Lecture notes on Topology and Geometry. address: URL: hqvu

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1 Lecture notes on Topology and Geometry Huỳnh Quang Vũ FACULTY OF MATHEMATICS AND INFORMATICS, UNIVERSITY OF NATURAL SCIENCES, VIETNAM NATIONAL UNIVERSITY, 227 NGUYEN VAN CU, DISTRICT 5, HO CHI MINH CITY, VIETNAM address: URL: hqvu

2 ABSTRACT. This is a set of lecture notes prepared for a series of introductory courses in Topology and Geometry for undergraduate students at the University of 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 delivered by a lecturer, namely myself, tailored to the need of my own students. I did not write it with self-study readers or with other lecturers in mind. Most statements are intended to be exercises. I provide proofs for some of the more difficult propositions, but even then there are still many details for students to fill in. More discussions will be carried out in class. I hope in this way 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 the Contents. A problem with a sign * is considered more difficult. A problem with a sign is an important one. This lecture notes will be continuously developed and I intend to keep it freely available on my web page. Although at this moment it is only a draft I do hope that it is useful to the readers. Your comments are very welcomed. May 14, 2005 September 23, 2008.

3 Contents Part 1. General Topology 1 Chapter 1. Theory of Infinite Sets The Cardinality of a Set The Axiom of Choice 8 Guide for Further Reading in General Topology 10 Chapter 2. Topological Spaces Topological Spaces Continuity Subspaces 18 Chapter 3. Connectivity Connected Space Path-connected Spaces 25 Further Reading 28 Chapter 4. Separation Axioms Separation Axioms 29 Chapter 5. Nets Nets 31 Chapter 6. Compact Spaces Compact Spaces 35 Chapter 7. Product Spaces Product Spaces Tikhonov Theorem 42 Further Reading 44 Chapter 8. Compactifications Alexandroff Compactification Stone-Cech Compactification 47 Chapter 9. Urysohn Lemma and Tiestze Theorem Urysohn Lemma and Tiestze Theorem 49 Further Reading 53 Chapter 10. Quotient Spaces Quotient Spaces 55 Chapter 11. Topological Manifolds Topological Manifolds 61 Further Reading 63 iii

4 iv CONTENTS Part 2. Algebraic Topology 65 Chapter 12. Homotopy and the fundamental groups Homotopy and the fundamental groups 67 Chapter 13. Classification of Surfaces Introduction to Surfaces Classification Theorem Proof of the Classification Theorem 71 Part 3. Differential Topology 75 Chapter 14. Differentiable manifolds Smooth manifolds Tangent spaces Derivatives 80 Chapter 15. Regular Values Regular Values Manifolds with Boundary 87 Chapter 16. The Brouwer Fixed Point Theorem The Brouwer Fixed Point Theorem 91 Chapter 17. Oriented Manifolds The Brouwer degree Orientation Brouwer Degree Vector Fields 98 Further Reading 99 Part 4. Differential Geometry 101 Guide for Reading 102 Chapter 18. Regular Surfaces Regular Surfaces The First Fundamental Form Orientable Surfaces The Gauss Map 107 Bibliography 109 Index 111

5 Part 1 General Topology

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7 CHAPTER 1 Theory of Infinite Sets In general topology we often work in very general settings, in particular we often deal with very large sets. Therefore we start with a deeper study of set theory The Cardinality of a Set Sets. We will not define what a set is. That means we only work on the level of the so-called naive set theory. Even so here we should be aware of certain problems in naive set theory. These problems are both educational and fascinating. Till the beginning of the 20th century, the set theory of German mathematician George Cantor, in which set is not defined, was widely used and thought to be a good basis for mathematics. Then certain critical problems were discovered relating to the liberal uses of the undefined notion of set EXAMPLE (Russell s Paradox). A famous version of it is the Barber paradox. It is as follows: In a village there is a barber. His job is to do hair cut to all villagers who cannot cut his hair himself. The question is who would cut the barber s hair? If he cut his hair, then he would have cut the hair of somebody who could do that himself, thus violating the terms of his job. On the other hand, if he does not cut this hair he also violates those terms, because he did not cut the hair of someone who could not do it himself. This means if we take the set of all villagers who had his hair cut by the barber then we can t decide whether the barber himself is a member of that set or not EXAMPLE. Consider the set S = {x/x x} (the set of all sets which are not a member of itself). Then whether S S or not is undecidable. These examples show that we are having too much liberty with the notion of set. Deeper study of the notion of set is needed. There are two axiomatic systems 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 general notion than set, called class, is used. For our purpose it is enough for us to be a bit careful when dealing with large sets, set of sets. In those occations we often replace the term set by the terms class or collection. See [Dug66, p. 32]. Indexed family. Suppose that A is a collection, I is a set and f : I A is a surjective map. The collection A together with the map f is called an indexed family. We often write f i = f (i), and denote the indexed family by { f i / i I}. Note that it can happen that f i = 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

8 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 {A i } i I be a family of sets indexed by a set I. The Cartesian product i I A i of this family is defined to be the collection of all maps f : I i I A i such that for each i I we have f (i) A i. An element of i I A i is often denoted by (a i ) i I, with a i A i 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 from one to the other. When A and B are equivalent we write A B Two intervals [a, b] and [c, d] on the real number line are equivalent via a linear map. Two intervals (a, b) and (c, d) are equivalent The interval ( 1, 1) is equivalent to R via the map x 1 1 x x The ball B(0, 1) = {(x 1, x 2,..., x n ) R n / x x x2 n < 1} is equivalent to R n Define the sphere S n to be the set {(x 1, x 2,..., x n+1 ) R n+1 / x x x2 n+1 = 1}. Then S n \ {(0, 0,..., 0, 1)} is equivalent to R n via the stereographic projection. Countable sets DEFINITION. A set is called countably infinite if it is equivalent to the set of positive 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. The elements of such a set can be indexed by the positive integers as a 1, a 2, a 3, EXAMPLE. The set of intergers Z is countable. The set of even integers is countable A union between a countable set and a finite set is countable A subset of a countable set is countable THEOREM. A union of a countable collection of countable sets is a countable set PROOF. The collection can be indexed as A 1, A 2,..., A i,... (if the collection is finite we can let A i be the same set for all i starting from a certain number). The elements of each set A i can be indexed as a i,1, a i,2,..., a i, j,.... This means the union i I A i can be mapped injectively to the index set Z + Z + by a i, j (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:

9 1.1. THE CARDINALITY OF A SET 5 (1, 1) (1, 2) (1, 3) (1, 4) (2, 1) (2, 2) (2, 3) (3, 1) (3, 2) (4, 1) (4, 2) (5, 1) Theorem can be reduced to the statement that Z + Z + is equivalent to Z +. Give another proof by checking that the map f : Z + Z + Z +, (m, n) 2 m 3 n is injective THEOREM. The set Q of rational numbers is countable. PROOF. Write Q = n=1 { p q / p, q Z, q > 0, p + q = n }. Another way is to observe that the map p q Q Q Q. (p, q) from Q to Z Z is injective. Inquiry minds would have noted that we did not define the set of integers or the set of real numbers. Such definitions are not easy. We contain ourself that those sets have some familiar properties THEOREM. The set of real numbers is uncountable. PROOF. The proof uses the Cantor diagonal argument. Suppose that the interval [0, 1] is countable and is enumerated as a sequence {a i / i Z + }. Suppose that a 1 = 0.a 11 a 12 a a 2 = 0.a 21 a 22 a a 3 = 0.a 31 a 32 a There are rational numbers whose decimal presentations are not unique, such as 1/2 = = To cover this case we should choose b n 0, 9. Choose a number b = 0.b 1 b 2 b 3... such that b 1 a 11, b 2 a 22,..., b n a nn,.... Then b a n for all n.

10 6 1. THEORY OF INFINITE SETS Cardinality. A genuine definition of cardinality of sets requires an axiomatic treatment of set theory, therefore here we contain ourselves that for each set A there exists an object called its cardinal A, and there is an order on the set of cardinals such 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 ℵ 0 (read aleph-0, aleph being the first character in the Hebrew alphabet): Z + = ℵ 0. The set R has cardinal c (continuum): R = c An infinite set contains a countably infinite subset ℵ 0 is the smallest infinite cardinal, and ℵ 0 < c. Georg Cantor put forward the The Continuum Hypothesis: There is no cardinal between ℵ 0 and c. Godel (1939) and Cohen (1964) have shown that the Continuum Hypothesis is independent from other axioms of set theory If A has n elements then 2 A = 2 n THEOREM. The cardinal of a set is strictly less than the cardinal of the set of its subsets, i.e. A < 2 A. This implies that there is no maximal cardinal. There is no universal set, a set which contains everything. PROOF. Let A and denote by 2 A the set of its subsets. (1) A 2 A : The map from A to 2 A : a {a} is injective. (2) A 2 A : Let φ be any map from A to 2 A. Let X = {a A/ a φ(a)}. Then there is no x A such that φ(x) = X (assuming the contrary there will be a contradiction), therefore φ is not surjective N is equivalent to the set of sequences of binary digits THEOREM (Cantor-Bernstein-Schroeder). If A is equivalent to a subset of B and B is equivalent to a subset of A then A and B are equivalent. ( A B B A ) A = B PROOF. Suppose that f : A B and g : B A are injective maps. Let A 1 = g(b), we will show that A A 1. Let A 0 = A and B 0 = B. Define B n+1 = f (A n ) and A n+1 = g(b n ). Then A n+1 A n. Furthermore via the map g f we have A n+2 A n, and A n \ A n+1 A n+1 \ A n+2. Using the following identities A = (A \ A 1 ) (A 1 \ A 2 ) (A n \ A n+1 )... ( A n ), A 1 = (A 1 \ A 2 ) (A 2 \ A 3 ) (A n \ A n+1 )... ( A n ), we see that A A 1. n=1 n=1

