2.1. METRIC SPACES 29 Definition 2.1.29. The function f is called uniformly continuous if it is continu- ous and, for all > 0, the δ > 0 can be chosen independently of x0. In precise mathematical notation, one has ( > 0)( δ > 0)( x X) ∀ ∃ ∀ 0 ∈ ( x x0 X d (x , x0) < δ ), d (f(x ), f(x)) < . ∀ ∈ { ∈ | X 0 } Y 0 Definition 2.1.30. A function f : X Y is called Lipschitz continuous on A X → ⊆ if there is a constant L R such that dY (f(x), f(y)) LdX (x, y) for all x, y A. ∈ ≤ ∈ Let fA denote the restriction of f to A X defined by fA : A Y with ⊆ → f (x) = f(x) for all x A. It is easy to verify that, if f is Lipschitz continuous on A ∈ A, then fA is uniformly continuous. Problem 2.1.31. Let (X, d) be a metric space and define f : X R by f(x) = → d(x, x ) for some fixed x X. Show that f is Lipschitz continuous with L = 1. 0 0 ∈ 2.1.3 Completeness Suppose (X, d) is a metric space. From Definition 2.1.8, we know that a sequence x , x ,... of points in X converges to x X if, for every δ > 0, there exists an 1 2 ∈ integer N such that d(x , x) < δ for all i N. i ≥ 1 n = 2 n = 4 0.8 n = 8 ) 0.6 t ( n f 0.4 0.2 0 1 0.5 0 0.5 1 − − t Figure 2.1: The sequence of continuous functions in Example 2.1.32 satisfies the Cauchy criterion. But, it does not converge to a continuous function in C[ 1, 1]. − 40 CHAPTER 2. METRIC SPACES AND TOPOLOGY In other words, x X is a limit point of A X if x A x , the closure of ∈ ⊂ ∈ − { } A x . The point x may or may not be in A. − { } Theorem 2.2.12. A subset of a topological space is closed if and only if it contains all its limit points. Definition 2.2.13. A subset A of a topological space X is dense in X if every x X ∈ is a limit point of the set A. This is equivalent to its closure A being equal to X. Definition 2.2.14. A topological space X is separable if it contains a countable subset that is dense in X. Example 2.2.15. Since every real number is a limit point of rational numbers, it follows that Q is a dense subset of R. This also implies that R, the standard metric space of real numbers, is separable. 2.2.2 Continuity Definition 2.2.16. Let X and Y be topological spaces. A function f : X Y is → 1 continuous if for each open subset O Y , the set f − (O) is an open subset of X. ⊆ 1 Recall that f − (B) is the set x X f(x) B . Continuity of a function { ∈ | ∈ } depends not only upon the function f itself, but also on the topologies specified for its domain and range! Theorem 2.2.17. Let X and Y be topological spaces and consider a function f : X Y . The following are equivalent: → 1. f is continuous 2. for every subset A X, f A f(A) ⊆ ⊆ 1 3. for every closed set C Y , the set f − (C) is closed in X. ⊆ Proof. (1 2). Assume f is a continuous function. We wish to show f A ⇒ ⊆ f(A) for every subset A X. To begin, suppose A is fixed and let y f A . ⊆ ∈ Then, there exists x A such that f(x) = y. Let O Y be a neighborhood ∈ ⊆ 1 of f(x). Preimage f − (O) is an open set containing x because f is continuous. 1 1 Since x A f − (O), we gather that f − (O) must intersect with A in some point ∈ ∩ 2.2. GENERAL TOPOLOGY* 41 1.2 1 | x | 1 0.8 1+ 0.6 ) = x ( f 0.4 0.2 0 2 0 2 4 − Argument x 1 Figure 2.3: The function f(x) = 1+ x is continuous. The set of integers Z is closed. | | Yet, the image of this set, f(Z) = 1/n : n N , is not closed. Thus, this is an { ∈ } example of a continuous function along with a set for which f Z ( f (Z). 