Banach-Tarski Paradox/Proof 2/Mistake

Source Work

1973: Thomas J. Jech: The Axiom of Choice:

Chapter $1$: Introduction

$1.3$ A paradoxical decomposition of the sphere:

Proof of Theorem $1.2$

Mistake

Now, it is easy to find some rotation $\alpha$ (not in $G$) such that $Q$ and $Q \cdot \alpha$ are disjoint, and so, using

$\overline C \approx \overline A \cup \overline B \cup \overline C$,

there exists $S \subset C$ such that $\overline S \approx \overline Q$. Let $p$ be some point in $\overline S - \overline C$. Obviously, $\ldots$

Correction

If $S \subset C$, as supposed, it follows that $\overline S \subset \overline C$ and so $\overline S - \overline C = \O$.

What is meant is:

Let $p$ be some point in $\overline C - \overline S$.

Sources

• 1973: Thomas J. Jech: The Axiom of Choice ... (previous) ... (next): $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Proof of Theorem $1.2$