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> Keep repeating this process, and from a certain perspective, you’ll have nothing left: The resulting set will cover so little of the original line segment tha

> Keep repeating this process, and from a certain perspective, you’ll have nothing left: The resulting set will cover so little of the original line segment that its length will be zero. But it is, in both an intuitive and a mathematical sense, “bigger” than just a single point. Its Hausdorff dimension is about 0.6. I don't follow. Can someone clarify what "it" is with a dimension of 0.6? Any point? Or all the points remaining after you remove specifically 1/3 from each remaining line segment a number (an infinite?) of times? Would it be different if we removed 1/4 repeatedly? Would it be different if the line was longer than 0 -> 1?

Sniffnoy 3y ago
The set is what has the dimension. And the set under discussion (the Cantor set) is the set remaining after repeating this process infinitely many times. That is to say, every time you repeat the process, the set gets smaller; so, if you take the intersection of all the finite stages, then you get what's left after repeating the process infinitely many times. That remaining set is the Cantor set being discussed. > Would it be different if we removed 1/4 repeatedly? That would result in what's known as a "fat Cantor set". If I'm not mistaken, it would have Hausdorff dimension 1, rather than than something intermediate like the usual Cantor set. > Would it be different if the line was longer than 0 -> 1? No, the length of the starting line segment is not material here.
- subroutine 3y ago
  > If I'm not mistaken, it would have Hausdorff dimension 1 What is the Hausdorff dimension for a line?
  - Sniffnoy 3y ago
    A line or line segment has Hausdorff dimension 1, while a point has Hausdorff dimension 0. In general, ordinary n-dimensional space has Hausdorff dimension n. That's why I referred to the Hausdorff dimension of the Cantor set -- equal to log(2)/log(3), or approximately 0.63 -- as "intermediate".
  - lmm 3y ago
    1 for a line, 2 for a square, 3 for a cube. Hausdorff dimension agrees with the usual kind for normal shapes.
- ouid 3y ago
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Sniffnoy 3y ago

The set is what has the dimension. And the set under discussion (the Cantor set) is the set remaining after repeating this process infinitely many times. That is to say, every time you repeat the process, the set gets smaller; so, if you take the intersection of all the finite stages, then you get what's left after repeating the process infinitely many times. That remaining set is the Cantor set being discussed. > Would it be different if we removed 1/4 repeatedly? That would result in what's known as a "fat Cantor set". If I'm not mistaken, it would have Hausdorff dimension 1, rather than than something intermediate like the usual Cantor set. > Would it be different if the line was longer than 0 -> 1? No, the length of the starting line segment is not material here.

subroutine 3y ago
> If I'm not mistaken, it would have Hausdorff dimension 1 What is the Hausdorff dimension for a line?
- Sniffnoy 3y ago
  A line or line segment has Hausdorff dimension 1, while a point has Hausdorff dimension 0. In general, ordinary n-dimensional space has Hausdorff dimension n. That's why I referred to the Hausdorff dimension of the Cantor set -- equal to log(2)/log(3), or approximately 0.63 -- as "intermediate".
- lmm 3y ago
  1 for a line, 2 for a square, 3 for a cube. Hausdorff dimension agrees with the usual kind for normal shapes.
ouid 3y ago
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