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Revision difference : Hausdorff paradox
Version 2 Version 1
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Let $S^2$ be the unit sphere in the Euclidean space $\mbb{R}^3$. Then 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 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 both of these parts are essentially congruent (we give a formal
version in a minute). This sounds paradoxical: version in a minute). This sounds paradoxical:
wouldn't that mean that half of the sphere's area is equal to only a 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 third? The ``paradox'' resolves itself if one takes into account that
one can choose the ``half'' and the ``third'' in a plausible group one can choose the ``half'' and the ``third'' in a plausible group
theoretical sense, yet the resulting subsets of the sphere are not theoretical sense, yet the resulting subsets of the sphere are not
measurable. measurable.
Let us now formally state the Theorem. Let us now formally state the Theorem.
\newtheorem*{thm}{Theorem} \newtheorem*{thm}{Theorem}
\begin{thm}[Hausdorff paradox~\cite{H}] \begin{thm}[Hausdorff paradox~\cite{H}]
There exists a disjoint \PMlinkescapetext{decomposition} of the unit sphere $S^2$ in the 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 Euclidean space $\mbb{R}^3$ into four subsets $A,B,C,D$, such that the
following conditions are met: following conditions are met:
\begin{enumerate} \begin{enumerate}
\item Any two of the sets $A$, $B$, $C$ and $B\cup C$ are congruent. \item Any two of the sets $A$, $B$, $C$ and $B\cup C$ are congruent.
\item $D$ is countable. \item $D$ is countable.
\end{enumerate} \end{enumerate}
\end{thm} \end{thm}
A cruical ingredient to the proof is the axiom of choice, so the A cruical ingedient to the proof is the axiom of choice, so that the
sets $A$, $B$ and $C$ are not constructible. The theorem itself is a sets $A$, $B$ and $C$ are not constructible. The theorem itself is a
cruical ingredient to the proof of the so-called \PMlinkid{Banach--Tarski cruical ingredient to the proof of the so-called \PMlinkid{Banach--Tarski
paradox}{4464}. paradox}{4464}.
\begin{thebibliography}{H} \begin{thebibliography}{H}
\bibitem[H]{H} \textsc{F.~Hausdorff}, Bemerkung über den Inhalt von \bibitem[H]{H} \textsc{F.~Hausdorff}, Bemerkung über den Inhalt von
Punktmengen, \emph{Math.\ Ann.}\ 75, 428--433, (1915). Punktmengen, \emph{Math.\ Ann.}\ 75, 428--433, (1915).
\end{thebibliography} \end{thebibliography}