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→Removing Points with Rigid Body Motions
===Removing Points with Rigid Body Motions===
===Removing Points with Rigid Body Motions===
First, lets examine why a circle is rigid-body equivalent to a circle with a point removed. Pick an irrational number <math>\alpha \ </math>, and let <math>R_\alpha \ </math> be the rigid motion which is rotation by <math>\alpha \ </math> degrees counterclockwise. Now we pick a point in the circle <math>P \ </math> and we form an infinite "necklace" <math>\left\{{P, R_\alpha(P), R_{2\alpha}(P), ... }\right\} \ </math>. It is clear that this necklace must be infinite, because if there was any <math>n \ </math> such that <math>R_{n\alpha}(P)=P \ </math>, then <math>\alpha \ </math> would not be irrational.
Since our necklace is a [[Countable Set|countable set]] and the circle is uncountable, there are points in the circle which are not in the necklace. So we partition the circle into two sets, the necklace, and everything left over. Observe that by performing the rigid motion <math>R_\alpha \ </math> on the necklace, we simply get <math>\left\{{R_\alpha(P), R_{2\alpha}(P), ... }\right\} \ </math>, that is, all the original points remain exactly where they were, except now the first point is gone. Re-combining the rotated necklace with those points not in the necklace gives us a circle with a point removed.
With this procedure, it becomes a simple matter to remove a point from the sphere or the interior of a ball - just pick a circle lying in the sphere or ball which contains the point you wish to remove, and partition the sphere or ball into three pieces - the necklace, the rest of the circle the necklace is on, and the everything in the sphere or ball which isn't in that circle.
Removing countably many points from a circle is a trickier matter.
===The Free Group===
===The Free Group===