11 ℵ 0 = c THE CARDINALITY OF A SET 7 Hint: That 2 N [0, 1] follows from For the reverse direction, observe that any real number can be written in binary form (or use the Continuum Hypothesis) ℵ 0 2 ℵ 0 = 2 ℵ 0. Hint: An injective map from 2 N 2 N to 2 N can be constructed as follows. Two binary sequences a 1 a 2... and b 1 b 2... correspond to the sequence a 1 b 1 a 2 b R 2 R, in other words c 2 = c. Hint: Use the results of and , or prove directly as in Note: In fact for all infinite cardinal ω we have ω 2 ω, see [Dug66, p. 52], [Lan93, p. 888]. More problems Check that ( i I A i ) ( j J B j ) = i I, j J A i B j Which of the following formulas are correct? (a) ( i I A i ) ( i I B i ) = i I(A i B i ). (b) i I( j J A i, j ) = i I( j J A i, j ) Let f be a function. Then: (a) f ( i A i ) = i f (A i ). (b) f ( i A i ) i f (A i ). If f is injective (one-one) then equality happens. (c) f 1 ( i A i ) = i f 1 (A i ). (d) f 1 ( i A i ) = i f 1 (A i ) 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 A set which contains an uncountable subset is uncountable If A is finite and B is inifinite then A B B. Hint: There is an infinitely countable subset of B The set of points in R n with rational coordinates is countable Is the set of functions f : {0, 1} Z countable? The set of functions f : A {0, 1} is equivalent to 2 A A real number α is called an algebraic number if it is a root of a polynomial with integer coefficients. Show that the set of algebraic numbers is countable A real number which is not algebraic is called transcendental. For example it is known that π and e are transcendental (whereas it is still not known whether π + e is transcendental or not). Show that the set of transcendental numbers is uncountable Show that [a, b] (a, b] (a, b) A countable union of continuum sets is a continuum set. Hint: n=1[n, n + 1] = [1, ).

12 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 suggestive notation for R EXAMPLE. (1) (R, ) is a totally ordered set. (2) Let S be a set. Then (2 S, ) is a partially ordered set but is not totally ordered if S has more than one element Let (S 1, 1 ) and (S 2, 2 ) be two ordered sets. Show that the following is an order on S 1 S 2 : (a 1, b 1 ) (a 2, b 2 ) if (a 1 < a 2 ) or ((a 1 = a 2 ) (b 1 b 2 )). This is called the dictionary order DEFINITION. A totally order on a set S is a well-order if every non-empty subset 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 THEOREM. The following statements are equivalent: (1) Axiom of Choice: Given an arbitrary set M, there exists a choice function giving with any subset of M an element of M. (2) Given a family of disjoint non-empty sets A i there exists a set A such that A contains exactly one element from each A i. (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 maximal element (i.e. a X, (b X b a) a = b). (6) Hausdorff Maximality Principle: A totally ordered subset of a partially ordered 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-Banach and Banach-Alaoglu Theorems in Functional Analysis, the existence of bases in a vector space, the existence of a Lebesgue unmeasureable set,.... Zorn Lemma is often a convenient form of the Axiom of Choice 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 {B i } is a totally ordered collection of members of B. Let A = i B i. Then A B and is an upper bound of {B i }. The following is based on the Axiom of Choice THEOREM (Zermelo, 1904). On any set there is a well-order.

13 1.2. THE AXIOM OF CHOICE 9 More problems The set Z + with the order m n m n is a partially ordered set Let (S 1, 1 ) and (S 2, 2 ) be two ordered sets. For a and b in S 1 S 2, define a b if (a 1 b) (a, b S 1 ), or (a 2 b (a, b S 2 \ S 1 )). Show that this is an order on S 1 S Find a partial order on R A totally ordered finite set is well-ordered A subset of a well-ordered set is well-ordered The set of rational numbers on the interval [0, 1] is not well-ordered (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.

14 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 although it was published in Its presentation is rather abstract. The book has no picture! Munkres book [Mun00] is famous. Its treatment is somewhat more modern than Kelley, with many examples, pictures and exercises. It also has a section on Algebraic Topology. Hocking and Young s book [HY61] contains many deep and difficult results. This book together with Kelley and Munkres contain many topics not discussed in our lectures, some are at more advanced levels than the level here. The more recent book by Roseman [Ros99] works mostly in R n. Its strength is that it is more down-to-earth and it contains many new topics such as space-filling curves, 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 General Topology you can have a look at a collection of open problems in [MR90]. All of the items in the References are available from me.

15 CHAPTER 2 Topological Spaces On sets equipped with topological structures we can study continuity of functions Topological Spaces DEFINITION. Let X be a set. A topology on X is a family τ of subsets of X satisfying: (1) The sets and X are elements of τ. (2) A union of elements of τ is an element of τ. (3) A finite intersection of elements of τ is an element of τ. Elements of τ 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 τ is called a topological space. We often call an element of a topological space a point EXAMPLE. (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 2 X EXAMPLE. Let X = {1, 2, 3}. The following are topologies on X: (1) τ 1 = {, {1}, {2, 3}, {1, 2, 3}}. (2) τ 2 = {, {1, 2}, {2, 3}, {2}, {1, 2, 3}} EXAMPLE (The Euclidean topology). In R n = {(x 1, x 2,..., x n )/x i R}, the Euclidean distance between two points x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) is d(x, y) = [ n i=1 (x i y i ) 2 ] 1/2. An open ball centered at x with radius r is the set B(a, r) = {y R n /d(y, x) < r}. A subset of R n is called open if it is either empty or is a union of open balls. This is the Euclidean topology of R n The finite complement topology on X consists of the empty set and all subsets of X whose complements are finite A finite intersection of open sets is open is equivalent to an intersection of two open sets is open 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 set containing x. Note that a neighborhood doesn t need to be open. 11

16 12 2. TOPOLOGICAL SPACES Bases of a topology. A collection B τ of open sets is called a basis for the topology τ of X if any non-empty member of τ is a union of members of B EXAMPLE. In the Euclidean space R n the set of balls with rational radii is a basis In the Euclidean space R n the set of balls with radii 1 2 m, m 1 is a basis A collection B of open sets is a basis if for each point x and each open set O containing x there is a U B containing x such that U is contained in O. A collection S τ of open sets is called a subbasis for the topology τ of X if the collection of finite intersections of members of S is a basis for τ EXAMPLE. Let X = {1, 2, 3}. The topology τ 2 = {, {1, 2}, {2, 3}, {2}, {1, 2, 3}} has a basis {{1, 2}, {2, 3} {2}} and a subbasis {{1, 2}, {2, 3}} The collection of open rays, that is sets of the forms (a, ) and (, a) is a subbasis of R with the Euclidean topology. Generating topologies. Given a collection of subsets of a set, when is it a basis for a topology? THEOREM. Let B be a collection of subsets of X. Then B { } is a basis for a topology on X if and only if the union of members of B is X and the intersection of two members of B is a union of members of B. The collection τ consists of the empty set and all unions of members of B is a topology on X. It is called the topology generated by B. PROOF. Verifying that τ is a topology is reduced to checking that the intersection of two members of τ is a member of τ. In deed ( i B i ) ( j B j ) = i, j(b i B j ). But then B i B j = i, j,k B k. Next we want to know, given a collection of subsets of a set, is there a smallest topology which contains these subsets? When is this collection a subbasis for a topology? THEOREM. Let S be a family of subsets of X whose union is X. Then S is a subbasis for a topology on X consisting of the empty set and unions of finite intersections of members of S. This topology is called the topology generated by S. A basis for this topology is the collection of finite intersections of members of S. By this theorem, just any collection of subsets covering the set generates a topology on that set EXAMPLE. Let X = {1, 2, 3, 4}. The set {{1}, {2, 3}, {3, 4}} generates the topology {, {1}, {3}, {1, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}. A basis for this topology is {{1}, {3}, {2, 3}, {3, 4}} (Order topology). Let (X, ) be a totally ordered set with at least two elements. The subsets of the forms {β X/ β < α} and {β X/ β > α} generate a topology on X, called the order topology EXAMPLE. The Euclidean topology on R is the ordering topology with respect to the usual order of real numbers.

