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Contents 1 Introduction and Set Theory 1 1.1 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Restrictions and Compositions of Functions . . . . . . . . . . . . 5 1.3 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2 Topological Spaces 11 2.1 Bases of Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.2 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.3 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . 20 3 New Spaces from Old 25 3.1 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.2 Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.3 Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.4 Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4 Separation Axioms 37 4.1 T0 , T1 , and T2 Axioms . . . . . . . . . . . . . . . . . . . . . . . . 37 4.2 Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.3 Complete Regularity . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.4 Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.5 Lindelöf Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 i ii CONTENTS 5 Convergence and Metric Spaces 51 5.1 First Countable Spaces . . . . . . . . . . . . . . . . . . . . . . . . 51 5.2 Convergence of Nets . . . . . . . . . . . . . . . . . . . . . . . . . 53 5.3 Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . 57 6 Compactness 63 6.1 Local Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . 68 6.2 Compactifications . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 6.3 Paracompactness . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 7 Connectedness 75 7.1 Basic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 7.2 Components and Local Connectedness . . . . . . . . . . . . . . . 78 7.3 Path Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 81 8 Homotopy And Fundamental Groups 85 8.1 Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . . 85 8.2 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 0 CONTENTS Chapter 1 Introduction and Set Theory This chapter should be mostly review, although the particular presentation may be new in many cases. While we do not do a thorough axiomatic treatment of sets, we will try to point out where naive set theory has difficulties and hint at how to solve them. A set is a collection of mathematical objects. Sets, in turn, are mathematical objects. Hence sets can have other sets as members. When X is a set and x is anything, we write x ∈ X to denote the fact that x is a member (also called an element) of X. The negation of this is written x 6∈ X. Sets are typically defined by giving some property that every element of the set satisfies and that nothing outside of the set satisfies. Hence, if P (x) is any property of and object x (in other words, P (x) can be either true or false for each x), then the set, A, of those things which satisfy property P (x) is written A = {x : P (x)}. In this case, x ∈ A if and only if P (x) is true. Another way of defining a set is simply by listing the elements in the set. Hence, with A = {a, b}, x ∈ A if and only if x = a or x = b. Notice that a = b is allowed in this. Similarly, with A = {a1 , a2 , · · · , an }, x ∈ A if and only if x = a1 or x = a2 or · · · or x = an . If X and Y are sets, we say that X is a subset of Y and write X ⊆ Y if 1 2 CHAPTER 1. INTRODUCTION AND SET THEORY x ∈ X implies x ∈ Y for all objects x. If X ⊆ Y and X 6= Y , we write X ( Y and say that X is a proper subset of Y . A particularly important set is the empty set ∅ = {x : x 6= x}. It is always the case that ∅ ⊆ X for any set X. We also have the intersection X ∩ Y = {x : x ∈ X and x ∈ Y } and the union X ∪ Y = {x : x ∈ X or x ∈ Y }. Also, we define X \ Y = {x ∈ X : x 6∈ Y }, the set difference. S More generally, if X is a set, we can define the union X = {x : x ∈ T A for some A ∈ X} and the intersection X = {x : x ∈ A for all A ∈ X}. This is typically done when X is a collection of sets, but the definitions work even in the general case. It turns out that inconsistencies arise unless some restriction is placed on those properties P (x) which can be used to define sets. In particular, forming the Russell ’set’ R = {x : x 6∈ x} leads quickly to a contradiction: is R ∈ R or is R 6∈ R? This is avoided by not allowing the construction of very ’large’. In spite of this, there are some important sets that are allowed. For example, if S X is a set, the power set P(X) = {A : A ⊆ X} is always allowed. Also, X T is allowed no matter what set X is, but X is only allowed if X is non-empty (why?). Whenever x and y are objects, we can define the ordered pair (x, y) = {{x}, {x, y}}. This has the property that (x, y) = (z, w) if and only if x = z and y = w. In fact, this is the only important property of ordered pairs from our point of view. This can be generalized by setting, say (x, y, z) = (x, (y, z)), so (x1 , y1 , z1 ) = (x2 , y2 , z2 ) if and only if x1 = x2 , y1 = y2 , and z1 = z2 . Clearly, this can be generalized to any n−tuple. If X and Y are sets, we define the cross product set X × Y = {(x, y) : x ∈ X, y ∈ Y }. In a similar way, we can define X × Y × Z by using ordered triples. Some important sets: 1. The set N of natural numbers, N = {1, 2, 3, · · · }. 2. The set R of real numbers. 3. The set Rn of n−tuples of real numbers. 3 4. The set Q of rational numbers. n+1 5. The set S n = {x ∈ R : kxk = 1}. the n−sphere. A function from X to Y is now a subset f ⊆ X × Y such that a) for each x ∈ X, there is a y ∈ Y with (x, y) ∈ f and b) if (x, y1 ), (x, y2 ) ∈ f , then y1 = y2 . When we think of functions actvely, we say that x ∈ X ’goes to’ y ∈ Y if (x, y) ∈ f . To be a function, then, each x ∈ X has to ’go to’ something in Y and only one such thing in Y . If (x, y) ∈ f , we write y = f (x). We write f : X → Y when f is a function from X to Y . The set {y ∈ Y : there is an x ∈ X with y = f (x) is called the image of f and denoted by im f . Similarly, we say that X is the domain of f . The collection of all functions from X to Y is a set denoted by Y X = {f |f : X → Y }. Now let I be a set and suppose there is a function whose domain is I that assigns each α ∈ I a non-empty set Aα . We call {Aα }α∈I an indexed family of sets. In this case, we define the union and intersections [ Aα = {x : x ∈ Aα for some α ∈ I}, α∈I \ Aα = {x : x ∈ Aα for all α ∈ I}. α∈I These are special cases of the union and intersection defined above. Theorem 1.1 Let {Aα }α∈I be an indexed family of subsets of X and suppose that A ⊆ X. Then a) A \ S α∈I Aα = T b) A \ T α∈I Aα = S c) A ∪ S α∈I Aα = S d) A ∩ T α∈I Aα = T α∈I A \ Aα , α∈I A \ Aα , α∈I A ∪ Aα , α∈I A ∩ Aα , Definition 1.2 Let f : X → Y and A ⊆ X and B ⊆ Y . Define a) f [A] = {y ∈ Y : there exists x ∈ X such that y = f (x)} 4 CHAPTER 1. INTRODUCTION AND SET THEORY b) f −1 [B] = {x ∈ X : f (x) ∈ B} Definition 1.3 Let f : X → Y . We say that f is a) one-to-one (or injective) if x1 , x2 ∈ X with f (x1 ) = f (x2 ) implies that x1 = x2 . b) onto (or surjective) if for each y ∈ Y , there is an x ∈ X with f (x) = y. c) a one-to-one correspondence (or bijective ) if f is both one-to-one and onto. Notice that f is onto exactly when im f = Y . If f is bijective, then there is an inverse function f −1 : Y → X defined by setting f −1 (y) = x exactly when f (x) = y. For any set, there is an identity function 1X : X → X defined by 1X (x) = x. This is clearly a one-to-one correspondence from X to itself. 1.1 Cardinality If X and Y are sets, we say that X and Y are equivalent (in cardinality) if there is a one-to-one correspondence from X to Y . We think of X and Y as being the ’same size’ when this happens. In particular, a set X is said to be countably infinite if X is equivalent to N, the set of natural numbers. In particular, Q is countably infinite. We say that X is countable if it is either finite or countably infinite. Any subset of a countable set is countable. Also, a countable union of countable sets is again countable. Finally, X is said to be uncountable if X is not countable. In particular, R is an uncountable set. Theorem 1.4 (Schröder-Bernstein) Suppose that f : X → Y and g : Y → X are one-to-one functions. Then, there exists a one-to-one correspondence between X and Y . Proof: Let C0 = X \g[Y ] and Cn+1 = g ◦f [Cn ]. Then let C = S∞ n=0 Cn . Notice that if x 6∈ C, then x ∈ g[Y ]. Since g is one-to-one, there is then a unique y ∈ Y such that g(y) = x. 1.2. RESTRICTIONS AND COMPOSITIONS OF FUNCTIONS 5 Define h : X → Y by h(x) = f (x) if x ∈ C and h(x) = g −1 (x) if x 6∈ C. By the last paragraph, h IS well-defined. We claim that h is one-to-one and onto. Suppose that x1 , x2 ∈ with h(x1 ) = h(x2 ). There are three cases: if x1 , x2 ∈ C, then f (x1 ) = f (x2 ), so x1 = x2 since f is one-to-one. If x1 , x2 6∈ C, then g −1 (x1 ) = g −1 (x2 ), so applying g gives x1 = x2 . Finally, if x1 ∈ C and x2 6∈ C (the other case is the same), f (x1 ) = g −1 (x2 ), so g ◦ f (x1 ) = x2 . But, if x1 ∈ C, then x1 ∈ Cn for some n and so g ◦ f (x1 ) ∈ Cn+1 , so x2 ∈ Cn+1 ⊆ C, a contradiction. Thus h is one-to-one. Next, if y ∈ Y , there are two cases. If y ∈ f [C], then there is an x ∈ C with h(x) = f (x) = y. Otherwise, let x = g(y). Since x ∈ g[Y ], x 6∈ C0 . But, now, if x ∈ Cn+1 for some n ≥ 0, there is a z ∈ Cn with g ◦ f (z) = x = g(y). Since g is one-to-one, y = f (z) ∈ f [C], a contradiction. Hence x 6∈ ∪∞ n=0 Cn = C, so h(x) = g −1 (x) = y. Hence h is onto. 1.2 Restrictions and Compositions of Functions Theorem 1.5 If f : X → Y , the following hold: S S a) If {Aα }α∈I is any indexed family of subsets of X, then f [ Aα ] = f [Aα ]. T T b) If {Aα }α∈I is any indexed family of subsets of X, then f [ Aα ] ⊆ f [Aα ]. In this case, equality may not hold. S c) If {Bα }α∈I is any indexed family of subsets of Y , then f −1 [ Bα ] = S −1 f [Bα ]. T d) If {Bα }α∈I is any indexed family of subsets of Y , then f −1 [ Bα ] = T −1 f [Bα ]. Now, suppose that f : X → Y and g : Y → Z are functions where the codomain of f is the same as the domain of g. We can then define the composition g ◦ f : X → Z by setting g ◦ f (x) = g(f (x)). It is easy to see that (g ◦ f )−1 [C] = f −1 [g −1 [C]]. Next, if f : X → Y and A ⊆ X, then we can define the restriction of f to A, f |A : A → Y by letting f |A (x) = f (x) for x ∈ A. The domain of this 6 CHAPTER 1. INTRODUCTION AND SET THEORY restriction is then A and the image is now f [A]. This is actually a special case of a composition, as we can see as follows: Define the inclusion map iA : A → X by iA (x) = x. Then f |A = f ◦ iA . In the case when f : X → Y is a one-to-one correspondence and f −1 is the inverse function, we have f ◦ f −1 = 1Y and f −1 ◦ f = 1X . 1.3 Products Given two sets X and Y , there are natural projections πX : X × Y → X and πY : X × Y → Y defined by πX (x, y) = x and πY (x, y) = y. We will generalize the idea of products so that infinitely many factors will be allowed. If {Aα } is an indexed family of sets, we can define the product set via Y α∈I Aα = {x : A → [ Aα |x(α) ∈ Aα for all α ∈ I}. α∈I An element of the product is called a choice function. It is quite common to write xα instead of x(α) and call xα the α coordinate of x. Y For each α0 ∈ I, there is a natural projection πα0 : Aα → Aα0 defined by πα0 (x) = x(α0 ). This generalizes the projections defined above for finite products. The Axiom of Choice states that if {Aα }α∈I is an indexed family of nonY empty sets, then the product Aα is non-empty. We will assume this axiom and use it without comment when needed. The most important property of products is the following: Theorem 1.6 Suppose that {Aα }α∈I is an indexed family of sets and that X is any set. Suppose also that there is an indexed family of functions fα : X → Aα . Y Then there is a unique function f : X → Aα such that fα = πα ◦ f for all α ∈ I. Proof: For x ∈ X and α ∈ I, define f (x)(α) = fα (x). It is easy to verify both that this works and that it is unique. 1.4. RELATIONS 1.4 7 Relations Let X be a set. A relation on X is a subset, R, of X × X. We generally write xRy in place of (x, y) ∈ R when dealing with relations. An equivalence relation on a set X is a relation R satisfying the following properties: a) For all x ∈ X, xRx (Reflexivity) b) Whenever xRy, we have yRx (Symmetry) c) If xRy and yRz, then xRz (Transitivity) If R is an equivalence relation on X and x ∈ X, we define the equivalence class of x by [x] = {y ∈ X : xRy}. It is then easy to show that [x] = [y] if and only if xRy and that [x] ∩ [y] = ∅ when xRy fails. The collection of all equivalence classes is denoted by X/R = {[x] : x ∈ X}. There is a natural quotient map qR : X → X/R defined by qR (x) = [x]. This map is always onto. We have the following property of such quotients: Theorem 1.7 Suppose that R is an equivalence relation on X and f : X → Y is such that x1 Rx2 implies that f (x1 ) = f (x2 ). Then, there is a unique function fR : X/R → Y such that f = fR ◦ qR . Proof: For x ∈ X, define fR to be the collection of ordered pairs fR = {([x], f (x)) : x ∈ X}. Then, if [x1 ] = [x2 ], we have that x1 Rx2 , so f (x1 ) = f (x2 ), so fR is a function. It is clear that dom f = X/R and that f = fR ◦ qR . Definition 1.8 A partition of a set X is a family P ⊆ P(X) such that a) S P =X b) If A, B ∈ P and A 6= B, then A ∩ B = ∅. From above, if R is an equivalence relation on X, the collection of equivalence classes X/R is a partition of X. Conversely, if P is a partition of X, we define xRy if there is an A ∈ P with x, y ∈ A. This gives an equivalence relation whose equivalence classes are exactly the elements of P . 8 CHAPTER 1. INTRODUCTION AND SET THEORY Another important type of relation is that of a partial order. For this, we usually use the symbol ≤ to denote the relation and we assume the following properties: a) For each x ∈ X, x ≤ x. b) If x ≤ y and y ≤ x, then x = y. c) If x ≤ y and y ≤ z, then x ≤ z. So a partial order is symmetric, transitive, but anti-symmetric. There are many types of partial orders, most of which we will not have occasion to investigate. We say that x0 ∈ X is a maximal element if y ∈ X with x0 ≤ y implies that y = x0 . If A ⊆ X, we say that x0 ∈ X is an upper bound for A if y ≤ x0 for all y ∈ A. Notice that it can happen that x0 is a maximal element of X while not being an upper bound for X. An upper bound for X is also called a largest element for X. Of course, similar definitions are made for the concepts of minimal elements, lower bounds, and smallest elements. But we will be interested in two types: linear orders and well-orderings. We say that a partial order ≤ is a linear order if, for each x, y ∈ X, we have x ≤ y or y ≤ x. In other words, every two elements of X are comparable. The usual order on the real line is a linear order. If X is a partially ordered set, we say that C ⊆ X is a chain if C is linearly ordered via ≤. In particular, every subset of a linearly ordered set is a chain. A well-ordering on a set X is a partial order ≤ with the property that every non-empty subset A ⊆ X has a smallest element in A. In particular, using A = {x, y}, we see that well ordered sets are linearly ordered. The usual order on the natural numbers, N, is a well-ordering, but that on the real line is NOT a well-ordering: the set (0, 1) does not have a smallest element. The following result gives two important equivalences of the Axiom of Choice. Both are heavily used in this course. Theorem 1.9 The following are equivalenct to the Axiom of Choice: a) Every set can be given a well-ordering. (The Well-Ordering Principle). 1.4. RELATIONS 9 b) Suppose that X is a partially ordered set. Suppose that every chain C in X has an upper bound in X. Then X has a maximal element. (Zorn’s Lemma). Theorem 1.10 There exists an uncountable well-ordered set (Ω0 , ≤) with the property that {x : x ≤ x0 } is countable for each x0 ∈ Ω0 . Proof: Let ≤ be a well-ordering of any uncountable set X. There are two possibilities: either the set {x : x ≤ y} is countable for each y ∈ X or there is some y where this set is uncountable. In the first case, let Ω0 = X. Otherwise, since ≤ is a well-ordering, in the second case, there is a smallest y with {x : x ≤ y} uncountable. Let Ω0 = {x : x < y}. Then Ω0 is a wellordered, uncountable set with the property that {x : x ≤ x0 } is countable for every x0 ∈ Ω0 . We call Ω0 the first uncountable ordinal. Notice that Ω0 does not have a largest element. We can extend it, however, by picking any ω1 6∈ Ω0 and setting Ω = Ω0 ∪ {ω1 } with the order extended by defining y ≤ ω1 for all y ∈ Ω0 . Then, Ω is a well-ordered, uncountable set with a largest element. The crucial propert of Ω0 is the following: Theorem 1.11 Suppose that A ⊆ Ω0 is a countable subset. Then A has an upper bound in Ω0 . Proof: For each x ∈ A, the set Ix = {y ∈ Ω0 : y ≤ x} is countable. Since A is S countable, so is x∈A Ix . Since Ω0 is uncountable, there is some y that is not in this union. Let y0 be the least such y. Then, for each x ∈ A, we must have x < y0 , so y0 is an upper bound for A. Exercises: 1. Let A = {a, b, c}. List all elements of P(A). 2. What happens with Y X when X = ∅? When Y = ∅? 3. Let X = {a, b, c} and Y = {x, y}. How many elements does Y X have? How about X Y ? 10 CHAPTER 1. INTRODUCTION AND SET THEORY 4. Let X be any set. Show that X {0,1} and X × X are equivalent via the function f (x) = (x(0), x(1)). 5. Let X be a set. Show that {0, 1}X and P(X)are equivalent via the function Φ(f ) = {x ∈ X : f (x) = 1}. 6. Let X be a set. Show that P(X) and X are not equivalent. Hint: Suppose that f : X → P(X) and consider A = {x ∈ X : x 6∈ f (x)} to show that f cannot be onto. 7. Prove Theorem 1.5. 8. Suppose that f : X → Y is one-to-one. Show that equality obtains in part b) of Theorem 1.5 9. Suppose that A = {1, 2} and let A1 and A2 be non-empty sets. Show that Y Aα is equivalent to A1 × A2 via the function f (x) = (x(1), x(2)). α∈A 10. Suppose that Aα = X for all α ∈ I. Show that Y Aα = X I . α∈I 11. By assuming the Axiom of Choice, show that the projections πα are all onto if every set Aα is non-empty. 12. Suppose that R is an equivalence relation on X and S is an equivalence relation on Y . Define a relation R × S on X × Y by setting (x1 , y1 )(R × S)(x2 , y2 ) if and only if x1 Rx2 and y1 Sy2 . Show that R × S is an equivalence relation on X × Y and that (X × Y )/(R × S) is equivalent to (X/R) × (Y /S) via the function f ([(x, y)]) = ([x], [y]). 