Added notes on equicontinuity.

preamble.sty

@@ -224,3 +224,8 @@
 
 % Real or Complex Numbers
 \newcommand{\RC}{\bracs{\real, \complex}}
+
+% Convex Stuff
+\newcommand{\conv}{\text{Conv}}
+\newcommand{\aconv}{\text{AbsConv}}
+

src/dg/derivative/higher.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[$n$-Fold Differentiability]
 \label{definition:n-differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
 
     Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable at $x_0$} if
     \begin{enumerate}
@@ -87,7 +87,7 @@
 
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric}
-    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed system that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
+    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
     Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
@@ -100,7 +100,7 @@
 
 \begin{proposition}[Power Rule]
 \label{proposition:multilinear-derivative}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a Hausdorff locally convex space, and
+    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
     \[
     T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
     \]

src/dg/derivative/sets.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Small]
 \label{definition:differentiation-small}
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
     \begin{enumerate}
         \item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
         \item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$.
@@ -16,7 +16,7 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family contains all finite sets, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
 \end{proposition}
 \begin{proof}
     Let $T \in B_\sigma(E; F)$ and suppose that there exists $V \in \cn_E(0)$ circled and $r \in \mathcal{R}_\sigma(E; F)$ such that $T|_V = r|_V$. For any $x \in V$, $\bracs{x} \in \sigma$, so $T(x) = \lim_{t \downto 0}T(tx)/t = 0$ as $F$ is separated.
@@ -25,7 +25,7 @@
 
 \begin{definition}[Derivative]
 \label{definition:derivative-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
@@ -37,7 +37,7 @@
 
 \begin{definition}[Differentiable]
 \label{definition:differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
 \end{definition}
 
 \begin{definition}
@@ -48,7 +48,7 @@
 
 \begin{proposition}[Chain Rule]
 \label{proposition:chain-rule-sets}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
+    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be covering ideals. If:
     \begin{enumerate}
         \item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \label{proposition:derivative-sets-real}
-    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
+    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be a covering ideal, then
     \begin{enumerate}
         \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
         \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is differentiable at $x_0$ if and only if

src/fa/duality/definitions.tex

@@ -44,7 +44,29 @@
     
 \end{proof}
 
-
-
+\begin{lemma}
+\label{lemma:duality-dense}
+    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality over $K$, then for any subspace $F_0 \subset F$, the following are equivalent:
+    \begin{enumerate}
+        \item $\dpn{E, F_0}{\lambda}$ is a duality. 
+        \item For any $y_0 \in F$, $\seqf{x_j} \subset E$, and $\eps > 0$, there exists $y \in F_0$ such that for each $1 \le j \le n$, $\dpn{x_j, y}{\lambda} = \dpn{x_j, y_0}{\lambda}$.
+        \item $F_0$ is $\sigma(F, E)$-dense in $F$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $E_0 = \text{span}\bracs{x_j|1 \le j \le n}$, and
+    \[
+    \phi: E_0 \to K \quad x \mapsto \dpn{x, y_0}{\lambda}
+    \]
+    
+    Since $\dpn{E, F_0}{\lambda}$ is a duality, there exists $\seqf{y_j} \subset F_0$ such that for all $x \in E_0$, 
+    \[
+    |\dpn{x, \phi}{E_0}| \le \sum_{j = 1}^n |\dpn{x, y_j}{\lambda}|
+    \]
+
+    Hence $\phi \in L(E_0, \sigma(E_0, F_0); K)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in L(E, \sigma(E, F_0); K)$ such that $\Phi|_{E_0} = \phi$. By \autoref{lemma:duality-dual}, there exists $y \in F_0$ such that $\dpn{x_j, y}{\lambda} = \dpn{x_j, y_0}{\lambda}$ for all $1 \le j \le n$. 
+
+    (3) $\Rightarrow$ (1): Let $x \in E$ such that $\dpn{x, y}{\lambda} = 0$ for all $y \in F_0$, then since $F_0$ is $\sigma(F, E)$-dense in $F$, $\dpn{x, y}{\lambda} = 0$ for all $y \in F$. Hence $x = 0$. 
+\end{proof}
 
 

src/fa/duality/index.tex

@@ -2,5 +2,6 @@
 \label{chap:duality}
 
 \input{./definitions.tex}
+\input{./polar.tex}
 
 

src/fa/duality/polar.tex

@@ -0,0 +1,35 @@
+\section{Polars}
+\label{section:polar}
+
+\begin{definition}[Real Polar]
+\label{definition:real-polar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \[
+    A^\circ = \bracsn{y \in F| \text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in A}
+    \]
+
+    is the \textbf{real polar} of $A$. 
+\end{definition}
+
+\begin{definition}[Absolute Polar]
+\label{definition:absolute-polar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \[
+    A^\square = \bracsn{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A}
+    \]
+
+    is the \textbf{absolute polar} of $A$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:polar-properties}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, then:
+    \begin{enumerate}
+        \item $\emptyset^\circ = \emptyset^\square = F$ and $F^\circ = F^\square = \bracs{0}$. 
+        \item For any $A, B \subset E$ and $\lambda \ne 0$, if $\lambda A \subset B$, then $B^\circ \subset \lambda^{-1}A^\circ$. 
+    \end{enumerate}
+    
+\end{proposition}
+
+
+

src/fa/lc/convex.tex

@@ -21,7 +21,7 @@
 \label{definition:convex-circled-hull}
     Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then the set
     \[
-    \Gamma(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset K, \seqf{x_j} \subset E, \sum_{j = 1}^n |t_j| \le 1 }
+    \aconv(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset K, \seqf{x_j} \subset E, \sum_{j = 1}^n |t_j| \le 1 }
     \]
 
     is the \textbf{convex circled hull} of $A$. 

src/fa/lc/spaces-of-linear.tex

@@ -3,18 +3,61 @@
 
 \begin{proposition}
 \label{proposition:lc-spaces-linear-map}
-    Let $T$ be a set, $E$ be a locally convex space defined by the seminorms $\seqi{[\cdot]}$, and $\mathfrak{S} \subset 2^T$ be an upward-directed family. For each $i \in I$ and $S \in \mathfrak{S}$, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, $E$ be a locally convex space over $K$, and $\cf \subset E^T$ be a subspace such that
+    \begin{enumerate}
+        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded. 
+    \end{enumerate}
+    
+    For each $S \in \sigma$ and continuous seminorm $\rho: E \to [0, \infty)$, let
     \[
-    [\cdot]_{S, i}: E^T \to [0, \infty) \quad f \mapsto \sup_{x \in S}[f(x)]_{S, i}
+    \rho_S: E^T \to [0, \infty] \quad f \mapsto \sup_{x \in S}\rho(f(x))
     \]
 
-    then the $\mathfrak{S}$-uniform topology on $E^T$ is defined by the seminorms
-    \[
-    \bracs{[\cdot]_{S, i}|S \in \mathfrak{S}, i \in I}
-    \]
-
-    and hence locally convex.
+    then the $\sigma$-uniform topology on $\cf$ is induced by seminorms of the form $\rho_S$, where $\rho$ is a continuous seminorm on $E$, and $S \in \sigma$. In which case, the $\sigma$-uniform topology on $\cf$ is locally convex. 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:set-uniform-pseudometric}.
+    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}

src/fa/lc/tensor.tex

@@ -19,7 +19,7 @@
         \item $E \otimes_\pi F$ is the linear span of $\iota(E \times F)$. 
         \item For any $U \subset E$ and $V \subset F$, let $U \otimes V = \bracs{u \otimes v|u \in U, v \in V}$, then the convex circled hulls
         \[
-        \fB = \bracsn{\Gamma(U \otimes V)| U \in \cn_E(0), V \in \cn_F(0)}
+        \fB = \bracsn{\aconv(U \otimes V)| U \in \cn_E(0), V \in \cn_F(0)}
         \]
 
         is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$.
@@ -42,9 +42,9 @@
 
     (5): By (4) of the \hyperref[tensor product]{definition:tensor-product}. 
 
-    (6): Let $U \in \cn_E(0)$ and $V \in \cn_F(0)$ be balanced. For any $\sum_{j = 1}^n x_j \otimes y_j \in E \otimes_\pi F$, then there exists $\lambda > 0$ such that $\seqf{x_j} \subset \lambda U$ and $\seqf{y_j} \subset \lambda V$. In which case, $\sum_{j = 1}^n x_j \otimes y_j \subset \lambda \Gamma (U \otimes V)$, so $\fB$ is a collection of convex, circled, and radial sets. Since $\fB$ defines a locally convex topology that satisfies (1) and (2), $\mathcal{S}$ contains the topology defined by $\fB$. 
+    (6): Let $U \in \cn_E(0)$ and $V \in \cn_F(0)$ be balanced. For any $\sum_{j = 1}^n x_j \otimes y_j \in E \otimes_\pi F$, then there exists $\lambda > 0$ such that $\seqf{x_j} \subset \lambda U$ and $\seqf{y_j} \subset \lambda V$. In which case, $\sum_{j = 1}^n x_j \otimes y_j \subset \lambda \aconv (U \otimes V)$, so $\fB$ is a collection of convex, circled, and radial sets. Since $\fB$ defines a locally convex topology that satisfies (1) and (2), $\mathcal{S}$ contains the topology defined by $\fB$. 
 
-    On the other hand, for any convex and circled set $W \in \cn_{E \otimes_\pi F}(0)$, there exists $U \in \cn_E(0)$ and $V \in \cn_F(0)$ such that $U \otimes V \subset W$. In which case, $W \supset \Gamma(U \otimes V)$, so $\fB$ is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$. 
+    On the other hand, for any convex and circled set $W \in \cn_{E \otimes_\pi F}(0)$, there exists $U \in \cn_E(0)$ and $V \in \cn_F(0)$ such that $U \otimes V \subset W$. In which case, $W \supset \aconv(U \otimes V)$, so $\fB$ is a fundamental system of neighbourhoods at $0$ for $E \otimes_\pi F$. 
 \end{proof}
 
 \begin{remark}
@@ -64,7 +64,7 @@
     then
     \begin{enumerate}
         \item $\rho$ is a continuous seminorm on $E \otimes_\pi F$.
-        \item $\rho$ is the gauge of $\Gamma(U \otimes V)$. 
+        \item $\rho$ is the gauge of $\aconv(U \otimes V)$. 
         \item For any $x \in E$ and $y \in F$, $\rho(x \otimes y) = p(x)q(Y)$. 
         \item $\rho$ is a norm if and only if $[\cdot]_U$ and $[\cdot]_V$ are norms. 
     \end{enumerate}
@@ -100,20 +100,20 @@
 
     so $\rho$ satisfies the triangle inequality. 
 
