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

src/topology/functions/set-systems.tex

@@ -33,7 +33,7 @@
         \item The topology induced by $\fV$ is finer than the $\mathfrak{S}$-topology on $T^X$.
         \item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then $\mathfrak{E}(\mathfrak{S}, \fU)$ is forms a fundamental system of entourages for $\fV$.
     \end{enumerate}
-    and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\mathfrak{S}$}/\textbf{$\mathfrak{S}$-uniform topology} on $X^T$.
+    The uniformity $\fV$ is the \textbf{$\mathfrak{S}$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\mathfrak{S}$}/\textbf{$\mathfrak{S}$-uniform topology} on $X^T$.
 \end{definition}
 \begin{proof}
     (1): Since $\Delta \subset E(S, U)$ for all $S \in \mathfrak{S}$ and $U \in \fU$, $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity on $X^T$.
@@ -71,3 +71,13 @@
     \]
     which is open in the product topology. The converse is given by \ref{definition:set-uniform}.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:set-uniform-complete}
+    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ such that $\bigcup_{S \in \mathfrak{S}}S = T$, and $(X, \fU)$ be a complete uniform space, then $X^T$ equipped with the $\mathfrak{S}$-uniformity is complete.
+\end{proposition}
+\begin{proof}
+    Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \mathfrak{S}$ with $x \in S$, then $\pi_x(\fF) \subset 2^X$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_x(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \mathfrak{S}}S$.
+
+    Let $S \in \mathfrak{S}$ be non-empty and $V_0, V$ be symmetric entourages of $X$ with $V_0 \circ V_0 \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_0)$-small set $F \in \fF$. Let $x \in S$, then $\pi_x(\fF) \to f(x)$ implies that $F_x = \pi_x^{-1}(V_0(f(x))) \in \fF$. Since $F \cap F_x \ne \emptyset$, $(f(x), g(x)) \in V_0 \circ V_0 \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.
+\end{proof}

src/topology/functions/uniform.tex

@@ -0,0 +1,39 @@
+\section{The Uniform Topology}
+\label{section:uniform-convergence}
+
+\begin{definition}[Uniform Topology]
+\label{definition:uniform-convergence}
+    Let $T$ be a set and $(X, \fU)$ be a uniform space. For each $U \in \fU$, let
+    \[
+    E(U) = \bracsn{(f, g) \in X^T| (f(x), g(x)) \in U \forall x \in X}
+    \]
+    then the \textbf{uniform topology} on $X^T$ is the topology induced by the uniformity generated by $\bracs{E(U)| U \in \fU}$.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:uniform-limit-continuous}
+    Let $X$ be a topological space and $Y$ be a uniform space, then:
+    \begin{enumerate}
+        \item $C(X; Y) \subset Y^X$ is closed with respect to the uniform topology.
+        \item If $X$ is a uniform space, then $UC(X; Y) \subset Y^X$ is closed with respect to the uniform topology.
+    \end{enumerate}
+    In particular, if $Y$ is complete, then the above spaces are complete.
+\end{proposition}
+\begin{proof}
+    Let $x \in X$ and $V$ be an entourage of $Y$, then there exists a symmetric entourage $W$ of $Y$ such that $W \circ W \circ W \subset V$.
+
+
+    (1): Let $f \in \ol{C(X; Y)}$, then there exists $g \in C(X; Y)$ with $(f, g) \in E(W)$. Let $U \in \cn(x)$ such that $g(U) \subset W(g(x))$, then for any $y \in U$,
+    \[
+    (f(x), g(x)), (g(x), g(y)), (g(y), f(y)) \in W
+    \]
+    so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
+
+    Therefore $f$ is continuous at $x$.
+
+    (2): Let $f \in \ol{UC(X; Y)}$, then there exists $g \in UC(X; Y)$ with $(f, g) \in E(W)$. Let $U$ be an entourage of $X$ such that $(g \times g)(U) \subset W$, then for any $(x, y) \in U$,
+    \[
+    (f(x), g(x)), (g(x), g(y)), (g(y), f(y)) \in W
+    \]
+    so $(f(x), f(y)) \in W \circ W \circ W \subset V$.
+\end{proof}

