Performed housekeeping for topological groups.

src/topology/groups/definition.tex

@@ -0,0 +1,131 @@
+\section{Group Topologies}
+\label{section:group-topologies}
+
+
+\begin{definition}[Topological Group]
+\label{definition:topological-group}
+    Let $G$ be a group and $\mathcal{T} \subset 2^G$ be a topology. If
+    \begin{enumerate}[label=(TG\arabic*)]
+        \item The composition map $G \times G \to G$ with $(g, h) \mapsto gh$ is continuous. 
+        \item The inversion map $G \to G$ with $g \mapsto g^{-1}$ is continuous. 
+    \end{enumerate}
+    
+    then the pair $(E, \mathcal{T})$ is a \textbf{topological group}. 
+\end{definition}
+
+\begin{definition}[Translation-Invariant Topology]
+\label{definition:translation-invariant-topology}
+    Let $G$ be a group and $\topo \subset 2^G$ be a topology, then $\topo$ is \textbf{left translation-invariant} if for every $U \in \topo$ and $g \in G$, $gU \in \topo$, and \textbf{right translation-invariant} if for every $U \in \topo$ and $g \in G$, $Ug \in \topo$. If $\topo$ is both left and right translation-invariant, then $\topo$ is \textbf{translation-invariant}.
+\end{definition}
+
+\begin{definition}[Translation-Invariant Uniformity]
+\label{definition:translation-invariant-uniformity}
+    Let $G$ be a group, $\fU$ be a uniformity on $G$, and $U \in \fU$, then $U$ is \textbf{left translation-invariant} if for every $z \in G$,
+    \[
+    U = zU = \bracs{(zx, zy)|(x, y) \in U}
+    \]
+
+    and \textbf{right translation-invariant} if for every $z \in G$, 
+    \[
+    U = Uz = \bracs{(xz, yz)|(x, y) \in U}
+    \]
+
+    The uniformity $\fU$ is \textbf{left/right translation-invariant} if it admits a fundamental system of left/right translation-invariant entourages. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:translation-invariant-symmetric}
+    Let $G$ be a group and $\fU \subset 2^{G \times G}$ be a left/right translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, left/right translation-invariant entourages.
+\end{lemma}
+\begin{proof}
+    Let $z \in G$, then the map $(x, y) \mapsto (zx, zy)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) = zU \cap zV$, and $U \cap V$ is left translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a left fundamental system of symmetric, translation-invariant entourages.
+\end{proof}
+
+\begin{definition}
+\label{definition:group-translation-invariant-uniformity}
+    Let $G$ be a topological group, then: 
+    \begin{enumerate}[label=(L\arabic*)]
+        \item There exists a unique left translation-invariant uniformity $\fU_L$ on $G$ that induces its topology. 
+        \item For each $U \in \cn_G(1)$, let $U_{L, V} = \bracsn{(x, y) \in G^2|x^{-1}y \in V}$, then $\fB_L = \bracs{U_{L, V}|V \in \cn_G(1)}$ is a fundamental system of entourages for $\fU_L$. 
+    \end{enumerate}
+
+    and
+    \begin{enumerate}[label=(R\arabic*)]
+        \item There exists a unique right translation-invariant uniformity $\fU_R$ on $G$ that induces its topology. 
+        \item For each $U \in \cn_G(1)$, let $U_{R, V} = \bracsn{(x, y) \in G^2|yx^{-1} \in V}$, then $\fB_R = \bracs{U_{R, V}|V \in \cn_G(1)}$ is a fundamental system of entourages for $\fU_R$. 
+    \end{enumerate}
+
+    The uniformities $\fU_L$ and $\fU_R$ are the \textbf{left} and \textbf{right uniformities} of $G$, respectively. The inversion map $G \to G$ with $g \mapsto g^{-1}$ is an isomorphism of the left and right uniformities. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[I.1.4]{SchaeferWolff}}}. ]
+    (L2): For each $V \in \cn_G(0)$, $U_{L, V}$ is left translation-invariant. 
+    \begin{enumerate}
+        \item[(FB1)] For each $V, V' \in \cn_G(1)$, $V \cap V' \in \cn_G(0)$, so $U_{L, V \cap V'} = U_{L, V} \cap U_{L, V'}$.
