Hausdorff paradoxHausdorffParadox2005-05-15 07:41:552006-12-24 13:22:57Theoremparadoxunit balldecomposition% this is the default PlanetMath preamble. as your knowledge
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Let $S^2$ be the unit sphere in the Euclidean space $\mbb{R}^3$. Then
it is possible to take ``half'' and ``a third'' of $S^2$ such that
both of these parts are essentially congruent (we give a formal
version in a minute). This sounds paradoxical:
wouldn't that mean that half of the sphere's area is equal to only a
third? The ``paradox'' resolves itself if one takes into account that
one can choose non-measurable subsets of the sphere which ostensively are ``half'' and a ``third'' of it, using geometric congruence as means of comparison.
Let us now formally state the Theorem.
\newtheorem*{thm}{Theorem}
\begin{thm}[Hausdorff paradox~\cite{H}]
There exists a disjoint \PMlinkescapetext{decomposition} of the unit sphere $S^2$ in the
Euclidean space $\mbb{R}^3$ into four subsets $A,B,C,D$, such that the
following conditions are met:
\begin{enumerate}
\item Any two of the sets $A$, $B$, $C$ and $B\cup C$ are congruent.
\item $D$ is countable.
\end{enumerate}
\end{thm}
A crucial ingredient to the proof is the \PMlinkid{axiom of choice}{310}, so the
sets $A$, $B$ and $C$ are not constructible. The theorem itself is a
crucial ingredient to the proof of the so-called Banach-Tarski
paradox.
\begin{thebibliography}{H}
\bibitem[H]{H} \textsc{F.~Hausdorff}, Bemerkung \"{u}ber den Inhalt von
Punktmengen, \emph{Math.\ Ann.}\ 75, 428--433, (1915), \texttt{\PMlinkexternal{http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?did=D28919}{http://dz-srv1.sub.uni-goettingen.de/sub/digbib/loader?did=D28919}} (in German).
\end{thebibliography}