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Finding Cluster Points

  1. Oct 5, 2008 #1
    The problem statement, all variables and given/known data
    Find all the cluster points of the subset of R given by {1/n + 1/m : n, m are positive integers}.


    Relevant equations
    Let S be a subset of a metric space E. Then p in E is a cluster point of S if any open ball with center p contains infinitely many points of S.


    The attempt at a solution
    Let S = {1/n + 1/m : n, m are positive integers}. Since 0 ≤ 1/n + 1/m ≤ 2, S is a subset of [0,2]. If p is a cluster point of S, it must be contained in [0,2]. I'm pretty sure that [0,2] is the set of all cluster points of S, but I'm having a hard time proving this. In particular, I'm having a hard time finding points of the form 1/n + 1/m around p contained in [0,2]:

    Suppose p is in [0,2]. Pick any open ball around p, say (p - r, p + r). If p < 2, how would I find a point of the form 1/n + 1/m greater than p but less than min{p+r, 2}?
     
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  3. Oct 5, 2008 #2

    Dick

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    You can't. Consider this, if e>0 then (e,2] contains only finitely many points of S. Doesn't it?
     
  4. Oct 5, 2008 #3
    I pondered about the possibility of not being able to find such a point. Does that mean that the cluster points of S are all in S?
     
  5. Oct 5, 2008 #4

    Dick

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    I don't think S has very many cluster points. And I don't think 'they' are in S.
     
  6. Oct 6, 2008 #5
    I think the cluster points of S are those of the form 1/n, where n is a positive integer. What do you think?
     
  7. Oct 6, 2008 #6

    Dick

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    Actually, I think you are right. But you are missing a cluster point that isn't in S.
     
    Last edited: Oct 6, 2008
  8. Oct 6, 2008 #7
    Are you talking about 0?
     
  9. Oct 6, 2008 #8

    Dick

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    None other.
     
  10. Oct 9, 2008 #9
    I'm stuck trying to prove there are no other cluster points: Suppose p in [0,2], where p is nonzero and not of the form 1/n. Suppose p < 1. Then we can find integers m, n such that 1/m + 1/n < p < 1/m + 1/(n-1). I'm unable to show that there are no points in S between 1/m + 1/n and 1/m + 1/(n-1). How would I do this? And is this the right approach?
     
  11. Oct 9, 2008 #10

    Dick

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    It's not hard. The proof is just a little slippery to phrase clearly. If p>0 is a cluster point then there are integer sequences ai and bi such that 1/ai+1/bi=pi->p. Let's also assume that all of pi are distinct. Let Mi=max(ai,bi) and mi=min(ai,bi). So 1/Mi+1/mi=pi->p and Mi>mi. Can Mi be bounded? Why not? Can mi be unbounded? Why not?
     
    Last edited: Oct 9, 2008
  12. Oct 9, 2008 #11
    If Mi and mi are both unbounded, then 1/Mi + 1/mi -> 0 ≠ p right? Hence, one of them must be bounded. You indirectly stated that Mi is unbounded and mi is bounded. I can't think of a reason why. But the conclusion will be that p = 1/n for some n right?
     
  13. Oct 9, 2008 #12

    Dick

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    If mi<=Mi<=N then there are only a finite different number of pi values. I used that p is a cluster point to make sure the sequence has an infinite number. The conclusion will be p=1/n, correct. But you aren't quite there yet. Now what would happen if mi were unbounded?
     
    Last edited: Oct 9, 2008
  14. Oct 9, 2008 #13

    Dick

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    Sort of right. If Mi=(1,2,1,3,1,4,...) and mi=(2,1,3,1,4,1,...) the limit is 1 but both are unbounded. That's what I meant by the proof being slippery and that's why I did the max, min thing.
     
  15. Oct 9, 2008 #14
    If Mi is bounded, then Mi <= N for some N and since mi <= Mi, mi must also be bounded. But as you stated, there will only be a finitely many pi values and hence p will turn out to be one of these pi. Right?
     
  16. Oct 9, 2008 #15

    Dick

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    No. We picked the sequence so that all of the pi were distinct so there are an infinite number. Hence Mi is NOT bounded. Pick a subsequence where Mi is increasing. Now you've already given the correct reason why mi is bounded (otherwise the limit is zero). That means 1/Mi->0 and 1/mi has only a finite collection of values. Since the sequence converges it must settle down to one of them. That's your 1/n.
     
  17. Oct 10, 2008 #16
    OK. I get it now. Thanks a lot.
     
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