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.
Best Answer
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.