+        \item[(UB1)] For each $V \in \cn_G(1)$, $1 \in V$, so $\Delta \subset U_{L, V}$. 
+        \item[(UB2)] For each $V \in \cn_G(1)$, by (TG1), there exists $W \in \cn_G(1)$ such that $WW \subset V$. In which case, $U_{L, W} \circ U_{L, W} \subset U_{L, V}$. 
+    \end{enumerate}
+
+    By \autoref{proposition:fundamental-entourage-criterion}, $\fB_L$ forms a fundamental system of entourages for a left translation-invariant uniformity $\fU_L$ on $G$. 
+
+    (L1): Let $\fV$ be a left translation-invariant uniformity on $G$. For each symmetric, left translation-invariant entourage $V \in \fV$, and $g \in G$, 
+    \[
+    V(g) = \bracsn{h \in G| (g, h) \in V} = g V(1) = \bracs{h \in G| g^{-1}h \in V(1)}
+    \]
+
+    so $V = U_{L, V(1)}$. Therefore $\fV \subset \fU_L$ by \autoref{lemma:translation-invariant-symmetric}. 
+
+    On the other hand, for each $W_0 \in \cn_G(1)$, there exists a symmetric, left translation-invariant entourage $W \in \fV$ such that $W(1) \subset W_0$. In which case, $W = U_{L, W(1)} \subset U_{L, W_0(1)}$, and $\fV \supset \fU_L$. 
+\end{proof}
+
+\begin{definition}[Left/Right Continuous]
+\label{definition:left-right-continuous}
+    Let $G$ be a topological group, $Y$ be a uniform space, and $f: G \to Y$, then $f$ is \textbf{left/right uniformly continuous} if it is uniformly continuous with respect to the left/right uniformity of $G$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:tvs-closure}
+    Let $G$ be a topological group, $A \subset G$, and $\fB \subset \cn_G(1)$ be a fundamental system of neighbourhoods, then
+    \[
+    \ol{A} = \bigcap_{U \in \fB}\bracs{AU| U \in \fB} = \bigcap_{U \in \fB}\bracs{UA|U \in \fB}
+    \]
+
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:uniformclosure}. 
+\end{proof}
+
+
+\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}]
+\label{proposition:tvs-set-operations}
+    Let $G$ be a topological group and $A, B \subset G$, then
+    \begin{enumerate}
+        \item If $A$ is open, then $AB$ is open.
+        \item If $A$ is closed and $B$ is compact, then $AB$ is closed.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For every $x \in B$, $Ab$ is open by translation invariance, so
+    \[
+    AB = \bigcup_{x \in B}(Ab)
+    \]
+
+    is open.
+
+    (2): Let $x \in \overline{AB}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap AB \ne \emptyset$, so $UB^{-1} \cap A \ne \emptyset$, and $\fB = \bracsn{UB^{-1}| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
+    \[
+    y \in \bigcap_{U \in \fF}\overline{UB^{-1}} = \bigcap_{U \in \fF}\overline{UB^{-1}}
+    \]
+
+
+    By \autoref{proposition:tvs-closure}, $\overline{UB^{-1}} \subset UUB^{-1}$, so
+    \[
+    y \in \bigcap_{U \in \fF}\overline{UB^{-1}} \subset \bigcap_{U \in \fF}[UUB^{-1}]
+    \]
+
+    Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
+    \[
+    y \in \bigcap_{U \in \fF}[UUB^{-1}] \subset \bigcap_{V \in \cn_G(1)}[xVB^{-1}] = \overline{xB^{-1}} = xB^{-1}
+    \]
+
+    so $x \in yB \subset AB$.
+\end{proof}

src/topology/groups/index.tex

@@ -0,0 +1,5 @@
+\chapter{Topological Groups}
+\label{chap:topological-group}
+
+\input{./definitions.tex}
+

src/topology/index.tex

@@ -5,4 +5,5 @@
 \input{./uniform/index.tex}
 \input{./functions/index.tex}
 \input{./metric/index.tex}
+\input{./groups/index.tex}
 \input{./notation.tex}

src/topology/uniform/definition.tex

@@ -25,7 +25,7 @@
 \label{definition:slice}
     Let $X, Y$ be sets, $U \subset X \times Y$, and $A \subset X$, then
     \[
-    U(A) = \bracs{y \in Y: (x, y) \in U, x \in A}
+    U(A) = \bracs{y \in Y| (x, y) \in U, x \in A}
     \]
 
     is the \textbf{slice} of $U$ at $A$.