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COMPLETE metric space

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Give an example of a decreasing sequence of closed balls in a COMPLETE metric space with empty intersection.
    Hint 1: use a metric on N topologically equivalent to the discrete metric so that {n≥k} are closed balls.
    Hint 2: In={n,n+1,n+2,...}. Consider the metric d(m,n)=∑1/2k where the sum is from k=m to k=n-1.


    2. Relevant equations
    N/A

    3. The attempt at a solution
    I really have no idea how to do this problem...how can we construct an example that would work?

    Any help/hints is greatly appreciated!
     
  2. jcsd
  3. Feb 2, 2010 #2

    HallsofIvy

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    Here's a major hint: use Hint 1 and Hint 2! Take your space to be N, the set of positive integers with metric [itex]d(m,n)= \sum_{k=m}^{n-1}[/itex] with m< n. (You will need to show that that is a metric.) d(m,n) is obviously an integer for any m, n so that is "equivalent to the discrete metric"- given [itex]0< \epsilon< 1[/itex] the "[itex]\epsilon[/itex] neighborhood" of any point n is just {n} itself and so every singleton set, and therefore every set, is closed. Therefore [itex]I_n= {n, n+1, n+2, ...} is a decreasing (every In+1 is a subset of In) sequence of closed sets that has empty intersection.
     
  4. Feb 3, 2010 #3
    Sorry, I don't get it...
    d(m,n)=∑1/2k where the sum is from k=m to k=n-1
    I am seeing a lot of fractions in this metric. Why is it obviously an integer for any m, n??

    Also, I don't understand why In={n,n+1,n+2,...} is a closed ball???

    Could somebody kindly help me, please?
    Thanks a million!!!
     
  5. Feb 3, 2010 #4

    Dick

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    The metric is not an integer. The points in the space are integers. Try this exercise. Can you show that the set {2,3,5,6...} is the closed ball of radius 1/4 around the point x=3?
     
  6. Feb 3, 2010 #5

    HallsofIvy

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    You are right, Dick, I don't know what I was thinking!
     
  7. Feb 4, 2010 #6
    Actually, can we backtrack a little?

    I am already confused when looking at hints 1 & 2 given in the question...even before trying to solve the problem...

    For hint 1: What does it mean to be a metric "on N"? How can {n>=k} ever be closed balls? This just doesn't make any sense to me...
    For hint 2: I just don't understand what In={n,n+1,n+2,...} is. How is In related to our problem?

    Also, in this problem, what is the "metric space" that we're talking about? A metric space is a "set" together with a metric. The metric is given in hint 2, but what is that "set" in our problem?

    Could somebody kindly explain these, please?
    To you, maybe this question is simple. But to me right now, it seems impossible. :(
    Any help is greatly appreciated!!
     
  8. Feb 4, 2010 #7

    Dick

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    The "set" is the positive integers. You have the metric ON the integers. Now I think you should do the exercise I proposed in post 4.
     
  9. Feb 4, 2010 #8
    Do you mean {2,3,4,5,6,...}?

    I assume so, and I'll try your exercise.

    For ANY n>3, n E N,
    d(3,n)=1/23+1/24+...+1/2n-1
    =(1/23)(1+1/2+...+1/2n-4)
    <(1/23)(1+1/2+...)
    =(1/23)(1+1/2+...)
    =(1/23)[1/(1-1/2)] (infinite geometric series)
    =2/23
    =1/4

    For ANY n>3, n E N, d(3,n)<1/4
    So the closed ball of radius 1/4 around the point x=3 = {2,3,4,5,6,...}.

    But how can I find a decreasing seqeunce of closed balls s.t....?
     
  10. Feb 4, 2010 #9

    Dick

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    Ok, you got me. I skipped the 4. So {2,3,4,5...} is B(3,1/4) (the closed ball around x=3 of radius 1/4). Now {3,4,5,6...} is B(4,1/8). B(3,1/4) CONTAINS B(4,1/8), right? Correct me if I slipped up again. Aren't we decreasing so far? Do you see how this continues?
     
