Problem Set 2 Answers

Problem Set 2 Answers - 7.17 Recall the Bolzano-Weierstrass...

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Unformatted text preview: 7.17 Recall the Bolzano-Weierstrass theorem for sequences: Let xi , i = 1, 2, 3, . . . be a bounded sequence in RN . Then xi contains a convergent subsequence. Let xν ∈ R, ν = 1, 2, 3, . . .. xν = 1 (−1)ν + ( 2 )ν . 1. Is xν bounded? Explain. 2. Is xν convergent? Explain. 3. If your answer to part “b” is “yes,” find the limit. Explain 4. If your answer to part “a” is “yes” and to part “b” is “no,” then find a convergent subsequence and its limit. Demonstrate convergence. Suggested Answer: 1. The sequence is completely contained in the interval [−2, 2]. It is bounded. 2. The sequence bounces between values near -1 and greater than +1 . It is not convergent. 3. Omit 1 4. For ν odd the subsequence converges to −1: − 2 , − 7 , − 31 , · · · . 8 32 1 1 For ν even the subsequence converges to +1: 1 + 1 , 1 + 16 , 1 + 64 , · · · . 4 1 Problems 7.19 - 7.25: Suggested Answers 7.19 Recall the Bolzano-Weierstrass theorem for sequences: Let xi , i = 1, 2, 3, . . .be a bounded sequence in RN . Then xi contains a convergent subsequence. Let xν ∈ R, ν = 1, 2, 3, . . .. xν = (−1)ν (10). That is, xν = −10 for ν odd and xν = 10 for ν even. 1. Is xν bounded? Explain. 2. Is xν convergent? Explain. 3. If your answer to part b is “yes” find the limit. Explain. If your answer to part a is “yes” and to part b is “no” then find a convergent subsequence and its limit. Demonstrate convergence. Suggested Answer (a) xν ∈ [−10, 10] so the sequence is bounded. (b) No. There is no single limit point since the sequence moves between -10 and 10 indefinitely. (c) The subsequence of xν for odd values of ν converges to -10; the subsequence for even values converges to 10. 7.20 Think of a vector of purchases of N goods as a point in RN , a = (a1 , a2 , . . . , aN ) where a1 is the amount of good 1 purchased, a2 is the amount of good 2 purchased, and so forth. Think of prices of the N goods as represented by a point in RN p = (p1 , p2 , . . . , pn , . . . , pN ) where p1 is the price of good 1, p2 is the price of good 2, and so forth. Prof. Debreu writes (paraphrasing N slightly): “The value of an action a relative to the price system p is i=1 pi ai , the [dot] product p · a.” Briefly explain this definition of the value of the purchase plan a = (a1 , a2 , . . . , aN ) relative to prices p = (p1 , p2 , . . . , pn , . . . , pN ). What does it mean in words? Why does it make sense? Your answer should run between ten and one hundred words. Suggested Answer: This is just adding up the cost of a shopping list. For each item, take unit price times number of units; that gives the gross value of the purchase of that item. Then add up the gross costs along the list of items. 7.21 Consider each of the following functions of R to R. In each case state whether the function is continuous at 0. 1. f (x) = x + 10 2. g(x) = −1 for x ≤ 0, g(x) = 1 for x > 0 3. h(x) = x2 Suggested Answer: (a) Yes, definitely. There’s no jump at 0. The function value progresses steadily from slightly less than 10 to slightly more than 10. (b) No, definitely not. The function value jumps at 0 from -1 to +1. 1 (c) Yes. The function value hovers consistently about 0. 7.22 Recall the following definitions, concerning subsets of RN : • a set is ‘closed’ if it contains all of its cluster points (limit points). • a set is ‘open’ if, for each point in the set, there is a small ball (neighborhood) centered at the point, contained in the set. • a set is ‘bounded’ if it can be contained in a cube of finite size, centered at the origin. • a set is ‘compact’ if it is both closed and bounded. 1. Is the following subset of R2 closed? open? bounded? compact? Explain your answer. T=45o line through the origin = {(x, y)|(x, y) ∈ R2 , x = y}. 2. Is the following subset of R2 closed? open? bounded? compact? Explain your answer. U = = ball of radius 10 centered at the origin, including its boundary {(x, y)|(x, y) ∈ R2 , x2 + y2 ≤ 100}. Suggested Answer: (a) The set T is closed. It is not bounded, open, or compact. Since the line extends indefinitely in both directions, it is not bounded or compact. Since at each point it does not contain the surrounding neighborhood in R2 , it is not open. (b) The set U is closed, bounded, compact, not open. U is closed since it contains its limit points. It is bounded since it is of finite extent: it can be contained in a finite cube centered at the origin. It is compact since it is closed and bounded. It is not open, since for points on the boundary, a small neighborhood of the point is not fully contained in U . 