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Mar 15, 2010 at 9:14 history edited Georges Elencwajg CC BY-SA 2.5
Corrected a false statement in a new section.
Mar 15, 2010 at 2:25 comment added Douglas Zare I think you are remembering something like an infinite binary tree doubled over a Cantor set.
Mar 15, 2010 at 1:28 comment added Ryan Budney Another argument that gives Petya's conclusion would be to triangulate your Cantor set complement (which can be done with countably-many triangles). So $\pi_1$ is countably-presented.
Mar 15, 2010 at 0:57 comment added Petya It seems to me that each loop could be approximated (in the same homotopy class) by a polygonal loop with finite number of rational vertexes. Hence, $\pi_1$ is countable.
Mar 15, 2010 at 0:45 history answered Georges Elencwajg CC BY-SA 2.5