src/topology/main/compact.tex

@@ -0,0 +1,80 @@
+\section{Compactness}
+\label{section:compact}
+
+\begin{definition}[Compact]
+\label{definition:compact}
+    Let $X$ be a topological space, then the following are equivalent:
+    \begin{enumerate}
+        \item For every family $\seqi{U}$ of open sets with $\bigcup_{i \in I}U_i = X$, there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_j = X$.
+        \item For every family $\seqi{E}$ of closed sets with $\bigcap_{j \in J}E_j \ne \emptyset$ for all $J \subset I$ finite, $\bigcap_{i \in I}E_i \ne \emptyset$.
+        \item Every filter in $X$ has a cluster point.
+        \item Every ultrafilter in $X$ converges.
+    \end{enumerate}
+    If the above holds, then $X$ is \textbf{compact}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For each $J \subset I$, let
+    \[
+    U_J = \bigcup_{j \in J}E_j^c
+    \]
+    then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.
+
+    Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
+    \[
+    \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
+    \]
+    is an open cover for $X$. By assumption, $U_J \subsetneq X$ for all $J \subset I$ finite. Thus $\mathbf{U}$ admits no finite subcover, contradiction.
+
+    (2) $\Rightarrow$ (3): Let $\fF \subset 2^X$ be a filter, then $\bracsn{\overline{E}| E \in \fF}$ satisfies the hypothesis of (2).
+
+    (3) $\Leftrightarrow$ (4): By \ref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
+
+    (3) $\Rightarrow$ (1): For each $J \subset I$, let
+    \[
+    E_J = \bigcap_{j \in J}U_j^c
+    \]
+    then for each $J, J' \subset I$, $E_J \cap E_{J'} = E_{J \cup J'}$. Assume for contradiction that $\mathbf{U}$ admits no finite subcover. Let
+    \[
+    \fB = \bracs{E_H|J \subset I \text{ finite}}
+    \]
+    then $\fB$ is a filter base consisting of closed sets. By assumption, there exists $x \in \bigcap_{i \in I}U_j^c$, so $\mathbf{U}$ is not an open cover, contradiction.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:compact-extensions}
+    Let $X$ be a topological space and $E, F \subset X$ be compact, then the following sets are compact:
+    \begin{enumerate}
+        \item $E \cup F$.
+        \item $G \subset E$ closed.
+        \item For any topological space $Y$ and $f \in C(E; Y)$, $f(E)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\seqi{U}$ be an open cover of $E \cup F$, then there exists $J_1, J_2 \subset I$ finite such that $\bigcup_{j \in J_1}U_j \supset E$ and $\bigcup_{j \in J_2}U_j \supset F$. In which case, $\bigcup_{j \in J_1 \cup J_2}U_j \supset E \cup F$.
+
+    (2): Let $\seqi{U}$ be an open cover of $G$. Since $G$ is closed, $\seqi{U} \cup \bracs{G^c}$ is an open cover of $E$. In which case, there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_j \cup G^c \supset E$, so $\bigcup_{j \in J}U_j \supset G$.
+
+    (3): Let $\seqi{U}$ be an open cover of $f(E)$, then $\bracs{f^{-1}(U_i)}_{i \in I}$ is an open cover for $E$. Thus there exists $J \subset I$ finite such that $\bigcup_{j \in J}f^{-1}(U_j) \supset E$, so $\bigcup_{j \in J}U_j \supset f(E)$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:compact-closed}
+    Let $X$ be a Hausdorff space and $E \subset X$ be compact, then $E$ is closed.
+\end{proposition}
+\begin{proof}
+    Let $x \in E^c$. For each $y \in E$, there exists $U_y \in \cn(y)$ and $V_y \in \cn(x)$ such that $U_y \cap V_y = \emptyset$. By compactness, there exists $E_0 \subset E$ finite such that $\bigcup_{y \in E_0}U_y \supset E$. In which case, $E \cap \bigcap_{y \in E_0}V_y = \emptyset$ and $\bigcap_{y \in E_0}V_y \in \cn(x)$. Thus $E^c$ is open.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:compact-hausdorff-normal}
+    Let $X$ be a Hausdorff space, $A, B \subset X$ be compact with $A \cap B = \emptyset$, then there exists $U \in \cn(A)$ and $V \in \cn(B)$ such that $U \cap V = \emptyset$.
+
+    In particular, if $X$ is compact and Hausdorff, then $X$ is normal.
+\end{proposition}
+\begin{proof}
+    For each $x \in A$ and $y \in B$, there exists $U_{x, y} \in \cn(x)$ and $V_{x, y} \in \cn(y)$ with $U_{x, y} \cap V_{x, y} = \emptyset$.
+
+    Fix $x \in A$. By compactness of $B$, there exists $B_0(x) \subset B$ finite such that $\bigcup_{y \in B_0(x)}V_{x, y} \in \cn(B)$. Let $U_x = \bigcap_{y \in B_0}U_{x, y} \in \cn(x)$ and $V_x = \bigcup_{y \in B_0(x)}V_{x, y} \in \cn(B)$, then $U_x \cap V_x = \emptyset$.
+
+    By compactness of $A$, there exists $A_0 \subset A$ finite such that $\bigcup_{x \in A_0}U_x \in \cn(A)$. Let $U = \bigcup_{x \in A_0}U_x \in \cn(A)$ and $V = \bigcap_{x \in A_0}V_x \in \cn(B)$, then $A \cap B = \emptyset$.
+\end{proof}

