Math 201 Topology I. Lecture notes of Prof. Hicham Gebran

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1 Math 201 Topology I Lecture notes of Prof. Hicham Gebran hicham.gebran@yahoo.com Lebanese University, Fanar, Fall

2 2 Introduction and orientation What is topology? Topology is the study of continuity in a general context. We can think of functions as deformations of objects. Continuity means that the deformation does not break the object, but it can stretch it, sequeeze it and twist it. This is why topology is known in popular terms as rubber-sheet geometry. In group theory or vector space theory, a fundamental concept is that of isomorphism. Two groups are isomorphic if they have the same algebraic structure. The analogous concept in topology is that of homeomorphism. A homeomorphism is a continuous bijection whose inverse is also continuous. For example, a circle and a square are homeomorphic as we shall see, that is, they are topologically the same. A fundamental problem of topology is the classification of certain spaces (manifolds) up to a homeomorphism. Also, topology allows one to go from the local to the global. Here are some examples that we will encounter in this course or in later courses. - Let f : X Y be locally constant. If X is connected (a topological property), then f is globally constant. - Let f : X Y be continuous (a local property). If X is compact (a topological property), then f is uniformly continuous (a global property). - Let f : X IR be locally bounded. If X is compact, then f is globally bounded. - Let F : Ω IR 2 IR 2 be an irrotational vector field. Then F has locally a potential function. If Ω is simply connected (a topological property), then F has globally a potential function. All questions, comments, remarks and suggestions are welcome. References James Munkres, Topology (Pearson, 2000). Colin Adams and Robert Franzosa, Introduction to topology, pure and applied (Pearson, 2007). Lynn Arthur Steen and J. Arthur Seebach, Counterexamples in Topology (Dover, 1978).

3 Contents 0 Some fundamental properties of the real line 5 1 Metric spaces and topological spaces Definitions and examples Open and closed sets and related concepts (neighborhood, interior, closure, boundary, convergence, density) Topological spaces Subspaces Product spaces Continuous functions and homeomorphisms 25 3 Compactness Definitions, examples and properties Compact subspaces of IR n Compact metric spaces Local compactness and the Alexandroff compactification Connectedness Definitions, examples and properties Components and local connectedness Complete metric spaces 53 3

4 4 CONTENTS

5 Chapter 0 Some fundamental properties of the real line We start by formulating two fundamental properties of the set of real numbers that we shall use in the sequel. This set is denoted by IR and it is naturally identified to a line. An upper bound of a subset E IR is a real number M such that x M for all x E. The set E is called bounded from above if it has an upper bound. A lower bound of a subset E IR is a real number m such that m x for all x E. The set E is called bounded from below if it has a lower bound. Definition 0.1 Let E IR be bounded from above. The number α is called the least upper bound of E if the following hold. i) α is an upper bound of E. ii) If β < α, then β is not an upper bound of E. In this case, we write α = sup E. If E is not bounded from above, we write sup E = +. Proposition 0.1 Let E IR be bounded from above. following hold. i) x α for all x E. ii) ε > 0 y E such that α ε < y. Then α = sup E if and only if the Definition 0.2 Let E IR be bounded from below. The number α is called the greatest lower bound of E if the following hold. i) α is a lower bound of E. ii) If β > α, then β is not a lower bound of E. In this case, we write α = inf E. If E is not bounded from below, we write inf E =. Proposition 0.2 Let E IR be bounded from below. following hold. i) x α for all x E. ii) ε > 0 y E such that α + ε > y. Then α = inf E if and only if the Proposition 0.3 Let A B IR. Then sup A sup B and inf A inf B. Exercise. Let A and B be two nonempty subsets of IR. If x y for all x A and all y B, then sup A inf B. Now we can state the two fundamental properties of IR. 5

6 6 CHAPTER 0. SOME FUNDAMENTAL PROPERTIES OF THE REAL LINE Theorem 0.1 Any nonempty subset of IR which is bounded from above has a least upper bound. Any nonempty subset of IR which is bounded from below has a greatest lower bound. Any proof of this theorem involves going back to the construction of the real numbers from the rational numbers. We do not address this issue here. Note that the above theorem is not true for the set of rational numbers Q. And this is one of the reasons we prefer to work in the IR rathre than in Q. Recall now that a subset E IR is an interval if and only if [x, y] E whenever x and y belong to E. An interval has thus one of the following eleven forms, {a}, ]a, b[, ]a, b], [a, b[, [a, b], ], a], ], a[, ]a, [, [a, [ and IR. An interval is called trivial if it is empty or a singleton. Theorem 0.2 Any nontrivial interval of the real line contains rational as well as irrational numbers. This theorem will be proved in the exercises. We end with a useful result. Proposition 0.4 Let a and b be given real numbers. If a b + ε for all ε > 0 then a b. Proof. Otherwise, taking ε = b a 2, we reach a contradiction.

7 Chapter 1 Metric spaces and topological spaces Topological spaces are generalizations of metric spaces which are themselves generalizations of IR and IR N. A metric space is a set equipped with a distance. The notion of distance give a meaning to the closeness of two ojects: two objects are close if their distance is small. So what is distance? 1.1 Definitions and examples Consider the usual Euclidean distance between two points in a plane equipped with an othonormal coordinate system (so we identify the plane with IR 2 ). Let d(x, y) denote the Euclidean distance bewtween two points x = (x 1, x 2 ) and y = (y 1, y 2 ). Note that by Pythagoras theorem d(x, y) = (x 1 y 1 ) 2 + (x 2 y 2 ) 2. It is clear (geometrically at least) that (i) d(x, y) 0 for all x, y IR 2. (ii) d(x, y) = 0 if and only if x = y. (iii) d(x, y) = d(y, x) for all x, y X (symmetry). (iv) d(x, z) d(x, y) + d(y, z) for all x, y, z X. This inequality is known as the triangle inequality: in the triangle xyz, the length of an edge is less than the sum of lengths of the other two edges. These properties of the Euclidean distance are used as an axiomatic definition of a distance over an arbitrary set. Definition 1.1 Let X be a non empty set. A distance or metric over X is a function d : X X IR that satisfies the following properties (i) d(x, y) 0 for all x, y X. (ii) d(x, y) = 0 if and only if x = y. (iii) d(x, y) = d(y, x) for all x, y X (symmetry). (iv) d(x, z) d(x, y) + d(y, z) for all x, y, z X (Triangle inequality). A metric space is a couple (X, d) where d is a distance on X. Examples. 1) The real line with the usual distance d(x, y) = x y. 2) The real line with the distance ρ(x, y) = x 3 y 3. 7

