Constructing a set with arbitrary strict Hausdorff dimension

Here’s the link to Lee’s Challenge #1:

https://leeconjecture.wordpress.com/2015/03/16/introduction-to-hausdorff-measures-and-fractals/

Consider a non-negative real number \beta; if \beta \in \mathbb{Z}, then constructing a set of strict Hausdorff dimension \beta is indeed trivial.

Suppose 0 < \beta < 1. Then, we may apply following theorem.

Theorem 1:
Suppose S_1, \dots, S_m are m separated similarities of a self-similar set F with the common ration r that satisfies 0 < r < 1. Then F has Hausdorff dimension equal to \log m / \log (1/r).

Theorem 1 suggests that Cantor set has Hausdorff dimension of \log 2 / \log 3. The proof of that Cantor set has a strict Hausdorff dimension is left to readers.

Let us define a \mathcal{C}_\delta as a Cantor set which was removed relative \delta from each interval for each stage of construction instead of 1/3. Then, Theorem 1 suggests the Hausdorff dimension of \mathcal{C}_\delta is \log 2 / \log (\frac{1-\delta}{2}). Since 0 < \delta < 1, we may choose appropriate \delta to construct \mathcal{C}_\delta of strict Hausdorff dimension \beta.

For \beta > 1, we may use the idea of n-dimensional Cantor dust. n-dimensional Cantor dust is constructed by n-cartesian product of Cantor sets. The n-dimensional Cantor dust has a strict Hausdorff dimension of n \log 2 / \log 3, and the proof is left to readers. (Key Idea: Use Theorem 1)

Therefore, for any \beta > 1, we may choose a natural number n satisfying \beta < n < \beta+1, and find appropriate \delta to render the strict Hausdorff dimension of \mathcal{C}_\delta to be \beta / n.

Thus, we may construct a set with arbitrary Hausdorff dimension.

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Constructing a set with arbitrary strict Hausdorff dimension

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