-    (2): Let $z \in \Gamma(U \otimes V)$, then there exists $\seqf{(x_j, y_j)} \subset U \times V$ and $\seqf{\lambda_j} \subset K$ such that $\sum_{j = 1}^n |\lambda_j| \le 1$ and $z = \sum_{j = 1}^n \lambda x_j \otimes y_j$. In which case, 
+    (2): Let $z \in \aconv(U \otimes V)$, then there exists $\seqf{(x_j, y_j)} \subset U \times V$ and $\seqf{\lambda_j} \subset K$ such that $\sum_{j = 1}^n |\lambda_j| \le 1$ and $z = \sum_{j = 1}^n \lambda x_j \otimes y_j$. In which case, 
     \begin{align*}
         \rho(z) &\le \sum_{j = 1}^n p(\lambda x_j)q(y_j) = \sum_{j = 1}^n |\lambda_j|p(x_j)q(y_j) \\
         &< \sum_{j = 1}^n |\lambda_j| \le 1
     \end{align*}
 
-    so $\Gamma(U \otimes V) \subset \bracs{\rho < 1}$.
+    so $\aconv(U \otimes V) \subset \bracs{\rho < 1}$.
 
     Let $z \in \bracs{\rho < 1}$, then there exists $\seqf{(x_j, y_j)} \subset E \times F$ such that $z = \sum_{j = 1}^nx_j \otimes y_j$ and $\sum_{j = 1}^n p(x_j)q(x_j) < 1$. Let $\eps > 0$ such that $\sum_{j = 1}^n(p(x_j) + \eps)(q(x_j) + \eps) < 1$, then
     \[
-    z = \sum_{j = 1}^n (p(x_j) + \eps)(q(x_j) + \eps) \cdot \underbrace{\frac{x_j}{p(x_j) + \eps}}_{\in \bracs{p < 1} = U} \otimes \underbrace{\frac{y_j}{q(x_j) + \eps}}_{\in \bracs{q < 1} = V} \in \Gamma(U \otimes V)
+    z = \sum_{j = 1}^n (p(x_j) + \eps)(q(x_j) + \eps) \cdot \underbrace{\frac{x_j}{p(x_j) + \eps}}_{\in \bracs{p < 1} = U} \otimes \underbrace{\frac{y_j}{q(x_j) + \eps}}_{\in \bracs{q < 1} = V} \in \aconv(U \otimes V)
     \]
 
-    and $\Gamma(U \otimes V) \supset \bracs{\rho < 1}$.
+    and $\aconv(U \otimes V) \supset \bracs{\rho < 1}$.
     
     (3): Let $x \in U$ and $y \in V$. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $\phi \in E^*$ and $\psi \in F^*$ such that $\dpn{x, \phi}{E} = p(x)$, $\dpn{y, \psi}{F} = q(x)$, $|\phi| \le p$, and $|\psi| \le q$. By (U1) of the \hyperref[projective tensor product]{definition:projective-tensor-product}, there exists $\Phi \in (E \otimes_\pi F)^*$ such that the following diagram commutes
     \[

src/fa/notation.tex

@@ -20,7 +20,7 @@
     $\widehat{E}$ & Hausdorff completion of TVS $E$. & \autoref{definition:tvs-completion} \\
     % ---- Locally Convex ----
     $\mathrm{Conv}(A)$ & Convex hull of $A$. & \autoref{definition:convex-hull} \\
-    $\Gamma(A)$ & Convex circled hull of $A$. & \autoref{definition:convex-circled-hull} \\
+    $\aconv(A)$ & Convex circled hull of $A$. & \autoref{definition:convex-circled-hull} \\
     $[\cdot]_A$ & Gauge of a radial set $A$. & \autoref{definition:gauge} \\
     $\rho_M$ & Quotient of seminorm $\rho$ by subspace $M$. & \autoref{definition:quotient-norm} \\
     $E \otimes_\pi F$ & Projective tensor product of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\

src/fa/tvs/spaces-of-linear.tex

@@ -3,9 +3,9 @@
 
 \begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
 \label{proposition:tvs-set-uniformity}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
     \begin{enumerate}
-        \item The $\mathfrak{S}$-uniformity on $F^T$ (\autoref{definition:set-uniform}) is translation invariant.
+        \item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
         \item The composition defined by
         \[
         T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
@@ -13,20 +13,21 @@
 
         is continuous.
     \end{enumerate}
+
     For any vector subspace $\cf \subset F^T$, the following are equivalent:
     \begin{enumerate}
-        \item[(3)] The $\mathfrak{S}$-uniform topology on $\cf \subset F^E$ is a vector space topology.
-        \item[(4)] For each $S \in \mathfrak{S}$ and $f \in \cf$, $f(S) \subset E$ is bounded.
+        \item[(T)] The $\sigma$-uniform topology on $\cf \subset F^E$ is a vector space topology.
+        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
+    (1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
 
-    (2): Let $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
+    (2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
 
-    (3) $\Rightarrow$ (4): Let $f \in \cf$, $S \in \mathfrak{S}$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\mathfrak{S}$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
+    (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
 
-    (4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, then for any $x \in X$,
+    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
     \begin{align*}
     \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
     &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
@@ -83,18 +84,18 @@
 
 \begin{definition}[Space of Bounded Linear Maps]
 \label{definition:bounded-linear-map-space}
-    Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
 
