Math 28 Spring 2010 Exam 2

Math 28 Spring 2010: Exam 2 Instructions: Each problem is scored out of 10 points for a total of 40 points. You may not use any outside materials(eg. notes or books). You have 50 minutes to complete this exam. Problem 1. Let (X, d) be a metric space and f : X → X be a continuous function. Prove that for every K ⊂ X compact, f (K) is compact. Proof. Let K ⊂ X be compact. Let (yn ) ⊂ f (K), then for each n ∈ N there exists and xn ∈ X such that yn = f (xn ) for some xn ∈ K. Since K ⊂ X is compact, then there exists a convergent subsequence (xnk ) → c. Since f is continuous f (xnk ) = (ynk ) is a subsequence of (yn ) which converges to y = f (c). Hence f (K) is compact. Problem 2. (a) Prove that if f : R → R is continuous on [a, b], then |f | is continuous on [a, b]. (b) Find a counterexample to the converse. Proof. Let f be continuous on [a, b] and let c ∈ [a, b]. Let ² > 0, then by the continuity of f there exists a δ > 0 such that if 0 < |x − c| < δ then |f (x) − f (c)| < ². Thus also ||f (x)| − |f (c)|| ≤ |f (x) − f (c)| < δ and hence |f | is also continuous. As a counter example, the function ( 1 x ∈ Q ∩ [a, b] f (x) = −1 x ∈ [a, b]\Q is discontinuous at every point in [a, b], but |f | is continuous on [a, b]. Problem 3. (a) Show that f (x) = ln x is not uniformly continuous on (0, ∞) (b) Show that f (x) = ln x is uniformly continuous on (1, 2) Proof. (a) Consider the sequences of points xn = 1 n and yn = 1 2n . They have images 1 1 f (xn ) − f (yn ) = ln( ) − ln( ) = ln 2. n 2n Additionally, xn − yn = 1 1 1 − = n 2n 2n and hence lim |xn − yn | = 0 n→∞ 1 bur for ² = ln 2 |f (xn ) − f (yn )| ≥ ². Thus, by the seqential crieterion of non-uniform continuity, ln x is not uniformly continuous on (0, ∞). (b) Since ln x is defined and continuous on the compact set [1, 2] we can apply the continuous extension theorem to see that ln x is uniformly continuous on (1, 2). Alternatively, applying the MVT to ln x for any x, y says that there exists a c ∈ (x, y) such that |ln x − ln y| = Since c ∈ (1, 2) we have that 1 c 1 |x − y| . c ≤ 1 and so |ln x − ln y| ≤ |x − y| for all x, y ∈ (1, 2). So given ² > 0 we may choose δ = ². Problem 4. Give an example with proof of each of the following. (a) A union of closed sets in R which is open. (b) An infinite set A ⊂ R with no limit points. (c) A non-empty set A ⊂ R with no isolated points. (d) A metric space which is not complete. (a metric space is complete if and only if every Cauchy sequence is convergent) Proof. £1 ¤ 1 (a) The union of closed sets ∪∞ n=1 n , 1 − n = (0, 1). Closed intervals in R are closed and open intervals in R are open, so we just have to prove that the union is in fact (0, 1). Let x ∈ (0, 1) then by the archimedean property there exists an N1 such that £ ¤ 1 such that x−1 < N2 . Let N = max(N1 , N2 ), then x ∈ N1 , 1 − N1 . The other inclusion is trivial since 0 < 1 n and 1 > 1 − 1 n 1 N1 < x and an N2 for all n. ¯ ¯ (b) Let A = { n1 | n ∈ N}. Every point is isolated since the sequence is decreasing and ¯ n1 − 1 n(n+1) and hence for ²n = ¯ 1 ¯ n+1 ¯ = 1 2n(n+1) µ ¶ ½ ¾ 1 1 V² n ∩A= . n n ¡ ¢ Since n1 → 0 is convergent every subsequence converges to the same value, so the only limit point is 0 which is not in the set. 2 (c) Let A = [0, 1]. For any x ∈ A, since A is an interval every V² (x) intersects A is an interval and hence has infinitely many points in A. (d) Consider the set (0, 1) with the standard metric on R, d(x, y) = |x − y| is a metric space. Let (xn ) = ( n1 ). Then (xn ) ⊂ (0, 1), but (xn ) → 0 6∈ (0, 1). Additionally, let ² > 0 and for n > m ≥ N > 1² ¯ ¯ ¯1 1 ¯¯ 1 ¯ <² |xn − xm | = ¯ − ¯ ≤ m n m so (xn ) is Cauchy but not convergent. 3