From 4b404d40a6192322ad96f5013862e876dd4f05aa Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Thu, 25 Jun 2026 13:58:38 -0400
Subject: [PATCH] Updated the A-A theorem to include convergence on a dense
 subset.

---
 src/topology/uniform/equicontinuous.tex | 21 ++++++++++++++-------
 1 file changed, 14 insertions(+), 7 deletions(-)

diff --git a/src/topology/uniform/equicontinuous.tex b/src/topology/uniform/equicontinuous.tex
index b8e638c..904cf8f 100644
--- a/src/topology/uniform/equicontinuous.tex
+++ b/src/topology/uniform/equicontinuous.tex
@@ -38,7 +38,12 @@
 
     then
     \begin{enumerate}[label=(C\arabic*)]
-        \item The uniform structures of pointwise and compact convergence on $\cf$ coincide. 
+        \item The following uniformities on $\cf$ coincide:
+        \begin{enumerate}[label=(\roman*)]
+            \item The uniform structure of pointwise convergence.
+            \item For any $D \subset X$ dense, the uniform structure of pointwise convergence on $D$. 
+            \item The uniform structure of compact convergence. 
+        \end{enumerate}
         \item The closure of $\cf$ in $Y^X$ with respect to the product topology is equicontinuous. 
     \end{enumerate}
 
@@ -55,19 +60,21 @@
     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 $\cf$.
+    (E1) $\Rightarrow$ (C1): It is sufficient to show that (ii) is finer than (iii).
 
-    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, 
+    Let $K \subset X$ be compact, and $U \in \fU$. Since $\cf$ is equicontinuous, for each $x \in X$, 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 $K \subset \bigcup_{j = 1}^n V_{x_j}$. Since $D$ is dense, there exists $\seqf{y_j} \subset D$ such that $y_j \in V_{x_j}$ for each $1 \le j \le n$. 
+    
+    Let $f, g \in \cf$ such that $(f(y_j), g(y_j)) \in E$ 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, 
     \[
-    (f(x), f(x_j)), (f(x_j), g(x_j)), (g(x_j), g(x)) \in U
+    (f(x), f(x_j)), (f(x_j), f(y_j)), (f(y_j), g(y_j)), (g(y_j), g(x_j)), (g(x_j), g(x)) \in U
     \]
 
-    so $(f(x), g(x)) \in U \circ U \circ U$. Therefore
+    so $(f(x), g(x)) \in U \circ U \circ U \circ U \circ U$. Therefore
     \[
-    \bigcap_{j = 1}^n E(\bracs{x_j}, U) \subset E(K, U \circ U \circ U)
+    \bigcap_{j = 1}^n E(\bracs{x_j}, U) \subset E(K, U \circ U \circ U \circ U \circ u)
     \]
 
-    so the uniform structures of pointwise and compact convergence coincide. 
+    and the uniformity of pointwise convergence on $D$ is finer than the uniformity of compact convergence. 
 
     (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.