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Computing lebesgue number for an open covering

  1. Mar 26, 2013 #1
    Munkres proof for the Lebesgue number lemma which is
    (If X is a compact metric space and A is an open covering then there exists δ>0 such that for each subset of X having diameter less than δ , then there exists an element of the covering A containing it)
    gives a way to compute δ using a finite subcollection that covers the compact metric space. However, this is not necessarily the smallest Lebesgue number. I wonder if there is another proof that involves evaluating the smallest Lebesgue number, if the latter exists.
     
  2. jcsd
  3. Mar 29, 2013 #2
    Useful nucleus, do you mean the largest Lebesgue number? Any positive number smaller than a Lebesgue number for a covering is also a Lebesgue number for the covering.
     
  4. Apr 1, 2013 #3
    lugita15, you are absolutely right, I meant largest Lebesgue number if exist. I did not have time to refelct on this question again, but will give it a try soon.
     
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