Results 1 to 4 of 4

Thread: Bounded set

  1. #1
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722

    Bounded set

    Could someone explain the the conclusion of this proof that the Julia set is bounded to me?

    Theorem: The Julia set J of a function f_c(z) = z^2 + c is compact for all c \in \mathbb{C}
    Proof: First, we show that J is bounded.
    Choose r = \textrm{max}(|c|,3) and let |z| \geq r.
    Then we have,
    |z^2| = |f_c(z) - c| \leq |f_c(z)| + |c|
    And so we have,
    |f_c(z)| \geq |z^2| - |c| \geq 3|z| - |z| = 2|z|
    Hence, since |z| \geq r then
    f_c^n(z) \geq 2^n |z| \rightarrow \infty \textrm{ as } n \rightarrow \infty (*)

    And so J(f_c) \subset B(0,r). (**)

    I found this in a journal, I don't really understand how the last line (**) happens. Surley (*) shows the function diverges... How can we say it's bounded?
    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 Deadstar View Post
    Could someone explain the the conclusion of this proof that the Julia set is bounded to me?

    Theorem: The Julia set J of a function f_c(z) = z^2 + c is compact for all c \in \mathbb{C}
    Proof: First, we show that J is bounded.
    Choose r = \textrm{max}(|c|,3) and let |z| \geq r.
    Then we have,
    |z^2| = |f_c(z) - c| \leq |f_c(z)| + |c|
    And so we have,
    |f_c(z)| \geq |z^2| - |c| \geq 3|z| - |z| = 2|z|
    Hence, since |z| \geq r then
    f_c^n(z) \geq 2^n |z| \rightarrow \infty \textrm{ as } n \rightarrow \infty (*)

    And so J(f_c) \subset B(0,r). (**)

    I found this in a journal, I don't really understand how the last line (**) happens. Surely (*) shows the function diverges... How can we say it's bounded?
    What (*) shows is that if z is any point outside B(0,r) then the iterates of z under the map f_c go off to infinity, which means that z does not belong to the Julia set. Therefore the Julia set is contained in B(0,r).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by Opalg View Post
    What (*) shows is that if z is any point outside B(0,r) then the iterates of z under the map f_c go off to infinity, which means that z does not belong to the Julia set. Therefore the Julia set is contained in B(0,r).
    Thanks! I had a feeling that was it...

    Could the '3' in the r = max... part be replaced by 'a' where a>2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Deadstar View Post
    Could the '3' in the r = max... part be replaced by 'a' where a>2?
    Yes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Closure of a totaly bounded set is totally bounded
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Apr 8th 2011, 08:42 PM
  2. Replies: 2
    Last Post: Sep 14th 2010, 07:15 AM
  3. Replies: 3
    Last Post: Mar 17th 2010, 07:12 PM
  4. Replies: 4
    Last Post: Oct 12th 2009, 09:43 PM
  5. Replies: 0
    Last Post: Jul 1st 2008, 03:47 AM

Search Tags


/mathhelpforum @mathhelpforum