Problem Set. Problems on Unordered Summation. Math 5323, Fall Februray 15, 2001 ANSWERS

Transcription

1 Problem Set Problems on Unordered Summation Math 5323, Fall 2001 Februray 15, 2001 ANSWERS

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3 1 Unordered Sums o Real Terms In calculus and real analysis, one deines the convergence o an ininite series (1.1) n=1 o reals numbers as ollows: For each postive integer n, deine n S n = a k, a n k=1 the n-th partial sum o the series. I the sequence { S n } converges to a real number S, the series converges and the sum is S. This deinition depends on the particular order o the terms a n i we reorder the terms, we get a dierent sequence o partial sums, and so possibly a dierent result. In this problem set, we want to consider how to add up ininitly many terms (including, perhaps, more than countably many terms) in a way that does not depend on the ordering. In our series (1.1) the sequence { a n } o terms amounts to a unction rom the index set N = { 1, 2, 3,... } to the real numbers R. In general we will consider an arbitary index set, and a unction : R. I F is a inite set, then the inite sum (x), x F makes sense, since the order o the terms does not matter in a inite sum. For brevity, we write = = (x). F F x F The emptyset is a inite set with = 0, or any unction. Note that i F 1, F 2 are inite and F 1 F 2 =, then = +, F 1 F 2 F F 2 since there are no duplicate terms in the sum on the let. Since we ll be using inite subsets o a lot, we introduce the notation Fin() = { F F is inite }. In the deinition o the sum o a series like (1.1), the idea is the we can add up initely many terms, and i we add in more and more terms we get closer and closer to the sum but we have a particular order in which to put in additional terms. In the ollowing deinition, we use the idea that adding in more terms should get closer to the sum, but we don t impose any order on which terms to take next. 1

4 Deinition 1.1. Let be a nonempty set and let : R. We say that is summable over i there is a real number S with the ollowing property: For every ε > 0, there is a inite set F 0 such that S < ε, whenever F F0 and F is inite. F Lemma 1.2. There is at most one number S that satisies the property in Deinition 1.1. Proo. Suppose that S 1 and S 2 have the property in the deintion. Let ε > 0 be arbitary. Then there is an F 1 Fin() such that F Fin(), F F 1 = S 1 < ε. F Similarily, there is an F 2 Fin() such that F Fin(), F F 2 = S 3 < ε. F I we set F = F 1 F 2 Fin(), then F F 1, F 2 and we have S 1 S 2 = S 1 + S 2 F F S 1 + F F S 2 < ε + ε = 2ε. Thus, we have S 1 S 2 < 2ε. Since ε was arbitary, we must have S 1 S 2 = 0, i.e., S 1 = S 2 I is summable, we will denote the unique number S in the deintion by or or (x), x and we introduce the notation S() = { : R is summable }. Problem 1. Suppose that : R. A. I itsel is inite, we ve given two deinitions o the symbol. Show these two deinitions are the same. B. I (x) = 0 or all x, then = 0. 2

5 Problem 2. The space S() o summable unctions is a vector space (under the pointwise operations) and the mapping S() R: is linear. In other words, A. I, g S(), then + g is summable and (1.2) ( + g) = + g. We want to show that the number + g satisies the condition in Deinition 1.1. Let ε > 0 be arbitrary. Since is summable, there is a inite set F 1 such that F Fin(), F F 1 = < ε. F Similarly, there is a inite set F 2 such that F Fin(), F F 2 = g g < ε. F Let F 3 = F 1 F 2, so F 3 is inite and F 3 F 1, F 2. Thus, i F F 3, we have [ ( + g) + ] g = + g g F F F F + g g F < ε + ε = 2ε. Since ε was arbitrary, this completes the proo. B. I S() and c R, c is summable and (1.3) c = c. We need to show the number on the right o (1.3) satisies the condition in Deinition 1.1. Let ε > 0 be arbitrary. Since is summable, there is a inite set F 0 such that F Fin(), F F 0 = < ε. F 3

