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This commit is contained in:
Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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22
src/fa/lc/inductive.tex
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@@ -13,12 +13,13 @@
\[
\mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
\]
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 \ref{proposition:tvs-0-neighbourhood-base}.
(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, $\frac{1}{2}U \in \mathcal{B}$.
@@ -48,6 +49,7 @@
}
\]
\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
\[
@@ -57,22 +59,24 @@
}
\]
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 $K$ (\ref{proposition:module-direct-limit}). Equip $E$ with the inductive topology (\ref{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).
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 \ref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \ref{definition:lc-inductive}, $S \in L(E; F)$.
(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 \ref{definition:lc-inductive}.
(5): By (5) of \autoref{definition:lc-inductive}.
\end{proof}
\begin{remark}
@@ -99,11 +103,13 @@
\[
(1 - \alpha)U + w = (1 - \alpha) U + \alpha x \subset V
\]
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.
@@ -128,9 +134,9 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $U \in \cn_{E_n}(0)$. By \ref{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, $V \cap E_n = U_n = U$.
(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 \ref{lemma:lc-induct-separate} and (1), there exists $V \in \cn_E(0)$ such that $V \cap E_n = U$, so $x \not\in V$.
(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 $k \in \natp$.
@@ -148,6 +154,7 @@
\[
\fU = \bracs{F + U|F \in \mathcal{F}, U \in \cn_E(0)}
\]
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 $U \in \cn_E(0)$.
@@ -156,6 +163,7 @@
\[
U = \text{Conv}\paren{\bigcup_{n \in \natp}(U_n \cap E_n)}
\]
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
@@ -168,11 +176,13 @@
\[
\underbracs{y - \sum_{k = 1}^n \lambda_kx_k}_{\in E_n} = \underbrace{z + \sum_{k = n + 1}^N \lambda_kx_k}_{\in F_n + U_n}
\]
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}