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Unformatted text preview: Math 301/ENAS 513, Fall 2007 Midterm Solution Triet Le October 29, 2007 1. (10 points). Prove that each bounded real sequence has a convergent subsequence. Discuss how this result is also true for bounded complex sequences. You can assume the followings: i. Every monotone and bounded real sequence converges. ii. Every real Cauchy sequence converges. iii. Every subsequence of a convergent sequence converges. iv. z n = a n + ib n → z = a + ib ⇔ a n → a and b n → b . v. { x n } ⊂ R converges if and only if lim inf n →∞ x n = lim sup n →∞ x n . Solution: Let { x n } be a bounded real sequence. I.e. there exists and M > 0 such that  a n  ≤ M for all n . One way to to prove this result is to use the properties of lim inf n →∞ x n = a to construct a subsequence that converges to a . The same also holds for lim sup n →∞ x n = b . The following proof is using the divide and conquer method. Let a = M and b = M . We have a ≤ a n ≤ b , for all n . We will construct a subsequence as follows: First, we divide the interval [ a , b ] into two intervals of equal lengths, [ a , ( a + b ) / 2] and [( a + b ) / 2 , b ]. Either of these two intervals must contain ∞many x n ’s. If [ a , ( a + b ) / 2] contains ∞many x n ’s, then we let [ a 1 , b 1 ] = [ a , ( a + b ) / 2]. Otherwise, let [ a 1 , b 1 ] = [( a + b ) / 2 , b ]. Pick any x n 1 ∈ [ a 1 , b 1 ] ∩ { x n } . Note that  b 1 a 1  = ( b a ) / 2. Suppose x n k ∈ [ a k , b k ] ∩ { x n } , for some k ≥ 1, is chosen with  b k a k  = ( b a ) / 2 k and [ a k , b k ] contains ∞many of x n ’s. Again, we divide [ a k , b k ] into two intervals of equal lengths, [ a k , ( b k + a k ) / 2] and [( a k + b k ) / 2 , b k ]. If [ a k , ( b k + a k ) / 2] contains ∞many of x n ’s, then we let [ a k +1 , b k +1 ] = [ a k , ( b k + a k ) / 2]. Otherwise, we let [ a k +1 , b...
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This note was uploaded on 07/18/2008 for the course MATH 301 taught by Professor Trietle during the Fall '07 term at Yale.
 Fall '07
 TrietLe
 Math

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