Concepts of Analysis. Homework 5 Solutions

Homework 5 Solutions Problem 1. (i) Define “countable set.” (ii) Determine (either prove or give a counterexample) whether the following statements are true: (a) The union of two uncountable sets is uncountable. (b) The intersection of two uncountable sets is uncountable. Solution. (ii) (a) True. If A ∪ B is countable, then there is a one-one mapping f : A ∪ B → N. The composite f ◦ i : A → A ∪ B → N is one-one, and hence A is countable. (b) False. A = (−∞, 1] and B = [1, ∞) are uncountable but A ∩ B = {1} is countable.  Problem 2. (i) Define “ lim an = l.” n→∞ 2n − 1 converges to l = 2. n+3 (ii) Prove, using the definition in (i), that an = Solution. (ii) Given ε > 0, let N = max{2, 7/ε − 2}. If n > N, then 7 2n − 1 −2 = < ε. n+3 N+3  Problem 3. Let an be the Fibonacci sequence, a1 = a2 = 1, an+2 = an + an+1 . (i) If rn = 1 an+1 , then prove that rn+1 = 1 + . an rn √ 1 1+ 5 (ii) Prove that r = lim rn exists, and r = 1 + . Conclude that r = . n→∞ r 2 an+1 , then Solution. (i) By definition, an+2 = an+1 + an . If rn = an rn+1 = an+2 an+1 + an an 1 = =1+ =1+ . an+1 an+1 an+1 rn (ii) After looking at the first few terms of the sequence (rn ) = (1, 2, 3/2, 5/3, 8/5, . . .), we conjecture that the subsequence of (rn ) given by the terms with even indexes, (r2n ), is decreasing and that with odd indexes, (r2n−1 ), is increasing. This is in fact true and can be proved by induction, using (i). To start, we have r1 < r3 and r4 < r2 . If we know that r2n < r2n−2 , then 1 1 r2n−1 = 1 + <1+ = r2n+1 , and similarly, if r2n−1 < r2n+1 , then r2n+2 < r2n . r2n−2 r2n Furthermore, the sequences (r2n ) and (r2n−1 ) are bounded because (rn ) is bounded on account of the identity proved in (i): 1 ≤ rn+1 = 1 + rn−1 ≤ 2. rn−1 + 1 Let re = lim r2n and ro = lim r2n−1 . Then, by the properties of limits and the identities n→∞ n→∞ r2n = 1 + 1 r2n−1 we have re = 1 + 1 ro and r2n+1 = 1 + and ro = 1 + 1 r2n 1 re 1 or re ro = re + 1 = ro + 1. Therefore, the limit lim rn = r exists (Why?) and r = re = ro satisfies r = 1 + , or r2 − r − 1 = 0. r √ 1+ 5 Since r is positive, it must equal the positive solution of the quadratic equation x2 − x − 1 = 0, that is, r = .  2 1 ( Problem 4. ) 1 1 (i) Find all the accumulation points of the set S = + | n and m in N n m (ii) Prove that p is an accumulation point of a set S ⊂ Rn if and only if every ball about p contains infinitely many points of S . 1 1 1 Solution. (i) The accumulation points of S are 0, 1, 1/2, 1/3, . . .. Indeed, each of the numbers = lim + , and 0 = n→∞ n n m 1 1 lim + . In each case the point is limit of a sequence whose terms are all in S and are distinct from the point itself. n→∞ n n We prove that those are the only accumulation points of S . Suppose that x , 0, 1, 1/2, · · · is an accumulation point of S . Then there is a sequence (sn ) in S such that sn converges to x and sn , x for all n. Each sn = 1/pn + 1/qn for some natural numbers pn and qn , with pn ≤ qn . If the sequence (qn ) is bounded above, then so is the sequence pn , and thus there are natural number p and q such that pn = q and qn = q for infinitely many n’s. This implies that sn = 1/p + 1/q for infinitely many n’s. Since a subsequence of a convergent sequence converges to the same limit, this forces sn = 1/p + 1/q = x for infinitely many n, which is a contradiction. Thus qn has a subsequence qnk which is strictly increasing, hence such that 1/qnk converges to 0. If the corresponding subsequence of natural numbers (pnk ) is bounded above, then it has a constant subsequence pnk l = p for some natural number p, which implies that snk l converges to 1/p, and thus that x = 1/p, again a contradiction. Therefore pnk  has a strictly increasing subsequence pnk l , which implies that snk l and thus sn , converges to 0, also a contradiction. Problem 5. (i) Let an be a bounded injective sequence of real numbers. Prove that if p is the only accumulation point of the set A = {an | n in N}, then the sequence an converges and lim an = p. n→∞ (ii) Show by a counterexample that this property is not true for unbounded sequences. Solution. (i) Suppose that an does not converge to p. Then there is ε > 0 such that for every natural number k, there is nk > k such that |ank − p| ≥ ε. The sequence (ank ) is a subsequence of (an ) and thus it is bounded. Therefore it has a subsequence (ank l ) which converges to a point q , p (because |ank l − p| > ε). Because the sequence (an ) is injective, all elements of this subsequence are distinct and therefore q is an accumulation point of the set A. Note. If “injective” is not assumed, then (i) may not be true. Let an = 1 is n is odd and an = 1/n if n is even. Then the set A = {an } = {1, 1/2, 1/4, 1/6, · · · } has only one accumulation point, namely 0, but the original sequence an does not converge to 0 (or to any other number). (ii) Let an = n if n is odd and an = 1/n if n is even. The sequence (an ) is unbounded and 0 is the only accumulation point of the set A = {an } = {1, 1/2, 3, 1/4, 5, 1/6, · · · }.  