13. Suppose that R, S are equivalence relations on X with R ⊆ S. Show that there is an equivalence relation, E on X/R defined by [x]R E[y]R if and only if xSy. Chapter 2 Topological Spaces Let X be a set. A collection τ ⊆ P(X) of subsets of X is called a topology if the following hold: a) ∅ ∈ τ and X ∈ τ . b) If U, V ∈ τ , then U ∩ V ∈ τ . c) If A ⊆ τ , then S A ∈ τ. A topological space is a set X with a topology τ on X. We call elements of τ open sets in X. Example: Let X be any set. The discrete topology on X is defined by setting τ = P(X). Hence, every subset of X is open. Example: Let X be a set. The indiscrete topology on X is defined by τ = {∅, X}. Example: Let X = {a, b}. Define τ = {∅, {a}, X}. Then τ is a topology on X. Example: Let X be a set. Define τ = {U ⊆ X : U = ∅ or X \ U is finite }. By DeMorgan’s laws and the fact that subsets of finite sets are finite, property c) is satisfied. Since unions of finite sets are finite, property b) is also satisfied. 11 12 CHAPTER 2. TOPOLOGICAL SPACES We call τ the co-finite topology on X. Notice that this topology is the same as the discrete topology if X is finite, but differs otherwise. Example: We can define the co-countable topology on a set X by replacing the word finite by the word countable in the previous definition. In this case, the topology is discrete exactly when X is countable. We will find many more examples of topologies, but we first have to find better ways of defining them. Before we do that, we define some important concepts. Definition 2.1 If τ1 and τ2 are both topologies on the same set, we say that τ1 is coarser (alternatively, weaker) than τ2 if τ1 ⊆ τ2 . In this case, we also say that τ2 is finer (alternatively, stronger) than τ1 . In particular, the discrete topology on X is the strongest topology on X and the indiscrete topology is the weakest. Suppose that X is a topological space. We say that A ⊆ X is closed in X if X \ A is open. Let F denote the collection of closed sets of X. Then, we have the following from an easy application of DeMorgan’s laws: a) ∅, X ∈ F.. b) If C, K ∈ F, then C ∪ K ∈ F. c) If A ⊆ F, then T A ∈ F. It is clear that we can define a topology by declaring which sets are closed instead of which sets are open as long as these three properties are satisfied. In fact, if F is a collection of subsets of X which satisfy these properties, then {U : X \ U ∈ F} is a topology on X. Now, let A ⊆ X be any subset. We define the interior of A by letting S A = {U : U ⊆ A, U ∈ τ }. In other words, A◦ is the union of all open subsets ◦ of A. Since unions of open sets are open, A◦ is an open set of A. In fact, because of our definition, it is the em largest open set contained in A. Another common notation for the interior of A is int(A). 13 In a similar way, we can define the closure of A, A = T {F : A ⊆ F, F ∈ F}. Then the closure is the intersection of all closed sets containing A as a subset. Since intersections of closed sets are closed, A is a closed set containing A. In fact, it is the smallest closed set containing A. Another notation for the closure of A is cl(A). The following holds: Theorem 2.2 For subsets A, B of the topological space X, we have a) A◦ ⊆ A. b) A ⊆ A. c) (A◦ )◦ = A◦ . d) (A ∩ B)◦ = A◦ ∩ B ◦ e) X \ A◦ = X \ A. f) X \ A = (X \ A)◦ . g) cl(cl(A)) = cl(A). h) A ∪ B = A ∪ B. i) A◦ = A if and only if A is open. j) A = A if and only if A is closed. Definition 2.3 We say a subset A ⊆ X is dense if A = X Proposition 2.4 Let A ⊆ X. Then the following are equivalent: a) A is dense in X b) Whenever U 6= ∅ is open, we have A ∩ U 6= ∅ This will be left as an exercise. Definition 2.5 If A ⊆ X and x ∈ A, we say that A is a neighborhood of x if x ∈ A◦ . 14 CHAPTER 2. TOPOLOGICAL SPACES It is clear that the intersection of two neighborhoods of x is again a neigh- borhood of x. Notice that neighborhoods do not have to be open sets, but we do have the following. Proposition 2.6 A set is open if and only if it is a neighborhood of each of its points. 2.1 Bases of Topologies The first difficulty in defining a topology is that, in general, topologies are too large to define easily. Our first goal is find an easier way to obtain topologies that makes them easier to work with also. This is done by considering collections of open sets that do not satisfy property c) in the definition of topology, but allow the full topology to be obtained by use of this property. Definition 2.7 Let X be a topological space with topology τ . A collection B ⊆ τ is called a base for the topology if whenever ∅ 6= U ∈ τ , there is a collection S {Bα } ⊆ B such that U = α Bα . In other words, B is a collection of open sets and all non-empty open sets can be obtained by taking unions from B. Theorem 2.8 Let X be a topological space. In order for B ⊆ P(X) to be a base for the topology τ , the following must hold. (i) Each B ∈ B must be open. In other words, B ⊆ τ . (ii) If U is a non-empty open set and x ∈ U , there exists a B ∈ B such that x ∈ B ⊆ U. Conversely, if B satisfies these properties, then B is a base for the topology of X. Proof: Since sets in a base have to be open, i) is clear. So suppose that U is a S non-empty open set and x ∈ U . If B is a base, then we can write U = α Bα for some indexed family {Bα }. But then x ∈ Bα for some α. Hence, x ∈ Bα ⊆ U . For the converse, we need to show that every non-empty open set is a union of elements from B. So let U be a non-empty open set. For each x ∈ U , there 2.1. BASES OF TOPOLOGIES is a Bx ∈ B such that x ∈ Bx ⊆ U . But then 15 S x Bx = U since containment is true both ways. In this result, we already have a topology on the set X and wish to determine if a collection of open sets is a base for that topology. In the next result, we want to define the topology by declaring a certain collection of sets to be a base. Of course, there are conditions on the collection for this to work. Theorem 2.9 Suppose that B ⊆ P(X) satisfies (i) S B = X, (ii) If B1 , B2 ∈ B and x ∈ B1 ∩ B2 , then there is a B3 ∈ B with x ∈ B3 ⊆ B1 ∩ B2 . Then there is a unique topology τ such that B is a base for τ . Notice that if (i) fails, we may simply replace B by B ∪ {X} Proof: Define τ to contain the empty set and all sets of the form S Bα where {Bα } is an indexed family of elements of B. Clearly, X ∈ τ and ∅ ∈ τ . This is also the only topology for which B could possibly be a base. Since unions of elements of τ are again simply unions of elements of B, τ is closed under unions. We need only show that τ is closed under intersections. First, suppose that B1 , B2 ∈ B. For each x ∈ B1 ∩ B2 , by assumption, there S is a Bx ∈ B such that x ∈ Bx ⊆ B1 ∩ B2 . Then, x Bx = B1 ∩ B2 . Hence, B1 ∩ B2 ∈ τ . S S Now suppose that U, V ∈ τ and write U = α Bα and V = β Bβ0 where S S Bα , Bβ0 ∈ B. Then U ∩ V = α β (Bα ∩ Bβ0 ). By the previous paragraph, each Bα ∩ Bβ0 ∈ τ . Since we already know τ is closed under unions, U ∩ V ∈ τ , showing that τ is, in fact, a topology on X. Example: Let X = R and set B = {(a, b) : a < b}. Since intersections of open intervals are open intervals, there is a unique topology making B a base. This is called the usual topology on R. Open sets in R are then unions of open intervals. Unless we state otherwise, we will always assume that R has this topology. 16 CHAPTER 2. TOPOLOGICAL SPACES Example: Again, let X = R, but this time let B = {[a, b) : a < b}. Again, intersections of these half-open intervals are again half-open intervals, so B defines a base for a topology. We call this the half-open topology and call R with this topology the Sorgenfry line.We will denote it by L. It can be important to know when two different bases manage to define the same topology. The next result gives an easy criterion for this determination. It actually is stronger than this, giving a criterion for determining when one topology is finer than another. Theorem 2.10 Suppose that B1 and B2 are bases for the topologies τ1 and τ2 on the set X, respectively. In order that τ1 ⊆ τ2 it is necessary and sufficient that whenever x ∈ B1 ∈ B1 there exists B2 ∈ B2 with x ∈ B2 ⊆ B1 . Definition 2.11 Let X be a topological space with topology τ . We say that S ⊆ τ is a sub-base for the topology of X if the collection of all finite intersections {S1 ∩ · · · Sn : S1 , · · · Sn ∈ S} ∪ {X} is a base for the topology of X. Proposition 2.12 Any collection of subsets of X defines a sub-base for some topology on X. Proof: Just show that the collection of finite intersections satisfies the condi- tions of Theorem 2.9. Example: Let X be any partially ordered set and define (←, x) = {y : y < x} and (x, →) = {y : x < y}. Let S = {(←, x) : x ∈ X} ∪ {(x, →) : x ∈ X}. Then S is a subbase for a topology, called the order topology on X. In particular, notice that the order topology of R agrees with the usual topology. Unless otherwise stated, Ω and Ω0 will be given their order topologies. Definition 2.13 Let X be a topological space and x ∈ X. A base of neighborhoods at x is a collection N of neighborhoods at x such that whenevewr A is any neighborhood at x, there is an N ∈ N such that x ∈ N ⊆ A. In particular, if B is a base for the topology, N = {B : x ∈ B ∈ B} is a base of neighborhoods at x. 2.2. METRIC SPACES 17 Definition 2.14 Let X be a topological space. We say that X is a) second countable if there is a countable base for the topology of X. b) first countable if every point in X has a countable neighborhood base. c) X is separable if there is a countable set D ⊆ X that is dense in X. Proposition 2.15 If X is second countable, then X is first countable and separable. We will learn later that the converse of this is not true. Proof: Let B be a countable base for the toplogy of X. Then for each x ∈ X, the collection {B : x ∈ B ∈ B} is a countable neighborhood base at x, so X is first countable. Now, write B = {Bn : n ∈ N}. For each n ∈ N, pick xn ∈ Bn and let D = {xn : n ∈ N}. We use the result from Exercise 23 to show that D is dense. So, let U be a non-empty open set in X. Since B is a base, there is a Bn ∈ B such that Bn ⊆ U . But then xn ∈ U ∩ D, showing U ∩ D is non-empty. Since D is clearly countable, X is separable. Proposition 2.16 Suppose that A, B ⊆ Ω0 are closed uncountable sets. Then A ∩ B 6= ∅. Proof: Pick any x1 ∈ A. Since B is uncountable and {y : y ≤ x1 } is countable, there is a y1 ∈ B with x1 < y1 . Similarly, there is an x2 ∈ A with x1 < y1 < x2 . Continuing in this way, we find sequences x1 < y1 < x2 < y2 < · · · where xn ∈ A and yn ∈ B for all n. Now let a = sup{xn : n ∈ N} = sup{yn ; n ∈ N}. By exercise 9, a ∈ A ∩ B. 2.2 Metric Spaces A very important class of topological spaces derive from spaces having a concept of distance. These are called metric spaces. Definition 2.17 Let X be a set. A metric on X is a function d : X × X → R such that 18 CHAPTER 2. TOPOLOGICAL SPACES a) For every x, y ∈ X, d(x, y) ≥ 0. b) We have d(x, y) = 0 if and only if x = y. c) For every x, y ∈ X, d(x, y) = d(y, x). d) For every x, y, z ∈ X, we have d(x, z) ≤ d(x, y) + d(y, z). A metric space is a set X together with a metric d on X. Example: Let X = R and set d(x, y) = |x − y|. n Example: Let X = R and set d(x, y) = n X ! 21 |xk − yk |2 k=1 This is called the usual metric on n R . n Example: Let X = R and set d1 (x, y) = n X |xk − yk | k=1 This is called the taxi-cab metric on Rn . n Example: Let X = R and set d∞ (x, y) = max{|xk − yk | : 1 ≤ k ≤ n} This is called the max metric or box metric on Rn . Definition 2.18 Let X be a metric space with metric d and let x ∈ X and r > 0. We define the open ball around x or radius r by Bd (x, r) = {y ∈ X : d(x, y) < r}. When the metric is understood, we shall often omit the subscript. Theorem 2.19 Let X b e a metric space with metric d. Define B = {B(x, r) : x ∈ X, r > 0}. Then B defines a base for a topology on X. We call this topology the metric topology. Moreover, for each x ∈ X, the collection {B(x, r) : r > 0} is a neighborhood base at x. 2.2. METRIC SPACES Proof: Clearly, for each x ∈ X, x ∈ B(x, 1), so X = 19 S x B(x, 1). This gives the first condition in Theorem 2.9. For the second, suppose that z ∈ B(x1 , r1 ) ∩ B(x2 , r2 ). Let r = min{r1 −d(z, x1 ), r2 −d(z, x2 ). If y ∈ B(x, r), then d(x1 , y) ≤ d(x1 , x)+d(z, y) < d(x1 , x)+r ≤ r1 . So, y ∈ B(x1 , r1 ). Similarly, y ∈ B(x2 , r2 ). Thus x ∈ B(x, r) ⊆ B(x1 , r1 ) ∩ B(x2 , r2 ) as required to invoke Theorem 2.9. Now assume that x ∈ X and x ∈ U with U open. By the definition of a base, there is a y ∈ X and an r1 > 0 such that x ∈ B(y, r1 ) ⊆ U . The subtlety here is that we want an open ball with center x instead of one with center y. But, if we let r = r1 − d(x, y), we have that B(x, r) ⊆ B(y, r1 ) ⊆ U , as required for a neighborhood base. Unless otherwise stated, we will always give a metric space the corresponding metric topology. Proposition 2.20 Every metric space is first countable. Proof: If X is a metric space and x ∈ X, then {B(x, n1 ) : n ∈ N} is a neighborhood base at x. Proposition 2.21 A separable metric space is second countable. Proof: Let D be a countable set with D = X. Write D = {xn : n ∈ B = {B(xn , k1 : k, n ∈ N}. Let N}. Then B is a countable collection of open sets in X. We show that B is a base for the topology of X. So, let x ∈ U where U is open in X. There is a k ∈ N such that B(x, k1 ) ⊆ U . Since D is dense, there is an 1 1 1 xn ∈ B(x, 2k ). But then x ∈ B(xn , 2k ) ⊆ B(x, k1 ) ⊆ U . Since B(xn , 2k ) ∈ B, this shows that B is a base for X. Hence, X is second countable. It is instructive to see where this argument breaks down for general topological spaces. The reader is encouraged to work through this. Definition 2.22 We say a topology on X is metrizable if it is induced by some metric on X. We say that two metrics are topologically equivalent if they induce the same topology on X. 20 CHAPTER 2. TOPOLOGICAL SPACES Example: The metrics d∞ , d2 , andd1 are all topologically equivalent on Rn Example: Suppose that d is a metric on X and define d(x, y) d(x, y) ≤ 1 d1 (x, y) = 1 d(x, y) ≥ 1 Then d1 is a metric on X that is topologically equivalent to d. The important thing about d1 is that it is bounded, yet defines the same topology as d. 2.3 Continuous Functions Definition 2.23 Let X and Y be two topological spaces with topologies τX and τY respecitvely. We say that a function f : X → Y is continuous at the point x ∈ X if, whenever V is open in Y with f (x) ∈ V , there is an open set U in X with x ∈ U and f [U ] ⊆ V . A function that is continous at every point of X is said to be continuous on X. The following shows that continuity can be tested using only elements from bases defining the respective topologies. Proposition 2.24 Suppose that BX and BY are bases for the topologies of X and Y respectively. Then f : X → Y is continous at x ∈ X if and only if whenever f (x) ∈ B ∈ BY , there is a B 0 ∈ BX with x ∈ B 0 and f [B 0 ] ⊆ B. Example: Let (X, d) and (Y ρ) be metric spaces. A function f : X → Y is continous at x ∈ X in the metric topologies if and only if for each ε > 0 there is a δ > 0 such that f [B(x, δ)] ⊆ B(f (x), ε). In more detail, whenever x0 ∈ X with d(x, x0 ) < δ, we can conclude that ρ(f (x), f (x0 )) < ε. This should be the familiar definition for the case where X and Y are Euclidean spaces. Theorem 2.25 Let X and Y be topological spaces and f : X → Y . The following are equivalent: a) f is continuous from X to Y . 2.3. CONTINUOUS FUNCTIONS 21 b) Whenever V is open in Y , f −1 [V ] is open in X. c) Whenever C is closed in Y , f −1 [C] is closed in X. d) For every set A ⊆ X, f [A] ⊆ f [A]. Proof: a) implies b): Suppose that f is continous and that V ⊆ Y is open. For each x ∈ f −1 [V ], f (x) ∈ V , so by continuity, there is an open set Ux such that x ∈ Ux and f [Ux ] ⊆ V . But this implies that x ∈ Ux ⊆ f −1 [V ]. Hence, S f −1 [V ] = x Ux is a union of open sets, so f −1 [V ] is open in X. b) implies c): Suppose that b) holds and let C ⊆ Y be closed. Then Y \ C is open in Y , so by assumption f −1 [Y \ C] is open in X. But f −1 [Y \ C] = X \ f −1 [C], so f −1 [C] is closed in X. c) implies d): Suppose that c) holds and that A ⊆ X. Then f [A] is closed in Y , so f −1 [f [A]] is closed in X. Notice that A ⊆ f −1 [f [A]]. Since A is the smallest closed set containing A, we have that A ⊆ f −1 [f [A]. This, in turn, implies f [A] ⊆ f [A]. d) implies a): Suppose that d) holds, x ∈ X, V ⊆ Y is open, and f (x) ∈ V . Let A = f −1 [Y \ V ] = X \ f −1 [V ].Then, f [A] ⊆ f [A] ⊆ Y \ V = Y \ V since Y \ V is closed. Since f (x) 6∈ Y \ V , this shows that x 6∈ A. Let U = X \ A. Then x ∈ U . Furthermore, U ⊆ X \ A = f −1 [V ]. Hence f [U ] ⊆ V . An easy, but important, consequence is the following: Theorem 2.26 Suppose that f : X → Y and g : Y → Z are continuous functions between topological spaces. Then g ◦ f : X → Z is continuous. Proof: Let V ⊆ Z be open. Then (g ◦ f )−1 [V ] = f −1 [g −1 [V ]] is open in X. Example: If X has the discrete topology, then any function f : X → Y into a topological space is continuous. Similarly, if Y has the indiscrete topology, then any function f : X → Y into Y is continuous. Definition 2.27 Let X and Y be topological spaces. A homeomorphism from X to Y is a continous, one-to-one correspondence f : X → Y such that the function f −1 : Y → X is also continuous. We say that two topological spaces are homeomorphic if there is a homeomorphism from one to the other. 22 CHAPTER 2. TOPOLOGICAL SPACES Since compositions and inverses of homeomorphisms are again homeomor- phisms, the following hold: (i) Every topological space X is homeomorphic to itself. (ii) If X is homeomorphic to Y , then Y is homeomorphic to X. (iii) If X is homeomorphic to Y and Y is homeomorphic to Z, then X is homeomorphic to Z. Hence, the property of being a homeomorphism is almost an equivalence relation. The only problem is that the collection of all topological spaces is too large to be a set, so we do not technically have a relation at all. However, for every set of topological spaces, we do get an equivalence relation. We generally regard homeomorphic spaces as being the same topologically. Definition 2.28 A function f : X → Y between topological spaces is said to be an open map if f [U ] is open in Y whenever U is open in X. Similarly, we say that f is a closed map if f [C] is closed for every closed set C ⊆. In particular, a continuous, one-to-one onto function is a homeomorphism if and only if it is also either an open map or a closed map. Note that if B is a base for the topology of X, it is enough to check that f [B] is open in Y for each B ∈ B to show that f is an open map by Theorem 1.5. Exercises: 1. Find all topologies on the three element set X = {a, b, c}. 