8 8 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES 3) The real line with the distance γ(x, y) = min(1, x y ). The first three properties of a distance are clear. We prove the triangle inequality. Observe that γ(x, y) 1 and γ(x, y) x y. Let x, y, z be three real numbers. We distinguish between two cases. Case 1. Either x y 1 or y z 1. Then γ(x, y) = 1 or γ(y, z) = 1. Therefore, γ(x, y) + γ(y, z) 1. On the other hand, γ(x, z) 1. Hence the triangle inequality in this case. Case 2. x y < 1 and y z < 1. Then γ(x, y) = x y and γ(y, z) = y z. Therefore, γ(x, z) x z x y + y z = γ(x, y) + γ(y, z). 4) More generally, let (X, d) be a metric space and set d(x, y) = min(1, d(x, y)). Then d is a distance on X. The proof is similar to the proof above. 5) Let a < b and X = C([a, b]) be the set of continuous functions f : [a, b] IR. Let d(f, g) = sup x [a,b] f(x) g(x). Then d is a distance on X. Write the details. 6) Again consider the set X = C([a, b]) and set d(f, g) = distance on X. Write the details. 7) Let X be a set. The discrete distance d on X is defined by { 0 if x = y d(x, y) = 1 if x y b a f(x) g(x) dx. Then d is a The first three properties are clear. Let x, y, z X. To prove the triangle inequality, we distinguish between two cases. Case 1. x = z. Then d(x, z) = 0 d(x, y) + d(y, z). Case 2. x z. Then either y x or y z. Therefore d(x, z) = 1 d(x, y) + d(y, z). Norms and distances on IR N Let N be a positive integer. The set of all N tuples of real numbers (x 1, x 2,..., x N ) is denoted by IR N. Otherwise stated, IR N is the cartesian product of N copies of IR. Elements of IR N are considered as vectors with N components. The origin of IR N is the vector (0,..., 0) denoted simply by 0. A norm on IR N is a way to measure the length of a vector (x 1,..., x N ). It is the analog of the absolute value of a real number. The norm of a vector x IR N is a number denoted by x that satisfies the following properties. (i) x 0. (ii) x = 0 if and only if x = 0. (iii) λx = λ x for any real number λ and any x IR N. (iv) x + y x + y for all x, y IR N. This inequality is also known as the triangle inequality. The most natural norm on IR N is the Euclidean norm. Let x = (x 1,..., x N ) IR N. The Euclidean norm of x is defined by ( N ) 1/2 x = x i 2. i=1 It is not difficult to show that x satisfies the three properties (i), (ii) and (iii) above. The triangle inequality is clear gemometrically but not trivial to prove algebraically. To prove it

9 1.1. DEFINITIONS AND EXAMPLES 9 algebraically, let us first observe that the Euclidean norm is just the square root of the inner product of x by itself. The inner product of two vectors x and y of IR N is x y = x 1 y x N y N = N x i y i. i=1 Let α be real number and consider the scalar product (x + αy) (x + αy). We have (x + αy) (x + αy) = x 2 + 2αx y + α 2 y 2 0. Thus we have a second order polynomial in α which is always nonnegative. Therefore its discriminant (x y) 2 x 2 y 2 0 (otherwise, the polynomial would have two distinct real roots and would therefore change sign). Thus, x y x y. This is known as the Cauchy-Schwartz inequality. From the last inequality we deduce that x + y 2 = (x + y) (x + y) = x 2 + 2x y + y 2 x x y + y 2 = ( x + y ) 2. Since both members of the above inequality are nonnegative, the triangle inequality follows. Other norms can also be defined on IR N. The sup norm or infinity norm of a vector x is defined by x = max( x 1,..., x N ). It is easy to check that this norm satisfies the four properties of a norm. The 1 norm is defined by N x 1 = x i. Here again it is not difficult to prove the four properties of a norm. More generally, let p 1. The p norm of x is ( N ) 1/p x p = x i p. i=1 It can be shown that this is indeed a norm (the triangle inequality is not trivial to prove). Note finally that all these norms coincide when N = 1. Now a norm on IR N defines a metric by setting d(x, y) = x y. Balls in metric spaces The open ball of center x and radius r is the set i=1 B(x, r) = {y X d(y, x) < r}. The closed ball of center x and radius r is the set B (x, r) = {y X d(y, x) r}.

10 10 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Examples. 1) The ball B(a, r) in IR (with the usual distance) is the interval ]a r, a + r[. In (IR 2, 2 ) the ball B(a, r) is the usual disc of center a and radius r. In (IR 3, 2 ), the ball B(a, r) is the usual geometric ball of center a and radius r. 3) In (IR 2, ) the ball B(0, 1) is the square of vertices (-1,-1), (1,-1) (1,1) and (-1,1). Draw a figure and explain. The closed ball is the same square with its boundary. 4) In (IR 2, 1 ), the ball B(0, 1) is the square of vertices (1,0), (0,-1) (-1,0) and (0,1). Draw a figure and explain. The closed ball is the same square with its boundary. 1.2 Open and closed sets and related concepts (neighborhood, interior, closure, boundary, convergence, density) Definition 1.2 Let X be a metric space. A subset O X is called open if for every x O there is an r > 0 such that B(x, r) O. Otherwise stated, a set is open if whenever it contains a point, it contains a whole ball around this point. This is why we call it open: if one member x can enter the set, some other members (the friends of x) can also enter the set. Examples. ]0, 1[ is open in IR but (0, 1] is not. Any interval of the form ]a, b[, ]a, [, ], a[ is open. Proposition 1.1 An open ball is an open set. Remark. This proposition is not a tautology because the adjective open has different meanings in open ball and open set. An open ball is a set of the form B(a, r) = {x d(x, a) < r}, whereas an open set is a set having the property that if a point belongs to this set, then a whole ball around this point is also in the set. So we have to prove that an open ball has this property. Proof. Let x B(a, r). Then d(x, a) < r. Set ρ = r d(x, a). We claim that B(x, ρ) B(a, r). Draw a figure. Indeed, let y B(x, ρ). Then d(x, y) < ρ = r d(x, a). Therefore d(x, a) + d(x, y) < r. By the triangle inequality, d(y, a) < r. This means that y B(a, r). The claim implies that B(a, r) is open. Proposition 1.2 Let X be a mteric space. Then the following properties hold. (i) X and are open. (ii) An arbitrary union of open sets is open. (iii) A finite intersection of open sets is open. Proof. (i) Let x X. Then indeed B(x, 1) X. Therefore X is open. The empty set is open because the condition x is never satsified. (ii) Let (O λ ) λ L be a collection of open subsets of X and let x O λ. Then x O µ for some µ L. Since O µ is open, there exists r > 0 such that B(x, r) O µ and so B(x, r) O λ. (iii) It is enough to prove that the intersection of two open sets is open because the result will follow by induction on the number of open sets. Let therefore O 1 and O 2 be open sets and let x O 1 O 2. Since O 1 is open and x O 1, there exists r 1 > 0 such that B(x, r 1 ) O 1. Similarly there exists r 2 > 0 such that B(x, r 2 ) O 2. It follows that B(x, r 1 ) B(x, r 2 ) O 1 O 2. But B(x, r 1 ) B(x, r 2 ) = B(x, r) where r = min(r 1, r 2 ). This means that O 1 O 2 is open.