-    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
+    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
 \end{definition}
 
 \begin{proposition}
 \label{proposition:multilinear-identify}
-	Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system that contains all singletons, and $k \in \natp$, then
+	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
 	\begin{enumerate}
         \item The map
         \[
-        I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
+        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
         \]
 
         defined by
@@ -105,7 +106,7 @@
         is an isomorphism.
         \item The map
         \[
-        I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
+        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
         \]
 
         defined by
@@ -117,33 +118,33 @@
     \end{enumerate}
     which allows the identification
     \[
-    \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
     \]
 
     under the map $I$ in (2).
 \end{proposition}
 \begin{proof}
-    (1): To see that $I$ is surjective, let $T \in B_{\mathfrak{S}}^{k+1}(E; F)$ and
+    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
     \[
-    I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
+    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
     \]
 
-    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
+    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
     \[
     T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
     \]
 
-    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
 
-    In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
+    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
 
-    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
+    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
 
-    On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
+    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
 
     (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
     \[
-    \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
     \]
 
     Thus (2) holds for all $k \in \natp$.

src/topology/functions/ideal.tex

@@ -0,0 +1,52 @@
+\section{Ideals}
+\label{section:set-ideal}
+
+
+\begin{definition}[Ideal]
+\label{definition:set-ideal}
+    Let $X$ be a set and $\sigma \subset 2^X$, then $\sigma$ is an \textbf{ideal} over $X$ if:
+    \begin{enumerate}[label=(I\arabic*)]
+        \item For any $E \in \sigma$ and $F \subset E$, $F \in \sigma$. 
+        \item For any $E, F \in \sigma$, $E \cup F \in \sigma$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Covering]
+\label{definition:set-ideal-covering}
+    Let $X$ be a set and $\sigma \subset 2^X$, then $\sigma$ \textbf{covers} $X$/is \textbf{covering} if $\bigcup_{E \in \sigma}E = X$. 
+\end{definition}
+
+\begin{definition}[Fundamental]
+\label{definition:set-ideal-fundamental}
+    Let $X$ be a set, $\sigma \subset 2^X$ be an ideal, and $\tau \subset \sigma$, then $\tau$ is \textbf{fundamental} with respect to $\sigma$ if for any $E \in \sigma$, there exists $F \in \tau$ such that $E \subset F$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:set-ideal-fundamental-criterion}
+    Let $X$ be a set and $\tau \subset 2^X$, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $E, F \in \tau$, there exists $G \in \tau$ such that $E \cup F \subset G$. 
+        \item There exists an ideal $\sigma \subset 2^X$ such that $\tau$ is fundamental with respect to $\sigma$. 
+    \end{enumerate}
+
+    If the above holds, then the ideal $\sigma$ in (2) is the \textbf{ideal generated by} $\tau$. 
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let
+    \[
+    \sigma = \bracs{F \subset X| \exists E \in \tau: F \subset E}
+    \]
+
+    then $\sigma$ satisfies (I1) by definition. For any $E, F \in \sigma$, there exists $E_0, F_0 \in \tau$ such that $E \subset E_0$ and $F \subset F_0$. By assumption, there exists $G \in \tau$ such that
+    \[
+    E \cup F \subset E_0 \cup F_0 \subset G
+    \]
+
+    so $\sigma$ satisfies (I2), and is an ideal. 
+
+    (2) $\Rightarrow$ (1): Let $E, F \in \tau$, then $E \cup F \in \sigma$. Since $\tau$ is fundamental, there exists $G \in \tau$ such that $E \cup F \subset G$. 
+\end{proof}
+
+
+
+

src/topology/functions/index.tex

@@ -1,5 +1,6 @@
 \chapter{Function Spaces}
 \label{chap:function-spaces}
 
+\input{./ideal.tex}
 \input{./set-systems.tex}
 \input{./uniform.tex}

src/topology/functions/set-systems.tex

@@ -1,58 +1,58 @@
-\section{Topology With Respect to Families of Sets}
+\section{Topologies With Respect to Ideals}
 \label{section:pointwise}
 
 \begin{definition}[Set-Open Topology]
 \label{definition:set-open}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \topo)$ be a topological space. For each $S \in \mathfrak{S}$ and $U \subset X$ open, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \topo)$ be a topological space. For each $S \in \sigma$ and $U \subset X$ open, let
     \[
     M(S, U) = \bracs{f \in X^T| f(S) \subset U}
     \]
 
     and
     \[
-    \ce(\mathfrak{S}, \topo) = \bracs{M(S, U)| S \in \mathfrak{S}, U \in \topo}
+    \ce(\sigma, \topo) = \bracs{M(S, U)| S \in \sigma, U \in \topo}
     \]
 
-    then the topology generated by $\ce$ is the \textbf{$\mathfrak{S}$-open topology} on $T^X$.
+    then the topology generated by $\ce$ is the \textbf{$\sigma$-open topology} on $T^X$.
 
