Cleanup

src/topology/uniform/cauchy.tex

@@ -29,7 +29,7 @@
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.
 \end{proposition}
 \begin{proof}
-    Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \ref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$.
+    Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \autoref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$.
 \end{proof}
 
 \begin{definition}[Cauchy Continuous]
@@ -60,12 +60,13 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \ref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for
+    Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \autoref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for
     \[
     \fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF
     \]
 
-    To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \ref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.
+
+    To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \autoref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.
 
     Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$.
 \end{proof}
@@ -75,9 +76,9 @@
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets.
 \end{proposition}
 \begin{proof}
-    Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \ref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \ref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
+    Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \autoref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \autoref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$.
 
-    Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \ref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets.
+    Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \autoref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets.
 \end{proof}
 
 \begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}]
@@ -90,9 +91,9 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \ref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$.
+    (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \autoref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$.
 
-    (2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \ref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.
+    (2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \autoref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$.
 
     (3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$.
 \end{proof}