Cleanup

src/topology/main/normal.tex

@@ -30,16 +30,17 @@
 \begin{proof}
     (1): Let $\seq{q_n}$ be an enumeration of $\rational \cap (0, 1)$. For each $n \in \natp$, let $Q_n = \bracs{0, 1} \cup \bracs{q_k|1 \le k \le n}$.
 
-    Let $U_1 = B^c$. By (2) of \ref{definition:topology-normal}, there exists $U_0 \in \cn(A)$ such that $A \subset U_0 \subset \ol{U_0} \subset B^c$. In which case, for $n = 0$,
+    Let $U_1 = B^c$. By (2) of \autoref{definition:topology-normal}, there exists $U_0 \in \cn(A)$ such that $A \subset U_0 \subset \ol{U_0} \subset B^c$. In which case, for $n = 0$,
     \begin{enumerate}
         \item[(i)] $U_1 = B^c$.
         \item[(ii)] For any $p, q \in Q_k$ with $p < q$, $\overline{U_p} \subset U_q$.
     \end{enumerate}
 
-    Suppose inductively that $\bracs{U_q|q \in Q_n}$ has been constructed, and (ii) holds for $n$. Let $p = \max\bracs{r \in Q_n|r < q_{n+1}}$ and $q = \min\bracs{r \in Q_n|r > r_{n+1}}$. By (2) of \ref{definition:topology-normal}, there exists $U_{q_{n+1}} \in \cn^o(\overline{U_p})$ such that
+    Suppose inductively that $\bracs{U_q|q \in Q_n}$ has been constructed, and (ii) holds for $n$. Let $p = \max\bracs{r \in Q_n|r < q_{n+1}}$ and $q = \min\bracs{r \in Q_n|r > r_{n+1}}$. By (2) of \autoref{definition:topology-normal}, there exists $U_{q_{n+1}} \in \cn^o(\overline{U_p})$ such that
     \[
     U_p \subset \overline{U_p} \subset U_{q_{n+1}} \subset \overline{U_{q_{n+1}}} \subset U_q
     \]
+
     and $\bracs{U_q|q \in Q_{n+1}}$ satisfies (ii) for $n + 1$.
 
     Now suppose that $\bracs{U_q|q \in \rational \cap [0, 1]}$ has been constructed and (ii) holds for all $n \in \nat$. For any $p, q \in \rational \cap [0, 1]$ with $p < q$, there exists $n \in \nat$ such that $p, q \in Q_n$. In which case, $\ol{U_p} \subset U_q$ by (ii). Thus (b) holds.
@@ -48,16 +49,19 @@
     \[
     f: X \to [0, 1] \quad x \mapsto \inf\bracs{q \in [0, 1] \cap \rational| x \in U_q}
     \]
+
     where $f(x) = 1$ if $x \not\in \bigcup_{q \in [0, 1] \cap \rational}U_q$. Since $A \subset \bigcap_{q \in [0, 1] \cap \rational}U_q$ and $U_1 = B^c$, $f|_A = 0$ and $f|_B = 1$.
 
     Let $\alpha \in [0, 1]$, then
     \[
     f^{-1}([0, \alpha)) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q < \alpha}}U_q
     \]
+
     is open. On the other hand, let $x \in X$, then $f(x) > \alpha$ if and only if there exists $q \in (\alpha, f(x)) \cap \rational$ such that $x \not\in U_q$. By (b) of (1), this is equivalent to the existence of $p \in (\alpha, f(x)) \cap \rational$ such that $x \not\in \ol{U_p}$. In which case,
     \[
     f^{-1}((b, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
     \]
+
     is open.
 \end{proof}
 
@@ -70,18 +74,21 @@
     \[
     R: BC(X; \real) \to BC(A; \real) \quad g \mapsto g|_A
     \]
+
     then $R \in L(BC(X; \real); BC(A; \real))$.
 
     For any $g \in C(A; [0, 1])$, let
     \[
     B = g^{-1}(\norm{g}_u \cdot [0, 1/3]) \quad C = g^{-1}(\norm{g}_u \cdot [2/3, 1])
     \]
+
     then $B, C \subset A$ are closed with $B \cap C = \emptyset$. Since $A$ is closed, $B, C \subset X$ are closed. By Urysohn's lemma, there exists $h \in C(X; [0, 1/3])$ such that $h|_C = 1/3$ and $h|_B = 0$. Thus $g - h|_A \in C(A; [0, 2/3])$.
 
     By linearity, this implies that for any $g \in BC(A; \real)$, there exists $h \in BC(A; \real)$ such that $\norm{h}_u \le \norm{g}_u/3$ and $\norm{g - h|_A}_u \le 2\norm{g}_u/3$.
 
-    Since $\real$ is complete, so is $BC(X; \real)$ by \ref{proposition:set-uniform-complete}. Using successive approximations (\ref{theorem:successive-approximation}), for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
+    Since $\real$ is complete, so is $BC(X; \real)$ by \autoref{proposition:set-uniform-complete}. Using \hyperref[successive approximations]{theorem:successive-approximation}, for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
     \[
     \norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = \norm{f}_u
     \]
+
 \end{proof}