2. What are the closed sets in the co-finite topology? The co-countable topology? 3. Suppose that τ1 and τ2 are both topologies on the set X. Show that τ1 ∩τ2 is a topology on X. What about τ1 ∪ τ2 ? 4. Suppose that {τα }α∈I is a collection of topologies on X. T α∈I τα is a topology on X. Show that 2.3. CONTINUOUS FUNCTIONS 23 5. Suppose that S ⊆ P(X) is any collection of subsets of X. Show that there is a weakest topology on X containing S. 6. Show that [a, b) is both open and closed in the half-open topology on R. 7. Prove Theorem 2.10. 8. Show that the half-open topology on R is finer than the usual topology. 9. Suppose that A ⊆ Ω0 is closed and {xn }n∈N is a sequence in A. Let x = sup{xn : n ∈ N}. Show that x ∈ A. 10. Suppose that A, B ⊆ Ω0 are closed, uncountable sets. Show that A ∩ B is uncountable. 11. Suppose that {An }∞ n=1 is a countable family of closed uncountable subsets T∞ of Ω0 . Show that n=1 An is uncountable. 12. Let B be a bse for the topology of X and suppose that A ⊆ X. Show that S A◦ = {B : B ∈ B, B ⊆ A}. In other words, we may obtain the interior of a set using only basic open sets in our definition. 13. What do the open balls look like in R2 with the taxi-cab metric? R3 ? 14. What do the open balls look like in R2 in the max metric? R3 ? 15. Show that the max metric and the taxi-cab metric define the same topology on R2 . 16. Let X be a metric space with metric d. The closed ball of radius r centered at x ∈ X is defined by B(x, r) = {y : B(x, y) ≤ r}. Show that B(x, r) is a closed set in the metric topology. 17. Let X = [0, 1]∪[2, 3] with distance d(x, y) = |x−y|. Show that the closure of B(2, 1) is NOT B(2, 1). 18. Let (X, d1 ) and (Y, d2 ) be metric spaces. Define d : (X ×Y )×(X ×Y ) → R by d((x1 , y1 ), (x2 , y2 )) = d1 (x1 , x2 ) + d2 (y1 , y2 ). Show that d is a metric on X × Y . 24 CHAPTER 2. TOPOLOGICAL SPACES 19. Let (X, d) be a metric space and give X × X the metric from the previous exercise (use d for both coordinates). Show that the function d : X × X → R is continuous on X × X. 20. Let τ1 and τ2 be topologies on the set X. When is the identity function 1X : (X, τ1 ) → (X, τ2 ) continuous? 21. Let (X, d) be a metric space and A ⊆ X. Define d(x, A) = inf{d(x, a) : a ∈ A}. Show that f : X → R defined by f (x) = d(x, A) is continuous. 22. Let (X, d) be a metric space and A ⊆ X. Show that d(x, A) = 0 if and only if x ∈ A. 23. Show that D is dense if and only if whenever U is a non-empty open set, we have D ∩U 6= ∅. If B is a base, this is equivalent to D ∩B 6= ∅ whenever B ∈ B is non-empty. Chapter 3 New Spaces from Old Our next concern is to find ways of creating new topological spaces from ones we have already found. 3.1 Subspaces So, suppose that A ⊆ X and that X is a topological space with topology τ . Define the subspace or relative topology on A by setting τA = {A ∩ U : U ∈ τ }. It is easily verified that τA is, in fact, a topology on the set A. We say that elements of τA are open in A. Unless otherwise stated, we will always give subsets of a topological space the subspace topology. Proposition 3.1 Let X be a topological space and A ⊆ X. The following hold: a) A subset V ⊆ A is open in A if and only if there is an open set U in X with V = A ∩ U . b) A subset C ⊆ A is closed in A if and only if there is a closed set K in X with C = A ∩ K. c) If B is a base for the topology of X, then {A ∩ B : B ∈ B} is a base for the topology of A. d) The inclusion map iA : A → X is continuous. 25 26 CHAPTER 3. NEW SPACES FROM OLD e) If f : X → Y is a continuous map, then the restriction f |A : A → Y is continuous. It should be pointed out that τA is the weakest topology on A that makes the inclusion map iA continuous. This follows since i−1 A [U ] = A ∩ U for U ⊆ X. This is one motivation for defining the topology on A as we have done. Notice that if A is an open subset of X and U is open in the subspace topology of A, then U is open in X. Similarly, if A is closed in X and C ⊆ A is closed in the subspace topology of A, we can conclude that C is cloased in X. It is often the case that we want to define a continuous function on a large set X by saying what it does on certain subsets. The following allows us to do this: Theorem 3.2 Suppose that X = A1 ∪ A2 is a topological space with A1 , A2 closed subsets. Suppose that f : X → Y is a function such that f1 = f |A1 and f2 = f |A2 are continuous on A1 and A2 , resp. Then, f is continuous on X. Proof: Let K ⊆ Y be closed in Y . Then f −1 [K] = f1−1 [K] ∪ f2−1 [K]. Since f1 and f2 are assumed to be continuous, f1−1 [K] is closed in A1 and f2−1 [K] is closed in A2 . Since A1 and A2 are closed in X, these are both closed sets in X, so f −1 [K] is closed in X. Hence f is continuous. Clearly, this can be extended to the case where there are finitely many closed subsets {Ai } of X. Also, if we use open sets instead of closed sets, we may use arbitrarily many sets and obtain the same result. Definition 3.3 Let X and Y be topological spaces. We define the em dijoint union of X and Y by X ⊕ Y = (X × {0}) ∪ (Y × {1}). A subset U is declared to be open in X ⊕ Y if {x ∈ X : (x, 0) ∈ U } and {y ∈ Y : (y, 1) ∈ U } are both open. Notice that the map iX : X → X ⊕ Y defined by iX (x) = (x, 0) is a homeomorphism of X onto the subspace X × {0}. Similarly, the map iY : Y → X ⊕ Y defined by iY (y) = (y, 0) is a homeomorphism of Y onto the subspace Y × {0}. Because of this, we will usually identify X and Y with their corresponding subspaces of X ⊕ Y . 3.2. QUOTIENT SPACES 27 Proposition 3.4 Suppose that X, Y and Z are topological spaces and that fX : X → Z and fY : Y → Z are continuous maps. Then the map f : X ⊕ Y :→ Z defined by setting f (x, 0) = fX (x) for x ∈ X and f (y, 0) = fY (y) for y ∈ Y is a continous function. 3.2 Quotient Spaces Suppose that X is a topological space and that R is an equivalence relation on X. There is a quotient map qR : X → X/R defined by qR (x) = [x], the equivalence class of x. Our goal is to put a topology on X/R that will make this map continuous. Of course, the indiscrete topology will work for this, but it doesn’t have anything to do with the topology of X. The problem is that it is too small. We want to find the finest topology on X/R which will make the quotient map continuous. Definition 3.5 Let X be a toplogical space and R an equivalence relation on X. Define the quotient topology on X/R by declaring U ⊆ X/R to be open in −1 X/R exactly when qR [U ] is open in X. Proposition 3.6 The quotient topology is, in fact, a topology on X/R. Furthermore, it is the finest topology on X/R making qR continuous. Proposition 3.7 Let X be a topological space and R and equivalence relation on X. Then, a function f : X/R → Y is continuous if and only if f ◦ qR is continuous. −1 −1 Proof: Just notice that for an open set U ⊆ Y , (f ◦ qR )−1 [U ] = qR [f [U ]] is open in X if and only if f −1 [U ] is open in X/R by the definition of the quotient topology. More generally, we have the following definition: Definition 3.8 A continuous map between topological spaces, f : X → Y is said to be a quotient map if f is surjective and U ⊆ Y is open in Y if and only if f −1 [U ] is open in X. 28 CHAPTER 3. NEW SPACES FROM OLD Example: Suppose that X and Y are topological spaces, R is an equivalence relation on X and f : X → Y is a continous function such that x1 Rx2 implies f (x1 ) = f (x2 ). Then the function fˆ : X/R → Y defined by fˆ([x]) = f (x) is well-defined and continuous. A very common situation is that of ’collapsing a set to a point.’ For this, let X be a topological space and A ⊆ X. Define a partition of X by P = {{x} : x 6∈ A} ∪ {A}. We let R be the corresponding equivalence relation on X. Then we will write X/A instead of X/R and say that X/A is the space obtained from X by collapsing A to a point. Example: Let X = [0, 1] and A = {0, 1}. Let S 1 = {(x, y) ∈ R2 : x2 + y2 = 1}. Define f : X → S 1 by f (t) = (cos t, sin t). Then f is continous and f (0) = (1, 0) = f (1), so there is an induced map on the quotient fˆ : X/A → S 1 . Notice that f |[0,1/2] and f |[1/2,1] are both homeomorphisms, so are closed. This shows that f is closed, so is a quotient map, so fˆ is a homeomorphism. Now let X and Y be topological spaces and f : A → Y be a continuous function where A ⊆ X. We define the attachment space of X onto Y via f to be (X⊕Y )/R where the equivalence classes for R are sets of the form {y}∪f −1 [y] for y ∈ Y . We denote this attachment space as X ⊕f Y . Definition 3.9 We say the equivalence relation R is upper semi-continuous if whenever y ∈ X/R with q −1 [y] ⊆ U open, there is a V ⊆ X/R open with y ∈ V and q −1 [V ] ⊆ U . Theorem 3.10 The quotient map q : X → X/R is a closed map if and only if R is upper semi-continuous. Proof: Let Y = X/R for convenience. Suppose q is closed. Suppose y ∈ Y and U is open in X with q −1 [y] ⊆ U . Then X \ U is closed in X, so V = Y \ q[X \ U ] is open in Y and y ∈ V . Notice that q −1 [V ] ⊆ U . Hence R is upper semicontinuous. Conversely, suppose that R is upper semi-continuous and suppose that C ⊆ X is closed. Let y 6∈ q[C]. Then q −1 [y] ⊆ U = X \ C, so there is an open set 3.3. PRODUCT SPACES 29 V ⊆ Y with y ∈ V and q −1 [V ] ⊆ X \ C. But then, V ∩ q[C] = ∅, showing that q[C] is closed in Y . 3.3 Product Spaces Definition 3.11 Let X and Y be topological spaces. We define the product topology on X × Y by declaring B = {U × V : U open in X, V open in Y } to be a base for the topology. Proposition 3.12 The above is a base for a topology on X × Y . Furthermore, it is the weakest topology such that both πX and πY are continuous. Proof: Since (U1 × V1 ) ∩ (U2 × V2 ) = (U1 ∩ U2 ) × (V1 ∩ V2 ), B satisfies the conditions in Theorem 2.9 to be a base for a topology on X × Y . Next, if U −1 is open in X, then πX [U ] = U × Y is open in X × Y , so πX is continuous. Similarly, πY is continuous. Conversely, suppose that τ is any topology making both πX and πY continuous. Then, if U is open in X and V is open in Y , −1 U × V = πX [U ] ∩ πY−1 [V ] will be in τ . Hence, τ contains B and so is stronger than the product topology. Proposition 3.13 Suppose that X, Y , and Z are topological spaces with f : Z → X × Y . Then f is continuous if and only if both πX ◦ f and πY ◦ f are continuous. Proof: If f is continuous, then πX ◦f and πY ◦f are compositions of continuous functions so are also continuous. For the converse, assume that the compositions are continuous. To show that f is continuous, it is enough to show that f −1 [U × V ] is open in Z for every basic open set U ×V . But f −1 [U ×V ] = (πx ◦f )−1 [U ]∩ (πY ◦ f )−1 [V ], so this follows. Example: We show that L is not second countable. If it were, then L × L would be second countable by exercise 14. In that case, every subspace of L × L would be second countable by exercise 4. But consider the subspace A = {(x, −x) : x ∈ L}. Every singleton is open in the subspace since A∩([x, x+ 30 CHAPTER 3. NEW SPACES FROM OLD 1) × [−x, −x + 1)) = {(x, −x)}. But an uncountable discrete space is definitely NOT second countable. Hence L cannot be second countable. This gives an example of a first countable, separable space which is not second countable. It also shows that L is not metrizable. We now want to extend the idea of product topologies to the case of infinite products. So, suppose that {Xα }α∈I is a collection of topological spaces. Q We want to put a topology on Xα . The natural goal would be to require each projection πα to be continuous. The product topology will be the weakest topology such that this is the case. Definition 3.14 The product topology on Q Xα is defined by declaring a sub- basic open set to be of the form πα−1 [Uα ] where Uα is open in Xα , α ∈ I. From this, a basic open set in the product topology is a set of the form Tn i=1 [Uαi ] where each Uαi is open in Xαi . Clearly, we may assume that πα−1 i the αi are all distinct. But now, let Uα = Xα unless α = α1 , · · · , αn , then Q Tn Uα = i=1 πα−1 [Uαi ]. In other words, products of open sets where all but i finitely many factors are trivial are open in the product topology and such sets form a base for the product topology. Q Note: if we allowed every product of the form Uα where Uα is open in Q Xα , we get a different topology on Xα known as the box topology. This topology is, in general, a finer topology than the product topology. Of the two, the product is the nicer, as we shall see. Proposition 3.15 The product topology on Q Xα is the weakest topology mak- ing all projections πα continuous. Furthermore, each πα is an open map. Proof: Since πα−1 [Uα ] is open in the product topology whenever Uα is open in Xα , each πα is continuous. Also, πα [B] is open for each basic open set, so πα Q is an open map. On the other hand, if Xα has a topology τ making each πα T continuous, then all sets of the form πα−1 [Uα ] with finitely many terms will be open. Hence every basic open set in the product topology would be open in τ . Thus τ is stronger than the product topology. 3.3. PRODUCT SPACES 31 Proposition 3.16 Suppose that Z is a topological space. Then, f : Z → Q Xα is continuous if and only if each πα ◦ f is continuous. Proof: Since each πα is continuous, the continuity of f would imply that of all πα ◦ f . For the converse, suppose that B = ∩ni=1 πα−1 [Uαi ] is a basic open set. i Then f −1 [B] = ∩ni=1 (παi ◦ f )−1 [Uαi ], showing f to be continuous. Lemma 3.17 If U is a non-empty open set in the product Q Xα , then {α : πα (U ) = Xα } is finite. Proof: Let x ∈ U and pick a basic open set Q Uα such that x ∈ Q Uα ⊆ U . Then {α : Uα 6= Xα } is finite and Uα ⊆ πα [U ]. Theorem 3.18 Suppose that {Xα } is a collection of topological space. Then Q Xα is second countable if and only if each Xα is second countable and {α : Xα is not indiscrete} is countable. This also holds with first countable in place of second countable everywhere. Proof: First suppose that Q Xα is second countable. By an exercise, each Xα is second countable. Let B = {Bn : n ∈ N} be a countable base for the product. Let An = {α : πα [Bn ] 6= Xα }. Then each An is finite, so A = ∪∞ n=1 An is countable. Now suppose that α 6∈ A and that Xα is not indiscrete. Then there is a non-empty open set Uα 6= Xα . Then U = πα−1 [Uα ] is open in the product and πα [U ] 6= Xα . Since α 6∈ A, this shows that no Bn is a subset of U . THis contradicts that B is a base for the product. Now, suppose that Xn is second countable for each n ∈ N. Next, let Bn = {Bnm : m ∈ N} be a countable base for Xn for each n ∈ N. Let Tk B = { i=1 πi−1 [Bimi ] : k ∈ N, Bimi ∈ Bi }. Then B is a countable base for Q Xn . Similar proofs hold for first countability. Theorem 3.19 Suppose that (Xn , dn ) for n ∈ N are metric spaces such that Q dn (xn , yn ) ≤ 1 for all xn , yn ∈ Xn . For x, y ∈ X = Xn , define d(x, y) = P n n dn (xn , yn )/2 . Then d is a metric on X inducing the product topology. Hence, countable products of metrizable spaces are metrizable. 32 CHAPTER 3. NEW SPACES FROM OLD Proof: Clearly the sum defining d(x, y) converges and gives a metric on X. We need to show that the metric topology induced by d is the same as the product topology on X. We first show that every open set in the product topology is open in the metric topology. It is sufficient to show that every sub-basic open set in the product is open in the metric topology. So, suppose that Un is open in Xn and that x ∈ πn−1 [Un ]. Then xn ∈ Un . Since Xn has the topology induced by the metric dn , there is an ε > 0 such that B(xn ; ε) ⊆ Un . Let δ = ε/2n . Then, if d(x, y) < δ, dn (xn , yn ) < ε, so yn ∈ Un , so y ∈ πn−1 [Un ]. Hence B(x, δ) ⊆ πn−1 [Un ], showing this subbasic open set is open in the metric topology. Conversely, consider a basic open set B(x, ε) in the metric topology. Choose P an N ∈ N such that n≥N 21n < ε/2. Next, let Un = B(xn ; ε/2). Then Un is open in Xn . Let U = U1 × · · · UN −1 × XN · · · . Then U is a basic open set in the product topology and x ∈ U . Furthermore, suppose that y ∈ U . Then, for n < N , dn (xn , yn ) < ε/2 and for n ≥ N , dn (xn , yn ) ≤ 1. Hence P P P d(x, y) = n dn (xn , yn )/2n < (ε/2) · n<N 21n + n≥N 21n < ε/2 + ε/2 = ε. So, U ⊆ B(x; ε). Hence B(x; ε) is open in the product topology. 