The Mandelbrot set is the subset of the complex plane on those points $c \in \mathbb{C}$ on which the iteration of the operation “square and add $c$” does not diverge.
This is a famous example of a fractal.
For $c \in \mathbb{C}$ a complex number, consider the function on the complex plane that squares its argument and adds $c$ to the result:
For $n \in \mathbb{N}$ write
for the $n$-fold composition of $f_c$ with itself ($f_c^0 \coloneqq id_{\mathbb{C}}$).
Starting with value $0 \in \mathbb{C}$, this defines a sequence of points in the complex plane $(f_c^n(0))_{i \in \mathbb{C}}$ for each complex number $c \in \mathbb{C}$.
The Mandelbrot set $Mdlbrt \subset \mathcal{C}$ is the subset of the complex plane on those values of $c$ for which the corresponding sequence $(f_c^n)_{n \in \mathbb{B}}$ is bounded
Globally, at low resolution, the Mandelbrot set looks like this:
(Mandelbrot space)
Regard the Mandelbrot set as a topological space
via the subspace topology $\tau_{sub}$ inherited from the Euclidean metric topology of $\mathbb{C} \simeq \mathbb{R}^2$.
The Mandelbrot space $(Mdlbrot, \tau_{sub})$ (def. 1) is a compact topological space.
We prove this below, after the following lemma:
(escape radius)
For ${\vert c\vert} \gt 2$ then the sequence $(f_c^n(0))_{n \in \mathbb{N}}$ is not bounded, hence the sequence of absolute values $( {\vert f_c^n(0)\vert} )_{n \in \mathbb{N}}$ diverges for ${\vert c \vert} \gt 2$.
In fact in this case the absolute values increase monotonically:
If ${\vert c\vert} \gt 2$ then for all $n \gt 0$ we have
So assume ${\vert c \vert} \gt 2$.
We prove the last statement by induction.
Observe that it is true for $n = 1$, where we have
Now assume that there is $n \in \mathbb{N}$ such that ${\vert f_c^n(0) \vert} \gt {\vert c\vert}$. Then it follows that
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 ${\vert c \vert} \gt 2$.
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 $c \in \mathbb{C} \backslash Mdlbrt$ not contained in the Mandelbrot set there is an open neighbourhood of $c$ which still does not intersect the Mandelbrot set.
Now that $c \notin Mdlbrt \subset \mathbb{R}^2$ means by definition that for every positive real number $r$ there is an $n \in \mathbb{N}$ such that $\vert f_c^n(0)\vert \gt r$.
Pick such an $n$ for $r = 2$. Let then
and consider the subset
This is clearly an open neighbourhood of $f_c^n(0)$. Hence by continuity of the function $f_c^n \colon \mathbb{C} \to \mathbb{C}$, the pre-image
is an open neighbourhood of $c \in \mathbb{C}$, and by lemma 1 this does not intersect the Mandelbrot set.