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Macroeconomics Exam Review 15

Macroeconomics Exam Review 15 - Solutions for Foundations...

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Choose any 𝑥 𝑛 =1 𝑆 𝑛 . That is, 𝑥 𝑆 𝑛 for each 𝑛 = 1 , 2 , . . . . Thus, for every 𝑟 > 0 and 𝑛 = 1 , 2 , . . . , there exists some 𝑥 𝑛 𝐵 𝑟 ( 𝑥 ) 𝑆 𝑛 We construct a subsequence as follows. For 𝑘 = 1 , 2 , . . . , let 𝑥 𝑘 be the first term in 𝑆 𝑘 which belongs to 𝐵 1 /𝑘 ( 𝑥 ). Then, ( 𝑥 𝑘 ) is a subsequence of ( 𝑥 𝑛 ) which converges to 𝑥 . We conclude that every sequence has a convergent subsequence. 1.119 Assume ( 𝑥 𝑛 ) is a bounded sequence in . Without loss of generality, we can assume that { 𝑥 𝑛 } ⊂ [0 , 1]. Divide 𝐼 0 = [0 , 1] into two sub-intervals [0 , 1 / 2] and [1 / 2 , 1]. At least one of the sub-intervals must contain an infinite number of terms of the sequence. Call this interval 𝐼 1 . Continuing this process of subdivision, we obtain a nested sequence of intervals 𝐼 0 𝐼 1 𝐼 2 . . . each of which contains an infinite number of terms of the sequence. Consequently, we can construct a subsequence ( 𝑥 𝑚 ) with 𝑥 𝑚 𝐼 𝑚 . Furthermore, the intervals get
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