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Proving totally bounded sets are bounded.

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the error in this proof and give an example in (ℝ,de) to illustrate where this proof breaks down.

    Proof that every totally bounded set in a metric space is bounded.

    The set S is totally bounded and can therefore be covered by finitely many balls of radius 1, say N balls of radius 1. Then S is a subset of any ball B(x,2N) provided X lies in S. Thus diam S≤4N so that S is bounded.

    I can't see the fault in the proof and therefore don't know where to start when looking for an example in (ℝ,de) that illustrates how the proof breaks down.
    Any suggestions?
     
  2. jcsd
  3. May 5, 2013 #2
    A totally bounded set is bounded. The former is a stronger property than the latter. The converse may not be true in any metric space. I am not sure what ##(R, d_e)## means.
     
  4. May 5, 2013 #3
    What I mean by (R,de) is the set of real numbes with the euclidean metric. I understand that totally bounded is a stronger property and that a proof of this exists but I don't know what is wrong with the proof given in my first post. I'm trying to find out why this proof is sufficient.
     
  5. May 5, 2013 #4

    micromass

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    I suppose it breaks down for ##S## empty...
     
  6. May 5, 2013 #5
    That's interesting and hadn't occurred to me. Is the empty set totally bounded because it is a finite subcover of itself?
     
  7. May 5, 2013 #6
    Would any ball cover the empty set?
     
  8. May 5, 2013 #7
    Yes surely. But does there exists, for every r, a finite A contained in the empty set such that U{B(a,r);a in A} contains the empty set?
     
  9. May 5, 2013 #8
    How could anything be contained in the empty set? The only thing it contains as a subset is itself.
     
  10. May 5, 2013 #9
    Sure. So in my above message A would be the empty set then U{B(a,r): a in A} would also be the empty set and therefore A is in the empty set which is covered by U{B(a,r): a in A} . Making the empty set totally bounded?
     
  11. May 5, 2013 #10
    I am not sure what you are trying do. What is U{B(a, r) : a in A}? This is a union over what?

    The empty set is totally bounded because of the reason given in #6, to which you said "yes surely".
     
  12. May 5, 2013 #11
    Ok cool.

    The definition of being totally bounded that I have been given is that X is totally bounded if for every r>0 there is a finite subset of X say A such that the union of balls U{B(a,r):a in A} contains X.

    So I'm just trying to get it into this form for my understanding
     
  13. May 5, 2013 #12
    Obviously, if A is empty, then the union is empty. And the empty set contains itself.
     
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