-    If $\cb \subset \topo$ generates $\topo$, then $\ce(\mathfrak{S}, \cb)$ generates the $\mathfrak{S}$-open topology.
+    If $\cb \subset \topo$ generates $\topo$, then $\ce(\sigma, \cb)$ generates the $\sigma$-open topology.
 \end{definition}
 
 
 \begin{definition}[Set Uniformity]
 \label{definition:set-uniform}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space. For each $S \in \mathfrak{S}$ and $U \in \fU$, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \fU)$ be a uniform space. For each $S \in \sigma$ and $U \in \fU$, let
     \[
     E(S, U) = \bracs{(f, g) \in X^T|(f(x), g(x)) \in U \forall x \in S}
     \]
 
     and
     \[
-    \mathfrak{E}(\mathfrak{S}, \fU) = \bracs{E(S, U)| S \in \mathfrak{S}, U \in \fU}
+    \mathfrak{E}(\sigma, \fU) = \bracs{E(S, U)| S \in \sigma, U \in \fU}
     \]
 
     then
     \begin{enumerate}
-        \item $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity $\fV$ on $X^T$.
-        \item The topology induced by $\fV$ is finer than the $\mathfrak{S}$-topology on $T^X$.
-        \item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then $\mathfrak{E}(\mathfrak{S}, \fU)$ is forms a fundamental system of entourages for $\fV$.
+        \item $\mathfrak{E}(\sigma, \fU)$ generates a uniformity $\fV$ on $X^T$.
+        \item The topology induced by $\fV$ is finer than the $\sigma$-open topology on $T^X$.
+        \item If $\mathfrak{E}(\sigma, \fU)$ forms a fundamental system of entourages for $\fV$.
     \end{enumerate}
-    The uniformity $\fV$ is the \textbf{$\mathfrak{S}$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\mathfrak{S}$}/\textbf{$\mathfrak{S}$-uniform topology} on $X^T$.
+    The uniformity $\fV$ is the \textbf{$\sigma$-uniformity}, and the topology induced by $\fV$ is the \textbf{topology of uniform convergence on the sets $\sigma$}/\textbf{$\sigma$-uniform topology} on $X^T$.
 \end{definition}
 \begin{proof}
-    (1): Since $\Delta \subset E(S, U)$ for all $S \in \mathfrak{S}$ and $U \in \fU$, $\mathfrak{E}(\mathfrak{S}, \fU)$ generates a uniformity on $X^T$.
+    (1): Since $\Delta \subset E(S, U)$ for all $S \in \sigma$ and $U \in \fU$, $\mathfrak{E}(\sigma, \fU)$ generates a uniformity on $X^T$.
 
     (2): Let $U \subset X$ be open, then for each $x \in U$, there exists $V_x \in \fU$ such that $x \in V_x(x) \subset U$. In which case, $U = \bigcup_{x \in U}V_x(x)$ and $M(S, U) = \bigcup_{x \in U}M(S, V_x)(x)$.
 
     (3): It is sufficient to verify
     \begin{enumerate}
-        \item[(FB1)] For any $S, S'  \in \mathfrak{S}$, there exists $T \in \mathfrak{S}$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
+        \item[(FB1)] For any $S, S'  \in \sigma$, there exists $T \in \sigma$ with $T \supset S, S'$. In which case, for any $U, U' \in \fU$,
         \[
         E(T, U \cap U') \subset E(S \cup S', U \cap U') \subset E(S, U) \cap E(S', U')
         \]
 
-        \item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \mathfrak{S}$.
-        \item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \mathfrak{S}$,
+        \item[(UB1)] For any $U \in \fU$, $\Delta \subset U$. Thus the diagonal in $X^T$ is in $E(S, U)$ for any $S \in \sigma$.
+        \item[(UB2)] For any $U \in \fU$, there exists $V \in \fV$ with $V \circ V \subset U$. Thus for any $S \in \sigma$,
         \[
         E(S, V) \circ E(S, V) \subset E(S, V \circ V) \subset E(S, U)
         \]
@@ -63,43 +63,43 @@
 
 \begin{proposition}
 \label{proposition:set-uniform-pseudometric}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be a non-empty family of sets, and $(X, \fU)$ be a uniform space whose uniformity is induced by the pseudometrics $\seqi{d}$. For each $i \in I$ and $S \in \mathfrak{S}$, let
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $(X, \fU)$ be a uniform space whose uniformity is induced by the pseudometrics $\seqi{d}$. For each $i \in I$ and $S \in \sigma$, let
     \[
     d_{i, S}: X^T \times X^T \quad (f, g) \mapsto \sup_{x \in S}d_i(f(x), g(x))
     \]
 
     then
     \begin{enumerate}
-        \item $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ is a family of pseudometrics induces the $\mathfrak{S}$-uniformity on $X^T$.
-        \item If $\mathfrak{S}$ is upward-directed with respect to inclusion, then
+        \item $\bracs{d_{i, S}| i \in I, S \in \sigma}$ is a family of pseudometrics induces the $\sigma$-uniformity on $X^T$.
+        \item The family
         \[
-        \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
+        \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
         \]
 
-        is a fundamental system of entourages for the $\mathfrak{S}$-uniformity on $X^T$.
+        is a fundamental system of entourages for the $\sigma$-uniformity on $X^T$.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $S \in \mathfrak{S}$ and $U \in \fU$, then there exists $r > 0$ and $J \subset I$ finite such that $\bigcap_{j \in J}E(d_j, r) \subset U$, so
+    (1): Let $S \in \sigma$ and $U \in \fU$, then there exists $r > 0$ and $J \subset I$ finite such that $\bigcap_{j \in J}E(d_j, r) \subset U$, so
     \[
     \bigcap_{j \in J}E(d_{j, S}, r) \subset E\paren{S, \bigcap_{j \in J}E(d_j, r)} \subset E(S, U)
     \]
 
-    and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \mathfrak{S}}$ contains the $\mathfrak{S}$-uniformity.
+    and the uniformity induced by $\bracs{d_{i, S}| i \in I, S \in \sigma}$ contains the $\sigma$-uniformity.
 
