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Sequence of closed sets

  1. Nov 27, 2008 #1
    I was reading about the Nested sphere theorem and a thought occurred. if you have a sequence of decreasing closed sets whose diameter goes to zero in the limit,
    we can show that the intersection of all these sets is a single point.

    my idea was to show this using nested sphere theorem if we can say that for any closed set we can find a smallest closed sphere containing this set as well as a biggest sphere contained in this set.
    that way we can trap our original sequence between two sequences of decreasing spheres.

    Is this if fact possible? if so/not, why?

    cheers!
     
  2. jcsd
  3. Nov 28, 2008 #2
    essentially, my question can be boiled down to:

    for any closed set, can we find a smallest closed sphere containing it? what about a smallest closed sphere contained IN it?
     
  4. Nov 28, 2008 #3

    Office_Shredder

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    Consider instead of something like the interval [0,1/n] we look at the union of ALL intervals of the form

    [tex]\bigcup [k,k+1/n] = K_n[/tex] (over R of course)

    then the limit of the intersection of these [tex] \bigcap K_n[/tex]... this is Z which has measure 0, even though each Kn has unbounded measure.

    EDIT: Obviously we could make this conform to limn->infinitym(Kn) = 0 by using just [tex][1,1+1/n] \cup [0,1/n] [/tex]
     
  5. Nov 28, 2008 #4
    thanks for the reply! it is true what you write, but i am having trouble making the connections...

    What does this imply?

    cheers!
     
  6. Nov 28, 2008 #5

    Office_Shredder

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    You said
    This isn't true. It's if you have a sequence of decreasing closed spheres that this happens... so when trying to extend the result, you get examples like I gave. The first one in fact isn't contained in any sphere. You'll need a boundedness condition to get anywhere
     
  7. Nov 28, 2008 #6
    of course! yes i apologize i mean in a complete metric space where the diameter of each set in the sequence is bounded.

    Thanks for pointing that out :D

    so how would i go about finding these spheres?
     
  8. Nov 30, 2008 #7
    is it even reasonable to say that any closed set in this space will have a closed sphere containing it?
    And if so, can we simply define a set this way and take the infimum along the radii?
     
  9. Dec 2, 2008 #8
    That can only be done if the closed set is bounded; from that, there is a ball containing it by the definition of boundedness. I want to say that there is a smallest closed ball containing a closed set, but I haven't figured out how to prove it yet. (However, it is true that for any point in the metric space, there is a smallest closed ball centered on it containing the closed set.)

    As for closed sets containing spheres, this does not necessarily hold even if the set is nonempty. For example, in the metric space Rn with the Euclidean metric, any one-point set is closed, but does not contain any ball of positive radius. It is true, however, that any nonempty open set in a metric space contains a ball; if U is nonempty and open and x is a point in U, there is an open ball B(x, r) centered on x with radius r contained in it (by the definition of the induced topology). I haven't tried to show that there is a largest open ball contained in a bounded open set, but I think that's true as well. (Again, it certainly is true that for any point in the open set, there is a largest open ball centered on it contained in the open set.)


    This thread is probably more appropriate in the Topology & Geometry forum.
     
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