\section{Inductive Limits}
\label{section:lc-inductive}

\begin{definition}[Inductive Locally Convex Topology]
\label{definition:lc-inductive}
    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 there exists a topology $\topo$ on $E$ such that:
    \begin{enumerate}
        \item $(E, \topo)$ is a locally convex space over $K$.
        \item For each $i \in I$, $T_i \in L(E_i; E)$.
        \item[(U)] For any topology $\mathcal{S}$ on $E$ satisfying (1) and (2), $\mathcal{S} \subset T$.
        \item For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T_i \in L(E_i; F)$ for all $i \in I$.
        \item The family
        \[
        \mathcal{B} = \bracs{U \subset E|U \text{ convex, circled, radial}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
        \]

        is a fundamental system of neighbourhoods for $E$ at $0$.
        \item If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then
        \[
        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
        \]

        is a fundamental system of neighbourhoods for $E$ at $0$. 
    \end{enumerate}

    The topology $\topo$ is the \textbf{inductive locally convex topology} on $E$ induced by $\seqi{T}$.
\end{definition}
\begin{proof}
    (1), (5): To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \autoref{proposition:tvs-0-neighbourhood-base}.
    \begin{enumerate}
        \item[(TVB1)] Every set in $\mathcal{B}$ is radial and circled by definition.
        \item[(TVB2)] For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
    \end{enumerate}

    Let $\topo$ be the vector space topology such that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, then $(E, \topo)$ is a locally convex space.

    (2): For each $i \in I$ and $U \in \mathcal{B}$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $T_i \in L(E_i; E)$.

    (U): Let $U \in \cn_{(E, \mathcal{S})}(0)$ be convex, circled, and radial. By (2), $T_i^{-1}(U) \in \cn_{E_i}(0)$ for all $i \in I$. Thus the convex, circled, and radial neighbourhoods of $0$ in $(E, \mathcal{S})$ is a subset of $\mathcal{B}$.

    (5): Let $i \in I$ and $U \in \cn_F(0)$ be convex, circled, and radial. Since $T \circ E_i \in L(E_i; F)$, $T_i^{-1}(T^{-1}(U)) \in \cn_{E_i}(0)$, so $T^{-1}(U) \in \mathcal{B} \subset \cn_E(0)$.

    (6): If $E$ is spanned by $\bigcup_{i \in I}T_i(E_i)$, then each set in $\fB$ is radial. Hence $\fB$ is a family of neighbourhoods of $E$ at $0$. 

    Let $U \in \cn_E(0)$ be convex, circled, and radial, then for each $i \in I$, $T_i^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_i[T_i^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \Gamma\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]} \in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.
\end{proof}

\begin{definition}[Locally Convex Direct Sum]
\label{definition:lc-direct-sum}
    Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, then there exists $(E, \seqi{\iota})$ such that:
    \begin{enumerate}
        \item $E$ is a locally convex space over $K$. 
        \item For each $i \in I$, $\iota_i \in L(E_i; E)$. 
        \item[(U)] For each $(F, \seqi{T})$ satisfying (1) and (2), there exists a unique $T \in L(E; F)$ such that the following diagram commutes:
        \[
        \xymatrix{
        A \ar@{->}[r]^{T} & B \\
        A_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} & 
        }
        \]
        
        \item The family
        \[
        \fB = \bracs{\Gamma\paren{\bigcup_{i \in I}\iota_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}
        \]

        is a fundamental system of neighbourhoods for $E$ at $0$. 
    \end{enumerate}

    
    The space $E = \bigoplus_{i \in I}E_i$ is the \textbf{locally convex direct sum} of $\seqi{E}$. 
\end{definition}
\begin{proof}
    Let $(E, \seqi{\iota})$ be the direct sum of $\seqi{E}$ as vector spaces, and equip it with the inductive topology induced by $\seqi{\iota}$, then $(E, \seqi{\iota})$ satisfies (1) and (2). 

    (U): By (U) of the \hyperref[direct sum]{definition:direct-sum}, there exists a unique $T \in \hom(E; F)$ such that the diagram commutes. In which case, by (4) of \autoref{definition:lc-inductive}, $T \in L(E; F)$. 

