Added notes on equicontinuity.

src/topology/uniform/equicontinuous.tex

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+\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}
+
+
+