How to prove Mandelbrot set is simply connected?

Solution 1:

What do you mean by "simply-connected" in the case of a compact subset $K$ of the plane that may or may not be locally connected?

Usually the term used is "full", which means that $K$ does not disconnect the complex plane (or, equivalently, its complement has no bounded component). This is probably the strongest interpretation for your question.

That $\mathcal{M}$ is full follows immediately from the definition of the Mandelbrot set as $$\mathcal{M} := \{c\in\mathbb{C}: |f_c^n(c)|\leq 2\text{ for all $n$} \},$$ using the maximum modulus principle. (Here $f_c(z) = z^2+c$ and $f_c^n$ denotes the $n$-th iterate of $f$.)

The usual proof that the Mandelbrot set is connected constructs - somewhat miraculously, if you are not used to these things - the Riemann map (conformal isomorphism) between the complement of the Mandelbrot set and the complement of the closed unit disk.

The proof is not so hard if you know complex analysis. However, there is also a topological proof, as pointed out by Jeremy Kahn:

http://www.math.brown.edu/~kahn/mconn.pdf