Timeline for Is the distance function from a point to the Mandelbrot set computable?
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Oct 13, 2010 at 4:47 | comment | added | Andrej Bauer | @Bjørn: "only because of (1)", well the general reason why you can extend computable maps from computably dense subsets is not the Lipshitz condition but computable pointwise continuity (actually, computable sequential continuity will do the job too), and that's a weaker condition. | |
| Oct 13, 2010 at 3:01 | vote | accept | CommunityBot | ||
| Oct 12, 2010 at 19:44 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |
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| Oct 12, 2010 at 19:43 | comment | added | Bjørn Kjos-Hanssen | @Andrej Bauer: Yes, (1) is a good restatement of the inequality above. (2) is also true but only because of (1). | |
| Oct 12, 2010 at 16:30 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |
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| Oct 12, 2010 at 13:23 | comment | added | Andrej Bauer | (1) The distance function is a Lipshitz map with Lipshitz constant 1, so it has a very easy modulus of continuity. (2) It does not matter whether we consider $\mathbb{Q}[i]$ or $\mathbb{C}$ as the domain of the distance function, since the former is computably dense in the latter. In other words, if we could compute the distance for points in $\mathbb{Q}[i]$ then a straightforward computable limit would give us the distance for points in $\mathbb{C}$. | |
| Oct 12, 2010 at 8:20 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |
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| Oct 12, 2010 at 8:14 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |
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| Oct 12, 2010 at 6:04 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |
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| Oct 12, 2010 at 5:46 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |
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| Oct 12, 2010 at 5:41 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 2.5 |