Question

What is a (relatively) simple proof that a complex number of modulus greater than two is not an element of the Mandelbrot set \(\mathcal M\)? In symbols, how could one prove the following, where I denote \(P_z:x \in \mathbb C\mapsto x^2+z\).
\[\left( c \in \mathbb C \wedge |c|>2 \right) \Rightarrow c \not\in \mathcal M := \{z \in \mathbb C :\ \forall n \in \mathbb N,\ \exists S \in \mathbb R,\ |P_c^n(0)|\le S \}\]If you aren't familiar, the Mandelbrot set \(\mathcal M \subset \mathbb C\) is the set of complex numbers \(z\) such that the infinite composition of \(P_z\) on 0 is bounded. I see this fact referenced a lot in texts regarding the Mandelbrot set, though I have never seen it proved in a way I could understand. Any links to proofs, or ones of your own?

Answer 1

Don't hesitate to ask if I can clarify anything for you... I know the above looks intimidating at first sight...

Answer 2

Suppose that
\[
|c| > 2\\

|v| \ge |c| >2\\

f(v) = v^2 + c\\
\]
Then
\[
|f(v)| > (|c| -1) |v|
\]

Proof:
\[
|f(v)| =|v^2 + c| \ge |v^2| -|c|=\\
|v| |v| - |c|\ge |c| |v| - |c| \ge |c| |v| - |v|= |v|(|c|-1)

\]
We are done.\\
Notice that
\[
|f(v)|\ge |c| \\

|f(f(v))| \ge (|c|-1)|f(v)| \ge (|c|-1)(|c|-1)|v| =(|c|-1)^2 |v|\\

|f^k (v)| \ge (|c|-1)^k |v|\\
\]
Since |c| - 1 > 1, then
\[
|f^k (v)| \to \infty,\, \text { if } k \to \infty
\]

Answer 3

Thank you so much for this! This is extremely appreciated!