Added compactness, Urysohn's lemma, and the Tietze extension theorem.

src/topology/main/normal.tex

@@ -0,0 +1,87 @@
+\section{Normal Spaces}
+\label{section:topology-normal}
+
+\begin{definition}[Normal]
+\label{definition:topology-normal}
+    Let $X$ be a topological space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $A, B \subset X$ closed with $A \cap B = \emptyset$, there exists $U \in \cn(A)$ and $V \in \cn(B)$ such that $U \cap V = \emptyset$.
+        \item For any $A \subset X$ closed and $U \in \cn^o(A)$, there exists $V \in \cn^o(A)$ such that $A \subset V \subset \ol{V} \subset U$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Since $U^c$ is closed, there exists $V \in \cn^o(A)$ and $W \in \cn^o(U^c)$ such that $V \cap W = \emptyset$. In which case, $A \subset V \subset \ol{V} \subset W^c \subset U$.
+
+    (2) $\Rightarrow$ (1): Since $B$ is closed with $A \cap B = \emptyset$, $B^c \in \cn^o(A)$. Let $V \in \cn^o(A)$ with $A \subset V \subset \ol{V} \subset B^c$, then $\ol{V}^c \in \cn^o(B)$ with $V \cap \ol{V}^c = \emptyset$.
+\end{proof}
+
+\begin{lemma}[Urysohn]
+\label{lemma:urysohn}
+    Let $X$ be a normal topological space, $A, B \subset X$ be closed with $A \cap B = \emptyset$, then
+    \begin{enumerate}
+        \item There exists $\bracs{U_q| q \in \rational \cap [0, 1]} \subset \cn(A)$ such that
+        \begin{enumerate}
+            \item[(a)] $U_1 = B^c$.
+            \item[(b)] For any $p, q \in \rational \cap [0, 1]$ with $p < q$, $\overline{U_p} \subset U_q$.
+        \end{enumerate}
+        \item There exists $f \in C(X; [0, 1])$ with $f|_A = 0$ and $f|_B = 1$.
+    \end{enumerate}
+\end{lemma}
+\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 $n = 0$,
+    \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
+    \[
+    U_p \subset \overline{U_p} \subset U_{q_{n+1}} \subset \overline{U_{q_{n+1}}} \subset U_q
+    \]
+    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.
+
+    (2): Let
+    \[
+    f: X \to [0, 1] \quad x \mapsto \inf\bracs{q \in [0, 1] \cap \rational| x \in U_q}
+    \]
+    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}
+
+\begin{theorem}[Tietze Extension Theorem]
+\label{theorem:tietze}
+    Let $X$ be a normal space, $A \subset X$ be closed, $U \in \cn(A)$, and $f \in BC(A; \real)$, then there exists $F \in BC(X; \real)$ such that $F|_A = f$ and $\norm{F}_u = \norm{f}_u$.
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    R: BC(X; \real) \to BC(A; \real) \quad g \mapsto g|_A
+    \]
+    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
+    \[
+    \norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = 1
+    \]
+\end{proof}