Added Goldstine's Theorem.

src/fa/duality/definitions.tex

@@ -78,4 +78,38 @@
     (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}
 
+\begin{theorem}[Goldstine]
+\label{theorem:goldstine-weak}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $A \subset E$ be non-empty, convex, circled, and $\sigma(E, F)$-compact, and $B$ be the closed unit ball of $E_A^*$, then $B \cap F$ is $\sigma(E_A^*, E_A)$-dense in $B$. 
+\end{theorem}
+\begin{proof}
+    For any $S \subset E$ or $S \subset F$, denote $S^\circ$ as the polar of $S$ with respect to $\dpn{E, F}{\lambda}$. For any $S \subset E_A$ or $S \subset E_A^*$, denote $S^\bullet$ as the polar of $S$ with respect to $\dpn{E_A, E_A^*}{E_A}$. 
+
+
+    Since $B_{E_A}(0, 1)$ is circled and $A = \ol{B_{E_A}(0, 1)}^{E_A}$ is compact in $E$, 
+    \[
+    B \cap F = \bracsn{\phi \in F| \text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall x \in A} = A^\circ
+    \]
+
+    is the polar of $A$ with respect to $\dpn{E, F}{\lambda}$. Now, as $A$ is convex, circled, and compact, the \hyperref[Bipolar Theorem]{theorem:bipolar} implies that
+    \begin{align*}
+    A^{\circ\bullet} &= \bracsn{x \in E_A|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ} \\
+    &= E_A \cap \bracsn{x \in E|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ} = E_A \cap A^{\circ\circ} = A
+    \end{align*}
+
+    Given that $B \cap F$ is a convex and circled subset of $E_A^*$, 
+    \[
+    \ol{B \cap F}^{\sigma(E_A^*, E_A)} = (B \cap F)^{\bullet\bullet} = A^{\circ\bullet\bullet} = A^\bullet = B
+    \]
+\end{proof}
+
+\begin{corollary}
+\label{corollary:weak-dense-unit-ball}
+    Let $E$ be a normed space over $K \in \RC$, then $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^*)$-dense in $\ol{B_{E^{**}}(0, 1)}$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{E^*}(0, 1)}$ is convex, circled, and $\sigma(E^*, E)$-compact. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $E \cap \ol{B_{E^{**}}(0, 1)}$ is $\sigma(E^{**}, E^*)$-dense in $\ol{B_{E^{**}}(0, 1)}$. 
+\end{proof}
+
+
 

src/fa/duality/mackey.tex

@@ -66,6 +66,22 @@
     On the other hand, let $\mathcal{T} \subset 2^E$ be a locally convex topology consistent with $\dpn{E, F}{\lambda}$. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, every $\mathcal{T}$-equicontinuous set is relatively $\sigma(F, E)$-compact. Therefore $\mathcal{T}$ is coarser than the topology of uniform convergence on relatively $\sigma(E, F)$-compact, convex, and circled sets. 
 \end{proof}
 
+\begin{corollary}
+\label{corollary:mackey-bounded}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $\topo$ be a topology on $E$ consistent with $\dpn{E, F}{\lambda}$, and $B \subset E$, then $B$ is bounded with respect to $\topo$ if and only if $B$ is bounded with respect to $\sigma(E, F)$. 
+\end{corollary}
+\begin{proof}
+    Assume without loss of generality that $\topo = \tau(E, F)$. Suppose that $B$ is $\sigma(E, F)$-bounded. Let $A \subset F$ be convex, circled, and $\sigma(F, E)$-compact. By continuity of dual pairing, 
+    \[
+    A = \bracsn{\phi \in F|\ |\dpn{x, \phi}{\lambda}| \le 1 \forall x \in B}
+    \]
+
+    is a convex, circled, and closed subset. Given that $B$ is $\sigma(E, F)$-bounded, $\sup_{x \in B}|\dpn{x, \phi}{\lambda}| < \infty$ for all $x \in E$. Thus $A$ is absorbing and hence a barrel. 
+
+    Since $B$ is compact, the auxiliary space $E_B$ is a Banach space. In particular, $E_B$ is barreled by \autoref{proposition:baire-barrel}. By continuity of the inclusion map, $A \cap E_B$ is a barrel in $E_B$. Therefore there exists $\lambda > 0$ such that $B \subset \lambda A \cap E_B \subset \lambda A$.
+\end{proof}
+
+
 \begin{definition}[Mackey Space]
 \label{definition:mackey-space}
     Let $E$ be a separated locally convex space over $K \in \RC$, then $E$ is a \textbf{Mackey space} if $E$ is equipped with the Mackey topology of $\dpn{E, E^*}{E}$. 