3.4 Weak Topologies We can generalize the construction of the product topology as follows. Suppose that {Xα } is a collection of topological spaces and X is a set. Suppose we are given functions fα : X → Xα . We can give X a topology be defining a subbase for X to be S = {fα−1 [Uα ] : Uα open in Xα }. We call this topology the weak topology induced by the functions {fα }. Proposition 3.20 The weak topology on X is the weakest topology making each fα continuous. Proposition 3.21 Let X have the weak topology induced by the functions {fα }. If X is a topological space, then f : Z → X is continuous if and only if fα ◦ f is continuous for each α. 3.4. WEAK TOPOLOGIES 33 Definition 3.22 We say a function f : X → Y is an embedding if it is a homeomorphism of X to f [X] where f [X] is regarded as a subspace of Y . Notice that every inclusion map is an embedding. Definition 3.23 We say a collection of functions fα : X → Xα separates points of X if, whenever x 6= y in X, there is an α such that fα (x) 6= fα (y). Recall, from Theorem 1.6 that a collection of functions fα : X → Xα induces Q a function f : X → Xα such that fα = πα ◦ f . The following shows when this map is an embedding into the product topological space. This is ultimately the basis for many important results. Theorem 3.24 Let {Xα } be a collection of topological spaces, X a topological Q space, and fα : X → Xα functions. In order for the induced map f : X → Xα to be an embedding it is necessary and sufficient for X to have the weak topology induced by {fα } and for this collection of functions to separate points of X. Proof: First assume that f is an embedding. The collection of all f [X]∩πα−1 [Uα ] is a subbase for the topology of f [X]. Hence the collection of f −1 [f [X]∩πα−1 [Uα ]] is a subbase for the topology of X. But this set is exactly fα−1 [Uα ], so X has the weak topology induced by {fα }. Also, if x 6= y in X, then f (x) 6= f (y), so there is an α such that πα ◦ f (x) 6= πα ◦ f (y). Hence fα (x) 6= fα (y), so {fα } separates points of X. For the converse, suppose that X has the weak topology induced by {fα } which separates points of X. Then, if x 6= y in X, there is an α so that fα (x) 6= fα (y). Since fα = πα ◦ f , this implies that f (x) 6= f (y), so f is oneto-one. Since each fα is continuous, the function f is also continuous. We only need to show that f : X → f [X] is an open map. But a subbasic open set in X is one of the form fα−1 [Uα ] whose image in f [X] is f [X] ∩ πα−1 [Uα ], which is a subbasic open set in f [X]. Because of this result, it is important to be able to identify when a space X has the weak topology induced by a family of functions. The crucial concept is that of separation of points and closed sets. 34 CHAPTER 3. NEW SPACES FROM OLD Definition 3.25 We say that a family of functions {fα } separates points from closed sets in X if, whenever C ⊆ X is closed and x ∈ / C, there is a function fα such that fα (x) 6∈ fα [C]. Theorem 3.26 Suppose that {fα : X → Xα } is a collection of continuous functions which separates points from closed sets in X. Then the collection of all sets of the form fα−1 [Uα ] where Uα is open in Xα is a base for the topology of X. In particular, X then has the weak topology induced by the functions {fα }. Proof: Certainly all sets of this form are open in X. So, suppose that U is open in X and x ∈ U . Let C = X \ U , so x 6∈ C. By assumption, there is an α such that fα (x) 6∈ fα [C]. Let Uα = Xα \ fα [C], which is an open set in Xα . Then x ∈ fα−1 [Uα ] ⊆ U , proving the result. Exercises: 1. Let (X, d) be a metric space with the metric topology. Let A ⊆ X and let dA : A × A → R be the restriction of d. Show that the subspace topology on A induced from X is the same as the metric topology induced by dA . 2. Prove Proposition 3.1. 3. Suppose that X is first countable and A ⊆ X. Show that A is first countable. 4. Suppose that X is second countable and A ⊆ X. Show that A is second countable. 5. Suppose that X and Y are topological spaces, A ⊆ X and that f : X → Y is continous with f |A a constant map. Show that there is a continuous function fˆ : X/A → Y such that fˆ ◦ qA = f where qA : X → X/A is the quotient map. 6. Suppose that X and Y are topological spaces and f : X → Y is an onto continuous function. Define an equivalence relation on X by setting x1 Rx2 if and only if f (x1 ) = f (x2 ). Show that the induced map fˆ : X/R → Y is a continuous, one-to-one, onto map. Show that fˆ is a homeomorphism if and only if f is a quotient map. 3.4. WEAK TOPOLOGIES 35 7. Suppose that f : X → Y is a continous, closed, onto map. Show that f is a quotient map. 8. Suppose that f : X → Y is a continuous, open, onto map. Show that f is a quotient map. 9. Let R, S be equivalence relations on X such that R ⊆ S. Show that there is a continous map qRS : X/R → X/S such that qRS ◦ qR = qS . Show that qRS is a quotient map. 10. Suppose that X and Y are topological spaces, A ⊆ X and f : A → Y is continuous. Suppose also that Z is a topological space and that the functions g : X → Z and h : Y → Z are continous with g|A = h ◦ f . Show that H : X ⊕A Y → Z defined by H([(x, 0)]) = g(x) for x ∈ X and H([(y, 0)]) = h(y) for y ∈ Y is well-defined and gives a continuous function H. 11. Suppose that (X, d) and (Y, ρ) are metric spaces. Define d∞ on X × Y by d∞ ((x1 , y1 ), (x2 , y2 )) = max{d(x1 , x2 ), ρ(y1 , y2 )}. Show that d∞ induces the product topology on X × Y . 12. As in the previous exercise, define d1 ((x1 , y1 ), (x2 , y2 )) = d(x1 , x2 ) + ρ(y1 , y2 ). Show that d1 also induces the product topology on X × Y . 13. Suppose that BX is a base for the topology of X and BY is a base for the topology of Y . Show that {B1 × B2 : B1 ∈ BX , B2 ∈ BY } is a base for the topology of X × Y . 14. Suppose that X and Y are both second countable spaces. Show that X ×Y is second countable. 15. Suppose that X and Y are first countable spaces. Show that X × Y is first countable. 16. Show that A × B = A × B for A ⊆ X and B ⊆ Y . 17. Suppose that X and Y are separable spaces. Show that X ×Y is separable. 36 CHAPTER 3. NEW SPACES FROM OLD 18. Suppose that f : X → Y is a continuous, onto function which is also either open or closed. Show that Y is homeomorphic to X/R where x1 Rx2 if and only if f (x1 ) = f (x2 ). 19. Suppose that f : X → Y is an onto, continuous, open map. Suppose that B is a base for the topology of X. Show that {f [B] : B ∈ B} is a base for the topology of Y . 20. Suppose that f : X → Y is an onto continuous map and D ⊆ X is dense. Show that f [D] is dense in Y . 21. Suppose that Q Xα is second countable. Show that each Xα is second countable. 22. Suppose that Q Xα is separable. Show that each Xα is separable for each α. 23. Let T = {0, 1, 2} with the topology τ = {∅, {0, 1}, T }. Let X be any topological space and A the collection of continuous functions f : X → T . Show that A separates points of X and separates points from closed sets. Conclude that X can be embedded into T A . Chapter 4 Separation Axioms The concept of a topology is a very general one. One difficulty that may arise is that a given topology may not have very many open sets. We will find it convenient to look at additional axioms that will guarantee a good collection of open sets. The most useful such axioms are the separation axioms. 4.1 T0 , T1 , and T2 Axioms Definition 4.1 We say that a topological space satisfies the T0 separation axiom (usually, we just say that X is T0 ) if, whenever x 6= y in X, there is an open set U such that either x ∈ U and y 6∈ U or vice versa. Definition 4.2 We say that a topological space X satisfies the T1 separation axiom if whenever x 6= y in X, there is an open set U such that x ∈ U and y 6∈ U . Note that there will also be an open set V such that y ∈ V and x 6∈ V . Theorem 4.3 For a topological space X, the following are equivalent: a) X is T1 , b) For each x ∈ X, the set {x} is closed in X, c) For any A ⊆ X, A = T {U : A ⊆ U open}. 37 38 CHAPTER 4. SEPARATION AXIOMS Proof: Suppose that X is T1 and let x ∈ X. Then, for each y 6= x, there is an S open set Uy such that y ∈ Uy and x 6∈ Uy . But then, X \ {x} = y Uy is open, so {x} is closed. Now suppose that each {x} is closed in X and let A ⊆ X. Then A = T x X \ {x} where x runs over the x 6∈ A. Thus, A is the intersection of open sets. Finally, suppose that the last condition holds and suppose x 6= y in X. Let A = {x}. By assumption, there is an open set U such that A ⊆ U and y 6∈ U . Hence, X is T1 . Theorem 4.4 Products and subspaces of T1 spaces are T1 . Proof: Suppose that X is T1 and A ⊆ X. Let x 6= y in A. Then, there is an open set U in X such that x ∈ U and y 6∈ U . But then A ∩ U is open in A and x ∈ A ∩ U while y 6∈ A ∩ U . Now suppose that {Xα } is a collection of T1 spaces and x 6= y in Q Xα . Then, there is a coordinate α such that xα 6= yα . Hence, there is an open set Uα in Xα such that xα ∈ Uα but yα 6∈ Uα . Then, with U = πα−1 [Uα ], we have Q x ∈ U and y 6∈ U . Hence, Xα is T1 . Perhaps the most important separation axiom is the next one. Definition 4.5 We say that a topological space satisfies the T2 separation axiom (or say that X is Hausdorff ) if whenever x 6= y in X, there are open sets U and V such that x ∈ U , y ∈ V and U ∩ V = ∅. Example: Let (X, d) be a metric space and suppose that x 6= y in X. Let ε = d(x, y)/2 > 0 and set U = B(x, ε) and V = B(y, ε). Then U and V are open and U ∩ V = ∅. Hence, metric spaces are always T2 . Theorem 4.6 Products and subspaces of T2 spaces are T2 . Proof: Suppose that X is T2 and A ⊆ X. Let a1 , a2 ∈ A with a1 6= a2 . Then, there are open sets U and V in X such that a1 ∈ U , a2 ∈ V and U ∩ V = ∅. 4.1. T0 , T1 , AND T2 AXIOMS 39 But then a1 ∈ A ∩ U , a2 ∈ A ∩ V where A ∩ U and A ∩ V are open sets in A with (A ∩ U ) ∩ (A ∩ V ) = ∅. Now suppose that {Xα } is a family of T2 spaces. Suppose that x = (xα ) 6= Q (yα ) = y in Xα . Then, there is an index α such that xα 6= yα . Since Xα is T2 , there are disjoint open sets Uα and Vα such that xα ∈ Uα and yα ∈ Vα . But Q then U = πα−1 [Uα ] and V = πα−1 [Vα ] are disjoint open sets in Xα with x ∈ U and y ∈ V . Theorem 4.7 A space X is T2 if and only if the diagonal ∆ = {(x, x) : x ∈ X} is closed in X × X. Proof: Suppose that X is T2 and (x, y) 6∈ ∆. Then x 6= y, so there are disjoint open sets U and V with x ∈ U and y ∈ V . But then (x, y) ∈ U × V and (U × V ) ∩ ∆ = ∅. Hence the complement of ∆ is open in X × X. Conversely, Suppose that ∆ is closed in X × X and suppose that x 6= y in X. Then (x, y) 6∈ ∆, so there is a basic open set U × V with (x, y) ∈ U × V ⊆ (X × Y ) \ ∆. But then, x ∈ U , y ∈ V and U ∩ V = ∅. Theorem 4.8 Suppose that f X → Y is continuous and Y is T2 . Then A = {(x1 , x2 ) : f (x1 ) = f (x2 )} is closed in X × X. Proof: Suppose that (x1 , x2 ) 6∈ A. Then f (x1 ) 6= f (x2 ). Then there are disjoint open sets U and V with f (x1 ) ∈ U and f (x2 ) ∈ V . But then x1 ∈ f −1 [U ], x2 ∈ f −1 [V ]. Furthermore, if (a, b) ∈ f −1 [U ] × f −1 [V ], then f (a) ∈ U and f (b) ∈ V . Since U and V are disjoint, f (a) 6= f (b), so (a, b) 6∈ A. Hence f −1 [U ] × f −1 [V ] ∩ A = ∅, so A is open in X × X. Theorem 4.9 Suppose that f, g : X → Y are continuous and Y is T2 . Then {x ∈ X : f (x) = g(x)} is closed in X. Proof: Let A = {x ∈ X : f (x) = g(x)}. Suppose that x0 6∈ A, so f (x0 ) 6= g(x0 ). Since Y is T2 , there are disjoint open sets U and V in Y such that f (x0 ) ∈ U and g(x0 ) ∈ V . Then x0 ∈ f −1 [U ] ∩ g −1 [V ] is open and if x ∈ f −1 [U ] ∩ g −1 [V ], then f (x) ∈ U and g(x) ∈ V , so f (x) 6= g(x), so x 6∈ A. Hence X \ A is open. 40 CHAPTER 4. SEPARATION AXIOMS Corollary 4.10 Suppose that f, g : X → Y are continuous and D ⊆ X is dense in the space X and that Y is T2 . Suppose that f (x) = g(x) for all x ∈ D. Then f = g. The following should be compared to the corresponding result for second countablility where the number of factors must be countable. Theorem 4.11 Let {Xα } be a collection of T2 spaces. Then Q Xα is separable if and only if each Xα is separable and card{α : cardXα 6= 1} ≤ card[0, 1]. Proof: Let A = {α : cardXα 6= 1}. For each α ∈ A, let Uα and Vα be disjoint open sets in Xα . First suppose that Q Xα is separable and let D be a countable dense set. Then πα [D] is dense in Xα for each α. In fact, if Uα is open in Xα , then πα−1 [Uα ] ∩ D 6= ∅, so πα [D] ∩ Uα 6= ∅. Next, for each α ∈ A, consider the subset πα−1 [Uα ] ∩ D = Dα of D. If α 6= β in A, then πα−1 [Uα ] ∩ πβ−1 [Vβ ] 6= ∅ so the intersection of this set with D is nonempty. Hence Dα ∩ πβ−1 [Vβ ] 6= ∅ while Dβ ∩ πβ−1 [Vβ ] = ∅. Thus Dα 6= Dβ . In other words, the map that takes α ∈ A to Dα ∈ P(D) is one-to-one. Hence, cardA ≤ cardP(D) = card[0, 1]. Conversely, suppose that each Xα is separable and cardA ≤ card[0, 1]. We may regard A as a subset of [0, 1]. Let Dα = {d1α , d2α , · · · } be a countable dense subset of Xα . For each finite sequence of natural numbers {n1 , · · · , nk } and each finite disjoint collection of closed intervals {F1 , · · · , Fk } in [0, 1] with Q rational endpoints, let d(n1 , · · · , nk , F1 , · · · , Fk ) ∈ Xα be defined by setting d if α ∈ Fi ni α d(n1 , · · · , nk , F1 , · · · , Fk )(α) = d1α if α 6∈ ∪Fi There are only countably many such sequences of natural numbers and intervals with rational endpoints, so the collection, D, of all d(n1 , · · · , nk , F1 , · · · , Fk ) is a countable subset of the product. Q Q Now, suppose that U = Uα is a basic open set in Xα with Uα = Xα unless α = α1 , · · · , αk . Since A ⊆ [0, 1], we may find disjoint closed 4.2. REGULARITY 41 intervals F1 , · · · , Fk with rational endpoints such that αi ∈ Fi if αi ∈ A. Next, we may find dni αi ∈ Uαi ∩ Dαi since Dα is dense in Xα . But then d(n1 , · · · , nk , F1 , · · · , Fk ) ∈ U ∩ D. Thus D is dense in the product. 4.2 Regularity Definition 4.12 We say that a topological space X is regular if whenever C ⊆ X is closed and x 6∈ C, there are disjoint open sets U and V such that x ∈ U and C ⊆ V . In other words, we can separate points from closed sets via open sets. We say that X is T3 if it is both regular and T1 . Since singletons are closed in T1 spaces, a T3 space is T2 . It is possible, however, that a regular space is not T1 and hence will not be T2 . Proposition 4.13 For a topological spaces X, the following are equivalent: a) X is a regular space. b) Whenever U ⊆ X is open and x ∈ U , there is an open set V such that x ∈ V ⊆ V ⊆ U. Theorem 4.14 Products and subspaces of regular spaces are regular. Proof: Suppose that X is regular and A ⊆ X. Suppose that x 6∈ C, which is closed in A. Then C = A ∩ K for some closed subset K ⊆ X. Since x 6∈ K, there are disjoint open sets U and V in X such that x ∈ U and K ⊆ V . But then A∩U and A∩V are disjoint open sets in A and x ∈ A∩U while C ⊆ A∩V . Theorem 4.15 Suppose that q : X → Y is an open, closed, continuous, onto map and that X is T3 . Then Y is T2 . Proof: Since q is closed, continuous and onto, q is a quotient map. Since X is T1 , singletons are closed in X, so singletons are closed in Y , so Y is T1 . Now suppose that y1 = q(x1 ) 6= q(x2 ) = y2 in Y . Then x1 ∈ X \ q −1 [y2 ]. Since X is T3 , there are disjoint open sets U and V so that x1 ∈ U and q −1 [y2 ] ⊆ 42 CHAPTER 4. SEPARATION AXIOMS V . By theorem 3.10, there is an open set W of Y such that y2 ∈ W and q −1 [W ] ⊆ V . Now, q[U ] is open in Y and y1 ∈ q[U ]. Finally, q[U ] ∩ W = ∅, showing that Y is T2 . 4.3 Complete Regularity The next separation axiom seems rather different than the others at first glance. It is, however, an important step. Definition 4.16 We say that a topological space X is completely regular if whenever C ⊆ X is closed and x 6∈ C, there is a continuous function f : X → [0, 1] such that f (x) = 0 and f [C] = {1}. We say that X is T3 12 if it is both T1 and completely regular. Theorem 4.17 Products and subspaces of completely regular spaces are completely regular. Proof: Suppose that X is completely regular and A ⊆ X. Suppose that C ⊆ A is closed in A and that x ∈ A with x 6∈ C. Then, we may write C = A ∩ K where K is closed in X. Notice that x 6∈ K, so there is a continuous function f : X → [0, 1] such that f (x) = 0 and f [K] = {1}. Then f |A : A → [0, 1] is continuous and f |A [C] = {1}. Hence subspaces of completely regular spaces are completely regular. Now suppose that {Xα } is a collection of completely regular spaces, C ⊆ Q X = Xα is closed and x ∈ X \ C. Then, there are finitely many α1 , · · · αn Tn and open sets Uαk ⊆ Xαk such that x ∈ U = k=1 πα−1 [Uαk ] ⊆ X \ C. Then, k xαk ∈ Uαk , so there is a continuous function fαk : Xαk → [0, 1] such that f (xαk ) = 1 and fαk [Xαk \ Uαk ] = {0}. Let f : X → [0, 1] be defined to be the product of fαk ◦ παk where 1 ≤ k ≤ n. Then f is continuous on X, f (x) = 1 and if y 6∈ U , some yαk 6∈ Uαk , so f (y) = 0. Hence f [X \ C] = {0}. Theorem 4.18 If a space X is completely regular then it has the weak topology induced by the collection of continuous functions into [0, 1]. 4.4. NORMALITY Proof: 43 If X is completely regular, the collection of all continuous functions from X into [0, 1] separates points from closed sets. Hence, by theorem 3.26, X has the weak topology induced by these functions. Theorem 4.19 A space can be embedded into a product of [0, 1] if and only if it is T3 21 . Proof: This now follows easily from theorem 3.24 and the previous results for subspaces and products of T3 21 spaces. In essence, this shows that T3 12 spaces are those with enough continuous real-valued functions to define the topology. These are then the natural spaces on which to do analysis of continuous functions. It is known that there are T3 spaces with more than two points where every continuous function is constant. Such spaces are not completely regular. 4.4 Normality Definition 4.20 We say that a topological space X is normal if whenever C and D are disjoint closed sets, there are disjoint open sets U and V such that C ⊆ U and D ⊆ V . We say that X is T4 if it is both T1 and normal. Clearly T4 spaces are T3 . It is not as clear, althought it is true, that T4 spaces are T3 21 . Theorem 4.21 Metric spaces are T4 . Proof: Let C and D be disjoint closed sets in X. Define f (x) = d(x, C) . d(x, C) + d(x, D) Then f : X → [0, 1] is defined for every x ∈ X since C and D are disjoint and gives a continuous function. Furthermore, f (x) = 0 if and only if x ∈ C and f (x) = 1 if and only if x ∈ D. Let U = f −1 [[0, 14 )] and V = f −1 [( 34 , 1]]. Then C ⊆ U , D ⊆ V and U and V are disjoint open sets in X. 44 CHAPTER 4. SEPARATION AXIOMS This proof shows, by the way, that metric spaces are T3 12 . Example: The Sorgenfry line L is normal. In fact, suppose that C and D are disjoint closed sets in L. For each x ∈ C, there is a εx > 0 such that [x, x + εx ) ∩ D = ∅. Similarly, for each y ∈ D there is a δy > 0 such that S S [y, y + δy ) ∩ C = ∅. Now, let U = [x, x + εx ) and V = [y, y + δy ). Clearly U and V are open with C ⊆ U and D ⊆ V . But, if U ∩ V 6= ∅, there would be x ∈ C and y ∈ D such that [x, x + εx ) ∩ [y, y + δy ) 6= ∅. But then either x ∈ [y, y + δy ) or y ∈ [x, x + εx ). Either one is a contradiction. Unfortunately, it is not the case that subspaces and products of T4 spaces are T4 . So, while this separation axiom is very nice in some ways, it is not so nice in others. Theorem 4.22 Suppose that X is a T4 space. Suppose that D ⊆ X is dense and that S ⊆ X is closed and discrete (in the subspace topology). Then 2|S| ≤ 2|D| . Proof: For each A ⊆ S, both A and S \ A are closed in X, so there are disjoint open sets U (A) and V (A) such that A ⊆ U (A) and S \ A ⊆ V (A). Consider the subset U (A) ∩ D of D. This gives us a function from P(S) to P(D). I claim that this map is one-to-one, showing the result. In fact, if A 6= B ⊆ S, we may assume that there is x ∈ A \ B. Then x ∈ U (A) ∩ V (B) so this open set has non-empty intersection with D, that is U (A) ∩ V (B) ∩ D 6= ∅. But, since U (B) ∩ V (B) = ∅, this shows that U (A) ∩ D 6= U (B) ∩ D, giving the result. Example: The product L × L is not a T4 space. In fact, let D = Q × Q, a countable dense subset and A = {(x, −x) : x ∈ L}, a discrete, closed subset of L × L. Then |A| = 2|D| , so 2|A| > 2|D| . Hence L × L cannot be T4 . Thus product of T4 spaces need not be T4 . On the other hand, we have the following crucial result about normal spaces. Theorem 4.23 (Urysohn’s Lemma) A topological space is normal if and only if whenever C and D are disjoint closed sets, there is a continuous function f : X → [0, 1] such that f [C] = {0} and f [D] = {1}. 