Chapter 9 The Topology of Metric Spaces 9.1 Abstract Topological Spaces We undertake a study of metric spaces because we wish to study, among other things the set of continuous functions defined on Rn,andRn is a simple instance of a metric space, which we shall shortly define. The topology of a space is of particular interest to us, because the topology of the space and the set of continuous function defined on that space are intimately connected, as we shall soon see. Definition 9.1 Let X be a set, and C a collection of subsets of X.Then(X, C) is called a topological space, and the elements of C are called the open sets of X, provided the following hold: 1. ∅∈C. 2. X ∈C. 3. If X, Y ∈Cthen X ∩ Y ∈C. 4. The union of any number (i.e. finite or infinite number) of elements of C is again an element of C. 1 2 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES Note that by induction, (3) implies that the intersection of any finite number of elements of C is an element of C,orsaidanotherway:C is closed under finite intersection, and from (4), C is closed under arbitrary union. Often, having exhibited the topological space (X, C), we will often refer to “an open set O in the topological space X”, understanding that that means O ∈C. Example 1 Let X = {0, 1}, that is, a set consisting of two elements. Then if we let C = {{0}, {1},X,∅},then(X, C) is a topological space. This is true because (1) and (2) can be verified by inspection. (3) and (4) require that certain subsets of X are elements of C, but we have chosen C to be all subsets of C, which make (3) and (4) hold automatically. This reasoning generalizes to the following example: Example 2 Let X be an arbitrary set, and C be the set of all subsets of X, including both ∅ and X.ThenaccordingtothereasoningofExample1,(X, C) is a topological space. This is called the discrete topology for X. Example 3 Let X be arbitrary, and let C = {∅,X}.Then(X, C) is a topological space, and the topology is called the trivial topology. Example 4 [The Usual Topology for R1.] Let X =(−∞, ∞), and let C consist of all intervals of the form (a, b), the arbitrary union of such intervals, and the intersection of any finite number of elements of C.Then(X, C) is a topological space, and the open sets are just the open sets we studied in Chapter 1. To see this, first note that since (1, 2) ∈Cand (3, 4) ∈C, ∅ =(1, 2) ∩ (3, 4) ∈C. Further, (x − 1,x+1)∈C x∈X which implies that X ∈C. To prove (3), suppose (a, b)and(c, d) are elements of C.Thenif (a, b) ∩ (c, d) = ∅,either(a, b) ⊂ (c, d)or(c, d) ⊂ (a, b)ora<c<b<dor c<a<d<b. In each of these four cases the intersection is an interval.1 The intersection of two arbitrary 1In Exercise 1 you are asked to compute the intersection explicitly, for each of the four cases. 9.1. ABSTRACT TOPOLOGICAL SPACES 3 unions of intervals is again a union of such intervals, and hence so is the intersection of finitely many such unions. Example 5 [The Right Order Topology] Let X =(−∞, ∞), and let C consist of all intervals of the form [a, ∞), arbitrary unions of such intervals, the empty set, and X.Then (X, C) is a topological space. We conclude the introduction with a simple and yet powerful theorem about open sets: Theorem 9.1 Let (X, C) be a topological space. Then O ⊂ X is an open set, that is, O ∈C, if and only if for every x ∈ O there exists a B ∈Csuch that x ∈ B ⊂ O. Proof: If O ∈C, then for every x ∈ O, there exists a B ∈C(O will play the role of B)so the theorem is trivially true. Conversely, if for every x ∈ O there exists a B ∈Csuch that x ∈ B ⊂ O, then indicating the dependence of B upon x by Bx, O ⊂ Bx ⊂ O {x:x∈O} and therefore O = Bx. {x:x∈O} Since arbitrary unions of sets in C are again in C, it follows that O ∈C. Continuous Functions on an Arbitrary Topological Space Definition 9.2 Let (X, C) and (Y,C) be two topological spaces. Suppose f is a function whose domain is X and whose range is contained in Y .Thenf is continuous if and only if the following condition is met: For every open set O in the topological space (Y,C),thesetf −1(O) is open in the topo- logical space (X, C). 