src/fa/lc/compact.tex

@@ -1,6 +1,27 @@
 \section{Compact Convex Sets}
 \label{section:compact-convex}
 
+
+\begin{theorem}[Mazur]
+\label{theorem:convex-hull-complete}
+    Let $E$ be a locally convex space over $K \in \RC$ and $A \subset E$ be compact, then
+    \begin{enumerate}
+        \item $\conv(A)$ is totally bounded in $E$. 
+        \item If $E$ is complete, then $\conv(A)$ is relatively compact. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $U \in \cn_E(0)$ be convex and circled, then since $A$ is compact, there exists $B \subset A$ finite such that $A \subset B + U$. As such, $\conv(A) \subset \conv(B + U) = \conv(B) + U$. Since $B$ is finite, $\conv(B)$ is compact. Thus there exists $C \subset \conv(B)$ finite such that
+    \[
+    \conv(A) \subset \conv(B + U) = \conv(B) + U \subset C + U
+    \]
+
+    which yields a finite covering of $\conv(A)$ using $U$. 
+
+    (2): By \autoref{proposition:compact-uniform}. 
+\end{proof}
+
+
 \begin{definition}[Extreme Point]
 \label{definition:extreme-point}
     Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$. 
@@ -83,3 +104,43 @@
     
 \end{proof}
 
+\begin{lemma}
+\label{lemma:compact-null-auxiliary}
+    Let $E$ be a Banach space over $K \in \RC$ and $A \subset E$ be compact, then:
+    \begin{enumerate}
+        \item There exists a null sequence $\seq{x_n} \subset E$ such that $A \subset \ol{\conv}(\seq{x_n})$.
+        \item There exists a compact, convex, and circled set $B \subset E$ such that $A$ is compact in $E_B$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma III.9.1]{SchaeferWolff}}}. ]
+    (1): Assume without loss of generality that $A \ne \emptyset$. Let $\bracsn{\lambda_n}_0^\infty \subset (0, \infty)$ such that $\sum_{n \in \natz}\lambda_n = 1$. For each $n \in \natz$, let $r_n = \lambda_{n+1}^2$ and $A_n \subset A$ be finite such that $A \subset \bigcup_{x \in A_n}B_E(x, r_n)$. 
+
+    Define $B_0 = \lambda_0^{-1}A_0$. For each $n \in \natp$, write $A_n = \bracsn{x_j}_1^k$. By definition of $A_{n-1}$, there exists $\bracsn{y_j}_1^k \subset A_{n-1}$ such that $d(x_j, y_j) < r_{n-1}$ for all $1 \le j \le k$. For every $1 \le j \le k$, let $z_j = (x_j - y_j)/\lambda_n$, and define $B_n = \bracsn{z_j}_1^k$. 
+
+    By the above construction, $A_n \subset \sum_{j = 0}^n \lambda_j B_j$ for all $n \in \natz$. As $\sum_{n \in \natz}\lambda_n = 1$, 
+    \[
+    A \subset \ol{\conv}\braks{\bigcup_{n \in \natz}A_n} \subset \ol{\conv}\braks{\bracs{0} \cup \bigcup_{n \in \natz}B_n}
+    \]
+
+    Finally, for each $n \in \natp$, $B_n$ is finite with $\norm{z}_E \le r_{n-1}/\lambda_{n} = \lambda_n$ for all $z \in B_n$. As $B_0$ is finite as well, any enumeration of $\bracs{0} \cup \bigcup_{n \in \natz}B_n$ yields a null sequence. 
+
+    (2): Using (1) and \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, assume without loss of generality that there exists a null sequence $\seq{x_n} \subset E$ such that $A$ is the closed convex hull of $\seq{x_n}$. 
+
+    Since $\seq{x_n} \subset E$ is a null sequence, there exists $\seq{\lambda_n} \subset [1, \infty)$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\lambda_n \to \infty$ as $n \to \infty$. 
+        \item $\lambda_n x_n \to 0$ as $n \to \infty$. 
+    \end{enumerate}
+
+    Let $B = \ol{\aconv}(\seq{\lambda_n x_n})$, then by \hyperref[Mazur's Theorem]{theorem:convex-hull-complete}, $B$ is a compact, convex, and circled subset of $E$ with $\seq{x_n} \subset B$. In addition, $\seq{x_n} \subset E_B$ with $\norm{x_n}_{E_B} \le \lambda_n^{-1}$ for all $n \in \natp$. Thus $\seq{x_n}$ is a null sequence in $E_B$ as well. 
+
+    Now, let $A'$ be the closed convex hull of $\seq{x_n}$ with respect to $E_B$. Since the inclusion $E_B \to E$ is continuous, $A'$ is a compact convex set in $E$ by \autoref{proposition:compact-extensions}. As such, $A' = A$ by \autoref{proposition:closure-of-image} and \autoref{proposition:compact-closed}. Therefore $A$ is a compact subset of $E_B$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:auxiliary-weak-dense}
+    Let $E$ be a separated locally convex space over $K \in \RC$, $A \subset E$ be compact, convex, and circled, and $
+\end{lemma}
+
+
+