    (4): By (6) of \autoref{definition:lc-inductive}.
\end{proof}

\begin{definition}[Inductive Limit]
\label{definition:lc-inductive-limit}
    Let $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of locally convex spaces over $K \in \RC$, then there exists $(E, \bracsn{T^i_E}_{i \in I})$ such that:
    \begin{enumerate}
        \item $E$ is a locally convex space over $K$.
        \item For each $i \in I$, $T^i_E \in L({E_i, E})$.
        \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:

        \[
        \xymatrix{
        E_i \ar@{->}[rd]_{T^i_E} \ar@{->}[r]^{T^i_j} & E_j \ar@{->}[d]^{T^j_E} \\
        & E
        }
        \]


        \item[(U)] For any pair $(F, \bracsn{S^i_F}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L({E, F})$ such that the following diagram commutes

        \[
        \xymatrix{
        E_i \ar@{->}[d]_{T^i_E} \ar@{->}[rd]^{S^i_F} &  \\
        E \ar@{->}[r]_{S} & F
        }
        \]


        for all $i \in I$.
        \item For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
        \item The family
        \[
        \mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, (T^i_E)^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
        \]

        is a fundamental system of neighbourhoods for $E$ at $0$.
    \end{enumerate}

    The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
\end{definition}
\begin{proof}
    Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\autoref{proposition:module-direct-limit}). Equip $E$ with the \hyperref[inductive topology]{definition:lc-inductive} induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).

    (U): By (U) of \autoref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \autoref{definition:lc-inductive}, $S \in L(E; F)$.

    (5): By (5) of \autoref{definition:lc-inductive}.
\end{proof}

\begin{remark}
\label{remark:tvs-limits}
    The projective topology behaves well across the constraints of topological vector spaces and locally convex spaces: the preimage of a vector space/locally convex topology is also a vector space/locally convex topology.

    On the inductive side, the story is not as simple: In principle, the locally convex inductive topology is smaller than the vector space inductive topology, which is smaller than the inductive topology. As such, the same construction must be performed three separate times, each time restricting to a smaller collection of sets.

    In addition to the neighbourhood construction given above, the inductive topology may also be constructed as the weak topology generated by all topologies satisfying certain properties. While this more non-constructive method is simpler, it does not directly provide an explicit fundamental system of neighbourhoods at $0$.
\end{remark}

\subsection{Strict Inductive Limits}
\label{subsection:lc-induct-strict}

\begin{lemma}[{{\cite[Lemma II.6.4]{SchaeferWolff}}}]
\label{lemma:lc-induct-separate}
    Let $E$ be a locally convex space over $K$, $M \subset E$ be a subspace, $U \in \cn_M(0)$ be convex and circled, then
    \begin{enumerate}
        \item There exists $V \in \cn_E(0)$ circled and radial such that $U = M \cap V$.
        \item For any $x \in E \setminus \ol M$, there exists $V \in \cn_E(0)$ circled and radial such that $U = M \cap V$ and $x \not\in U$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    (1): Let $W \in \cn_E(0)$ be circled and radial with $W \cap M \subset U$, and $V = \text{Conv}(W \cup U)$.
    \begin{itemize}
        \item For any $u \in U$, $w \in W$, and $t \in [0, 1]$, there exists $\alpha \in (0, 1)$ such that $x = \alpha^{-1}w \in W$. In which case,
        \[
        (1 - \alpha)U + w = (1 - \alpha) U + \alpha x \subset V
        \]

        so $V \in \cn(0)$.
        \item For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $u \in U$, $w \in W$, and $t \in [0, 1]$,
        \[
        \lambda (1 - t)u + \lambda tw = (1 - t)\lambda u + t \lambda w \in V
        \]

        as $U$ and $W$ are both circled.
    \end{itemize}
    so $V \in \cn_E(0)$ is convex and circled.

    For any $u \in U$, $w \in W$, and $t \in [0, 1]$, if $(1 - t)u + tw \in M$, then $u \in M \cap U \subset M \cap W$, so $(1 - t)u + tw \in W$.

    (2): Since $x \not\in \ol M$, there exists $W \in \cn_E(0)$ circled and radial with $x + W \cap M = \emptyset$. Let $V$ as in (1), then $x \not\in W + M \supset W + U = V$.
\end{proof}

\begin{proposition}[{{\cite[II.6.4, II.6.5, II.6.6]{SchaeferWolff}}}]
\label{proposition:lc-strict-inductive-countable}
    Let $(\seq{E_n}, \seq{\iota_n^m|m, n \in \natp, m \le n})$ be a strict inductive system of locally convex spaces over $K \in \RC$ and $(E, \seq{\iota_n})$ be its direct limit, then:
    \begin{enumerate}
        \item For each $n \in \natp$, the topology of $E_n$ is induced by $\iota_n$, which allows the identification of $E_n$ as a subspace of $E$.
        \item If $E_n$ is separated for all $n \in \natp$, then $E$ is also separated.
        \item If $\iota^m_n(E_m) \subset E_n$ is closed for all $m, n \in \natp$ with $m \le n$, then the following are equivalent for any $B \subset E$:
        \begin{enumerate}
            \item[(a)] $B$ is bounded.
            \item[(b)] There exists $n \in \natp$ such that $B \subset E_n$ is bounded.
        \end{enumerate}

        \item If $E_n$ is complete for each $n \in \natp$, then $E$ is also complete.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Let $U \in \cn_{E_n}(0)$. By \autoref{lemma:lc-induct-separate}, there exists $\bracs{U_m| m \in \natp, m \ge n} \subset 2^E$ such that $U_n = U$, $U_m \in \cn_{E_m}(0)$ and $U_{m} = U_{m + 1} \cap E_m$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_m$, then $V \cap E_m = U_m$ for all $m \ge n$. In particular, $V \cap E_n = U_n = U$.

