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# B a n is bounded below if and only if lim inf n a n c

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(b) { a n } is bounded below if and only if lim inf n →∞ a n > -∞ . (c) The sequence { a n } converges, or diverges to , or diverges to -∞ if and only if lim inf n →∞ a n = lim sup n →∞ a n , in which case we also have lim n →∞ a n = lim inf n →∞ a n = lim sup n →∞ a n . (d) Let S be the set of all real numbers x such that there is a subsequence of { a n } converging to x . Then lim inf n →∞ a n = g. l. b. S, lim sup n →∞ a n = l. u. b. S. A standard way of proving that a sequence { a n } converges is to prove that lim inf n →∞ a n lim sup n →∞ a n . The converse inequality being always true, one would be done. In conjunction with all this, here are a few other exercises to think about. Could one of them show up on the exam? Only time will tell. (i) Prove: Let { p n } be a sequence in a metric space and let S = { p n : n N } be the range of the sequence. Prove: If p is an accumulation point of S , then there is a subsequence of { p n } converging to p . Is the converse true? That is, if there is a subsequence of { p n } converging to p , must p be an accumulation point of S ? 4

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Suppose S has a single accumulation point p . Is it true or false that lim n →∞ p n = p ? (ii) Let { p n } be a sequence in a metric space X and this time let X consist of the set of all points q S such that there exists a subsequence of { p n } converging to q . Prove: S is closed. (iii) Define lim sup , lim inf as Rosenlicht does in Exercise 18 of Chapter 3. Prove the following additional properties for a sequence { a n } of real numbers. Assume it bounded if you want to avoid working with or -∞ . (a) There exists a subsequence { a n k } of { a n } such that lim k →∞ a n k = lim sup n →∞ a n . Similarly, there exists a subsequence { a m k } of { a n } such that lim k →∞ a m k = lim inf n →∞ a n . (b) For every subsequence { a n k } of { a n } we have lim inf n →∞ a n lim inf k →∞ a n k lim sup k →∞ a n k lim sup n →∞ a n . (c) If a n b n for all n , and lim n →∞ b n = b , then lim sup n →∞ a n b . Similarly, if a n b n for all n , and lim n →∞ b n = b , then lim inf n →∞ a n b . (d) lim sup n →∞ a n = - lim inf n →∞ ( - a n ) . (iii) Define a sequence { a n } of real numbers as follows. Consider the decimal representation of n , n = d 1 . . . d k , where d 1 , . . . , d k ∈ { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } ; d 1 6 = 0 . Just to be sure this is understood, if n = 1234, then d 1 = 1 , d 2 = 2 , d 3 = 3 , d 4 = 4; if n = 99, then d 1 = 9 , d 2 = 9. In other words, d 1 , . . . , d k are the digits of n . Now define: If k , the number of digits of k is odd, set a n = - 1. If k is even, let a = d 1 . . . d k 2 , b = d k 2 +1 . . . d k and set a n = b/a . It is possible for b to be zero, but a 6 = 0. Again to make sure we understand how this works, here are some terms of this sequence: a 1 = a 2 = a 3 = a 4 = a 5 = a 6 = a 7 = a 8 = a 9 = - 1; a 10 = 0 , a 11 = 1 , a 12 = 2 , a 13 = 3 , a 19 = 9 , a 20 = 0 , a 21 = 1 2 , a 125 = - 1 , a 4364 = 64 43 , etc. Determine the set of all numbers x R such that there exists a subsequence of { a n } converging to x .
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