Added notes on equicontinuity.

src/topology/functions/set-systems.tex

@@ -1,4 +1,4 @@
-\section{Topology With Respect to Ideals}
+\section{Topologies With Respect to Ideals}
 \label{section:pointwise}
 
 \begin{definition}[Set-Open Topology]
@@ -132,3 +132,21 @@
 
     Let $S \in \sigma$ be non-empty and $V_0, V$ be symmetric entourages of $X$ with $V_0 \circ V_0 \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_0)$-small set $F \in \fF$. Let $x \in S$, then $\pi_x(\fF) \to f(x)$ implies that $F_x = \pi_x^{-1}(V_0(f(x))) \in \fF$. Since $F \cap F_x \ne \emptyset$, $(f(x), g(x)) \in V_0 \circ V_0 \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:compact-uniform-open}
+    Let $X$ be a topological space, $\kappa \subset 2^X$ be the collection of all precompact sets in $X$, and $(Y, \fU)$ be a uniform space, then the $\kappa$-open topology and $\kappa$-uniform topology on $C(X; Y)$ coincide. 
+\end{proposition}
+\begin{proof}
+    By \autoref{definition:set-uniform}, the $\kappa$-uniform topology is finer than the $\kappa$-open topology. 
+
+    Let $K \in \kappa$ be compact, $U \in \fU$ be symmetric, and $f \in C(X; Y)$. For each $x \in K$, there exists $V_x \in \cn_X(x)$ such that $f(V_x) \subset U(f(x))$. Since $K$ is compact, there exists $\seqf{x_j} \subset X$ such that $K \subset \bigcup_{j = 1}^n V_{x_j}$. 
+
+    Let $g \in \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j)))$ and $y \in K$, then there exists $1 \le j \le n$ such that $y \in V_{x_j}$. In which case, since $f(y), g(y) \in U(f(x_j))$, $(f(y), g(y)) \in U \circ U$. Therefore
+    \[
+    f \in  \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j))) \subset [E(K, U \circ U)](f)
+    \]
+
+    and the $\kappa$-open topology is finer than the $\kappa$-uniform topology.
+\end{proof}
+

src/topology/uniform/compact.tex

@@ -0,0 +1,74 @@
+\section{Compact Uniform Spaces}
+\label{section:compact-uniform}
+
+\begin{definition}
+\label{definition:totally-bounded-uniform}
+    Let $(X, \fU)$ be a uniform space, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $U \in \fU$, there exists $\seqf{x_j} \subset U$ such that $X = \bigcup_{j = 1}^n U(x_j)$. 
+        \item Every ultrafilter on $X$ is Cauchy.
+        \item For every filter $\fF_0 \subset 2^X$, there exists $\fF \supset \fF_0$ such that $\fF$ is Cauchy. 
+    \end{enumerate}
+
+    If the above holds, then $(X, \fU)$ is \textbf{totally bounded}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $\fF \subset 2^X$ be an ultrafilter and $U \in \fU$, then there exists $U$-small sets $\seqf{E_j} \subset 2^X$ such that $X = \bigcup_{j = 1}^n E_j$. By (3) of \autoref{definition:ultrafilter}, there exists $1 \le j \le n$ such that $E_j \in \fF$, so $\fF$ is Cauchy.
+
+    (2) $\Rightarrow$ (3): By the \hyperref[ultrafilter lemma]{lemma:ultrafilter}. 
+
+    $\neg (1) \Rightarrow \neg (3)$: Let $U \in \fU$ be symmetric such that for all $X_0 \subset X$ finite, $\bigcup_{x \in X_0}U(x) \subsetneq X$. Let $\fF_0$ be the filter generated by
+    \[
+    \bracs{X \setminus \bigcup_{x \in X_0}U(x) \bigg | X_0 \subset X \text{ finite}}
+    \]
+
+    then for any $U$-small set $E \subset X$ and $x \in E$, $E \cap [X \setminus U(x)] = \emptyset$. Therefore no filter finer than $\fF_0$ contains a $U$-small set. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:totally-bounded-image}
+    Let $X, Y$ be uniform spaces and $f \in UC(X; Y)$. If $X$ is totally bounded, then so is $f(X)$. 
+\end{proposition}
+\begin{proof}
+    Let $U$ be an entourage of $Y$, then there exists $\seqf{x_j} \subset X$ such that $\bigcup_{j = 1}^n[(f \times f)^{-1}(U)](x_j) = X$. In which case, $f(x) \subset \bigcup_{j = 1}^n U(f(x_j))$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:compact-uniform}
+    Let $X$ be a uniform space, then the following are equivalent: 
+    \begin{enumerate}
+        \item $X$ is compact. 
+        \item $X$ is complete and totally bounded. 
+    \end{enumerate}
+
+    In particular, $X$ is totally bounded if and only if its Hausdorff completion is compact. 
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By (1) of \autoref{definition:compact}, $X$ is totally bounded. Let $\fF$ be a Cauchy filter, then by (4) of \autoref{definition:compact}, $\fF$ has at least one accumulation point. Since $\fF$ is Cauchy, the accumulation points and limit points of $\fF$ coincide by (2) of \autoref{proposition:cauchyfilterlimit}. 
+
+    (2) $\Rightarrow$ (1): By (2) of \autoref{definition:totally-bounded-uniform}, every ultrafilter is Cauchy, and hence converges, which satisfies (4) of \autoref{definition:compact}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:totally-bounded-product}
+    Let $\seqi{X}$ be totally bounded uniform spaces, then $\prod_{i \in I}X_i$ is totally bounded. 
+\end{proposition}
+\begin{proof}
+    Let $J \subset I$ be finite and $\seqj{U}$ such that $U_j$ is an entourage of $X_j$ for each $j \in J$. Since each $X_j$ is totally bounded, there exists $Y_j \subset X_j$ finite such that $X_j = U_j(Y_j)$. Let $Y = \prod_{j \in J}Y_j$ and $\phi_y \in \pi_J^{-1}(y)$ for each $y \in Y$, then for any $x \in X$, there exists $y \in Y$ such that $\pi_j(x) \in U_j(\pi_j(y))$ for each $j \in J$. Therefore
+    \[
+    X = \bigcup_{x \in Y}\braks{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)}(\phi_x)
+    \]
+\end{proof}
+
+
+
+\begin{proposition}
+\label{proposition:uniform-continuous-compact}
+    Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$.
+\end{proposition}
+\begin{proof}
+    Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. 
+    
+    Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. 
+\end{proof}