    (2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_n$. Since $E_n$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By \autoref{lemma:lc-induct-separate} and (1), there exists $V \in \cn_E(0)$ such that $V \cap E_n = U$, so $x \not\in V$.

    (3), $\neg (b) \Rightarrow \neg (a)$: If $B \not\subset E_n$ for all $n \in \natp$, then there exists a subsequence $\bracsn{n_k}_0^\infty \subset \natp$ and $\seq{x_k} \subset B$ such that $x_k \in E_{n_{k}} \setminus E_{n_{k - 1}}$ for all $k \in \natp$.

    Since $E_{n_k} \subset E_{n_{k+1}}$ is closed for all $k \in \natp$, there exists $\seq{U_k} \subset 2^E$ such that:
    \begin{enumerate}
        \item For each $k \in \natp$, $U_k \in \cn_{E_{n_k}}(0)$.
        \item For each $k \in \natp$, $U_k = U_{k+1} \cap E_{n_k}$.
        \item For each $k \in \natp$, $n^{-1}x_k \not\in U_k$.
    \end{enumerate}

    then $V = \bigcup_{k \in \natp}U_k \in \cn_E(0)$ with $V \cap E_{n_k} = U_k$ for all $k \in \natp$. For any $n \in \natp$, $x_k \not\in nU_k = nV \cap E_{n_k}$. Therefore $B$ is not bounded.

    (3), $(b) \Rightarrow (a)$: Let $U \in \cn_E(0)$, then $U \cap E_n \in \cn_{E_n}(0)$, so there exists $\lambda \in K$ with $\lambda (U \cap E_n) \supset B$.

    (4): Let $\fF \subset 2^E$ be a Cauchy filter and
    \[
    \fU = \bracs{F + U|F \in \mathcal{F}, U \in \cn_E(0)}
    \]

    then $\fU$ is also a Cauchy filter, which converges if and only if $\fF$ does.

    Since each $E_n$ is complete, it is sufficient to show that there exists $n \in \natp$ such that $F + U \cap E_n \ne \emptyset$ for all $F \in \fF$ and $U \in \cn_E(0)$.

    Suppose for contradiction that for every $n \in \natp$, there exists $F_n \in \fF$ and $U_n \in \cn_E(0)$ such that $E_n \cap F_n + U_n = \emptyset$. Assume without loss of generality that for every $n \in \natp$, $U_n$ is convex and circled with $U_n \supset U_{n+1}$. Let
    \[
    U = \text{Conv}\paren{\bigcup_{n \in \natp}(U_n \cap E_n)}
    \]

    then since each $U_n$ is circled, so is $U$. Thus $U \cap E_n \supset U_n \cap E_n \in \cn_{E_n}(0)$, and $U \in \cn_E(0)$.

    Now, suppose that $(F_n + U) \cap E_n \ne \emptyset$. Let $y \in (F_n + U) \cap E_n$, then there exists $N \in \natp$, $\bracs{x_k}_1^N \subset E$, $\bracs{\lambda_k}_1^N \subset [0, 1]$, and $z \in F_n$ such that
    \begin{itemize}
    \item For each $1 \le k \le N$, $x_k \in U_k \cap E_k$.
    \item $\sum_{k = 1}^N \lambda_k = 1$.
    \item $y = z + \sum_{k = 1}^N \lambda_k x_k$.
    \end{itemize}
    In which case, since $U_{k} \supset U_{k+1}$ for all $k \in \natp$,
    \[
    \underbrace{y - \sum_{k = 1}^n \lambda_kx_k}_{\in E_n} = \underbrace{z + \sum_{k = n + 1}^N \lambda_kx_k}_{\in F_n + U_n}
    \]

    which is impossible. Therefore $(F_n + U) \cap E_n = \emptyset$ for all $n \in \natp$.

    Finally, since $\fF$ is a Cauchy filter, there exists $F \in \fF$ such that $F - F \subset U$. Let $z \in F$, then there exists $n \in \natp$ such that $z \in E_n$. In which case, for any $y \in F \cap F_n$,
    \[
    z = y + (z - y) \in y + (F - F) \subset y + U \subset F_n + U
    \]

    which contradicts the fact that $(F_n + U) \cap E_n = \emptyset$.
\end{proof}