Incremental update.

src/topology/main/index.tex

@@ -11,6 +11,8 @@
 \input{./src/topology/main/hausdorff.tex}
 \input{./src/topology/main/regular.tex}
 \input{./src/topology/main/normal.tex}
+\input{./src/topology/main/unity.tex}
 \input{./src/topology/main/compact.tex}
-\input{./src/topology/main/metric.tex}
+\input{./src/topology/main/sigma-compact.tex}
+\input{./src/topology/main/lch.tex}
 \input{./src/topology/main/baire.tex}

src/topology/main/interiorclosureboundary.tex

@@ -51,6 +51,15 @@
     Since $f$ is continuous, $f^{-1}(\ol{f(A)})$ is closed and contains $A$.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:closure-finite-union}
+    Let $X$ be a topological space and $\seqf{E_j} \subset 2^X$, then $\bigcup_{j = 1}^n \ol{E_j} = \ol{\bigcup_{j = 1}^n E_j}$.
+\end{proposition}
+\begin{proof}
+    Since $\bigcup_{j = 1}^n \ol{E_j}$ is closed, $\bigcup_{j = 1}^n \ol{E_j} \supset \ol{\bigcup_{j = 1}^n E_j}$. On the other hand, $\ol{\bigcup_{j = 1}^n E_j} \supset E_j$ for each $1 \le j \le n$. Thus $\ol{\bigcup_{j = 1}^n E_j} \supset \ol{E_j}$ for each $1 \le j \le n$, and $\ol{\bigcup_{j = 1}^n E_j} \supset \bigcup_{j = 1}^n\ol{E_j}$.
+\end{proof}
+
+
 
