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→The Banach-Tarski Construction
Make sure you understand the notation: <math>\mathbb{F} \ </math> is a set of rotations, <math>W \ </math> is a set of points, and <math>\mathbb{F}(W) \ </math> is the set of points resulting from applying every rotation in <math>\mathbb{F} \ </math> to every point in <math>W \ </math>. Similarly we will be defining <math>U,D,L,R \ </math> as sets of rotations and <math>U(x),\ </math> etc. as the point <math>x \ </math> rotated by all the points in <math>U \ </math>.
Make sure you understand the notation: <math>\mathbb{F} \ </math> is a set of rotations, <math>W \ </math> is a set of points, and <math>\mathbb{F}(W) \ </math> is the set of points resulting from applying every rotation in <math>\mathbb{F} \ </math> to every point in <math>W \ </math>. Similarly we will be defining <math>U,D,L,R \ </math> as sets of rotations and <math>U(x),\ </math> etc. as the point <math>x \ </math> rotated by all the points in <math>U \ </math>.
Let's define <math>U_* = U(W), D_* = D(W), L_*(W), R_*(W) \ </math>. Obviously, <math>U_* \cup D_* \cup L_* \cup R_* = \mathbb{S}^2 - B \ </math>. But we remember <math>b'(U)\cup D = F \ </math>. This immediately yields:
<math>b'(U_*)\cup D_* = b'(U(W)) \cup D(W) = (b'(U) \cup D)(W) = \mathbb{F}(W) = \mathbb{S}^2-B \ </math>.
Since we also had <math>a'(U) \cup r = F_2 \ </math>, we have <math>b'(L_*)\cup R_* = \mathbb{S}^2 - B \ </math>.
Hence <math>U_* \cup D_* \cong \mathbb{S}^2 - B \ </math> and <math>L_* \cup D_* \cong \mathbb{S}^2 - B \ </math>, a paradoxical decomposition of <math>\mathbb{S}^2-B \ </math>. Since we've already shown that <math>\mathbb{S}^2-B \ </math> is rigid-body equivalent to <math>\mathbb{S}^2 \ </math>, this proves the Banach-Tarski paradox.
==Interpretation==
==Interpretation==