Suppose I is an interval on a Cantor tree with m children I′, each of length epsilon . I have that sum_(hat I) is a child of |hat I|^S=|I|^S Rightarrow s=(log(m))/(-log epsilon) . Is s the "local dimension" of the tree?

Paul Reilly

Paul Reilly

Answered question

2022-09-05

Suppose I is an interval on a Cantor tree with m children I′, each of length ϵ . I have that I ^  is a child of  I | I ^ | s = | I | s s = log ( m ) log ε . Is s the "local dimension" of the tree?
Is there a relationship between the the Hausdorff dimension of a Cantor tree and its local dimension?

Answer & Explanation

Gracelyn Paul

Gracelyn Paul

Beginner2022-09-06Added 17 answers

Step 1
There are many different definitions of local dimension of a set, but the one I've seen the most often occurring together with the ordinary Hausdorff dimension is the following:
Let C be the Cantor set obtained by at each step removing some closed interval from each interval from the last step. The size of this interval need not be fixed neither between steps nor for the intervals in each step and it need not be centered. Now pick any point ξ C . Then ξ can be written as the intersection of all intervals from the construction containing ξ . Let this sequence of intervals be { I j } where I j is an interval from the nth step of the construction. Let I j 1 , I j 2 , ..., I j m be the m children of I j . Then the local Hausdorff dimension of C at ξ is defined as
sup { s : ( | I j 1 | | I j | ) s + . . . + ( | I j | m | I j | ) s < }
In your case, since all children have the same length, this would simplify to
sup { s : ( m | I j 1 | | I j | ) s < }
If defined in this way, the Hausdorff dimension of the set (for nice enough sets) is the supremum of its local dimensions.
The local dimension is thus always defined at a point, and not in a certain step, which implies that we must use the information in infinitely many steps to be able to say something at all.

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