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Fix typo reported in #339
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rzach committed Oct 8, 2023
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It is instructive to compare and contrast the arguments in
this section with Russell's paradox:
\begin{enumerate}
\item Russell's paradox: let $S = \Setabs{x}{x \notin x}$. Then $x
\in S$ if and only if $x \notin S$, a contradiction.
\item Russell's paradox: let $S = \Setabs{x}{x \notin x}$. Then $S
\in S$ if and only if $X \notin S$, a contradiction.

\emph{Conclusion:} There is no such set~$S$. Assuming the existence of a
``set of all sets'' is inconsistent with the other axioms of set
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