11 1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS 11 Remark 1.1 An arbitrary intersection of open sets need not be open. For example, is not open. n 1 ] 1 n, 1 [ = {0} n Definition 1.3 A neighborhood of a point x in a metric space X is a set V which contains an open set containing x. The set of neighborhoods of a point x is usually denoted by U(x). Remark 1.2 Note that the whole space is a neighborhood of x, therefore U(x) is not empty. Note also that if V U(x) and V U then U U(x). Example. [ 1, 1] is a neighborhood of any point x ] 1, 1[. However it is not a neighborhood of 1 nor of 1. Proposition 1.3 A set is open if and only if it is a neighborhood of all of its points. Proof. Suppose first that A is open and let x A. Then A contains an open set (itself) containing x. This means that A is a neighborhood of x. Conversely, suppose that A is a neighborhood of all its points. This means that for every x A, there exists an open set O x such that x O x A. Then, we can write A = x A O x. Therefore A is open as a union of open sets. Remark 1.3 For some mathematicians, a neighborhood of a point is an open set containing that point. This makes a little difference because in practise we can always assume that a neighborhood is open. Definition 1.4 The interior of a set A is the union of all open sets that are contained in A. It is therefore the biggest open set contained in A. It is denoted by A or inta. Examples. a) int [0,1]=]0,1[. Indeed, ]0,1[ is open and contained in [0,1]. Therefore it is contained in the interior of [0,1]. So we have ]0, 1[ int[0, 1] [0, 1]. Therefore int[0, 1] is either ]0,1[, ]0,1] [0,1[ or [0,1]. But the only open set among these sets is ]0,1[. Hence the result. Similarly, int[0, 1[= int]0, 1] =]0, 1[. b) Q =. This is because any open interval contains points outside Q and so Q cannot contain any open interval and therefore cannot contain any nonempty open set. Similarly int(ir\q) =. c) Similarly, Z =. Proposition 1.4 A set is open if and only if it is equal to its interior. Proof. Suppose first A = A. But A is open. Therefore A is open. Conversely, suppose that A is open. But A contains A. Since A is the biggest open set containing A, We have A A. But A A. Hence the equality. Definition 1.5 Let X be a metric space. A subset F X is called closed if its complement is open.

12 12 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Examples. 1) [a, b] is closed because its complement ], a[ ]b, [ is open. 2) [a, [ and ], a] are closed. 3) In a metric space singletons are closed. Indeed, let {a} be a singleton (one point set) and let x X {a}. Then x a and so d(x, a) > 0. We claim that B(x, d(x, a)) X\{a}. Indeed, let y B(x, d(x, a)). Then d(x, y) < d(x, a). This implies that y a because otherwise we would have d(x, a) < d(x, a). This means that y X\{a}. Therefore etc. Proposition 1.5 A closed ball is a closed set. This proposition is not a tautology. Proof. Let B (a, r) be a closed ball in a metric space X. We have to prove that X\B (a, r) is open. Let x X\B (a, r). Then d(x, a) > r. Set ρ = d(x, a) r. We claim that B(x, ρ) X\B (a, r). Indeed, let y B(x, ρ). Then d(x, y) < ρ = d(x, a) r. Therefore r < d(x, a) d(x, y). By the triangle inequality, d(x, a) d(x, y) d(y, a). Hence d(y, a) > r and so y X\B (a, r). This proves the claim. The claim implies that X\B (a, r) is open. Proposition 1.6 In a metric space X, the following properties hold. (i) X and are closed. (ii) An arbitrary intersection of closed sets is closed. (iii) A finite union of closed sets is closed. Proof. Take complements and use the properties of open sets. Corollary 1.1 In a metric space every finite set is closed. Remark 1.4 An arbitrary union of closed sets need not be closed. For example for each n 2, the set [ 1 n, 1 1 n ] is closed. However is not closed. n=2 [ 1 n, 1 1 ] =]0, 1[ n Remark 1.5 A set which is not open is not necessarily closed. For example ]0,1] is neither open nor closed. Also a set can be both open and closed. Indeed, in any metric space X, and X are closed an open. Here is another example. Let d be the discrete distance on a set X containing more than one point. Then B(a, 1) = {a}. Hence {a} is open. But we know that {a} is also closed. In fact, every subset of X is open (because a subset is a union of singletons). But if every subset of X is open then every subset is also closed. A subset which is both closed and open is sometimes called clopen. Definition 1.6 The closure of a set A is the intersection of all closed sets that contain A. It is therefore the smallest closed set containing A. It is usually denoted by Ā and sometimes by cl(a). Examples. a) ]0, 1[ = [0, 1]. Indeed, [0,1] is a closed set containing ]0,1[, therefore it contains ]0, 1[. Thus ]0, 1[ ]0, 1[ [0, 1]. Therefore ]0, 1[ is either ]0,1[, ]0,1], [0,1[ or [0,1]. However, the only closed set among these four sets is [0,1]. Hence the result. Similarly, ]0, 1] = [0, 1[ = [0, 1]. b) We shall see below that Q = IR. We say that Q is dense in IR. We shall also prove that the irrationals are dense in IR. c) We shall see in the exercises that if A is the open disk x 2 + y 2 < 1, then the closure of A is the closed disk x 2 + y 2 1.

