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Sep 11, 2010 at 17:41 vote accept Joseph O'Rourke
Sep 11, 2010 at 6:43 answer added Bill Thurston timeline score: 10
Sep 5, 2010 at 15:59 comment added Joseph O'Rourke @Victor: I was trying to formulate a way to incorporate the Cantor set myself, without success. So your clear and precise construction is welcome indeed! At this point I cannot fathom what $X$ is in this case.
Sep 5, 2010 at 10:51 comment added Victor Protsak Here is a construction worth thinking about. Let $C$ be a nowhere dense closed subset of the unit circle $S=\partial D, U$ be its complement, so that $U$ is a union of at most countable family of open intervals, and σ be a fixed-point free involution on the set of components of $U.$ Then σ generates an orientation-preserving equivalence relation $\sim$ on $S$ and we form the quotient $X=D/\sim$ of the closed unit disk by $\sim.$ If, in addition, σ preserves the length of the intervals then $X$ carries a natural metric and if $C$ is finite, $X$ is a closed surface. What if $C$ is a Cantor set?
Sep 4, 2010 at 23:18 history edited Joseph O'Rourke CC BY-SA 2.5
The bit-complement example makes no sense.
Sep 4, 2010 at 22:51 comment added Joseph O'Rourke @Victor: Good observation by you and Sergei---I had intended the example to be "some weird pairwise identification" to quote Sergei, but inadvertently specified $aa^{−1}$!
Sep 4, 2010 at 17:35 answer added Sergei Ivanov timeline score: 5
Sep 4, 2010 at 17:18 comment added Victor Protsak The points with complementary binary representations have coordinates $x$ and $1-x,$ so the corresponding identification coincides with the identification obtained from a digon with the word $aa^{-1}.$
Sep 4, 2010 at 13:53 history asked Joseph O'Rourke CC BY-SA 2.5