Mandelbrot set



The Mandelbrot set is the subset of the complex plane on those points cc \in \mathbb{C} on which the iteration of the operation “square and add cc” does not diverge.

This is a famous example of a fractal.


For cc \in \mathbb{C} a complex number, consider the function on the complex plane that squares its argument and adds cc to the result:

AAf cAA z AAAA z 2+c. \array{ \mathbb{C} &\overset{\phantom{AA}f_c\phantom{AA}}{\longrightarrow}& \mathbb{C} \\ z &\overset{\phantom{AA}\phantom{AA}}{\mapsto}& z^2 + c } \,.

For nn \in \mathbb{N} write

f c nfffnfactors f_c^n \coloneqq \underset{n\,\text{factors}}{\underbrace{ f \circ \cdots \circ f \circ f }}

for the nn-fold composition of f cf_c with itself (f c 0id f_c^0 \coloneqq id_{\mathbb{C}}).

Starting with value 00 \in \mathbb{C}, this defines a sequence of points in the complex plane (f c n(0)) i(f_c^n(0))_{i \in \mathbb{C}} for each complex number cc \in \mathbb{C}.

The Mandelbrot set Mdlbrt𝒞Mdlbrt \subset \mathcal{C} is the subset of the complex plane on those values of cc for which the corresponding sequence (f c n) n𝔹(f_c^n)_{n \in \mathbb{B}} is bounded

Mdlbrt{c:(f c n(0)) nis bounded}. Mdlbrt \coloneqq \left\{ c \in \mathbb{C} \;\colon\; (f_c^n(0))_{n \in \mathbb{N}}\, \text{is bounded} \right\} \;\subset\; \mathbb{C} \,.

Globally, at low resolution, the Mandelbrot set looks like this:


Topological properties


(Mandelbrot space)

Regard the Mandelbrot set as a topological space

(Mdlbrt,τ sub) (Mdlbrt, \tau_{sub})

via the subspace topology τ sub\tau_{sub} inherited from the Euclidean metric topology of 2\mathbb{C} \simeq \mathbb{R}^2.


The Mandelbrot space (Mdlbrot,τ sub)(Mdlbrot, \tau_{sub}) (def. 1) is a compact topological space.

We prove this below, after the following lemma:


(escape radius)

For |c|>2{\vert c\vert} \gt 2 then the sequence (f c n(0)) n(f_c^n(0))_{n \in \mathbb{N}} is not bounded, hence the sequence of absolute values (|f c n(0)|) n( {\vert f_c^n(0)\vert} )_{n \in \mathbb{N}} diverges for |c|>2{\vert c \vert} \gt 2.

In fact in this case the absolute values increase monotonically:

If |c|>2{\vert c\vert} \gt 2 then for all n>0n \gt 0 we have

|f c n+1(0)|>|f c n(0)|. {\vert f_c^{n+1}(0)\vert } \gt {\vert f_c^n(0)\vert} \,.

So assume |c|>2{\vert c \vert} \gt 2.

We prove the last statement by induction.

Observe that it is true for n=1n = 1, where we have

|f c 2(0)| =|c 2+c| |c 2||c| =|c|(|c|1)>1 >|c| =|f c(0)|. \begin{aligned} {\vert f_c^2(0) \vert} & = {\vert c^2 + c \vert } \\ & \geq {\vert c^2\vert } - {\vert c \vert } \\ & = {\vert c\vert}\underset{\gt 1}{\underbrace{({\vert c\vert}-1)}} \\ & \gt {\vert c\vert } \\ & = {\vert f_c(0) \vert } \end{aligned} \,.

Now assume that there is nn \in \mathbb{N} such that |f c n(0)|>|c|{\vert f_c^n(0) \vert} \gt {\vert c\vert}. Then it follows that

|f c n+1(0)||f c n(0)| =|(f c n(0)) 2+c|f c n(0) |f c n(0)| 2|c||f c n(0)| =|f c n(0)||c||f c n(0)| >|c|1 >1. \begin{aligned} \frac{ {\vert f_c^{n+1}(0)\vert} }{ \vert f_c^n(0) \vert } & = \frac{ {\vert (f_c^n(0))^2 + c \vert} }{ f_c^n(0) } \\ & \geq \frac{ {\vert f_c^n(0)\vert}^2 - {\vert c \vert} }{ {\vert f_c^n(0)\vert} } \\ & = {\vert f_c^n(0) \vert} - \frac{ {\vert c \vert} }{ {\vert f_c^n(0) \vert} } \\ & \gt \vert c \vert - 1 \\ & \gt 1 \end{aligned} \,.

Here the first inequality is due to the triangle inequality, the second is due to the induction assumption, and the last one is due to the initial assumption that |c|>2{\vert c \vert} \gt 2.


that the Mandelbrot space is compact (prop. 1)

By lemma 1 the Mandelbrot set is a bounded subset of 2d Euclidean space. Hence by the Heine-Borel theorem is is now sufficient to show that it is a closed subset, this will imply that it is compact.

The subset is closed if for every point c\Mdlbrtc \in \mathbb{C} \backslash Mdlbrt not contained in the Mandelbrot set there is an open neighbourhood of cc which still does not intersect the Mandelbrot set.

Now that cMdlbrt 2c \notin Mdlbrt \subset \mathbb{R}^2 means by definition that for every positive real number rr there is an nn \in \mathbb{N} such that |f c n(0)|>r\vert f_c^n(0)\vert \gt r.

Pick such an nn for r=2r = 2. Let then

ϵ|f c n(0)|2. \epsilon \coloneqq {\vert f_c^n(0) \vert} - 2 \,.

and consider the subset

U c,n{z|((|f c n(0)|ϵ)<(|z|<|f c n(0)|+ϵ))}. U_{c,n} \coloneqq \left\{ z \in \mathbb{C} \;\vert\; \left( \left({\vert f_c^n(0) \vert} - \epsilon \right) \lt \left( {\vert z\vert} \lt {\vert f_c^n(0) \vert} + \epsilon \right) \right) \right\} \,.

This is clearly an open neighbourhood of f c n(0)f_c^n(0). Hence by continuity of the function f c n:f_c^n \colon \mathbb{C} \to \mathbb{C}, the pre-image

(f c n1) 1(U c,n) (f_c^{n-1})^{-1}(U_{c,n})

is an open neighbourhood of cc \in \mathbb{C}, and by lemma 1 this does not intersect the Mandelbrot set.

Revised on May 16, 2017 13:58:08 by Urs Schreiber (