In order to define the Mandelbrot set $\mathbb M$, one looks at the sequences $$ s = f(0),(f\circ f)(0), (f\circ f \circ f)(0),\ldots, $$ where $f:\mathbb C \to \mathbb C,~z\mapsto z^2+c$ for different $c\in\mathbb C$. Now if $c$ is such that the resulting sequence $s$ is bounded, then $c\in\mathbb M$, otherwise $c\notin \mathbb M$. Does anyone know if there are characterizations or (even better) visualizations of the following related sets? I'd be grateful for a reference.

$$\begin{aligned} \mathbb M_\ell & :=\{c\in \mathbb C, s\text{ is a periodic sequence with period $\ell$.}\}\\ \mathbb M_p & := \bigcup_{\ell \in \mathbb N}\mathbb M_{\ell}\\ \mathbb M_{np} & := \mathbb M\setminus \mathbb M_p\\ \mathbb M_{\text{con}} & := \{c \in \mathbb C, s\text{ converges.}\} \end{aligned}$$


1 Answer 1

  • The sets $M_l$ are finite (they have $2^l$ elements). They are "in the center" of the connected components of the interior of the Mandelbrot set, and are not on its boundary.

  • The set $M_p$ is therefore countable, consisting of points all in the interior of $M$; but its closure contains the boundary of $M$.

  • $M_{np}$ is too big to have a simple description: after all, you only removed countably many points from $M$.

  • Presumably, $M_{con}$ is exactly the interior of the main cardioid (plus the parameter $c=\frac{1}{4}$, together with a countable collection of points corresponding to parameters for which the sequence $s$ is eventually constant. Those countable parameters are located at certain tips on the boundary of the Mandelbrot set.

  • $\begingroup$ Thanks a lot for your answer. Could you please explain how you find the cardinality of $M_l$ and what the reason for your guess for $M_{con}$ is? $\endgroup$
    – frog
    Nov 29, 2019 at 18:39
  • $\begingroup$ mrob.com/pub/muency/enumerationoffeatures.html $f_c^p(0) = 0$ is a degree $2^{p-1}$ polynomial in $c$ so it has $2^{p -1}$ roots. Some of these roots may have a lower period too, maybe you want to count them in only one $M_l$ (with $l$ as low as possible). I'm not sure how to take into account potential multiplicity of the roots. $\endgroup$
    – Claude
    Dec 3, 2019 at 11:40

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