13 1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS 13 Proposition 1.7 A set is closed if and only if it is equal to its closure. Proof. Suppose first that A = Ā. Since Ā is closed, A is closed. Conversely, suppose that A is closed. Since A contains A and Ā is the smallest closed set containing A, we have Ā A. But A Ā. Hence the equality. In a metric space, we can define the notion of convergencce. A sequence (x n ) converges to x in a metric space, if x n can be made arbitrarily close to x for all n large enough. This idea of convergence is therefore captured by the following definition. Definition 1.7 Let (x n ) be a sequence in a metric space (X, d). We say that (x n ) converges to a point x if the following condition holds. ε > 0 n 0 IN n n 0 d(x n, x) < ε. If you studied calculus, this condition is equivalent to the condition that the sequence of real numbers (d(x n, x)) conveges to 0 in IR. Proposition 1.8 Let X be a metric space. Let A X and let x X. Then, the following conditions are equivalent. (i) x Ā. (ii) Every neighborhood of x intersects A. (iii) There is a sequence in A which converges to x. Proof. Let B denote the set of points t satisfying property (ii), i.e., satisfying that every neighborhood of t meets A. Proving that (i) (ii) is equivalent to proving that Ā = B So let us prove first that Ā B. We claim that B is closed. Let t / B, then there is an open neighborhood U of t not intersecting A. Consider an element y U. Since U is a neighborhood of y not intersecting A, we deduce that y / B. This means that U B and accordingly B is open. Therefore B is closed as claimed. Observe now that B contains A. Since Ā is the smallest closed set containing A, we get Ā B. Let us prove next that B Ā. Let t / Ā. Then t Ā. But Ā is an open set not intersecting A since Ā A A A =. This means that t / B. (ii) (iii). For every positive integer n the open ball B(x, 1 n ) intersects A. Choose accordingly for each n, an element x n in the intersection. This defines a sequence of points in A which clearly converges to x. (iii) (ii). Let U be a neighborhood of x. Since (x n ) converges to x, we have that x n U for n large enough. This means that U intersects A. Corollary 1.2 Let X be a metric space. Let A X and let (x n ) be sequence in A which converges to some x. Then x Ā. Definition 1.8 Let X be a metric space. A subset A X is called dense if Ā = X. Corollary 1.3 Let X be a metric space and A X. Then the following are equivalent (i) A is dense in X. (ii) Every nonempty open set of X meets A. (iii) For every x X, there exists a sequence in A that converges to x.

14 14 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Corollary 1.4 The set of rational numbers and the set of irrational numbers are dense in IR. Proof. Let O be an open nonempty subset of IR. Then O contains an open interval I. But any open inerval of IR contains rational and irrational numbers. Therefore O meets Q as well as IR\Q. Other examples. a) IR = IR\{0} is dense in IR. b) IR\Z is dense in IR. This is because Z has an empty interior. Definition 1.9 Let A be a subset of a metric space X. A point x X is called a limit point of A if every neighborhood of x meets A at a point different from x. The point x is called an isolated point of A if x A and x is not a limit point of A. It is clear that a limit point of A belongs to Ā and an isolated point of A belongs to A. So a point in Ā is either a limit point of A or is isolated in A. Examples. a) Let A =]0, 1[ {2}. Then any point in [0, 1] is a limit point of A, whereas 2 is an isolated point of A. b) Every finite set in a metric space consists of isolated points. Indeed, let A = {x 1,..., x n } be a finite set of n distinct points. For each i = 1,..., n,.let r i = min j i d(x i, x j ). Then B(x i, r i ) A = {x i }. c) Let A = { 1 n n = 1, 2, 3,...}. Then 0 is a limit point of A, and every point in A is isolated. The following proposition will be proved in the exercises. Proposition 1.9 Let A be a subset of a metric space X and let x be a limit point of A. Then every neighborhood of x contains infinitely many points of A. Definition 1.10 Let A be a set in a metric space X. The boundary of A denoted by A is the set A = Ā\ A. Examples. 1) [0, 1] = [0, 1]\]0, 1[= {0, 1}. Similarly, ]0, 1] = [0, 1[= {0, 1} 2) Q = Q\ Q = IR. 3) We shall see in the exercises that if A is the open disk x 2 + y 2 < 1, then A is the circle x 2 + y 2 = 1. Remark 1.6 The boundary of a set is always a closed set. Why? Remark 1.7 A = A is clopen. 1.3 Topological spaces Now that we have enough examples of metric spaces, we can move to the next level of abstraction. If you studied carefully the previous section and did the exercises, you observe that the majority of concepts and results in metric spaces rely only on the notion of an open set. Therefore we can extend the results and concepts of the previous to a still more general situation. The properties of open sets in a metric space were recorded in Proposition 1.2 which will be our starting point toward more generality.