17 2.1. TOPOLOGICAL SPACES 13 Metric spaces. Metric space is a motivation for generalization to topological space. 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} In the theory of metric spaces, a set U X is said to be open if x U, ɛ > 0, B(x, ɛ) U. Show that the family of balls is a basis for a topology on (X, d). When we speak about topology on a metric space we mean this topology. Comparing topologies. Let τ 1 and τ 2 are two topologies on X. If τ 1 τ 2 we say that τ 2 is finer (or stronger) than τ 1 and τ 1 is coarser (or weaker) than τ EXAMPLE. On a set the trivial topology is the coarsest topology and the discrete topology is the finest one The topology generated by a collection of subsets is the coarsest topology containing those subsets. More problems Is the collection of all open half-planes (meaning not consisting the separating lines) a subbasis for the Euclidean topology of R 2? Show that on R 2 the basis consists of interiors of circles, the basis consists of interiors of squares, and the basis consists of interiors of ellipses generate the same topology Show that two bases generate the same topology if and only if each member of one basis is a union of members of the other basis (R n has a countable basis). The set of balls each with rational radius whose center has rational coordinates forms a basis for the Euclidean topology of R n On R n consider the following metrics, which are generated by norms. Find the unit ball for each metric. Show that these metrics generate the same topology. (a) d 1 (x, y) = n i=1 x i y i (b) d 2 (x, y) = max 1 i n x i y i (c) d 3 (x, y) = ( n i=1 (x i y i ) 2 ) 1/ Let d 1 and d 2 be two metrics on X. If there are α, β > 0 such that for all x, y X, αd 1 (x, y) < d 2 (x, y) < βd 1 (x, y) then the two metrics are said to be equivalent. This is indeed an equivalence relation. Two equivalent metrics generate the 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 R n generate equivalent metrics, so all norms on R n generate the same topology, exactly the Euclidean topology Is the Euclidean topology on R 2 the same as the ordering topology on R 2 with respect to the dictionary order? An open set in R is a countable union of open intervals The family of intervals of the form [a, b) generates a topology on R. Is it the Euclidean topology?

18 14 2. TOPOLOGICAL SPACES Describe all sets that have only one topology * 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 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.

19 2.2. CONTINUITY Continuity Continuity DEFINITION. Let X and Y be topological spaces. We say a map f : X Y is continuous at x X if for any neighborhood U of f (x) there is a neighborhood V of x such that f (V) U. Equivalently, f is continuous at x if for any neighborhood U of f (x) the inverse image f 1 (U) is a neighborhood of x. We say that f is continuous on X if it is continuous everywhere on X THEOREM. A map is continuous if and only if the inverse image of an open set is an open set. PROOF. ( ) Suppose that f : X Y is continuous. Let U be an open set in Y. Let x f 1 (U), and let y = f (x). Since f is continuous at x and U is a neighborhood of y, the set f 1 (U) is a neighborhood of x. Thus x is an interior point 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 neighborhood U of y. Then f 1 (U ) is an open set containing x, therefore the set f 1 (U) will be a neighborhood of x A map is continuous if and only if the inverse image of a closed set is a closed set Let (X, d 1 ) and (Y, d 2 ) be two metric spaces. In the theory of metric spaces, a map f : (X, d 1 ) (Y, d 2 ) is continuous at x X if and only if ɛ > 0, δ > 0, d 1 (y, x) < δ d 2 ( f (y), f (x)) < ɛ. Show that this is a special case of the notion of continuity in topological spaces REMARK. Therefore from now on we can use any results in previous courses involving continuity in metric spaces. Homeomorphism. A map from one topological space to another is said to be a homeomorphism if it is a bijection, is continuous and its inverse map is also continuous. Two spaces X and Y are said to be homeomorphic, written X Y, if there is a homeomorphism 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 A homeomorphism is both an open map and a closed map PROPOSITION. If f : (X, τ X ) (Y, τ Y ) is a homeomorphism then it induces a bijection between τ X and τ Y. PROOF. The map is a bijection. f : τ X O τ Y f (O)

20 16 2. TOPOLOGICAL SPACES In the category of topological spaces and continuous maps, when two spaces are homeomorphic they are the same. An embedding (or imbedding) (phép nhúng) from the topological space X to the 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. We say that X can be embedded in Y EXAMPLE. The Euclidean line R can be embedded in the Euclidean plane R EXAMPLE. Suppose that f : R R 2 is continuous under the Euclidean topology. Then R can be embedded onto the graph of f. Topologies generated by maps Let (X, τ) 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 Let X be a set and (Y, τ) be a topological space. Let f : X Y be a map. Find the coarsest and finest topologies on X such that f is continuous Let X be a set and (Y, τ) be a topological space. Let f i : X Y, i I be a collection of maps. Find the coarsest topology on X such that all maps f i, i I are continuous. 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 If f : X Y and g : Y Z are continuous then g f is continuous Suppose that f : X Y and B is a basis for the topology of Y. Show that f is continuous if and only if the inverse image of any element of B is an open set in X A continuous bijection is a homeomorphism if and only if it is an open map Any two closed intervals in R are homeomorphic. Also (a, b) R S n \ {(0, 0,..., 0, 1)} R n via the stereographic projection Any two balls in the Euclidean R n are homeomorphic The unit ball B(0, 1) in the Euclidean R n is homeomorphic to R n. Hint: Consider the map x 1 1 x x (a) A square and a circle are homeomorphic. (b) The region bounded by a square and the region bounded a the circle are homeomorphic If f : X Y is a homeomorphism and Z X then X \ Z and Y \ f (Z) are homeomorphic 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.

21 2.2. CONTINUITY On the Euclidean plane R 2, show that: (a) R 2 \ {0, 0} and R 2 \ {1, 1} are homeomorphic. (b) R 2 \ {{0, 0}, {1, 1}} and R 2 \ {{1, 0}, {0, 1}} are homeomorphic.

22 18 2. TOPOLOGICAL SPACES 2.3. Subspaces Relative Topology. Let (X, τ) be a topological space and A X. The relative topology, or the subspace topology of A is {A O/ O τ}. With this topology we say 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 restriction of an open set in X to A A subset of a subspace A X is closed in A if and only if it is a restriction of a closed set in X to A Suppose that X is a topological space and Z Y X. Then the relative topology on Z with respect to Y is the same as the topology on Z with respect to X REMARK. Let Y be a subspace of a topological space X. It is not true that if a set 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 we are 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 A set is open if and only if it is equal to its interior. In other words, all of its points are interior. The intersection A of all closed set of X containing A is called the closure of A in 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 subset A of X if any neighborhood of x contains a point of A 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 acummulation point) of the subset A of X if any neighborhood of x contains a point of A other than x A set is closed if and only if it contains all its limit points 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 say that it is a boundary point of A A point is in the boundary of A if and only if every of its neighborhoods has non-empty intersections with A and the complement of A. More problems Show that A is the disjoint union of A and A Show that X is the disjoint union of A, A, and X \ A.

23 Verify the following properties. (a) X\ A= X \ A. (b) X \ A = X \ A. (c) If A B then A B SUBSPACES The set {x Q/ 2 x 2} is both closed and open in Q R The map ϕ : [0, 1) S 1 R 2 given by t e 2πit is a bijection but is not a homeomorphism. Hint: Compare the subinterval [1/2, 1) and its image via ϕ Consider Q R. Then Q= and Q = R Find the closures, interiors and the boundaries of the interval [0, 1) under the Euclidean, discrete and trivial topologies of R Find the closures, interiors and the boundaries of N R In a metric space X, a point x X is a limit point of the subset A X if and only if there is a sequence in A \ {x} converging to x. Note: This is not true in general topological spaces In R n with the Euclidean topology, the boundary of the ball B(x, r) is the sphere {y/ d(x, y) = r}. The closed ball B (x, r) = {y/ d(x, y) r} is the closure of B(x, r) 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 Suppose that A Y X. Then A Y = A X Y. Furthermore if Y is closed in X then A Y = A X Let O n = {k Z + /k n}. Then { } {O n /n Z + } is a topology on Z +. Find the closure of the set {5}. Find the closure of the set of even positive integers Consider R with the finite complement topology. Find the closures, interiors and the boundaries of the subsets {1, 2} and N 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 Show that these spaces are not homeomorphic to each other: Z, Q, R with Euclidean topology, and R with the finite complement topology.

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25 CHAPTER 3 Connectivity 3.1. Connected Space Connected space. A topological space is said to be connected if is not a union of two non-empty disjoint open subsets PROPOSITION. A topological space X is connected if and only if its only subsets which are both closed and open are the empty set and X. When we say that a subset of a topological space is connected we implicitly mean that the subset under the subspace topology is a connected space In a topological space a set containning one point is connected PROPOSITION. Let X be a topological space and A and B are two connected subsets. If A B then A B is connected. Thus the union of two non-disjoint connected subsets is a connected set. PROOF. Suppose that C is subset of A B that is both open and closed. Suppose that C. Then either C A or C B. Without loss of generality, assume that C A. Note that C A is both open and closed in A. Since A is connected, C A = A. Then C B, hence C B = B, so C = A B. The same proof gives a more general result: PROPOSITION. Let X be a topological space and let A i, i I be connected subsets. If i I A i then i A i is connected THEOREM (Continuous image of connected set is connnected). If f : X Y is continuous and X is connected then f (X) is connected If two spaces are homeomorphic and one space is connected then the other space is also connected. Connected Component THEOREM. Let X be a topological space. Define a relation on X whereas two points are related if both belong to a connected subset of X. Then this is an equivalence relation PROPOSITION. An equivalence class under the above equivalence relation is connected. PROOF. Consider the equivalence class [p] represented by a point p. By the definition, q [p] if and only if there is a connected set containing both p and q. Thus [p] = q [p] O q where O q is a connected set containing both p and q. By 3.1.4, [p] is connected. 21

