Question

What are the Hausdorff and box dimensions of the set $\left\{0,1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \ldots\right\}$ ? 

   What are the Hausdorff and box dimensions of the set $\left\{0,1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \ldots\right\}$ ?
Fractal Geometry: Mathematical Foundations and Applications
Fractal Geometry: Mathematical Foundations and Applications
Kenneth Falconer 2nd Edition
Chapter 3, Problem 11 ↓

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The set in question is $\left\{0,1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \ldots\right\}$. This set consists of 0, 1, and the reciprocals of the squares of natural numbers. We can express this set as $\{0, 1\} \cup \left\{\frac{1}{n^2} : n \in \mathbb{N}, n \geq  Show more…

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What are the Hausdorff and box dimensions of the set $\left\{0,1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \ldots\right\}$ ?
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Key Concepts

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Box (Minkowski) Dimension
The box dimension, also known as the Minkowski dimension, quantifies the scaling behavior of the number of small boxes (or balls) needed to cover the set as the size of the boxes shrinks. In contrast to the Hausdorff dimension, the box dimension for sets with accumulation points can be positive, reflecting the rate at which points become denser at the accumulation point even when the set is countable.
Hausdorff Dimension
The Hausdorff dimension is a way of measuring the ‘size’ of a set that takes into account its fine-scale structure by looking at coverings with arbitrarily small sets and then summing the sizes raised to a power. In many cases, including for any countable set, the Hausdorff dimension is zero because you can cover the set with arbitrarily small intervals that make the associated sums as small as desired.
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The purpose of this problem is to show that there exists a set whose fractal dimension does not exist. Let A be the following subset of [0, 1], where we think of representing each point of [0, 1] by its base ten series expansion(s): A = {x = ∑_{i=1}^∐ (d_i / 10^i) : d_i ∈ {0, 1, ..., 9} and d_i = 0 whenever there exists n ∈ N ∪ {0} such that 2^{2n} ≤ i ≤ 2^{2n+1} - 1}. For ε_n = 10^{-2^n}, n = 0, 1, 2, ..., show lim_{n → ∞} (ln N(A, ε_{2n}) / ln(ε_{2n}^{-1})) = 2/3, lim_{n → ∞} (ln N(A, ε_{2n+1}) / ln(ε_{2n+1}^{-1})) = 1/3. One possible approach: (a) (3 points) Argue that, for n ≥ 1, N(A, ε_{2n}) ≈ 10^{∑_{k=0}^{n-1} 2^{2k+2} - 2^{2k+1}}, N(A, ε_{2n+1}) ≈ 10^{1 + ∑_{k=0}^{n-1} 2^{2k+2} - 2^{2k+1}}. You do not need to give a completely formal argument here, but you should provide some justification for why (3) is true. Hint: try to get an intuitive understanding of which points in [0, 1] belong to A. Do this at different levels by decomposing [0, 1] into the collections I_n, n = 0, 1, 2, ..., of 10^n equally spaced intervals, [0, 1/10^n], [1/10^n, 2/10^n], ..., [(10^n - 1)/10^n, 1]. (b) (3 points) Taking (3) as equalities, show (2).
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