src/topology/main/continuity.tex

@@ -32,3 +32,18 @@
 
     Global continuity, $(2) \Leftrightarrow (3)$: Follows from local continuity.
 \end{proof}
+
+\begin{lemma}[Gluing Lemma for Continuous Functions]
+\label{lemma:gluing-continuous}
+    Let $X, Y$ be topological spaces, $\seq{U_i} \subset 2^X$ be open sets, and $\seqi{f}$ with $f_i \in C(U; Y)$ for all $i \in I$. If:
+    \begin{enumerate}
+        \item[(a)] $\bigcup_{i \in I}U_i = X$.
+        \item[(b)] For each $i, j \in I$, either $U_i \cap U_j = \emptyset$, or $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$.
+    \end{enumerate}
+    then there exists a unique $f \in C(X; Y)$ such that $f|_{U_i} = f_i$ for all $i \in I$.
+\end{lemma}
+\begin{proof}
+    By \ref{lemma:glue-function}, there exists a unique $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
+
+    Let $x \in X$. By assumption (a), there exists $i \in I$ such that $x \in U_i$. Since $f_i \in C(U_i; Y)$, for any $V \in \cn_Y(f(x))$, there exists $U \in \cn_{U_i}(x)$ such that $f_i(U) \subset V$. As $U_i$ is open in $X$, $U \in \cn_X(x)$. Thus $f(U) = f_i(U) \subset V$, and $f$ is continuous at $x$.
+\end{proof}

src/topology/main/filters.tex

@@ -4,7 +4,7 @@
 
 \begin{definition}[Filter]
 \label{definition:filter}
-    Let $X$ be a set, a \textbf{filter} $\fF \subset X^2$ is a non-empty family of sets such that:
+    Let $X$ be a set, a \textbf{filter} $\fF \subset 2^X$ is a non-empty family of sets such that:
     \begin{enumerate}
         \item[(F1)] For any $E \in \fF$ and $X \supset F \supset E$, $F \in \fF$.
         \item[(F2)] For any $E, F \in \fF$, $E \cap F \in \fF$.
@@ -13,6 +13,7 @@
 \end{definition}
 
