Hausdorff Paradox/Proof 2

Theorem

There exists a disjoint decomposition of the sphere $\mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C, B \cup C$ are all congruent and $Q$ is countable.

Proof

Lemma $1$

Let $G$ be the free product of the groups $G_1 = \set {e_1, \phi}$ and $G_2 = \set {e_2, \psi, \psi^2}$.

Let $U := \mathbb D^3 \subset \R^3$ be a unit ball in real Euclidean space of $3$ dimensions.

Let $\phi$ and $\psi$ be represented by the axes of rotation $a_\phi$ and $a_\psi$ passing through the center of $U$ such that:

$\phi$ is a rotation by $180 \degrees$, that is $\pi$ radians about $a_\phi$

$\psi$ is a rotation by $120 \degrees$, that is $\dfrac {2 \pi} 3$ radians about $a_\psi$

Hence consider $G$ as the group of all rotations generated by $\phi$ and $\psi$.

The identity of $G$ is then the identity mapping $\mathbf I_3$.

Then $a_\phi$ and $a_\psi$ can be determined in such a way that distinct elements of $G$ represent distinct rotations generated by $\phi$ and $\psi$.

$\Box$

Lemma $2$

The group $G$ can be partitioned into $3$ disjoint sets:

$G = A \cup B \cup C$

such that:

\(\text {(1)}: \quad\)

\(\ds A \circ \phi\)

\(=\)

\(\ds B \cup C\)

\(\text {(2)}: \quad\)

\(\ds A \circ \psi\)

\(=\)

\(\ds B\)

\(\text {(3)}: \quad\)

\(\ds A \circ \psi^2\)

\(=\)

\(\ds C\)

$\Box$

Let $Q$ be the set of all fixed points of $\Bbb S^2$ of all rotations $\alpha \in G$.

Each such $\alpha$ has $2$ fixed points.

Hence $Q$ is countable.

The set $\Bbb S^2 \setminus Q$ is a disjoint union of all orbits $P_x$ of $G$:

$P_x = \set {x \alpha: \alpha \in G}$

By the Axiom of Choice, there exists a set $M$ which contains exactly $1$ element in each $P_x$ where $x \in \Bbb S^2 \setminus Q$.

Let:

\(\ds A\)

\(=\)

\(\ds M \circ \mathscr A\)

\(\ds B\)

\(=\)

\(\ds M \circ \mathscr B\)

\(\ds C\)

\(=\)

\(\ds M \circ \mathscr C\)

It then follows from Lemma $2$ that:

$A$, $B$ and $C$ are disjoint

$A$, $B$ and $C$ are congruent to each other

$B \cup C$ is congruent to each of $A$, $B$ and $C$

and moreover:

$S = A \cup B \cup C \cup Q$

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Source of Name

This entry was named for Felix Hausdorff.

Sources

• 1973: Thomas J. Jech: The Axiom of Choice ... (previous) ... (next): $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere