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HW4 Solutions

# HW4 Solutions - Math 131A Analysis Summer Session A...

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Math 131A Analysis Summer Session A Homework 4 Solutions 1. Suppose that ( s n ) is a bounded sequence such that all convergent subsequences have limit s R . Prove ( s n ) converges to s . Proof. Since ( s n ) is bounded, any subsequence of ( s n ) is bounded. By Corollary 11.4, ( s n ) contains a bounded monotonic subsequence that has limit lim sup s n and one that has limit lim inf s n . Since bounded monotonic sequences converge and by hypothesis every convergent subsequence has limit s , lim sup s n = s = lim inf s n . Thus lim n →∞ s n = s . 2. (a) Let S be a nonempty subset of R and let - S = {- s : s S } . Prove inf S = - sup( - S ). (Handle the cases inf S = -∞ and inf S > -∞ separately.) (b) Let ( s n ) be any sequence. Prove lim inf s n = - lim sup( - s n ). Proof. (a) Suppose inf S R . Then sup - S R since inf S s for all s R implies - s ≤ - inf S for all s R . Therefore, sup - S ≤ - inf S < . To show - sup - S is a lower bound for S , note - s sup - S for all s S by definition. Thus - sup - S s for all s S . Let a be any lower bound for S . Thus for all s S , a s or equivalently, - s ≤ - a . So - a is an upper bound for the set - S . Hence, sup - S ≤ - a by definition of sup and we get a ≤ - sup - S . Suppose inf S = -∞ . Then S is not bounded below which means - S is not bounded above. For if there exists M > 0 such that - s M for all s S , then - M s for all s S which contradicts S not having a lower bound. Thus sup - S = which is equivalent to - sup - S = -∞ . Recall sup - S > M for any M > 0 if and only if - sup - S < C for any C < 0. (b) Letting S N = { s n : n > N } , we see - S N = {- s n : n > N } . By (a) inf S N = - sup - S N . Thus lim inf s n = lim N →∞ inf S N = lim N →∞ - sup - S N = - lim N →∞ sup - S N = - lim sup - s n 1

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3. Let ( a n ) and ( b n ) be bounded sequences and c R such that c 0. Prove (a) lim sup( a n + b n ) lim sup a n + lim sup b n (b) lim inf( a n + b n ) lim inf a n + lim inf b n (c) lim sup( ca n ) = c lim sup a n (d) lim inf( ca n ) = c lim inf a n Proof. (a) Claim for every N N , sup { a n + b n : n > N } ≤ sup { a n : n > N } + sup { b n : n > N } Since ( a n ) and ( b n ) are bounded, sup { a n + b n : n > N } , sup { a n : n > N } and sup { b n : n > N } are real for every N and are bounded monotonic sequences, hence they converge: If for some positive constants M 1 and M 2
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HW4 Solutions - Math 131A Analysis Summer Session A...

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