HW 4 &amp; 5 Solutions

# HW 4 &amp; 5 Solutions - Math 117 F 09 HW#4,5(sample...

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Math 117, F 09, HW#4,5 (sample solutions) P.57, #35. For abbreviation’s sake, set S = { a n | n N } . We deﬁne a subsequence by recursion. First, since x is an accumulation point of S , the neighborhood ( x - 1 , x + 1) contains inﬁnitely many points of S . Hence, there is a positive integer n 1 such that | a n 1 - x | < 1. We claim that there are positive integers n 1 < n 2 < n 3 < ··· such that | a n k - x | < 1 k for all k . We have already chosen n 1 . Now suppose that n 1 < ··· < n k have been chosen. Since x is an accumulation point of S , the neighborhood ( x - 1 k +1 , x + 1 k +1 ) contains inﬁnitely many points of S . Hence, there are inﬁnitely many positive integers n such that | a n - x | < 1 k +1 . On the other hand, there are only ﬁnitely many positive integers less than or equal to n k . Therefore, among the inﬁnitely many positive integers n such that | a n - x | < 1 k +1 , there must be some (inﬁnitely many, actually) which are greater than n k . Choose one, and label it n k +1 . Thus n k +1 > n k and | a n k +1 - x | < 1 k +1 . This establishes the recursion. At this point, we have chosen a subsequence { a n k } k =1 of the original sequence { a n } n =1 . It remains to prove that

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HW 4 &amp; 5 Solutions - Math 117 F 09 HW#4,5(sample...

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