Adjusted organisation in the TVS chapter.

src/fa/tvs/bounded.tex

@@ -9,7 +9,7 @@
         \item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
     \end{enumerate}
 
-    If the above holds, then $B$ is \textbf{bounded}. The collection $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
+    If the above holds, then $B$ is \textbf{bounded}. The collection $\mathfrak{B}(E) = \mathfrak{B}(E, \topo)$ is the set of all bounded sets of $E$.
 \end{definition}
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$. 
@@ -19,7 +19,7 @@
 
 \begin{proposition}[{{\cite[I.5.1]{SchaeferWolff}}}]
 \label{proposition:bounded-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded:
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \in \mathfrak{B}(E)$, then the following sets are bounded:
     \begin{enumerate}
         \item Any $C \subset B$.
         \item The closure $\ol{B}$.