1
$\begingroup$

Each point in Cantor Set can be built according to the well-known "delete the middle third" rule but also as a real number in the unit interval:

In arithmetical terms, the Cantor set consists of all real numbers of the unit interval $[0,1]$ that do not require the digit $1$ in order to be expressed as a ternary (base 3) fraction.

(Source: Cantor Set on Wikipedia)

It is also known that when writing numbers from an infinite sequence of digits in a given base, infinitely repeating the highest digit leads to an alternate way of writing a number having a finite number of digits in the same base. Thus, I expect $0.0222\dots_3=0.1_3$.

Now, I am unsure whether $0.0222\dots_3$ actually belongs to Cantor set for obvious reasons: thinking at the number as $0.0222\dots_3$ makes me think it belongs to the set, as it describes a perfectly valid path on the picture below (path being: left, then always right); on the other hand, the number $0.1$ obviously does not belong to the set. Then, does $0.0222\dots_3$ belong to the set?

https://en.wikipedia.org/wiki/Cantor_set

$\endgroup$
3
  • $\begingroup$ Things break when you take them to infinity. Like $\bigcup_{n=2}^{\infty} [\frac1n,1-\frac1n]=(0,1)$. $\endgroup$
    – player3236
    Sep 9, 2020 at 8:18
  • $\begingroup$ The Wiki page says on it that: "... So removing the line segment $(1/3,2/3)$ from the original interval $[0, 1]$ leaves behind the points $1/3$ and $2/3$. Subsequent steps do not remove these (or other) endpoints, since the intervals removed are always internal to the intervals remaining. ..." $\endgroup$
    – Peter Foreman
    Sep 9, 2020 at 8:27
  • $\begingroup$ @PeterForeman Thank you for pointing that. My question actually focused on the quoted statement (about ternary expansion). The answer below is very good since it explains about the word "require" in the sentence. $\endgroup$
    – Thomas Baruchel
    Sep 9, 2020 at 8:29

1 Answer 1

Reset to default
6
$\begingroup$

Yes, the number $0.022222..._3 = 0.1_3 = \frac{1}{3}$ lies in the Cantor set. In each step of the Cantor set you take out the open interval in the middle, so you get

Step $0$: $[0,1]$

Step $1$: $[0,\frac{1}{3}]\cup[\frac{2}{3},1]$

...

and after Step 1, you never change anything near $\frac{1}{3}$, so it remains inside for all steps and thus is in there at the end as well.

This is also consistent with the Wikipedia comment, the important detail is: that do not require the digit 1.

In arithmetical terms, the Cantor set consists of all real numbers of the unit interval [0,1] that do not require the digit 1 in order to be expressed as a ternary (base 3) fraction.

$0.1_3$ does not require the digit $1$, since $0.022222..._3 = 0.1_3$ and thus (according to this rule) is part of the cantor set.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .