# The Mandelbrot Set

1. Nov 16, 2008

### Troels

Don't know if this is the right subforum, but here goes:

The Mandelbrot set is formally defined as the values of a complex constant c for which the Julia Sets of the iterative function:

$$f(z)=z^2+c$$

are connected (i.e. consists of a single figure)

Recall that the Julia sets are the launch points $$z_0$$ of the iteration that gives rise to bound orbits in the complex plan for some given constant c

However, according to some less formal sources (e.g. wikipedia) the Mandelbrot set can also be constructed simply by analyzing the same iterative function, only this time keep the launch point constant (0) and vary the constant - or parameter - c instead. Those values that give bound orbits are the Mandelbrot Set

This seems like a peculiar accident, but that cannot be so. So my question is: Why does a bound orbit for the iterative function

$$f(z)=z^2+c$$

launched from $$z_0=0$$ for a given value of c, imply that the corrosponding Julia Set is connected?

Last edited: Nov 16, 2008
2. Nov 17, 2008

### Troels

Okay, I've given this a bit more thought and the following occured to me:

If the orbit launches from z=0 is bounded then z=0 is of course in the julia set.

And all the Connected julia sets I have seen contain the point z=0, but I haven't seen a dis-connected Julia Set that contains the point z=0. Of course, that is an argument from ignorance on my part.

So I guess that the question now is: Does there exist a proof that states that a Julia Set contains the point z=0 if and only if it is connected? If so, then the issue is solved.