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