 
+
 \begin{definition}[Filter Base]
 \label{definition:filterbase}
     Let $X$ be a set, $\fF \subset 2^X$ be a filter, and $\fB \subset \fF$, then $\fB$ is a \textbf{filter base} for $\fF$ if for every $F \in \fF$, there exists $E \in \fB$ such that $E \subset F$.
@@ -91,17 +92,91 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     If $\fF' \supset \fB_0$ is a filter, then $\fF \supset \fB$ by (F2), and $\fF' \supset \fF$ by (F1). Thus $\fF$ is the smallest filtetr containing $\fB$.
 \end{proof}
 
-\begin{definition}[Convergence]
+\begin{lemma}
-\label{definition:filterconvergence}
+\label{lemma:filter-extend}
-    Let $X$ be a topological space, $\fB \subset 2^X$ be a filter base, and $x \in X$, then \textbf{$\fB$ converges} to $x$ if $\cn(x) \subset \fF(\fB)$.
+    Let $X$ be a set, $\fF, \mathfrak{G} \subset 2^X$ be filter subbases. If for any $E \in \fF$ and $F \in \mathfrak{G}$, $E \cap F \ne \emptyset$, then there exists a filter $\fU \supset \fF \cup \mathfrak{G}$.
+\end{lemma}
+\begin{proof}
+    Let $\seqf{E_j} \subset \fF$ and $\seqf[m]{F_j} \subset \mathfrak{G}$, then $\bigcap_{j = 1}^n E_j \in \fF$ and $\bigcap_{j = 1}^m F_j \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By \ref{definition:generatedfilter}, there exists a filter containing $\fF \cup \mathfrak{G}$.
+\end{proof}
 
-    If $A \subset X$ and $\fB \subset 2^A$, then $\fB$ converges to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
+\begin{definition}[Ultrafilter]
+\label{definition:ultrafilter}
+    Let $X$ be a set and $\fF \subset 2^X$ be a filter, then the following are equivalent:
+    \begin{enumerate}
+        \item $\fF$ is maximal with respect to inclusion.
+        \item For any $E \subset X$, either $E \in \fF$ or $E^c \in \fF$.
+        \item For any $\seqf{F_j} \subset X$ such that $\bigcup_{j = 1}^n F_j \in \fF$, there exists $1 \le j \le n$ such that $F_j \in \fF$.
+    \end{enumerate}
+    If the above holds, then $\fF$ is an \textbf{ultrafilter}.
 \end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $E \subset X$ with $E^c \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By \ref{lemma:filter-extend}, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E} \in \fU \subset \fF$.
+
+    (2) $\Rightarrow$ (1): Let $E \subset X$ with $E \not \in \fF$, then $E^c \in \fF$. Since $E \cap E^c = \emptyset$, there exists no filter containing $\fF \cup \bracs{E}$. Thus $\fF$ is maximal.
+
+    (2) $\Rightarrow$ (3): Suppose that $F_n \not\in \fF$, then $F_n^c \in \fF$ and $F_n^c \subset \bigcup_{j = 1}^{n - 1}F_n \in \fF$. Proceeding inductively, there must exist $1 \le j \le n$ such that $F_n \in \fF$.
+
+    (3) $\Rightarrow$ (2): $E \cup E^c = X \in \fF$.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:ultrafilter}
+    Let $X$ be a set and $\fF \subset 2^X$, then
+    \begin{enumerate}
+        \item There exists an ultrafilter $\fU \supset \fF$.
+        \item $\fF$ is the intersection of all ultrafilters containing it.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let
+    \[
+    \mathbf{U} = \bracs{\fU \supset \fF| \fU \subset 2^X \text{ is a filter}}
+    \]
+    and order it by inclusion. For any chain $\mathbf{C} \subset \mathbf{U}$, let $\fB = \bigcup_{\fF \in \textbf{C}}\fF$, then $\fB$ is a filter subbase, and the filter $\fU \in \mathbf{U}$ generated by $\fB$ is an upper bound for $\mathbf{C}$. By Zorn's lemma, $\textbf{U}$ has a maximal element.
+
+    (2): Let $E \subset X$. If there exists $F \in \fF$ with $F \cap E^c = \emptyset$, then $E \supset F$ and $E \in \fF$. Otherwise, $\fF \cup \bracs{E^c}$ is a filter subbase, and there exists an ultrafilter containing it by (1). In which case, there exists an ultrafilter $\fU \supset \fF$ with $E \not\in \fU$.
+\end{proof}
+
+
+\begin{definition}[Convergence]
+\label{definition:filter-convergence}
+    Let $X$ be a topological space, $\fF \subset 2^X$ be a filter with base $\fB \subset 2^X$, and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\fF \supset \cn(x)$.
+        \item For each ultrafilter $\fU \supset \fF$, $\fU \supset \cn(x)$.
+        \item There exists a fundamental system of neighbourhoods $\cb(x) \subset \cn(x)$ such that for every $E \in \cb$, there exists $F \in \fB$ with $F \subset E$.
+    \end{enumerate}
+    If the above holds, then $x$ is a \textbf{limit point} of $\fB$, and $\fB$ \textbf{converges} to $x$.
+
+    If $A \subset X$ and $\fB \subset 2^A$, then $\fB$ \textbf{converges} to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
+\end{definition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): By (2) of \ref{lemma:ultrafilter}.
+
+    (1) $\Rightarrow$ (3): $\cn(x)$ is a fundamental system of neighbourhoods at $x$.
+
+    (3) $\Rightarrow$ (1): For any $U \in \cn(x)$, there exists $B \in \cb(x)$ and $F \in \fB$ with $F \subset B \subset U$. In which case, $U \in \fF$.
+\end{proof}
 
