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src/topology/uniform/cauchy.tex

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+\section{Cauchy Filters}
+\label{section:cauchy-filter}
+
+\begin{definition}[Small]
+\label{definition:small}
+    Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is \textbf{$V$-small} if $A \times A \subset V$.
+\end{definition}
+
+\begin{lemma}[{{\cite[Proposition 2.3.1]{Bourbaki}}}]
+\label{lemma:small-intersect}
+    Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that:
+    \begin{enumerate}
+        \item $A, B$ are $V$-small.
+        \item $A \cap B \ne \emptyset$.
+    \end{enumerate}
+    then $A \cup B$ is $V \circ V$-small.
+\end{lemma}
+\begin{proof}
+    Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$.
+\end{proof}
+
+\begin{definition}[Cauchy Filter]
+\label{definition:cauchyfilter}
+    Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
+\end{definition}
+
+\begin{proposition}[Cauchy Criterion]
+\label{proposition:cauchycriterion}
+    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 $V(x) \times V(x) \subset V$.
+\end{proof}
+
+\begin{definition}[Cauchy Continuous]
+\label{definition:cauchy-continuous}
+    Let $X, Y$ be uniform spaces and $f: X \to Y$, then $f$ is \textbf{Cauchy continuous} if for any Cauchy filter base $\fB \subset 2^X$, $f(\fB) \subset 2^Y$ is Cauchy.
+\end{definition}
+
+
+\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}]
+\label{proposition:imagecauchy}
+    Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous.
+\end{proposition}
+\begin{proof}
+    Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^X$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$.
+\end{proof}
+
+\begin{definition}[Minimal Cauchy Filter]
+\label{definition:minimalcauchy}
+    Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion.
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}]
+\label{proposition:minimalcauchyexistence}
+    Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then:
+    \begin{enumerate}
+        \item There exists a unique minimal Cauchy filter $\fF_0$ such that $\fF_0 \subset \fF$.
+        \item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$.
+    \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
+    \[
+    \fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF
+    \]
+
+    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.
+
+    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}
+
+\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}]
+\label{proposition:cauchyinterior}
+    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 $V \in \fU$, 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.
+\end{proof}
+
+\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}]
+\label{proposition:cauchyfilterlimit}
+    Let $(X, \fU)$ be a uniform space, then:
+    \begin{enumerate}
+        \item For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter.
+        \item For each Cauchy filter $\fF \subset 2^X$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$.
+        \item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$.
+    \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 $\fF$.
+
+    (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 $\cn(x)$, $\fF \supset \cn(x)$.
+
+    (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}