 \begin{definition}[Dense]
 \label{definition:dense}

src/topology/main/lch.tex

@@ -0,0 +1,123 @@
+\section{Locally Compact Hausdorff Spaces}
+\label{section:lch}
+
+\begin{definition}[Locally Compact Hausdorff Space]
+\label{definition:lch}
+    Let $X$ be a Hausdorff space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $x \in X$, there exists $K \in \cn(x)$ compact.
+        \item For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of compact sets.
+        \item For any $x \in X$, $\cn(x)$ admits a fundamental system of neighbourhoods consisting of precompact sets.
+    \end{enumerate}
+    If the above holds, then $X$ is a \textbf{locally compact Hausdorff (LCH)} space.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By \ref{proposition:compact-closed}, $K$ itself is closed, so $\overline{U \cap K} \subset K$ is a closed subset of a compact set, and compact by \ref{proposition:compact-extensions}.
+
+    (2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By \ref{proposition:compact-closed}, $K$ is closed, so $\overline{K^o} \subset K$ is compact by \ref{proposition:compact-extensions}.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:lch-compact-neighbour}
+    Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$.
+\end{lemma}
+\begin{proof}
+    For each $x \in K$, there exists $V_x \in \cn^o(x)$ be precompact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \ref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
+    \[
+    K \subset \bigcup_{j = 1}^n V_{x_j} \subset U
+    \]
+    By \ref{proposition:closure-finite-union},
+    \[
+    \ol{\bigcup_{j = 1}^n V_{x_j}} = \bigcup_{j = 1}^n \overline{V_{x_j}} \subset U
+    \]
+    so $V = \bigcup_{j = 1}^n V_{x_j} \in \cn^o(K)$ is precompact.
+\end{proof}
+
+\begin{lemma}[Urysohn's Lemma (LCH), {{\cite[Lemma 4.32]{Folland}}}]
+\label{lemma:lch-urysohn}
+    Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $f \in C_c(X; [0, 1])$ such that $\supp{f} \subset U$.
+\end{lemma}
+\begin{proof}
+    By \ref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
+    \[
+    K \subset V \subset \ol{V} \subset W \subset \ol{W} \subset U
+    \]
+    As $\ol{W}$ is compact, it is normal by \ref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \ref{proposition:compact-closed}.
+
+    By Urysohn's lemma (\ref{lemma:urysohn}), there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_K = 1$ and $f|_{\ol{W} \setminus V} = 0$. Let
+    \[
+    F: X \to [0, 1] \quad x \mapsto \begin{cases}
+    f(x) &x \in W \\
+    0 &x \in X \setminus \ol{V}
+    \end{cases}
+    \]
+    then by the gluing lemma for continuous functions (\ref{lemma:gluing-continuous}), $F \in C_c(X; [0, 1])$ with $F|_{K} = 1$ and $\supp{f} \subset U$.
+\end{proof}
+
+\begin{theorem}[Tietze Extension Theorem (LCH)]
+\label{theorem:lch-tietze}
+    Let $X$ be a LCH space, $K \subset X$ be compact, $U \in \cn^o(K)$, and $f \in C(K; \real)$, then there exists $F \in C_c(U; \real)$ such that $F|_K = f$.
+\end{theorem}
+\begin{proof}
+    By \ref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \ref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \ref{proposition:compact-closed}.
+
+    By the Tietze extension theorem (\ref{theorem:tietze}), there exists $F \in C(\ol{W}; \real)$ such that $F|_K = f$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\eta \in C_c(X; [0, 1])$ such that $\eta|_K = 1$ and $\supp{\eta} \subset V$. In which case, define
+    \[
+    \ol F: X \to \real \quad x \mapsto \begin{cases}
+        \eta(x) \cdot f(x) &x \in V \\
+        0 &x \in X \setminus \supp{\eta}
+    \end{cases}
+    \]
+    then by the gluing lemma for continuous functions (\ref{lemma:gluing-continuous}), $\ol F \in C_c(X; \real)$ with $\ol F|_K = F|_K = f$ and $\supp{F} \subset \supp{\eta} \subset V \subset U$.
+\end{proof}
+
+\begin{proposition}[{{\cite[Proposition 4.39]{Folland}}}]
+\label{proposition:lch-sigma-compact}
+    Let $X$ be a LCH space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is $\sigma$-compact.
+        \item There exists an exhaustion of $X$ by compact sets.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $\seq{K_n} \subset 2^X$ be compact such that $\bigcup_{n \in \natp}K_n = X$, and $U_0 = \emptyset$.
+
+    Assume inductively that $\bracs{U_j}_0^n$ has been constructed such that:
+    \begin{enumerate}
+        \item[(a)] For each $0 \le k \le n$, $U_k$ is a precompact open set.
+        \item[(b)] For each $0 \le k < n$, $\overline{U_k} \subset U_{k+1}$.
+        \item[(c)] For each $1 \le k \le n$, $U_k \supset \bigcup_{j = 1}^k K_j$.
+    \end{enumerate}
+    By \ref{lemma:lch-compact-neighbour}, there exists $U_{n+1} \in \cn^o(\overline{U_n} \cup K_{n+1})$ precompact. In which case, by (c),
+    \[
+    U_{n+1} \supset \ol{U_n} \cup K_{n+1} \supset \bigcup_{j = 1}^n K_j \cup K_{n+1} = \bigcup_{j = 1}^{n+1}K_j
+    \]
+    Thus $\bracs{U_j}_0^{n+1}$ satisfies (a), (b), and (c), and $\seq{U_n}$ is an exhaution of $X$ by compact sets.
+\end{proof}
+
+\begin{proposition}[{{\cite[Proposition 4.41]{Folland}}}]
+\label{proposition:lch-partition-of-unity}
+    Let $X$ be a LCH space, $K \subset X$ be compact, and $\seqi{U}$ be an open cover of $K$, then there exists a $C_c$ partition of unity on $K$ subordinate to $\seqi{U}$.
+\end{proposition}
+\begin{proof}
+    Since $K$ is compact, assume without loss of generality that $\seqi{U} = \seqf{U_j}$.
+
+    For every $x \in K$, there exists $1 \le j \le n$ and $N_x \in \cn(x)$ compact such that $x \in N_x \subset U_j$. By compactness of $K$, there exists $\seqf[m]{x_j} \subset K$ such that $K = \bigcup_{j = 1}^m N_{x_j}$.
+
+    For each $1 \le j \le n$, let
+    \[
+    F_j = \bigcup_{\substack{1 \le k \le m \\ N_{x_k} \subset U_j}}N_{x_k}
+    \]
+    then $F_j \subset U_j$ is compact, and $\bigcup_{j = 1}^n F_j \supset K$.
+
+    By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\seqf{f_j} \subset C_c(X; [0, 1])$ such that for each $1 \le j \le n$, $f_j|_{F_j} = 1$, and $\supp{f_j} \subset U_j$.
+
+    By Urysohn's lemma again, there exists $f_{j + 1} \in C(X; [0, 1])$ such that $f_{j+1}|_{K} = 0$ and $\bracs{f_{j+1} = 0} \subset \bigcup_{j = 1}^n \supp{f_j}$. Let $F = \sum_{j = 1}^{n+1}f_j$, then $F(x) > 0$ for all $x \in X$. For each $1 \le j \le n$, let $g_j = f_j/F$, then $g_j \in C_c(X; [0, 1])$ with $\supp{g_j} \subset U_j$. In addition, since $f_{j+1}|_K = 0$,
+    \[
+    \sum_{j = 1}^n g_j|_K = \frac{\sum_{j = 1}^n f_j}{F} = \frac{\sum_{j = 1}^n f_j}{\sum_{j = 1}^n f_j} = 1
+    \]
+    Therefore $\seqf{g_j}$ is the desired partition of unity.
+\end{proof}
+
+
+LOCALLY FINITE EXHAUSTION

