Fixed equicontinuous formulation.

src/fa/lc/barrel.tex

@@ -25,14 +25,14 @@
 
 \begin{summary}
 \label{summary:barreled-space}
-    The following types of locally convex spaces are barreled:
+    The following types of locally convex spaces are barrelled:
     \begin{enumerate}
         \item Every locally convex space with the Baire property.
         \item Every Banach space and every Fréchet space. 
-        \item Inductive limits of barreled spaces. 
+        \item Inductive limits of barrelled spaces. 
         \item Spaces of type (LB) and (LF). 
-        \item The locally convex direct sum of barreled spaces. 
-        \item Products of barreled spaces.
+        \item The locally convex direct sum of barrelled spaces. 
+        \item Products of barrelled spaces.
     \end{enumerate}
 \end{summary}
 \begin{proof}
@@ -46,7 +46,7 @@
 
 \begin{proposition}
 \label{proposition:baire-barrel}
-    Let $E$ be a locally convex space over $K \in \RC$. If $E$ is a Baire space, then $E$ is barreled. 
+    Let $E$ be a locally convex space over $K \in \RC$. If $E$ is a Baire space, then $E$ is barrelled. 
 \end{proposition}
 \begin{proof}[Proof, {{\cite[II.7.1]{SchaeferWolff}}}. ]
     Let $D \subset E$ be a Barrel, then $E = \bigcup_{n \in \natp}nD$ is a countable union of closed sets. Since $E$ is Baire, there exists $n \in \natp$, $U \in \cn_E(0)$ circled, and $x \in E$ such that $x + U \in nB$. In which case, 
@@ -59,9 +59,9 @@
 
 \begin{proposition}
 \label{proposition:barrel-limit}
-    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then the inductive locally convex topology on $E$ induced by $\seqi{T}$ is barreled.  
+    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_i \in \hom(E_i; E)$ for all $i \in I$, then the inductive locally convex topology on $E$ induced by $\seqi{T}$ is barrelled.  
 \end{proposition}
 \begin{proof}[Proof, {{\cite[II.7.2]{SchaeferWolff}}}. ]
-    Let $D \subset E$ be a barrel, then for each $i \in I$, $T_i^{-1}(D) \subset E_i$ is also a barrel, and thus a neighbourhood of $0$ in $E_i$. By (5) of \autoref{definition:lc-inductive}, $D$ is a neighbourhood of $0$ in $E$, so $E$ is barreled. 
+    Let $D \subset E$ be a barrel, then for each $i \in I$, $T_i^{-1}(D) \subset E_i$ is also a barrel, and thus a neighbourhood of $0$ in $E_i$. By (5) of \autoref{definition:lc-inductive}, $D$ is a neighbourhood of $0$ in $E$, so $E$ is barrelled. 
 \end{proof}
 

src/fa/norm/normed.tex

@@ -73,21 +73,14 @@
 \label{theorem:uniform-boundedness}
     Let $E, F$ be normed vector spaces and $\mathcal{T} \subset L(E; F)$. If
     \begin{enumerate}
-        \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
-        \item $E$ is a Banach space.
+        \item[(B)] $E$ is a Banach space.
+        \item[(E2)] For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
     \end{enumerate}
 
     then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
 \end{theorem}
 \begin{proof}
-    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
-
-    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
-    \[
-    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
-    \]
-
-    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
+    By the \autoref{theorem:banach-steinhaus} theorem, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
 \end{proof}
 
 

src/fa/tvs/equicontinuous.tex

@@ -16,6 +16,15 @@
     (5) $\Rightarrow$ (1): Let $V \in \cn_F(0)$, then $U = \bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. Thus for any $x, y \in E$ with $x - y \in U$, $Tx - Ty \in V$ for all $T \in \alg$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:equicontinuous-bounded}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, then for any ideal $\sigma \subset \mathfrak{B}(E)$, $\alg$ is a bounded subset of $B_\sigma(E; F)$. 
+\end{proposition}
+\begin{proof}
+    Let $S \in \sigma$ and $U \in \cn_F(0)$, then there exists $V \in \cn_E(0)$ such that $\bigcup_{T \in \alg}T(V) \subset U$. Since $S$ is bounded, there exists $\lambda \in K$ such that $S \subset \lambda V$. Therefore $\bigcup_{T \in \alg}T(S) \subset \lambda U$, and $\alg$ is bounded in $B_\sigma(E; F)$. 
+\end{proof}
+
+
 \begin{proposition}[{{\cite[IV.4.3]{SchaeferWolff}}}]
 \label{proposition:equicontinuous-linear-closure}
     Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, and $\alg'$ be the closure of $\alg$ in $F^E$ with respect to the product topology, then $\alg'$ is equicontinuous and hence $\alg' \subset L(E; F)$. 
@@ -29,7 +38,7 @@
     Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
     \begin{enumerate}
         \item[(B)] $E$ is a Baire space. 
-        \item[(B')] $E$ is barreled and $F$ is locally convex.
+        \item[(B')] $E$ is barrelled and $F$ is locally convex.
     \end{enumerate}
 
     and that
@@ -86,7 +95,7 @@
     Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
     \begin{enumerate}
         \item[(B)] $E$ is Baire.
-        \item[(B')] $E$ is barreled and $G$ is locally convex. 
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
     \end{enumerate}
 
     and that

src/measure/vector/variation.tex

@@ -119,7 +119,5 @@
     \]
 
     by \hyperref[Fubini's theorem]{theorem:fubini-tonelli}.
-    
-    % TODO: Actually link Fubini once it's there.
 \end{proof}
 

src/topology/metric/metric.tex

@@ -36,8 +36,6 @@
 
     Let $J \subset I$ be countable such that for each $k \in K$, there exists $j \in J$ with $V_n \subset U_j$, then $X = \bigcup_{k \in K}V_k = \bigcup_{j \in J}V_j$.
 
-    % TODO: This stuff may be moved to more general spaces.
-
     (2) $\Rightarrow$ (3): Let $n \in \nat$, then $\bracs{B(x, 1/n)|x \in X}$ is an open cover of $X$, so there exists $\seq{x_{n, k}} \subset X$ such that $X = \bigcup_{k \in \natp}B(x_k, 1/n)$. Let $D = \bracs{x_{n, k}| n, k \in \natp}$, then for any $x \in X$ and $n \in \natp$, there exists $k \in \natp$ such that $x \in B(x_{n, k}, 1/n)$, so $x_{n, k} \in B(x, 1/n)$. Therefore $D$ is dense in $X$, and $X$ is separable. 
 
     (3) $\Rightarrow$ (1): Let $\seq{x_n} \subset X$ be a countable dense subset. Let $x \in X$ and $k \in \natp$, then there exists $x_n \in \natp$ such that $d(x, x_n) < 1/(2k)$. In which case, $x \in B(x_n, 1/(2k)) \subset B(x_n, 1/k)$. Therefore $\bracs{B(x_n, 1/k)|n, k \in \natp}$ forms a countable basis for $X$. 

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}