Fixed equicontinuous formulation.

src/topology/uniform/equicontinuous.tex

@@ -14,13 +14,13 @@
     \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 For any upward-directed set $A$ with $|A| \le |\cn_X(x)|$, $\angles{x_\alpha}_{\alpha \in A} \subset X$ 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 $\alpha \ge \alpha_0$.  
 
-    $\neg (1) \Rightarrow \neg (3)$: Direct $\cn_X(x)$ under reverse inclusion. If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \cn_X(x)$, 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 $V \in \cn_X(x)$. 
+    $\neg (1) \Rightarrow \neg (3)$: 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}
 
 
@@ -38,7 +38,7 @@
 
     then
     \begin{enumerate}[label=(C\arabic*)]
-        \item The product uniformity and the compact uniformity on $\cf$ coincide. 
+        \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}
 
@@ -49,7 +49,7 @@
 
     then
     \begin{enumerate}[label=(C\arabic*), start=2]
-        \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. 
+        \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).
@@ -67,7 +67,7 @@
     \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. 
+    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. 
 
@@ -80,7 +80,7 @@
 
     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 compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
+    (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}