6 Feb
2016
6 Feb
'16
11:17 a.m.
With the Mandelbrot set, the fractal is the boundary. For every point on the boundary, there's a neighborhood containing points outside of the set, so it seems to me that if the fractal is dense, its complement must be dense, too. On Sat, Feb 6, 2016 at 11:03 AM, David Wilson <davidwwilson@comcast.net> wrote:
Is there a fractal set M a la the Mandelbrot set where both M and its complement are dense on the plane?
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