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X is an accumulation point show there is subsequence that converges to x

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose x is an accumulation point of {an: n is a member of integers}. Show there is a subsequence of (an) that converges to x.

    3. The attempt at a solution
    I'm a little stuck on this one. I know that since x is an accumulation point then every neighborhood around x, (x-e,x+e) contains infinitely many points of an. I guess I just don't know how to construct the subsequence.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Feb 16, 2010 #2
    Suppose the subsequence we want is [tex]a_{n_1}, a_{n_2}, \ldots[/tex] with [tex]n_1 < n_2 < \cdots[/tex]. If you could ensure that [tex]x-1/k < a_{n_k} < x+1/k[/tex] for all k you'd be done (why?). Can you think of how you can ensure such a sequence exists?
     
  4. Feb 17, 2010 #3
    Can we say the sequence is {x-1/k2} for integers>0 ? I'm still a little confused.

    I know that the interval (x-1/k, x+1/k) will be contained in a neighborhood of x, and thus has infinitely many points. I'm not sure why we are using 1/k though. I guess we are in essence using the definition of an accumulation point repeatedly and narrowing the interval and making it converge to x?
     
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