26 22 3. CONNECTIVITY DEFINITION. Under the above equivalence relation, the equivalence classes are called the connected components of the space. Thus a space is a disjoint union of its connected components A connected component is a maximal connected subset under the set inclusion PROPOSITION. The closure of a connected set is a connected set. PROOF. Suppose that A is connected. Let B A be both open and closed in A and 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 The proof above actually showed that if A is connected and A B A then B is connected A connected component must be closed. Connected sets in the Euclidean real number line PROPOSITION. A connected set in R under the Euclidean topology must be an interval. PROOF. Suppose that A X is connected. Suppose that x, y A and x < y. If x < 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 THEOREM. The interval (0, 1) as a subset of the Euclidean real number line is connected. A proof of this fact must deal with some fundamental properties of real numbers. PROOF. 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 in C 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 both open 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 then there are points in D which are bigger than s. If D = we let E = C \ (0, x) and consider t = inf E THEOREM. An interval in the Euclidean real number line is connected. Therefore a subset of the Euclidean real number line is connected if and only if it is an interval. PROOF. 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

27 3.1. CONNECTED SPACE 23 More problems S 1 is connected in the Euclidean topology of the plane. Hint: The circle is a continuous image of an interval (Intermediate Value Theorem). If f : R R is continuous under the Euclidean topology then the image of an interval is an interval If f : R R is continuous then its graph is connected in the Euclidean plane Let X be a topological space and A Y X. If A is connected in Y then A is connected in X, and vice versa Let X be a topological space and let A i, i I be connected subsets. If A i A j for all i j then i A i is connected Let X be a topological space and let A i, i Z + be connected subsets. If A i A i+1 for all i 1 then i=1 A i is connected. Hint: Note that the conclusion is stronger than that n i=1 A i is connected for all n Z Is an intersection of connected sets connected? Let X be a topological space. A map f : X Y is called a discrete map if Y 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 What are the connected components of N R under the Euclidean topology? What are the connected components of Q R under the Euclidean topology? 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 in C * What are the connected components of Q 2 R 2 under the Euclidean topology? If a space has finitely many components then each component is both open and closed Let X = {0} { 1 n /n Z+ } R where R has the Euclidean topology. Find the open connected components of X If X and Y are homeomorphic then they have the same number of connected components, i.e. there is a bijection between the sets of connected components of the two sets The Euclidean space R n is connected No interval on the Euclidean real number line is homeomorphic to S The Euclidean spaces R and R 2 are not homeomorphic. Hint: Delete a point from R. Use and

28 24 3. CONNECTIVITY That R 2 and R 3 are not homeomorphic is not easy. It is a consequence of the following difficult theorem: THEOREM (Invariance of Domain). If two subsets of the Euclidean R n are homeomorphic and one set is open then the other is also open. As a consequence, R m and R n are not homeomorphic if m n. This result allows us to talk about topological dimension Show that R with the finite complement topology and R 2 with the finite complement topology are homeomorphic.

29 3.2. PATH-CONNECTED SPACES Path-connected Spaces Let X be a topological space and a, b X. A path in X from a to b is a continuous 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 and b in X there is a path in X from a to b The Euclidean R n is path-connected A ball in the Euclidean R n is path-connected. Let f be a path from a to b. Then the inverse path of f is defined to be the path f 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 composition of f with g is the path f (2t), 0 t 1 2 h(t) = 1 g(2t 1), 2 t 1. The path h is often denoted as f g Show that f g is continuous. Hint: If f (t) = f (2t) and g(t) = g(2t 1) then h 1 (U) = f 1 (U) g 1 (U) Let X be a topological space. Define a relation on X whereas a point x is related to a point y there is a path in from x to y. Then this is an equivalence relation An equivalence class under the above equivalence relation is path-connected. Each equivalence class is called a path-connected component Let X be a topological space and A and B are two path-connected subsets. If A B then A B is path-connected A path-connected component is a maximal path-connected subset under the set inclusions THEOREM. A path-connected space is connected. PROOF. The proof is based on the fact that the image of a path is a connected set. Let X be path-connected. Let a, b X. There is a path from a to b. The image of this path is a connected subset of X. That means any two points of X are in the same connected component of X. Therefore X has only one connected component. Generally, the reverse statement of is not correct. However we have: THEOREM. An open, connected subset of the Euclidean space R n is pathconnected. PROOF. Let A be open, connected in R n. Let B be a path-connected component of A. Then B must be both open and closed in A. Indeed, if x B then there is a ball B(x, r) A. Then B B(x, r) is still path-connected, so B B(x, r), thus B is open. Similarly we can show that A \ B is open If f : X Y is continuous and X is path-connected then f (X) is pathconnected.

30 26 3. CONNECTIVITY Locally connected and locally path-connected spaces. A topological space is said to be locally connected if every neighborhood of a point contains a connected neighborhood of that point. A topological space is said to be locally path-connected if every neighborhood of a point contains a path-connected neighborhood of that point EXAMPLE. All open sets in the Euclidean space R n are locally connected and locally path-connected. Generalize we get: THEOREM. A connected, locally path-connected space is path-connected. PROOF. Suppose that X is connected and locally path-connected. Let C be a path-connected component of X. If x X has a path-connected neighborhood U such that U C, then U C is path-connected, and so U C. This means that C is both open and closed in X, hence C = X. Topologist s Sine Curve. The closure of the graph of the curve y = sin 1 x, x > 0 is often called the Topologist s Sine Curve. 1 sin(1/x) FIGURE 3.1. Topologist s Sine Curve. Denote A = {(x, sin 1 x )/0 < x 1} and B = {0} [ 1, 1]. Then A B = and the Topologist s Sine Curve is X = A B PROPOSITION. The Topologist s Sine Curve is connected. PROOF. By the set A is connected. And A X A PROPOSITION. The Topologist s Sine Curve is not path-connected. PROOF. Suppose that there is a path γ(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 t 0 = sup{t [0, 1]/x(t) = 0}. Then x(t 0 ) = 0, t 0 < 1, and for all t > t 0 we have x(t) > 0. Thus t 0 is the moment when the path γ departs from B. We can see that the path jumps immediately when it departs from B. Thus we will show that γ(t) cannot be continuous at t 0, by showing that for any δ > 0 there are

31 3.2. PATH-CONNECTED SPACES 27 t 1, t 2 (t 0, t 0 +δ) such that y(t 1 ) = 1 and y(t 2 ) = 1, therefore y(t) is not continuous at t 0. To find t 1, note that the set {x(t)/t 0 t t 0 + δ} is an interval [0, x 0 ] where x 0 > 0. There exists an x 1 (0, x 0 ) such that sin 1 x 1 = 1: we just need to take 1 x 1 = π 2 +k2π with sufficiently large k. There is t 1 (t 0, t 0 + δ) such that x(t 1 ) = x 1. Then y(t 1 ) = sin 1 x(t 1 ) = 1. More problems R 2 \ {one point} is path-connected under the Euclidean topology R 2 \ A where A is a finite set is path-connected under the Euclidean topology. Hint: Bound A in a rectangle. For any point in R 2 \ A there is a straight line through it that does not intersect A, by an argument using cardinalities of sets * R 2 \ Q 2 is path-connected under the Euclidean topology. Indeed if A is countable then R 2 \ A is path-connected. Hint: Let a R 2 \ A. Let l a be a line passing through a and does not intersect A. If b R 2 \ A let l b be a line passing through b and does not intersect A, but does intersect l a 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 The Topologist s Sine Curve is not locally path-connected The Topologist s Sine Curve is not locally connected * (a) Classify the alphabetical characters up to homeormophisms, that is, which of the following characters are homeomorphic to each other as subsets of the 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 to modify a letter part by part. (b) Note that the result depends on the font you use! Indeed, what if instead of a 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

32 28 3. CONNECTIVITY Further Reading Jordan Curve Theorem. The following is an important and deep result of plane topology THEOREM (Jordan Curve Theorem). A simple, continuous, closed curve separates the plane into two disconnected regions. More concisely, if C is a subset of the Euclidean plane homeomorphic to the circle then R 2 \ C has two connected components. Space Filling Curves. A rather curious and surprising result is: THEOREM. 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 be simple. Such a curve is called a Peano curve. It could be constructed as a limit of an iteration of piecewise linear curves.

33 CHAPTER 4 Separation Axioms 4.1. Separation Axioms DEFINITION. The following are called separation axioms. T 0 : A topological space is called a T 0 -space if for any two different points there is an open set containing one of them but not the other. T 1 : A topological space is called a T 1 -space if for any two points x y there is an open set containing x but not y and an open set containing y but not x. T 2 : A topological space is called a T 2 -space or Hausdorff if for any two points x y there are disjoint open sets U and V such that u U and v V. T 3 : A T 1 -space is called a T 3 -space or regular (chính t ăc) if for any point x and a closed set F not containing x there are disjoint open sets U and V such that x U and F V. T 4 : A T 1 -space is called a T 4 -space or normal if for any two disjoint closed sets F and G there are disjoint open sets U and V such that F U and G V. These are called separation axioms because they involve separating certain sets from one another by open sets PROPOSITION. A space is T 1 if and only if a set containing exactly one point is a closed set (T 4 T 3 T 2 T 1 T 0 ). If a space is T i then it is T i 1, for 1 i Any space with the discrete topology is normal Give an example of a space which is T 0 but not T The real number line under the finite complement topology is T 1 but is not T A metric space is regular PROPOSITION. A metric space is normal. PROOF. We introduce the notion of distance between two sets in a metric space X. If A and B are two subsets of X then we define the distance between A and B 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 a continuous function with respect to x. Now suppose that A and B are disjoint closed sets. Let U = {x/d(x, A) < d(x, B)} and V = {x/d(x, A) > d(x, B)}. Then A U, B V, U V =, and both U and V are open A subspace of a Hausdorff space is Hausdorff. 29

34 30 4. SEPARATION AXIOMS PROPOSITION. A T 1 -space X is regular if and only if given a point x and an open set U containing x there is an open set V such that x V V U. PROOF. Suppose that X is regular. Since X \ U is closed and disjoint from C there is an open set V containing x and an open set W containing X \ U such that V and W are disjoint. Then V (X \ W), so V (X \ W) U. Now suppose that X is T 1 and the condition is satisfied. Given a point x and a closed set C disjoint from x. Let U = X \C. Then there is an open set V containing x such that V V U. Then V and X \ V separate x and C A T 1 -space X is normal if and only if given a closed set C and an open set U containing C there is an open set V such that C V V U. More problems If a finite set is a T 1 -space then the topology is the discrete topology * There are examples of a T 2 -space which is not T 3, and a T 3 -space which is not T 4, but they are rather difficult, see and [Mun00, p. 197] * Let X be normal, f : X Y is surjective, continuous, and closed. Prove that Y is normal.

