# Why is the bailout value of the Mandelbrot set 2?

For the past few days I've been studying the Mandelbrot set, and many say that if the iterations of a point stay within a magnitude of 2, the point converges. A very natural question of "why is the bailout value 2?" came to me. Is there a proof that the value will diverge if the iterations do not stay within the bailout value of 2?

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It is not correct that "the point converges". Merely that it remains bounded. Which is what you need when defining the Mandelbrot set. –  GEdgar Jun 19 '13 at 12:25

First assume that $|c| \le 2$. Let $f(z) = z^2 + c$ and assume that $|z_n| = 2+a$, where $a > 0$. Then $$|z_{n+1}| =|f(z_n)| = |z_n^2 + c| \ge |z_n|^2 - |c| \ge |z_n|^2 - 2 = (2+a)^2-2 = 2+2a+a^2 > 2+2a.$$ By induction $|z_{n+k}| > 2+ak \to \infty$ as $k\to\infty$, so $z_n$ escapes to infinity.
If $|c| > 2$, a similar analysis shows that $z_n \to \infty$ so we don't even have to consider such values of $c$.
You should be careful with your final statement - it is true that $z_n\to\infty$ for the orbit of the critical point, but not for all orbits. (You did not say precisely what your $z_n$ is.) –  mathstribble Jun 21 '13 at 12:50
Hint: Consider the map $x\mapsto x^2-|c|$, with $|c|>2$. Show that the orbit of zero tends to infinity under this map, and conclude that the orbit of zero under $z^2+c$ does also. –  mathstribble Jun 26 '13 at 15:26