src/fa/lc/tensor.tex

@@ -28,7 +28,7 @@
     
     The space $E \otimes_\pi F$ is the \textbf{projective tensor product} of $E$ and $F$, and the mapping $\iota \in L^2(E, F; E \otimes_\pi F)$ is the \textbf{canonical embedding}.
 
-    The space $E \widetilde{\otimes}_\pi F$ denotes the Hausdorff completion of $E \otimes_\pi F$.
+    The space $E \wh{\otimes}_\pi F$ denotes the Hausdorff completion of $E \otimes_\pi F$.
 \end{definition}
 \begin{proof}
     Let $E \otimes_\pi F = E \otimes F$ be the \hyperref[tensor product]{definition:tensor-product} of $E$ and $F$ as vector spaces. Let $\mathscr{T} \subset 2^{2^X}$ be the collection of all locally convex topologies satisfying (1) and (2), and let $\mathcal{S}$ be the projective topology on $E \otimes_\pi F$ generated by $\mathscr{T}$. 

src/fa/norm/ap.tex

@@ -0,0 +1,45 @@
+\section{The Approximation Property}
+\label{section:approximation-property}
+
+\begin{definition}[Approximation Property]
+\label{definition:approximation-property}
+    Let $E$ be a separated locally convex space over $K \in \RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item The closure of $E^* \otimes E$ in $L_c(E; E)$ contains the identity map.
+        \item $E^* \otimes E$ is dense in $L_c(E; E)$. 
+        \item For each locally convex space $F$ over $K$, $E^* \otimes F$ is dense in $L_c(E; F)$. 
+        \item For each locally convex space $F$ over $K$, $F^* \otimes E$ is dense in $L_c(F; E)$. 
+    \end{enumerate}
+
+    If the above holds, then $E$ has the \textbf{approximation property}. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $T \in L_c(E; E)$ and $A \subset E$ be precompact, then $T(A)$ is also precompact by \autoref{proposition:totally-bounded-image}. Let $U \in \cn_E(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in U$ for all $x \in T(A)$. In which case, $STx - Tx \in U$ for all $x \in A$. 
+
+    (1) $\Rightarrow$ (3): Let $T \in L_c(E; F)$ and $A \subset E$ be precompact, and $U \in \cn_F(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in T^{-1}(U)$ for all $x \in A$. In which case, $TS \in E^* \otimes F$ and $TSx - Tx \in U$ for all $x \in A$. 
+
+    (1) $\Rightarrow$ (4): Let $T \in L_c(F; E)$ and $A \subset F$ be precompact, then $T(A)$ is also precompact. Let $U \in \cn_E(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in U$ for all $x \in T(A)$. Thus $STx - Tx \in U$ for all $x \in A$.  
+\end{proof}
+
+\begin{proposition}
+\label{proposition:approximation-property-associated}
+    Let $E$ be a locally convex space over $K \in \RC$. If there exists a fundamental system of convex and circled neighbourhoods $\fB \subset \cn_E(0)$ such that for each $V \in \fB$, $\wh E_V$ has the approximation property, then $E$ has the approximation property. 
+\end{proposition}
+\begin{proof}
+    Let $V \in \fB$, $\pi_V: E \to \wh E_V$ be the canonical projection, and $A \subset E$ be precompact, then $\pi_V(A)$ is precompact as well. Since $\wh E_V$ has the approximation property, there exists $T \in E_V^* \otimes \wh E_V$ such that $Tx - x \in \pi_V(V)$ for all $x \in \pi_V(A)$. As $E_V$ is dense in $\wh E_V$, there exists $S \in E_V^* \otimes E_V$ such that $Sx - Tx \in \pi_V(V)$ for all $x \in \pi_V(A)$. In which case, $Sx - x \in 2\pi_V(V)$ for all $x \in \pi_V(A)$, and $S \circ \pi_V(x) - \pi_V(x) \in 2\pi_V(V)$. 
