AcademicGround >> The Cantor Set
7/27/04 2:38 AM  
jgibson
Edited: 27Jul04 Member Since: 04/30/2001 Posts: 4189 
Let E(sub 0) be the interval [0,1]. Remove the segment (1/3,2/3) and let E(sub 1) be the union of the intervals [0,1/3] and [2/3,1].
Remove the middle thirds of these intervals, and let E(sub 2) be the union of these intervals:
[0,1/9], [2/9,3/9], [6/9,7/9]. [8/9,1].
Continuing on this way, we obtain sets E(sub n) such that:
E(sub 3) is a subset of E(sub 2) and E(sub 2) is a subset of E(sub 1).
The set (lim n > inf) E(sub n) = E is called the Cantor Set. The Cantor Set consists of the numbers that remain in [0,1] after all those numbers have been deleted.
Here is my question:
If you associate to each sequence a = {ALPHA sub n}, in which ALPHA(sub n) is 0 or 2, the real number
x(a) = (sum, n=1 to inf, (alpha(subn)/ 3^n)
How would I prove that the set of all x(a) is precisely the Cantor Set? I'm not looking for a direct proof, just some hints.

7/28/04 4:08 AM  
jgibson
Edited: 28Jul04 Member Since: 04/30/2001 Posts: 4192 
Once I figure it out, I will. I'm taking it to a math prof tomorrow. He says he'll give me some hints. 
7/29/04 2:18 AM  
Dogbert
Edited: 29Jul04 Member Since: 01/01/2001 Posts: 13466 
First a smartass comment: The set lim_{n>8} E_{n} isn´t really well defined. The Cantor set is defined as the intersection of all E_{n}. Your question is basically about proving two inclusions: 1. The first inclusion is that the Cantor set is a subset of the set of all x(a). If c is in [0 1/3] make alpha_{1}=0 otherwise alpha_{1}=2. Now look at the interval ([0,1/3] or [2/3,1]) it is in. If it is in the left part of the interval set alpha_{2}=0 otherwise alpha_{2}=2. I hope you get the idea. Now you simply have to show that the partial sums converge to c. 2. The second inclusion is that the set of all x(a) is a subset of the Cantor set. We should note two points. First the Cantor set is closed, as the intersection of closed sets. Secondly, any sequence x(a) (a sequence of partial sums) is a cauchy sequence and must therefore converge. What remains is to show is that this is a convergent sequence <i>in</i> the Cantor set. That is, we have to show that all partial sums lie within the Cantor set. One can do that by induction. Just note that each partial sum of any such sequence lies on the left boundary of the first part or the right boundary of the third part and that this implies that the same thing happens with the next higher partial sum. I hope you find my confused writing to be useful. 
7/29/04 4:00 AM  
jgibson
Edited: 29Jul04 Member Since: 04/30/2001 Posts: 4195 
It gives me something to chew on. Thanks. 
8/4/04 2:22 PM  
jgibson
Edited: 04Aug04 Member Since: 04/30/2001 Posts: 4207 
Anyone else.... 
8/5/04 7:15 PM  
Rastus
Edited: 05Aug04 Member Since: 01/01/2001 Posts: 17413 
The terminology is throwing me a little, Mr. Gibson. I think you will find it illuminating, however, to consider the Cantor Set as expressed in base 3, or a "ternary" representation. What you will find with the successive exclusions is that, in ternary notation, the numbers that get excluded are precisely those with a "1" in each decimal place. As such, any decimal whose digits consist of 0's or 2's become elements of the cantor set. The key to this observation, however, is by the conversion to base 3. 
8/5/04 7:25 PM  
jgibson
Edited: 05Aug04 Member Since: 04/30/2001 Posts: 4212 
Thanks for the input, rastus. I already noticed this behavior. Here are some random thoughts I came up with. My prof said I am very close to a proof, he said I am just not stating it as concisely as as I could. SOme of my random thoughts. I am thinking of doing a different proof, not necessarily by induction. For g is an element of E, show there is an x(a) such that x(a) = g. Look at this: x(a)  g <= (1/3)^n In essence, if we can get the remainder term (1/3)^n to equal 0, x and g will be equal to each other and therefore we have the proof. Every time to we go one level down (n+1), we have to make sure we pick the correct subinterval. If g is in an odd (left) subinterval, we will make alpha 0. (i.e. if g is an element of [0,1/3] make alpha 1 = 0) If g is in an even (right) subinterval, we will make alpha 2. (i.e. if g is an element of [2/3,1] make alpha 1 = 2) Because we picked alpha carefully, we are now in the subinterval En+1, hence x(a)  g <= (1/3)^n+1 or the distance bewteen x and g is decreasing. So we can say for every integer n the distance between x and g <= (1/3)^n. Therefore we can get the distance between them to become less and less with every nth iteration. In fact, if we let n approach infinity, the distance between them will in essence be zero. This means that x and g are the same real number! But this all begins with the idea that we are assuming that we can pick the first n alphas such that x and g are within (1/3)^n of each other. This means that x and g have to be in the same subinterval. 
8/6/04 3:57 AM  
Rastus
Edited: 06Aug04 Member Since: 01/01/2001 Posts: 17422 
The notation is getting me again.
Let me start over. I want to make sure the problem's written correctly (that I can understand) before I attempt to solve.
a = a1,a2,..., where each ai=0,2
x(a) = a1/3 + a2/9 + ...
We want to prove that the set of all x(a)'s is precisely the cantor set. Yet?
OK, I hate to do this to you, but let's write any decimal between zero and one in ternary notation. Therefore, we can have a decimal that consists entirely of 0's, 1'ss and 2's.
The first step of the Cantor process is to eliminate the open segment (0.1, 0.2). The effect of this is to create two numbers of the form {0.0a1a2a3...} or {0.2a1a2a3...}. Further division of the cantor sequence operates on successive decimal places. (Obviously, this is where the terniary notation is so critical).
We can, therefore, view a sequence of Cantor Sets:
C1 = [0,1], C2 = {0.0a1a2a3...} U {0.2a2a2a3...} C3 = {0.00a1a2a3...} U {0.02a1a2a3...} U {0.20a1a2a3...} U {0.22a1a2a3...} let's distinguish between two types of variables: ai = 0,1,2 ci = 0,2 Then C1 = [0,1] C2 = {0.c1a1a2...} C3 = {0.c1c2a1a2...} With this construction, it becomes immediately evident that the limit as n > inf of Cn is not only the Cantor Set but also the sequance of 0's and 2's you wanted to find to begin with. qed 
8/11/04 2:45 AM  
jgibson
Edited: 11Aug04 Member Since: 04/30/2001 Posts: 4225 
Alright, i turned the paper in and he said "Well done." This is the first time I've ever heard him give praise, so it must be a good sign. I won't post the whole paper on the net, but if you are interested in looking at it let me know. 
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