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Existence of a limit point implies existence of inifintely many limit points?

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following statement is true or not:

    the statement:
    Let [itex](X,d)[/itex] be a non-empty metric space and [itex]A[/itex] is a non-empty subset of [itex]X[/itex]. Then if [itex]A'[/itex] is not empty, then [itex]A'[/itex] is infinte.


    2. Relevant equations
    Definition of limit point and its negation.


    3. The attempt at a solution
    I tried to prove by contradiction in this way: (to prove the statement is true)
    Suppose [itex]p_{1},\dots,p_{N}[/itex] are limit points of [itex]A[/itex]. Since [itex]A[/itex] has a limit point, it is infinte. Now observe that if [itex]p'\in A,p\notin\{p_{1},\dots,p_{N}\}[/itex] then [itex]p'[/itex] is not a limit point of [itex]A[/itex], i.e., [itex]A\cap B(p';r_{p'})=\{p'\}[/itex] for some [itex]r_{p'}>0[/itex]. Thus [itex]A[/itex] has a kind of gap inside it.

    But from here, I cannot go further. It almost seems that [itex]A'[/itex] finite is consistent.
     
  2. jcsd
  3. Mar 13, 2012 #2

    Dick

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    Go with that. I can think of a metric space with one limit point. Can you?
     
    Last edited: Mar 14, 2012
  4. Mar 14, 2012 #3
    Sorry, I tried but couldn't think of one. Could you give me at least a hint? Like, what kind of metric space is it with what kind metric??
     
  5. Mar 14, 2012 #4

    HallsofIvy

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    Well, one obvious way to avoid "an infinite number of limit points" is to start with a space that only contains a finite number of points!
     
  6. Mar 14, 2012 #5
    Humm.. But, you know, if a set has only finite points then there is no limit point of this set for whatever metric space is talked about. Isn't it?
     
  7. Mar 14, 2012 #6

    Dick

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    Nothing fancy. Pick a subset of the real numbers. They should be discrete but approach a single limit point. Like 0.
     
  8. Mar 14, 2012 #7
    Wow, this one right? [itex]\bigcup_{n=1}^{\infty}\{-\frac{1}{n},\frac{1}{n}\}[/itex].
     
  9. Mar 14, 2012 #8

    Dick

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    Looks generally ok to me. But I think you want to include 0 in your subset. What do you think?
     
    Last edited: Mar 14, 2012
  10. Mar 15, 2012 #9
    Ah... I think it's okay to disprove the statement. I showed that A' is not empty but finite anyway. Thanks!
     
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