Performed housekeeping for topological groups.

src/fa/tvs/definition.tex

@@ -6,18 +6,13 @@
 \label{definition:tvs}
     Let $E$ be a vector space over $K \in \bracs{\real, \complex}$ and $\topo \subset 2^E$ be a topology. If
     \begin{enumerate}
-        \item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
-        \item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
+        \item[(TVS1)] The addition map $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
+        \item[(TVS2)] The scalar multiplication map $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
     \end{enumerate}
 
     then the pair $(E, \topo)$ is a \textbf{topological vector space}.
 \end{definition}
 
-\begin{definition}[Translation-Invariant Topology]
-\label{definition:translation-invariant-topology}
-    Let $E$ be a vector space and $\topo$ be a topology on $E$, then $\topo$ is \textbf{translation-invariant} if for any $U \in \topo$ and $y \in E$, $U + y \in \topo$.
-\end{definition}
-
 \begin{lemma}
 \label{lemma:tvs-translation-invariant}
     Let $E$ be a TVS over $K \in \RC$, then the topology of $E$ is translation-invariant.
@@ -26,26 +21,8 @@
     Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open.
 \end{proof}
 
-\begin{definition}[Translation-Invariant Uniformity]
-\label{definition:translation-invariant-uniformity}
-    Let $E$ be a vector space, $\fU$ be a uniformity on $E$, and $U \in \fU$, then $U$ is \textbf{translation-invariant} if for every $z \in E$,
-    \[
-    U = \bracs{(x + z, y + z)|(x, y) \in U}
-    \]
 
-    and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
-\end{definition}
-
-\begin{lemma}
-\label{lemma:translation-invariant-symmetric}
-    Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
-\end{lemma}
-\begin{proof}
-    Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
-\end{proof}
-
-
-\begin{proposition}[{{\cite[I.1.4]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:tvs-uniform}
     Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then:
     \begin{enumerate}
@@ -56,73 +33,7 @@
     The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
 \end{proposition}
 \begin{proof}
-    (2): Firstly, for any $V \in \fB_0$, $U_V$ is translation-invariant.
-    \begin{enumerate}
-        \item[(FB1)] For any $V, V' \in \fB_0$, there exists $W \in \fB_0$ such that $W \subset V \cap V'$. In which case, $U_V \cap U_{V'} \supset U_W \in \fB$.
-        \item[(UB1)] For any $V \in \fB_0$, $0 \in V$, so $\Delta \subset U_V$.
-        \item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
-    \end{enumerate}
-
-    By \autoref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
-
-    (1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
-
-    Let $W \in \cn(0)$, then by \autoref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
-
-    Let $V \in \mathfrak{V}$. Using \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
-\end{proof}
-
-\begin{proposition}
-\label{proposition:tvs-closure}
-    Let $E$ be a TVS over $K \in \RC$, $A \subset E$, and $\fB \subset \cn(0)$ be a fundamental system of neighbourhoods, then
-    \[
-    \ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
-    \]
-
-\end{proposition}
-\begin{proof}
-    Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
-
-    Assume without loss of generality that $\fB$ consists of symmetric entourages. By \autoref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \autoref{proposition:uniformclosure} implies that
-    \[
-    \ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
-    \]
-
-\end{proof}
-
-\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}]
-\label{proposition:tvs-set-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then:
-    \begin{enumerate}
-        \item If $A$ is open, then $A + B$ is open.
-        \item If $A$ is closed and $B$ is compact, then $A + B$ is closed.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    $(1)$: For every $x \in B$, $A + x$ is open by \autoref{definition:translation-invariant-topology}, so
-    \[
-    A + B = \bigcup_{x \in B}(A + x)
-    \]
-
-    is open.
-
-    $(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
-    \[
-    y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
-    \]
-
-
-    By \autoref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
-    \[
-    y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
-    \]
-
-    Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
-    \[
-    y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
-    \]
-
-    so $x \in y + B \subset A + B$.
+    By \autoref{definition:group-translation-invariant-uniformity}. 
 \end{proof}
 
 \begin{definition}[Balanced/Circled]

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 $A$.