124 CHAPTER 6. METRIC SPACES 6.2 Convergence of Sequences You are familiar with the notion of a convergent sequence of real numbers. It is defined as follows. The sequence x1, x2, . , xn,... of real numbers is said to converge to the real number x if given any ε > 0 there exists an integer n0 such that for all n ≥ n0, |xn − x| < ε. It is obvious how this definition can be extended from R with the euclidean metric to any metric space. 6.2.1 Definitions. Let (X, d) be a metric space and x1, . , xn,... a sequence of points in X. Then the sequence is said to converge to x ∈ X if given any ε > 0 there exists an integer n0 such that for all n ≥ n0, d(x, xn) < ε. This is denoted by xn → x. The sequence y1, y2, . , yn,... of points in (X, d) is said to be convergent if there exist a point y ∈ X such that yn → y. The next Proposition is easily proved, so its proof is left as an exercise. 6.2.2 Proposition. Let x1, x2, . , xn,... be a sequence of points in a metric space (X, d). Further, let x and y be points in (X, d) such that xn → x and xn → y. Then x = y. For convenience we say that a subset A of a metric space (X, d) is closed (respectively, open) in the metric space (X, d) if it is closed (respectively, open) in the topology τ induced on X by the metric d. 6.2. CONVERGENCE OF SEQUENCES 125 The following proposition tells us the surprising fact that the topology of a metric space can be described entirely in terms of its convergent sequences. 6.2.3 Proposition. Let (X, d) be a metric space. A subset A of X is closed in (X, d) if and only if every convergent sequence of points in A converges to a point in A. (In other words, A is closed in (X, d) if and only if an → x, where x ∈ X and an ∈ A for all n, implies x ∈ A.) Proof. Assume that A is closed in (X, d) and let an → x, where an ∈ A for all positive integers n. Suppose that x ∈ X \ A. Then, as X \ A is an open set containing x, there exists an open ball Bε(x) such that x ∈ Bε(x) ⊆ X \ A. Noting that each an ∈ A, this implies that d(x, an) > ε for each n. Hence the sequence a1, a2, . , an,... does not converge to x. This is a contradiction. So x ∈ A, as required. Conversely, assume that every convergent sequence of points in A converges to a point of A. Suppose that X \ A is not open. Then there exists a point y ∈ X \ A such that for each ε > 0, Bε(y) ∩ A =6 Ø. For each positive integer n, let xn be any point in B1/n(y) ∩ A. Then we claim that xn → y. To see this let ε be any positive real number, and n0 any integer greater than 1/ε. Then for each n ≥ n0, xn ∈ B1/n(y) ⊆ B1/n0 (y) ⊆ Bε(y). So xn → y and, by our assumption, y∈ / A. This is a contradiction and so X \ A is open and thus A is closed in (X, d). 126 CHAPTER 6. METRIC SPACES Having seen that the topology of a metric space can be described in terms of convergent sequences, we should not be surprised that continuous functions can also be so described. 6.2.4 Proposition. Let (X, d) and (Y, d1) be metric spaces and f a mapping of X into Y . Let τ and τ 1 be the topologies determined by d and d1, respectively. Then f :(X, τ ) → (Y, τ 1) is continuous if and only if xn → x ⇒ f(xn) → f(x); that is, if x1, x2, . , xn,... is a sequence of points in (X, d) converging to x, then the sequence of points f(x1), f(x2), . , f(xn),... in (Y, d1) converges to f(x). Proof. Assume that xn → x ⇒ f(xn) → f(x). To verify that f is continuous it suffices to show that the inverse image of every closed set in (Y, τ 1) is closed in (X, τ ). So let A be closed in (Y, τ 1). Let x1, x2, . , xn,... be a sequence of points in −1 f (A) convergent to a point x ∈ X. As xn → x, f(xn) → f(x). But since each f(xn) ∈ A and A is closed, Proposition 