Help with hausdorff dimension and countableness.

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The discussion centers on the relationship between Hausdorff dimension and countability, specifically addressing the claim that a subset S of R with a Hausdorff dimension greater than 0 must be uncountable. The participants clarify that while a set with Hausdorff dimension greater than 0 is indeed uncountable, the converse does not hold; a set with Hausdorff dimension 0 can still be uncountable. The example provided involves modifying the Cantor set, demonstrating that it is possible to construct uncountable sets with varying Hausdorff dimensions.

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  • Understanding of Hausdorff dimension and its implications in set theory.
  • Familiarity with the Cantor set and its properties.
  • Basic knowledge of countability in mathematical sets.
  • Concepts of limits and logarithmic functions, particularly ln2 and ln n.
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  • Research the proof that establishes the relationship between Hausdorff dimension and countability.
  • Explore variations of the Cantor set and their respective Hausdorff dimensions.
  • Study the implications of sets with Hausdorff dimension 0 in measure theory.
  • Investigate counterexamples in set theory that illustrate the nuances of dimension and countability.
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Mathematicians, students of real analysis, and anyone interested in advanced set theory and its applications in topology and measure theory.

icantadd
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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?

Thanks,

jon
 
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icantadd said:
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?

Thanks,

jon
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.
 
HallsofIvy,

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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