15 1.3. TOPOLOGICAL SPACES 15 Definition and examples Definition 1.11 Let X be a set. A family T of subsets of of X is called a topology on X provided the following conditions hold. (i), X T. (ii) An arbitrary union of elements of T belongs to T (we say that T is stable under arbitray unions). (iii) A finite intersection of elements of T belongs to T (we say that T is stable under finite intersections). The elements of T are called open sets of X. A topological space is a couple (X, T ). Remarks. 1) If T is a topology on X, then T P(X). 2) To prove property (iii), it is enough to prove that the intersection of two elements of T is also an element of T because then the result follows by induction. 3) Suppose that T satisfies property (i). To prove (ii), we may assume that the sets are nonempty, because the empty sets do not change the union and so we can remove them. To prove (iii), we may assume that none of the sets is empty. Because otherwise the intersection would be empty and so belongs to T by (i). Examples. 1) As we observed, a metric space is a topological space. But the converse is not true as we shall see. The topology defined from a distance is called the topology generated or induced by a distance. Let (X, T ) be a topological space. We say that T is metrizable if T is generated by a distance. 2) Let X be a set. The collection {, X} is a topology called the indiscrete topology. This topology is not metrizable (see below). 3) Let X be a set. The collection P(X) of all subsets of X is a topology called the discrete topology. This topology is metrizable because it is generated by the discrete distance. Observe also that any topology on X, satisfies {, X} T P(X). The trivial topology is the smallest (coarsest) topology and the discrete topology is the biggest (finest) topology on a set. 4) Let X be a set and A X. Then {, A, X} is a topology on X. This topology is not metrizable. 5) Let X = {a, b, c} be a three point set. Then {, {a}, {a, b}, X} is a topology on X. This topology is not metrizable. 6) Let X be a set. Let T be the collection of all subsets of U X such that X\U is finite or equal to X. Then T is a topology on X called the finite complement topology. We prove this. (i) T because X\ = X and X T because X\X = is finite. (ii) Let (O α ) be a collection of nonempty elements of T. Then X\ α O α = α (X\O α ) This set is finite because it is an intersection of finite sets. (iii) Let O 1 and O 2 be nonempty elements of T. Then X\(O 1 O 2 ) = (X\O 1 ) (X\O 2 ) is finite as a union of two finite sets. 7) Let X be a set. Let T be the collection of all subsets of U X such that X\U is countable or equal to X. Then T is a topology on X called the countable complement topology. Prove this as an exercise. 8) Let X be a set and let a X. Let T be the collection of all subsets of U X that are either empty or contain a. Then T is a topology called the particular point topology.

16 16 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Closed sets and related concepts All the concepts that we learned in the previous section can be defined only in terms of open sets. A set is closed if its complement is open. A neighborhood of a point is a set containing an open set containing the point. The interior of a set is the biggest open set contained in the set. The closure of a set is the smallest closed set containing the set. The boundary of a set is the difference between the closure and the interior. Most of the results that we proved in the metric case still hold in a general topological space. To be more precise, Proposition 1.10 Propositions 1.3, 1.4, 1.6, 1.7 and the results in exercises 4,8,12,13,14,19,20 hold in an arbitrary topological space. Let us now define the concept of convergence in a general topological space. In a metric space, a sequence (x n ) is convergent to x if This is equivalent to ε > 0 n 0 IN n n 0 d(x n, x) < ε. ε > 0 n 0 IN n n 0 x n B(x, ε). Since a neighborhood of a point contains an open ball about this point, this condition is equivalent to the condition that very neighborhood U of x, contains x n for all n large enough. In terms of quantifiers U U(x) n 0 IN n n 0 x n U. Now this condition involves only neighborhoods, so it depends only on the topology of the space and we take it as our definition of convergence in an arbitrary topological space. Remark 1.8 In a general topological space, weird and unfamiliar situations can happen. For example a sequence can converge to many points. Consider for instance IR equipped with the finite complement topology and consider the sequence x n = n. Then (x n ) converges to every x IR. Indeed, let U be neighborhood of x. Then IR\U is finite. Choose n 0 > max(ir\u). Then x n = n U for all n n 0. This situation does not happen in metric spaces and more generally in Hausdorff spaces (see below). Proposition 1.8 and Corollary 1.3 remain partially true. More precisely, Proposition 1.11 Let X be a topological space and let A X. conditions. Consider the following (i) x Ā. (ii) Every neighborhood of x intersects A. (iii) There is a sequence in A which converges to x. Then (i) (ii) and (iii) (i). Remark 1.9 It is not true that (i) (iii) in a an arbitrary topological space. Here is an example. Equipp [0,1] with the countable complement topology. Let A =]0, 1]. Then 0 Ā because every neigborhood of 0 meets A (since {0} is not open, any neigborhood of zero contains a point different from 0). However no sequence of A converges to 0 in this topology because every convergent sequence is eventually constant (constant from a certain rank). Indeed, let x n a and let F = {x n x n a}. Then F is countable and so [0, 1]\F is a open neighborhood of a. Therefore x n [0, 1]\F starting from a certain rank. This means that starting from a certain rank, x n / F and so x n = a.

17 1.3. TOPOLOGICAL SPACES 17 Corollary 1.5 Let X be a topological space and A X. Consider the following conditions. (i) A is dense in X. (ii) Every nonempty open set of X meets A. (iii) For every x X, there exists a sequence in A that converges to x. Then (i) (ii) and (iii) (i). Finally Proposition 1.9 does not hold in an arbitrary topological space. In the exercises your are asked to give a counterexample. Hausdorff spaces Definition 1.12 A topological space X is called a Hausdorff space if every two distinct points of X can be separated by two disjoint open sets. This means that if x y then there exist two disjoint open sets U and V such that x U and y V. A Hausdorff space is also called a T 2 space. Proposition 1.12 A metrizable space is a Hausdorff space. Proof. Choose a distance d that generates the topology of the space. Let x y and let r = d(x, y)/2 > 0. Then the open balls B(x, r) and B(y, r) are disjoint open sets containing x and y respectively. Corollary 1.6 A space which is not Hausdorff is not metrizable. Now you can see why some topologies defined above are not metrizable. The finite complement topology on an infinite space X is not Hausdorff and so not metrizable. In fact there are no disjoint non empty open sets. For suppose that U V = with U, V. Then X\U and X\V are finite. Therefore X\U X\V is finite but X\U X\V = X\(U V ) = X. So X is finite contrary to our assumption. (The finite complement topology on a finite space coincides with the discrete topology so it is metrizable). There are however Hausdorff spaces which are not metrizable. You will encouter them in your graduate studies. Proposition 1.13 A finite set in a Hausdorff space is closed. Proof. It is enough to prove that singletons are closed. Consider a singletons {x}. Let y X {x}. We can separate x and y with disjoint sets U and V respectively. Then V X U X {x}. This means that X {x} is a neighborhood of y. Since y was arbitrary, this means that X {x} is a neighborhood of all its points and so it is open. Remark 1.10 A topological space in which singletons are closed is called a T 1 space. Therefore a T 2 space is a T 1 space. The converse is not true. For example, let X be an infinite set equipped with the finite complement topology. Then X is a T 1 space which is not T 2. Proposition 1.14 In a Hausdorff space, a sequence converges to at most one point. Proof. Suppose that there is a sequence (x n ) that converges to two distinct points x and y. Let U and V be two neighborhoods of x and y respectively. Then x n U for all n n 1 and x n V for all n n 2. Then x n U V for all n max(n 1, n 2 ). Contradiction.

