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Metric Space and Balls

  1. Oct 7, 2006 #1
    I need to find a complete metric space and a sequence of nested, closed balls such that their infinite intersection is empty. How is this possible?
     
  2. jcsd
  3. Oct 7, 2006 #2

    quasar987

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    Out of curiosity, what does "nested"means? I haven't been able to find a definiiton on wiki.
     
  4. Oct 7, 2006 #3
    [tex]B_1\supset B_2\supset ... B_n\supset ...[/tex]
     
  5. Oct 7, 2006 #4

    StatusX

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    Are you sure you're not asked to prove it's not possible?
     
  6. Oct 7, 2006 #5

    HallsofIvy

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    You want an example that violates the "finite intersection property": if a collection of a certain type of sets has the property that the intersection of any finite subcollection is non-empty, then the intersection of all the sets is non-empty. That is true for compact sets so you want an example where the closed balls are not compact. Since, in Rn, closed and bounded sets are compact, you want either an example that is not in R- say subsets of the rational numbers with the usual topology- or unbounded closed intervals. I'm not clear on what you mean by "closed balls"- intervals of the form [p- d, p+ d]? Would [itex][x, \infty][/itex] be a "closed ball"?
     
  7. Oct 7, 2006 #6
    I emailed my professor about the same thing; he said a closed ball must have a nonzero radius (I was trying to get away with the empty set). I'm not sure [x,infty] would be a ball since it has no 'center'.

    Also, I need a complete space, so rationals won't work.
     
  8. Oct 7, 2006 #7

    StatusX

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    Ok, I deleted that last post, forget what I said.

    If one of those sets were compact, all subsequent sets would be, and then by the finite intersection property of compact sets the intersection would be non-empty. A set in a metric space is compact iff it is complete and totally bounded. In your example, the sets are complete and bounded. In R^n, boundedness is equivalent to total boundedness, so you'll have to look to a different metric space. Try an infinite dimensional space, like a Hilbert space.
     
  9. Oct 11, 2006 #8
    I tried l_p spaces but I'm not sure what to do.
     
  10. Oct 12, 2006 #9

    matt grime

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    We've decided it does it not work with a finite dimensional normed space, so what can we use about infinite dimensional normed spaces that distinguishes them from finite ones? Well, it can only be the infinite dimensionality, right? Can you think what to do with that?
     
  11. Oct 12, 2006 #10
    Let's say in the space of all sequences whose sum is finite, I can have a ball centered at (1, 0, 0, ...) and another inside at (0, 1, 0, ...), and so on, but I don't see why their intersection would be empty.
     
  12. Oct 13, 2006 #11

    matt grime

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    Nope. You've no need to pick balls centered on different points. since you didn't insert the radii of the balls above, no one can state what the intersection would be, or if there even nested.

    Can you think of anyway of getting sets B_r (nested, but ignore that for now) so that their intersection is definitely empty? Is there anyway you can think of to use r to define which elements may lie in the set (nothing to do with the radius), so that the intersection is definitely empty? How about an exampe in the integers to help you: what is the intersection of

    {0,1,2,.....}
    {1,2,3,......}
    {2,3,4,.....}
    {3,4,5,.....}
    etc
     
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