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Convergent sequence and limit points

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if [itex]E \subset X[/itex] and if p is a limit point of E, then there is a sequence [itex]\{p_{n}\}[/itex] in [itex]E[/itex] such that [itex]p=\lim_{n\to\infty}\{p_{n}\}[/itex] (I presume that there is an invisible "[itex]p_{n} \rightarrow p[/itex] implies that" at the beginning of the sentence).

    2. Relevant equations

    -

    3. The attempt at a solution

    We want to show that every neighbourhood around p contains infinitely many [itex]p_{n}[/itex]'s.

    By the definition of convergence we can choose infinitely many N's so that infinitely many [itex]p_{n}[/itex]'s can be contained in a neighbourhood of p with a correspondent radius.

    Is this valid?
     
  2. jcsd
  3. Sep 17, 2012 #2

    micromass

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    Notation issue: when writing [itex]p=\lim_n p_n[/itex], we don't write brackets around the [itex]p_n[/itex]

    Nope, the sentence is fine how it is. Why do you think that should be added?

    No, this is not valid. And I think you're missing the entire point of the exercise. The exercise wants you to actually construct a sequence [itex](p_n)_n[/itex]. You act like the [itex]p_n[/itex] are given, but they are not. You must show that such a sequence actually exist.
     
  4. Sep 17, 2012 #3
    What if the actual question were p_n converges to p implies that p is limit point of range of {p_n}?
     
  5. Sep 17, 2012 #4

    micromass

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    That is not even true. Take [itex]p_n=2[/itex] for all n. Then [itex]p_n\rightarrow 2[/itex], but 2 is not a limit point of [itex]\{p_n~\vert~n\}[/itex].

    What is true is: if [itex]p_n\rightarrow p[/itex] and if [itex]p_n\neq p[/itex] for all n, then p is a limit point of [itex]\{p_n~\vert~n\}[/itex].
     
  6. Sep 17, 2012 #5
    Then what about the latter?
     
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