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

src/topology/uniform/metric.tex

@@ -10,11 +10,13 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
         \item[(PM1)] For any $x \in X$, $d(x, x) = 0$.
         \item[(PM2)] For any $x, y \in X$, $d(x, y) = d(y, x)$.
         \item[(PM3)] For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     If $d$ satisfies the above and
     \begin{enumerate}
         \item[(M)] For any $x, y \in X$ with $x \ne y$, $d(x, y) > 0$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $d$ is a \textbf{metric}.
 \end{definition}
 
@@ -62,7 +64,8 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
         \item For each $U \subset X$, $U$ is open if and only if for every $x \in U$, there exists $J \subset I$ finite and $r > 0$ such that $\bigcap_{j \in J}B_j(x, r) \subset U$.
         \item For each $i \in I$, $d_i \in UC(X \times X; [0, \infty))$.
         \item[(U)] For any other uniformity $\mathfrak{V}$ satisfying (4), $\mathfrak{U} \subset \mathfrak{V}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     The uniformity $\fU$ is the \textbf{pseudometric uniformity} induced by $\seqi{d}$, and the topology induced by $\fU$ is the \textbf{pseudometric topology} on $X$ induced by $\seqi{d}$.
 \end{definition}
 \begin{proof}
@@ -112,7 +115,8 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
         \item[(a)] $U_{0} = X \times X$.
         \item[(b)] For each $n \in \natz$, $U_n$ is symmetric.
         \item[(c)] For each $n \in \natz$, $U_{n + 1} \circ U_{n+1} \subset U_n$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then there exists a pseudometric $d: X \times X \to [0, 1]$ such that
     \[
     U_{n+1} \subset E(d, 2^{-n}) \subset U_{n-1}
@@ -141,7 +145,8 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
         \]
 
         As this holds for all such $\seqf{x_j}$ and $\seqf[m]{y_j}$, $d(x, z) \le d(x, y) + d(y, z)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     so $d$ is a pseudometric.
 
     For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1} \subset E(d, 2^{-n})$.
@@ -250,7 +255,8 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
         \item[(a)] For each $1 \le k \le n$, $V_k$ is symmetric.
         \item[(b)] For each $1 \le k \le n$, $V_k \subset U_k$.
         \item[(c)] For each $1 \le k < n$, $V_{k+1} \circ V_{k+1} \subset V_{k}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Let $W = V_n \cap U_{n+1}$, then by \autoref{lemma:symmetricfundamentalentourage}, there exists $V_{n+1} \in \fU$ symmetric such that $V_{n+1} \circ V_{n+1} \subset W$. Thus $\bracs{V_k|1 \le k \le n + 1} \subset \fU$ satisfies (a), (b), and (c) for $n + 1$.
 
     Let $V_0 = X \times X$, then by \autoref{lemma:uniform-sequence-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for each $n \in \natp$,