18 18 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Comparison of topologies Definition 1.13 Let T and T be two topologies on a set X. If T T, we say that T is finer or stronger or just bigger than T ; we also say that T is coarser or weaker or just smaller than T. We say that T and T are comparable if either T T or T T. Examples. a) Let X = {a, b, c} and consider the following topologies T 1 = {, {a}, {a, b}, X}, T 2 = {, {a}, {b}, {a, b}, X} and T 3 = {, {a}, {c}, {a, c}, X}. Then T 2 is finer than T 1. However, T 1 and T 3 are not comparable. Also T 2 and T 3 are not comparable. b) The finite complement topology on IR is weaker than the usual topology. Suppose that we have two distances on the same set. topology, we can ask if these topologies are comparable. Since each distance generates a Definition 1.14 Let X be a set. Let d be distance on X that generates a topology T. Let d be distance on X that generates a topology T. We say that d and d are topologically equivalent if T = T, that is, if they generate the same topology. We give some examples. Proposition 1.15 Consider on IR N the three norms x 1 = ( N N ) 1/2 x i, x 2 = x i 2, x = max( x 1,..., x N ). i=1 i=1 Let d 1, d 2 and d be respectively the distances associated with these norms. Then d 1, d 2 and d are topologically equivalent Proof. We prove that d 2 and d are equivalent. In the exercises, you are asked to prove that d 1 and d are equivalent. It is not difficult to see that x x 2 N x. It follows that Therefore, d (x, y) d 2 (x, y) Nd (x, y). B d2 (x, r) B d (x, r) and B d (x, r N ) B d2 (x, r). Now let O be an open set for the topology generated by d and let x O. Then there exists r > 0 such that B d (x, r) O. But B d2 (x, r) B d (x, r) and so B d2 (x, r) O. This means that O is open for the topology generated by d 2. Conversely, let O be an open set for the topology generated by d 2 and let x O. Then there r exists r > 0 such that B d2 (x, r) O. But B d (x, r N ) B d2 (x, r) and so B d (x, N ) O. This means that O is open for the topology generated by d. Remark 1.11 Let d and d be two two distances on the same set X. Then d generates a finer topology than d if and only if the following condition holds: x X r > 0 δ > 0 B d (x, δ) B d (x, r). Therefore d and d are topologically equivalent if and only the following two conditions hold x X r > 0 δ > 0 B d (x, δ) B d (x, r) x X r > 0 δ > 0 B d (x, δ) B d (x, r).

19 1.3. TOPOLOGICAL SPACES 19 Proposition 1.16 Let (X, d) be a metric space. Set ρ(x, y) = min(1, d(x, y)). Then ρ and d are topologically equivalent. Proof. We know that ρ is a metric. Let x X. Observe that B d (x, r) B ρ (x, r) since ρ(x, y) d(x, y). By the previous remark, d generates a finer topology than ρ. Conversely, let x X and r > 0 be given. Let δ = min(1, r). Then B ρ (x, δ) B d (x, r). Indeed, let y B ρ (x, δ). Then ρ(x, y) < δ 1. It follows that ρ(x, y) = d(x, y) (because ρ(x, y) 1) and so d(x, y) < δ r; that is, y B d (x, r). Basis for a topology Recall that in a metric space, we defined open sets in terms of balls. Now the collection of balls is not a topology but it generates a topology. We will give now a more general definition. Definition 1.15 A subset B P(X) is called a basis provided the following hold. 1. For all x X, there exists B B such that x B. 2. If B 1, B 2 B and x B 1 B 2, then there exists B 3 B such that x B 3 B 1 B 2. The elements of B are called basis elements. The topology T generated by B is defined as follows: O T if whenever x O, there exists B B such that x B O. We need to check that T defined in this way is indeed a topology. 1. The empty set satisfies the condition vacuously. The first condition of a basis implies that X T. 2. Let (O α ) α A be a collection of elements of T. Let x O α. Then there is β A such that x O β. Since O β T, there exists B B such that x B O β and so B O α. This means that O α T. 3. Let O 1, O 2 T and let x O 1 O 2. Then there exists B 1, B 2 B such that x B 1 O 1 and x B 2 O 2. The second condition of a basis implies that there exists B 3 B such that x B 3 B 1 B 2. Therefore x B 3 B 1 B 2 O 1 O 2. Remark 1.12 Observe that B T. Examples. 1) Let X be a metric space. The collection of balls B(x, r) (x X, r > 0) is a basis for the metric topology. In particular, the collection of open disks and the collection of open squares are bases for the usual topology of IR 2. The first condition of a basis is satisfied. Let now x B(a 1, r 1 ) B(a 2, r 2 ). Let r = min(r 1 d(x, a 1 ), r 2 d(x, a 2 )). Then B(x, r) B(a 1, r 1 ) B(a 2, r 2 ). 2) Let X be a set. The collection of singletons of X is basis for the discrete topology on X. 3) The collection of all open intervals ]a, b[ of IR is a basis for the usual topology of IR ( indeed the collection of these intervals coincide with the collection of balls ]x r, x + r[ which generate the usual topology of IR). 4) Let X and Y be two topological spaces. Let B be the collection of all sets of the from U V where U is open in X and V is open in Y. Then B is a basis. The topology generated by this basis is called the product topology of X Y. Proposition 1.17 Let B be a basis for a topology T on a set. Then T equals the collection of all unions of elements of B. Proof. Since B T and T is stable under unions, a union of elements of B belongs to T. Conversely, given O T, for each x O, there exists B x B such that x B x O. Then O = x B x.

