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Math Help - Hausdorff dimension of subsets of R, and cardinality

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    Hausdorff dimension of subsets of R, and cardinality

    If a subset of the unit interval has the cardinality of the continuum, will it necessarily have non-zero Hausdorff dimension?

    While I am at it, is the Sierpinski triangle a continuous curve, like the Sierpinski carpet?

    Thanks.
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    Senior Member Tinyboss's Avatar
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    This is a guess, not a deliberately vague hint: what happens if you do a "Cantor set" where at the n-th iteration, you remove the middle (n-1)/n of each remaining interval?
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    Tinyboss, thanks, but, although my guess is that you would still end up with a continuum number of points (although it would no longer be self-similar), unless I see where this is supposed to lead me, I'm not tempted to pursue this idea further. Please enlighten me as to what it would have to do with either of my questions.
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    Senior Member Tinyboss's Avatar
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    I wonder about it because a generalized Cantor set can have any Hausdorff dimension in (0,1) depending on what fraction of each interval is removed. In the construction above, that fraction gets arbitrarily close to 1, so maybe (vague intuition here--I'm not in a calculating mood) the Hausdorff dimension will be zero.
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    MHF Contributor Drexel28's Avatar
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    I wondered this same question a year ago or so. To be more precise I asked "Does there exist an uncounbtable subset of the reals with zero Hausdorff dimension". I searched for a few days in books and couldn't find anything. I then moved to the internet and the only thing I could come up with is something similar to what Tinyboss suggested, by taking limiting values of modified Cantor sets.
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    Thanks, Tinyboss and Drexel28. I now see where the argument is going, but I am not sure it would pan out: if we had two parts each time, with the scaling factor being
    (1-(n-1)/n)/2 = 1/2n, each dimension would be 1/log_2(2n). Taking the fractal as a limit, then its dimension does indeed tend to zero. However, taking the limit process also to the number of points left, the limit of the portion of the line you are taking out is also tending to taking the entirety of each segment remaining from the last step out, so that it is not clear that you end up with an uncountable number of points. In any case, the proof that the Cantor set has an uncountable number of points no longer works for this new construction. Hence, although the construction is interesting, I am not sure that it would produce a set of uncountable points with dimension zero.

    PS. Any thoughts on my second question? (about Sierpinski)
    Last edited by nomadreid; January 15th 2011 at 10:06 PM. Reason: PS added, and forgot factor 2
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    Senior Member Tinyboss's Avatar
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    I don't know precisely how the Sierpinski triangle curve is defined, but I assume it is the limit of a family of continuous curves, and so it's continuous if the convergence is uniform.
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    There's the rub: the Sierpinski triangle is not defined as a limit of curves. It is formed in a similar fashion to the Cantor set, except in the Sierpinski triangle you start with an equilateral triangle and at each iterative step you remove from each closed triangle the middle open triangle (oriented upside down to the original triangle), leaving three closed equilateral triangles (each of the same size as the one removed, except that they are closed). It's also called the Sierpinski gasket, or Sierpinski Sieve, and has a Hausdorff dimension of log(3)/log(2). This does not automatically make it a curve at all, but somewhere (?) I read that it was a curve. However, whether it is a continuous curve is another question.
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  9. #9
    Senior Member Tinyboss's Avatar
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    I believe the original construction was as a curve, and that the construction by removing center triangles has the same limit set.
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