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===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.

First, let's examine why a circle is rigid-body equivalent to a circle with a point removed.  We'll work in the plane, for now.

 

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.

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.

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>

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Removing countably many points from a sphere is similar, except now we are using the longitudes of the points, with respect to some axis that doesn't pass through any of the points of <math>A \ </math>.  A tricky technical matter arises: how do we know such an axis exists?

Let's now move back to three dimensional space.  Removing countably many points from a sphere is similar, except now we are using the longitudes of the points, with respect to some axis that doesn't pass through any of the points of <math>A \ </math>.  A tricky technical matter arises: how do we know such an axis exists?

This is simple!  Since <math>A \ </math> is countable, so is the union of <math>A \ </math> and all the antipodes of <math>A \ </math>, which means there are points not in this union.  By design, neither are the antipodes of such points.

This is simple!  Since <math>A \ </math> is countable, so is the union of <math>A \ </math> and all the antipodes of <math>A \ </math>, which means there are points not in this union.  By design, neither are the antipodes of such points.