src/topology/uniform/equicontinuous.tex

@@ -0,0 +1,69 @@
+\section{Equicontinuity}
+\label{section:equicontinuity}
+
+\begin{definition}[Equicontinuous]
+\label{definition:equicontinuous}
+    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^X$, and $x \in X$, then $\cf$ is \textbf{equicontinuous at $x$} if for every $U \in \fU$, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $y \in V$ and $f \in \cf$. 
+    
+    The set $\cf \subset C(X; Y)$ is \textbf{equicontinuous} if it is equicontinuous at every point in $x$. 
+\end{definition}
+
+\begin{definition}[Uniformly Equicontinuous]
+\label{definition:uniformly-equicontinuous}
+    Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is \textbf{uniformly equicontinuous} if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$. 
+\end{definition}
+
+
+\begin{theorem}[Arzelà-Ascoli]
+\label{theorem:arzela-ascoli}
+    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, and $\cf \subset C(X; Y)$. If
+    \begin{enumerate}[label=(E\arabic*)]
+        \item $\cf$ is equicontinuous.
+    \end{enumerate}
+
+    then
+    \begin{enumerate}[label=(C\arabic*)]
+        \item The product uniformity and the compact uniformity on $\cf$ coincide. 
+    \end{enumerate}
+
+    In addition, if $\cf$ satisfies (E1) and
+    \begin{enumerate}[label=(E\arabic*), start=1]
+        \item For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$.
+    \end{enumerate}
+
+    then
+    \begin{enumerate}[label=(C\arabic*), start=1]
+        \item $\cf$ is a precompact subset of $Y^X$ with respect to the compact uniformity. 
+    \end{enumerate}
+
+    Conversely, if $X$ is a LCH space, then (C2) implies (E1) + (E2).
+\end{theorem}
+\begin{proof}
+    (E1) $\Rightarrow$ (C1): By \autoref{proposition:compact-uniform-open}, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$ and thus $\cf$.
+
+    Let $K \subset X$ be compact, and $U \in \fU$. Since $\cf$ is equicontinuous, for each $x \in K$, there exists $V_x \in \cn_X(x)$ such that $g(V_x) \subset U(g(x))$ for all $g \in \cf$. By compactness of $K$, there exists $\seqf{x_j} \subset K$ such that $K \subset \bigcup_{j = 1}^n V_{x_j}$. Let $f, g \in \cf$ such that $(f(x_j), g(x_j)) \in E$ for all $1 \le j \le n$. For any $x \in K$, there exists $1 \le j \le n$ such that $x \in V_{x_j}$. In which case, 
+    \[
+    (f(x), f(x_j)), (f(x_j), g(x_j)), (g(x_j), g(x)) \in U
+    \]
+
+    so $(f(x), g(x)) \in U \circ U \circ U$. Therefore
+    \[
+    \bigcap_{j = 1}^n E(\bracs{x_j}, U) \subset E(K, U \circ U \circ U)
+    \]
+
+    so the product uniformity and the compact uniformity coincide. 
+
+    (E1) + (E2) $\Rightarrow$ (C2): By \autoref{proposition:totally-bounded-product}, $\prod_{x \in X}\cf(x)$ is totally bounded. Since $\cf \subset \prod_{x \in X}\cf(x)$, $\cf$ is also totally bounded. As the pointwise and compact uniformities coincide, $\cf$ is totally bounded with respect to the compact uniformity. By \autoref{proposition:product-complete} and \autoref{proposition:compact-uniform}, $\prod_{x \in X}\ol{\cf(x)}$ is compact and hence complete. Therefore the closure of $\cf$ with respect to the compact uniformity is compact. 
+
+    (C2) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
+    \[
+    (g(x), f_j(x)), (f_j(x), f_j(y)), (f_j(y), g(y)) \in U \circ U \circ U
+    \]
+
+    Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous. 
+
+    (C2) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
+\end{proof}
+
+
+

src/topology/uniform/index.tex

@@ -4,6 +4,8 @@
 \input{./definition.tex}
 \input{./uc.tex}
 \input{./metric.tex}
+\input{./compact.tex}
+\input{./equicontinuous.tex}
 \input{./cauchy.tex}
 \input{./complete.tex}
 \input{./completion.tex}

src/topology/uniform/uc.tex

@@ -164,14 +164,3 @@
 
 \end{proof}
 
-
-\begin{proposition}
-\label{proposition:uniform-continuous-compact}
-    Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$.
-\end{proposition}
-\begin{proof}
-    Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. 
-    
-    Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. 
-\end{proof}
-