Show that Mandelbrot set is contained within the closed disc of r=2 Show that the Mandelbrot set is contained within the closed disc of radius 2 around
the origin.
How do I show this?
 A: Suppose the point in question $c$ is outside the disc, at $|c| = 2+\epsilon$, where $\epsilon$ is real and $\epsilon >0$.
Then we will show by induction that when $z_0 = 0$ and $z_{n+1} = z_n^2 + c$, 
with $|c|=2+\epsilon$, for $\epsilon > 0$, that for all integer $n > 0$
$$
|z_n| \geq 2+ (2^n-1) \epsilon 
$$
The basis is trivial:  $|z_1| = |c| = 2+(2^1-1)\epsilon$.
Now assume  $|z_n| \geq 2+(2^n-1)\epsilon$.  Then 
$$
|z_n|^2 = \left( 2+(2^n-1)\epsilon \right)^2 = 4 + 2^{n+2}\epsilon -4\epsilon+ \epsilon^2
>  4 + 2^{n+1}\epsilon + \epsilon
$$
because for $n\geq 1$, $2^{2n+2} - 4 >2^{2n+1} + 1$.
And now using the triangle equality in the form $|a+b| \geq |a|-|b|$ we have
$$
|z_{n+1}| =  \left| z_n^2 + c \right| \geq |z_n|^2 - |c| \geq 4 + 2^{n+1}\epsilon + \epsilon  - 2 - \epsilon = 2 + (2^{n+1}-1) \epsilon 
$$
So $|z_{n+1}| \geq 2 + (2^{n+1}-1) \epsilon$ and induction is established.
Finally, take $|c|= 2+ \epsilon$, and consider any $L$ however large:
Choosing $n > \log_2( L/\epsilon + 1)$ we have  $|z_{n+1}| > L$.
So if $c$ is outside the disc fof radius 2, the sequence is unbounded and $c$ is outside the Mandelbrot set.
A: A complex number $c$ is in the Mandelbrot set if repeated squaring and adding $c$ doesn't diverge to infinity. If you're further away than $2$ from the origin, then squaring increases the absolute value more than adding $c$ could hope to make up for, even if the two operations are directly opposed.
