Added bornological spaces.

.vscode/project.code-snippets

@@ -137,6 +137,11 @@
     "prefix": "cal",
     "body": ["\\mathcal{$1}$0"]
   },
+  "Mathscr": {
+  "scope": "latex",
+    "prefix": "scr",
+    "body": ["\\mathscr{$1}$0"]
+  },
   "Mathfrak": {
   "scope": "latex",
     "prefix": "fk",

src/fa/lc/bornologic.tex

@@ -1,15 +1,15 @@
-\section{Bornologic Spaces}
+\section{Bornological Spaces}
-\label{section:bornologic}
+\label{section:bornological}
 
-\begin{definition}[Bornologic Space]
+\begin{definition}[Bornological Space]
-\label{definition:bornologic-space}
+\label{definition:bornological-space}
     Let $E$ be a locally convex space, then the following are equivalent:
     \begin{enumerate}
         \item For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_E(0)$.
         \item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous. 
     \end{enumerate}
     
-    If the above holds, then $E$ is a \textbf{bornologic space}.
+    If the above holds, then $E$ is a \textbf{bornological space}.
 \end{definition}
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $B \subset E$ be bounded, then there exists $R > 0$ such that $\rho(B) \subset [0, R)$. In which case, 
@@ -23,8 +23,8 @@
 \end{proof}
 
 \begin{proposition}
-\label{proposition:bornologic-bounded}
+\label{proposition:bornological-bounded}
-    Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
+    Let $E$ be a bornological space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
     \begin{enumerate}
         \item $T$ is continuous.
         \item $T$ is sequentially continuous. 
@@ -34,32 +34,57 @@
 \begin{proof}
     (2) $\Rightarrow$ (3): Let $B \subset E$ be a bounded set, $\seq{x_n} \subset E$, and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$ as $n \to \infty$, then $\lambda_n x_n \to 0$ as $n \to \infty$. By sequential continuity of $T$, $T(\lambda_n x_n) \to 0$ as $n \to \infty$ as well. Thus $T(B)$ is also bounded. 
 
-    (3) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornologic, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
+    (3) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornological, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
 \end{proof}
 
 \begin{proposition}
-\label{proposition:metrisable-bornologic}
+\label{proposition:metrisable-bornological}
-    Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
+    Let $E$ be a metrisable locally convex space, then $E$ is bornological.
 \end{proposition}
 \begin{proof}
-    Let $U \subset E$ be convex and balanced such that $U$ absorbs every bounded set of $E$. Let $\seq{U_n} \subset \cn^o(0)$ be a decreasing countable fundamental system of neighbourhoods at $0$. If $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$, then there exists $\seq{x_n}$ such that $x_n \in U_n \setminus nA$ for all $n \in \natp$. In which case, $x_n \to 0$ as $n \to \infty$, so $\seq{x_n}$ is bounded. By assumption, there exists $n \in\natp$ such that $nA \supset \seq{x_n}$, which contradicts the fact that $\seq{x_n} \cap A = \emptyset$. 
+    Let $A \subset E$ be a convex and circled set that absorbs every bounded set of $E$, and $\seq{U_n} \subset \cn_E(0)$ be a decreasing fundamental system of neighbourhoods at $0$. If $A \not\in \cn_E(0)$, then $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$. For each $n \in \natp$, let $x_n \in U_n \setminus nA$, then $x_n \to 0$ as $n \to \infty$, and $\seq{x_n}$ is bounded. However, since $A$ absorbs every bounded set of $E$, there exists $n \in \natp$ such that $nA \supset \seq{x_n}$, which contradicts the assumption that $A \not\in \cn_E(0)$. 
 \end{proof}
 
 \begin{proposition}
-\label{proposition:bornologic-limit}
+\label{proposition:bornological-limit}
-    Let $\seqi{E}$ be bornologic 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 $E$ equipped with the inductive topology is bornolgic. 
+    Let $\seqi{E}$ be bornological 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 $E$ equipped with the inductive topology is bornolgic. 
 \end{proposition}
 \begin{proof}
-    Let $\rho: E \to [0, \infty)$ be a seminorm on $E$ that is bounded on all bounded sets. For each $i \in I$ and $B \subset E_i$ bounded, $T_i(B)$ is bounded by \autoref{proposition:bornologic-bounded}, and $\rho \circ T_i(B)$ is bounded by assumption. Thus for every $i \in I$, $\rho \circ T_i$ is continuous, so $\rho$ is continuous by (4) of \autoref{definition:lc-inductive}. 
+    Let $\rho: E \to [0, \infty)$ be a seminorm on $E$ that is bounded on all bounded sets. For each $i \in I$ and $B \subset E_i$ bounded, $T_i(B)$ is bounded by \autoref{proposition:bornological-bounded}, and $\rho \circ T_i(B)$ is bounded by assumption. Thus for every $i \in I$, $\rho \circ T_i$ is continuous, so $\rho$ is continuous by (4) of \autoref{definition:lc-inductive}. 
 \end{proof}
 
 
 \begin{proposition}
-\label{proposition:bornologic-continuous-complete}
+\label{proposition:bornological-continuous-complete}
-    Let $E$ be a bornologic space and $F$ be a complete separated locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
+    Let $E$ be a bornological space and $F$ be a complete separated locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:bornologic-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
+    By \autoref{proposition:bornological-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
 \end{proof}
 
+\begin{definition}[Associated Bornological Space]
+\label{definition:associated-bornological}
+    Let $(E, \mathcal{T})$ be a separated locally convex space over $K \in \RC$, then there exists a locally convex topology $\mathcal{T}_B \subset 2^E$ such that:
+    \begin{enumerate}
+        \item $B(E, \mathcal{T}_B) \supset B(E, \mathcal{T})$. 
+        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$. 
+        \item $(E, \mathcal{T}_B)$ is a bornological space. 
+        \item Let $\mathcal{B} \subset B(E, \topo)$ be the collection of all convex, circled, bounded, and closed subsets of $E$, ordered by inclusion. For each $B \in \mathcal{B}$, let $(E_B, \rho_B)$ be the normed space associated with $B$ and $\iota_B: E_B \to E$ be the canonical inclusion, then $\mathcal{T}_B$ is the inductive topology induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
+    \end{enumerate}
+
+    The space $(E, \mathcal{T}_B)$ is the \textbf{bornological space associated with} $E$. 
+\end{definition}
+\begin{proof}
+    Let $\mathscr{T}$ be the collection of all topologies satisfying (1), then $\mathcal{T} \in \mathscr{T}$, so $\mathscr{T} \ne \emptyset$. Let $\mathcal{T}_B$ be the projective topology induced by $\mathscr{T}$. 
+
+    (1): Since $\mathcal{T}_B \supset \mathcal{T}$, $B(E, \mathcal{T}) \supset B(E, \mathcal{T}_B)$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S} \subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}} \subset U$. In which case, since $B(E, \mathcal{T}) = B(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
+    
+    (U): By (U) of the \hyperref[projective topology]{definition:tvs-initial}.
+
+    (3): Let $\rho: E \to [0, \infty)$ be a seminorm that is bounded on bounded sets, then the topology induced by $\rho$ satisfies (1). By (U), $\rho$ is continuous with respect to $\mathcal{T}_B$, so $(E, \mathcal{T}_B)$ is a bornological space. 
+
+    (4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in B(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B} \in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B} \circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_B \supset \mathcal{S}$. 
+
+    On the other hand, since $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_B: E_B \to (E, \mathcal{T}_B)$ is continuous. Therefore by (U) of the \hyperref[inductive topology]{definition:lc-inductive}, $\mathcal{S} \supset \mathcal{T}_B$. 
+\end{proof}
 

src/fa/lc/convex.tex

@@ -199,3 +199,32 @@
 
     $(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
 \end{proof}
+
+
+
+\begin{definition}[Associated Normed Space]
+\label{definition:lc-associated-normed-space}
+    Let $E$ be a separated locally convex space and $A \subset E$ be convex and circled. Let $E_0 = \bigcup_{n \in \natp}nA$, $\rho_0: E_0 \to [0, \infty)$ be the gauge of $A$, and $(E_A, \rho_A)$ be the quotient of $E_0$ by $\bracs{\phi = 0}$, equipped with the quotient norm of $\rho_0$, then
+    \begin{enumerate}
+        \item $(E_A, \rho_A)$ is a normed space. 
+    \end{enumerate}
+
+    If $A$ is radial, then $E_0 = E$ and the map $\pi_A: E \to E_A$ is the \textbf{canonical projection}, and
+    \begin{enumerate}[start=1]
+        \item If $A \in \cn_E(0)$, then $\pi_A \in L(E; E_A)$.
+    \end{enumerate}
+
+    If $(E_0, \rho_0)$ is separated, then $(E_0, \rho_0) = (E_A, \rho_A)$, and the map $\iota_A: E_A \to E$ is the \textbf{canonical inclusion}. In particular, if $A$ is bounded, then
+    \begin{enumerate}[start=2]
+        \item $(E_0, \rho_0)$ is separated. 
+        \item $\iota_A \in L(E_A; E)$. 
+    \end{enumerate}
+
+    The space $(E_A, \rho_A)$ is the \textbf{normed space associated with} $A$. 
+\end{definition}
+\begin{proof}
+    (3): Let $x \in E_0 \setminus \bracs{0}$. Since $E$ is separated, there exists $U \in \cn_E(0)$ such that $x \not\in U$. As $A$ is bounded, there exists $\lambda > 0$ such that $\lambda U \supset A$. In which case, $x \not\in \lambda^{-1}A$, and $E_0$ is separated. 
+
+    (4): Let $U \in \cn_E(0)$, then there exists $\lambda > 0$ such that $\lambda U \supset A$, so $\iota_A^{-1}(U) \supset \lambda^{-1}A \in \cn_{E_A}(0)$, and $\iota_A$ is continuous by \autoref{proposition:tvs-convex-morphism}.
+\end{proof}
+

src/fa/tvs/bounded.tex

@@ -3,13 +3,13 @@
 
 \begin{definition}[Bounded]
 \label{definition:bounded}
-    Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
+    Let $(E, \topo)$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
     \begin{enumerate}
         \item For every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. 
         \item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
     \end{enumerate}
 
-    If the above holds, then $B$ is \textbf{bounded}. The collection $B(E)$ is the set of all bounded sets of $E$.
+    If the above holds, then $B$ is \textbf{bounded}. The collection $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
 \end{definition}
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$. 

src/fa/tvs/complete-metric.tex

@@ -111,7 +111,7 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    Suppose that $T(E)$ is not meagre. Let $r_0 > 0$ and $r > 0$ such that $B_E(0, r) + B_E(0, r) \subset B_E(0, r_0)$, then since $B_E(0, r)$ is absorbing,
+    Suppose that $T(E)$ is not meagre. Let $r_0 > 0$ and $r > 0$ such that $B_E(0, r) + B_E(0, r) \subset B_E(0, r_0)$, then since $B_E(0, r)$ is radial,
     \[
     E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
     \]

src/fa/tvs/definition.tex

@@ -137,7 +137,7 @@
 \label{proposition:tvs-good-neighbourhood-base}
     Let $E$ be a topological vector space over $K \in \RC$, then
     \begin{enumerate}
-        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of circled and absorbing sets.
+        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of circled and radial sets.
         \item The fundamental system of neighbourhoods in $(1)$ can be taken to be open or closed.
     \end{enumerate}
 \end{proposition}

src/fa/tvs/projective.tex

@@ -20,6 +20,7 @@
 
         is a fundamental system of neighbourhoods for $E$ at $0$.
     \end{enumerate}
+    
     The uniformity $\fU$ and its topology are the \textbf{projective uniformity/topology} induced by $\seqi{T}$.
 \end{definition}
 \begin{proof}

src/measure/vector/vector.tex

@@ -34,7 +34,7 @@
     \sup_{A \in \cm}|\dpn{\phi, x_A}{E^{**}}| = \sup_{A \in \cm}|\dpn{\mu(A),\phi}{E}| < \infty
     \]
 
-    By \autoref{proposition:bornologic-continuous-complete} and \autoref{proposition:metrisable-bornologic}, $E^*$ is a Banach space. The \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness} implies that
+    By \autoref{proposition:bornological-continuous-complete} and \autoref{proposition:metrisable-bornological}, $E^*$ is a Banach space. The \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness} implies that
     \[
     \sup_{A \in \cm}\norm{\mu(A)}_{E} = \sup_{A \in \cm}\norm{x_A}_{E^{**}} < \infty
     \]