src/topology/main/sigma-compact.tex

@@ -0,0 +1,16 @@
+\section{$\sigma$-Compact Spaces}
+\label{section:sigma-compact}
+
+\begin{definition}[$\sigma$-Compact]
+\label{definition:sigma-compact}
+    Let $X$ be a topological space, then $X$ is \textbf{$\sigma$-compact} if there exits $\seq{K_n} \subset 2^X$ compact such that $X = \bigcup_{n \in \natp}K_n$.
+\end{definition}
+
+\begin{definition}[Exhaustion by Compact Sets]
+\label{definition:compact-exhaustion}
+    Let $X$ be a topological space and $\seq{U_n} \subset 2^X$, then $\seq{U_n}$ is an \textbf{exhaustion of $X$ by compact sets} if:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $U_n$ is open and precommpact.
+        \item For each $n \in \natp$, $\ol{U_n} \subset U_{n+1}$.
+    \end{enumerate}
+\end{definition}

src/topology/main/unity.tex

@@ -0,0 +1,14 @@
+\section{Partitions of Unity}
+\label{section:partition-of-unity}
+
+
+\begin{definition}[Partition of Unity]
+\label{definition:partition-of-unity}
+    Let $X$ be a topological space and $E \subset X$, then a \textbf{partition of unity} on $E$ is a family $\seqi{f} \subset C(X; [0, 1])$ such that:
+    \begin{enumerate}
+        \item For each $x \in X$, there exists $U \in \cn(x)$ such that $\bracs{i \in I|f_i|_U \ne 0}$ is finite.
+        \item $\sum_{i \in I}f_i|_E = 1$.
+    \end{enumerate}
+
+    For any open cover $\mathcal U$ of $X$, $\seqi{f}$ is \textbf{subordinate} to $\mathcal U$ if for every $i \in I$, there exists $U \in \mathcal U$ such that $\supp{f_i} \subset \mathcal U$.
+\end{definition}

src/topology/uniform/metric.tex

@@ -22,7 +22,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    (1) $\Rightarrow$ (2): Let $r > 0$, then there exists $V \in \fU$ symmetric such that for any $(x, x'), (y, y') \in V$, $\abs{d(x, y) - d(x', y')} < r$. In particular, for any $(x, y) \in V$, $(x, x), (x, y) \in V$. Thus $d(x, y) < d(x, x) + r = r$. Therefore $V \subset U_r$, and $U_r \in \fU$.
+    (1) $\Rightarrow$ (2): Let $r > 0$, then there exists $V \in \fU$ symmetric such that for any $(x, x'), (y, y') \in V$, $\abs{d(x, y) - d(x', y')} < r$. In particular, for any $(x, y) \in V$, $(x, x), (x, y) \in V$. Thus $d(x, y) < d(x, x) + r = r$, $V \subset U_r$, and $U_r \in \fU$.
 
     (2) $\Rightarrow$ (1): Let $r > 0$, then for any $(x, x'), (y, y') \in U_{r/2}$, $\abs{d(x, y) - d(x', y')} < r$ by the triangle inequality.
 \end{proof}
@@ -229,14 +229,14 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
     \]
     so $d$ induces the uniformity on $\fU$.
 
-    (3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity},
+    (3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity}, if
+    \[
+    U_{n, r} = \bracs{(x, y) \in X \times X| d_n(x, y) < r}
+    \]
+    then
     \[
     \fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r > 0}
     \]
-    where
-    \[
-    U_{j, r} = \bracs{(x, y) \in X \times X| d_n(x, y) < r}
-    \]
     is a fundamental system of entourages for $\fU$. Since for any $r > 0$, there exists $q \in \rational \cap (0, r)$,
     \[
     \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r \in \rational, r > 0}