Updated Schaefer & Wolff citations.

src/fa/tvs/completion.tex

@@ -29,6 +29,6 @@
     By continuity and the density of $\iota(E)$ in $E$, $\wh E$ with these operations forms a TVS, and $T$ is linear.
 \end{proof}
 
-\begin{remark}[{{\cite[Section 1.1]{SchaeferWolff}}}]
+\begin{remark}[{{\cite[Section I.1]{SchaeferWolff}}}]
 	The Hausdorff completion works in general with arbitrary valuated fields. Though the completion yields a TVS over the completion of the field, the field need not to be complete.
 \end{remark}