How can I express the endpoints of the Cantor set construction in closed form?

[Dragonfall]

How can I write down a closed form expression for the endpoints used in the construction of the Cantor set? i.e., 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, etc.

 

[StatusX]

You don't see any patterns there? Try writing down more terms. Or is it that you're trying to prove some expression is correct?

 

[Dragonfall]

I have to show that these numbers are dense in the Cantor set. I found another way without using the closed form.

 

[AKG]

The closed form is k/3ⁿ for all n, and for all 0 < k < 3ⁿ.

 

[HallsofIvy]

The closed form is k/3ⁿ for all n, and for all 0 < k < 3ⁿ.

No, it isn't. when n= 1, the first "cut", the enpoints are 1/3 and 2/3 so that the intervals left are [0, 1/3], [2/3, 1] but when you remove the middle third of those, the endpoint are 0, 1/9, 2/9, 1/3= 3/9, 2/3= 6/9, 7/9, 8/9, 1. In your notation, k/3ⁿ, with n= 2, k is not 4 or 5. It gets worse with higher values of n.

Dragonfall, think in terms of base 3. Writing a number between 0 and 1 in base 3, the first "digit" (trigit?) may be 0, 1, or 2. If 0, the number is between 0 and 1/3; if 1, between 1/3 and 2/3; if 2, between 2/3 and 1. When you remove the middle third you remove all numbers have a "1" as first digit. Now, all numbers between 0 and 1/3 must have .00, .01, or .02 as first two digits, all numbers between 2/3 and 1 must have .20, .21, or .22 as first two digits. When you remove the middle third of each you remove all numbers that have 1 as the second digit. Do you see what happens in the limit?

 

[AKG]

Sorry, just wasn't thinking.

 

[Dragonfall]

Dragonfall, think in terms of base 3. Writing a number between 0 and 1 in base 3, the first "digit" (trigit?) may be 0, 1, or 2. If 0, the number is between 0 and 1/3; if 1, between 1/3 and 2/3; if 2, between 2/3 and 1. When you remove the middle third you remove all numbers have a "1" as first digit. Now, all numbers between 0 and 1/3 must have .00, .01, or .02 as first two digits, all numbers between 2/3 and 1 must have .20, .21, or .22 as first two digits. When you remove the middle third of each you remove all numbers that have 1 as the second digit. Do you see what happens in the limit?

Yes, I was able to prove that the Cantor set contains those and only those numbers whose triadic expansion contains only 0 or 2. I did not use this fact for the denseness; however, but I did use it to prove that any number in [0,2] can be written as a sum of 2 'Cantor numbers'.

For denseness, I simply showed that since anything between two endpoints is either entirely in or out of the Cantor set, and if there exists a non-endpoint such that some closed ball of radius e about it contains no endpoint, then the Cantor set has at least length 2e, which is impossible.