20 20 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Accumulation points and subsequences Recall that a sequence (x n ) of a topological space is convergent to a limit x if every neighborhood of x contains x n for all n large enough, that is except for a finite set of indices. Not all sequences are convergent. A weaker notion than that of limit is that of accumulation point or cluster point. Definition 1.16 Let (x n ) be a sequence of a topological space. We say that x is an accumulation point or a cluster point of (x n ) if every neighborhood of x contains x n for infinitely many indices n. In symbols U U (x) n IN m n such that x m U. This is equivalent to saying that the set {n IN x n U} is infinite. Compare with the condition of convergence U U (x) n 0 IN m n 0 x m U. Examples. 1) It should be clear that a limit of a convergent sequence is an accumulation point of the sequence. The converse however is not true as testified by the next example. 2) Let x n = ( 1) n. Then (x n ) is not convergent in IR but has two accumulation points 1 and 1. 3) The sequence (1, 2, , 2 1 2, , 2 1 3,...) has two accumulation points 1 and 2. Proposition 1.18 Let A be the set of accumulation points of a sequence (x n ). Then A = {x k k n} = n=1 {x n, x n+1,...}. n=1 Proof. Let F n = {x n, x n+1,...}. Then x A U U (x) n IN m n x m U n IN U U (x) U F n n IN x F n x F n. Definition 1.17 Let (x n ) be a sequence of some set and let n 1 < n 2 < < n i < be an increasing sequence of positive integers. Then the sequence (y i ) defined by y i = x ni is called a subsequence of the sequence (x n ). Examples. a) Let x n = n. Then (x 2n+1 ) is a subsequence of (x n ). b) Let x n = 2 n. Then the sequence 1, 4, 16, 64,..., i.e. (x 2n ) is a subsequence of (x n ). Proposition 1.19 Let (x n ) be a sequence of a topological space X. (i) If (x n ) has a subsequence converging to l, then l is an accumulation point of (x n ). (ii) Conversely, if X is metrizable and l is an accumulation of (x n ) then there exists a subsequence of (x n ) that converges to l.

21 1.4. SUBSPACES 21 Proof. (i). Let (x nk ) be a subsequence of (x n ) that converges to l. Let U be a neighborhood of l. Then, there exists k 0 such that x nk U for all k k 0. This means that x m U for all m {n k0, n k0 +1, n k } and this set of integers is infinite. (ii). In the condition U U (x) n IN m n such that x m U, take first U = B(l, 1) and n = 1. Then there exists n 1 1 such that x n1 B(l, 1). Next take U = B(l, 1 2 ) and n = n Then there exists n 2 n such that x n2 B(l, 1 2 ). At the kth step, take U = B(l, 1 k ) and n = n k Then there exists n k n k such that x nk B(l, 1 k ). This procedure defines a subsequence (x n k ) that converges to l. 1.4 Subspaces Proposition 1.20 Let (X, T ) be a topological space and let Y X. The collection T Y = {Y O O T } is a topology on Y called the subspace topology. Therefore, in the subspace topology, the open sets of Y are the intersection of Y with the open subsets of X. Proof. (i) Observe that = Y and Y = Y X. It follows that, Y T Y. (ii) Let (O α ) be a collection of elements of T Y. Then there exists a collection (U α ) of elements of T such that O α = Y U α. Then O α = (U α Y ) = ( U α ) Y. Since U α T, it follows that O α T Y. (iii) Let O 1, O 2 T Y. Then there exists U 1, U 2 T such that O 1 = Y U 1 and O 2 = Y U 2. Then O 1 O 2 = Y (U 1 U 2 ) T Y. Examples. 1) Consider [0, [ as a subspace of IR with the usual topology. Then although [0, 1[ is not open in IR, it is open in [0, [ because for example [0, 1[= [0, [ ] 1, 1[. 2) Consider Z as a subspace of IR with the usual topology. Then each singleton {n} is open in Z because {n} =]n 1, n + 1[ Z. It follows that that every subset of Z is open in the subspace topology. This implies that the subspace topology and the discrete topology on Z coincide. Corollary 1.7 Let X a be a topological space, let Y be a subspace of X and let U Y. If U is open in Y and Y is open in X, then U is open in X. Proof. If U is open in Y, then U = Y V where V is open in X. Since Y is also open in X, then so is Y V. Remark 1.13 Let X be a topological space and let Z Y X. The the topology on Z inherited from X is the same as the topology on Z inhereted from Y. Indeed if O is open in X, then O Z = (O Y ) Z. Proposition 1.21 Let Y be a subspace of a topological space X and let A Y then A is closed in Y if and only if A = Y F where F is closed in X.

22 22 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES Proof. We have to show first that if F is closed in X, then Y F is closed in Y, that is, its complement in Y is open in Y. Now Y \(Y F ) = Y (X\F ). This set is open in Y because X F is open in X. Now conversely, suppose that A is closed in Y. Then Y A is open in Y and so Y \A = Y O where O is open in X. Then A = Y (X\O). But X O is closed in X. Hence the conclusion. Example. ]0,1] is closed in ]0, [ since ]0, 1] =]0, [ [0, 1]. Corollary 1.8 Let X a be a topological space, let Y be a subspace of X and let B Y. If B is closed in Y and Y is closed in X, then B is closed in X. Let (X, d) be a metric space and let Y X. Let d Y denote the restriction of d to Y Y. Then d Y is a metric on Y which therefore generates a topology on Y. How does this topology relate to the subspace topology? It turns out that the two topologies coincide. If x Y, we denote by B Y (x, r) the ball in Y of center and radius r. It coincides with Y B(x, r). Proposition 1.22 The metric topology induced on Y coincides with the subspace topology of Y. Proof. Let A be open relative to the subspace topology. Then, A = Y O where O is open in X. Let x A. Then x O. Since O is open there exists δ > 0 such that B(x, δ) O. Then B(x, δ) Y O Y = A, that is, B Y (x, δ) A. This means that A is open relative to the metric topology on Y. Conversely, Suppose that A is open relative to the metric topology of Y. For each x A, set δ x = d Y (x, Y A). Then δ x > 0 because d Y (x, Y A) = 0 x Y A = Y A since Y A is closed in Y. We claim that if y Y and d(x, y) < δ x then y A. For if y / A, then d(x, y) δ x by definition of δ x. It follows that A = x A B Y (x, δ x ). Let now O = x A B(x, δ x ). Then O is open in X and O Y = A. 1.5 Product spaces If X 1, X 2..., X n are sets, the set of n tuples (x 1,..., x n ) where each x i X i is denoted by X 1 X 2 X n or by n i=1 X i. Suppose now that X 1,..., X n are topological space. If U i is an open set of X i for each i = 1,..., n, their product U 1 U 2 U n will be called an open rectangle. Now, it is easy to check that the collection B of all open rectangles is a basis (indeed, first, any x n i=1 X i is contained in the open rectangle n i=1 X i; second, the intersection of two open rectangles is an open retangle). The topology generated by this basis is called the product topology of X 1 X 2 X n. This means that a set O X 1 X 2 X n is open if whenever x = (x 1,..., x n ) O, there exist open sets U i X i such that x n i=1 U i O. Proposition 1.23 The product of open sets is open. The product of closed sets is closed. Proof. The first part is rather trivial because a product of open sets is a basis element and therefore belongs to the product topology. For the second part, it is enough to prove that the product of two closed sets is closed (then the result follows by induction). Let A X 1 and B X 2 be closed sets. Then X 1 X 2 A B = (X 1 A 1 ) X 2 X1 (X 2 B) which is the union of two open sets. In the exercises, you are asked to prove the following proposition.