4.4. NORMALITY 45 Proof: If such a function exists, then U = f −1 [[0, 41 )] and V = f −1 [( 34 , 1]] are disjoint open sets containing C and D. So this condition implies normality. The converse is more difficult because we have to construct a continuous function on X ’from scratch’. First, let U1 = X \ D. Then C ⊆ U1 , so there is an open set U0 such that C ⊆ U0 ⊆ U0 ⊆ U1 . Next, there is an open set U 12 such that U0 ⊆ U 12 ⊆ U 12 ⊆ U1 . Continuing in this fashion we can find open sets Uq for every dyadic rational q = m/2n such that if q < r, we have Uq ⊆ Ur . Now, define f : X → [0, 1] by f (x) = inf{q : x ∈ Uq } if x ∈ U1 while f (x) = 1 otherwise. Clearly, if x ∈ C, f (x) = 0 and if x ∈ D, f (x) = 1. It remains to prove that f is continuous. To see this, notice that f −1 [[0, a)] = [ Uq q<a and f −1 [(b, 1]] = [ X \ Uq b<q are both open sets in X. The second equality follows since if f (x) > b, there are dyadic rationals with b < q < r < f (x), so x 6∈ Ur , which implies that x 6∈ Uq . Conversely, if x 6∈ Uq , then f (x) ≥ q > b. It follows that T4 spaces are T3 12 . This also shows that every T4 space can be embedded in a product of [0, 1]. The next result is another crucial result, allowing the extension of continuous functions from subspaces to the whole space. Theorem 4.24 (Tietze’s Extension Theorem). A topological space X is normal if and only if whenever C ⊆ X is closed and f : C → [0, 1] is continuous, there is a continuous F : X → [0, 1] with F |C = f . Proof: It is easier to use [−1, 1] in place of [0, 1] for this proof. So suppose that f : A → [−1, 1] is continuous. Let C = f −1 [−1, 1 31 ] and D = f −1 [ 13 , 1]. Then C and D are disjoint closed sets in A and hence in X. But Urysohn’e Lemma, there is a function g1 : X → [− 13 , 31 ] such that g1 [C] = − 13 and g1 [D] = Then, for every a ∈ A, we have |f (a) − g1 (a)| ≤ 2 3 and |g1 (x)| ≤ 1 3 1 3. for all x ∈ X. 46 CHAPTER 4. SEPARATION AXIOMS Now consider the function ( 23 )(f − g1 ) : A → [−1, 1]. By the previous paragraph, there is a function g2 : X → [− 13 , 13 ] such that |( 32 (f − g1 ) − g2 | ≤ on A and |g2 | ≤ 1 3 2 3 on X. But then |f − (g1 − 23 g2 )| ≤ ( 23 )2 on A. Continue in this way to find a sequence gn : X → [− 13 , 13 ] of continuous functions such that |f − n X 2 2 ( )k−1 gk | ≤ ( )n 3 3 k=1 Notice that the sum in this is convergent on X and the limit will be a continuous F : X → [−1, 1]. By construction, f = F on A. The same result is true if we replace [0, 1] by (0, 1). In fact, suppose that f : A → (−1, 1) is continuous. Use Tietze’s Extension theorem to extend to F : X → [−1, 1]. Now, A and B = f −1 [{−1, 1}] are disjoint closed sets in X, so there is a continuous function g : X → [0, 1] such that g[A] = {1} and g[B] = {0}. Now let F 0 (x) = F (x)g(x). Then F 0 : X → (−1, 1) is continous and extends f . 4.5 Lindelöf Spaces Definition 4.25 An open cover of a topological space X is a collection of open S sets {Uα }α∈A in X such that X = α∈A Uα . A subcover of {Uα }α∈A is a cover {Uβ }β∈B where B ⊆ A. We say the subcover is finite if the set B is finite. Similarly, the subcover is said to be countable if B is countable. Definition 4.26 A topological space X is said to be Lindelöf if every open cover of X has a countable subcover. Proposition 4.27 In the definition of Lindelöf, is is sufficient to consider only basic open sets. Proof: Suppose that B is a base for the topology of X and that U is an open cover of X. For each x ∈ X, there is a Ux ∈ U and a Bx ∈ B such that x ∈ Bx ⊆ Ux . But then {Bx }x∈X is a cover of X by basic open sets. If {Bxn } is a countable subcover, then {Uxn } is a countable subcover of U. In particular, second countable spaces are Lindelöf. 4.5. LINDELÖF SPACES 47 Theorem 4.28 Closed subspaces and continuous images of Lindelöf spaces are Lindelöf. Proof: Suppose that X is Lindelöf and A ⊆ X is closed. Suppose also that {Vα } is an open cover of A in the subspace topology. Then, there are open sets {Uα } in X with Vα = A ∩ Uα . Then, the collection of open sets {X \ A} ∪ {Uα } is an open cover of X. Since X is Lindelöf, there is a countable subcover, say {X \ A} ∪ {Uαn }. Notice that it does no harm to assume that X \ A is in this subcover. But then {Vαn } is a countable subcover of {Vα } for A. Theorem 4.29 A regular Lindelöf space is normal. Proof: Suppose that C and D are disjoint closed subsets of the Lindelöf space X. For each x ∈ C, there is an open set Ux with x ∈ Ux and D ∩ Ux = ∅. Similarly, for each y ∈ D, there is a Vy open with y ∈ Vy and C ∩ Vy = ∅. Since closed subspaces of Lindelöf spaces are Lindelöf, there are countably many S∞ S∞ {Un } and {Vn } with C ⊆ n=1 Un and D ⊆ n=1 Vn . Now, let W1 = U1 , Sn Z1 = V1 \ W1 , W2 = U2 \ Z1 , Z2 = V2 \ (W1 ∪ W2 , Wn+1 = Un+1 \ k=1 Zk , Sn+1 Zn+1 = Vn+1 \ k=1 Wk . S∞ S∞ Finally, let W = n=1 Wn and Z = n=1 Zn . Then W and Z are open, C ⊆ W and D ⊆ Z. Finally, W ∩ Z = ∅. Example: The Sorgenfry line L is Lindelöf. Proof: Suppose that {[aα , bα )} is a cover of L by basic open sets. Let A = S α (aα , bα ). Since R is second countable, the subset A of R is second countable, S∞ and so it is Lindelöf. Hence, we can write A = n=1 (an , bn ). Now, suppose that x ∈ L \ A. Then x = aα for some α and we may pick a rational qx ∈ (aα , bα ) ⊆ A. Suppose that y 6= x in L \ A. Write y = aβ . Then, for qx = qy , we would need that (aα , bα ) ∩ (aβ , bβ ) 6= ∅. But then, either x = aα ∈ (aβ , bβ ) ⊆ A or y = aβ ∈ (aα , bα ). Either case is a contradiction. Since there are only countably many rational numbers, the set L \ A must be countable. Thus L \ A can be covered by countably many of the original open sets. Since A is already covered by countably many, we have that L is covered by countably many of the basic open sets. 48 CHAPTER 4. SEPARATION AXIOMS Since L × L is not normal, but is T3 , the product of Lindelöf spaces need not be Lindelöf. Finally, we give an incredible result which identifies purely topological properties which imply metrizability, at least in the case of separablility. Theorem 4.30 (Urysohn Metrization Theorem) The following are equivalent for a topological space X. a) X is a T3 , second countable space. b) X is separable and metrizable. c) X can be embedded into a countable product of [0, 1]. Proof: First notice that a countable product of [0, 1] is metrizable and second countable, so every subspace will be metrizable and second countable, hence separable. Hence c) implies b). Next, a separable metric space is second countable and T4 , hence T3 , so b) implies a). So, now assume that X is T3 and second countable. Then X is Lindelöf, so we know that X is T4 . Let B be a countable base for the topology of X and consider the collection {(B1 , B2 ) : B1 , B2 ∈ B, B1 ⊆ B2 }. This is a countable 0 collection of pairs, so we may re-write this set as {(Bn , Bn ) : n ∈ N}. 0 Now, for each pair, (Bn , Bn ), Urysohn’s lemma gives a continuous function 0 fn : X → [0, 1] such that fn [Bn ] = {0}. and fn [X \ Bn ] = {1}. Notice that the collection {fn } separates points from closed sets. In fact, if C ⊆ X is closed and x 6∈ C, then there is a B ∈ B such that x ∈ B and B ∩ C = ∅. But, since X is T3 , there is a B 0 ∈ B such that x ∈ B 0 and B 0 ⊆ B. Then, the pair (B 0 , B) corresponds to some n ∈ N and fn (x) = 0 while fn [C] = {1}. Q This now shows that the function f : X → n [0, 1] such that fn = πn ◦ f is an embedding. Hence a) implies c). Exercises: 1. Suppose that fn : X → R are continuous functions which converge uni- formly to f : X → R. Show that f is continuous on X. 4.5. LINDELÖF SPACES 49 2. Prove Proposition 4.13 3. Suppose that X is a T3 space and x 6= y in X. Show that there are open sets U and V with x ∈ U , y ∈ V and U ∩ V = ∅. 4. Suppose that X is a T3 space that A ⊆ X is an infinite set. Show that there are open sets Un such that Un ∩A 6= ∅ for all n ∈ N and Un ∩Um = ∅ for all n 6= m. 5. A subset A of a topological space is called a retract of X if there is a continuous map r : X → A such that r(a) = a for all a ∈ A. Suppose that X is T2 and that A is a retract of X. Show that A is closed in X. 6. Show that A ⊆ X is a retract if and only if whenever f : A → Y is a continuous function, there is an F : X → Y which extends f . 7. We call a space Y an absolute retract if whenever A ⊆ X is a closed subspace of a T4 space and f : A → Y is continuous, there is an extension F : X → Y . Note the Tietze’s Extension theorem says exactly that [0, 1] is an absolute retract. Show that if Y is an absolute retract and f : Y → X is an embedding as a closed subspace of a T4 space, then f [Y ] is a retract of X. 8. Show that a product of absolute retracts is again an absolute retract. 9. Suppose that A ⊆ R is uncountable. Show that there is an x ∈ R such that for every ε > 0, the set A ∩ [x, x + ε) is uncountable. 10. Suppose that A ⊆ X is a closed subset. Show that the quotient map q : X → X/A is a closed map. 11. Suppose that X is a T3 space and A ⊆ X is a closed subset. Show that X/A is T2 . 12. Suppose that both X and Y are T1 (resp, T2 , T3 , T3 12 , T4 ). Show that X ⊕ Y is T1 (resp, T2 , T3 , T3 12 , T4 ). 50 CHAPTER 4. SEPARATION AXIOMS 13. A space X is said to be completely normal if every subspace of X is normal. Show that X is completely normal if and only if whenever A, B ⊆ X with A ∩ B = ∅ = B ∩ A, there are disjoint open sets U and V with A ⊆ U and B ⊆ V . Hint: for one direction consider the subspace X \ (A ∩ B). Chapter 5 Convergence and Metric Spaces 5.1 First Countable Spaces Definition 5.1 A sequence {xn } in a topological space X is said to converge to the point x ∈ X if, whenever U is an open set with x ∈ U , there is an N ∈ N such that n ≥ N implies that xn ∈ U . We will write xn → x. Notice that it is enough to check this condition for open sets U in a neighborhood base at x. In particular, if (X, d) is a metric space, then xn → x if and only if for each ε > 0, there is an N ∈ N such that n ≥ N implies that xn ∈ B(x; ε), in other words, n ≥ N implies that d(xn , x) < ε. Theorem 5.2 Suppose that X is a first countable space. Then the following hold. a) Suppose that A ⊆ X. Then x ∈ A if and only if there is a sequence xn ∈ A with xn → x. b) A subset A ⊆ X is closed if and only if whenever xn is a sequence in A and xn → x ∈ X, then x ∈ A. 51 52 CHAPTER 5. CONVERGENCE AND METRIC SPACES c) A subset U ⊆ X is open if and only if whenever xn is a sequence with xn → x ∈ U , then there is an N inN such that n ≥ N implies xn ∈ U . d) Suppose that Y is a topological space and f : X → Y . Then f is continuous if and only if whenever xn → x ∈ X, we have that f (xn ) → f (x) ∈ Y . Proof: a) Suppose that x ∈ A and let N = {Un : n ∈ N} be a countable neighborhood base at x. We may assume that Un+1 ⊆ Un for all n by replacing Un by U1 ∩ · · · Un . For each n, A ∩ Un 6= ∅, so pick xn ∈ A ∩ Un . By construction n ≥ N implies that xn ∈ UN , so xn → x. Conversely, suppose that xn ∈ A and xn → x ∈ X but that x 6∈ A. Then x ∈ U = X \ A, so by definition of convergence there is an N ∈ N such that n ≥ N implies that xn ∈ U . But this contradicts that xn ∈ A. b) This follows immediately from a). c) This follows by letting A = X \ U and applying b). d Suppose that f : X → Y is continuous, that xn → x ∈ X and that f (x) ∈ V , open in Y . Then x inf −1 [V ] is open in X, so there is an N ∈ N such that n ≥ N implies that xn ∈ f −1 [V ]. But then n ≥ N implies that f (xn ) ∈ V , so f (xn ) → f (x). Notice that this direction does not require that X be first countable. Conversely, suppose that whenever xn → x ∈ X, we can conclude that f (xn ) → f (x) ∈ Y . We need to show that f is continuous. So, suppose that V ⊆ Y is open and consider the set U = f −1 [V ]. We use the criterion in c to show that U is open in X. But, if xn → x ∈ U , then f (xn ) → f (x) ∈ V . Hence, there is an N ∈ xn ∈ f −1 N such that n ≥ N implies that f (xn ) ∈ V . But then [V ] = U for n ≥ N . Hence, U is open in X and so f is continuous. One way to think of this theorem is that convergence of sequences is enough to define the topology of any first countable space. Unfortunately, this is not the case for general topological spaces and the following example shows. Example: Let X = [0, 1][0,1] as a product space with uncountably many factors. Let A be the collection of f : [0, 1] → [0, 1] such that {x : f (x) 6= 0} is a finite set. Let g be the function which is identically 1. The claim is that g ∈ A but 5.2. CONVERGENCE OF NETS 53 that no sequence fn from A converges to g. For the first, suppose that U is Q a basic open set with g ∈ U . Write U = x∈[0,1] Ux where Ux = [0, 1] except for finitely many x = x1 , · · · xn . Let f (x) = 1 if x = x1 , · · · xn and f (x) = 0 otherwise. Then, clearly f ∈ A ∩ U . Thus A ∩ U 6= ∅, showing that g ∈ A. However, if fn is a sequence in A, we can let Fn = {x : fn (x) 6= 0}. Then each Fn is a finite set, so F = ∪n Fn is a countable set. Pick any x 6∈ F . Then πx : [0, 1][0,1] → [0, 1] is continuous, πx (fn ) = 0 for all n ∈ N, but πx (g) = 1. Hence, πx (fn ) does not converge to πx (g). This fn does not converge to g. Example: Another illustrative example is found in Ω, the first uncountable ordinal. Recall that subbasic open sets are those of the form (←, ω0 ) = {ω : ω < ω0 } and (ω0 , →) = {ω : ω0 < ω}. Recall that ω1 is the largest element of Ω and that Ω0 = {ω ∈ Ω : ω < ω1 }. Then ω1 ∈ Ω0 , but if {ωn } is a sequence in Ω0 , there is an upper bound in Ω0 , so ωn does not converge to ω1 . 5.2 Convergence of Nets Definition 5.3 We say that Λ is a directed set if it has a relation ≤ such that a) λ ≤ λ for each λ ∈ Λ b) If λ1 ≤ λ2 and λ2 ≤ λ3 , then λ1 ≤ λ3 for λi ∈ Λ c) If λ1 , λ2 ∈ Λ, then there is a λ3 ∈ Λ such that λ1 , λ2 ≤ λ3 . Notice that ≤ does not have to be a partial order on Λ since we do not assume anti-symmetry. A very important directed set for our purposes is Nx = {U ∈ τ : x ∈ U } where x ∈ X, a topological space and we define U ≤ V if and only if V ⊆ U . Definition 5.4 A net in a set X is a function x : Λ → X from a directed set Λ to X. We often denote the image of λ ∈ Λ by xλ . If X is a topological space and x0 ∈ X, we say that the net (xλ ) converges to x0 if whenever U is an open set with x ∈ U , there is a λ0 ∈ Λ such that λ0 ≤ λ implies that xλ ∈ U . In this case, we write xλ → x. 54 CHAPTER 5. CONVERGENCE AND METRIC SPACES If (xλ ) is a net and λ0 ∈ Λ, we call a set of the form {xλ : λ0 ≤ λ} a tail of the net (xλ ). Convergence of a net to x means that every neighborhood around x contains a tail of the net. Notice that sequences are particular types of nets and our notion of convergence for ners generalizes that for sequences. The following should be compared to theorem 5.2) in both the statement and the proof. Theorem 5.5 Let X be a topological space. Then, the following hold. a) Let A ⊆ X. Then x ∈ A if and only if there is a net (xλ ) in A such that xλ → x. b) A subset A ⊆ X is closed if and only if whenever (xλ ) is a net in A with xλ → x ∈ X, we can conclude that x ∈ A. c) A subset U ⊆ X is open if and only if whenever (xλ ) is a net with xλ → x ∈ U , then some tail of (xλ ) is contained in U . d) Let f : X → Y be a function between topological spaces. Then f is continuous if and only if whenever (xλ ) is a net in X with xλ → x ∈ X, then f (xλ ) → f (x) ∈ Y . Proof: a) Suppose that (xλ ) is a net in A with xλ → x ∈ X. Let U be an open set with x ∈ U . Then, there is a λ0 ∈ Λ such that λ ≥ λ0 implies that xλ ∈ U . Hence A ∩ U 6= ∅. This shows that x ∈ A. Conversely, suppose that x ∈ A. Let Λ = {U : x ∈ U open} with U1 ≤ U2 if and only if U2 ⊆ U1 . Then, for each U ∈ Λ, A ∩ U 6= ∅, so there is a xU ∈ A ∩ U . Now, (xU ) is a net in A which convwerges to x. In fact, if x ∈ U is open, just take λ0 = U . b) This follows immediately from a). c) Again, this follows immediately from b) by letting A = X \ U . d) Suppose that f is continuous and that xλ → x ∈ X. Let f (x) ∈ V , an open set in Y .Then x ∈ f −1 [V ], an open set in X, so there is a λ0 ∈ Λ such 5.2. CONVERGENCE OF NETS 55 that λ ≥ λ0 implies that xλ ∈ f −1 [V ]. But then f (xλ ) ∈ V , so f (xλ ) → f (x) in Y . Converely, we use c) to show that f −1 [V ] is open in X when V is open in Y . In particular, suppose that xλ → x ∈ f −1 [V ]. Then f (xλ ) → f (x) ∈ V , so there is a λ0 ∈ Λ such that λ ≥ λ0 implies that f (xλ ) ∈ V . But then the tail {xλ : λ ≥ λ0 } is contained in f −1 [V ]. Thus, f −1 [V ] is open in X and so f is continuous. Theorem 5.6 Suppose that X is a T2 space and (xλ ) is a net in X that converges to both x and y. Then x = y. Proof: If not, there are disjoint open sets U and V with x ∈ U and y ∈ V . There are then λ0 , λ1 ∈ Λ such that λ ≥ λ0 implies that xλ ∈ U and λ ≥ λ1 implies that xλ ∈ V . Since there is a λ ≥ λ0 , λ1 , this is a contradiction. Theorem 5.7 Let {Xα } be a collection of topological spaces and let X = Q Xα . Then, a net (xλ ) in X converges if and only if πα (xλ ) → πα (x) for every index α. Proof: Since each πα is continuous, one implication is automatic. Now suppose [Uαi ] be a basic open set that πα (xλ ) → πα (x) for every α. Let U = ∩ni=1 πα−1 i containing x. Then παi (x) ∈ Uαi for each 1 ≤ i ≤ n. Hence, there are λi ∈ Λ such that λ > λi implies that παi (xλ ) ∈ Uαi . Since Λ is a directed set, there is a λ0 ≥ λi for all 1 ≤ i ≤ n. But then, if λ > λ0 , we have that παi (xλ ) ∈ Uαi for all i, which shows that xλ ∈ U . Thus, xλ → x. Example: Let’s reconsider example 5.1. In that example, A is the set of functions which are 0 except on a finite subset of [0, 1] and g is the constant function 1. It was shown there that g ∈ A but that no sequence from A converges to g. Now, let Λ consist of all finite subseteq of [0, 1] and define F1 ≤ F2 if F1 ⊆ F2 . For F ∈ Λ, let fF be the function which is 1 on F and 0 elsewhere. Q We claim that the net (fF ) converges to g. In fact, suppose that U = Uα is a basic open set containing g. Then F0 = {α : Uα 6= [0, 1]} ∈ Λ and if F0 ≤ F , we clearly have that fF ∈ U . 