Math 301 Midterm Solutions

Math 301 Midterm Solutions - Math 301/ENAS 513 Fall 2007...

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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 0 = - M and b 0 = M . We have a 0 a n b 0 , for all n . We will construct a subsequence as follows: First, we divide the interval [ a 0 , b 0 ] into two intervals of equal lengths, [ a 0 , ( a 0 + b 0 ) / 2] and [( a 0 + b 0 ) / 2 , b 0 ]. Either of these two intervals must contain -many x n ’s. If [ a 0 , ( a 0 + b 0 ) / 2] contains -many x n ’s, then we let [ a 1 , b 1 ] = [ a 0 , ( a 0 + b 0 ) / 2]. Otherwise, let [ a 1 , b 1 ] = [( a 0 + b 0 ) / 2 , b 0 ]. Pick any x n 1 [ a 1 , b 1 ] ∩ { x n } . Note that | b 1 - a 1 | = ( b 0 - a 0 ) / 2. Suppose x n k [ a k , b k ] ∩ { x n } , for some k 1, is chosen with | b k - a k | = ( b 0 - a 0 ) / 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 k +1 ] = [( b k + a k ) / 2 , b k ]. We then pick x n k +1 [ a k +1 , b k +1 ] ∩ { x n } such that n k +1 > n k . And continue. Note that | b k +1 - a
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