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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?

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    $\begingroup$ The usual Cantor set has zero measure, but the set you described has positive measure. See this link: en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set $\endgroup$ – Chee Han May 26 '15 at 7:25
  • $\begingroup$ Abbott uses term "length". It's the same as measure, right? $\endgroup$ – zark novik May 26 '15 at 7:29
  • $\begingroup$ I would avoid using that term but in one dimension yes I think you're right. $\endgroup$ – Chee Han May 26 '15 at 7:30
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    $\begingroup$ If you remove a constant fraction from the middle of each remaining interval than you get much the same effect, but a nice feature of removing a third is the binary/ternary explanation. $\endgroup$ – Henry May 26 '15 at 7:31
  • $\begingroup$ OK, I replaced measure with length. $\endgroup$ – zark novik May 26 '15 at 7:31
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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.

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  • $\begingroup$ Why wouldn't this work with 2? Could I not remove the middle 1/2, leaving a quarter on each side? Or is it because I can't break it into 2 equal connected components? $\endgroup$ – Ben Kushigian Dec 21 '15 at 2:27
  • $\begingroup$ @BenKushigian You can; I was thinking of the construction in which you take $[0, 1/2]$ and $[1/2, 1]$ (in which case, you just get the unit interval again). $\endgroup$ – user296602 Dec 21 '15 at 2:28
  • $\begingroup$ Ahh, makes perfect sense. That's a pretty boring Cantor set. $\endgroup$ – Ben Kushigian Dec 21 '15 at 2:28
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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).

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  • $\begingroup$ Work in base $8$, and you get the numbers that have no $3$'s or $4$'s. In general, if you remove the middle $k$th, you can work in base $2k$. $\endgroup$ – user296602 Dec 21 '15 at 2:25
  • $\begingroup$ Yes, exactly, you get a more complicated characterization depending on what you remove - only slightly more complicated in this case, but not as nice as the base-3 one in any case. $\endgroup$ – Carl Mummert Dec 21 '15 at 2:26

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