56 CHAPTER 5. CONVERGENCE AND METRIC SPACES Definition 5.8 A subnet of a net (xλ ) is the composition x◦φ where φ : M → Λ is a function from a directed set M into Λ such that m1 ≤ m2 in M implies that φ(m1 ) ≤ φ(m2 ) in Λ (we say that φ is increasing) and such that whenever λ0 ∈ Λ, there is an m ∈ M such that λ0 ≤ φ(m) (we say that φ is cofinal). Definition 5.9 Let (xλ ) be a net in a topological space X and let x ∈ X. We say that x is a cluster point of (xλ ) if whenever x ∈ U , an open set and λ0 ∈ Λ, there is a λ ≥ λ0 such that xλ ∈ U . We say that xλ is in U frequently. Theorem 5.10 Let (xλ ) be a net in a topological space X and let x ∈ X. Then the following are equivalent: a) x is a cluster point of (xλ ) b) There is a subnet of (xλ ) which converges to x c) x ∈ Proof: T λ∈Λ {xµ : µ ≥ λ}. First assume that x is a cluster point of (xλ ). Define M = {(λ, U ) : x ∈ U open , xλ ∈ U }. Define (λ1 , U1 ) ≤ (λ2 , U2 ) if and only if λ1 ≤ λ2 and U2 ⊆ U1 . Then M is a directed set. Now define φ(λ, U ) = λ. Clearly φ is increasing. Also, φ(λ, X) ≥ λ, so φ is also cofinal. We show that the subnet x ◦ φ converges to x. For this, suppose that x ∈ U open. By assumption, there is a λ ∈ Λ such that xλ ∈ U . Hence (λ, U ) ∈ M . Now suppose that (µ, V ) ≥ (λ, U ). Then, we have that x ◦ φ(µ, V ) = xµ ∈ V ⊆ U , as required. Next, suppose that the subnet x ◦ φ converges to x, λ ∈ Λ and let U be open with x ∈ U . Then, there is a m0 ∈ M such that whenever m ≥ m0 , we have x ◦ φ(m) ∈ U . Next, there is an m1 ∈ M such that φ(m1 ) ≥ λ. Finally, since M is a directed set, there is an m ∈ M such that m ≥ m0 , m1 . But then x ◦ φ(m) ∈ U and φ(m) ≥ φ(m1 ) ≥ λ, so U ∩ {xµ : µ ≥ λ} 6= ∅. Hence x ∈ {xµ : µ ≥ λ}. This holds for all λ ∈ Λ. T Finally, suppose that x ∈ λ∈Λ {xµ : µ ≥ λ}. Then, for each open set U such that x ∈ U and for each λ ∈ Λ there is a µ ≥ λ such that xµ ∈ U . Hence x is a cluster point of (xλ ). 5.3. COMPLETE METRIC SPACES 5.3 57 Complete Metric Spaces Definition 5.11 Let (X, d) be a metric space. A sequence {xn } in X is said to be a Cauchy sequence if whenever ε > 0, there is a N ∈ N such that whenever n, m ≥ N , we have d(xn , xm ) < ε. We say that (X, d) is a complete metric space if every Cauchy sequence in X converges. Proposition 5.12 In any metric space, any convergent sequence is Cauchy. Example: The metric space R with the usual metric is complete. However, the homeomorphic space (0, 1) is not complete in the usual metric. This shows that completeness is a metric property and not a topological property. Example: If (X, d) is complete if and only if (X, d0 ) is complete where d0 (x, y) = max{d(x, y), 1}. Hence every complete metric space has a complete, bounded metric. Definition 5.13 A subset A of a metric space (X, d) is said to be bounded if there is an M ∈ R such that d(x, y) ≤ M for all x, y ∈ A. in this case, we define the diameter of A by diam(A) = sup{d(x, y) : x, y ∈ A}. Proposition 5.14 Cauchy sequences are bounded. Proof: There is an N ∈ R such that whenever n, m ≥ N , we have d(xn , xm ) ≤ 1. Let M = 1 + max{d(xn , xm ) : n, m ≤ N }. Then, for every n, m ∈ N, d(xn , xm ) ≤ M . In particular, if n < N and m ≥ N , we have d(xn , xm ) ≤ d(xn , xN ) + d(xN , xm ) ≤ 1 + d(xn , xN ) ≤ M . Proposition 5.15 If a Cauchy sequence has a convergent subsequence, then the Cauchy sequence converges. Proof: Suppose that the subsequence xnk converges to x. Let ε > 0 and choose K ∈ N so that k ≥ K implies that d(xnk , x) < ε/2. Now choose N ∈ N so that n, m ≥ N implies that d(xn , xm ) < ε/2. Then, if n ≥ max{nK , N }, we have d(xn , x) ≤ d(xn , xnK ) + d(xnK , x) < ε/2 + ε/2 = ε. 58 CHAPTER 5. CONVERGENCE AND METRIC SPACES Theorem 5.16 A subspace of a complete metric space is complete (in the restricted metric) if and only if it is closed. Proof: Suppose that A ⊆ X is complete and suppose that xn → x0 ∈ X. Then {xn } is Cauchy in X, so is Cauchy in A, so it converges in A. Since metric spaces are T2 , this implies that x0 ∈ A, so A is closed. Conversely, if A is closed and {xn } is a Cauchy sequence in A, then it is a Cauchy sequence in X, so it converges to some point x0 ∈ X. But since A is closed, x0 ∈ A, so {xn } converges in A, so A is complete. Theorem 5.17 Suppose that X is a complete metric space and f : X → X satisfies d(f (x), f (y) ≤ α · d(x, y) for all x, y ∈ X with α < 1. Then, there is a unique x0 ∈ X such that f (x0 ) = x0 . Proof: Pick any x1 ∈ X and define, inductively, xn+1 = f (xn ). Let A = d(x1 , x2 ). Then, d(xn+1 , xn+2 ) ≤ α · d(xn , xn+1 ) ≤ · · · ≤ αn d(x1 , x2 ) = αn A. Thus, for m < n, we have d(xm , xn ) ≤ d(xm , xm+1 ) + d(xm+1 , xm+2 ) + · · · + d(xn−1 , xn ) ≤ αm−1 A + αm A + · · · + αn−2 A ≤ Aαm−1 /(1 − α). Since αm−1 → 0 as m → ∞, this shows that {xn } is a Cauchy sequence. Suppose that xn → x0 . Then xn+1 = f (xn ) → f (x0 ), but xn+1 → x0 also. Thus f (x0 ) = x0 . To show uniqueness, suppose that f (y0 ) = y0 also. Then d(x0 , y0 ) = d(f (x0 ), f (y0 )) ≤ α · d(x0 , y0 ). Since α < 1, this implies that d(x0 , y0 ) = 0, so x0 = y0 . Theorem 5.18 Suppose that (Xn , dn ) is a sequence of complete metric spaces Q∞ with dn (xn , yn ) ≤ 1 for all xn , yn ∈ Xn . Let X = n=1 Xn with the metric d(x, y) = ∞ X dn (xn , yn ) 2n n=1 Then (X, d) is a complete metric space. Proof: To reduce confusion, we will let the index for sequences in X be a superscript rather than a subscript. Hence, we assume that x(n) is a Cauchy sequence (n) (m) in X. Then, for each subscript, we have that dk (xk , xk ) ≤ 2k d(x(n) , x(m) ). 5.3. COMPLETE METRIC SPACES 59 Thus, the sequence of k th coordinates is a Cauchy sequence in Xk . Let this sequence converge to x ∈ X . Let x ∈ X have the k th coordinate x . Then k k k πk (xn ) converges to πk (x) for all n, so xn converges to x. Example: Let (X, d) and (Y, ρ) be metric spaces and let Cb (X, Y ) denote all continuous bounded functions from X to Y . For f, g ∈ Cb (X, Y ), define d∞ (f, g) = sup{ρ(f (x), g(x)) : x ∈ X}. Then d∞ gives a metric for Cb (X, Y ). We claim that d∞ is complete if and only if ρ is complete. So suppose that ρ is complete and that {fn } is a Cauchy sequence in Cb (X, Y ). For each x ∈ X, the sequence {fn (x)} is a Cauchy sequence in Y since ρ(fn (x), fm (x)) ≤ d∞ (fn , fm ). Hence, this sequence converges in Y . Let fn (x) → f (x) define f : X → Y . We need to show that f ∈ Cb (X, Y ) and that fn → f using the d∞ metric. Let ε > 0 and choose N ∈ N so that n, m ≥ N implies that d∞ (fn , fm ) < ε/3. For every x ∈ X, there is an m ≥ N so that ρ(f (x), fm (x)) < ε/3. But then, whenever n ≥ N , we have ρ(f (x), fn (x)) ≤ ρ(f (x), fm (x)) + ρ(fm (x), fn (x)) < 2ε/3. This holds for every x ∈ X and n ≥ N , so d∞ (f, fn ) ≤ 2ε/2 < ε for n ≥ N . Hence, if f ∈ Cb (X, Y ), then fn → f . But now, if ε > 0 and x0 ∈ X are given, choose N ∈ N so that n ≥ N implies that d∞ (f, fn ) < ε/3. Next, choose δ > 0 so that d(x, x0 ) < δ implies that ρ(fN (x), fN (x0 )) < ε/3. But then, for such x, ρ(f (x), f (x0 )) ≤ ρ(f (x), fN (x))+ ρ(fN (x), fN (x0 ) + ρ(fN (x0 ), f (x0 )) < ε. Hence f is continuous. Finally, choose N so that n ≥ N implies that d∞ (f, fn ) < 1 and let M ∈ R be such that x1 , x2 ∈ X implies that ρ(fN (x1 ), fN (x2 )) ≤ M . Then, ρ(f (x1 ), f (x2 )) ≤ M + 2 for all x1 , x2 ∈ X, so f is bounded. Definition 5.19 Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y is said to be an isometry into Y if ρ(f (x), f (y)) = d(x, y) for all x, y ∈ X. If, in addition, f can be found to be onto, we say that X and Y are isometric. Theorem 5.20 Let (X, d) be a metric space. Then, there is a complete metric space (Y, ρ) along with an isometry Φ : X → Y into Y such that Φ[X] is dense in Y . Furthermore (Y, ρ) is unique in the sense that if (Z, η) is another complete 60 CHAPTER 5. CONVERGENCE AND METRIC SPACES metric space and g : X → Z is an isometry with g[Z] dense in Z, then Y and Z are isometric via an isometry h : Y → Z such that h ◦ Φ = g. Proof: First, fix x0 ∈ X. For each a ∈ X, define φa (x) = d(x, x0 ) − d(x, a). Then φa is a continuous function from X into R. Also, we have for each x ∈ X that |φa (x)| ≤ d(x0 , a), so φa is a bounded function. Now define Φ : X → Cb (X, R) by Φ(a) = φa . We show that Φ is an isometry. In other words, d∞ (Φ(a), Φ(b)) = d∞ (φa , φb ) = d(a, b) for all a, b ∈ X. Now let x ∈ X. We have that |φa (x) − φb (x)| = |(d(x, x0 ) − d(x, a)) − (d(x, x0 ) − d(x, b))| = |d(x, a) − d(x, b)| ≤ d(a, b). Thus, d∞ (φa , φb ) ≤ d(a, b). On the othe rhand, when x = a, |φa (x) − φb (x)| = d(a, b), so the inequality is not strict. Now, let Y = Φ[X]. Then Y is closed in the complete metric space Cb (X, R) so is complete. Clearly Φ[X] is dense in Y . This shows the existence in the theorem. Now suppose that g : X → Z is an isometry, Z is complete and g[X] is dense in Z. Let y ∈ Y . By density, there is a sequence {xn } in X such that Φ(xn ) → y. Then {Φ(xn )} is Cauchy in Y and since Φ is an isometry, {xn } is Cauchy in X. But this implies that {g(xn )} is Cauchy in Z. Hence, there is a z ∈ Z such that g(xn ) → z. We want to define h(y) = z. To do this, we have to show that h is well defined. 0 0 0 So, suppose that Φ(xn ) → y also. Then, d(xn , xn ) = d∞ (Φ(xn ), Φ(xn )) ≤ 0 0 0 d∞ (φ(xn ), y) + d∞ (y, Φ(xn )) → 0. Thus η(g(xn ), g(xn )) → 0, so η(g(xn ), z) ≤ 0 0 η(g(xn ), g(xn )) + η(g(xn ), z) → 0. Hence {g(xn )} also converges to z. We leave it as an exercise to show that h is an isometry between Y and Z and that h is onto. Theorem 5.21 (Baire Category Theorem) Suppose that (X, d) is a complete metric space and {Un } is a sequence of dense open sets in X. Then ∩∞ n=1 Un is dense in X. Proof: Let D = ∩∞ n=1 Un . Let U be any non-empty open set in X. We wish to show that D ∩ U is non-empty. 5.3. COMPLETE METRIC SPACES 61 Since U1 is dense, U ∩ U1 is non-empty, so there is an x1 ∈ U ∩ U1 . Since U ∩ U1 is open, there is a ε1 < 1 such that B(x1 ; ε1 ) ⊆ U ∩ U1 . Now, B(x1 ; ε1 ) is open and U2 is dense, so B(x1 ; ε1 ) ∩ U2 is non-empty. As before, there is a x2 and a ε2 < 1 2 such that B(x2 ε2 ) ⊆ B(x1 ; ε1 ) ∩ U2 . Continue in this way to find a sequence {xn } and εn < 1 2n such that B(xn+1 ; εn+1 ) ⊆ B(xn , εn ) ∩ Un+1 ⊆ U ∩ U1 ∩ · · · Un+1 . Now, if ε > 0, find N so that xn , xm ∈ (xN , εN ), so d(xn , xm ) 1 2N −1 < 21N < ε. Then, if n, m ≥ N , we have that < ε, so the sequence {xn } is Cauchy. Since X is complete, xn → x for some x ∈ X. But now, if m ≥ n, xm ∈ B(xn ; εn ) ⊆ Un . Since B(xn ; εn ) is closed, x ∈ Un . This holds for every n ∈ N, so x ∈ U ∩ ∩∞ n=1 Un , as desired. Example: In particular, we can see that Q is not the intersection of a countable number of open sets in R. For, if it were, Q = ∩∞ n=1 Un , then each Un would be dense and open. Also, for each x ∈ Q, R \ {x} is dense and open. So we would then have ∩∞ n=1 Un ∩ ∩x∈Q R \ {x} = ∅ is an intersection of a countbable family of dense open sets, but is no longer desne. Example: We can no show explicitly that L × L is not normal. Let A = {(x, −x) : x ∈ Q} and B = {(x, −x) : x 6∈ Q}. Then A and B are disjoint closed subsets of L × L. Suppose that U and V are open with A ⊆ U and B ⊆V. For each n ∈ Bn = N define An = {x ∈ R : (x + 1/n, −x + 1/n) ∈ U } and ◦ ∪∞ k=n An . Q ⊆ Bn for every n. In fact, if x ∈ Q, then there is an ε > 0 such that [x, x+ε)×[−x, −x+ε) ⊆ U . Claim: Choose N ≥ n such that 2/N < ε. Then, if y ∈ (x − 1/N, x + 1/N ), we have (y + 1/N, −y + 1/N ) ∈ U , so y ∈ AN . Thus, (x − 1/N, x + 1/N ) ⊆ AN , so x ∈ A◦N . But this shows that Bn is a dense open set. By the previous example, ∩∞ n=1 Bn 6= Q. Pick some x ∈ ∩∞ n=1 Bn \ Q. Then (x, −x) ∈ B, so there is an ε > 0 such that [x, x + ε) × [−x, −x + ε) ⊆ V . Now, pick 1/N < ε and n ≥ N such that x ∈ A◦n . Then (x + 1/n, −x + 1/n) ∈ U ∩ V , so there are no disjoint 62 CHAPTER 5. CONVERGENCE AND METRIC SPACES open sets containing A and B. Exercises: 1. Show that h as defined in the proof of theorem 5.20 is an isometry and is onto. 2. We say that a metric space (X, d) is totally bounded if for each ε > 0, there are finitely many ε-balls that cover X. Show that a totally bounded metric space is bounded. 3. Show that a totally bounded metric space is separable. 4. Show that every sequence in a totally bounded space has a Cauchy subsequence. 5. Show that every subspace of a totally bounded space is totally bounded. 6. Suppose that A is a dense, totally bounded subspace of the metric space X. Show that X is totally bounded. 7. Show that Q is not a Gδ in R. 8. We say that a subset of a topological space X is nowhere dense if (A)◦ = ∅. Show that an open set is dense if and only if X \ U is nowhere dense. 9. We say that a subset A of X is of first category if it is the union of countably many nowhere dense subsets. Show that a complete metric space is not of first category in itself. Chapter 6 Compactness Definition 6.1 A collection A of subsets of X is said to have the finite intersection property if whenever A1 · · · , An ∈ A, then ∩nk=1 Ak 6= ∅. Definition 6.2 A topological space is said to be compact if every open cover of X has a finite subcover. Theorem 6.3 For a topological space X, the following are equivalent: a) X is compact. b) Whenever C is a collection of closed sets with the finite intersection propT erty, then C 6= ∅. c) Every net has a cluster point. d) Every net has a convergent subnet. Proof: Suppose that X is compact and that C is a collection of closed sets of X with the finite intersection property. Then, {X \ C}C∈C is a collection of open sets such that no finite sub-collection covers X by DeMorgan’s theorem. Hence, This collection cannot cover X, so, again by DeMorgan, ∩C∈C C 6= ∅. Hence a) implies b). The converse is evidence again by an argument using DeMorgan. By our general results on convergence of nets, we know that c) and d) are equivalent. 63 64 CHAPTER 6. COMPACTNESS Next, suppose that (xλ ) is a net and let Fλ = {xα : α ≥ λ}. Then C = {Fλ } is a collection of closed sets. Since Λ is a directed set, given any λ1 · · · , λn , there is a λ ≥ λ1 · · · λn , the collection C has the finite intersection property. If we assume b), the intersection of all Fλ is non-empty. But this shows that the net (xλ ) has a cluster point. Finally, assume c) and let C be a collection of closed sets with the finite intersection propery. Let Λ consist of all finite sets F = {F1 · · · , Fn } ⊆ C ordered by inclusion. For each such F, there is an xF ∈ F1 ∩ · · · Fn by assumption. Now, if x is a cluster point of the net (xF ), we must have x ∈ F = F for each F ∈ C. Thus, the intersection of all elements of C is non-empty, so b) follows. Theorem 6.4 A closed subspace of a compact space is compact. A compact subspace of a T2 space is closed. Proof: Suppose that A ⊆ X is a closed subspace of the compact space X. Suppose that {Vα } is an open cover of A in the subspace topology. Then, we can write Vα = A ∩ Uα where each Uα is open in X. Then {X \ A} ∪ {Uα } is an open cover of X. Since X is compact, this has a finite subcover, {X\A, Uα1 , · · · , Uαn }. Then Vα1 · · · , Vαn } is a cover of A. Now suppose that X is T2 and A ⊆ X is compact. Let x 6∈ A. We find an open set containing x that is disjoint from A. This will show that A is closed. For each y ∈ A, the T2 property gives disjoint open sets Uy and Vy such that x ∈ Uy and y ∈ Vy . Now, {A ∩ Vy }y∈A is an open cover of A so there are y1 , · · · yn so that A ⊆ Vy1 ∪ · · · ∪ Vyn . Let U = Uy1 ∩ · · · Uyn . Then x ∈ U and U ∩ A = ∅. Theorem 6.5 A compact T2 space is T4 . Proof: Let X be compact and T2 We first show that X is T3 . Let x 6∈ C where C is closed in X. For each y ∈ C, there are disjoint open sets Uy and Vy such that x ∈ Uy and y ∈ Vy . But then {Vy }y∈Y is an open cover of C, which is compact since it is closed in a compact space. Hence, there are y1 , · · · , yn such Sn T that C ⊆ k=1 Vyk = V . Let U = Uyk . Then x ∈ U , C ⊆ V and U ,V are disjoint open sets in X. 65 Now we show that X is T4 . Suppose that C and D are disjoint closed sets in X. For each x ∈ C, there are disjoint open sets Ux and Vx such that x ∈ Ux and D ⊆ Vx . Again, the collection {Ux } covers the compact set C, so there are Tn Sn x1 , · · · , xn such that C ⊆ k=1 Uxk = U . Let V = k=1 Yxk . Then C ⊆ U , D ⊆ V and U ,V are disjoint open sets in X. Theorem 6.6 The continuous image of a compact space is compact. Proof: Suppose that X is compact and f : X → Y is continuous and onto. if {Uα } is an open cover of Y , then {f −1 [Uα ]} is an open cover of X. By S compactness, there are α1 , · · · , αn such that X = f −1 [Uαk ]. But then Y = S Uαk , so we have a finite subcover of the original cover. Theorem 6.7 (Alexander Sub-base Theorem) Suppose that S is a subbase for a topological space X. In order for X to be compact, it is necessary and sufficient that every cover of X by elements of S have a finite subcover. Proof: Clearly, if X is compact, every cover by elements of S has a finite subcover. Conversely, suppose that every cover by elements of S has a finite subcover, but that X is not compact. In particular, there is an open cover with no finite subcover. Let O denote the collection of all open covers without finite subcovers. If U1 , U2 ∈ O are open covers, we say U1 ≤ U2 if U1 ⊆ U2 ; in other words, U1 has fewer open sets than U2 . In this way, O becomes a partially ordered set. S Now, suppose that C ⊆ O is a chain. Let U = C. Then U is an open cover of X since every element of C is an open cover. Also, U has no finite subcover. In fact, if U1 , U2 , · · · Un are in U, then there are Ui ∈ C such that Ui ∈ Ui for each 1 ≤ i ≤ n. But, since C is a chain, there is a U0 ∈ C such that Ui ⊆ U0 for each 1 ≤ i ≤ n. But then U1 , U2 , · · · , Un cannot be a cover since U0 has no finite subcover. Thus, U ∈ O is an upper bound for C. By Zorn’s lemma, there is a maximal element V ∈ O. Then V is an open cover of X with no finite subcover, but if U is an open set, U 6∈ V, then V ∪ {U } has a finite subcover. Now consider S ∩ V. There is no finite subcover from this set, so by our assumption, S ∩ V is not a cover of X. 66 CHAPTER 6. COMPACTNESS Suppose that x 6∈ S (S ∩ V). Since V is a cover, there is a V ∈ V such that x ∈ V . Also, since S is a subbase for X, there are S1 , · · · , Sn ∈ S such Tn that x ∈ k=1 Sk ⊆ V . But no Sk is in V, so by the maximality of V, there are S V11 , V21 , · · · Vnk k such that Sk ∪V11 ∪· · ·∪Vnk k = X. But then ∩nk=1 Sk ∪ ij Vij =. S But this implies that V ∪ ij Vij = X, giving a finite subcover of V. This contradiction shows that X must be compact. Theorem 6.8 (Tychonoff ’s Theorem) The product of topological spaces is compact if and only if each factor is compact. Proof: Since each projection πα is continuous and surjective, compactness of ΠXα implies the compactness of each Xα . Conversely, suppose that U is a cover of ΠXα by subbasic open sets. For each S α, let Uα = {U : πα−1 [U ] ∈ U}. If Uα is not a cover of Xα , pick xα 6∈ Uα . If no S Uα is a cover of the corresponding Xα , then x = (xα ) ∈ ΠXα , but x 6∈ U. But we assumed that U is a cover of ΠXα , so this means some Uα is an open cover of Xα . But if U1 , · · · , Un is a finite cover from Uα , then πα−1 [U1 ], · · · , πα−1 [Un ] is a finite subcover from U. By the Alexander subbase theorem, ΠXα is compact. In particular, any cube, [0, 1]A is compact and T2 , so is T4 . We now turn to an important example. Recall that Ω0 is a well-ordered uncountable set such that whenever x ∈ Ω0 , the set {y ∈ Ω0 : y ≤ x} is countable. We let ω1 6∈ Ω0 and define Ω = ω0 ∪ {ω1 } with x < ω1 for all x ∈ Ω0 . Now, if x ∈ Ω0 , we define the successor of x, x+ to be the smallest element of (x, →). If y = x+ for some x, we say that y is a successor ordinal. Otherwise, we say that y is a limit ordinal. We give both Ω and Ω0 the order topology: sub-basic open sets are of the form (←, x) = {y : y < x} or (x, →) = {y : x < y}. Since both Ω and Ω0 are linearly ordered, this means that basic open sets are of the form (←, x), (x, →), or (x, y) = {a : x < z < y}. In particular, a neighborhood of x contains a set of the form (y, x] = (y, x+). Theorem 6.9 The space Ω is compact. 