INTRODUCTORY ANALYSIS I Homework #13 (The lucky one)

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1 INTRODUCTORY ANALYSIS I Homework #13 (The lucky one) 1. Let X be a metric space. Let {p n } be a sequence of points in X. Let p X. Prove: The sequence {p n } converges to p in X if and only if the sequence of real numbers {d(p n, p)} converges to 0 in R. (d denotes the distance function of X). In symbols: p n = p d(p n, p) = Definition. A sequence {x n } of real numbers diverges to iff for every real number A there exists N such that n N implies x n A. In this case we write x n =. Prove: If x n > 0 for all n N, then x n = if and only if 3. The Fibonacci Sequence. Define {f n } as follows: 1 x n = 0. f 1 = 1, f 2 = 2, f n = f n 1 + f n 2 if n 3. (a) If f n exists (notice that it says if ), then = 1 +. f n 2 Do this by setting x n = f n and noticing that x n+1 = 1 + 1/x n. (b) Show that there exist real numbers a, b, r, s such that f n = ar n + bs n for alln N, and use this result to prove that = 1 +. f n 2 Hint: Consider first the recurrence relation f n+2 = + f n. Determine all numbers r such that f n = r n satisfies the relation. You should find precisely two; those are your r and s. Verify that if a, b are any real numbers, {ar n + bs n } again satisfies the recurrence relation and determine a, b so that arbs = 1, ar 2 + bs 2 = 2. Prove that now {ar n + bs n } must be the sequence of Fibonacci numbers. 1

2 4. Let {x n } be a sequence of real numbers and assume the sequence either converges to x R or diverges to or to. In other words, assume that x n = L, where L R {, }. Let {y n } be the sequence defined by Prove: y n = L. y n = 1 n n x n, n = 1, 2,.... k=1. We define a somewhat strange sequence in this exercise. Let n N. Then we can write n = 2 a 3 b c m, where a, b, c are non-negative integers and m is not divisible by 2, 3, or. To fix ideas, here are some integers and their a, b, c, m values. n Decomposition a b c m We now define x n as follows: If a = 0 or if b = 0, we set x n = 0. Otherwise, we set x n = ( 1) c a/b. Describe the set of its of convergent subsequences of {x n }. Prove your assertions. 6. Let {p n } be a sequence in a metric space X. Let p X. Prove: If then p n = p. p 2n = p, and p 2n+1 = p, 7. Let Q = {r 1, r 2,...} be an enumeration of the rational numbers. Let ɛ > 0 and for n N let I n = (r n ɛ 2 n, r n + ɛ 2 n ). Let U = I n ; let F = U c. n=1 Prove or disprove: F is a non-empty, perfect set. 8. Let A, B be subsets of R k. Prove: 2

3 (a) If A, B are compact, then A + B is compact. (b) If A is compact and B is closed, then A + B is closed. (c) Its is possible to have A, B closed, but A + B is not closed. 9. Let X be a metric space, {p n } a sequence in X, r R, r > 0. Assume that d(p n, p m ) r for all n, m N such that n m. Prove: The sequence {p n } does not have convergent subsequences. In other words, no subsequence of {p n } can converge. 10. This is perhaps a difficult, but important exercise. If you do it you will have a proof of the following extremely important characterization of compactness in metric spaces. Theorem A. Let X be a metric space. A subset C of X is compact if and only if it satisfies the following property: Every sequence of points of C has a subsequence converging to a point of C. In other words, C is compact if and only if for every sequence {p n } with p n C for all n N, there exists p C and integers n k for k N, 1 n 1 < n 2 <, such that p n k = p. k Using this characterization, it is possible to give a different, probably simpler, proof of the Heine Borel theorem. We have to start with a definition. Let X be a metric space, A X and let G = {G λ : λ Λ} be an open covering of A. We say that G has a Lebesgue number iff there exists a real number r > 0 such that for every p A, there is G λ G satisfying N r (a) G λ. This could be a somewhat weird notion, but it is not as obscure as one might think. Here are a few examples. Try to understand them. Example 1. Let X = R (as usual), A = [0, 1], the covering being G = {(x 1 10, x + 1 ) : x [0, 1]}. This is, I hope, a very easy case. Any 10 number r (0, 1/10] will do. For example, r = 1/10. For every x [0, 1], we have {(x 1 10, x ) G and N 1/10(x) is not only included in {(x 1 10, x + 1 ), but actually equal to it. 10 Example 2. Let X = R 2, let A = {(x 1, x 2 ) R 2 : x x 2 2 1} (closed unit disc in R 2 ). Let U 0 = {(x 1, x 2 ) R 2 : x x 2 2 < 1/16}, and for n N let ( U n = {(x 1, x 2 ) R 2 1 : n 2 < x2 1 + x n) 1 2 }. Exercise, Part 1 Prove that {U n : n = 0, 1, 2...} is an open covering of A (you can declare that the sets are obviously open, if you think they are obviously open) and that it has a Lebesgue number. 3

