Why Cantor set removes one third?

Question

I found the derivation of Cantor-like set in Understanding Analysis by Abbott. There he removes one fourth, and most properties (length, cardinality, compactness, uncountableness) are preserved (except dimension).

That's why I wonder: why we remove one third to build Cantor set? Is it because Cantor himself did this? Or is there some property that would be lost if we use some other fraction? Well, Abbott showed that dimension would be different, but I don't think we care about that number much, do we?

Answer 1

You're correct that removing $1/3$ isn't terribly important; rather, one reason to use it is that $3$ is the smallest number after $2$, and the construction of taking the first stage to be $[0, 1/3] \cup [1 - 1/3, 1]$ isn't interesting with $2$ replacing $3$. The direct connection with base $3$ is nice for proving things such as uncountability, but is not an essential feature.

There is a good reason to study Cantor sets with different ratios, however - they're frequently helpful in constructing easy to work with examples of various dimensions. For example, if we take a different construction with intervals $[0, \lambda], [1 - \lambda, 1]$ replacing $[0, 1/3]$, $[2/3, 1]$, then we get a set with Hausdorff dimension

$$\frac{\log 2}{\log 1 / \lambda}$$

As an example application, this gives way to construct a $1$ dimensional set as a countable union of strictly lower dimensional sets.

These are just relatively simple examples within the larger scheme of iterated function systems, which frequently can be used to define fractals with easily found Hausdorff dimension.

Answer 2

On nice thing about removing the middle third is the theorem:

The middle-thirds Cantor set in $[0,1]$ consists of exactly the points in $[0,1]$ that can be expressed using a base-3 expansion with no $1$s.

That, in turn, makes it easy to show that points such as $1/4$ are in the middle-thirds Cantor set despite not being the endpoint of any interval in the construction. It also makes the uncountability easy to show, by showing there is a bijection with $2^{\mathbb N}$.

If you removed the middle fourth of the interval (so the first step leaves you with $[0,3/8] \cup [5/8, 1]$, for example) then there is not such an elegant characterization of the points that are left in the final set (although there must certainly be some characterization along similar lines).