35 CHAPTER 5 Nets 5.1. Nets In metric spaces we can study continuity of functions via convergence of sequences. In general topological spaces, we need to use a notion more general than sequences, called nets. Roughly speaking in general topological spaces, sequences (countable indices) might not be enough to describe the neigborhood systems at a point, 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 x i = n(i), we often denote n as (x i ) i I EXAMPLE. Nets on I = N with the usual order are exactly sequences EXAMPLE. Let X be a topological space and x X. Let I be the family of open neighborhoods of x. Define an order on I by U V U V. Then I becomes a directed set. It will be clear in a moment that this is the most important example of directed sets concerning nets. Convergence. A net (x i ) 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 (x i ), and we often write x i x EXAMPLE. Convergence of nets on I = N is exactly convergence of sequences EXAMPLE. Let X = {x 1, x 2, x 3 } with topology {, X, {x 1, x 3 }, {x 2, x 3 }, {x 3 }}. The net (x 3 ) converges to x 1, x 2, and x 3. The net (x 1, x 2 ) converges to x PROPOSITION. A point x X is a limit point for a set A X if and only if there is 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 to topological spaces by simply replacing sequences by nets. PROOF. ( ) Suppose that x is a limit point of A. Consider the directed set I consisting of all the open neighborhoods of x with the partial order U V if U V. For any open neighborhood U of x there is an element x U U A, x U x. Consider the net (x U ) U I. It is a net in A \ {x} convergent to x. Indeed, given an open neighborhood U of x, for all V U, x V V U. ( ) Suppose that there is a net (x i ) i I in A \ {x} convergent to x. Let U be an open neighborhood of x. There is an i I such that for j i we have x j U, in particular x i U (A \ {x}). 31

36 32 5. NETS REMARK (When can nets be replaced by sequences?). By examining the above proof we can see that the term net can be replaced by the term sequence if there is a countable collection F of neighborhood of x such that any neighborhood of x contains a member of F. In other words, the point x has a countable neighborhood basis. A space having this property at every point is said to be a first countable space. A metric space is such a space. Similar to the case of metric spaces we have: THEOREM. Let X and Y be topological spaces. Then f : X Y is continuous if and only if whenever a net n in X is convergent to x then the net f n is convergent to f (x). In more familiar notations, f is continuous if and only if for all nets (x i ) and all points x in X, x i x f (x i ) f (x). The proof is simply a repeat of the proof for metric spaces. PROOF. ( ) 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 (x i ) is convergent to x, there is an i 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 of X \ f 1 (U). There is a net (x i ) in X \ f 1 (U) convergent to x. Since f is continuous, f (x i ) Y \ U is convergent to f (x) U. That contradicts the assumption that U is open A map f is continuous at x if and only if for all nets (x i ), x i x f (x i ) f (x) PROPOSITION. If a space is Hausdorff then a net has at most one limit. PROOF. Suppose that a net (x i ) is convergent to two different points x and y. Since the space is Hausdorff, there are disjoint open neighborhoods U and V of x and y. There is i I such that for γ i we have x γ U, and there is j I such that for γ j we have x γ U. Since there is a γ I such that γ i and γ j, the point x γ will be in U V, a contradiction The converse statement of is also true. A space is Hausdorff if and only 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 (U x, V y ) of open neighborhoods of x and y, under set inclusion. Take a net n such that n(u x, V y ) is a point in U x V y. Then this net converges to both x and y. More problems 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 with n((0, a)) = a/2. Is n convergent? Suppose that τ 1 and τ 2 are two topologies on X. Furthermore suppose that τ 1 τ 2 for all nets x i and all points x, x i x x i x. Show that τ 1 τ 2. In other words, if convergence in τ 1 implies convergence in τ 2 then τ 1 is finer than τ 2.

37 5.1. NETS 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 on A, by: (a) using nets. (b) not using nets.

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39 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 cover is said to be an open cover if each member of the cover is an open subset of X. A subset 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 EXAMPLE. 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 A is a compact space under the topology restricted from X REMARK. Let I be an open cover of A. Each O I is an open set of A, so it is the restriction of an open set U O of X. Thus we have a collection of open sets {U O / O I} whose union contains A. On the other hand if we have a collection I of open sets of X whose union contains A then the collection {U O/ O I} is an open cover of A Any subset of R with the finite complement topology is compact. Hint: Any open set covers a subset of R except finitely many points THEOREM. In a Hausdorff space compact sets are closed. PROOF. Let A be a compact set in a Hausdorff space X. We show that X \ A is open. Let x X\A. For each a A there are disjoint open sets U a containing x and V a containing a. The family {V a /a A} covers A, so there is a finite subcover {V ai /1 i n}. Let U = n i=1 U ai and V = n i=1 V ai. Then U is an open neighborhood of x disjoint from V, a neighborhood of A THEOREM (Continuous image of compact set is compact). If X is compact and f : X Y is continuous then f (X) is compact PROPOSITION. If X is compact and A X is closed then A is compact. PROOF. Add X \ A to an open cover of A REMARK. When A is a compact subset of X and X is a subspace of Y then A is also a compact subset of Y, because the topology of X restricted to A and the topology 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: PROPOSITION. If X is compact, Y is Hausdorff, f : X Y is bijective and continuous, then f is a homeomorphism PROPOSITION. In a compact space an infinite set has a limit point. 35

40 36 6. COMPACT SPACES PROOF. Let A be an infinite set in a compact space X. Suppose that A has no limit point. Let x X, then there is an open neighborhood U x of x that contains at most one point of A. The family of such U x 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 of subsets of a set is said to have the finite intersection property if the intersection of every finite subcollection of A is non-empty THEOREM. A space is compact if and only if any family of closed subsets with the finite intersection property has non-empty intersection. Compact sets in metric spaces. Most of the following results have been studied in a course in Real Analysis or Functional Analysis. It is very useful to have a look back at notes for those courses In a metric space, if a set is compact then it is closed and bounded If X is a compact space and f : X (R, Euclidean) is continuous then f has a maximum value and a minimum value A metric space is compact if and only if every sequence has a convergent subsequence In the Euclidean space R n a subset is compact if and only if it is closed and bounded. More problems Find a cover of the interval (0, 1) with no finite subcover A discrete compact topological space is finite (An extension of Cantor Lemma in Calculus). Let X be compact and X A 1 A 2 A n... be a ascending sequence of closed, non-empty sets. Then n=1 A n A compact Hausdorff space is normal. Hint: Use The proof of contains the following: In a Hausdorff space a point and a compact set disjoint from it can be separated by open sets. More generally: In a Hausdorff space two disjoint compact sets can be separated by open sets. Hint: We already have that for each x in the first set there are an open set U x containing x and an open set V x containing the second set and disjoint from U x. Since the first set is compact, it is covered by n i=1 U xi. Consider n i=1 V xi In a regular space a compact set and a disjoint closed set can be separated by open sets In a topological space a finite unions of compact subsets is compact.

41 6.1. COMPACT SPACES In a Hausdorff space an intersection of compact subsets is compact The set of n n-matrix with real coefficients, denoted by M(n; R), could be naturally considered as a subset of the Euclidean space R n2 by considering entries of a matrix as coordinates, via the map (a i, j ) (a 1,1, a 2,1,..., a n,1, a 1,2, a 2,2,..., a n,2, a 1,3,..., a n 1,n, a n,n ). The Orthogonal Group O(n) is defined to be the group of matrices representing orthogonal linear maps of R n, that are linear maps that preserve inner product. Thus O(n) = {A M(n; R)/ A A T = I n }. 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 form cos(ϕ) sin(ϕ) sin(ϕ) cos(ϕ). This is a rotation in the plane around the origin with an angle ϕ. Thus SO(2) is the group 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 R n could be rotated to the unit vector e 1 = (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, R 1 where R 1 SO(n 1). (g) Show that S O(n) is connected. The General Linear Group GL(n; R) is the group of all invertible n n-matrices with real coefficients. (h) Show that GL(n; R) is not compact. (k) Find the number of connected components of GL(n; R).