6 Then, i F F 0, we have F c c = c F c ε. Since ε was arbitrary, c ε can be as small as we like, so this completes the proo. Problem 3. Let : R. A. The unction is summable over i and only i the ollowing Cauchy Condition is satisied: For every ε > 0, the is a inite set F 0 so that F, F Fin(), F, F F 0 = < ε. F F First suppose that is summable. We show that the Cauchy condition holds. Let ε > 0 be given. Then there is a inite set F 0 such that F Fin(), F F 0 = < ε. F Suppose that F, F are inite sets that contain F 0. Then, F = + F F F + F F < ε + ε = 2ε. Since ε was arbitrary, the Cauchy Condition holds. For the second part o the proo, suppose the Cauchy condition holds. We must show that is summable. Applying the Cauchy condition with ε = 1/n or n N gives us a sequence o inite sets F n so that (A) F, F Fin(), F, F F n = F 1 < F n. I we set E n = F 1 F 2 F n, then E 1 E 2 E 3 and E n F n. We claim that the sequence { E n } is a Cauchy sequence o real numbers. To see this, let ε > 0 be given. Choose N N so that 1/N < ε. I m, n N then E m, E n E N, so by (A), we have E n E m < 1/N < ε. Thus, { E n } is Cauchy. 4

7 Since a Cauchy sequence o real numbers converges, there is a number S so that E n S as n. We claim that S =. To see this, let ε > 0 be arbitrary, and choose N 1 N so that 1/N 1 < ε. Then, as above, (B) F, F Fin(), F, F E N1 = < ε. F F On the other hand, we can ind N 2 so that (C) n N 2 = S < ε. E n Let N = max(n 1, N 2 ). Suppose that F E N. Then we have F E N < ε, by (B) and we have E N S < ε, by (C). Thus, S = F F + S E N E N + S F E N E N < ε + ε = 2ε. Since ε was arbitrary, we conclude that is summable. B. The Cauchy Condition is equivalent to the ollowing Alternate Cauchy Condition: For all ε > 0 there is a inite set G such that F Fin(), F G = = < ε. F First suppose the Cauchy condition holds. Let ε > 0 be arbitrary. Then there is a inite set F 0 so that F, F Fin(), F, F F 0 = < ε. F F Suppose that H Fin() and that H F 0 =. Then both F 0 and H F 0 contain F 0, so we have = H H + F 0 F 0 = < ε. H F 0 F 0 Thus, the Alternate Cauchy Condition holds. For the second part o the proo, assume that Alternate Cauchy Condition holds, and we want to show the Cauchy Condition holds. To do this, let ε > 0 be given and let G be as in the Alternate Cauchy Condition. Suppose 5

8 that F is inite and F G. Then we can write F = G H, where H G =, so H < ε. Now suppose that F 1, F 2 G and write F i = G H i (i = 1, 2) as above. Then we have F 1 F 2 = G H 1 G H 2 = + G H 1 G H 2 = H 1 H 2 + H 1 H 2 < ε + ε = 2ε. Since ε was arbitrary, we conclude the Cauchy Condition holds. I A, we can consider the unction A : A R: a (a), the restriction o to A. I A is summable over A, we say is summable over A and write the sum as just A (instead o A A). Problem 4. Let : R and let A be a nonempty subset o. I is summable over, then is summable over A. We will use the Cauchy Condition to show that is summable over A. To show the Cauchy Condition holds on A, let ε > 0 be arbitrary. Let B = \ A, so is the disjoint union o A and B. Since is summable over, is satisies the Cauchy Condition on, so there is a set G Fin() so that (A) F 1, F 2 Fin(), F 1, F 2 G = < ε. F 1 F 2 The set G is the disjoint union o G A and G B. Suppose that H 1, H 2 Fin(A) and that H 1, H 2 G A. Then H 1 (G B) G and H 2 (G B) G. Thus, we have H 1 (G B) H 2 (G B) < ε, by (A). But H 1 (G B) H 2 (G B) = + H 1 G B H 2 G B = H 1 H 2 6

9 and so we have < ε. H 1 H 2 We conclude that satisies the Cauchy Condition on A, and so is summable on A. Problem 5. Let : R and suppose that is the disjoint union o two sets A and B. Then is summable over i and only i is summable over both A and B and, in this case, (1.4) = +. A B First suppose that is summable over both A and B. We want to show that is summable over and that (1.4) holds. Let ε > 0 be arbitrary. Since is summable over A, there is a set F A Fin(A) so that F Fin(A), F F A = < ε. F A Similarly, there is a set F B Fin(B) so that F Fin(B), F F B = < ε. F B Set F = F A F B. I F Fin() and F F, we can write F and the disjoint union o F A and F B, with F A F A and F B F B. Then we have [ + ] = F A B A B (F A) (F B) = + A B F A F B + A B F A < ε + ε = 2ε. F B We conclude that is summable over and (1.4) hold. For the second part o the proo, we assume that is summable over. But then, by the previous problem, is summable over A and B and hence (1.4) holds by the irst part o the proo. 7

