Added barreled spaces.

src/fa/lc/convex.tex

@@ -118,15 +118,16 @@
         \item $[\cdot]$ is continuous.
         \item $[\cdot]$ is continuous at $0$.
         \item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
+        \item $\bracs{x \in E| [x] \le 1} \in \cn_E(0)$.
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    $(4) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
+    $(5) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
     \[
-    x - y \in \bracs{x \in E|[x] < r} = r\bracs{x \in E|[x] < 1} \in \cn_E(0)
+    x - y \in \bracs{x \in E|[x] \le r} = r\bracs{x \in E|[x] \le 1} \in \cn_E(0)
     \]
     
-    then $[x - y] < r$.
+    then $[x - y] \le r$.
 \end{proof}
 
 
@@ -147,7 +148,7 @@
 
 \begin{definition}[Gauge/Minkowski Functional]
 \label{definition:gauge}
-    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be radial, then the mapping
     \[
     [\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
     \]
@@ -157,11 +158,12 @@
         \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
         \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
         \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
+        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
     \end{enumerate}
     In particular,
-    \begin{enumerate}
-        \item[(4)] If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
-        \item[(5)] If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
+    \begin{enumerate}[start=4]
+        \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
+        \item If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
     \end{enumerate}
 \end{definition}
 \begin{proof}
@@ -171,6 +173,13 @@
     \]
 
     then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
+
+    (4): Let $x \in \bracs{\rho \le 1}$, then $\lambda x \in A$ for all $\lambda \in (0, 1)$. Therefore
+    \[
+    x \in \overline{\bracs{\lambda x|\lambda \in (0, 1)}} \subset A
+    \]
+
+    so $x \in \overline{A}$. 
 \end{proof}
 
 \begin{definition}[Locally Convex Space]