23 1.5. PRODUCT SPACES 23 Proposition 1.24 Let (X 1, d 1 ),..., (X n, d n ) be n metric spaces. Let X = X 1 X n. For x = (x 1,..., x n ) and y = (y 1,..., y n ), define ρ 1 (x, y) = n d i (x i, y i ) i=1 ( n ) 1/2 ρ 2 (x, y) = d i (x i, y i ) 2 i=1 ρ (x, y) = max i=1...,n d i(x i, y i ). Then ρ 1, ρ 2 and ρ are metrics on X that generate the product topology on X.

24 24 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES

25 Chapter 2 Continuous functions and homeomorphisms Let us start by defining continuity in metric spaces. Let f : (X, d X ) (Y, d Y ) be a function between two metric space and let x X. We say that f is continuous at x if f(y) can be made arbitrarily close to f(x) for y sufficiently close to x. In metric spaces the concept of a distance gives a meaning to closeness. This leads to the following definition. We can think of ε as the accuracy in the approximation. Definition 2.1 Let f : (X, d X ) (Y, d Y ) be a function between two metric spaces and let x X. f is said to be continuous at x if the following condition holds. ε > 0 δ > 0 such that d X (x, y) < δ d Y (f(x), f(y)) < ε. Remark 2.1 In the definition above, one can replace < by. Examples. 1) A constant function is continuous. 2) The identity function is continuous. 3) Let (X, d) be a metric space and a X. The function x d(x, a) is continuous at every point because d(x, a) d(y, a) d(x, y). More generally, if A X, then the function x d(x, A) is continuous at every point because d(x, A) d(y, A) d(x, y). 4) All the elementary functions of calculus are continuous on their domains. The elementary functions are the rational functions (ratio of polynomials), the logarithm, the exponential, the trigonometric and inverse trigonometric functions. Now there are several equivalent characterizations of continuity and we shall bring out a condition which is independent of the metrics so that it will serve as a definition of continuity in general topological spaces. Proposition 2.1 Let f : (X, d X ) (Y, d Y ) be a function between two metric spaces and let x X. Then the following are equivalent. (i) f is continuous at x. (ii) ε > 0, δ > 0 such that f(b(x, δ)) B(f(x), ε). (iii) V U(f(x)), U U(x) such that f(u) V. (iv) If {x n } is a sequence of X that converges to x then {f(x n )} converges to f(x). 25

26 26 CHAPTER 2. CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS (v) If {x n } is a sequence of X that converges to x then there exists a subsequence (x nk ) of (x n ) such that {f(x nk )} converges to f(x). Proof. We prove that (i) (ii) (iii) (iv) (v) (i). (i) (ii). (ii) is just a reformulation of (i). Indeed, let ε and δ be as in the definition of continuity. Let y f(b(x, δ)). Then y = f(z) for some z B(x, δ). This means that d(z, x) < δ and so by the condition of continuity, d(f(z), f(x)) < ε, that is, d(y, f(x)) < ε. This means that y B(f(x), ε). (ii) (iii). Let U be a neighborhood of f(x). Then there exists a ball B(f(x), ε) V. Take now U = B(x, δ). Then U is a neighborhood of x. Now (ii) can be written as f(u) B(f(x), ε) V. (iii) (iv). Let (x n ) be a sequence of X that converges to x. Let V be a neighborhood of f(x). Then there exists a neighborhood U of x such that f(u) V. Now convergence means that x n U for all n large enough. Therefore f(x n ) f(u) V for n large enough. This means that (f(x n )) converges to f(x). (iv) (v). Is clear. (v) (i). Suppose that f is not continuous at x. Then there exists ε > 0 such that for all δ > 0 there exists y with d(y, x) < δ and d(f(y), f(x)) ε. Taking δ = 1 n we get a sequence (y n) converging to x and such that no subsequence of (f(y n )) converges to f(x), a contradiction. The only condition which does not involve metrics is condition (iii) and this will be our definition of continuity in a general topological space. Definition 2.2 A function f : X Y between two topological space is said to be continuous at x X, if for every neighborhood V of f(x), there exists a neighborhood U of x such that f(u) V. Remark 2.2 We can replace neighborhood by open neighborhood. Remark 2.3 If f is continuous at x and (x n ) is a sequence converging to x, then the sequence (f(x n )) converges to f(x). This follows from the definitions. The converse is however not true, i.e. if for every sequence (x n ) converging to x we have (f(x n )) converges to f(x), it does not follow that f is continuous at x. The converse is true if X is metrizable. Definition 2.3 A function f : X Y between two topological spaces is called continuous if it is continuous at every point of X. Theorem 2.1 Let f : X Y be a function between two topological spaces. Then the following are equivalent. (i) f is continuous. (ii) The inverse image under f of any open subset of Y is open in X. (iii) The inverse image under f of any closed subset of Y is closed in X. Proof. (i) (ii). Let O Y be open and let x f 1 (O) (so that f(x) O). Since O is a neighborhood of f(x), there exists according to the previous proposition, a neighborhood U of x such that f(u) O. Therefore U f 1 (f(u)) f 1 (O). This means that f 1 (O) is open. (ii) (i). Let x X and let V be an open neighborhood of f(x). Set U = f 1 (V ). Then U is a neighborhood of x (being an open set containing x) and f(u) = f(f 1 (V )) V. According to the previous proposition, f is a continuous at x. Since x was arbitrary, this proves that f is continuous. The equivalence (ii) (iii) follows from the fact that f 1 (Y \O) = X\f 1 (O).