+    
+    Write $S = \sum_{j = 1}^n \phi_j \otimes y_j$. For each $1 \le j \le n$, choose any representative $x_j \in \pi_V^{-1}(y_j)$, then for any $x \in A$, 
+    \[
+        \pi_V \braks{x - \sum_{j = 1}^n x_j\dpn{x, \phi_j \circ \pi_V}{E}} = S \circ \pi_V(x) - \pi_V(x) \in -2\pi_V(V) = 2\pi_V(V)
+    \]
+
+    Finally, since $\ker(\pi_V) = \bigcap_{\lambda > 0}\lambda V \subset V$, $x - \sum_{j = 1}^n x_j\dpn{x, \phi_j \circ \pi_V}{E} \in -3V = 3V$. Therefore if $R = \sum_{j = 1}^n (\phi_j \circ \pi_V) \otimes x_j \in E^* \otimes E$, then $Rx - x \in 3V$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:approximation-property-hilbert}
+    Every subspace of a product of Hilbert spaces has the approximation property. Every subspace of a projective limit of Hilbert spaces has the approximation property. 
+\end{corollary}
+
+
+

src/fa/norm/compact.tex

@@ -0,0 +1,28 @@
+\section{Compact Operators}
+\label{section:compact-operator}
+
+\begin{definition}[Compact Operator]
+\label{definition:compact-operator}
+    Let $E, F$ be locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then $T$ is \textbf{compact} if there exists $U \in \cn_E(0)$ such that $T(U)$ is relatively compact in $F$. 
+
+    The set $\mathcal{K}(E; F)$ is the \textbf{space of compact operators} from $E$ to $F$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:compact-normed-complete}
+    Let $E$ be a normed space over $K \in \RC$ and $F$ be a complete Hausdorff topological vector space over $K$, then $\mathcal{K}(E; F)$ is a closed subspace of $L_b(E; F)$. 
+\end{proposition}
+\begin{proof}
+    Let $T \in \ol{\mathcal{K}(E; F)}$ and $U \in \cn_F(0)$ be circled, then there exists $S \in \mathcal{K}(E; F)$ such that $Sx - Tx \in U$ for all $x \in B_E(0, 1)$. Since $S$ is compact, there exists $Y \subset F$ finite with $S(B_E(0, 1)) \subset Y + U$. In which case, $T(B_E(0, 1)) \subset Y + U + U = Y + 2U$. Therefore $T(B_E(0, 1))$ is totally bounded, and as $F$ is complete, relatively compact in $F$. 
+\end{proof}
+
+\begin{theorem}[Schauder]
+\label{theorem:compact-adjoint}
+    Let $E, F$ be normed vector spaces over $K \in \RC$ with $F$ being complete and $T \in L(E; F)$, then $T$ is compact if and only if $T^* \in L(F^*; E^*)$ is compact. 
+\end{theorem}
+\begin{proof}
+    ($\Rightarrow$): Let $\cf \subset F^*$ be bounded, then $\cf$ is equicontinuous and hence relatively compact in the weak* topology by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Since $F$ is complete, $\ol{T(B_E(0, 1))}$ is compact. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is relatively compact with respect to the topology of uniform convergence on $\ol{T(B_E(0, 1))}$. Thus $T^*(\cf) = \bracsn{\phi \circ T|\phi \in \cf}$ is relatively compact with respect to the topology of uniform convergence on $B_E(0, 1)$. In other words, $T^*(\cf)$ is relatively compact in $E^*$, and $T^*$ is a compact operator. 
+
+    ($\Leftarrow$): By \autoref{proposition:operator-space-completeness}, $E^*$ is complete. The preceding case then implies that $T^{**} \in L(E^{**}; F^{**})$ is compact. As such, its restriction to $E$, being identified with $T$, is also compact. 
+\end{proof}
+