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43 CHAPTER 7 Product Spaces 7.1. Product Spaces Finite products of spaces. Let X and Y be two topological space, and consider the Cartesian product X Y. The product topology on X Y is the topology generated by the collection F of sets of the form U V where U is an open set in X and V is an open set in Y. The collection F is a subbasis for the product topology. Actually, since the intersection of two members of F is also a member of F, the collection F is a basis for the product topology. Thus every open set in the product topology is a union of products of open sets of X with open sets of Y Note that, as sets: (a) (A B) (C D) = (A C) (B D). (b) (A B) (C D) (A C) (B D) = (A B) (A D) (C B) (C D) If τ 1 is a basis for X and τ 2 is a basis for Y then τ 1 τ 2 is a basis for the product topology on X Y. Similarly the product topology on n i=1 (X i, τ i ) is defined to be the topology generated by n i=1 τ i If each τ i is a basis for X i then n i=1 τ i is a basis for the product topology on ni=1 (X i, τ i ) EXAMPLE (Euclidean topology). Recall that R n = } R R {{ R}. The n times Euclidean topology on R is generated by open intervals. An open set in the product topology of R n is a union of products of open intervals. Since a product of open intervals is an open rectangle, and an open rectangle is a union of open balls and vice versa, the product topology is exactly the Euclidean topology. Infinite products. Let {(X i, τ i ), i I} be a family of topological spaces. The product topology on i I X i is the topology generated by the collection F consisting of sets of the form i I U i, where U i τ i and U i = X i for all except finitely many i I Check that F is really a basis for a topology. For j I the projection p j : i I X i X j is defined by p j ((x i )) = x j. It is the projection to the j coordinate PROPOSITION (Product topology is the topology such that projections are continuous). (a) If i I X i has the product topology then p i is continuous for all i I. 39

44 40 7. PRODUCT SPACES (b) The product topology is the coarsest topology on i I X i such that all the maps p i are continuous. This is the topology generated by the all the maps p i, i I, as in PROOF. (a) Note that if O j X j then p 1 j (O j ) = i I U i with U i = X i for all i except j, and U j = O j. (b) The topology generated by all the maps p i is the topology generated by sets of the form p 1 i (O i ) with O i τ i. A finite intersections of these sets is exactly a member of the basis of the product topology as in the definition Show that each projection map p i is a an open map, mapping an open set to an open set. Hint: Only need to show that the projection of an element of the basis is open (Map to product space is continuous if and only if each component map is continuous). Show that f : Y i I X i is continuous if and only if f i = p i f is continuous for every i I PROPOSITION (Convergence in product topology is coordinate-wise convergence). A net n : J i I X i is convergent if and only if all of its projections p i n are convergent. PROOF. ( ) Suppose that each p i n is convergent to a i, we show that n is convergent to a = (a i ) i I. A neighborhood of a contains an open set of the form U = i I O i with O i are open sets of X i and O i = X i except for i K, where K is a finite subset of I. For each i K, p i n is convergent to a i, therefore there exists an index j i J such that for j j i we have p i (n( j)) O i. Take an index j 0 such that j 0 j i for all i K. Then for j j 0 we have n( j) U. More problems Check that in topological sense R 3 = R R R = R 2 R. Hint: Look at their bases If X is homeomorphic to X 1 and Y is homeomorphic to Y 1 then X Y is homeomorphic to X 1 Y Give an example to show that in general the projection p i is not a closed map Fix a point x = (x i ) i I X i. Define the inclusion map f : X i i I X i by x j if j i y ( f (y)) j = y if j = i. The inclusion f is a homeomorphism to its image (an embedding of X i ). The image of the inclusion map is X i = j I A j where A j = {x j } if j i and A i = X i. Thus X i is a parallel copy of X i in i I X i. The spaces X i have x as common point. Hint: Use to prove that the inclusion map is continuous (a) If each X i, i I is Hausdorff then i I X i is Hausdorff. (b) If i I X i is Hausdorff then each X i is Hausdorff. Hint: (b) Use

45 7.1. PRODUCT SPACES (a) If i I X i is path-connected then each X i is path-connected. (b) If X i, i I is path-connected then i I X i is path-connected. Hint: (b) Let (x i ) and (y i ) be in i I X i. Let γ i (t) be a continuous path from x i to y i. Let γ(t) = (γ i (t)). Use (a) If i I X i is connected then each X i is connected. (b) If X and Y are connected then X Y is connected. (c)* If each X i, i I is connected then i I X i is connected. Hint: (b) Use (c) Fix a point x i I X i. 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 i I X i, such that the union of these connected subsets is i I X i (a) If i I X i is compact then each X i is compact. (b) * Prove that if X and Y are compact then X Y is compact, without using the Axiom of Choice. Hint: Use 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 finite subcover {(U x ) i (V x ) i / 1 i n x }. Take U x = n x i=1 (U x) i. The family {U x / x X} covers X so there is a subcover {U x j / 1 j n}. The family {(U x j ) i (V x j ) i / 1 i n x j, 1 j n} is a finite subcover of X Y (a) If O i is an open set in X i for all i I then is i I O i open? (b) If F i is a closed set in X i for all i I then is i I F i closed? (Zariski Topology). Let F = R or F = C. A polynomial in n variables on F is a function from F n to F which is a finite sum of terms of the form ax m 1 1 xm 2 2 xm n n, where a, x i F and m i 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 F n / p S, p(x) = 0}. Such a set is called an algebraic set. (a) Show that if we define that a set in F n is closed if it is algebraic, then this gives a topology on F n, called the Zariski topology. (b) Show that the Zariski topology on F is exactly the finite complement topology. (c) Show that if both F and F n have the Zariski topology then all polynomials on F n are continuous. (d) Is the Zariski topology on F n the product topology? Note: The Zariski topology is the fundamental topology in Algebraic Geometry.

46 42 7. PRODUCT SPACES 7.2. Tikhonov Theorem THEOREM (Tikhonov theorem). A product of compact spaces is compact. Applications of this theorem include the Banach-Alaoglu Theorem in Functional Analysis and the Stone-Cech compactification. Tikhonov theorem is equivalent to the Axiom of Choice, and needs a difficult proof. 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. PROOF. Let X i be compact for all i I. We will show that X = i I X i is compact by showing that if a collection of closed subsets of X has the finite intersection property then it has non-empty intersection. Let F be a collection of closed subsets of X that has the finite intersection property. We will show that A F A. Note: Have a look at the following argument, which suggests that proving the Tikhonov theorem might not be easy. If we take the closures of the projections of the collection F to the i-coordinate then we get a collection {p i (A), A F} of closed subsets of X i having the finite intersection property. Since X i 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 is not true, see the figure. In what follows we will overcome this difficulty by enlarging the collection F to the extent that the intersections of the closures of the projections of the collection will give common elements of the collection. (1) We show that there is a maximal collection F of subsets of X that contains F and still has the finite intersection property. We will use Zorn Lemma for this purpose. (This is a routine step; it is easier for the reader to carry it out instead of reading.) Let K be the collection of collections G of subsets of X such that G contains F and has the finite intersection property. On K we define an order by the usual set inclusion. Now suppose that L is a totally ordered subcollection of K. Let M = A L 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 intersection property. Suppose that M i M, 1 i n. Then M i A i for some A i L. There is an i 0 such that A i0 contains all A i. Then M i A i0 for all 1 i n, and since A i0 has the finite intersection property, we have n i=1 M i. (2) Since F is maximal, it is closed under finite intersections.

47 7.2. TIKHONOV THEOREM 43 (3) We will show that [ p i (A)] A. i I A F To show that i I[ A F p i(a)] is non-empty means to show that A F p i(a) is non-empty for each i I. Since F has the finite intersection property, the collection {p i (A), A F} also has this property, and so is the collection {p i (A), A F}. Furthermore {p i (A), A F} is a collection of closed subsets of the compact set X i. So A F p i(a) is non-empty. (4) Let x i I[ A F p i(a)], we will show that x A, A F. Since A is closed we need to show that any neighborhood of x has non-empty intersection with A. It is suffice to prove this for neighborhoods of the form i D p 1 i (U i ) where U i is open in X i, and D is a finite subset of I. Let A F. We have p i (x) = x i p i (A). For each i D, since U i is a neighborhood of x i we have U i p i (A). Therefore p 1 i (U i ) A. Then {p 1 i (U i )} F has the finite intersection property. Since F is maximal we must have p 1 i (U i ) F. Therefore [ i D p 1 i (U i )] A. More problems Let [0, 1] have the Euclidean topology. If I is an infinite set then [0, 1] I is called the Hilbert cube. The Hilber cube is compact Using the characterization of compact subsets of Euclidean spaces, prove the Tikhonov theorem for finite products of compact subsets of Euclidean spaces Let (X, d X ) and (Y, d Y ) be metric spaces. (a) Prove that the product topology on X Y is given by the metric A F d((x 1, y 1 ), (x 2, y 2 )) = max{d X (x 1, x 2 ), d Y (x 2, y 2 )}. (b) Using the characterization of compact metric spaces in terms of sequences, prove the Tikhonov theorem for finite products of compact metric spaces.

48 44 7. PRODUCT SPACES Further Reading Strategy for a proof of Tikhonov theorem based on net. The proof that we will outline here is based on further developments of the theory of nets and a characterization of compactness in terms of nets DEFINITION (Subnet). Suppose that I and I are directed sets, and h : I I is 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 take n i Z + such that n i < n i+1 then (x ni ) is a subsequence of (x n ). In this case the map h : Z + Z + is given by h(i) = n i. Thus a subsequence of a sequence is a subnet of that sequence. On the other hand there are subnets of sequences which are not subsequences. Suppose that x 1, x 2, x 3,... be a sequence. Then x 1, x 1, x 2, x 2, x 3, x 3,... is a subnet of that sequence, but not a subsequence. In this case the map h is given by h(2i 1) = h(2i) = i. A net (x i ) i I is called eventually in A X if there is j I such that i j x i A DEFINITION (Universal net). A net n in X is universal if for any subset A of X either n is eventually in A or n is eventually in X \ A PROPOSITION. If f : X Y is continuous and n is a universal net in X then f (n) is a universal net PROPOSITION. 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. PROOF OF TIKHONOV THEOREM. Let X = i I X i where each X i is compact. Suppose that (x j ) j J is a universal net in X. By the net (x j ) is convergent if and only if the projection (p i (x j )) is convergent for all i. But that is true since (p i (x j )) is a universal net in the compact set X i.