4 Example 3. Let X = R and let A = (0, 1). For x A, let U x = (0, 2x). Exercise, Part 2. Show that {U x : x (0, 1)} is an open covering of A that does not have a Lebesgue number. Example 4. If A is a compact subset of a metric space X, then every open cover has a Lebesgue number. Exercise, Part 3. Prove this result. As you can imagine, we sould start with the statement Let {G λ : λ Λ} be an open covering of A. This is one of the rare instances in which we say this for a set we know is compact. Then, of course, there is a finite subcovering. However, that last part could be irrelevant. What you can do is construct a new open covering of A by showing that for every p A there exists r p > 0 such that N rp (p) G λ for some λ Λ. The new covering is {N rp/2(p) : p A} and once one has a finite subcovering of this last covering, something that may look like r = min i r pi /2 could work as a Lebesgue number. Exercise, part 4. Complete the blanks in the argument below to obtain a proof of Theorem A. Assume first that C is sequentially compact; i.e., it satisfies the property that every sequence of points of C has a subsequence converging to some point of C. We begin showing that every open covering of C must have a Lebesgue number. Assume not. Then there exists an {G λ : λ Λ} such that for each r > 0 there exists p r C such that N r (p r ) G λ for all λ Λ. Take r = 1/n, n N; writing p n for p 1/n, the sequence {p n } is a sequence of points in C such that. By hypothesis, the sequence has a convergent subsequence {p nk }, converging to a point p. Because {G λ : λ Λ} is, there exists λ such that p. Because is open, there exists r > 0 such that N r (p). Because {p nk }, there exists K such that k implies d(p nk, p) < r/2. Let k and such that 1/n k < r/2. Then N 1/nk (p nk ) N r (p) ; in fact,. The inclusion N 1/nk (p nk ) contradicts the assumption that N 1/n (p n ) is not included in any. This contradiction is due to the assumption that did not have a Lebesgue number. We conclude that has a Lebesgue number, and since it was an arbitrary of C, we proved that every of C has a Lebesgue number. To prove now that C is indeed compact, let G = {G λ : λ Λ} be. Assume, for a contradiction, that G contains no finite subcovering of C. By what we proved, G has a Lebesgue number r > 0. We construct inductively sequences {p n }, {λ n } in C and in Λ, respectively, as follows. Let p 1 C (C because ). Because, there exists λ 1 Λ such that p 1 G λ1. Assume we have determined for some n N points p 1,..., p n in C, indices λ 1,..., λ n in Λ such that N r (p k ) G λk for k = 1,..., n. This has been done for n =. Now C n k=1 G λ k, because we are assuming that. Thus there exists 4

5 p n+1. Because r is a Lebesgue number, there exists λ n+1 Λ such that. The sequences have been properly constructed and satisfy: 1. N r (p n ) G λn for n = 1, 2, p n+1 / n k=1 G λ k for n = 1, 2,.... It also satisfies d(p n, p m ) r if n m. In fact, assume n < m. Then N r (p n ) G λn while p m /, because., contradicting the se- It follows that the sequence {p n } has no quential compactness of C. Conversely, assume that C is compact. Theproof that C is sequentially compact will be done in class.