Various additions.

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} \\

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]