-    On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\mathfrak{S}$-uniformity contains the induced uniformity.
+    On the other hand, for any $i \in I$ and $r > 0$, $E(d_j, r/2) \in \fU$ by \autoref{definition:pseudometric-uniformity}. Therefore $E(S, E(d_j, r/2)) \subset E(d_{j, S}, r)$, so the $\sigma$-uniformity contains the induced uniformity.
 
-    (2): If $\mathfrak{S}$ is upward-directed with respect to inclusion, then by \autoref{definition:set-uniform},
+    (2): By \autoref{definition:set-uniform},
     \[
-    \bracs{E(S, U)| U \in \fU, S \in \mathfrak{S}}
+    \bracs{E(S, U)| U \in \fU, S \in \sigma}
     \]
 
-    Following the same steps in (1),
+    is a fundamental system of entourages for the $\sigma$-uniformity. Following the same steps in (1),
     \[
-    \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \mathfrak{S}}
+    \bracs{\bigcap_{j \in J}E(d_{j, S}, r)|J \subset I \text{ finite}, r > 0, S \in \sigma}
     \]
 
-    is a fundamental system of entourages for the $\mathfrak{S}$-uniformity.
+    is a fundamental system of entourages for the $\sigma$-uniformity.
 \end{proof}
 
 
@@ -109,7 +109,7 @@
     Let $T$ be a set and $X$ be a topological space, then the following topologies on $X^T$ coincide:
     \begin{enumerate}
         \item The product topology on $X^T$.
-        \item The $\mathfrak{S}$-open topology, where $\mathfrak{S} = \bracs{\bracs{x}| x \in X}$.
+        \item The $\sigma$-open topology, where $\sigma$ is the collection of all finite sets.
         \item (If $X$ is a uniform space) The $\mathfrak{F}$-uniform topology, where $\fF = \bracs{F| F \subset X \text{ finite}}$.
     \end{enumerate}
     This topology is the \textbf{topology of pointwise convergence} on $X^T$.
@@ -125,10 +125,28 @@
 
 \begin{proposition}
 \label{proposition:set-uniform-complete}
-    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ such that $\bigcup_{S \in \mathfrak{S}}S = T$, and $(X, \fU)$ be a complete uniform space, then $X^T$ equipped with the $\mathfrak{S}$-uniformity is complete.
+    Let $T$ be a set, $\sigma \subset 2^T$ be a covering ideal, and $(X, \fU)$ be a complete uniform space, then $X^T$ equipped with the $\sigma$-uniformity is complete.
 \end{proposition}
 \begin{proof}
-    Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \mathfrak{S}$ with $x \in S$, then $\pi_x(\fF) \subset 2^X$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_x(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \mathfrak{S}}S$.
+    Let $\fF \subset 2^{X^T}$ be a Cauchy filter. Let $x \in T$. If there exists $S \in \sigma$ with $x \in S$, then $\pi_x(\fF) \subset 2^X$ is a Cauchy filter. By completeness of $X$, there exists $f: T \to X$ such that $\pi_x(\fF) \to f(x)$ for all $x \in \bigcup_{S \in \sigma}S$.
 
-    Let $S \in \mathfrak{S}$ be non-empty and $V_0, V$ be symmetric entourages of $X$ with $V_0 \circ V_0 \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_0)$-small set $F \in \fF$. Let $x \in S$, then $\pi_x(\fF) \to f(x)$ implies that $F_x = \pi_x^{-1}(V_0(f(x))) \in \fF$. Since $F \cap F_x \ne \emptyset$, $(f(x), g(x)) \in V_0 \circ V_0 \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.
+    Let $S \in \sigma$ be non-empty and $V_0, V$ be symmetric entourages of $X$ with $V_0 \circ V_0 \subset V$. Since $\fF$ is Cauchy, there exists an $E(S, V_0)$-small set $F \in \fF$. Let $x \in S$, then $\pi_x(\fF) \to f(x)$ implies that $F_x = \pi_x^{-1}(V_0(f(x))) \in \fF$. Since $F \cap F_x \ne \emptyset$, $(f(x), g(x)) \in V_0 \circ V_0 \subset V$ for all $g \in F$. As this holds for all $x \in S$, $E \subset E(S, V)(f) \in \fF$. Thus $\fF \to f$.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:compact-uniform-open}
+    Let $X$ be a topological space, $\kappa \subset 2^X$ be the collection of all precompact sets in $X$, and $(Y, \fU)$ be a uniform space, then the $\kappa$-open topology and $\kappa$-uniform topology on $C(X; Y)$ coincide. 
+\end{proposition}
+\begin{proof}
+    By \autoref{definition:set-uniform}, the $\kappa$-uniform topology is finer than the $\kappa$-open topology. 
+
+    Let $K \in \kappa$ be compact, $U \in \fU$ be symmetric, and $f \in C(X; Y)$. For each $x \in K$, there exists $V_x \in \cn_X(x)$ such that $f(V_x) \subset U(f(x))$. Since $K$ is compact, there exists $\seqf{x_j} \subset X$ such that $K \subset \bigcup_{j = 1}^n V_{x_j}$. 
+
+    Let $g \in \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j)))$ and $y \in K$, then there exists $1 \le j \le n$ such that $y \in V_{x_j}$. In which case, since $f(y), g(y) \in U(f(x_j))$, $(f(y), g(y)) \in U \circ U$. Therefore
+    \[
+    f \in  \bigcap_{j = 1}^n M(V_{x_j} \cap K, U(f(x_j))) \subset [E(K, U \circ U)](f)
+    \]
+
+    and the $\kappa$-open topology is finer than the $\kappa$-uniform topology.
+\end{proof}
+

