The Mandelbrot Set Membership

Solution 1:

For Q1: Consider first $|c| > 2$. Then we have that by using the triangle inequality $$ |c^2| = |c^2 + c - c| \leq |c^2 + c| + |c| $$ or $$ |c^2 + c| \geq |c|(|c| - 1) > |c|$$ So we have that $|z_2| > |c|$.

Next, claim: if $|z_k| > \max(|c|,2)$ for some $k$, then $|z_k|\nearrow \infty$. To prove this claim consider $$ |z_k|^2 = |z_{k+1} - c| \leq |z_{k+1}| + |c| $$ so we have that $$ |z_k|^2 - 2|c| \leq |z_{k+1}| - |c| $$ Now using the assumption on $|z_k|$, we have that $$ 2(|z_k| - c) \leq |z_{k+1}| - |c|$$ so the sequence of (positive) numbers $|z_n| - |c|$ grows at least geometrically. Hence it will diverge to infinity.

Now, if $|c| > 2$, we have that $|z_2| > \max(|c|,2)$ and so by the above claim diverges. If $|c| \leq 2$, however, the hypothesis of the above claim is that $|z_k| > 2$, which is precisely what we want.

For Q2: as stated, you can't. Since for any $m \geq 2$, we have the same property that if $|z_k|> m$ then $|z_n|$ goes to infinity. What you want to prove instead is that 2 is the smallest number with this property. To do so let $m <2$ be some number. And let $-c= z_2 = 2$. We have that $|c| = |z_2| = 2 > m$ by definition. But then

$$ z_k^2 + c = 4 - 2 = 2 = z_{k+1}$$

so that $z_2 = z_3 = \ldots = z_k = z_{k+1} = z_{k+2} = \ldots = 2$ so the sequence does not diverge. Hence if $m < 2$ we can find a $z_k$ for which the property does not hold. And this counterexample shows that $2$ is the smallest number for which the property can hold.