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Fix mistake in proof 09GK
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aisejohan committed Jan 31, 2018
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Expand Up @@ -747,7 +747,7 @@ \section{Algebraic extensions}
is finite (details omitted). Moreover, the fact that $E$ is algebraic
over $F$ implies that $E = \bigcup_{P \in S} r(P, E)$.
It is clear that $S$ has cardinality bounded by $\max(\aleph_0, |F|)$
because the cardinality of a finite product of copies of $F$ has
because the cardinality of a countable product of copies of $F$ has
cardinality at most $\max(\aleph_0, |F|)$.
Thus so does $E$.
\end{proof}
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