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

src/fa/tvs/complete-metric.tex

@@ -21,12 +21,14 @@
     \begin{enumerate}
         \item[(a)] $\eta(y - Tx) \le \gamma \eta(y)$.
         \item[(b)] $\rho(x) \le C \eta(y)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
     \begin{enumerate}
         \item $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)/(1 - \gamma)$.
         \item $y = \limv{N}\sum_{n = 1}^N Tx_n$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     In particular,
     \[
     T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
@@ -38,12 +40,14 @@
     \begin{enumerate}
         \item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
         \item[(II)] $\eta\paren{y - \sum_{n = 1}^N Tx_n} \le \eta(y)\gamma^N$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     By assumption, there exists $x_{N+1} \in E$ such that:
     \begin{enumerate}
         \item[(i)] $\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n} \le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n} \le \gamma^{N+1}$.
         \item[(ii)] $\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n} \le C\eta(y)\gamma^N$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Combining (I) and (ii) shows that $\sum_{n = 1}^N \rho(x_n) \le C \eta(y) \sum_{n = 0}^N \gamma^n$. Therefore there exists $\seq{x_n} \subset E$ such that (I) and (II) holds for all $N \in \natp$.
 
     By (I), $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)\sum_{n \in \natz}\gamma^n = C \eta(y)/(1 - \gamma)$. By (II), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n} = \limv{N}\eta(y)\gamma^N = 0$.
@@ -55,7 +59,8 @@
     \begin{enumerate}
         \item[(a)] For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))} \supset B_F(0, \delta(r))$.
         \item[(b)] $E$ is complete.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then for every $s > r$, $T(B_E(0, s)) \supset B_F(0, \delta(r))$.
 \end{proposition}
 \begin{proof}
@@ -65,12 +70,14 @@
         \item[(ii)] $s_1 = r$.
         \item[(iii)] For all $n \in \natp$, $\overline{T(B_E(0, s_n))} \supset B_F(0, \delta_n)$.
         \item[(iv)] $\rho_1 = \rho$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Let $y_0 \in B(0, r)$ and $x_0 = 0$. Let $N \in \natp$ and suppose inductively that $\bracs{x_n}_1^N \subset E$ has been constructed such that:
     \begin{enumerate}
         \item[(I)] For each $0 \le n \le N - 1$, $\rho(x_{n+1} - x_n) < s_n$.
         \item[(II)] For each $0 \le n \le N$, $\eta(Tx_n - y) \le \rho_{n+1}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     By density of $T(x_N + B_E(0, s_N))$ in $Tx_N + B_F(0, \rho_N)$, there exists $x_{N+1} \in T(x_N + B_E(0, s_N))$ such that $\eta(Tx_{N+1} - y) \le \rho_{N+2}$.
 
     By (I), $\seq{x_N}$ is a Cauchy sequence, so
@@ -88,7 +95,8 @@
     \begin{enumerate}
         \item[(a)] For any $r > 0$, there exists $C \ge 0$ such that for any $y \in T(E)$, there exits $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y)$.
         \item[(b)] $E$ is complete.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $T(E)$ is closed.
 \end{proposition}
 \begin{proof}