Added the Banach-Steinhaus theorem.

src/topology/uniform/equicontinuous.tex

@@ -13,7 +13,6 @@
     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
@@ -24,6 +23,7 @@
     then
     \begin{enumerate}[label=(C\arabic*)]
         \item The product uniformity and the compact uniformity 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
@@ -32,11 +32,11 @@
     \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. 
+    \begin{enumerate}[label=(C\arabic*), start=3]
+        \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. 
     \end{enumerate}
 
-    Conversely, if $X$ is a LCH space, then (C2) implies (E1) + (E2).
+    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 $\cf$.
@@ -53,32 +53,18 @@
 
     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. 
+    (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. 
 
-    (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 $y \in W$,  
+    (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. 
 
-    (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}. 
+    (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}. 
 \end{proof}
 
 
-\begin{corollary}
-\label{corollary:arzela-locally-compact}
-    Let $X$ be a LCH space, $Y$ be a uniform space, and $\cf \subset C(X; Y)$ such that:
-    \begin{enumerate}[label=(E\arabic*)]
-        \item $\cf$ is equicontinuous.
-        \item For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$.
-    \end{enumerate}
-
-    then $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. 
-\end{corollary}
-\begin{proof}
-    By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is a precompact subset of $Y^X$. By \autoref{proposition:lch-compactly-generated}, $C(X; Y)$ is a closed subset of $Y^X$. Therefore $\cf$ is a precompact subset of $C(X; Y)$. 
-\end{proof}
-
-
-