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Convergent Subsequences in Compact Metric Space

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that [tex](x_n)[/tex] is a sequence in a compact metric space with the property that every convergent subsequence has the same limit [tex]x[/tex]. Prove that [tex]x_n \to x[/tex] as [tex]n\to \infty[/tex]


    2. Relevant equations
    Not sure, most of the relevant issues pertain to the definitions of the space. In this case I believe the following is relevant:
    In a compact metric space every sequence must have a convergent subsequence, defining it as sequentially compact.

    I'll add more.

    3. The attempt at a solution
    My basic hang up is this: Does every subsequence have to be convergent? If so...you can.

    Suppose not:
    Assume there exists a subsequence [tex]s_n[/tex] in [tex](x_n)[/tex] s.t. [tex]s_m =\displaystyle\sum\limits_{k=1}^{m} x_k \to y \neq x[/tex]

    Then there would exist an [tex]m>a>0[/tex]

    Under the assumption that [tex]s_a = \displaystyle\sum\limits_{k=a}^{\infty} x_k \to x[/tex]

    Where [tex]s_m = \displaystyle\sum\limits_{k=1}^{m+a} x_k[/tex] Would not be convergent.

    Then the sequence [tex](x_n)[/tex] is not Cauchy, which implies it's not a Complete Metric Space.
     
  2. jcsd
  3. Aug 30, 2010 #2

    lanedance

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    how about contradiction? may need some tightening but general idea is there

    assume xn doesn't tend to x, then either there it does not converge or it tends to y not equal to x

    if it tends to y its clearly a contradication

    now if it does not converge it still visits x infinitely often, but it must also be in the neighbourhood of some other point infinitely often (compactness) which is also a contradiction
     
  4. Aug 30, 2010 #3
    That's essentially the proof I gave. Although I've revised it a little to be more true for subsequences rather than partial sums.

    Let [tex](n_1>n_2>n_3...) \in \mathbb{N}[/tex]

    Suppose [tex]lim x_n = x[/tex]

    given any [tex]\epsilon > 0[/tex] we must find an [tex]N[/tex] s.t. [tex]j \ge N[/tex] then [tex]|x_n_j - x| < \epsilon[/tex] [tex]\forall n \ge N[/tex]

    [tex]n_j \ge j[/tex] [tex]\forall j[/tex] by induction and [tex]n_1 \ge 1[/tex] because [tex]n_1 \in \mathbb{N}[/tex]

    if [tex]n_j \ge j \to n_{j+1} \ge n_j \ge j \to n_{j+1} > j +1[/tex] so if [tex]j\ge N[/tex] [tex]n_j \ge N[/tex] then [tex]|x_n_j -x| > \epsilon[/tex]

    And thus every subsequence needs to converge to the same value.
     
  5. Aug 30, 2010 #4
    This appears to be a (confusingly written) proof that if [tex]x_n \to x[/tex] then every subsequence of [tex](x_n)[/tex] also converges to [tex]x[/tex]. Unfortunately that statement is the converse of what you are required to prove.

    Begin with the hypothesis that every subsequence of [tex](x_n)[/tex] converges to [tex]x[/tex], and suppose that [tex]x_n \not\to x[/tex]. What does it mean for [tex]x_n[/tex] not to converge to [tex]x[/tex]? (What is the negation of the statement that [tex](x_n)[/tex] is eventually in every neighborhood of [tex]x[/tex]?)

    (Use words! If your argument isn't essentially correct, symbols and abbreviations don't make it any clearer; they just make it harder to understand what needs fixing.)
     
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