Are the Mandelbrot set's bulb's countably infinite?

My daughter asked me this question, after I pointed out that some Julia sets are a Cantor dust.

For a point not in the Mandelbrot set, the corresponding Julia set is a Cantor dust with an uncountably infinite number of disconnected points. There are also an infinite number of baby Mandelbrot's, each with an infinite number of bulbs.
Can you prove that the set of bulb's is countably infinite?

I think the answer is that every bulb can be associated with an external ray, and it seems that the known landing rays are all associated with rational numbers; the set of rational numbers is countably infinite. Perhaps if a landing ray is a real number, it would be associated with an infinitely small baby Mandelbrot, which is no longer a baby Mandelbrot because it has zero area? But another possible answer may also be because bulbs have area>0, and perhaps you can't have an uncountably infinite number of things with positive area.

  • $\begingroup$ A bit offtopic , But I think you might be intrested in a question I asked myself about fractals : math.stackexchange.com/questions/204885/zoom-out-fractals Perhaps you have seen it before. $\endgroup$ – mick Jan 29 '13 at 22:34
  • $\begingroup$ Thanks for the comment mick! I guess I'll have to answer your question about zoom out fractals; of course the answer involves tetration :). Kneser's Chi-star function, used in proof of the existence of a tetration solution is exactly such a zoom out fractal! But the chi-star is soooo complicated that it will take me a couple of days to put together a nice post with pictures as an answer to your question. $\endgroup$ – Sheldon L Jan 30 '13 at 12:33
  • $\begingroup$ Im curious about this chi-star thing ! $\endgroup$ – mick Dec 22 '14 at 22:39

I don't know if you have a precise definition of "bulb", but it's reasonable to expect that any bulb ought to contain a sufficiently small ball. Any ball contains a point with rational coordinates, because the latter are dense. Assuming bulbs are disjoint, this lets us define an injection from bulbs to rational points, which are countable, therefore there cannot be uncountably many bulbs.

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  • $\begingroup$ thanks! that is exactly the argument I was looking for. $\endgroup$ – Sheldon L Nov 25 '12 at 13:38

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