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Endpoints of Cantor Set

by Dragonfall
Tags: cantor, endpoints
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Dragonfall
#1
Sep21-06, 10:08 AM
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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.
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StatusX
#2
Sep21-06, 10:51 AM
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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
#3
Sep21-06, 04:21 PM
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I have to show that these numbers are dense in the Cantor set. I found another way without using the closed form.

AKG
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Sep22-06, 12:38 AM
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Endpoints of Cantor Set

The closed form is k/3n for all n, and for all 0 < k < 3n.
HallsofIvy
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Sep22-06, 04:54 AM
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Quote Quote by AKG
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
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Sep22-06, 09:07 AM
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Sorry, just wasn't thinking.
Dragonfall
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Sep22-06, 07:52 PM
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Quote Quote by HallsofIvy
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.


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