7.23 Let A, B ⊂ R. A = [−1, 12], the closed interval from -1 to 12. B = [7, 18], the closed interval from 7 to 18. 1. Describe A sufficient. B. Representing A B as one or more intervals is 2. Describe A sufficient. B. Representing A B as one or more intervals is 3. Describe A\B . Representing A\B as one or more intervals is sufficient. Suggested Answer: (a) A ∪ B = [−1, 18] (b)A ∩ B = [7, 12] (c) A \ B = [−1, 7) 7.24 Let S ⊂ R2 , S compact (closed and bounded). ν = 1, 2, 3, ..., be a sequence in S . xν ∈ S , all ν . 2 Let xν = (xν , xν ), 1 2 Show that A B is convex. Let x, y A BThen x, y A and B. Then by convexity of A and B we have that x+(1-)y A and B. Then x+(1-)y A B. Show that A+B is convex. x, y A+B means that there are xa, ya A, and xb, yb B so that xa + xb = x A+B and ya + yb y A+B. Then x + (1-)y = xa + xb + (1-)ya +(1-)yb = xa+(1-)ya+xb+(1-)yb. But a x +(1-)ya A, xb+(1-)yb B, by convexity of A and B. So x + (1-)y A+B. Show that A is convex. Let x, y A . We wish to show that x+(1-)y A . The only difficulty arises if x or y A. Suppose x A. But then x is the limit of a sequence x A . Then the sequence x+(1-)y A and approaches x+(1-)y as its limit, so x+(1-)y A . 8.1 1 9.1, 9.2 Suggested Answers 9.1 The Brouwer Fixed-Point Theorem can be stated in the following way: Let S ⊂ RN be compact and convex. Let f : S → S be a continuous function. Then there is x∗ ∈ S so that f (x∗ ) = x∗. Show how a fixed point would fail to exist when the assumptions of the Brouwer Fixed-Point theorem are not fulfilled, as specified in the following cases: i. Suppose S is not convex. Let S = [1, 2] ∪ [3, 4]; S ⊂ R. That is, S is the union of two disjoint closed intervals in R. Find continuous f : S → S so that there is no fixed point x∗ fulfilling the theorem. ii. Suppose f is not continuous. Let S = [1, 4]; S ⊂ R. Let f (x ) = 4 − x for x < 2, x − 1 for x ≥ 2. Show that although f : S → S there is no fixed point of f in S . iii. Suppose S is not compact. Let S = R and f (x) = x + 1. Note that f : S → S and f is continuous. Show that there is no fixed point of f in S . Suggested Answer :(i) Suppose S is not convex. Let S = [1, 2] ∪ [3, 4]; S ⊂ R. That is, S is the union of two disjoint closed intervals in R. Find continuous f : S → S so that there is no fixed point x* fulfilling the theorem. Let f(x) = 5 - x. f : [1, 2] → [3, 4]; f : [3, 4] → [1, 2]. f : S → S and f continuous but there is no fixed point. (ii) Suppose f is not continuous. Let S = [1, 4]; S ⊂ R. Let f (x) = 4 − x for x < 2, f (x) = x − 1 for x ≥ 2. Show that f : S → S but there is no fixed point of f in S. For 1 ≤ x < 2, we have 3 ≥ f (x) > 2, 2 ≤ x ≤ 4, we have 1 ≥ f (x) ≤ 3, f (x) = x. So f : S → S, but there is no fixed point. (iii) Suppose S is not compact. Let S = R, f (x) = x + 1. Show that f : S → S but there is no fixed point of f in S. To have a fixed point would require that f (x) = x + 1 = x, or equivalently that 0 = 1. This is impossible so there is no fixed point. 1 9.2 Recall the Intermediate Value Theorem: Let [a, b] be a closed interval in R and h a continuous real-valued function on [a, b] so that h(a) < h(b). Then for any real k so that h(a) < k < h(b) there is x [a, b] so that h(x) = k . Recall the Brouwer Fixed-Point Theorem: Let S ⊂ RN be compact and convex. Let f : S → S be a continuous function. Then there is x∗ ∈ S so that f (x∗ ) = x∗. Consider the special case S = [0, 1], the unit interval in R, and let f be a continuous function from S into itself. Using the Intermediate Value Theorem, prove the Brouwer Fixed-Point Theorem for this case. You may find the function g (x) = x−f (x) useful. Suggested Answer : Define g (x) = x − f (x) . Then g (0) ≤ 0 and g (1) ≥ 0. Then by the Intermediate Value Theorem there is x∗ ∈ [0, 1] so that g (x∗) = 0. But then g (x∗ ) = 0 = x∗ − f (x∗) so x∗ = f (x∗ ) and x∗ is the required fixed point. 2 ...
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This note was uploaded on 11/30/2011 for the course ECON 311 taught by Professor Zambrano during the Fall '08 term at Cal Poly.

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