67 Proof: Suppose that {Uα } is an open cover of Ω. Then, there is an α0 such that ω1 ∈ Uα0 . Since ω1 is not a successor ordinal, there is an x < ω1 such that (x, ω1 ] ⊆ Uα0 . Let x1 be the smallest such x. If x1 6= 0, then x1 6∈ Uα0 . Now pick Uα1 such that x1 ∈ Uα1 and the smallest x2 such that (x2 , x1 ] ⊆ Uα1 . If x2 6= 0, then x2 6∈ Uα1 . Continuing in this manner, we either obtain an infinite decreasing sequence {xn } or we find some xn = 0. The first case violates the well-ordered property of Ω and the second gives a finite subcover for Ω. On the other hand, Ω0 is not compact. In fact, it is not even Lindelöf since the cover {[0, α)} where α ∈ Ω0 is an open cover with no countable subcover. Lemma 6.10 Suppose that A and B are closed, uncountable subsets of Ω0 . Then A ∩ B is uncountable. Proof: Suppose not. Then there is an x0 ∈ Ω0 such that x ≤ x0 for all x ∈ A ∩ B. Since A is uncountable and {x : x ≤ x0 } is countable, there is an a1 ∈ A with x0 < a1 . Now, since B is uncountable and {x : x ≤ a1 } is countable, there is a b1 ∈ B with a1 < b1 . Continue in this manner to obtain an ∈ A, bn ∈ B such that x0 < a1 < b1 < a2 < b2 < · · · . Then y = sup{an } = sup{bn } and y ∈ A ∩ B with x0 < y. This is a contradiction. Theorem 6.11 Suppose that f : Ω0 → and there exists x0 ∈ Ω0 and α0 ∈ R is continuous. Then f is bounded R such that f (x) = α0 for all x ≥ x0 . In other words, f is constant on some tail of Ω0 . Proof: We first show that f is a bounded function. Suppose not. Then, there is an x1 ∈ Ω1 such that |f (x1 )| > 1. Now, the set [0, x1 ] is compact by the same proof that shows Ω is compact, so f is bounded on [0, x1 ]. Hence, there is an x2 > x1 such that |f (x2 )| > 2. Continuing in this manner, we obtain an increasing sequence {xn } such that |f (xn )| > n. But now, if y = sup{xn }, f cannot be continuous at y ∈ Ω0 . Suppose now that f [Ω0 ] ⊆ [−M, M ] ⊆ R. Claim: There is an α0 ∈ [−M, M ] such that f −1 [(α0 − ε, α0 + ε)] is uncountable for every ε > 0. 68 CHAPTER 6. COMPACTNESS Suppose not. Then for each α ∈ [−M, M ], there is an εα > 0 such that f −1 [(α−εα , α+εα )] is countable. Let Uα = (α−εα , α+εα ). Then {Uα }α∈[−M,M ] is an open cover of the compact set [−M, M ], so there is a finite subcover, say with α1 , · · · , αn ∈ [−M, M ]. This would imply that Ω0 = ∪nk=1 f −1 [Uαn ] is countbale. This is a contradiction. We now show that the α0 of the claim is unique. In fact, suppose α1 also has the property that f −1 [(α1 − ε, α1 + ε)] is uncountable for every ε > 0. If α0 6= α1 , let ε = |α1 − α0 |/3. Then f −1 [α0 − ε, α0 + ε] and f −1 [α1 − ε, α1 + ε] are two disjoint, uncountable closed sets in Ω0 in contradiction to the lemma above. Now, for each n ∈ N, the closed sets f −1 [α0 − n1 , α0 + n1 ] and f −1 [R \ (α0 − 2 n , α0 + 2 n )] are disjoint and the first is uncountable. This means the second is not uncountable, which means it must be countable. But f −1 [R \ {α0 }] = −1 ∪∞ [R \(α0 − n2 , α0 + n2 )], is countable. Choose may now choose any x0 ∈ Ω0 n=1 f larger than any element of this set to finish the proof. N∗ = N ∪ {ω0 } with n < ω0 for all n ∈ N given the order ∗ topology. Let T = Ω × N . Then T is a compact T2 space, so is normal. Let Example: Let T ∗ = T \ {(ω1 , ω0 )}. Then T ∗ is a T3 12 space. We show that T ∗ is not normal. Hence, subspaces of T4 spaces need not be T4 . ∗ In fact, let A = {ω1 } × N and B = Ω × {ω0 }. Then A ∩ T ∗ and B ∩ T ∗ are disjoint closed subsets of T ∗ . Suppose that U and V are open in T ∗ with A ⊆ U and B ⊆ V . Then, for each n ∈ N, the point (ω1 , n) ∈ U , so there is an αn ∈ Ω0 such that αn < x implies (n, x) ∈ U . Let α = sup{αn } ∈ Ω0 . Let β > α in Ω0 . Then (β, ω0 ) ∈ V , so there is an n0 ∈ N such that n > n0 implies that (β, n) ∈ V . But then, any such (β, n) ∈ U ∩ V , so U and V cannot be disjoint. 6.1 Local Compactness Definition 6.12 A topological space X is said to be locally compact if whenever x ∈ U open, there is an open V such that x ∈ V ⊆ V ⊆ U such that V is 6.1. LOCAL COMPACTNESS 69 compact. In other words,a space is locally compact if the compact neighborhoods form a neighborhood base at every point. Notice that a locally compact space is automatically regular. In fact, more is true. Proposition 6.13 A locally compact Hausdorff space is completely regular. Proof: Suppose that x0 6∈ C closed. Pick an open set V such that x0 ∈ V ⊆ V ⊆ X \ C and such that V is compact. Then V is a compact Hausdorff space, so is T4 . Also, x0 6∈ V \ V , so there is a continuous function g : V → [0, 1] such that g(x0 ) = 1 and g = 0 on V \ V . Now, define f : X → [0, 1] by setting f (x) = g(x) if x ∈ V and f (x) = 0 if x 6∈ V . Notice that f is then well defined and is continuous on the closed sets V and X \ V . But then f is continuous on X = V ∪ (X \ V ). Also, f (x0 ) = 1 and f (x) = 0 for x ∈ C. Notice that we actually get a function that is 0 off the compact set V , i.e. a function with compact support. This is important in some applications. Proposition 6.14 Let X be locally compact, K ⊆ U where K is compact and U is open. Then, there is an open set V with V compact and K ⊆ V ⊆ V ⊆ U . Proof: For each x ∈ K, there is an open Vx such that x ∈ Vx ⊆ Vx ⊆ U with Vx compact. But then {Vx } is an open cover of K, so there are x1 , · · · , xn with S S K ⊆ Vxk = V . But then V is open and V = Vxk is compact with V ⊆ U . Proposition 6.15 A T2 space is locally compact if and only if each point has a compact neighborhood. Proof: One direction is clear. Now suppose that each point has a compact neighborhood and suppose that x ∈ U open. Also suppose that x ∈ V with K = V is compact. Then x ∈ U ∩ V and U ∩ V is an open set in K, which is a compact, T2 space. By regularity of K, there is an open set W such that x ∈ W ∩K ⊆ ClK (W ∩K) ⊆ U ∩V . But then, x ∈ U ∩V ∩W and ClK (W ∩K) is 70 CHAPTER 6. COMPACTNESS a compact, hence closed subset of X with U ∩ V ∩ W ⊆ ClK (W ∩ K). But then, U ∩ V ∩ W ⊆ ClK (W ∩ K) ⊆ U ∩ V ⊆ U and U ∩ V ∩ W ⊆ K, so U ∩ V ∩ W has compact closure in X. 6.2 Compactifications Definition 6.16 Let X be a topological space. We say that a pair (Y, f ) is a compactification of X if Y is a compact T2 space, f : X → Y is an embedding of X to a dense subspace f [X] of Y . Notice that if X has a compactification, then X is T3 12 . Example: The pair ([0, 1], i) is a compactification of (0, 1) where i : (0, 1) → [0, 1] is the inclusion map. Example: The pair (S 1 , e) is a compactification of (0, 1) where e : (0, 1) → S 1 is given by e(x) = (cos(2πx), sin(2πx)). Example: Let f : X → Y be any embedding of X into a compact T2 space. Then (f [X], f ) is a compactification of X. Theorem 6.17 Let X be a locally compact T2 space. Define X∞ = X ∪ {∞} and define U ⊆ X∞ to be open if either U ⊆ X is open in X or ∞ ∈ U and X \U is compact in X. Then (X∞ , i) is a compactification where i is the inclusion map. We call (X∞ , i) the one point compactification of X. Definition 6.18 Let (Y, f ) and (Z, g) be compactifications of the space X. We say (Y, f ) ≤ (Z, g) if there is a function h : Z → Y such that f = h ◦ g. Notice that h is then an extension of the homeomorphism f ◦ g −1 : g[X] → f [X]. By density, this means h is unique if it exists. We say (Y, f ) and (Z, g) are equivalent compactifications if (Y, f ) ≤ (Z, g) and (Z, g) ≤ (Y, f ) In general, it is common to identify X and f [X] for a compactification (Y, f ). We call the set Y \ X = Y \ f [X] the remainder of X in Y . 6.2. COMPACTIFICATIONS 71 Lemma 6.19 Suppose that f : X → Y is continuous, where X is T2 . Suppose that A ⊆ X is dense and f |A is a homeomorphism. Then f [X \ A] ⊆ Y \ f [A]. Proof: Suppose not. Then there are x 6∈ A and a ∈ A such that f (a) = f (x). Hence, there are disjoint open sets U and V such that x ∈ U and a ∈ V . Now, A ∩ V is an open set in A, so by the fact that f |A is a homeomorphism, there is an open set W of Y such that f [A ∩ V ] = W ∩ f [A]. Now, f (x) = f (a) ∈ W ∩ f [A], so x ∈ f −1 [W ]. By density of A, A ∩ U ∩ f −1 [W ] 6= ∅. But, if a0 is in this set, f (a0 ) ∈ W ∩ f [A] = f [A ∩ V ]. Since a0 ∈ U and U ∩ V = ∅, this contradicts that f is one-to-one on A. In particular, if h : Z → Y as in the definition of (Y, f ) ≤ (Z, g), then h[Z \ g[X]] = Y \ f [X]. One inclusion follows from the lemma. The other follows since h must be onto (Z is compact, so h[Z] is closed in Y and contains f [X]). Hence, h takes the remainder of X in Z to the remainder of X in Y . Definition 6.20 Let X be T3 12 and let F denote the collection of all continuous f : X → [0, 1]. We know that X has the weak topology from the set F, so e : X → [0, 1]F is an embedding where e(x)f = f (x) for f ∈ F. Then, if we let βX = e[X], we have that (βX, e) is a compactification of X. This is called the Stone-Cech compactification of X. The Stone-Cech compactification has a very useful property: if f : X → [0, 1] is a continuous map, there is an F : βX → [0, 1] such that F ◦ e = f . In other words, every continuous function from X into [0, 1] has an extension to βX. Example: From this, we see that [0, 1] is not equivalent to β(0, 1) since, for example, f (x) = sin(1/x) has no extension to [0, 1]. More is true about βX: Theorem 6.21 Suppose X is T3 12 and f : X → Y is continuous where Y is compact and T2 . Then there is a continuous F : βX → Y such that F ◦ e = f . Proof: Let G denote the collection of all continuous g : Y → [0, 1]. There is an embedding eY : Y → [0, 1]G such that πg ◦ eY = g for all g ∈ G. Since Y is compact, the image of this map is closed in [0, 1]G . 72 CHAPTER 6. COMPACTNESS For each g ∈ G, the map g ◦ f : X → [0, 1] is continous, so has an extension Gg : βX → [0, 1] such that Gg ◦e = g◦f . Hence, there is a map G : βX → [0, 1]G such that πg ◦ G = Gg . Now, for x ∈ X, πg ◦ G ◦ e(x) = Gg ◦ e = g ◦ f (x) = πg ◦eY ◦f (x). Since this happens for all g ∈ G, we have that G◦e = eY ◦f . Thus, G[e[X]] ⊆ eY [Y ]. Since e[X] is dense in βX, this shows that G[βX] ⊆ eY [Y ]. But now F = e−1 Y ◦ G : βX → Y satisfies the conditions of the theorem. As a consequence, we find that certain compactifications are surprisingly large: Example: We know that [0, 1][0,1] is separable, so there is a countable dense subset D ⊆ [0, 1][0,1] . Let f : N → D be any onto map. Then f is continuous so there is an extension F : β N → [0, 1][0,1] . The image of this map is closed since β N is compact and dense because it contains D. Hence, F is an onto map. In particular, card(β N) ≥ card([0, 1][0,1] = 2card([0,1] = card(P([0, 1])). Another consequence of this result is that whenever (Y, eY ) is a compactification of X, then (βX, e) ≥ (Y, eY ). In other words, (βX, e) is the ’largest’ compactification of X. Furthermore, the only property of βX we used in the above proof is the fact that all continuous functions from X to [0, 1] can be extended to βX. This means that any other compactification with this property shares the maximality with βX and so is equivalent to βX. Example: We have shown that every continuous function f : Ω0 → R is constant on some tail. But if we then define f (ω1 ) to be this constant, we get and extension of f to Ω. Hence, βΩ0 = Ω. 6.3 Paracompactness Definition 6.22 Suppose that U = {Uα }α∈I is an open cover of X. Another open cover V = {Vα }α∈I is said to be a shrinking of U if Vα ⊆ Uα for each αinI. In particular, the index set of the two covers is the same. Definition 6.23 Suppse that U is an open cover of X. Another open cover V is said to be a refinement of U if whenever V ∈ V, there is a U ∈ U such that V ⊆ U. 6.3. PARACOMPACTNESS 73 To elucidate in the case of indexed covers, the cover V = {Vβ }β∈J is a refinement of U = {Uα }α∈I if there is a function φ : J → I such that Vβ ⊆ Uφ(β) for each β ∈ J. We write V ≺ U. Exercises: 1. Show that a finite union of compact sets is compact. 2. Suppose that X is compact, U ⊆ X is open, and C is a collection of closed sets such that ∩C ⊆ U . Show there are C1 , · · · , Cn ∈ C such that ∩nk=1 Ck ⊆ U . 3. Suppose that A ⊆ X and B ⊆ Y are compact sets and that A × B ⊆ U , which is open in X × Y . Show that there are open sets A ⊆ V and B ⊆ W such that V × W ⊆ U . 4. Suppose that X is compact. Show that the projection πY : X × Y → Y is a closed map. 5. Suppose that X is compact, Y is T2 and that f : X → Y is continuous, one-to-one, and onto. Show that f is a homeomorphism. 6. Suppose that X is compact and A ⊆ X is dense. Show that A is locally compact if and only if A is open in X. 7. Suppose that X is T2 and that f : X → Y is a closed, onto map with f −1 (y) compact for each y ∈ Y . Show that Y is T2 . 8. Suppose that X is locally compact, T2 and that f : X → Y is continuous, closed, open, and onto. Show that Y is locally compact. 9. A space X is said to be countably compact if every countable open cover has a finite subcover. Show that X is countably compact if and only if every sequence has a cluster point, i.e, that every sequence has a convergent sub-net. 10. Show that a space is countably compact if and only if whenever C1 ⊇ C2 ⊇ · · · are non-empty closed sets, then ∩∞ n=1 Cn 6= ∅. 74 CHAPTER 6. COMPACTNESS 11. Show that Ω0 is countably compact. 12. Show that the continuous image of a countably compact space is countably compact. 13. A space X is said to be sequentially compact if every sequence has a convergent subsequence. Show that every sequentially compact space is countably compact. 14. Show that every first countable countably compact space is sequentially compact. 15. Show that a metric space is compact if and only if it is sequentially compact. 16. Show that a metric space is compact if and only if it is both complete and totally bounded. 17. Show that every compact subset of the Sorgenfry line L is countable. 18. Show that equivalence of compactifications is an equivalence relation. 19. Suppose that X is T4 and that A and B are closed disjoint subsets of X. Show that ClβX A and ClβX B are disjoint. Chapter 7 Connectedness 7.1 Basic Results Definition 7.1 We say a topological space X is connected if the only subsets of X that are both open and closed are the emptyset and X. Otherwise, we say X is disconnected. We say that ∅ and X are the trivial open and closed sets in X. Proposition 7.2 For a topological space X, the following are equivalent: a) X is disconnected. b) There are disjoint, non-empty open sets U and V such that X = U ∪ V . In this case, we call {U, V } a disconnection of X. c) There are disjoint, non-empty closed sets C and D such that X = C ∪ D. d) There is a non-constant continuous function f : X → {0, 1}, the two-point discrete space. Proof: In fact, if U is both open and closed, then V = X \ U is open and closed and X = U ∪ V . In this case, defining f (x) = 0 if x ∈ U and f (x) = 1 if x ∈ V defines a continous function to {0, 1}. 75 76 CHAPTER 7. CONNECTEDNESS Proposition 7.3 Let X be a topological space and A ⊆ X. Then A is connected in the subspace topology if and only if whenever U and V are open sets in X such that A ⊆ U ∪ V and A ∩ U ∩ V = ∅, then A ⊆ U or A ⊆ V . Proof: Consider the disjoint sets A ∩ U and A ∩ V . These are disjoint sets that are open in A. Furthermore, every pair of disjoint open sets in A arise in this way from some U and V with A ∩ U ∩ V = ∅. For connectedness, we need either A ∩ U = A or A ∩ V = A. Example: The closed unit interval [0, 1] is connected. In fact, suppose that U ⊆ [0, 1] is both open and closed. We may assume 0 ∈ U (otherwise use [0, 1] \ U . Let A = {x : [0, x) ⊆ U }. Let x0 = sup A. Then [0, x0 ) = ∪x∈A [0, x) ⊆ U , so x0 ∈ A. Notice that x0 > 0 since 0 ∈ U and U is open. Also, x0 ∈ U since U is closed (otherwise, there would be an ε > 0 such that (x0 − ε, x0 + ε) ∩ U = ∅, which violates the definition of x0 .). But now, if x0 < 1, there is a ε > 0 such that (x0 − ε, x0 + ε) ⊆ U , which contradicts x0 = sup A. But now, [0, 1] = [0, x0 ) ∪ {x0 } ⊆ U . Hence [0, 1] is connected. Proposition 7.4 Suppose that X is connected and f : X → Y is continuous and onto. Then Y is connected. Proof: f −1 Suppose not. If U ⊆ Y is a non-trivial open and closed subsets, then [U ] is a non-trivial (by the fact that f is onto) open and closed subset of X. In other words, continuous images of connected sets are connected. The following turns out to be a very useful way to show that sets are connected: show they are unions of connected subsets that overlap. Theorem 7.5 Suppose {Aα }α∈I are connected subsets of the space X. Suppose also that ∩α∈I Aα is non-empty. Then ∪α∈I Aα is connected. Proof: Let A = ∪α∈I Aα and let x0 ∈ ∩α∈I Aα . Suppose that U and V are open subsets of X such that A ⊆ U ∪ V and A ∩ U ∩ V = ∅. Without loss of generality, assume x0 ∈ U . Then, for each α ∈ I, Aα ⊆ U ∪V and Aα ∩U ∩V = ∅. Since x0 ∈ Aα ∩ U , the connectedness of Aα implies that Aα ⊆ U . But then A = ∪α∈I Aα ⊆ U . Hence, A is connected. 