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Totally bounded & Heine Borel?

  1. Jun 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Why doesn't a set being totally bounded imply the set has the Heine Borel property?

    Another related question is what happens if a cover consist of open balls that cover the set and more of it? i.e. A=(-1,2) U (1,3) covers (0,1) but really covers more than (0,1). Is A considered a legitimate cover? I have a feeling it is.



    3. The attempt at a solution
    They both refer to the set being covered by a finite number of sets.

    Is it because for example, Take the set (0,1). One cover for it is an infinite cover consisting of infinitely many open balls of the type B(x,1/n), n>1. There is no finite subcover of this cover that covers (0,1). But such a cover is not allowed to show totally boundedness as the radius has to be fixed.
     
  2. jcsd
  3. Jun 10, 2007 #2

    matt grime

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    What is the Heine Borel property precisely?

    The difference is that for compactness, which I assume is H-B, the cover is given to you: you have no choice over the size of the sets, or where they are centred. For totally bounded, you get to choose how to lay down the cover once the radius has been fixed.
     
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