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I am reading Abbott's Understanding Analysis. He uses the following reasoning to conclude that the Cantor set has dimension 0.631.

He asks the reader to consider the following informal definition of dimension:

"Without attempting a formal definition of dimension (of which there are several), we can nevertheless get a sense of how one might be defined by observing how the dimension affects the result of magnifying each particular set by a factor of 3."

A single point undergoes no change at all; the line segment triples in length; the triple-magnified square contains 9 copies of the original square; similarly, the triple-magnified cube contains 27 copies of the original cube. Hence, the shapes have the respective dimensions of $0$, $1$, $2$, and $3$.

"Notice that, in each case, to compute the 'size' of the new set, the dimension appears as the exponent of the magnification factor."

Using this definition, he goes on to say that the Cantor set has dimension $0.631$ by the following reasoning:

Consider the set $[0,1]$, and magnify it three times to produce $[0,3]$. When beginning the construction of the Cantor set on the set $[0,3]$, we come upon $C_1 = [0,1] \cup [2,3]$, which contains two copies of the original set (and therefore will produce two copies of the Cantor set, with the following step in the construction being equivalent to the first step of the construction of the Cantor set in both $[0,1]$ and $[2,3]$. Therefore, if the Cantor set has dimension $x$, then $2 = 3^x$, and $x = 0.63$.


Could someone clarify this reasoning? Something seems funky about magnifying the original set $[0,1]$, removing the middle third, and then noting that there are two copies of the Cantor set, that is, a copy in each of the respective sets $[0,1]$ and $[2,3]$, and therefore concluding that the dimension $x$ of the Cantor set satisfies the equation $2 = 3^x$. To try to demonstrate the dimension of the Cantor set, was he not suppose the magnify the Cantor set by $3$ (and not the set $[0,1]$)? Sorry if my question is imprecise. I do not have a specific question. I am just unconvinced by the reasoning (it seems floppy) and would like someone to clarify it.

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Abbott's exposition is very intentionally inexact–the goal is provide some intuition rather than an exact mathematical definition. The hope is, I think, that the reader will struggle with the ideas a little, see the basic outline, and be convinced by the result even if they don't fully understand it. I will note that Abbott's rather loose definition of dimension is, essentially, the similarity dimension. This can be made rigorous as follows:

Definition: Let $E\subseteq \mathbb{R}^n$ be bounded. For any positive number $c \in \mathbb{R}$ and any vector $\mathbf{b}$, let $$ cE := \{ cx : x \in E\} \qquad\text{and}\qquad E + \mathbf{b} = \{x+\mathbf{b} : x\in E\}. $$ That is $cE$ is a similar copy of $E$ scaled by a factor of $c$, and $E+\mathbf{b}$ is a congruent copy of $E$ translated by the vector $\mathbf{b}$. Suppose that there is some $c > 0$ and some finite collection $\{\mathbf{b}_j\}_{j=1}^{N} \subseteq \mathbb{R}^n$ such that $$ E = \bigcup_{j=1}^{N} cE + \mathbf{b}_j \qquad\text{and}\qquad (cE + \mathbf{b}_j) \cap (cE + \mathbf{b}_k) = \emptyset\quad \text{when $j\ne k$}.$$ We say that $E$ is a self-similar set, and the similarity dimension of $E$ is given by $$ \dim_s(E) = -\frac{\log(N)}{\log(c)}.$$

In other words, if $E$ is exactly made up of $N$ copies of itself (after scaling by a factor of $c$), then the dimension of $E$ is given by $-\log_c(N)$. If you compare this to Abbott's exposition, you should see the agreement. It should be noted that this is a simplified version of the similarity dimension (a more general definition does not restrict us to $\mathbb{R}^n$, and can handle scalings of different sizes).

To apply this to the Cantor set, I am going to use an argument different from Abbott's, starting with a different description of what the Cantor set it.

Definition: The Cantor set is the set of real numbers $C$ in the interval $[0,1]$ which possess the property that if $x \in C$, then $x$ has a ternary expansion consisting of only 0s and 2s.

For example, $\frac{2}{3} \in C$, since $$ \frac{2}{3} = 0.122\overline{2}_3 = 0.2_3$$ Note that $\frac{2}{3}$ does have a representation containing a 1, but this isn't a problem. Contrast this with $$ \frac{1}{2} = 0.111\overline{1}_3, $$ which does not possess a ternary expansion consisting of only 0s and 2s.

It can be shown that this definition of the Cantor set coincides exactly with the usual notion, which involves removing intervals. I'm not going to go through the details here, but note that the set numbers $x$ satisfying $0.1_3 < x < 0.2_3$ (i.e. the set of numbers with a 1 right after the "decimal point" in their ternary expansions) is the interval $(\frac{1}{3}, \frac{2}{3})$, which is the first interval removed in the usual construction of the Cantor set.

I claim that $$ C = \left( \frac{1}{3} C + 0 \right) \cup \left( \frac{1}{3}C + \frac{2}{3}\right). $$ Observe that if $x = 0.a_1a_2a_3\dotso$ is the ternary expansion of some number, then $$ \frac{1}{3}x = 0.0a_1a_2a_3\dotso \qquad\text{and}\qquad \frac{1}{3}x + \frac{2}{3} = 0.2a_1a_2a_3\dotso.$$ From this, it is not too hard to see that the claim is true. All of the numbers in the Cantor set with a zero in their first "decimal" place are in the set $\frac{1}{3}C$, while all of the points in the Cantor set with a two in their first "decimal" place are in the set $\frac{1}{3}C + \frac{2}{3}$. Moreover, these two sets are disjoint.

Hence we have expressed the Cantor set as the union of 2 copies of itself, after scaling by a factor of $\frac{1}{3}$. Therefore $$ \dim_s(C) = -\frac{\log(2)}{\log(\frac{1}{3})} = \frac{ \log(2) }{ \log(3) } \approx 0.631. $$

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He's not noting that there are two copies of the Cantor set inside $[0,1] \cup [2,3]$; he's noting that the version of the Cantor set obtained by beginning from $[0,3]$ instead of $[0,1]$ is made up of two copies of the original Cantor set. Put more precisely:

The Cantor set is constructed by beginning with $C_0 = [0,1]$ and obtaining $C_{n + 1}$ by removing the middle third of each interval in $C_n$. Then the Cantor set is the set $C = \bigcap_iC_i$.

The tripled Cantor set is constructed by beginning with $\hat{C}_0 = [0,3]$ and obtaining $\hat{C}_{n + 1}$ by removing the middle third of each interval in $\hat{C}_n$. Then the tripled Cantor set is the set $\hat{C} = \bigcap_i\hat{C}_i$.

Now, $\hat{C}$ is clearly three times as "wide" as $C$ (it stretches across three times as much space). Because the process is recursive on the intervals, $\hat{C}\cap[0,1]$ is identical to $C$, and $\hat{C} \cap [2,3]$ is a translation of $C$. Therefore, $\hat{C}$ consists of two copies of $C$.

Note that the part that goes "$\hat{C}$ is clearly three times as 'wide' as $C$" is very loose - what does "wide" mean here? That's not an accident - this is a purposefully loose "definition" of dimension, because the actual, precise definition is difficult to understand without a warm-up.

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