  11. Feb 4, 2010 #10
    Yes, identical to my proof above we can show in general that {n E N: d(k,n)≤1/2k-1} = {k-1,k,k+1,k+2,...} = Ik-1.
    And I1,I2,I3,... is a decreasing sequence of closed balls with empty intersection.

    But how can we prove that "d(m,n)=∑1/2k where the sum is from k=m to k=n-1" is actually a metric? It is clearly nonnegative, but I am struggling to prove the other three properties...

    Thanks for helping!
     
  12. Feb 4, 2010 #11

    Dick

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    Well, it's symmetric by definition if you read between the lines. The d(m,n) they gave you is only valid for m<=n. Otherwise you define d(n,m)=d(m,n). The only property you really have to worry about is the triangle inequality. And the only case you really have to worry about there is d(m,o)<=d(m,n)+d(n,o) where m<n<o. Are you sure you really have to PROVE it's a metric, or did you impose that duty on yourself?
     
  13. Feb 5, 2010 #12
    How about the property d(m,n)=0 iff m=n? How can we prove this property for our metric? The sum as defined in our metric is from k=m to k=n-1, so it looks like it's undefined when m=n?


    About the triangle inequality, how many cases do we need?
    For the case d(m,o)<=d(m,n)+d(n,o) where m<n<o, isn't it trivial? I mean, if m<n<o, then we will just have equality in the triangle inequality, right?

    I think we have to prove it because we should show that our example satisfies all the conditions stated in the question.

    Thanks!
     
  14. Feb 5, 2010 #13

    Dick

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    I think you have to interpret their definition. In the case m=n I think they mean that there are no numbers to sum. So d(m,m)=0. For the triangle inequality I would say there are three cases to consider. You can always pick m<o since the inequality is symmetrical in m and o. So where m is in the inequality is what splits the cases. I'd say m<n<o, n<m<o and n<o<m. I'm glad you found m<n<o trivial, yes, in that case both sides are equal. The other cases are even more trivial.
     
  15. Feb 5, 2010 #14
    So I guess it's best to modfiy the function d given in the hint a little and define d as a piecewise function with 3 cases.

    I don't see why only these 3 cases are enough. There are other cases that are possible, right? Also, some of m,n,o may be equal as well, right?
    Also, you said we should look at m<o, but the 3rd one, i.e. n<o<m, does not satifiy m<o?

    Thanks!
     
  16. Feb 5, 2010 #15

    Dick

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    Ok, I meant m<o and the three cases to be n<m<o, m<n<o and m<o<n. I.e whether n smallest, greatest or in the middle. If m>o just swap the places of m and o. If that still doesn't make sense, you can check all six cases if you want to. It just gets kind of repetitious.
     
  17. Feb 5, 2010 #16
    We can just check the 3 cases where m<o, and say by symmetry the cases for m><o are true. Is this because we have verified that d(m,n)=d(n,m) for all n,m E N is satified beforehand, so that d(m,o)<=d(m,n)+d(n,o) and d(o,m)<=d(o,n)+d(n,m) are the same thing?
     
  18. Feb 5, 2010 #17

    Dick

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    Sure.
     
  19. Feb 5, 2010 #18
    How can we prove that (N,d) with the metric d as defined above is a complete metric space? (i.e. every Cauchy sequence in N converges (in N))
     
  20. Feb 5, 2010 #19

    Dick

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    Prove it's a discrete metric space. I.e. prove that there is a ball around every point that contains only the single point. That means all Cauchy sequences are eventually constant.
     
  21. Feb 5, 2010 #20
    d(m,n)=∑1/2k where the sum is from k=m to k=n-1
    But with our metric, I don't think it's possible? What will be the "radius" of the ball for which that is true for EVERY point?

    If there is a ball around every point that contains only the single point, why does it imply that all Cauchy sequences are eventually constant? (my textbook also says something similar in another example, but it is not so obvious to me why this is true...)

    Thank you.
     
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