If you do this with the other critical orbit it looks sort of the same but the minbrot has vanished.

I have been quiet for a couple of days. (I think) I understand what is happening in Gerrit's observation. But I spent a whole day trying to prove it, only to discover my math skills are quite rusty. It has been 40 years since topology and complex functional analysis. I just wasted another hour this morning. So, I will present "conjecture". Please help fill in the details if you are so inclined.

The "official" definition of the Mandelbrot set is \( M = \{ c | J_c \) is connected \( \} \). That "is connected" part is difficult to work with, so we use the equivalent definition \( M_z = \{ c | z \in K_c \} \) where \( K_c \) is the filled in Julia set. \( M=M_0 \), and we have a natural generalization to alternate starting points. However \( K_c \) has hard coded assumptions that the other critical point is \( \infty \) and it is attractive. This generalization can be extended to polynomials, but it breaks down for rational functions.

The "connected Julia" definition may generalize nicely, but that one hurts my head. Let's agree that any generalization should include \( M_z = \{ c | z \in J_c \} \). (For polynomials, \( K_c \supseteq J_c \), and the "extra" points are usually uninteresting.)

For \( z^2 + c \). If\( |c| > 2 \), then then \( J_c \) is Fatou dust. If \( |z| > 2 \) and \( |c| < 2 \) then \( J_c \) is contained in the \( |z|<=2 \) disk. Now I want to swap planes, I keep tripping up on this part. Fix a starting point z, with \( |z| > 2 \), for the parameter plane calculation. Then pick a "pixel" value c. Does z converge for c? Only if z is in the 0-measure set \( J_c \). So unless you engineered the choice of z and c, the answer is no. The parameter plane image is empty, the brot disappears. Yes, this is a long way to say what everyone already knows. However, the \( |c| < 2 \) condition is often ignored in escape-checks. And this sets up a contrast with the rational function case. based on the nature of the Julia set.

Now suppose f is a rational function. Any of them discussed in the thread will work. Although degree numerator <= degree denominator may be required. For these rational functions we know there exists c where \( J_c = \bar{C} \). Now f(z,c) is continuous, so there is an open set around the c with this property. (Yes, this needs proof, but it makes intuitive sense and all images so far support the statement). So the parameter space image, will have an open set (measure > 0) around c with this property. The filled-in Julia set definition of M-set fails because \( \infty \) does not attract. Whatever we do to replace it should include\( \{ c | z \in J_c \} \), in particular this c is included. Any of the alternate, non-escape time, coloring methods in this thread seem to do a good job of exposing these points. You might be suspicious that the starting point z has not been mentioned yet. For which starting values of z is this c an interesting point? When is \( z \in J_c \)? Always because \( J_c = \bar{C} \). The starting point does not matter!\( \)

These images show distinct brot and antibrot components. I think it makes sense that the brot disappears without careful selection of the start value, while the antibrot seems unchanged with any start value.