Problem 6. (i) Define the concept “bounded sequence.” (ii) Prove that a set S ⊂ R is bounded if and only if every sequence of points in S has a convergent subsequence. Solution. (i) A sequence (xn ) is bounded if there is a number M such that |xn | ≤ M for all n. (ii) Assume that S ⊂ R is bounded. If (xn ) is a sequence in S , then (xn ) is bounded and thus it has a convergent subsequence. Assume that S is not bounded. Then for any integer n there is an xn in S such that |xn | > n. Since any convergent sequence is bounded, the sequence (xn ) cannot have a convergent subsequence.  Problem 7. (i) Define the concept “ f is a continuous function at the point p.” (ii) Let f : R → R be the function given by f (x) = x if x is rational, and f (x) = −x if x is irrational. Prove that f is continuous only at p = 0. Solution. (i) f is continuous at p if for every ε > 0 there is δ > 0 such that if 0 < |x − p| < δ and x is in the domain of f , then | f (x) − f (p)| < ε. The following equivalent version of continuity will be used bellow: f if continuous at p if and only if f (xn ) → f (p) for any sequence (xn ) in domain f such that xn → p. (ii) f is continuous at 0. Given ε > 0, take δ = ε. If |x| < δ, then | f (x) − f (0)| = | f (x)| = |x| < δ = ε. If p , 0, then f is not continuous at p. Indeed, if p is not rational there is a sequence pn of rational √ numbers such that pn → p, and if p is rational, there is a sequence of irrational numbers pn → p (for example, pn = p + 2/n). In either case, lim f (pn ) = −p, which is , p for p , 0.  n→∞ 2 Problem 8.   (i) If lim f (x) and lim g(x) do not exist, can lim f (x) + g(x) or lim( f · g)(x) exist? x→a x→a x→a x→a (ii) If lim f (x) exists and lim f (x) + g(x) exists, must lim g(x) exist?  x→a x→a  x→a   (iii) If lim f (x) exists and lim g(x) does not exist, can lim f (x) + g(x) exist? x→a x→a x→a (iv) If lim f (x) exists and lim f (x)g(x) exists, does it follow that lim g(x) exists? x→a x→a x→a Solution. (i) Yes. Let f (x) = 1 if x is rational and f (x) = −1 if x is not rational, and let g = − f . Then f (x) + g(x) = 0 for all x, and f (x)g(x) = −1 for all x. (ii) Yes. Write g(x) = ( f (x) + g(x)) − f (x). Each of the terms f (x) + g(x) and f (x) on the right side has limit when x → a, so their difference also has limit when x → a. (iii) No. Apply (ii). (iv) No. Let f (x) = 0 for all x and let g(x) = 1 if x is rational and g(x) = −1 if x is not rational. Then, for any a, f (x) and f (x)g(x) = 0 both have limit when x → a, but g(x) does not have limit when x → a.  Problem 9. (i) Define the concepts “closed set” and “closure of a set.” (ii) Prove that the closure of a set S ⊂ Rn is the smallest closed subset of Rn which contains S . Solution. (i) A set is closed if it contains all its boundary points. The closure of a set S is S = S ∪ ∂S . (ii) By definition, the closure of S is S = S ∪ ∂S , hence S contains S . We show that S is closed. Let p be point in of ∂S , the boundary of S . Then every ball around p intersects S and its c complement S . Hence there is a sequence (xn ) in S such that |xn − p| < 1/n for all n. Because xn is in S , there is yn in S such that |yn − xn | < 1/n. Then, by the triangle inequality, |yn − p| ≤ |yn − xn | + |xn − p| < 2/n. Thus p is in S because the (yn ) is in S and converges to p. We show that S is the smallest closed set containing S , that is, if T is closed and S ⊂ T , then S ⊂ T . Assume this was not the case. Then there is p in S which is also in T c . In particular, p is not in S . Moreover, since T c is open, there is a ball around p which is completely contained in T c . This ball cannot intersect S because S ⊂ T , and this p cannot be a boundary point of S . Hence p is not in S ∪ ∂S = S , a contradiction.  3