 \begin{definition}[Accumulation Point]
-\label{definition:accumulationpoint}
+\label{definition:accumulation-point}
-    Let $X$ be a topological space, $\fB \subset 2^X$ be a filter base, and $x \in X$, then $x$ is an \textbf{accumulation point} of $\fB$ if $x \in \bigcap_{E \in \fB}\ol{E}$.
+    Let $X$ be a topological space, $\fF \subset 2^X$ be a filter with base $\fB$, and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $x \in \bigcap_{E \in \fB}\overline{E}$.
+        \item $x \in \bigcap_{E \in \fF}\overline{E}$.
+        \item There exists a fundamental system of neighbourhoods $\cb(x) \subset \cn(x)$ such that for every $E \in \cb$ and $f \in \fB$, $E \cap F \ne \emptyset$.
+        \item There exists a filter $\fU \supset \fB$ that converges to $x$.
+    \end{enumerate}
+    If the above holds, then $x$ is a \textbf{cluster/accumulation point} of $\fB$. In particular, if $\fF$ is an ultra filter, then (6) implies that the limit points and cluster points of $\fF$ coincide.
 \end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (3): Let $U \in \cn(x)$, then $U \cap E \ne \emptyset$ for all $E \in \fB$.
+
+    (3) $\Rightarrow$ (4): By \ref{lemma:filter-extend}, there exists a filter $\fU \supset \cb(x) \cup \fB$. Since $\cb(x)$ and $\fB$ are bases for $\cn(x)$ and $\fF$, respectively, $\fU \supset \cn(x) \cup \fF$.
+
+    (4) $\Rightarrow$ (1): Let $U \in \cn(x)$, then since $\fU \supset \fB \cup \cn(x)$, $U \cap E \ne \emptyset$ for all $E \in \fB \subset \fF \subset \fU$. Thus $x \in \bigcap_{E \in \fF}\ol{E}$.
+\end{proof}
 
 
 \begin{definition}[Limit]

src/topology/main/index.tex

@@ -10,5 +10,7 @@
 \input{./src/topology/main/product.tex}
 \input{./src/topology/main/hausdorff.tex}
 \input{./src/topology/main/regular.tex}
+\input{./src/topology/main/normal.tex}
+\input{./src/topology/main/compact.tex}
 \input{./src/topology/main/metric.tex}
 \input{./src/topology/main/baire.tex}

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}