10 The result o the last problem is easily extended inductively to the case where is partitioned into a inite number o subsets A 1, A 2..., A n. In this case is summable over i and only i it is summable over each o the A i and, i this is so, = A 1 + A A n. A simple, but useul, corollary to the last problem is the ollowing. Corollary 1.3. Suppose that A and that : R is zero outside A. Then is summable over i and only i it is summable over A and, in this case, = A Problem 6. Let : R be summable. Then, or every ε > 0, there is a inite set F 0 such that A F 0 = < ε. A This diers rom the deinition o summablitity because A is allowed to be ininite. Let ε > 0 be given. Since is summable, there is a inite set F 0 such that (A) < ε F or all inite F F 0. Suppose that A is any subset o that contains F 0. Since is summable over, it is summable over A, so there is a inite set G 0 such that (B) < ε F A or all inite sets F with A F G 0. Let F = F 0 G 0. Then F A and F F 0, G 0. Thus, both (A) and (B) hold. Then we have A = + A F F A + F F This completes the proo. < ε + ε = 2ε. 8

11 Our next step is to extend one direction o Problem??? to arbitary partitions o. Recall that a partition P o is a collection o nonempty subsets o such that the elements o P are pairwise disjoint and P = P =. P P It ollows that every point o is in exactly one o the elements o P Problem 7. Let S() and let P be any partition o. Since is summable over every subset o, we can deine a unction g : P R: P g(p ) by g(p ) = P, Then, the unction g is summable over P and P g =. Let ε > 0 be given. By the last problem, there an F 0 Fin() so that (A) A F 0 = < ε. A Every point x is contained in exactly one element o P, call it P (x). Deine F 0 by F 0 = { P (x) x F 0 }. Then F 0 is a inite subcollection o P and certainly F 0 (the union o the elements o F) contains F 0. Suppose that F is a inite subcollection o P that contains F 0. Then F F 0 F 0. I we label the elements o F by F = { P 1, P 2,..., P k }, we have F g = P F g = g(p 1 ) + g(p 2 ) + + g(p k ) = P 1 + P P k = P 1 P k = F. But, (A) implies that < ε, F 9

12 because F F 0. Thus, we have g < ε. F or all inite subcollections F that contain F 0. This completes the proo. Note that the converse o this theorem is alse. That is, it is possible that is summable over every element o P, so that g is deined, and that g is summable over P, but is not summable over. For an example like this, let = N and deine : N R by (n) = ( 1) n. Let P be the partition P = { { 1, 2 }, { 3, 4 }, { 5, 6 },... }. Then is summable over each element o P and g is summable over P, but is not summable over N. To see this, note that i was summable over N, it would be summable over every subset o N. In particular, it would be summable over { 2, 4, 6, 8,... }, which is not the case (why?). 2 Unordered Sums o Postive Terms Suppose that : [0, ]. I F Fin(), the inite sum F makes sense (it is, i takes the value at some point o F ). For unctions with non-negative values, the inite sums have two monotonicity properties. First, F 1, F 2 Fin(), F 1 F 2 = F 1 F 2. Secondly, i g : [0, ] is another unction, we have g, F Fin() = F F g. For any unction : [0, ], we deine = = (x) x by which may be. { = sup } F Fin, F Problem 8. Suppose that, g : R. 10

13 A. I A and B are subsets o, (2.1) A B = A B By deinition, (A) (B) { = sup } F A, F inite A F { = sup } F B, F inite. B F The set on the right-hand side o (B) is larger than the set in (A), and so has a larger sup. This establishes (2.1). B. For any subset A o, (2.2) g = g. A A Suppose g. For any inite set F A, we have F g g. F A Thus, g is an upper bound or the set A { F F A, F inite }, and hence is bigger than or equal to the sup o this set, which is. A Thus, (2.2) holds. Problem 9. In this problem, we establish that A. I, g : [0, ] then is a close to linear as it can be. ( + g) = + g. 11