src/fa/norm/index.tex

@@ -7,3 +7,5 @@
 \input{./separable.tex}
 \input{./multilinear.tex}
 \input{./hilbert.tex}
+\input{./compact.tex}
+\input{./ap.tex}

src/fa/norm/separable.tex

@@ -56,4 +56,31 @@
     then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. 
 \end{proof}
 
+\begin{lemma}
+\label{lemma:compact-embed}
+    Let $E$ be a normed vector space over $K \in \RC$ and $A \subset [0, 1]$ be closed, then $C(A; E)$ embeds isometrically into $C([0, 1]; E)$. 
+\end{lemma}
+\begin{proof}
+    First note that if $0 \not\in A$ or $1 \not\in A$, $C(A; E)$ embeds isometrically into $C(A \cup \bracs{0, 1}; E)$ through extension by $0$. Thus assume without loss of generality that $A$ contains the endpoints $0$ and $1$. 
+
+    Let $U = [0, 1] \setminus A$, then there exists $\seq{(a_n, b_n)} \subset [0, 1]^2$ such that  $U = \bigsqcup_{n \in \natp}(a_n, b_n)$. For each $f \in C(A; E)$, let
+    \[
+    Tf: [0, 1] \to E \quad x \mapsto \begin{cases}
+        f(x) &x \in A \\
+         \frac{b_n - x}{b_n - a_n}f(a_n) + \frac{x - a_n}{b_n - a_n}f(b_n) &x \in (a_n, b_n) \subset [0, 1]
+    \end{cases}
+    \]
+
+    then the mapping $f \mapsto Tf$ is an isometric embedding into $E^{[0, 1]}$ with respect to the uniform norm.
+    
+    Since $U$ is open and $Tf$ is affine on each component of $U$, $Tf$ is continuous on $U$. It remains to show that $Tf$ is continuous on $A$. Let $x \in A$ and $\eps > 0$, then there exists $\delta > 0$ such that $\norm{f(y) - f(x)}_E < \eps$ for all $y \in (x -\delta, x + \delta) \cap A$. Now, a case analysis: 
+    \begin{enumerate}
+        \item If there exists $y \in (x - \delta, x) \cap A$, then for any $z \in U \cap (y, x)$, there exists $n \in \natp$ such that $(a_n, b_n) \subset (y, x)$ and $z \in (a_n, b_n)$. In which case, since $\norm{f(a_n) - f(x)}_E < \eps$ and $\norm{f(b_n) - f(x)}_E < \eps$, $\norm{Tf(z) - Tf(x)}_E < \eps$. Thus $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x)$.
+        \item Otherwise, $x = 0$ or $Tf|_{(x - \delta, x)}$ is an affine function. Either way, there exists $y \in (x - \delta, x)$ such that $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. 
+    \end{enumerate}
+
+    Thus there exists $y \in (x - \delta, x)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. Similarly, there exists $y' \in (x, x + \delta)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (x, y') \cap [0, 1]$. Therefore $Tf$ is continuous at $x$. Since this holds for all $x \in U$ and $x \in A$, $Tf \in C([0, 1]; E)$. 
+\end{proof}
+
+
 

