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royden2 - Royden Real Analysis 3rd ed Chapter 2 Problem...

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Royden, Real Analysis 3rd ed. Chapter 2 Problem 2-11 a. If | x n l | ² for n N, we get immediately | x n x m | | x n l | + | x m l | 2 ε (1) for n, m N. b. Equation (1) implies | x n | | x N | + 2 ε for all n N. Hence { x n } is bounded by max 1 i N | x i | + 2 ε . c. Assume lim x n m = l. Then for any ε > 0 , there is a number L such that | x n m l | ε for m L. (2) Moreover, we may assume WLOG that n L N. By (1) and (2), | x n l | | x n x n L | + | x n L l | 2 ε for n n L . Since ε can be arbitrarily small, lim x n = l. d. By part a it is enough to know that every Cauchy sequence { x n } converges. By Problem 2-9.b { x n } has a convergent subsequence and the conclusion follows then from part c . Problem 2-13 (i) only if part. Let l = lim x n . By de fi nition l = inf a n = lim a n , where a n =sup k n x k is a monotone sequence. Hence we have, for any ε > 0 , l a n l + ε (1) holds for n no less than a certain number N. By de fi nition of a N , x k a N l + ε for k N and there exists n ( ε ) n such that x n ( ε ) a n ε l ε . 1
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Combining together, we have for any ε > 0 and n, l ε x m l + ε holds for some m n. Now let ε vary like 2 1 , 3 1 , . . . , we can get a sequence { x n k } such that l k 1 x n k l + k 1 and n 1 < n 2 < n 3 < · · · , which implies lim k x n k = l. (ii) if part. The given conditions imply l ε a n l + ε holds for all m n. Hence lim x n = lim a m = l. Problem 2-15 For each n, b n inf k n x k x n sup k n x n a n . Letting n → ∞ , lim x n = sup n b n = lim b n lim a n = inf n a n = lim x n . (1) For any ε > 0 , we knew from Problem 2-13 that x k l + ε for k N 1 . (2) Using the de fi nition of lim x n = l, we have similarly x k l ε for k N 2 . (3) Combining together, l ε x k l + ε for k N = N 1 N 2 . That is lim x n = l. Conversely, (2) and (3) will hold, which imply lim x n l and lim x n l.
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