solutions-assignment1

# solutions-assignment1 - MATH 255 ASSIGNMENT 1, short...

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MATH 255 ASSIGNMENT 1, short solutions 1. [10 points] Let ( x n ) be a bounded sequence of real numbers. Prove that lim inf n →∞ x n = sup { t : { n : x n < t } is ﬁnite } . Solution. Denote α = lim inf n →∞ x n , β = sup { t : { n : x n < t } is ﬁnite } . Let t be such that { n : x n < t } is ﬁnite. Such t must exist since ( x n ) is a bounded sequence. Let n t = max { n : x n < t } . Then, for n > n t , x n t . It follows that for m n t , y m = inf { x n : n m } ≥ t. Since y m is increasing sequence, α = lim inf n →∞ x n t. Hence, α is an upper bounded for the set { t : { n : x n < t } is ﬁnite } , and so α β. (1) Let ± > 0. Then the set { n : x n < β + ± } is inﬁnite (why?) and for all m , y m = inf { x n : n m } ≤ β + ± (why?). Since y m is an increasing sequence, α = lim m →∞ y m β + ±, and since ± > 0 is arbitrary, α β. (2) The relations (1) and (2) imply the statement. 2. [10 points] Let ( x n ) be a sequence of real numbers. Prove that lim sup n →∞ ( - x n ) = - lim inf n →∞ x n . 1

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Solution. Suppose that ( x n ) is not bounded from below. Then lim inf n →∞ x n = -∞ . Since ( - x n ) is not bounded from above, lim sup n →∞ ( - x n ) = , and the statement holds. Suppose now that ( x n ) is bounded from below. This implies that ( - x n ) is bounded from above. Let y m = sup {- x n : n m } , ˆ y m = inf { x n : n m } . Note that if
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solutions-assignment1 - MATH 255 ASSIGNMENT 1, short...

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