src/fa/notation.tex

@@ -17,6 +17,8 @@
     $L_s(E; F)$ & $L(E; F)$ with strong operator topology. & \autoref{definition:strong-operator-topology} \\
     $L_w(E; F)$ & $L(E; F)$ with weak operator topology. & \autoref{definition:weak-operator-topology} \\
     $L_b(E; F)$ & $L(E; F)$ with topology of bounded convergence. & \autoref{definition:bounded-convergence-topology} \\
+    $L_c(E; F)$ & $L(E; F)$ with topology of precompact convergence. & \autoref{definition:compact-operator-topology} \\
+    $\mathcal{K}(E; F)$ & Space of compact operators from $E$ to $F$. & \autoref{definition:compact-operator} \\ 
     $\widehat{E}$ & Hausdorff completion of TVS $E$. & \autoref{definition:tvs-completion} \\
     % ---- Locally Convex ----
     $\mathrm{Conv}(A)$ & Convex hull of $A$. & \autoref{definition:convex-hull} \\
@@ -24,7 +26,7 @@
     $[\cdot]_A$ & 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} \\
-    $E \,\widetilde{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
+    $E \,\wh{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\
     $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ 
     % ---- Order Structures ----

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

@@ -120,11 +120,19 @@
 
 \begin{definition}[Bounded Convergence Topology]
 \label{definition:bounded-convergence-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
+    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}, or the \textbf{uniform topology}. 
 
     The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
 \end{definition}
 
+\begin{definition}[Topology of Precompact Convergence]
+\label{definition:compact-operator-topology}
+    Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{K} \subset 2^E$ be the collection of precompact subsets of $E$, then the $\mathfrak{K}$-uniform topology on $L(E; F)$ is the \textbf{topology of precompact convergence}. 
+
+    The space $L_c(E; F)$ denotes $L(E; F)$ equipped with the topology of precompact convergence.
+\end{definition}
+
+
 \begin{proposition}
 \label{proposition:operator-space-completeness}
     Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:

src/topology/dst/index.tex

@@ -0,0 +1,4 @@
+\chapter{Polish Spaces and Analytic Sets}
+\label{chap:polish-spaces}
+
+\input{./polish.tex}