1 x0. Moreover, f(x0) f (f − (O)) O and f(x0) f(A). Thus, O intersects ∈ ⊆ ∈ with f(A) in the point f(x0). Since O is an arbitrary neighborhood of f(x), we deduce that f(x) f(A) by Theorem 2.2.9. Collecting these results, we get that ∈ any y f A is also in f(A). ∈ (2 3). For this step, we assume that f A f(A) for every subset A X. ⇒ ⊆ ⊆ 1 1 Let C Y be a closed set and let A = f − (C). Then, f(A) = f (f − (C)) C. ⊆ ⊆ If x A, we get ∈ f(x) f A f(A) C = C. ∈ ⊆ ⊆ 1 So that x f − (C) = A and, as a consequence, A A. Thus, A = A is closed. ∈ ⊆ (3 1). Let O be an open set in Y . Let Oc = Y O; then Oc is closed in Y . ⇒ − 1 c By assumption, f − (O ) is closed in X. Using elementary set theory, we have 1 c c 1 X f − (O ) = x X f(x) / O = x X f(x) O = f − (O). − { ∈ | ∈ } { ∈ | ∈ } 1 That is, f − (O) is open. Theorem 2.2.18. Suppose X and Y are two metrizable spaces with metrics dX and d . Consider a function f : X Y . The function f is continuous if and only if it Y → is d-continuous with respect to these metrics. 42 CHAPTER 2. METRIC SPACES AND TOPOLOGY 6 ) x 4 ( f 2 A 0 0 1 2 3 4 5 Argument x Figure 2.4: Given a function with a discontinuity and a set A, the image of the closure, f A , need not be a subset of the closure of the image, f(A), as seen in the example above. Proof. Suppose that f is continuous. For any x X and > 0, let O = 1 ∈ y BdY (f(x1), ) and consider the set 1 Ox = f − (Oy) which is open in X and contains the point x . Since O is open and x O , 1 x 1 ∈ x there exists a d-open ball BdX (x1, δ) of radius δ > 0 centered at x1 such that B (x , δ) O . We also see that f(x ) O for any x B (x , δ) be- dX 1 ⊂ x 2 ∈ y 2 ∈ dX 1 cause A O implies f(A) O . It follows that d (f(x ), f(x )) < for all ⊆ x ⊆ y Y 1 2 x B (x , δ). 2 ∈ dX 1 Conversely, let Oy be an open set in Y and suppose that the function f is d- 1 continuous with respect to d and d . For any x f − (O ), there exists a d-open X Y ∈ y ball BdY (f(x), ) of radius > 0 centered at f(x) that is entirely contained in Oy. By the definition of d-continuous, there exits a d-open ball BdX (x, δ) of radius δ > 0 centered at x such that f (B (x, δ)) B (f(x), ). Therefore, every dX ⊂ dY 1 1 x f − (O ) has a neighborhood in the same set, and that implies f − (O ) is ∈ y y open. Definition 2.2.19. A sequence x1, x2,... of points in X is said to converge to x X ∈ if for every neighborhood O of x there exists a positive integer N such that x O i ∈ for all i N. ≥ 2.2. GENERAL TOPOLOGY* 43 A sequence need not converge at all. However, if it converges in a metrizable space, then it converges to only one element. Theorem 2.2.20. Suppose that X is a metrizable space, and let A X. There ⊆ exists a sequence of points of A converging to x if and only if x A. ∈ Proof. Suppose x x, where x A. Then, for every open set O containing n → n ∈ x, there is an N, such that x O for all n > N. By Theorem 2.2.9, this implies n ∈ that x A. Let d be a metric for the topology of X and x be a point in A. For ∈ each positive integer n, consider the neighborhood B x, 1 . Since x A, the set d n ∈ A B x, 1 is not empty and we choose x to be any point in this set. It follows ∩ d n n that the sequence x1, x2,... converges to x. Notice that the “only if” proof holds for any topological space, while “if” requires a metric. Theorem 2.2.21. Let f : X Y where X is a metrizable space. The function f is → continuous if and only if for every convergent sequence x x in X, the sequence n → f(xn) converges to f(x). Proof. Suppose that f is continuous. Let O be a neighborhood of f(x). Then 1 f − (O) is a neighborhood of x, and so there exists an integer N such that x n ∈ 1 f − (O) for n N. Thus, f(x ) O for all n N and f(x ) f(x). ≥ n ∈ ≥ n → To prove the converse, assume that the convergent sequence condition is true.