Prove that the Mandelbrot Set Is A Closed Set

Question

The Problem: Suppose we define the Mandelbrot Set as the following

For $c \in \mathbb{C}$ , $\mathbb{M}$ = $({c:|c| \leq 2}) \cap ({c: |c^2 + c| \leq 2}) \cap ({c: |(c^2+c)^2 + c| \leq 2}) \cap ...$

Carefully argue that each set in this intersection of sets is a closed subset of the Complex Plane. By this, show that the Mandelbrot Set is closed.

The attempt - So for each $i \in \mathbb{N} $, we can write this set as the following:

$\mathbb{M}= (c \in \mathbb{C} : |Q_{c}^{n} (0)| \leq 2)$, for $i \geq 1$, which $Q_{c} (z) = z^2+c$. Now if we are going to show the set is closed, we can show that the complement of each set is open, which it is for each $i$, $\mathbb{M}^{c}= (c \in \mathbb{C} : |Q_{c}^{n} (0)| > 2)$, for $i \geq 1$.

To show each set is open, we can find an ε-neighborhood of any point, $z_{0} \in \mathbb{M^c}$ for which $N(z_{0}, ε) \subseteq \mathbb{M} ^{c}$.

I can define $ε = max ({2, |z_{0}|})$ and that is all I got so far.

I am not sure if I am on the right track. However, there was a hint to this problem which I do not know what it means (Hint: If $F : \mathbb{C} \mapsto \mathbb{R}$, is a continuous function, then for every $b \in \mathbb{R}$, the set $(c \in \mathbb{C} : F(c) \leq b)$. Is this the neighborhood I was supposed to be defining?

Thank you very much for your help!

Answer 1

Each of your sets has the form $$ \{ c \in \mathbb C \mid \lvert p(c) \rvert \le 2 \} $$ where $p$ is some polynomial.

Now you ought to know that every polynomial is continuous, and that $z\mapsto \lvert z \rvert$ is continuous, and that the composition of two continuous functions is continuous.

Thus, apply the hint with the function $F(x) = \lvert p(x)\rvert$ and $b=2$ to see that each of your sets are closed. Since any (possibly infinite) intersection of closed sets is closed, the conclusion follows.

(And entirely without having to do manual $\varepsilon$-$\delta$ proofs yourself. It's all in connecting building blocks you should already have at this point).