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 \ </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.

Make sure you understand the content of <math>W \ </math>.  If the first few values of <math>A = \left\{{6,2\sqrt{3},\frac{e}{\pi}, ... }\right\} \ </math>, then <math>W \ </math> contains, among other things, <math>6-2\sqrt{3}, (2\sqrt{3}-\frac{e}{\pi})/653, \ </math> etc.  Now, since <math>W \ </math> is countable, and the angles of <math>\mathbb{S}^1 \ </math> are uncountable, there is an angle in <math>\mathbb{S}^1 \ </math> which is not in <math>W \ </math>.  Pick one such angle, and call it <math>\theta \ </math>.  This is our magic rotation angle.

Make sure you understand the content of <math>W \ </math>.  If the first few values of <math>A = \left\{{6,2\sqrt{3},\frac{e}{\pi}, ... }\right\} \ </math>, then <math>W \ </math> contains, among other things, <math>6-2\sqrt{3}, (2\sqrt{3}-\frac{e}{\pi})/653, \ </math> etc.  Now, since <math>W \ </math> is countable, and the angles of <math>\mathbb{S}^1 \ </math> are uncountable, there is an angle in <math>\mathbb{S}^1 \ </math> which is not in <math>W \ </math>.  Pick one such angle, and call it <math>\theta \ </math>.  This is our magic rotation angle.