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This is a weakened version of Is the measure induced by the Mandelbrot set computable on rational rectangles? ;

Given a (computable, or rational) rectangle in the complex plane, is it computable whether:

  1. The rectangle is contained within the Mandelbrot set?
  2. The rectangle is disjoint from the Mandelbrot set?
  3. other

The above-referenced question asked whether the measure of the intersection of a (computable) rectangle and the Mandelbrot set is computable.

Given a (computable) complex number, it is probably computable whether that number is in the Mandelbrot set, but I'm not even sure of that. If that is insoluble, it does not necessarily make my question (for an open rectangle) insoluble.

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  • $\begingroup$ I don not know the answer to your question but I would like to point out that the phrase "whether that number is in the Mandlebrot set" may no be precise enough, since the complement of the set is computable, while the set itself may not be. $\endgroup$ – Daron Jun 10 '14 at 22:13
  • $\begingroup$ Membership in Mandelbrot set is not computable. This is in the book by Blum, Cucker, Shub and Smale, "Complexity and real computations". $\endgroup$ – Moishe Kohan Jun 10 '14 at 22:42
  • $\begingroup$ @Daron: The complement of a decidable set is decidable. The complement of a set that's merely recursively enumerable may not be, though. $\endgroup$ – Henning Makholm Jun 10 '14 at 23:57

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