49 CHAPTER 8 Compactifications 8.1. Alexandroff Compactification A compactification of a space X is a compact space Y such that X is homeomorphic to a dense subset of Y EXAMPLE. A compactification of the Euclidean (0, 1) is the Euclidean [0, 1] EXAMPLE. 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 EXAMPLE. The Euclidean space R n is locally compact THEOREM (Alexandroff compactification). Let X be a locally compact Hausdorff 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 is X \ C where C is a compact set in X. Then this is the only topology on X such that 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 compactification of X. PROOF. (1) Suppose that there is a topology on X such that X is compact, 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 then C = X \ O is a closed subset of the compact space X, therefore C is compact in X. But C X, so C must be compact in X too. Thus the open sets of X must be open sets of X or complements of compact sets of X. (2) We check that we really have a topology. Let C i be compact sets in X. Consider i(x \ C i ) = X \ i C i. Note that since X is Hausdorff, i C i is compact. If O is open in X and C is compact in X then O (X \ C) = X \ (C (X \ O)). If C i, 1 i n are compact sets then n i=1 X \ C i = X \ ( n i=1 C i ). Finally O (X \ C) = O (X \ C). With this topology X is a subspace of X. (3) We show that X is compact. Let {O i, i I} be an open cover of X. Then one O i0 will cover. Therefore the complement of O i0 in X will be a compact set C in X. Then {O i, i I, i i 0 } is an open cover of C. From this cover there is a finite cover. This finite cover together with O i0 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

50 46 8. COMPACTIFICATIONS Since X is locally compact there is a compact set C containing an open 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 PROPOSITION. Let X and Y be two locally compact Hausdorf spaces. If X and Y are homeomorphic then X and Y are homeomorphic. PROOF. A homeomorphism from X to Y induces a homeomorphism from X to Y. We can also use REMARK. Since a one-point compactification is unique up to homeomorphisms we often use the word the one-point compactification EXAMPLE. The one-point compactification of the Euclidean line R is S 1. The one-point compactification C { } of the complex plane C is homeomorphic to S 2, is often called the extended complex plane or the Riemann sphere. More problems Find the one-point compactification of the Euclidean (0, 1) (2, 3) What is the one-point compactification of Z with the Euclidean topology? Describe the topology of the compactification What is the one-point compactification of { 1 n /n Z+ } under the Euclidean topology? Describe the topology of the compactification What is the one-point compactification of the Euclidean open ball B(0, 1)? Find the one-point compactification of the Euclidean space R n What is the one-point compactification of the Euclidean annulus {(x, y) R 2 /1 < x 2 + y 2 < 2}? 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 containing x, 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 We could have noticed that the notion of local compactness as we have defined is not apparently a local property. For a property to be local, every neighborhood of any point must contain a neighborhood of that point with the given property (as in the cases of connectedness and path-connectedness). Show that for Hausdorff spaces local compactness is indeed a local property The Topologist s Sine Curve (3.2) is a compactification of the Euclidean interval (0, 1). space. Note: This is not a one-point compactification. There can be many ways to compactify a Define a topology on R {± } such that it is a compactification of the Euclidean R.

51 8.2. STONE-CECH COMPACTIFICATION Stone-Cech Compactification A space is said to be completely regular (sometimes called a T space) if it is a T 1 -space and for each point x and each closed set A with x A there is a continuous 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 it could be separated by continuous functions A completely regular space is regular. Let X be a completely regular space. Let F be the set of all continuous maps from X to [0, 1]. Let Φ be a function from X to [0, 1] F = {[0, 1]/ f F}} defined by for x X, Φ(x) f = f (x) for each f F. Since [0, 1] F is compact, and Φ(X) [0, 1] F, we have Φ(X) is compact. The compact space Φ(X) is called the Stone-Cech compactification of the completely regular space X THEOREM. If X is completely regular then Φ : X Φ(X) Φ(X) is a homeomorphism. In other words Φ is an embedding. PROOF. (1) Φ is injective: If Φ(x) = Φ(y) then Φ(x) f = Φ(y) f for all f F, so f (x) = f (y) for all f F. Since X is completely regular we must have x = y. (2) Φ is continuous: Let (x i ) i I be a net converging to x X. We show that Φ(x i ) Φ(x). Recall that the product topology on [0, 1] F has the property that convergence is coordinate-wise convergence (see 7.1.9), so Φ(x i ) Φ(x) if and only if Φ(x i ) f = f (x i ) Φ(x) f = f (x) for all f F. (3) Φ 1 is continuous: Suppose that Φ(x i ) Φ(x). This implies f (x i ) f (x) for all f F. Suppose that (x i ) i I does not converge to x. There is a neighborhood U of x such that for all i I there is a j I such that j i and x j U. Choose f on X such that f (x) = 0 and f (X \ U) = {1} then f (x i ) does not converge to f (x), a contradiction The Stone-Cech compactification of a completely regular space is Hausdorff PROPOSITION. If f is a bounded continuous real function on a completely regular space X then there is a unique continuous extension of f to the Stone-Cech compactification of X. PROOF. (1) We can reduce the problem to the case when f is bounded between 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 (y i ) be a net in [0, 1] F convergent to y. Then the net (g(y i )) is convergent to g(y), since convergence in product space is coordinate-wise convergence (7.1.9). In some sense the Alexandroff compactification is the smallest compactification of a space and the Stone-Cech compactification is the largest one. For more discussions, see [Mun00, p. 237]

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53 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 THEOREM (Urysohn Lemma). If X is normal, F is closed, U is open, and F U, then there exists a continuous map f : X [0, 1] such that f F 0 and f X\U 1. PROOF OF URYSOHN LEMMA. Recall : Since X is normal, if F is closed, U is 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 U 1 = U. n = 0: F U 0 U 0 U 1. n = 1: U 0 U 1 2 U 1 2 U 1. n = 2: U 0 U 1 4 U 1 4 U 2 4 = U 1 2 U 2 4 U 3 4 U 3 4 U 4 4 = U 1. Inductively we have a family of open sets: F U 0 U 0 U 1 U 1 U 2 U 2 U 3 U 3 2 n 2 n 2 n 2 n 2 n 2 n U 2 n 1 2 n U 2 n 1 2 n U 2 n 2 n = U 1. Let I = {r = m 2 /m, n N; 0 m 2 n }. We have a family of open n sets {U r /r I} having the property r < s U r U s. (2) We can check that I is dense in [0, 1]. (3) Define a function f : X [0, 1], inf{r I/x U r } if x U f (x) = 1 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<a U r. We have f (x) < a if and only if r I, r < a, f (x) < r, which in turns is if and only if r I, r < a, x U r. (5) Checking that f 1 ((b, + )) is open. First we show that {x X/ f (x) > b} = r>b X \ U r. If f (x) > b then r I, r > b, f (x) > r, so x U r, hence x X \ U r. Conversely, if there is r I, r > b such that x U r, then f (x) r > b. Now we show that r>b X \ U r = r>b X \ U r. Indeed, if r I, r > b then there is s I, r > s > b. Then U s U r, therefore r>b X \ U r r>b X \ U r Let A and B be two disjoint closed subsets of a normal space X. Show that there is a continuous function f from X to [0, 1] such that f (x) is 0 on A and is 1 on B. 49

54 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 It is much easier to prove Urysohn Lemma for metric space, using the function d(x, A) f (x) = d(x, A) + d(x, B). An application of Urysohn Lemma is: THEOREM (Tiestze Extension Theorem). Let X be a normal space. Let F be closed in X. Let f : F R be continuous. Then there is a continuous map g : X R such that g F = f. PROOF. First consider the case when f is bounded. (1) The general case can be reduced to the case when inf F f (x) = 0 and sup F f (x) = 1. We will restrict our attention to this case. (2) Let f 0 = f. By Urysohn Lemma, there is a continuous function g 1 : X [0, 1 3 ] such that 0 x f0 1 g 1 (x) = ([0, 1 3 ]) 1 3 x f0 1([ 2 3, 1]). Let f 1 = f 0 g 1. Then sup x F f 1 (x) = 2 3, inf x F f 1 (x) = 0 and sup x X g(x) = 1 3. (3) Inductively, once we have such a ( function ) f n : F R, n 0 we will obtain a function g n+1 : X [0, 1 2 n] 3 3 such that ( ) 0 x fn 1 ([0, 1 2 n]) 3 3 g n+1 (x) = ( ) 1 2 n 3 3 x fn 1 ([ ( ) 2 n+1. 3, 1]) Let f n+1 = f n g n+1. Then sup x F f n+1 (x) = ( 2 n+1 3) and supx X g n+1 (x) = ) n. 1 3 ( 2 3 (4) The series n=1 g n converges uniformly to a continuous function g(x). (5) Since f n = f n i=1 g i the series n=1 g n F converges uniformly to f. Therefore g F = f. (6) Note that with this construction inf x X g(x) = 0 and sup x X g(x) = 1. Now consider the case when f is not bounded. (1) Suppose that f is neither bounded from below nor bounded from above. Let h be a homeomorphism from (, ) to (0, 1). Then f 1 = h f is bounded by 0 and 1, therefore it can be extended, as in the previous case, to a continuous function g 1 such that inf x X g 1 (x) = inf x F f 1 (x) = 0 and sup x X g 1 (x) = sup x F f 1 (x) = 1. If the range of g 1 does not include 0 or 1 then g = h 1 g 1 will be the desired function. It may happens that the range of g 1 includes 0 or 1. In this case let C = g 1 1 ({0, 1}). Note that C F =. By Usrysohn Lemma, there is a continuous function k : X [0, 1] such that k C = 0 and k F = 1. Let g 2 = kg 1 + (1 k) 1 2 then g 2 F = g 1 F and the range of g 2 is a subset of (0, 1). Then g = h 1 g 2 will be the desired function.

