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Homework Help: A problem about compactness

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Serge Lang Undergraduate Analysis Chapter Ⅷ §1 Exe4

    Let{Xn} be a sequence in a normed vector space E such that {Xn} converges to v. Let S be the set consisting of all v and Xn.
    Show that S is compact.

    2. Relevant equations
    None

    3. The attempt at a solution
    I guess that maybe it is useful to consider it from the aspect of the definition of compactness,i.e. every sequence of elements of S has a convergent subsequence whose limit is in S. But I coudn't convince that why there must be such a convergent subsequence in every sequence, you know, some sequences are not the given ones that converge.
     
    Last edited: Feb 13, 2007
  2. jcsd
  3. Feb 13, 2007 #2

    quasar987

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    Here's a plan for you.

    1° Show that the only accumulation point in S is v.

    2° Consider a sequence {a_n} in S. Convince yourself that there can be one of two possibilities. Either {a_n} contains a finite number distinct elements or an infinity of distinct element.

    3° Show that in the first case, there is at least one element of the sequence that appears an infinity of times (this defines a converging subsequence).

    4° Show that in the later case, the sequence is convergent and converges to v. (defining a "trivially converging" subsequence)
     
  4. Feb 13, 2007 #3
    Thanks.
    but I still have some problems.
    I can't show that
    If {a_n} contains an infinity of distinct element, the sequence is convergent and converges to v.

    Is the set S bounded?
     
  5. Feb 13, 2007 #4

    StatusX

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    It's probably easiest to go back to the definition in terms of open covers. If U_alpha is an open cover, v is in one of the open sets, and this open set contains all but finitely many of the x_n.
     
  6. Feb 13, 2007 #5

    quasar987

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    This point raised by StatusX (namely, that If U_alpha is an open cover, v is in one of the open sets, and this open set contains all but finitely many of the x_n), is also the key to your problem:

    The fact that v is the only accumulation point of S means that

    [tex]\forall\epsilon>0, \ \ |S\setminus (B_{\epsilon}(v)\setminus\{v\})|<\infty[/tex]

    In words: for any radius epsilon as small as we like, there is only a finite number of elements of S outside the open ball of radius epsilon centered on v.

    It follows immediately that for any sequence {a_n} with an infinity of distinct elements and for all epsilon>0, there exist an N>0 such that for all n>N, a_n [itex]\in B_{\epsilon}(v)\setminus\{v\}[/itex] and that is precisely the definition of convergence.

    (If it weren't so, i.e. if there was an epsilon such that there is only a finite number of elements inside the epsilon-ball, then there is an infinity of them outside. But there is only a finite number of distinct elements outside the epsilon-ball, so our sequence cannot have an infinity of distinct elements: contradiction.)
     
    Last edited: Feb 13, 2007
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