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Hausdorff Dimension

  1. Mar 18, 2011 #1

    I had a question about understanding some basic thing about the Hausdorff dimension. Specifically, I'm trying to understand why the two dimensional Hausdorff dimension of a 1-d line is zero.

    In terms of the two dimensional Lebesgue measure, I can see that I can cover the line by a countable union of rectangles, where each rectangle has length of one coordinate = 0.

    For example, Suppose the line is [0,1] and lies on the x axis. Then I can cover it with rectangles with their length in the y dimension zero, and in the end, I would have to sum up the measure of each rectangle, and each rectangle would have

    m( A_{i} x B_{i} ) = m(A_{i}) m_(B_{i}) = 0. (where A_{i} is in X and B_{i} is in Y)

    So I can see how a two dimensional Lebesgue measure of a one dimensional line is zero.

    However for Hausdorff measures, I have to use circles to cover the line. I'm not able to visualize this. Can I make the radius so small that it only covers a single number? But in that case I would need an uncountable number of circles? I'm confused since the definition of a Hausdorff measure only lets me control the radius of the n-balls and not anything else, so I can't understand how to play around with the radius to make the two Hausdorff measure zero when covering a line in 1 dimension.
  2. jcsd
  3. Mar 19, 2011 #2


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    It is a theorem that the n-dimensional Hausdorff measure is related to the n-dimensional Lebesgue measure. Thus if H is the n-dimensional Hausdorff measure, and if L is the Lebesgue measure, than there is a constant factor c such that H=cL.

    In particular, if something has measure zero for Lebesgue measure, then it has measure zero for Hausdorff measure. Maybe you could look up that proof?
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