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Cauchy sequence with a convergent subsequence

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Theorem: In a metric space X, if (xn) is a Cauchy sequence with a subsequence (xn_k) such that xn_k -> a, then xn->a.

    2. Relevant equations

    3. The attempt at a solution
    1) According to this theorem, if we can show that ONE subsequence of xn converges to a, is that enough? Or do we need to show that EVERY subseuqence of xn converges to a in order to claim that xn->a?

    2) How can we prove the theorem?
    By definition, xn is Cauchy iff for all ε>0, there exists N s.t. if n,m≥N, then d(an,am)<ε.
    By definition, xn->a iff for all ε>0, there exists M s.t. if n≥M, then d(xn,a)<ε.
    I know the definitions, but I don't see how to link these all together to prove the theorem...

    Any help is greatly appreciated!
  2. jcsd
  3. Feb 10, 2010 #2
    You can only assume it for one subsequence. Otherwise you could just take the subsequence n_k=k and be done.

    The general idea of this proof is that for a_nk to converge to a means that for large enough N for all k>N a_nk is close to a. Now the issue here is that a_nk doesn't include all elements of a_n, but to say that a_n is Cauchy means precisely that for large enough indices the elements are close to each other.

    To actually perform the proof let us choose some [itex]\epsilon > 0[/itex]. Note that since [itex]a_{n_k} \to a[/itex] there exists some K>0 such that for all k>K we have [itex]d(a_{n_k},a) < \epsilon[/itex]. Now since [itex](a_n)[/itex] is Cauchy there exists some N>0 such that if n,m>N, then [itex]d(a_n,a_m) < \epsilon[/itex]. This is the necessary setup.

    There exists some k' such that [itex]n_{k'} > N[/itex] and [itex]k' > K[/itex]. By the Cauchy condition if [itex]n>N[/itex], then,
    [tex]d(a_{n_{k'}},a_n) < \epsilon[/tex]
    Similarily by the convergence we have,
    [tex]d(a_{n_{k'}},a}) < \epsilon[/tex]
    Add these and use the triangle inequality to get,
    [tex]d(a_n,a) < 2\epsilon[/tex]
    Since [itex]\epsilon[/itex] was arbitrary this shows that [itex](a_n)[/itex] converges to a.
  4. Feb 12, 2010 #3
    There exists some k' such that [itex]n_{k'} > N[/itex] and [itex]k' > K[/itex] <---can you please explain this part?

    Thank you!!
  5. Feb 12, 2010 #4
    Well [itex]1 \leq n_1 < n_2 < \cdots[/itex] so [itex]n_i \geq i[/itex] (n_1 must be at least 1, n_2 must be at least one more, n_3 must be at least one more than n_2, etc.). Thus [itex]n_{N+1} > N[/itex] so simply let [itex]k' = \max(N+1,K+1)[/itex], then
    [tex]n_{k'} \geq n_{N+1} > N \qquad k' \geq K+1 > K[/tex]
    Basically what this says is the sequence (n_k) will eventually become greater than N (it's unbounded) and there exists an integer larger than K.
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