequence, but not a subsequence. In this case the map h is given by h(2i 1) = h(2i) = i.

    A net (xi)iI is called eventually in A X if there is j I such that i j xi A.7.2.6. D (Universal net). A net n in X is universal if for any subset A of Xeither n is eventually in A or n is eventually in X \ A.7.2.7. P. If f : X Y is continuous and n is a universal net in X thenf (n) is a universal net.

    7.2.8. P. The following statements are equivalent.(a) X is compact.(b) Every universal net in X is convergent.(c) Every net in X has a convergent subnet.

    The proof of the last two propositions above could be found in [Bre93].Then we finish the proof of Tikhonov theorem as follows.

    P T . Let X =

    iI Xi where each Xi is compact. Sup-pose that (x j) jJ is a universal net in X. By 7.1.9 the net (x j) is convergent if andonly if the projection (pi(x j)) is convergent for all i. But that is true since (pi(x j))is a universal net in the compact set Xi.

  • CHAPTER 8

    Compactifications

    8.1. Alexandroff Compactification

    A compactification of a space X is a compact space Y such that X is homeo-morphic to a dense subset of Y .

    8.1.1. E. A compactification of the Euclidean (0, 1) is the Euclidean [0, 1].

    8.1.2. E. A one-point compactification of the open Euclidean interval (0, 1)is the circle S 1.

    A space X is called locally compact if every point has a compact neighborhood.

    8.1.3. E. The Euclidean space Rn is locally compact.

    8.1.4. T (Alexandroff compactification). Let X be a locally compactHausdorff space which is not compact. Let be not in X and let X = X {}.Define a topology on X as follows: an open set in X is an open set in X or isX \ C where C is a compact set in X. Then this is the only topology on X suchthat X is compact, Hausdorff and contains X as a subspace.

    With this topology X is dense in X.The space X is called the one-point compactification or the Alexandroff com-

    pactification of X.

    P. (1) Suppose that there is a topology on X such that X is com-pact, Hausdorff and contains X as a subspace. Let O be open in X. If < O then O X. Then O = O X must be open in X. If O thenC = X \ O is a closed subset of the compact space X, therefore C iscompact in X. But C X, so C must be compact in X too. Thus theopen sets of X must be open sets of X or complements of compact setsof X.

    (2) We check that we really have a topology.Let Ci be compact sets in X. Consider

    i(X \ Ci) = X \ i Ci.

    Note that since X is Hausdorff,

    i Ci is compact.If O is open in X and C is compact in X then O (X \ C) = X \

    (C (X \ O)).If Ci, 1 i n are compact sets then ni=1 X \Ci = X \ (ni=1 Ci).Finally O (X \C) = O (X \C).With this topology X is a subspace of X.

    (3) We show that X is compact. Let {Oi, i I} be an open cover of X.Then one Oi0 will cover . Therefore the complement of Oi0 in X willbe a compact set C in X.

    Then {Oi, i I, i , i0} is an open cover of C. From this cover there isa finite cover. This finite cover together with Oi0 is a finite cover of X

    .(4) To check that X is Hausdorff, we only need to check that and x X

    can be separated by open sets.

    45

  • 46 8. COMPACTIFICATIONS

    Since X is locally compact there is a compact set C containing anopen neighborhood O of x. Then O and X \C separates x and.

    (5) Since X is not compact and X is compact, X cannot be closed in X,therefore the closure of X in X is X.

    8.1.5. P. Let X and Y be two locally compact Hausdorf spaces. If X andY are homeomorphic then X and Y are homeomorphic.

    P. A homeomorphism from X to Y induces a homeomorphism from X

    to Y. We can also use 6.1.8.

    8.1.6. R. Since a one-point compactification is unique up to homeomor-phisms we often use the word the one-point compactification.

    8.1.7. E. The one-point compactification of the Euclidean line R is S 1.The one-point compactification C {} of the complex plane C is homeomor-

    phic to S 2, is often called the extended complex plane or the Riemann sphere.

    More problems.

    8.1.8. Find the one-point compactification of the Euclidean (0, 1) (2, 3).8.1.9. What is the one-point compactification of Z with the Euclidean topology?Describe the topology of the compactification.

    8.1.10. What is the one-point compactification of { 1n/n Z+} under the Euclideantopology? Describe the topology of the compactification.

    8.1.11. What is the one-point compactification of the Euclidean open ball B(0, 1)?Find the one-point compactification of the Euclidean space Rn.

    8.1.12. What is the one-point compactification of the Euclidean annulus {(x, y) R2/1 < x2 + y2 < 2}?8.1.13. A locally compact Hausdorff space is regular.

    Hint: Suppose that a point x and a closed set C are disjoint. Since X\C is an open set containingx, it contains a compact neighborhood A of x. So A contains an open neighborhood U of x. Since X

    is Hausdorff, A is closed. Consider U and X \ A.

    8.1.14. We could have noticed that the notion of local compactness as we havedefined is not apparently a local property. For a property to be local, every neigh-borhood of any point must contain a neighborhood of that point with the givenproperty (as in the cases of connectedness and path-connectedness).

    Show that for Hausdorff spaces local compactness is indeed a local property.

    8.1.15. The Topologists Sine Curve (3.2) is a compactification of the Euclideaninterval (0, 1).

    Note: This is not a one-point compactification. There can be many ways to compactify a

    space.

