18.100pset4

18.100pset4 - 18.100B/C: Fall 2008 Solutions to Homework 4...

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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 4 1. We proceed by contradiction. Choose R > 0 and suppose that there are infinitely many n such that | a n | ≤ R . Let ( a n k ) ∞ k =1 be the subsequence of ( a n ) consisting of those points. We have that a n k ∈ [- R,R ] for all k and that [- R,R ] is compact, so some subsequence of ( a n k ) converges. But this subsequence is also a subsequence of ( a n ), so ( a n ) has a subsequence that converges. This is a contradiction, so for all R > 0 there are at most finitely many n such that | a n | ≤ R . Thus | a n | → ∞ , as desired. The same statement does not hold of a sequence of rationals in the metric space Q : consider, for example, the sequence a 1 = 1 ,a 2 = 1 . 4 ,a 3 = 1 . 41 ,a 4 = 1 . 414 ,... of rational approximations to √ 2. As a sequence in the metric space R , every subsequence of ( a n ) approaches √ 2. Since √ 2 6∈ Q , no subsequence of ( a n ) converges in Q . However, the sequence is bounded and so certainly | a n | 6→ ∞ . 2.(a) Since a n 6→ 0, we know there exists some r > 0 such that ( a n ) contains infinitely many terms satisfying a n 6∈ B r (0), i.e., infinitely many terms such that | a n | > r . Since ( a n ) is a Cauchy sequence, we know that for all ε there exists some N such that all terms after the sequence differ pairwise by at most ε . In particular, choose ε = r 2 . Then there is some N such that m,n ≥ N implies | a m- a n | < r 2 . Now, choose some k ≥ N such that | a k | > r . We have by the triangle inequality that | a m- | + | a m- a k | ≥ | a k- | . Thus for all m ≥ N we have | a m | + r 2 > | a m- | + | a m- a k | ≥ | a k- | > r and so | a m | > r 2 for all m ≥ N . Setting ω = r 2 gives us our desired result. In addition, because | a k | > r and | a m- a k | < r 2 for all m ≥ N we actually must have that a m has the same sign as a k . (b) We wish to show that for all ε > 0, if we choose any m,n sufficiently large then 1 a m- 1 a n < ε . We have from part (a) that there is some ω > 0 such that | a n | > ω for all n ≥ N . We may also assume without loss of generality that a n > 0 for all n ≥ N and so actually that a n > ω . Choose any ε > 0 and set r = ω 2 ε 1+ ωε . Then there is some N such that m,n ≥ N , | a m- a n | < r . Observe that ω- r = ω 1+ ωε > 0, so the preceding sentence implies that a n > a m- r > 0 and a m > a n- r > 0....
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.

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18.100pset4 - 18.100B/C: Fall 2008 Solutions to Homework 4...

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