6.2.3 then implies that f(x) ∈ A. Thus x ∈ f −1(A). So we have shown that every convergent sequence of points from f −1(A) converges to a point of f −1(A). Thus f −1(A) is closed, and hence f is continuous. Conversely, let f be continuous and xn → x. Let ε be any positive real number. −1 Then the open ball Bε(f(x)) is an open set in (Y, τ 1). As f is continuous, f (Bε(f(x)) is an open set in (X, τ ) and it contains x. Therefore there exists a δ > 0 such that −1 x ∈ Bδ(x) ⊆ f (Bε(f(x))). As xn → x, there exists a positive integer n0 such that for all n ≥ n0, xn ∈ Bδ(x). Therefore f(xn) ∈ f(Bδ(x)) ⊆ Bε(f(x)), for all n ≥ n0. Thus f(xn) → f(x). The Corollary below is easily deduced from Proposition 6.2.4. 6.2. CONVERGENCE OF SEQUENCES 127 6.2.5 Corollary. Let (X, d) and (Y, d1) be metric spaces, f a mapping of X into Y , and τ and τ 1 the topologies determined by d and d1, respectively. Then f :(X, τ ) → (Y, τ 1) is continuous if and only if for each x0 ∈ X and ε > 0, there exists a δ > 0 such that x ∈ X and d(x, x0) < δ ⇒ d1(f(x), f(x0)) < ε. Exercises 6.2 1. Let C[0, 1] and d be as in Example 6.1.5. Define a sequence of functions f1, f2, . , fn,... in (C[0, 1], d) by sin(nx) f (x) = , n = 1, 2, . , x ∈ [0, 1]. n n Verify that fn → f0, where f0(x) = 0, for all x ∈ [0, 1]. 2. Let (X, d) be a metric space and x1, x2, . , xn,... a sequence such that xn → x and xn → y. Prove that x = y. 3. (i) Let (X, d) be a metric space, τ the induced topology on X, and x1, x2, . , xn,... a sequence of points in X. Prove that xn → x if and only if for every open set U ∋ x, there exists a positive integer n0 such that xn ∈ U for all n ≥ n0. (ii) Let X be a set and d and d1 equivalent metrics on X. Deduce from (i) that if xn → x in (X, d), then xn → x in (X, d1). 4. Write a proof of Corollary 6.2.5. 5. Let (X, τ ) be a topological space and let x1, x2, . , xn,... be a sequence of points in X. We say that xn → x if for each open set U ∋ x there exists a positive integer n0, such that xn ∈ U for all n ≥ n0. Find an example of a topological space and a sequence such that xn → x and xn → y but x =6 y. 6. (i) Let (X, d) be a metric space and xn → x where each xn ∈ X and x ∈ X. Let A be the subset of X which consists of x and all of the points xn. Prove that A is closed in (X, d). 1 R (ii) Deduce from (i) that the set {2} ∪ {2 − n : n = 1, 2,... } is closed in . 1 R (iii) Verify that the set {2 − n : n = 1, 2,... } is not closed in . 128 CHAPTER 6. METRIC SPACES 7. (i) Let d1, d2, . , dm be metrics on a set X and a1, a2, . am positive real numbers. Prove that d is a metric on X, where d is defined by m d(x, y) = X aidi(x, y), for all x, y ∈ X. i=1 (ii) If x ∈ X and x1, x2, . , xn,... is a sequence of points in X such that xn → x in each metric space (X, di) prove that xn → x in the metric space (X, d). 8. Let X, Y, d1, d2 and d be as in Exercises 6.1 #4. If xn → x in (X, d1) and yn → y in (Y, d2), prove that hxn, yni → hx, yi in (X × Y, d). 9. Let A and B be non-empty sets in a metric space (X, d). Define ρ(A, B) = inf{d(a, b): a ∈ A, b ∈ B}. [ρ(A, B) is referred to as the distance between the sets A and B.] (i) If S is any non-empty subset of (X, d), prove that S = {x : x ∈ X and ρ({x},S) = 0}. (ii) If S is any non-empty subset of (X, d) prove that the function f :(X, d) → R defined by f(x) = ρ({x},S), x ∈ X is continuous. 10. (i) For each positive integer n let fn be a continuous function of [0, 1] into itself and let a ∈ [0, 1] be such that fn(a) = a, for all n. Further let f be a ∗ ∗ continuous function of [0, 1] into itself. If fn → f in (C[0, 1], d ) where d is the metric of Example 6.1.6, prove that a is also a fixed point of f.