14 I F is a inite set, then F ( + g) = + + g. F F Taking the sup over F, we conclude that ( + g) + g. To complete the proo, it will suice to show the opposite inequality (A) + g ( + g) I ( + g) =, this inequality is true, so we are reduced to considering the case (( + g)) <. Since, g + g we have, g ( + g) <. Let ε > 0 be arbitrary. Then there is a inite set F 1 so that ε < F 1. Similarly, there is a inite set F 2 so that Let F = F 1 F 2. Then we have g ε < F 2 g. (2.3) ε < F 1 F (2.4) g ε < g g F 2 F Adding these inequalities (ignoring the middle term) gives + g 2ε < + g = ( + g) ( + g). F F F Thus, we have Since ε was arbitrary, this implies (A) + g 2ε < ( + g). 12

15 B. I c [0, ) and : [0, ], then c = c where we re using the deinition 0 = 0. I c = 0, both sides o the equation in the problem are zero. So consider the case c > 0. For inite F, we have F c = c F (which is the distributive law). Thus, we have { } { F c F Fin = c } { } F Fin() = c F F F Fin(),, rom which the equation in the problem ollows by an elementary property o sup s. Problem 10. A. Suppose that A and that : [0, ] is zero outside o A. Then =. A Since A, we have A, so it will suice to establish the opposite inequality. Suppose that F is inite. Since = 0 outside A, we have F = F A. Thus, Sup ing over F yields and the proo is complete. F =. F A A B. Suppose that : [0, ]. Let A be a subset o and deine χ A, the characteristic unction o A, by { 1, x A χ A (x) = 0, x / A. A 13

16 Then, we have A = χ A. Since χ A is zero outside A, so A χ A = A. χ A = A χ A. But, = χ A on A, C. Let : [0, ] and suppose that is the disjoint union o A and B. Then = +. A B This result is easily extended to a partition o into initely many subsets. Since is the disjoint union o A and B, χ A +χ B = 1. Thus, = χ A +χ B and so = χ A + χ B. Now apply the previous part to the problem. Problem 11. Let : [0, ] and let P be a partition o. Deine g : P [0, ] by g(p ) = P. Then P g =. I F = { P 1, P 2,... P k } is a inite subcollection o P, then g = g(p 1) + g(p 2 ) + + g(p k ) F = P 1 P 2 = P 1 P 2 P k = F. P k Thus, i F is a inite subcollection o P, g =. F F 14

17 Taking the sup over inite subcollection F P, we conclude that P g. To get the opposite inequality, let F be a inite subset o and set F = { P (x) x F } where P (x) is the unique element o P containing x. Then F is a inite subcollection o P and F F. Thus, F F = g g. F P Supping over F gives and the proo is complete. g, P Problem 12. Let : [0, ] and suppose that is countably ininite. Let { x n } n=1 be any enumeration o. Then (2.5) = (x n ), n=1 where the right-hand side is the limit o the partial sums, which may be ininity. Let F n = { x 1, x 2,..., x n }. Then, s n = n (x k ), k=1 the nth partial sum o the series on the right o (2.5) is the same as F n. Thus, s n = F n. Since the partial sums s n converge upward to the sum o the series s, we conclude that s. For the opposite inequality, suppose that F is inite. Find the maximum value o n such that x n F. Say this maximum value is N. Then F F N, so = s N s. F F N Supping over F gives s and the proo is complete. 15

18 I : [0, ], we deine the carrier o, denoted carr(), by From our previous results, we have carr() = { x (x) 0 }. = carr(). I carr() is inite, the right hand side o this equation would be a inite sum. I carr() is countably ininite, we can apply the last problem. Problem 13. Suppose that : [0, ] and that Then carr() is countable. <. Let E n = { x (x) 1/n } or n N. Then carr() = (To see this, note is (x) > 0 then (x) 1/n or some n.) We claim that each o the sets E n is inite. To prove this, suppose, or a contradiction, that E N is ininite. Then, or any k N, we can ind a subset F k o E N consisting o k elements. We then have n=1 E n k N F k Thus, k/n. Since k is arbitrary, this would imply =, contrary to our hypothesis. We conclude that each E n is inite. Then carr() is a countable union o countable sets, and hence countable. 16