| RE: .2
That's exactly what I was thinking of saying. However, I chickened
out, due to the realization I couldn't personally back it up
mathematically. Another thing that made me balk was: even if it is
connected, what's to stop it from having an "island" not belonging to
the set? From the images that have been generated, it is easy to say
that no such "island" exists inside of the set, but how can we REALLY
know, unless it can be proven mathematically? I think it might be
possible to prove it, but don't ask me--I don't have a doctorate in
math.
I wonder if that is the point of the question in .0? Sounds like a
great challenge!!
Andy
|
| Hubbard and Douady proved that the boundary of the Mandelbrot set
is a deformed circle which never "crosses" itself. While there
appear to be disconnected islands, these are attached to the main
set by the boundary curve. The curve becomes tangent to itself
at points, but never meets itself transversally.
This proof involves a potential theoretic limiting argument - one
neat result is that the contours one sees in pictures are equipotentials
that would result if the set were a charged conductor.
The proof is in the paper
A. Douady, J. Hubbard, "Iteration des polynomes quadratiques
complexes", Comptes Rendus Paris Acad. Sci. 294, (1982) pp 123-126.
They also explained some other neat stuff about the order of branching
along the boundaries of the various parts of the set.
They acknowledged that computer pictures were "decisive" in
formulating their proofs.
- Jim
|