Why are these two definitions of the Mandelbrot set equivalent?

Solution 1:

Let $S_1$ be a circle of radius $100$ in the complex plane centered at the origin. Clearly everything outside of $S_1$ diverges to $\infty$ under iteration of $P_c$, so the filled Julia set lies entirely inside of $S_1$.

Now take the preimage $S_2$ of $S_1$ under $P_c$. This will be a smaller curve (close to a circle, with a radius of approximately $\sqrt{100} = 10$), which again must contain the entire filled Julia set. Iterating this process, we obtain a sequence $S_1,S_2,\ldots$ of closed curves, each of which contains the filled Julia set in its interior. In fact, since any point outside of the filled Julia set goes to $\infty$ under iteration of $P_c$, the intersection of the interiors of the curves $S_n$ is precisely the filled Julia set. (See this picture for an example. The curves separating the different shades of orange are the iterated preimages of some large circle.)

Unfortunately, this reasoning is not quite correct, because the preimage of a closed curve under $P_c$ is not always a single closed curve. Sometimes it is one closed curve, and sometimes it is two closed curves with disjoint interiors. If we repeatedly take the preimages of $S_1$, we may find that $S_n$ is a union of a very large number of closed curves! Note, however, that these curves still contain the filled Julia set entirely inside of them.

Now here is the key bit: the preimage of a curve will have one component if and only if $0$ lies in the the interior of the preimage. This is because $0$ is the critical point of the map $P_c$. Thus the preimage of a curve is either a single curve that surrounds $0$, or two curves, neither of which surrounds $0$. (In the latter case, the two curves are actually negatives of one another, i.e. symmetric across the origin.) Therefore, there are exactly two cases:

1. The point $0$ lies in the filled Julia set. In this case, each preimage $S_n$ must be a single curve, so the intersection of the interiors is connected.

2. The point $0$ lies outside the filled Julia set. In this case, some preimage $S_n$ does not encircle $0$, so it must have two components. Then each successive preimage will have twice as many curves as the last, and the resulting filled Julia set is homeomorphic to the Cantor set. (See this picture for an example. The curves separating the different shades of blue are the iterated preimages of some large circle. You ought to be able to see the first step at which the curve separates into two components.)