From 227436a9c23558d3bbb6ae54cd5ef084bcab2836 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Mon, 4 May 2026 17:54:03 -0400
Subject: [PATCH] Added saturated ideals.

---
 src/fa/lc/spaces-of-linear.tex                | 43 +++++++++++++++++++
 src/topology/main/interiorclosureboundary.tex |  4 +-
 2 files changed, 46 insertions(+), 1 deletion(-)

diff --git a/src/fa/lc/spaces-of-linear.tex b/src/fa/lc/spaces-of-linear.tex
index 67b6c3d..e2fb48c 100644
--- a/src/fa/lc/spaces-of-linear.tex
+++ b/src/fa/lc/spaces-of-linear.tex
@@ -18,3 +18,46 @@
 \begin{proof}
     By \autoref{proposition:tvs-set-uniformity} and \autoref{proposition:set-uniform-pseudometric}.
 \end{proof}
+
+\begin{definition}[Saturated Ideal]
+\label{definition:saturated-ideal}
+    Let $E$ be a locally convex space over $K \in \RC$ and $\sigma \subset 2^E$ be an ideal, then $\sigma$ is \textbf{saturated} if:
+    \begin{enumerate}
+        \item For each $\lambda \in K$ and $S \in \sigma$, $\lamdba S \in \sigma$. 
+        \item For each $S \in \sigma$, $\ol{\aconv}(S) \in \sigma$. 
+    \end{enumerate}
+
+    For any ideal $\sigma \subset 2^E$, the smallest saturated ideal $\ol \sigma$ containing it is the \textbf{saturated hull} of $\sigma$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:locally-convex-saturated}
+    Let $E$, $F$ be locally convex spaces over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $\ol \sigma$ be its saturated hull, then the $\sigma$-uniformity and $\ol \sigma$-uniformity on $L(E; F)$ coincide. 
+\end{lemma}
+\begin{proof}
+    Let $\tau \subset \ol \sigma$ be the collection of sets such that for each $S \in \tau$ and $U \in \cn_F(0)$,
+    \[
+    N(S, U) = \bracs{(S, T) \in L(E; F)| (S - T)(S) \subset U}
+    \]
+    
+    is an entourage in the $\sigma$-uniformity. 
+
+    For each $S \in \tau$, $U \in \cn_F(0)$, and $\lambda \in K$ with $\lambda \ne 0$, 
+    \begin{align*}
+    N(\lambda S, U) &= \bracs{(S, T) \in L(E; F)| (S - T)(\lambda S) \subset U} \\
+    &= \bracs{(S, T) \in L(E; F)| (S - T)(S) \subset \lambda^{-1}U}
+    \end{align*}
+
+    is another entourage in the $\sigma$-uniformity. If $\lambda = 0$, then $N(\lambda S, U) = L(E; F)$, which is also an entourage. 
+
+    Now, let $S \in \tau$ and $U \in \cn_F(0)$ be convex and circled, then by \autoref{proposition:closure-of-image},
+    \begin{align*}
+        N(\ol{\aconv}(S), U) &= \bracs{(S, T) \in L(E; F)| (S - T)(\ol{\aconv}(S)) \subset U} \\
+        &\supset \bracs{(S, T) \in L(E; F)| \overline{(S - T)(\aconv(S))} \subset U} \\
+        &= \bracs{(S, T) \in L(E; F)| \ol{\aconv}{(S - T)(S)} \subset U}
+    \end{align*}
+
+    so $N(\ol{\aconv}(S), U)$ contains an entourage in the $\sigma$-uniformity. 
+
+    Since $\tau$ is a saturated ideal that contains $\sigma$, $\tau = \ol \sigma$. Therefore the $\sigma$-uniformity and $\ol \sigma$-uniformity on $L(E; F)$ coincide. 
+\end{proof}
diff --git a/src/topology/main/interiorclosureboundary.tex b/src/topology/main/interiorclosureboundary.tex
index 95372ea..c5af445 100644
--- a/src/topology/main/interiorclosureboundary.tex
+++ b/src/topology/main/interiorclosureboundary.tex
@@ -9,6 +9,7 @@
         \item There exists $U \in \fB$ with $U \subset A$.
         \item There exists $U \in \cn(x)$ with $U \subset A$.
     \end{enumerate}
+
     The set of all points satisfying the above is the \textbf{interior} $A^o$ of $A$, which is the largest open set contained in $A$.
 \end{definition}
 \begin{proof}
@@ -28,7 +29,7 @@
         \item For every $B \supset A$ closed, $x \in B$.
         \item For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
         \item For every $U \in \fB$, $U \cap A \ne \emptyset$.
-        \item There exists a filter $\fF \subset 2^A$ that converges to $\fF$.
+        \item There exists a filter $\fF \subset 2^A$ that converges to $x$.
     \end{enumerate}
 
     The set $\ol{A}$ of all points satisfying the above is the \textbf{closure} of $A$ in $X$. By (1), it is the smallest closed set containing $A$.
@@ -69,6 +70,7 @@
         \item For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
         \item For every $U \subset X$ open, $\overline{A \cap U} \supset U$.
     \end{enumerate}
+
     If the above holds, then $A$ is a \textbf{dense} subset of $X$.
 \end{definition}
 \begin{proof}