1873_solutions

# Lminf x n n n solution this exercises follows at once

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Unformatted text preview: Ý Since lmsup x n  nÝ 2  lmsup x n , nÝ there are at most finitely many integers n for which x n  lmsup x n  2 nÝ and we see, in the same way, that there are at most finitely many integers n for which y n  lmsup y n  . 2 nÝ Thus, for all but at most finitely many integers n we have  lmsup y n  z n  x n  y n lmsup x n  2 2 nÝ nÝ which is impossible since  lmsup y n   lmsup z n . lmsup x n  2 2 nÝ nÝ nÝ Exercises on the Cantor Intersection Theorem 1. Suppose that H n is a sequence (not necessarily contracting) of closed bounded sets and that for every positive integer n we have n  Hi i1 176 . Prove that Ý  Hi . i1 For each n we define n  Hi Kn  i1 Since the sequence K n is a contracting sequence of nonempty closed bounded sets, we know from the Cantor intersection theorem that Ý  Kn . n1 We observe finally that, since Ý Ý n1 n1  Hn   Kn the intersection of all of the sets H n must also be nonempty. 2. Suppose that H is a closed bounded set of real numbers and that U n is an expanding sequence of open sets. a. Explain why the sequence of sets H U n is a contracting sequence of closed bounded sets. For each n, it follows at once from the inclusion U n U n1 that H U n 1 H Un. b. Use the Cantor intersection theorem to deduce that if H for every n then Un Ý H Un . n1 Since H is closed and bounded and since H Un  H R Un for each n, the sets H U n must be closed and bounded. The desired result therefore follows at once from the Cantor intersection theorem. c. Prove that if Ý  Un H n1 then there exists an integer n such that H We suppose that Un. Ý  Un. H n1 Since the set Ý  H Un  H n1 Ý  Un n1 is empty we deduce from part b that there is a value of n for which H integer n we have H U n . Un  and for any such 3. Suppose that U n is a sequence of open sets (not necessarily expanding) and that H is a closed bounded set and that Ý H  Un. n1 Prove that there exists a positive integer N such that 177 N  Un. H n1 For each n we define n Vn   Ui. i1 The sequence V n is expanding and, since Ý Ý n1 n1  Un   Vn H it follows from Exercise 2 that there exists a positive integer n for which n Vn  H  Ui. i1 4. The Cantor intersection theorem depends upon the completeness of the real number system. Where in the proof of the theorem is the completeness used? We use the completeness to guarantee that a nonempty closed bounded set has a least member and then we use the completeness again to guarantee that the sequence of all the least members of the given sets has a supremum. Exercises on Complete Metric Spaces 1. Prove that the metric space R k with the Ý-metric is complete. Hint: From the inequality x x Ý kx Ý Rk that holds for every point x in you can show easily that a sequence x n in R k is a Cauchy sequence if and only if x n is a Cauchy sequence in R k with the Ý-metric. The assertion this this exercise also follows at once from the one that appears in Exercise 2. 2. Prove that if S is...
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