55 9.1. URYSOHN LEMMA AND TIESTZE THEOREM 51 (2) If f is bounded from below then similarly to the previous case we can use a homeomorphism h : [a, ) [0, 1), and we let C = g 1 1 ({1}). The case when f is bounded from above is similar. More problems Show that the Tiestze extension theorem implies the Urysohn lemma The Tiestze extension theorem is not true without the condition that the set F is closed Show that the Tiestze extension theorem can be extended to maps to the Euclidean R n Let X be a normal space and F be a closed subset of X. Then any continuous map f : F S n can be extended to an open set containing F. Hint: Use Let X be a topological space, and Y be a subspace of X. We say that Y is a retract of X if there is a continuous map r : X Y such that r Y = id Y. In other words the identity map id Y can be extended to X If Y is a retract of X then any map from Y to a topological space Z can be extended to X (a) Show that S 1 is a retract of R 2 \ {(0, 0)}. (b) Show that a subset containing two points cannot be a retract of R 2. Note: This shows that the Tiestze extension theorem cannot be automatically generalized to maps to general topological spaces. (c)* Could S 1 be a retract of R 2 or D 2? Note: This question could be answered using basic tools in homotopy theory (in Algebraic Topology) or smooth functions (Differential Topology) * Show that if X is Hausdorff and Y is a retract of X then Y is closed in X. Hint: Use nets Prove the following version of the Urysohn Lemma, as stated in [Rud86]. Suppose that X is a locally compact Hausdorff space, V is open in X, K V, and K is compact. Then there is a continuous function f : X [0, 1] such that f (x) = 1 for x K and supp( f ) V, where supp( f ) is the closure of the set {x X/ f (x) 0}, called the support of f. Hint: Use and (Niemytzki space). * Let H = {(x, y) R 2 / y 0} be the upper half-plane. Equip H with the topology generated by the Euclidean open disks (i.e. open balls) in K = {(x, y) R 2 / y > 0}, together with sets of the form {p} D where p is a point on the line L = {(x, y) R 2 / y = 0} and D is an open disk in K tangent to L at p. This is called the Niemytzki space. (a) Check that this is a topological space. (b) What is the subspace topology on L? (c) What are the closed sets in H? (d) Show that H is Hausdorff. (e) Show that H is regular.

56 52 9. URYSOHN LEMMA AND TIESTZE THEOREM (f) Show that H is not normal. Hint: Any real function on L is continuous. The cardinality of the set of such function is c c. A real function on H is continuous if and only if it is continuous on the dense subset of points with rational coordinates, so the cardinality of the set of such functions is c ℵ 0. Since c c > c ℵ 0, the space H cannot be normal, by Tiestze Extention Theorem.

57 FURTHER READING 53 Further Reading Metrizability. A space is said to be metrizable if its topology can be generated by a metric THEOREM (Urysohn Metrizability Theorem). A regular space with a countable basis is metrizable. The proof uses the Urysohn Lemma [Mun00].

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59 CHAPTER 10 Quotient Spaces Quotient Spaces Suppose that X is a topological space, Y is a set, and f : X Y is a surjective map. We want to find a topology on Y so that f becomes a continuous map. Define V Y to be open in Y if f 1 (V) is open in X. This is a topology on Y, called the quotient topology generated by f, denoted by Y/ f This is the finest topology on Y such that f is continuous. Let X be a topological space and be an equivalence relation on X. Let p : X X/ be the projection x [x]. The set X/ with the quotient topology generated by p is called a quotient space. If A X then there is an equivalence relation on X: x x if x A, and x y if x, y A. The quotient space X/ is also denoted by X/A THEOREM. Suppose that f : X Y is surjective and Y has the quotient topology corresponding to f. Let Z be a topological space. Then g : Y Z is continuous if and only if g f is continuous. f X Y g f The following result will provide us a tool to identify quotient spaces THEOREM. Suppose that X is compact, Y is Hausdorff, f : X Y is continuous and onto. If f factors through the quotient, i.e. f (x 1 ) = f (x 2 ) if x 1 x 2, then it induces a homeomorphism from X/ to Y. In more details, define h : X/ Y by h([x]) = f (x), then f = h p. X p f Z g X/ If f is continuous then h is a homeomorphism. Y h PROOF. By and EXAMPLE (Gluing the two end-points of a line segment gives a circle). More precisely [0, 1]/0 1 is homeomorphic to S 1 : [0, 1] p [0, 1]/0 1 h f S 1 Here f is the map t (cos(2πt), sin(2πt)). 55

60 QUOTIENT SPACES (Gluing a pair of opposite edges of a square gives a cylinder). Let X = [0, 1] [0, 1]/ where (0, t) (1, t) for all 0 t 1, then X is homeomorphic to the cylinder [0, 1] S 1. The homeomorphism is induced by the map (s, t) (s, cos(2πt), sin(2πt)) (Gluing opposite edges of a square gives a torus). Let X = [0, 1] [0, 1]/ where (s, 0) (s, 1) and (0, t) (1, t) for all 0 s, t 1, then X is homeomorphic to the torus T 2 = S 1 S 1. FIGURE The torus. The torus T 2 is homeomorphic to a subspace of R 3 (we say that the torus can be embedded in R 3 ). The subspace is the surface of revolution obtained by revolving a circle around a line not intersecting it. FIGURE The torus embedded in R 3. Suppose that the circle is on the Oyz-plane, the center is on the y-axis and the axis for the rotation is the z-axis. Let a be the distance from the center of the circle to the z-axis, b be the radius of the circle (a > b). Let S be the surface of revolution, then the embedding is given by [0, 2π] [0, 2π] T 2 = ([0, 2π]/0 2π) ([0, 2π]/0 2π) f S h where f (s, t) = ((a + b cos(s)) cos(t), (a + b cos(s)) sin(t), b sin(s)) (Gluing the boundary circle of a disk together gives a sphere). More precisely D 2 / D 2 is homeomorphic to S 2. We only need to construct a continuous map from D 2 onto S 2 such that after quotient out by the boundary D 2 it becomes injective, see Figure

61 10.1. QUOTIENT SPACES 57 FIGURE FIGURE The Mobius band embedded in R (The Mobius band). Gluing a pair of opposite edges of a square in opposite directions gives the Mobius band. More precisely the Mobius band is X = [0, 1] [0, 1]/ where (0, t) (1, 1 t) for all 0 t 1. The Mobius band could be embedded in R 3. It is homeomorphic to a subspace of R 3 by rotating a straight segment around the z-axis while also turning that segment up side down. The embedding can be induced by the map (s, t) ((a + t cos(s/2)) cos(s), (a + t cos(s/2)) sin(s), t sin(s/2)), with 0 s 2π and 1 t 1. t a s s/2 FIGURE The embedding of the Mobius band in R (The projective plane). Identifying opposite points on the boundary of a disk we get a topological space called the the projective plane RP Show that RP 2 is homeomorphic to the sphere with antipodal points identified, that is, S 2 /x x.

62 QUOTIENT SPACES The real projective plane cannot be embedded in R 3. It can be embedded in R (The projective space). More generally, identifying antipodal points of S n, or homeomorphically, identifying antipodal boundary points of D n gives us the (real) projective space RP n (The Klein bottle). Identifying one pair of opposite edges of a square and the other pair in opposite directions gives a topological space called the Klein bottle. More precisely it is [0, 1] [0, 1]/ with (0, t) (1, t) and (s, 0) (1 s, 1). FIGURE The Klein bottle. This space cannot be embedded in R 3, but it can be immersed in R 3. An immersion (phép nhúng chìm) is a local embedding. More concisely, f : X Y is an immersion if each point in X has a neighborhood U such that f U : U f (U) is a homeomorphism. FIGURE The Klein bottle immersed in R REMARK. The above topological spaces are obtained by identifying parts of the boundary circle of a disk. They are two-dimensional surfaces. For the general theory, see 13. In the above examples, a question might be raised: QUESTION. If the identifications are carried out in steps rather than simultaneously, will the result be different? Let R 1 and R 2 be two equivalence relations on the space X such that R 1 R 2 is also an equivalence relation. This is the requirement that ( ([x] R1 [y] R2 ) ([x] R2 [y] R1 ) ) [x] R1 [x] R2. Consider X/R 1. On this space we define an equivalence relation R 2 induced from R 2 as follows: [x] R1 R 2 [y] R1 if ([x] R1 [y] R2 ) ([x] R2 [y] R1 ).