Results 1 to 2 of 2

Thread: Cantor set question

  1. #1
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303

    Cantor set question

    Let $\displaystyle C$ represent the Cantor set.

    Q: Show that there exists $\displaystyle x_{1},y_{1}\in{C}$ satisfying $\displaystyle x_{1}+y_{1}=s\in[0,2]$.

    I found some work that goes like this...

    $\displaystyle C_{1}=[0,1/3]\cup[2/3,1]$. Then we have, $\displaystyle [0,1/3]+[0,1/3]=[0,1/3]\cup[0,2/3]+[2/3,1]=[2/3,4/3]\cup[2/3,1]
    +[2/3,1]$. So, $\displaystyle C_{1}+C_{1}=[0, 2/3]\cup[2/3, 4/3]\cup[4/3, 2]=[0, 2]$. Thus, for all $\displaystyle s\in[0, 2]$, we can find an $\displaystyle x_{1} , y_{1}$ $\displaystyle C_{1}$ satisfying $\displaystyle x_{1} + y_{1} = s$

    I am not sure that I follow exactly what is being done. Can someone help me to understand the work above. Do I think of these intervals as only lengths?

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Danneedshelp View Post
    Let $\displaystyle C$ represent the Cantor set.

    Q: Show that there exists $\displaystyle x_{1},y_{1}\in{C}$ satisfying $\displaystyle x_{1}+y_{1}=s\in[0,2]$.

    I found some work that goes like this...

    $\displaystyle C_{1}=[0,1/3]\cup[2/3,1]$. Then we have, $\displaystyle \color{blue}[0,1/3]+[0,1/3]=[0,1/3]\cup[0,2/3]+[2/3,1]=[2/3,4/3]\cup[2/3,1]
    +[2/3,1]$. So, $\displaystyle C_{1}+C_{1}=[0, 2/3]\cup[2/3, 4/3]\cup[4/3, 2]=[0, 2]$. Thus, for all $\displaystyle s\in[0, 2]$, we can find an $\displaystyle x_{1} , y_{1}$ $\displaystyle C_{1}$ satisfying $\displaystyle x_{1} + y_{1} = s$

    I am not sure that I follow exactly what is being done. Can someone help me to understand the work above. Do I think of these intervals as only lengths?
    $\displaystyle C_1+C_1$ means the set of all numbers that are equal to an element of $\displaystyle C_1$ plus an element of $\displaystyle C_1$. But $\displaystyle C_1 = [0,1/3]\cup[2/3,1]$. By adding two numbers from the interval [0,1/3] you can get anything in the interval [0,2/3]. By adding a number from the interval [0,1/3] and a number from the interval [2/3,1] you can get anything in the interval [2/3,4/3]. By adding two numbers from the interval [2/3,1] you can get anything in the interval [4/3,2]. Putting those facts together, you see that by adding two numbers from the set $\displaystyle C_1$ you can get anything in the interval [0,2]. That is what is meant by the line $\displaystyle C_{1}+C_{1}=[0, 2/3]\cup[2/3, 4/3]\cup[4/3, 2]=[0, 2]$. (The previous line, which I have highlighted in blue, seems totally confused.) Therefore, for any $\displaystyle s\in[0,2]$, we can find $\displaystyle x_1,\ y_1 \in C_1$ with $\displaystyle x_1+y_1=s$.

    That is just the first stage of this proof. You need to go on to show that $\displaystyle C_n+C_n = [0,2]$ for each of the sets $\displaystyle C_n$ used in the construction of the Cantor set. So there exist $\displaystyle x_n,\ y_n \in C_n$ with $\displaystyle x_n+y_n=s$. Finally, you need to show that a subsequence of the sequence $\displaystyle (x_n)$ converges to a point x in the Cantor set. The corresponding subsequence of the $\displaystyle y_n$s will then converge to a point y in the Cantor set, with x+y=s.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. a question about the cantor set
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: Jul 21st 2011, 03:51 PM
  2. Integral Question with Cantor Set
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 9th 2011, 12:46 PM
  3. Question on the cantor space
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Feb 11th 2010, 02:01 PM
  4. Cantor Set - quick question
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 9th 2009, 12:16 PM
  5. A Cantor set
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Oct 29th 2008, 02:12 PM

Search Tags


/mathhelpforum @mathhelpforum