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}

src/topology/notation.tex

@@ -17,7 +17,7 @@
     $U \circ V$ & Composition of $U, V \subset X \times X$. & \autoref{definition:composition} \\
     $U(A)$ & Slice of $U \subset X \times Y$ at $A \subset X$: $\{y \mid \exists x \in A,\, (x,y) \in U\}$. & \autoref{definition:slice} \\
     $E(S, U)$ & Entourage of the form $\{(f,g) \in X^T \mid (f(x),g(x)) \in U\ \forall x \in S\}$. & \autoref{definition:set-uniform} \\
-    $\mathfrak{E}(\mathfrak{S}, \mathfrak{U})$ & $\mathfrak{S}$-uniformity, generated by $\{E(S,U) \mid S \in \mathfrak{S},\ U \in \mathfrak{U}\}$. & \autoref{definition:set-uniform} \\
+    $\mathfrak{E}(\sigma, \mathfrak{U})$ & $\sigma$-uniformity, generated by $\{E(S,U) \mid S \in \sigma,\ U \in \mathfrak{U}\}$. & \autoref{definition:set-uniform} \\
 
     % Function Spaces
     $\mathrm{supp}(f)$ & Support of $f$. & \autoref{definition:support} \\

src/topology/uniform/compact.tex

@@ -0,0 +1,74 @@
+\section{Compact Uniform Spaces}
+\label{section:compact-uniform}
+
+\begin{definition}
+\label{definition:totally-bounded-uniform}
+    Let $(X, \fU)$ be a uniform space, then the following are equivalent:
+    \begin{enumerate}
+        \item For every $U \in \fU$, there exists $\seqf{x_j} \subset U$ such that $X = \bigcup_{j = 1}^n U(x_j)$. 
+        \item Every ultrafilter on $X$ is Cauchy.
+        \item For every filter $\fF_0 \subset 2^X$, there exists $\fF \supset \fF_0$ such that $\fF$ is Cauchy. 
+    \end{enumerate}
+
+    If the above holds, then $(X, \fU)$ is \textbf{totally bounded}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $\fF \subset 2^X$ be an ultrafilter and $U \in \fU$, then there exists $U$-small sets $\seqf{E_j} \subset 2^X$ such that $X = \bigcup_{j = 1}^n E_j$. By (3) of \autoref{definition:ultrafilter}, there exists $1 \le j \le n$ such that $E_j \in \fF$, so $\fF$ is Cauchy.
+
+    (2) $\Rightarrow$ (3): By the \hyperref[ultrafilter lemma]{lemma:ultrafilter}. 
+
+    $\neg (1) \Rightarrow \neg (3)$: Let $U \in \fU$ be symmetric such that for all $X_0 \subset X$ finite, $\bigcup_{x \in X_0}U(x) \subsetneq X$. Let $\fF_0$ be the filter generated by
+    \[
+    \bracs{X \setminus \bigcup_{x \in X_0}U(x) \bigg | X_0 \subset X \text{ finite}}
+    \]
+
+    then for any $U$-small set $E \subset X$ and $x \in E$, $E \cap [X \setminus U(x)] = \emptyset$. Therefore no filter finer than $\fF_0$ contains a $U$-small set. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:totally-bounded-image}
+    Let $X, Y$ be uniform spaces and $f \in UC(X; Y)$. If $X$ is totally bounded, then so is $f(X)$. 
+\end{proposition}
+\begin{proof}
+    Let $U$ be an entourage of $Y$, then there exists $\seqf{x_j} \subset X$ such that $\bigcup_{j = 1}^n[(f \times f)^{-1}(U)](x_j) = X$. In which case, $f(x) \subset \bigcup_{j = 1}^n U(f(x_j))$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:compact-uniform}
+    Let $X$ be a uniform space, then the following are equivalent: 
+    \begin{enumerate}
+        \item $X$ is compact. 
+        \item $X$ is complete and totally bounded. 
+    \end{enumerate}
+
+    In particular, $X$ is totally bounded if and only if its Hausdorff completion is compact. 
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By (1) of \autoref{definition:compact}, $X$ is totally bounded. Let $\fF$ be a Cauchy filter, then by (4) of \autoref{definition:compact}, $\fF$ has at least one accumulation point. Since $\fF$ is Cauchy, the accumulation points and limit points of $\fF$ coincide by (2) of \autoref{proposition:cauchyfilterlimit}. 
+
+    (2) $\Rightarrow$ (1): By (2) of \autoref{definition:totally-bounded-uniform}, every ultrafilter is Cauchy, and hence converges, which satisfies (4) of \autoref{definition:compact}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:totally-bounded-product}
+    Let $\seqi{X}$ be totally bounded uniform spaces, then $\prod_{i \in I}X_i$ is totally bounded. 
+\end{proposition}
+\begin{proof}
+    Let $J \subset I$ be finite and $\seqj{U}$ such that $U_j$ is an entourage of $X_j$ for each $j \in J$. Since each $X_j$ is totally bounded, there exists $Y_j \subset X_j$ finite such that $X_j = U_j(Y_j)$. Let $Y = \prod_{j \in J}Y_j$ and $\phi_y \in \pi_J^{-1}(y)$ for each $y \in Y$, then for any $x \in X$, there exists $y \in Y$ such that $\pi_j(x) \in U_j(\pi_j(y))$ for each $j \in J$. Therefore
+    \[
+    X = \bigcup_{x \in Y}\braks{\bigcap_{j \in J}(\pi_j \times \pi_j)^{-1}(U_j)}(\phi_x)
+    \]
+\end{proof}
+
+
+
+\begin{proposition}
+\label{proposition:uniform-continuous-compact}
+    Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$.
+\end{proposition}
+\begin{proof}
+    Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. 
+    
+    Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. 
+\end{proof}