src/topology/dst/polish.tex

@@ -0,0 +1,65 @@
+\section{Polish Spaces}
+\label{section:polish}
+
+
+
+
+\begin{definition}[Polish Space]
+\label{definition:polish-space}
+    Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:polish-space-extension}
+    The following spaces are Polish:
+    \begin{enumerate}
+        \item Closed subspace of a Polish space.
+        \item Open subspace of a Polish space. 
+        \item Countable products of Polish spaces. 
+        \item Countable disjoint union of Polish spaces. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $X$ be a Polish space with complete metric $d$ and $A \subset X$ be a closed subset, then $A$ is second countable. By \autoref{proposition:complete-closed}, $A$ is complete with respect to $d$. 
+
+    (2): Let $X$ be a Polish space with complete metric $d$ and $U \subset X$ be open. Assume without loss of generality that $U \subsetneq X$ and $d(X \times X) \subset [0, 1]$. Define
+    \[
+    d_U: U \times U \to [0, \infty] \quad (x, y) \mapsto d(x, y) + \abs{\frac{1}{d(x, U^c)} - \frac{1}{d(y, U^c)}}
+    \]
+
+    then $d_U$ is a metric on $U$. Since $d \le d_U$, the topology induced by $d_U$ is finer than the topology induced by $d$. On the other hand, the mapping $x \mapsto d(x, U^c)$ is continuous, so the topology induced by $d_U$ is coarser than the topology induced by $d$. Therefore $d_U$ induces the subspace topology of $U$. 
+
+    Now, let $\seq{x_n} \subset U$ be a Cauchy sequence with respect to $d_U$, then there exists $N \in \natp$ such that $d_U(x_m, x_n) \le 1$ for all $m, n \ge N$. Thus
+    \[
+    \frac{1}{d(x_n, U^c)} \le \frac{1}{d(x_N, U^c)} + d(x_n, x_N) + \abs{\frac{1}{d(x_n, U^c)} - \frac{1}{d(x_N, U^c)}} \le \frac{1}{d(x_N, U^c)} + 1
+    \]
+
+    and $\delta = \inf_{n \in \natp}d(x_n, U^c) > 0$. Since $d \le d_U$, $\seq{x_n}$ is Cauchy with respect to $d$ as well. Thus as $\bracsn{x \in X|d(x, U^c) \ge \delta} \subset U$ is a closed subset of $X$, there exists $x \in U$ such that $x_n \to x$ as $n \to \infty$. Therefore $U$ is complete with respect to $d_U$. 
+
+    (3): By \autoref{proposition:separable-product}, \autoref{proposition:product-complete}, and \autoref{theorem:uniform-metrisable}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:polish-subspace}
+    Let $X$ be a Polish space and $Y \subset X$, then $Y$ is Polish if and only if it is $G_\delta$ in $X$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.1.5]{CohnMeasure}}}. ]
+    ($\Rightarrow$): Suppose that $Y$ is Polish. Let $d_X: X^2 \to [0, 1]$ and $d_Y: Y^2 \to [0, 1]$ be complete metrics on $X$ and $Y$, respectively. For each $n \in \natp$, let $\mathcal{U}_n \subset 2^X$ be the collection of subsets of $X$ such that for each $U \in \mathcal{U}_n$,
+    \begin{enumerate}[label=(\roman*)]
+        \item $U$ is a non-empty open subset of $X$. 
+        \item $\sup_{x, y \in U}d_X(x, y) \le 1/n$. 
+        \item $\sup_{x, y \in U \cap Y}d_Y(x, y) \le 1/n$. 
+    \end{enumerate}
+    
+    Let $U_n = \bigcup_{U \in \mathcal{U}_n}U$, then $Y \subset \ol{Y} \cap \bigcap_{n \in \natp}U_n$. On the other hand, let $x \in \ol{Y} \cap \bigcap_{n \in \natp}U_n$. Let $V_1 \in \mathcal{U}_1 \cap \cn_X(x)$ and $x_1 \in V_1$. For each $n \in \natp$ with $n \ge 2$, let $V_n \in \mathcal{U}_n \cap \cn_X(x)$ with $V_n \subset V_{n-1}$ and $x_n \in V_n \cap Y$, then by (ii) and (iii), $\seq{x_n}$ is Cauchy with respect to $d_X$ and $d_Y$. In particular, there exists $y \in Y$ such that $x_n \to y$ with respect to $d_Y$ as $n \to \infty$. Since $d_Y$ induces the subspace topology on $Y$, $x_n \to y$ with respect to $d_X$ as $n \to \infty$ as well. Therefore $x = y$, and $Y \supset \ol{Y} \cap \bigcap_{n \in \natp}U_n$. 
+
+    As every closed subset of $X$ is $G_\delta$, $Y = \ol{Y} \cap \bigcap_{n \in \natp}U_n$ is also $G_\delta$. 
+
+    ($\Leftarrow$): Suppose that $Y$ is $G_\delta$ in $X$. Let $\seq{U_n} \subset 2^X$ be open sets such that $Y = \bigcap_{n \in \natp}U_n$. In which case, $Y$ is homeomorphic to the diagonal
+    \[
+    \Delta = \bracs{x \in \prod_{n \in \natp}U_n \bigg | x_m = x_n \forall m, n \in \natp}
+    \]
+
+    For each $n \in \natp$, $U_n$ is Polish by (2) of \autoref{proposition:polish-space-extension}. As a closed subspace of a product of Polish spaces, $\Delta$ is Polish by (1) and (3) of \autoref{proposition:polish-space-extension}. Therefore $Y$ is also Polish. 
+\end{proof}
+

src/topology/metric/metric.tex

@@ -68,11 +68,6 @@
     (2) $\Rightarrow$ (3): Since $\fF$ is Cauchy, there exists $\seq{E_n} \subset \fF$ such that for each $n \in \natp$, $E_n \supset E_{n+1}$ and $\sup_{y, z \in E_n}d(y, z) \le 1/n$. For each $n \in \natp$, let $x_n \in E_n$, then there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{n}x_n$. In which case, $x \in \bigcap_{n \in \natp}\overline{E_n}$. For each $n \in \natp$, $\sup_{y, z\in E_n}d(y, z) \le 1/n$, so $B_X(x, 2/n) \supset E_n$. Therefore $\fF \to x$. 
 \end{proof}
 
-\begin{definition}[Polish Space]
-\label{definition:polish-space}
-    Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. 
-\end{definition}
-
 
 
 \begin{theorem}[Banach's Fixed Point Theorem]