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Math Help - Advanced Calculus Problem

  1. #1
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    Advanced Calculus Problem

    Let (S, d) be a metric space. Suppose {sn belongs to S} be a sequence converging to s, where s belongs to S. Suppose {tn belongs to S} is a sequence of points so that d(s_n, t_n) < 1/n for all n. Prove that \left( {t_n} \right) \to s .

    -------------------------------------------------------------------

    I know this may come as a silly question but should I prove that tn is bounded?

    Because I'm stuck on finishing this. Here's what I have so far.

    We know that {sn} is a converging sequence and that it converges to some s in S so we write for each epsilon > 0, there exists an N belonging to the set of natural numbers such that n > N implies that |sn - s| < epsilon (or lim sn = s).

    From Kenneth Ross, Definition 13.2: A sequence (sn) in a metric space (S,d) converges to s in S if the limit (n -> infinity) d(sn, s) = 0. A sequence (sn) in S is a Cauchy sequence if for each epsilon > 0, there exists an N such that m,n > N implies d(sn,s) < epsilon.

    We know that {tn} is a sequence in S and that the d(sn,tn) < 1/n for all n in the set of natural numbers. So we know that at some point s, both {sn} and {tn} meet but the sequence {sn} stops at s so...this where I get a little stuck.

    I know that the lim d(sn, tn) = 0 but I don't think that's suffice to show that they converge (or am I wrong)? What's getting me stuck is whether or not I have to prove that tn is bounded.

    Can you lend a hint?
    Last edited by MissMousey; November 7th 2011 at 07:43 AM.
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  2. #2
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    Re: Advanced Calculus Problem

    Quote Originally Posted by MissMousey View Post
    Let (S, d) be a metric space. Suppose {sn belongs to S} be a sequence converging to s, where s belongs to S. Suppose {tn belongs to S} is a sequence of points so that d(sn, tn) < 1/n for all n. Prove that {tn} converges to s.
    Don't over-think the problem.
    d(S,t_n)\le d(S,s_n)+d(s_n,t_n).
    Now make each of the last two terms <\frac{\epsilon}{2}
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  3. #3
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    Re: Advanced Calculus Problem

    Quote Originally Posted by Plato View Post
    Don't over-think the problem.
    d(S,t_n)\le d(S,s_n)+d(s_n,t_n).
    Now make each of the last two terms <\frac{\epsilon}{2}
    Why d(S, t_n)?
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  4. #4
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    Re: Advanced Calculus Problem

    Quote Originally Posted by MissMousey View Post
    Why d(S, t_n)?
    Well, that is what you are asked to prove:
    \left( {t_n } \right) \to S
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  5. #5
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    Re: Advanced Calculus Problem

    Quote Originally Posted by Plato View Post
    Well, that is what you are asked to prove:
    \left( {t_n } \right) \to S
    But I'm asked to show that \left( {t_n} \right) \to s , not the set S.
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  6. #6
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    Re: Advanced Calculus Problem

    Quote Originally Posted by MissMousey View Post
    But I'm asked to show that \left( {t_n} \right) \to s , not the set S.
    That was just my mistake in using a upper case. Change it to lower case.
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  7. #7
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    Re: Advanced Calculus Problem

    Quote Originally Posted by Plato View Post
    That was just my mistake in using a upper case. Change it to lower case.
    Sorry for the late response but one more question. Why are you able to set d(s_n,t_n) to be <\frac{\epsilon}{2}? Because the problem says that d(s_n,t_n) < 1/n, which I'm assuming is equal to {\epsilon}. I know that we're trying to show that d(s,t_n) < {\epsilon}, but with what they provided me in the problem, can I just disregard their d(s_n,t_n) < 1/n?
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  8. #8
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    Re: Advanced Calculus Problem

    Quote Originally Posted by MissMousey View Post
    Sorry for the late response but one more question. Why are you able to set d(s_n,t_n) to be <\frac{\epsilon}{2}? Because the problem says that d(s_n,t_n) < 1/n, which I'm assuming is equal to {\epsilon}. I know that we're trying to show that d(s,t_n) < {\epsilon}, but with what they provided me in the problem, can I just disregard their d(s_n,t_n) < 1/n?
    The idea is we can make d(s_n,s)<\tfrac{\epsilon}{2} as we can make d(s_n,t_n)<\tfrac{1}{n}<\tfrac{\epsilon}{2}
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  9. #9
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    Re: Advanced Calculus Problem

    Ah, ok. Great. Thank you very much for your help.
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