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So I'm familiar with the simple methods of calculating the Hausdorff dimension of a fractal, but when I try to apply them for this case I get into trouble. What I mean with the simple method is using the scaling factor and the amount of repetitions to exactly calculate the Hausdorff dimension.

The fractal below is constructed by taking all horizontal line segments, dividing each into three equally long line segments, raise the one in the middle by it's own length and connecting it to its former neighbors with two vertical line segments.

enter image description here

If I decided during my construction steps not to add those vertical lines, it would be clear that the whole fractal would be 1-dimensional. Each vertical line segment is 1-dimensional as well. But if I were to assume that the whole fractal was 1-dimensional, then it would be of infinite length.

There are more cases of fractals I can think of where I run into the same questions. For example a fractal where each point has position (1/2)i on the line interval [0,1] for all i in ℕ. Would this be a 0-dimensional fractal of infinite size?

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    $\begingroup$ Yes, your set has Hausdorff dimension one. It is the union of a one dimensional self-similar set consisting of three parts scaled by the fact $1/3$ together with countably many line segments. Since Hausdorff dimension is stable under countable unions, the result has Hausdorff dimension 1. Note that even a bonded differentiable curve can have infinite length so there's no problem there. $\endgroup$ – Mark McClure Jun 7 '18 at 15:17
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After $n$ subdivisions, there are $3^n$ horizontal pieces, each of which is $3^{-n}$ in length. The $n^\text{th}$ division adds $2\cdot3^{-n}\cdot3^{n-1}=\frac23$ in vertical pieces to the length. So looking at the resolution of $3^{-n}$, the curve takes at most $\left(1+\frac23n\right)3^n$ disks of that size to cover.

Thus, the Hausdorff dimension is $$ \lim_{n\to\infty}\frac{\log\left(\left(1+\frac23n\right)3^n\right)}{\log\left(3^n\right)}=1 $$

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First, an important result:

Theorem: Let $\{X_n\}_{n\in\mathbb{N}}$ be a countable collection of (possibly empty) subsets of some metric space $\mathscr{X}$. Then $$ \dim_H\left( \bigcup_{n\in\mathbb{N}} X_n \right) = \sup_{n\in\mathbb{N}} \dim_H(X_n), $$ where $\dim_H(X)$ is the Hausdorff dimension of $X$.

This can be summarized (as per Mark McClure's comment) as "the Hausdorff dimension is stable under countable unions." I'll not prove the result here, but there is a proof given in answer to another question.

The set shown in the question consists (broadly speaking) of two parts: a collection of vertical line segments (each of which has dimension 1), and some "dust" at the ends of the segments. The intervals are easily dealt with: there is one vertical segment at each triadic rational number between 0 and 1 (i.e. if $x = \frac{k}{3^n}$ for some $n\in\mathbb{N}$ and $0 \le k \le 3^n$, then there is a vertical segment based at the point $(x,0)$—if we are a bit more careful, we could work out the length of each such segment and thereby determine the total length of all of the segments, but the question was about dimension, not rectifiability). The set of triadic rationals is countable, so the total collection of vertical segments is countable. Hence the union of all of the segments form a set of Hausdorff dimension 1.

The "dust" is a bit more difficult to contend with, but not impossible. Again following Mark McClure's lead, we can note that the dust is a self-similar set of Hausdorff dimension 1. To make the argument more precise in a way that I find comfortable (as I am a guy that likes to work with iterated function systems), define $$\varphi_1(x,y) := \frac{1}{3}(x,y), \quad \varphi_2(x,y) := \frac{1}{3}(x,y) + \left(\frac{2}{3},0\right) \quad\text{and}\quad \varphi_3(x,y) := \frac{1}{3}(x,y) + \left(\frac{1}{2},\frac{1}{2}\right). $$ The "dust" is the attractor or fixed point of this iterated function system (IFS). That is, call the dust $\mathscr{D}$ and define $$ \Phi(X) := \bigcup_{j=1}^{3} \varphi_j(X) $$ for any $X \subseteq \mathbb{R}^2$, then $\mathscr{D}$ has the property that $$ \Phi(\mathscr{D}) = \mathscr{D}. $$ Via some abstract nonsense (essentially, an application of the Banach fixed point theorem applied to $\Phi$ acting on the space of compact subsets of $\mathbb{R}^2$, see for example Hutchinson, 1981), we can be sure that this IFS has a compact attractor. Moreover, it is not too hard to check that the IFS satisfies the open set condition, from which it follows (via some more abstract nonsense, see the above cited paper) that $\dim_H(\mathscr{D}) = s$, where $s$ is the unique real solution the the Moran equation $$ 3\left(\frac{1}{3}\right)^s = 1. $$ It is not too hard to see that $s = 1$, and so $\dim_H(\mathscr{D}) = 1$.

Therefore, since the set in question is a countable union of sets which each have Hausdorff dimension 1, it follows from the stability of the Hausdorff dimension under countable unions that the Hausdorff dimension of this set is 1.

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