Minor adjustments.

src/fa/tvs/definition.tex

@@ -69,7 +69,7 @@
     Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
 \end{proof}
 
-\begin{proposition}[{{\cite[I.1.2]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:tvs-0-neighbourhood-base}
     Let $E$ be a vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ such that:
     \begin{enumerate}
@@ -88,7 +88,7 @@
         \item[(3)] $(E, \topo)$ is a TVS.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[I.1.2]{SchaeferWolff}}}. ]
     \textbf{Forward:} By \autoref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
 
     \textbf{Converse:} For each $V \in \fB$, let $U_V = \bracs{(x, y) \in E|x - y \in V}$, then $U_V$ is symmetric and translation-invariant by (TVB1). Let