Endpoints of Cantor Set

1. Sep 21, 2006

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.

2. Sep 21, 2006

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?

3. Sep 21, 2006

Dragonfall

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

4. Sep 22, 2006

AKG

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

5. Sep 22, 2006

HallsofIvy

Staff Emeritus
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?

6. Sep 22, 2006

AKG

Sorry, just wasn't thinking.

7. Sep 22, 2006

Dragonfall

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.

Last edited: Sep 22, 2006