Show that Mandelbrot set is contained within the closed disc of r=2 [closed]

Solution 1:

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.