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Math Help - Cauchy Sequence

  1. #1
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    Cauchy Sequence

    Problem Statement:
    Suppose that { p_n} is a Cauchy sequence and that there is a subsquence { p_{n_i}} and a number p such that p_{n_i} \rightarrow p. Show that the full sequence converges, too; that is p_n \rightarrow p.

    My Solution:
    It will suffice to prove that { p_n} converges to p if and only if every subsequence of { p_n} converges to p.
    ( \Rightarrow)Suppose p_n \rightarrow p. Consider { p_{n_i}} \subseteq { p_n}.
    Let \varepsilon > 0. For some n \geq N, d(p_n,p) < \varepsilon. Then if i \geq N, n_i \geq i \geq N. so d(p_{n_i},p) < \varepsilon.
    ( \Leftarrow){ p_n} is a subset of itself, so p_n \rightarrow p.
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  2. #2
    Member Mauritzvdworm's Avatar
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    it should look more like this

    given \epsilon>0, integers n,m and N we have
    \|p_n-p_m\|<\epsilon whenever n,m\geq N

    we know that \|p_{n_i}\| converges to p and convergent sequences are also Cauchy sequences, so we have the inequality
    \|p_{n_r}-p_{n_s}\|<\frac{\epsilon}{2} whenever r,s\geq N

    then play around with the inequality, you should find the desired result
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  3. #3
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    Is this solved then? I realize I made the mistake of thinking that showing my iff statement is true proves the whole thing. But it's only ONE subsequence, not all of them...
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  4. #4
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    OK, sorry. Looking at this again, I do not understand. I apologize, the sequences are particularly confusing to me.
    I understand why |p_n-p_m| < \varepsilon but i don't see why |p_{n_r}-p_{n_s}| < \frac{\varepsilon}{2}.
    I always have trouble seeing these things, but i don't see why i can't just take |p_{n_r}-p_{n_s}| < \varepsilon for all \varepsilon > 0 as p_{n_i} \rightarrow p and since p_{n_s},p_{n_r} \in { p_n}, then p_n must converge to p as well.
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  5. #5
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    Quote Originally Posted by DontKnoMaff View Post
    i don't see why |p_{n_r}-p_{n_s}| < \frac{\varepsilon}{2}.
    That's just the definition of Cauchy sequence (and all convergent seq. are Cauchy), although I think the one that works is |p-p_{n_k}|<\varepsilon/2. Now think triangle inequality applied to |p-p_n| with those bounds.
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  6. #6
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    ok i think i get it. Could someone check me to make sure though?

    Take \varepsilon > 0. take N s.t. n_k,n > N implies that d(p_{n_k},p),d(p_n,p_{n_k}) < \frac{\varepsilon}{2}. Hence d(p_n,p) \leq d(p_{n_k},p)+d(p_{n_k},p_n) \leq \varepsilon Thus { p_n} converges to p.
    Last edited by DontKnoMaff; October 4th 2010 at 08:46 PM.
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