Polished A-A and added new lines for broken enumerates.

src/fa/tvs/metric.tex

@@ -82,7 +82,8 @@
     \begin{enumerate}
         \item[(a)] For each $n \in \natp$, $U_n$ is circled, radial, and contains $0$.
         \item[(b)] For each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $n \in \natp$,
     \[
     U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
@@ -110,7 +111,8 @@
 
         so $\rho(\lambda x) \le \rho(x)$.
         \item[(PN3)] Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_M$ and $y \in U_N$. Assume without loss of generality that $\rho_M + \rho_N < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_P = \rho_M + \rho_N$. In which case, $U_P \supset U_M + U_N$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \autoref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
     \begin{enumerate}
         \item[(PN4)] Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda} \ge \alpha$, $x \in \lambda U_n$. Therefore for any $\lambda \in K$ with $\abs{\lambda} \le \alpha^{-1}$, $\lambda x \in U_n$, and $\rho(x) \le 2^{-n}$.