# Bounded set

• Feb 13th 2010, 11:34 AM
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?
• Feb 13th 2010, 01:18 PM
Opalg
Quote:

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).
• Feb 13th 2010, 02:41 PM
Quote:

Originally Posted by Opalg
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?
• Feb 14th 2010, 01:35 AM
Opalg
Quote: