\section{Bornological Spaces}
\label{section:bornological}

\begin{definition}[Bornological 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{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, 
    \[
    B \subset \bracs{\rho < R} = R\bracs{\rho < 1}
    \]

    By assumption, $\bracs{\rho < 1} \in \cn_E(0)$, so $\rho$ is continuous by \autoref{lemma:continuous-seminorm}.

    (2) $\Rightarrow$ (1): Let $\rho$ be the \hyperref[gauge]{definition:gauge} of $U$, then for any $B \subset E$ bounded, there exists $R > 0$ such that $B \subset RU$. In which case, $\rho(B) \subset [0, R]$. 
\end{proof}

\begin{proposition}
\label{proposition:bornological-bounded}
    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. 
        \item $T$ is bounded.
    \end{enumerate}
\end{proposition}
\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 bornological, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
\end{proof}

\begin{proposition}
\label{proposition:metrisable-bornological}
    Let $E$ be a metrisable locally convex space, then $E$ is bornological.
\end{proposition}
\begin{proof}
    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:bornological-limit}
    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: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:bornological-continuous-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: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 $\mathfrak{B}(E, \mathcal{T}_B) \supset \mathfrak{B}(E, \mathcal{T})$. 
        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$. 
        \item $(E, \mathcal{T}_B)$ is a bornological space. 
        \item Let $\mathcal{B} \subset \mathfrak{B}(E, \topo)$ 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 $\mathcal{T}_B \supset \mathcal{T}$, $\mathfrak{B}(E, \mathcal{T}) \supset \mathfrak{B}(E, \mathcal{T}_B)$. 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 $\mathfrak{B}(E, \mathcal{T}) = \mathfrak{B}(E, \mathcal{S})$ 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 $B \in \mathfrak{B}(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, 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 $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$, 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}

