Question About Filled Julia and Julia Sets

Question

Question: Let $Q_{c}(z) = z^{2} +c $ which $ c \in \mathbb{C}$ and suppose that $z_{0} \in K _{c}$ for the filled Julia Set, $K_{c}$ of $Q_{c}$. Suppose further that $z_{1} = Q_{c}(z_{0})$ and it belongs to the Julia Set, $J_{c}$ of $Q_{c}$. Also, assume that $Q_{c} (K_{c}) = K_{c}$. Show that $z_{0} \in J_{c}$

My Attempt: Since $z_{0}$ is in the Filled Julia Set, Also, what I think is that $z_{1} = Q_{c} (z_{0})$ means is that $z_{1}$ is a attracting critical point. I guess the only way that $z_{0} \in J_{c}$ is only when $z_{0}$ is a repelling periodic point. Should I use Cauchy's Inequality? I am not sure what else to think about this problem.

Can you give me some hints on this problem please?

Thank you very much for your help!

Answer 1

First, it's not completely clear what you mean by the Julia set $J_c$; there are several equivalent formulations. Given your notation, I would not be surprised if you're studying one of Bob Devaney's books that uses the following defintions:

• The filled Julia set $K_c$ of $Q_c(z)=z^2+c$ is the set of all complex points whose orbit under iteration of $Q_c$ is bounded.

Then,

• The Julia set $J_c$ of $Q_c$ is the boundary of the filled Julia set, i.e. the closure of $K_c$ minus its interior.

If, for example, $c=0$, then $Q_0(z)=z^2$. It's not too hard to see that, if $|z|\leq1$, then the orbit of $z$ is bounded by $1$. (You can't square a number less than or equal to $1$ in absolute value and get a number whose absolute value is larger than $1$). On the other hand, if $|z|>1$, then $|z^2|>|z|$. If we keep squaring, we generate a sequence that diverges to $\infty$. So for this example, the filled Julia set is exactly the solid unit disk, i.e. $$K_c = \left\{z\in\mathbb C:|z|\leq1\right\}.$$ The Julia set is the boundary of the unit disk, namely the unit circle: $$J_c = \left\{z\in\mathbb C:|z|=1\right\}.$$

For large $c$ values (outside of the Mandelbrot set, in fact), $K_c$ has empty interior so that $J_c=K_c$. For values in the interior of the Mandelbrot set, though, it appear that the above example is typical.

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Now, you are essentially trying to show that $J_c$ is backward invariant, i.e. if $z_0\in J_c$ whenever $Q_c(z_0)\in J_c$. (Your problem, as stated, actually assumes more than is necessary.) An outline of an approach might look like so:

• Pick an open set $U$ containing $z_0$. You need to find a point in $U$ whose orbit is bounded and another point in $U$ whose orbit is unbounded. As $U$ is arbitrary, you could then conclude that $z_0$ is on the boundary of $K_c$.
• Consider the image of $U$ under $Q_c$, i.e. $Q_c(U)$. This is an open set by the open mapping theorem.
• Now, since $Q_c(U)$ is an open set intersecting the boundary of $K_c$, it contains a point $Q_c(z_1)$ whose orbit is bounded and another point $Q_c(z_2)$ whose orbit is unbounded, where $z_1,z_2\in U$.
• The, of course, we can can draw the same conclusions on $z_1$ and $z_2$, so we are done.

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Again, there are other characterizations of $J_c$. Another common one is: The Julia set $J_c$ is the closure of the set of repelling, periodic points of $Q_c$. It's worth mentioning that a similar argument works for this characterization as well. If $Q_c(z_0)\in J_c$, then every neighborhood of $Q_c(z_0)$ contains a repelling periodic point. Again using the open mapping theorem, every neighborhood of $z_0$ will contain a repelling periodic point.