\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{proposition}
\label{proposition:equicontinuous-net}
    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^X$, and $x \in X$, then the following are equivalent:
    \begin{enumerate}
        \item $\cf$ is equicontinuous at $x$. 
        \item For $\angles{x_\alpha}_{\alpha \in A} \subset X$ with $x_\alpha \to x$, $\angles{f_\alpha}_{\alpha \in A} \subset \cf$, and $U \in \fU$, there exists $\alpha_0 \in A$ such that $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  
        \item There exists a fundamental system of neighbourhoods $\fB \subset \cn_X(x)$ at $x$ such that for any $\angles{x_V}_{V \in \fB} \subset X$ with $x_\alpha \to x$, $\angles{f_V}_{V \in \fB} \subset \cf$, and $U \in \fU$, there exists $V_0 \in \fB$ such that $(f_V(x_V), f_V(x)) \in U$ for all $V \subset V_0$.  
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1) $\Rightarrow$ (2): Since $\cf$ is equicontinuous at $x$, there exists $V \in \cn_X(x)$ such that $(f_\alpha(y), f_\alpha(x)) \in U$ for all $y \in V$ and $\alpha \in A$. Given that $x_\alpha \to x$, there exists $\alpha_0 \in A$ such that $x_\alpha \in V$ for all $\alpha \ge \alpha_0$, so $(f_\alpha(x_\alpha), f_\alpha(x)) \in U$ for all $\alpha \ge \alpha_0$.  

    $\neg (1) \Rightarrow \neg (3)$: If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \fB$, there exists $f_V \in \cf$ and $x_V \in V$ with $(f_V(x_V), f_V(x)) \not\in U$. In which case, $x_V \to x$ but $(f_V(x_V), f_V(x)) \not\in U$ for all $V \in \cn_X(x)$. 
\end{proof}


\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 uniform structures of pointwise and compact convergence on $\cf$ coincide. 
        \item The closure of $\cf$ in $Y^X$ with respect to the product topology is equicontinuous. 
    \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=2]
        \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the uniform structure of compact convergence. 
    \end{enumerate}

    Conversely, if $X$ is a LCH space, then (C3) 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 uniform structures of pointwise and compact convergence coincide. 

    (E1) $\Rightarrow$ (C2): Let $\cf'$ be the closure of $\cf$ in $Y^X$ with respect to the product topology. Let $x \in X$ and entourage $U$ of $Y$. Using \autoref{proposition:goodentourages}, assume without loss of generality that $U$ is closed. Since $\cf$ is equicontinuous, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $f \in \cf$ and $y \in V$. For any element $g \in \cf'$, $(g(x), g(y)) \in \ol U = U$ for all $y \in V$. Therefore $\cf'$ is also equicontinuous. 

    (E1) + (E2) $\Rightarrow$ (C3): Using (C2), assume without loss of generality that $\cf$ is closed in $Y^X$ with respect to the product topology. In which case, $\cf$ is a closed subset of $\prod_{x \in X}\ol{\cf(x)}$ with respect to the product topology. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff} and \autoref{proposition:compact-extensions}, $\cf$ is compact in the product topology. By (C1), $\cf$ is also compact in the compact uniform topology.

    (C3) $\Rightarrow$ (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. 

    (C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the uniform structure of compact convergence, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
\end{proof}