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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!

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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.


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.

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.

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  • $\begingroup$ Okay. Sorry I did not clarify about the Julia Set. The Julia Set I am studying is exactly what you described, which is the boundary of the unit disk. Now to your questions, if I have an open set, which I need to find two points which the orbit of one point is bounded and the orbit of the other is unbounded respectively, then the two points which are in $Q_{c} (U)$ is when $|Q_{c}(z')| < 1$ and when $|Q_{c}(z')| > 1$. To be honest, I am not sure what I think about the second question. Well since $Q_{c} (z_{0}) \in J_{c}$, it means that the magnitude of $Q_{c} (z_{0})$ equals 1. $\endgroup$ – user211962 Aug 4 '15 at 2:31
  • $\begingroup$ @Overachiever I completed the argument more explicitly, rather than just providing hints. To address your specific questions, it seems that you have one point of fundamental confusion. Your original question does not concern the boundary of the unit disk. That's just a specific case, namely when $c=0$; most Julia sets are much more complicated than this. So there's no reason to suppose that $|Q_c(z_0)|=1$ when $Q_c(z_0)\in J_c$. Rather, you should draw conclusions on points in a neighborhood of $z_0$ from assumptions involving corresponding points in a neighborhood of $Q_c(z_0)$. $\endgroup$ – Mark McClure Aug 4 '15 at 11:07

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