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Sequence Convergence

  1. Mar 25, 2013 #1
    If ##p## is a limit point of ##E## then ##\exists \ (p_n) \ s.t. (p_n) \rightarrow p##

    For the sequence construction, can I just define ##(p_n)## as such:

    ##For \ q \in E, \ \ define \ (p_n) := \left\{\Large{\frac{d(p,q)}{n}} \right\}_{n=1}^\infty##​
     
  2. jcsd
  3. Mar 25, 2013 #2

    pwsnafu

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    Are you assuming E is subset of R?

    Try E = {1}.

    Edit: I also think your definition of limit point is wrong.
     
  4. Mar 25, 2013 #3
    Thanks. I am not looking for the def. of ##l.p.##, rather the theorem that states we can alway construct a seq. converging to any limit point.

    Rudin uses a different proof, but I just thought about this one and wanted to see if it is correct.
     
  5. Mar 25, 2013 #4
    I assume you're working in ##\mathbb{R}^n##? This is true in metric spaces. It's not true in general. Your approach is correct, though your proof is not correct as written. You need to be more rigorous about how you're selecting elements for your subsequence.
     
  6. Mar 28, 2013 #5

    Bacle2

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    Notice your sequence is a collection of numbers, which are not necessarily points in your space (i.e., outside of the reals, that I can think of ). For example, in R^n, for n>1, the sequence of numbers d(p,q)/n is not a collection of points in your space.
     
    Last edited: Mar 28, 2013
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