    8.1.16. Define a topology on R {} such that it is a compactification of theEuclidean R.

  • 8.2. STONE-CECH COMPACTIFICATION 47

    8.2. Stone-Cech Compactification

    A space is said to be completely regular (sometimes called a T3 12 -space) if itis a T1-space and for each point x and each closed set A with x < A there is acontinuous map f from X to [0, 1] such that f (x) = 0 and f (A) = {1}.

    Thus in a completely regular space a point and a closed set disjoint from itcould be separated by continuous functions.

    8.2.1. A completely regular space is regular.

    Let X be a completely regular space. Let F be the set of all continuous mapsfrom X to [0, 1]. Let be a function from X to [0, 1]F = {[0, 1]/ f F}} definedby for x X, (x) f = f (x) for each f F.

    Since [0, 1]F is compact, and (X) [0, 1]F , we have (X) is compact.The compact space (X) is called the Stone-Cech compactification of the com-

    pletely regular space X.

    8.2.2. T. If X is completely regular then : X (X) (X) is ahomeomorphism. In other words is an embedding.

    P. (1) is injective: If (x) = (y) then (x) f = (y) f for allf F, so f (x) = f (y) for all f F. Since X is completely regular wemust have x = y.

    (2) is continuous: Let (xi)iI be a net converging to x X. We showthat (xi) (x). Recall that the product topology on [0, 1]F has theproperty that convergence is coordinate-wise convergence (see 7.1.9), so(xi) (x) if and only if (xi) f = f (xi) (x) f = f (x) for all f F.

    (3) 1 is continuous: Suppose that (xi) (x). This implies f (xi) f (x) for all f F. Suppose that (xi)iI does not converge to x. There is aneighborhood U of x such that for all i I there is a j I such that j iand x j < U.

    Choose f on X such that f (x) = 0 and f (X \U) = {1} then f (xi) doesnot converge to f (x), a contradiction.

    8.2.3. The Stone-Cech compactification of a completely regular space is Haus-dorff.

    8.2.4. P. If f is a bounded continuous real function on a completelyregular space X then there is a unique continuous extension of f to the Stone-Cechcompactification of X.

    P. (1) We can reduce the problem to the case when f is boundedbetween 0 and 1.

    (2) A continuous extension of f , if exists, is unique.(3) Define g : [0, 1]F [0, 1] by g(y) = y f . Then g is an extension of f .(4) We check that g is continuous. Let (yi) be a net in [0, 1]F convergent to

    y. Then the net (g(yi)) is convergent to g(y), since convergence in productspace is coordinate-wise convergence (7.1.9).

    In some sense the Alexandroff compactification is the smallest compactifica-tion of a space and the Stone-Cech compactification is the largest one. For morediscussions, see [Mun00, p. 237]

  • CHAPTER 9

    Urysohn Lemma and Tiestze Theorem

    9.1. Urysohn Lemma and Tiestze Theorem

    We consider real functions, i.e. maps to the Euclidean R.

    9.1.1. T (Urysohn Lemma). If X is normal, F is closed, U is open, andF U, then there exists a continuous map f : X [0, 1] such that f |F 0 andf |X\U 1.

    P U L. Recall 4.1.11: Since X is normal, if F is closed, Uis open, F U then there is V open such that F V V U.

    (1) We construct a family of open sets in the following manner.Let U1 = U.

    n = 0: F U0 U0 U1.n = 1: U0 U 1

    2 U 1

    2 U1.

    n = 2: U0 U 14 U 1

    4 U 2

    4= U 1

    2 U 2

    4 U 3

    4 U 3

    4 U 4

    4= U1.

    Inductively we have a family of open sets:

    F U0 U0 U 12n U 1

    2n U 2

    2n U 2

    2n U 3

    2n U 3

    2n

    U 2n12n U 2n1

    2n U 2n

    2n= U1.

    Let I = {r = m2n /m, n N; 0 m 2n}. We have a family of opensets {Ur/r I} having the property r < s Ur Us.

    (2) We can check that I is dense in [0, 1].(3) Define a function f : X [0, 1],

    f (x) =

    inf{r I/x Ur} if x U1 if x < U .We prove that f is continuous.

    (4) Checking that f 1((, a)) is open. We show that {x X/ f (x) < a} =r b} = r>b X \ Ur.If f (x) > b then r I, r > b, f (x) > r, so x < Ur, hence x X \ Ur.Conversely, if there is r I, r > b such that x < Ur, then f (x) r > b.Now we show that

    r>b X \Ur = r>b X \Ur. Indeed, if r I, r > b

    then there is s I, r > s > b. Then Us Ur, therefore r>b X \ Ur r>b X \ Ur.

    9.1.2. Let A and B be two disjoint closed subsets of a normal space X. Show thatthere is a continuous function f from X to [0, 1] such that f (x) is 0 on A and is 1on B.

    49

  • 50 9. URYSOHN LEMMA AND TIESTZE THEOREM

    The sets A and B are said to be separated by a continuous function.As a corrolary: A normal space is completely regular.

    9.1.3. It is much easier to prove Urysohn Lemma for metric space, using the func-tion

    f (x) =d(x, A)

    d(x, A) + d(x, B).

    An application of Urysohn Lemma is:

    9.1.4. T (Tiestze Extension Theorem). Let X be a normal space. Let Fbe closed in X. Let f : F R be continuous. Then there is a c