2 Doubling a Sphere
2.1 Sphere
Let \(L = \{ (x,y,z) : x^2 + y^2 + z^2 \leq 1\} \) be the unit sphere. We define \(L' = L \setminus \{ (0,0,0)\} \) as the unit sphere without its center.
Two points \(a\) and \(b\) belong to the same orbit if and only if there exists a \(\rho \) in \(G\) such that \(\rho (a) = b\).
The set of all orbits is countable.
We can select a representative point from each orbit.
This follows directly from the Axiom of Choice.
Let \(M\) denote the set of all chosen representative points.
We can now reach every point in \(L'\) by applying a rotation from \(G\) to a specific element in \(M\).
2.2 Duplicating a Part of the Unit Sphere
We desire that each point in \(L'\) is reached by only one rotation in \(G\). Therefore, we decompose \(L'\) according to the rotations that map to a point. A point that is mapped by multiple rotations may belong to multiple sets.
Let \(Y\) be a set and \(f: Y \rightarrow Y\) a function. A point \(y \in Y\) is called a fixed point if it satisfies \(f(y) = y\).
Let \(D\) denote the set of all points in \(L'\) that are fixed points of the rotations in \(G\).
\(G\) is countable.
Each rotation in \(G\) has exactly one rotation axis.
The rotation axes lie on countably many lines.
Therefore, almost every point in \(L'\) can be reached by a specific rotation. We now consider a decomposition of \(L' \setminus D\) and will address the fixed points later.
\(X = \bigcup \limits _{k=1}^{\infty } A^{-k} M\). Thus, \(X\) is the set of all elements of \(M\) that consist solely of rotations by \(A^{-1}\).
\(L' \setminus D = P_1 \cup P_2 \cup P_3 \cup P_4\)
The set of representative points is countable.
We now have a decomposition that allows us to duplicate the sphere, excluding its center and the points on the rotation axes.
2.3 Fixed Points and the Center
Two sets \(C\) and \(D\) are called equidecomposable if \(C\) can be partitioned into finitely many pieces, which can be rearranged through rotations and translations to form \(D\).
\(L' \setminus D\) and \(L'\) are equidecomposable.
Since the points in \(D\) lie on countably many axes, we can find a line \(l\) through the origin that does not intersect \(D\). Furthermore, there exists an angle \(\theta \) such that a rotation by \(\theta \) around \(l\) does not map any point in \(D\) onto another point in \(D\). Define \(E = D \cup \rho (D) \cup \rho ^2(D) \cup \rho ^3(D) \cup \dots \). It follows immediately that \(L' = E \cup (L' \setminus E)\), which is equidecomposable with \(\rho (E) \cup (L' \setminus E)\). From the definition of \(E\), we obtain \(\rho (E) = E \setminus D\), so that we get \(\rho (E) \cup (L' \setminus E) = (E \setminus D) \cup (L' \setminus E) = L' \setminus D\). Thus, it follows that \(L'\) is equidecomposable with \(L' \setminus D\).
Now that we know that \(L' \setminus D\) is equidecomposable with \(L'\), the fixed points of the rotations in \(G\) are no longer an issue. Hence, we now focus on the center.
\(\nexists p,q\in \mathbb {Z}\) such that \(\sqrt{2} \cdot \frac{p}{q} = \pi \)
Assume that there exist \(p,q\in \mathbb {Z}\) such that \(\sqrt{2} \cdot \frac{p}{q} = \pi \). By the definition of \(\cos ^{-1}\) on the interval \([0,2\pi )\), we have \(\cos ^{-1}(-1) = \pi \) and thus
This is equivalent to \(-1 = \cos (\sqrt{2} \cdot \frac{p}{q})\). Using the Taylor series expansion of cosine, we would obtain:
This contradicts the properties of rational approximations, proving the claim.
A circle is equidecomposable with a circle minus a single point.
We consider the unit circle \(S^1 = \{ (x,y) : x^2 + y^2 = 1\} \) minus the point \((1,0)\). We use the unit circle to simplify notation, but the following argument can be applied to any circle. Define \(A = \{ (\cos n, \sin n) : n \in \mathbb {Z}\} \). Since \(\pi \) is irrational, we have \((\cos n, \sin n) \neq (\cos m, \sin m)\) for all distinct \(n, m \in \mathbb {Z}\). Thus, \(A\) is countably infinite and does not contain \((1,0)\). Let \(B = (S^1 \setminus \{ (1,0)\} ) \setminus A\). Now, rotate \(A\) about the origin by one unit. Denote the rotated set by \(A'\). This rotation maps \((\cos 1, \sin 1)\) to \((1,0)\), which is exactly the missing point. Since every point originally in \(A\) is still present in \(A'\), we obtain \(A' = A \cup \{ (1,0)\} \), and thus \(S^1 \setminus \{ (1,0)\} = A \cup B\) is equidecomposable with \(A' \cup B = S^1\).
A sphere minus its center is equidecomposable with the complete sphere.
Constructing a circle within the sphere that includes the center of the sphere, together with Lemma 27, shows that the sphere minus its center is equidecomposable with the complete sphere.
2.4 The Final Proof
Having gathered all the necessary details, we can now establish the full paradox.
A sphere is equidecomposable with two copies of itself.
In the second subsection, we showed that the sphere minus its center and the points on the rotation axes is equidecomposable with two copies of itself. Using Lemma 25, it follows that the sphere minus its center is equidecomposable with two copies of itself. Finally, by Theorem 29, the complete sphere is equidecomposable with two copies of itself.