4 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES Informally, we say: The inverse image under f of every open set in Y is an open set in X. If f : X → Y is continuous we occasionally call f a mapping from X to Y . Note that whether or not a particular function f is continuous depends upon the topologies, that is, what the open sets are, of both the domain and range. This is an important property of continuity. Later, when we specialize our study of topological spaces to Metric Spaces, we shall see that our − δ definition of continuity and the topological definition of continuity are the same. A useful way to test continuity of a function is given by the following theorem. Theorem 9.2 f :(X, C) → (Y,C) is continuous on X if and only if for every x ∈ X and every open set V containing f(x) there exists an open set U containing x such that f(U) ⊂ V. Proof: If f : X → Y is continuous on X, then since the inverse of every open set in Y is open in X, for any V ∈C, f −1(V )isopeninX and provides the U. Thus the necessity of the condition is proved. Conversely, suppose that V is open in Y ,andy ∈ V . Then for any x ∈ X such that f(x)=y, by hypothesis there exists an open set U in X, containing x,andsuchthat f(U) ⊂ V. Label the above U by Ux to indicate its dependence upon x,andlet O = Ux. {x:f(x)∈V } Then O is open because it is the union of open sets, and clearly2 O = f −1(V ). Corollary 9.3 Let f : R1 →R1 be any function where R1 =(−∞, ∞) with the usual topology (see Example 4), that is, the open sets are open intervals (a, b) and their arbitrary unions. Then in R1, f is continuous in the − δ sense if and only if f is continuous in the topological sense. 2Provide the details. See Exercise 2. 9.1. ABSTRACT TOPOLOGICAL SPACES 5 1 1 Proof: Suppose f is −δ continuous, and x0 ∈R.LetV be an open set in R , and suppose v = f(x0) ∈ V .SinceV is open, it is the union of open intervals, and hence v ∈ (a, b)for some a, b.Thenlet =min{v − a, b − v}, and note that (v − , v + ) ⊂ (a, b). Since f is assumed continuous in the − δ sense, there exists a δ>0 such that |x − x0| <δimplies |f(x) − f(x0)| <implies x ∈ V. Letting U =(x0 − δ, x0 + δ) in Theorem 9.2 shows that f is continuous in the topological sense. Conversely, suppose f is continuous in the topological sense. Let x be arbitrary, and y = f(x). Let >0. Then since V =(y − , y + ) is an open set in the range, by hypothesis thereexistsanopensetU, containing x, in the domain of f,andsuchthatf(U) ⊂ V. But U open implies there exists an interval (a, b), containing x, which is contained in U.That is, z ∈ (a, b) implies f(z) ∈ V .Nowletδ =min{x − a, b − x} and observe that if |z − x| <δ, then z ∈ (a, b) and hence f(z) ∈ V. But f(z) ∈ V implies |y − f(z)| <.Thus, f is − δ continuous at x. Exercise 1 Compute the four intersections in Example 1. The following two exercise indicate the degree with which continuity is connected to the topology of the spaces involved: Exercise 2 In the proof of Theorem 9.2, why is O = f −1(V )? The following exercises show the intimate connection between topology and continuity. Exercise 3 Let X be an arbitrary set with the discrete topology: C is the set of all subsets of X.Let(Y,C) be an arbitrary topological space. Then f :(X, C) → (Y,C) is continuous, for any f ! Exercise 4 Let Y be an arbitrary set endowed with the trivial topology: C = {∅,Y},and (X, C) an arbitrary topological space. Then f :(X, C) → (Y,C) is continuous, for any f. Exercise 5 Let X =(−∞, ∞)andtheopensetsofX be half-open, half-closed intervals of the form [a, b), for a<b,and their arbitrary unions. Let Y =(−∞, ∞), and the open sets 6 CHAPTER 9. THE TOPOLOGY OF METRIC SPACES of Y be arbitrary. Then prove that f(x)=[x] is continuous(!) [Hint: what does the range of f consist of?] Limit Points and the Derived Set Definition 9.3 Let (X, C) be a topological space, and A ⊂ X. Then x ∈ X is called a limit point of the set A provided every open set O containing x also contains at least one point a ∈ A,witha = x. Definition 9.4 Let (X, C) be a topological space, and A ⊂ X.