Changes

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.

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.  Let <math>A = \left\{{ a_1, a_2, ... }\right\} \ </math> be countable set of points in the circle, where it is understood that the <math>\alpha \ </math> measure the degree of the angle the point can be found at.  We construct a set <math>W = \left\{{ \frac{a_m-a_k}{n}:(m,k,n)\in \mathbb{N}^3 }\right\} \ </math>, which is also a countable set, since it is in one-to-one correspondence with the three dimensional lattice.

Removing countably many points from a circle is a trickier matter.  Let <math>A = \left\{{ a_1, a_2, ... }\right\} \ </math> be countable set of points in the circle, where it is understood that the <math>\alpha \ </math> measure the degree of the angle the point can be found at.  We construct a set <math>W = \left\{{ \frac{a_m-a_k}{n}:(m,k,n)\in \mathbb{N}^3 }\right\} \ </math>, which is also a countable set, since it is in one-to-one correspondence with the three dimensional lattice.

We form the partition <math>\mathbb{S}^1 = T \cup (\mathbb{S}^1-T) \ </math> and then note <math>R_\theta(T) \cup (\mathbb{S}^1 - T) = \mathbb{S}^1 - A \ </math> and we have shown that <math>\mathbb{S}^1 \cong \mathbb{S}^1 - A \ </math>

We form the partition <math>\mathbb{S}^1 = T \cup (\mathbb{S}^1-T) \ </math> and then note <math>R_\theta(T) \cup (\mathbb{S}^1 - T) = \mathbb{S}^1 - A \ </math> and we have shown that <math>\mathbb{S}^1 \cong \mathbb{S}^1 - A \ </math>

===The Free Group===

===The Free Group===