src/topology/uniform/equicontinuous.tex

@@ -0,0 +1,69 @@
+\section{Equicontinuity}
+\label{section:equicontinuity}
+
+\begin{definition}[Equicontinuous]
+\label{definition:equicontinuous}
+    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^X$, and $x \in X$, then $\cf$ is \textbf{equicontinuous at $x$} if for every $U \in \fU$, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $y \in V$ and $f \in \cf$. 
+    
+    The set $\cf \subset C(X; Y)$ is \textbf{equicontinuous} if it is equicontinuous at every point in $x$. 
+\end{definition}
+
+\begin{definition}[Uniformly Equicontinuous]
+\label{definition:uniformly-equicontinuous}
+    Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is \textbf{uniformly equicontinuous} if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$. 
+\end{definition}
+
+
+\begin{theorem}[Arzelà-Ascoli]
+\label{theorem:arzela-ascoli}
+    Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, and $\cf \subset C(X; Y)$. If
+    \begin{enumerate}[label=(E\arabic*)]
+        \item $\cf$ is equicontinuous.
+    \end{enumerate}
+
+    then
+    \begin{enumerate}[label=(C\arabic*)]
+        \item The product uniformity and the compact uniformity on $\cf$ coincide. 
+    \end{enumerate}
+
+    In addition, if $\cf$ satisfies (E1) and
+    \begin{enumerate}[label=(E\arabic*), start=1]
+        \item For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$.
+    \end{enumerate}
+
+    then
+    \begin{enumerate}[label=(C\arabic*), start=1]
+        \item $\cf$ is a precompact subset of $Y^X$ with respect to the compact uniformity. 
+    \end{enumerate}
+
+    Conversely, if $X$ is a LCH space, then (C2) implies (E1) + (E2).
+\end{theorem}
+\begin{proof}
+    (E1) $\Rightarrow$ (C1): By \autoref{proposition:compact-uniform-open}, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$ and thus $\cf$.
+
+    Let $K \subset X$ be compact, and $U \in \fU$. Since $\cf$ is equicontinuous, for each $x \in K$, there exists $V_x \in \cn_X(x)$ such that $g(V_x) \subset U(g(x))$ for all $g \in \cf$. By compactness of $K$, there exists $\seqf{x_j} \subset K$ such that $K \subset \bigcup_{j = 1}^n V_{x_j}$. Let $f, g \in \cf$ such that $(f(x_j), g(x_j)) \in E$ for all $1 \le j \le n$. For any $x \in K$, there exists $1 \le j \le n$ such that $x \in V_{x_j}$. In which case, 
+    \[
+    (f(x), f(x_j)), (f(x_j), g(x_j)), (g(x_j), g(x)) \in U
+    \]
+
+    so $(f(x), g(x)) \in U \circ U \circ U$. Therefore
+    \[
+    \bigcap_{j = 1}^n E(\bracs{x_j}, U) \subset E(K, U \circ U \circ U)
+    \]
+
+    so the product uniformity and the compact uniformity coincide. 
+
+    (E1) + (E2) $\Rightarrow$ (C2): By \autoref{proposition:totally-bounded-product}, $\prod_{x \in X}\cf(x)$ is totally bounded. Since $\cf \subset \prod_{x \in X}\cf(x)$, $\cf$ is also totally bounded. As the pointwise and compact uniformities coincide, $\cf$ is totally bounded with respect to the compact uniformity. By \autoref{proposition:product-complete} and \autoref{proposition:compact-uniform}, $\prod_{x \in X}\ol{\cf(x)}$ is compact and hence complete. Therefore the closure of $\cf$ with respect to the compact uniformity is compact. 
+
+    (C2) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
+    \[
+    (g(x), f_j(x)), (f_j(x), f_j(y)), (f_j(y), g(y)) \in U \circ U \circ U
+    \]
+
+    Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous. 
+
+    (C2) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
+\end{proof}
+
+
+

src/topology/uniform/index.tex

@@ -4,6 +4,8 @@
 \input{./definition.tex}
 \input{./uc.tex}
 \input{./metric.tex}
+\input{./compact.tex}
+\input{./equicontinuous.tex}
 \input{./cauchy.tex}
 \input{./complete.tex}
 \input{./completion.tex}

src/topology/uniform/uc.tex

@@ -164,14 +164,3 @@
 
 \end{proof}
 
-
-\begin{proposition}
-\label{proposition:uniform-continuous-compact}
-    Let $X$ be a compact uniform space, $Y$ be a uniform space, and $f \in C(X; Y)$, then $f \in UC(X; Y)$.
-\end{proposition}
-\begin{proof}
-    Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. 
-    
-    Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. 
-\end{proof}
-