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}
+