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Help with hausdorff dimension and countableness.

  1. May 9, 2008 #1
    My question, because I keep seeing this on the internet, is that if S is a subset of R and Hausdorff dimension greater than 0, it is uncountable... is this true.

    It seems not to be. If one were to modify the Cantor third set and remove some length of 1/n from the middle of the sets at each iteration, one would achieve a set with Hausdorff dimension: 2 = n^d => ln2/ln n =d, and as 1/n -> 1 , n -> infinity, and ln2/ln n -> 0. Yet the set is still uncountable.

    Perhaps I have missed something, or perhaps most of the time the relationship holds. Does anyone know either way? or if there are only special cases where this happens, what they are?


  2. jcsd
  3. May 9, 2008 #2


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    Your example does NOT disprove the theorem you cite. The theorem says that if a set has Hausdorff dimension larger than 0, then it is uncountable. It does NOT say that if the Hausdorff dimension is 0, it MUST be uncountable. Your example is a counterexample to the converse of the stated theorem.
  4. May 9, 2008 #3

    You are absolutely correct, thank you. If a set is countable then it must have H. dimension >0. But it does not hold that if Hausdorff dimension = 0, that the set must be countable.

    Yet, I still do not understand the why of it. A countable set is a qualification of smallness, as is H.dimension = 0, but I cannot come to see how having a Hausdorff dimension greater than zero implies that the set is uncountable. I can't find a proof of it, I can't think of how to go about proving it.
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