
#1
Sep2106, 10:08 AM

P: 994

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
Sep2106, 10:51 AM

HW Helper
P: 2,566

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
Sep2106, 04:21 PM

P: 994

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




#4
Sep2206, 12:38 AM

Sci Advisor
HW Helper
P: 2,589

Endpoints of Cantor Set
The closed form is k/3^{n} for all n, and for all 0 < k < 3^{n}.




#5
Sep2206, 04:54 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898

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
Sep2206, 09:07 AM

Sci Advisor
HW Helper
P: 2,589

Sorry, just wasn't thinking.




#7
Sep2206, 07:52 PM

P: 994

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 nonendpoint 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. 


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