7.1. BASIC RESULTS Example: Suppose a < b in 77 R. Then [a, b] is a continuous image of [0, 1], so is connected. But then, [a, b) = ∪n∈N [a, b − n1 ] is connected. Similarly, (a, b] and (a, b) are connected. Also, [a, ∞) = ∪n∈N [a, a + n], (−∞, b] and 1 2 2 R are 2 connected. As another example, the set S = {(x, y) ∈ R : x + y = 1} is the image of [0, 1] under f (t) = (cos 2πt, sin 2πt), so is connected. Theorem 7.6 (Intermediate Value Theorem) Suppose that X is connected and f : X → R is continuous. Suppose that x, y ∈ X and c ∈ R such that f (x) < c < f (y). Then, there exists a z ∈ X such that f (z) = c. Proof: Otherwise U = f −1 (−∞, c) and V = f −1 (c, ∞) would give a non-trivial disconnection of X. Theorem 7.7 Suppose that A ⊆ X is a connected subset and that A ⊆ B ⊆ A. Then B is connected. In particular, the closure of a connected set is connected. Proof: Suppose that U and V are open in X such that B ⊆ U ∪ V and B ∩ U ∩ V = ∅. Then A ⊆ U ∪ V and A ∩ U ∩ V = ∅, so by connectedness of A, we can assume that A ⊆ U . But then A ⊆ X \ V , which is closed, so A ⊆ X \ V . But then B ⊆ A ⊆ U . Hence B is connected. Example: Let X = ({0} × [0, 1]) ∪ {(x, sin( x1 )) : x ∈ (0, 1]}. Then X is the closure of the set {(x, sin( x1 ) : x ∈ (0, 1]}, which is the continuous image of 2 (0, 1]. Hence, X is a connected set in R . The space X is called the Topologist’s Sine Curve. We now turn to products of connected spaces. Theorem 7.8 Let X and Y be connected spaces. Then X × Y is connected. Proof: We may assume that both X and Y are non-empty. Let (x0 , y0 ) ∈ X×Y . Then, the set {x0 } × Y is connected and for each y ∈ Y , the set X × {y} is connected. Since (x0 , y) ∈ ({x0 }×Y )∩(X ×{y}), so Ay = ({x0 }×Y )∪(X ×{y}) is connected. But now, X × Y = ∪y∈Y Ay and (x0 , y0 ) ∈ Ay for all y, so X × Y is connected. 78 CHAPTER 7. CONNECTEDNESS Theorem 7.9 Let {Xα }α∈I be any collection of connected spaces. Then Πα∈I Xα is connected. Proof: Let a ∈ X = ΠXα . For each finite set F ⊆ I, the set XF = {x ∈ X : xα = aα , α 6∈ F } is homeomorphic to the space Πα∈F Xα so it is connected. S Also, x ∈ XF for all finite F ⊆ I. Thus, F ⊆I XF is a connected subset of X. But, this union is dense in X (open sets restrict on at most finitely many coordinates), so the closure, X, is connected. 7.2 Components and Local Connectedness Definition 7.10 Let X be a topological space and x ∈ X. The component of x S in X is the set {C : x ∈ C, C connected}. A subset of X is called a component of X if it is the component in X of some x ∈ X. Notice that the component of x is a connected set since x is an element of corresponding intersection. Hence, the component is the largest connected set in X containing x. Theorem 7.11 Every component of X is closed in X. Distinct components of X are disjoint sets. Hence, the collection of components of X forms a partition of X into closed subsets. Proof: Since components are maximal connected sets and since the closure of a connected set is connected, each component must be closed. Again, if C and D are components with C ∩ D 6= ∅, then C ∪ D is connected and contains both C and D. Hence, C = C ∪ D = D by maximality. This shows that the collection of components is a partition of X. Definition 7.12 We say that a space X is totally disconnected if all the components of x are singletons. Example: The product X = {0, 1}N of countably many 2-point spaces is totally disconnected since the projection of any connected subset must be 7.2. COMPONENTS AND LOCAL CONNECTEDNESS 79 a point. Since this space is compact and infinite, it cannot be discrete. In particular, no singleton in X is open. There is another important description of this same space. a+2b Let [a, b] be any closed interval. Define E(I) = [a, 2a+b 3 ] ∪ [ 3 , b]. So E(I) is obtained from I by removing the middle third of the interval. Now, if A = ∪nk=1 Ik is a disjoint union of closed intervals, define I(A) = ∪nk=1 E(Ik ). Notice that I(A) is a disjoint union of closed intervals and that I(A) ⊆ A. Also notice that if x is an endpoint of some interval in A, then x is also an endpoint of some interval in I(A). Now, define inductively, E0 = [0, 1], En+1 = E(En ). Then En is a disjoint union of 2n closed intervals, each of length 1/3n . Finally, let E = ∩∞ n=1 En . We call E the Cantor middle thirds set. Define a function f : X → E by setting f ((εn )) = P∞ 2εn n=1 3n . We leave it as an e xercise to show that f is a continuous, one-to-one function from X onto E. Hence, f is a homeomorphism. In particular, 1 4 = f ((0, 1, 0, 1, 0, 1, · · · )) ∈ E. Theorem 7.13 Let X be a compact Hausdorff space and x ∈ X. Then the component of x is the intersection of all subsets containing x which are both open and closed. Proof: Let C be the component of x. Suppose that U is an open and closed set with x ∈ U . Then, C ∩ U is an open and closed subset of C. Since C is connected, and x ∈ U ∩ C, we must have C ⊆, C ∩ U , i.e, C ⊆ U . In particular, T if A = {U : x ∈ U, U open and closed}, then C ⊆ A = K. We now show that K is connected. Since C is maximal connected, this will show K ⊆ C, yielding C = K, as claimed. Notice that K is closed since it is the intersection of closed sets. So, suppose that K = K1 ∪ K2 with K1 , K2 disjoint closed sets in X. Since X is T4 , there are disjoint open sets V1 , V2 with Ki ⊆ Vi for i = 1, 2. But now, T A ⊆ V1 ∪ V2 , and X is compact, so there are finitely many Uj in A such that K ⊆ ∩nj=1 Uj ⊆ V1 ∪ V2 . Let U = ∩j Uj so that U is open and closed and K ⊆ U ⊆ V1 ∪ V2 . Suppose that x ∈ V1 . 80 CHAPTER 7. CONNECTEDNESS But now, x ∈ U ∩ V1 and U ∩ V1 is open in X. But also, U ∩ V1 = U \ V2 is closed in X. So, x ∈ K ⊆ U ∩ V1 . But then K2 ⊆ K \ V1 = ∅. Thus, K is connected. We actually showed a bit more than was claimed. In fact, the proof shows that if C is a component of X and if C ⊆ U with U open, then there is a set V which is both open and closed with C ⊆ V ⊆ U . In particular, the closed and open subsets form a base for the topology of any totally disconnected, compact Hausdorff space. Theorem 7.14 Suppose C1 ⊇ C2 ⊇ C3 ⊇ · · · is a decreasing sequence of T∞ compact, connected, T2 spaces. Then n=1 Cn is connected (and compact). Proof: Each of the spaces Cn is closed in C1 , so T Cn is also. By the finite intersection property, the intersection is also non-empty. Now, suppose that we T can decompose Cn = K ∪ L where K and L are disjoint, non-empty, closed sets. Since C1 is a T4 space, there are disjoint open sets U and V with K ⊆ U T and L ⊆ V . Then Cn ⊆ U ∪ V . Now, by compactness, there is a Cn with Cn ⊆ U ∪ V . But, since U ∩ Cn and V ∩ Cn are non-empty, this contradicts the connectedness of Cn . In some senses, the opposite extreme from total disonnectedness is local connectedness: Definition 7.15 We say a space X is locally connected if whenever x ∈ U with U open, there is a connected neighborhood V of x such that x ∈ V ⊆ U . Theorem 7.16 A space X is locally connected if and only if every component of every open set is open. Proof: If every component of every open set is open, suppose that x ∈ U and U is open. Let V be the component of U containing x. Then by assumption V is open, so it is a neighborhood of x and x ∈ V ⊆ U . Hence X is locally connected. Conversely, suppose that X is locally connected and that C is a component of the open set U . Let x ∈ C. Then x ∈ U , so there is a connected neighborhood 7.3. PATH CONNECTEDNESS 81 V of x with x ∈ V ⊆ U . But now, since C is the component of x and since V is connected, x ∈ V ⊆ C. Hence, C contains a neighborhood of each of its points, so C is open. Example: It is possible to be connected and not locally connected. Let X = ([0, 1] × {0}) ∪ ({0} ∪ (({1/n : n ∈ N}) × [0, 1]). Then X is connected. But the open set U = X \ ([0, 1] × {0}) has {0} × [0, 1] as a component, but this component is not open. Alternatively, notice that (0, 1) ∈ U , but there is no connected neighborhood of (0, 1) contained in U . 7.3 Path Connectedness Definition 7.17 Let X be a topological space and x, y ∈ X. A path from x to y is a map p : [0, 1] → X such that p(0) = x and p(1) = y. We say that X is path connected if for every x, y ∈ X there is a path from x to y. We say that X is locally path connected if each point has a base of path connected neighborhoods. Proposition 7.18 Let X be a topological space and define the relation x y if there is a path from x to y in X. Then is an equivalence relation on X. The equivalence classes are called path components of X. Proof: If x ∈ X, then the constant path p(t) = x for all t ∈ [0, 1] is a path from x to x, so x x. Suppose x y and that p : [0, 1] → X is a path from x to y. Define q(t) = p(1 − t), so q : [0, 1] → X is a path from y to x, so y x. Finally, assume that x y and y z. Let p be a path from x to y and q a path from y to z. Define r : [0, 1] → X by p(2t) 0 ≤ t ≤ 21 r(t) = q(2t − 1) 1 ≤ t ≤ 1 2 Then r is continuous because its restriction is continuous on the closed sets [0, 12 ] and [ 12 , 1] and r( 12 ) is well defined. Then r is a path from r(0) = x to r(1) = z, so x z. 82 CHAPTER 7. CONNECTEDNESS Proposition 7.19 A path connected space is connected. Proof: Fix any x ∈ X. For each y ∈ X, there is a path py from x to y. But then, the image p[0, 1] is a connected set containing both x and y, so x and y are in the same component of X. Hence, X has exactly one component, so is connected. Proposition 7.20 A connected, locally path connected space is path connected. Proof: By local path connectivity, the path components are all open. But the complement of a path component is the union of all the other path components. Hence the complement is also open. Connectivity then shows there can be only one path component. Example: The topologist’s sine curve, X, is connected but not path con- nected. In fact, suppose that p : [0, 1] → X is a path from (0, 0) to (1, sin 1). Let f = π1 ◦ p, so f (0) = 0 and f (1) = 1. By continuity, there is a t1 ∈ (0, 1) such that f (t1 ) = 2 π. Then, there is a t2 ∈ (0, t1 ) such that f (t2 ) = Continue in this manner to find t1 > t2 > t3 > · · · with f (tn ) = 2 3π . 2 (2n−1)π . Let x0 = inf{tn : n ≥ 1} = lim tn ∈ [0, 1]. Now let g = π2 ◦ p. Then = (−1)n−1 . Then continuity of g is violated at x0 . g(tn ) = sin (2n−1)π 2 Exercises: 1. Show that a space is connected if and only if every continuous function into the discrete space {0, 1} is constant. 2. Suppose that {Cn }∞ n=1 are connected sets such that Cn ∩ Cn+1 6= ∅ for all S∞ n ≥ 1. Show that n=1 Cn is connected. 3. Suppose that X is connected and locally connected and C ⊆ X is a connected, closed subspace. Suppose that U is a component of X \ C. Show that X \ U is connected. 4. Suppose that f : X → Y is continuous, closed and f −1 (y) is connected for all y ∈ Y . Suppse that C ⊆ Y is connected. Show that f −1 [C] is connected. 7.3. PATH CONNECTEDNESS 83 5. Suppose that X is a compact T2 space and define xRy when there is a connected C with x, y ∈ C. Show that X/R is a compact totally disconnected T2 space and that the quotient q : X → X/R satisfies the conditions of the previous problem. 6. Show that products and continuous images of path connected spaces are path connected. 7. Show that products and subspaces of totally disconnected spaces are totally disconnected. 8. Show that R2 \ Q2 is connected. 9. Find two disjoint connected sets A and B such that A ∪ B is connected. 10. Show that a connected, T4 space with at least two points is uncountable. 11. Show that a T3 12 space X is connected if and only if the Stone-Cech compactification βX is connected. 12. Show that β R \ R is not connected. 13. Show that β R \ R has exactly two components. 2 2 14. Show that β R \ R is connected. 84 CHAPTER 7. CONNECTEDNESS Chapter 8 Homotopy And Fundamental Groups 8.1 Fundamental Group Definition 8.1 Let x, y ∈ X be two points in a topological space. Define Px,y to be the collection of paths p : [0, 1] → X from x to y, i.e, p(0) = x, p(1) = y. We say that two paths, p, q ∈ Px,y are homotopic if there is a continuous H : [0, 1] × [0, 1] → X such that H(t, 0) = p(t), H(t, 1) = q(t) for all t ∈ [0, 1] and H(0, s) = x, H(1, s) = y for all s ∈ [0, 1]. We say that H is a homotopy from p to q. We write p ' q or H : p ' q if we want to make H explicit. Definition 8.2 Let p ∈ Px,y and q ∈ Py,z be paths. Define n p(2t) 0≤t≤1/2 p ∗ q(t) = q(2t−1) 1/2≤t≤1 Then p ∗ q ∈ Px,z . Also, for p ∈ Px,y , define p̂(t) = p(1 − t), so p̂ ∈ Py,x . Finally, if x ∈ X, define the constant path cx (t) = x for all t. Proposition 8.3 Homotopy of paths in Px,y is an equivalence relation. Definition 8.4 Define π1 (X; x, y) = Px,y / '. We simplify if x = y and write π1 (X, x) = π1 (X; x, x). 85 86 CHAPTER 8. HOMOTOPY AND FUNDAMENTAL GROUPS Proposition 8.5 Suppose p, p0 ∈ Px,y and q, q 0 ∈ Py,z . Suppose also that p ' p0 and q ' q 0 . Then p ∗ q ' p0 ∗ q 0 . In other words, we can define an operation π1 (X; x, y) × π1 (X; y, z) → π1 (X; x, z) by [p] ∗ [q] = [p ∗ q]. The proposition says that this operation is well defined. Proposition 8.6 For p ∈ Px,y , q ∈ Py,z , r ∈ Pz,w , we have p∗(q ∗r) ' (p∗q)∗r. Also, for p ∈ Px,y , we have cx ∗ p ' p ' p ∗ cy and cx ' p ∗ p̂ and p̂ ∗ p ' cy . Theorem 8.7 The set π1 (X, x) with the operation [p] ∗ [q] = [p ∗ q] is a group with identity [cx ] and inverse [p]−1 = [p̂]. Furthermore, if x and y are in the same path component of X, then π1 (X, x) is isomorphic to π1 (X, y). Proposition 8.8 Let f : X → Y be a function. Then f∗ : π1 (X, x) → π1 (Y, f (x)) defined by f∗ [p] = [f ◦ p] is well defined and is a homomorphism of groups. Proposition 8.9 We have (1X )∗ = 1π1 (X,x) , and (f ◦ g)∗ = f∗ ◦ g∗ 8.2 Homotopy Definition 8.10 Let f, g : X → Y be two continous functions. We say that f and g are homotopic and write f ' g is there is a continous H : X × [0, 1] → Y such that H(x, 0) = F (x) and H(x, 1) = g(x) for all x ∈ X. In this case, we say that H is a homotopy from f to g. We can also write H : f ' g to explicitly mention the homotopy H. If, in addition A ⊆ X and f |A = g|A , we say that the homotopy H is relative to A if H(a, t) = f (a) = g(a) for all 0 ≤ t ≤ 1. In this case, we write H : f ' g (rel A). Definition 8.11 A topological pair (X, A) is a topological space X with a subspace A ⊆ X. A map f : (X, A) → (Y, B) is a continous function f : X → Y with f [A] ⊆ B. A pointed topological space (X, x) is a topological pair (X, {x}). 8.2. HOMOTOPY 87 Notice that this is consistent with our previous definition of homotopy of paths if we require the homotopy to be relative to {0, 1} ⊆ [0, 1]. n Example: Let Y ⊆ R be a convex subset (i.e, if x, y ∈ Y and 0 ≤ t ≤ 1, then tx + (1 − t)y ∈ Y ). Then any two maps f, g : X → Y are homotopic via the homotopy H : X × [0, 1] → Y defined by H(x, t) = tg(x) + (1 − t)f (x). Notice that if f |A = g|A , then H is a homotopy relative to A. Corollary 8.12 If X ⊆ Rn is a convex set and x ∈ X, then π1 (X, x) is the trivial group. Definition 8.13 If X and Y are topological spaces and y ∈ Y , we denote the constant map cy : X → Y where cy (x) = y for all x ∈ X. Example: If X ⊆ Rn is convex, x0 ∈ X and f : X → Y is any map, then f is homotopic to cy where y = f (x0 ). A homotopy showing this is given by H(x, t) = f ((1 − t)x + tx0 ). Proposition 8.14 If X and Y are topological spaces and A ⊆ X, then homotopy relative to A is an equivalence relation on C(X, Y ), the collection of continuous functions from X to Y . Proposition 8.15 Suppose that f1 ' f2 : X → Y and g1 ' g2 : Y → Z. Then g1 ◦ f1 ' g2 ◦ f2 : X → Z. Proof: Suppose that H : f1 ' f2 . Then g1 ◦ H : X × [0, 1] → Z is a homotopy g1 ◦ f1 ' g1 ◦ f2 . Also, if G : g1 ' g2 , and we define G1 (x, t) = G(f2 (x), t), then G1 : g1 ◦ f2 ' g2 ◦ f2 . Definition 8.16 We say that two spaces X and Y are homotopically equivalent if there are maps f : X → Y and g : Y → X such that f ◦g ' 1Y and g ◦f ' 1X . Proposition 8.17 Homotopy equivalence is an equivalence relation on the collection of topological spaces. equivalent. Also, homeomorphic spaces are homotopically 88 CHAPTER 8. HOMOTOPY AND FUNDAMENTAL GROUPS The easiest situation is that of contractible spaces. Definition 8.18 A space X is said to be contractible if X is homotopy equivalent to a singleton {y}. Theorem 8.19 For a space X, the following are equivalent: a) X is contractible b) The identity map 1X is homotopic to a constant map cx0 : X → X for some x0 ∈ X. c) There is an x0 ∈ X and a homotopy H : X × [0, 1] → X such that H(x, 0) = x and H(x, 1) = x0 for all x ∈ X. d) Whenever Y is a space and f, g : Y → X are continuous, then f ' g. Proposition 8.20 Convex subsets of Rn are contractible. Exercises 1. Show that the map f : π1 (X × Y, (x, y)) → π1 (X, x) × π1 (Y, y) given by f [p] = (π1∗ [p], π2∗ [p]) is an isomorphism. 2. Let p ∈ Px,y , q ∈ Py,z , r ∈ Pz,w . Give a homotopy H : p∗(q ∗r) ' (p∗q)∗r. 3. We say that a pair (X, A) with A ⊆ X has the homotopy extension property if whenever f : X → Y and H : A × [0, 1] → Y are maps with H(a, 0) = f (a) for a ∈ A, then there is an extension G : X×[0, 1] → Y such that G(x, 0) = f (x) for x ∈ X and G(a, t) = H(a, t) for a ∈ A, t ∈ [0, 1]. Show that the pair (X, A) has the homotopy extension property if and only if (X × {0}) ∪ (A × [0, 1]) is a retract of X × [0, 1]. 4. Show that ([0, 1], {0, 1}) has the homotopy extension property. 5. Suppose that H : [0.1] × [0, 1] → X is a map. Let p(t) = H(t, 0), q(t) = H(1, t), r(t) = H(0, t), s(t) = H(t, 1). Show that p ∗ q ' r ∗ s as paths in X.