From 3d9c47bda16c6ae427f99b8b7718357b3330b842 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Sat, 27 Jun 2026 14:12:40 -0400
Subject: [PATCH] Minor housekeeping.

---
 src/fa/duality/mackey.tex     | 16 ++++++++++++++++
 src/fa/tvs/equicontinuous.tex |  2 +-
 2 files changed, 17 insertions(+), 1 deletion(-)

diff --git a/src/fa/duality/mackey.tex b/src/fa/duality/mackey.tex
index 31648e5..aacf0c4 100644
--- a/src/fa/duality/mackey.tex
+++ b/src/fa/duality/mackey.tex
@@ -66,3 +66,19 @@
     On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets. 
 \end{proof}
 
+\begin{definition}[Mackey Space]
+\label{definition:mackey-space}
+    Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:barreled-mackey}
+    Let $E$ be a separated barreled space over $K \in \RC$, then $E$ is a Mackey space. 
+\end{proposition}
+\begin{proof}
+    Let $\cf \subset E^*$ be a $\sigma(E^*, E)$-compact set and $U \in \cn_{K}(0)$ be a barrel, then $V = \bigcap_{\phi \in \cf}\phi^{-1}(U)$ is convex, circled, and closed. For each $x \in E$, $\cf(x) = \bracs{\dpn{x, \phi}{E}|\phi \in \cf}$ is bounded. Thus $V$ is absorbing and hence a barrel. Since $E$ is barreled, $V \in \cn_E(0)$. Therefore the Mackey topology is contained in the topology of $E$, and $E$ is a Mackey space. 
+\end{proof}
+
+
+
+
diff --git a/src/fa/tvs/equicontinuous.tex b/src/fa/tvs/equicontinuous.tex
index 2aef80e..89542e8 100644
--- a/src/fa/tvs/equicontinuous.tex
+++ b/src/fa/tvs/equicontinuous.tex
@@ -43,7 +43,7 @@
 
     and that
     \begin{enumerate}
-        \item[(B2')] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
+        \item[(B2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
     \end{enumerate}
 
     then