Fixed diagram.

src/fa/lc/convex.tex

@@ -135,7 +135,7 @@
 \label{definition:seminorm-topology}
     Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
     \begin{enumerate}
-        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric.
+        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudometric.
         \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
         \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
     \end{enumerate}

src/fa/norm/separable.tex

@@ -31,31 +31,6 @@
     (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
 \end{proof}
 
-\begin{proposition}
-\label{proposition:separable-banach-borel-sigma-algebra}
-    Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
-    \begin{enumerate}
-        \item Open sets in $E$ with respect to the strong topology. 
-        \item $\bracs{B(x, r)|x \in E, r > 0}$.
-        \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. 
-        \item Open sets in $E$ with respect to the weak topology. 
-    \end{enumerate}
-
-    
-\end{proposition}
-\begin{proof}
-    (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.
-
-    (4) $\subset$ (1): Every weakly open set is strongly open. 
-
-    (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
-    \[
-    \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E
-    \]
-
-    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 $C([0, 1]; E)$. 
@@ -82,5 +57,59 @@
     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}
 
+\begin{theorem}[Banach-Mazur]
+\label{theorem:banach-mazur}
+    Let $E$ be a separable normed vector space over $K \in \RC$, then there exists an isometric embedding $\iota \in L(E; C([0, 1]; K))$. 
+\end{theorem}
+\begin{proof}
+    Let $B$ be the closed unit ball of $E^*$, equipped with the weak* topology. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, the linear mapping
+    \[
+    E \to C(B; K) \quad x(\phi) = \dpn{x, \phi}{E}
+    \]
+
+    is an isometric embedding. By \autoref{proposition:separable-dual}, $B$ is a compact metric space. The \hyperref[Alexandroff-Hausdorff Theorem]{theorem:cantor-universality} then provides a continuous surjection $f: 2^{\natp} \to B$. Thus the composition map
+    \[
+    C(B; K) \to C(2^{\natp}; K) \quad g \mapsto g \circ f
+    \]
+
+    is a linear isometric embedding. Let $\mathcal{C} \subset [0, 1]$ be the Cantor set, then $\mathcal{C}$ is homeomorphic to $2^{\natp}$ through \autoref{proposition:cantor-space-embedding}. Hence $C(2^{\natp}; K)$ is isometrically isomorphic to $C(\mathcal{C}; K)$. 
+
+    Finally, \autoref{lemma:compact-embed} provides yet another linear isometric embedding $C(\mathcal{C}; K) \to C([0, 1]; K)$. Composing the above maps as follows
+    \[
+    \xymatrix{
+E \ar@{->}[r] & C(B; K) \ar@{->}[r] & C(2^{{\mathbb N}^+}; K) \ar@{->}[r] & C(\mathcal{C}; K) \ar@{->}[r] & C([0, 1]; K)
+}
+    \]
+
+    yields the desired embedding. 
+\end{proof}
+
+
+
+\begin{proposition}
+\label{proposition:separable-banach-borel-sigma-algebra}
+    Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
+    \begin{enumerate}
+        \item Open sets in $E$ with respect to the strong topology. 
+        \item $\bracs{B(x, r)|x \in E, r > 0}$.
+        \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. 
+        \item Open sets in $E$ with respect to the weak topology. 
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.
+
+    (4) $\subset$ (1): Every weakly open set is strongly open. 
+
+    (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
+    \[
+    \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E
+    \]
+
+    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}
+
 
 

src/measure/radon/c0.tex

@@ -125,7 +125,7 @@
     
     As such a $\phi \in C_0(X; E)$ exists for all $\eps > 0$ and $\seqf{A_j}$, $\norm{I_\mu}_{C_0(X; E)^*} \ge \norm{\mu}_{\text{var}}$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
 
-    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak*-topology and
+    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak* topology and
     \[
     T: C_0(X; E) \to C_0(X \times B; K) \quad (Tf)(x, \phi) = \dpn{f(x), \phi}{E}
     \]

src/measure/sets/borel.tex

@@ -8,6 +8,32 @@
 \end{definition}
 
 
+
+\begin{lemma}
+\label{lemma:monotone-borel-characterisation}
+    Let $X$ be a topological space, then $\cb_X$ is the smallest subset of $2^X$ that:
+    \begin{enumerate}
+        \item contains open and closed subsets of $X$. 
+        \item is closed under countable intersections.
+        \item is closed under countable disjoint unions. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 8.2.4]{CohnMeasure}}}. ]
+    Let $\cf \subset 2^X$ be the smallest subset of $2^X$ satisfying the lemma. Since $\cb_X$ satisfies the lemma, $\cf \subset \cb_X$. On the other hand, let
+    \[
+    \cf_0 = \bracs{A \subset X| A \in \cf, A^c \in \cf}
+    \]
+
+    then by definition, $\cf_0$ is closed under complements. Let $\seq{A_n} \subset \cf_0$, then
+    \[
+    \bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}A_n \setminus \bigcup_{k = 1}^{n-1} A_k
+     = \bigsqcup_{n \in \natp}A_n \cap \bigcap_{k = 1}^{n - 1}A_k^c
+    \]
+
+    Since $\cf_0 \subset \cf$ is closed under complements, $\seq{A_n^c} \subset \cf$ as well. By (2) and (3), $\bigcup_{n \in \natp}A_n \in \cf$ and $\bigcap_{n \in \natp}A_n^c \in \cf$. Thus $\bigcup_{n \in \natp}A_n \in \cf$ as well. By (1), $\cf_0$ is a $\sigma$-algebra that contains all open subsets of $X$, so $\cf \supset \cf_0 \supset \cb_X$.  
+\end{proof}
+
+
 \begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$]
 \label{definition:borel-sigma-algebra-extended}
     The family

src/topology/dst/analytic.tex

@@ -0,0 +1,34 @@
+\section{Analytic Sets}
+\label{section:analytic-sets}
+
+\begin{definition}[Analytic Set]
+\label{definition:analytic-set}
+    Let $X$ be a Polish space and $A \subset X$, then $A$ is \textbf{analytic} if there exists a Polish space $Z$ and $f \in C(Z; X)$ such that $f(Z) = A$. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:analytic-sets}
+    Let $X$ be a Polish space, then:
+    \begin{enumerate}
+        \item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcup_{n \in \natp}A_n$ is analytic. 
+        \item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcap_{n \in \natp}A_n$ is analytic. 
+        \item For any $A \in \cb_X$, $A$ is analytic. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.2.1-8.2.3]{CohnMeasure}}}. ]
+    For each $n \in \natp$, let $Z_n$ be a Polish space and $f_n \in C(Z_n; X)$ such that $A_n = f_n(Z_n)$. 
+
+    (1): By \autoref{proposition:polish-space-extension}, $Z = \bigsqcup_{n \in \natp}Z_n$ is a Polish space. Let $f \in C(Z; X)$ be the gluing of $\seq{f_n}$, then $\bigcup_{n \in \natp}A_n = f(Z)$.
+
+    (2): By \autoref{proposition:polish-space-extension}, $Z = \prod_{n \in \natp}Z_n$ is a Polish space. For each $m, n \in \natp$, $\bracs{f_m \circ \pi_m = f_n \circ \pi_n}$ is closed. Thus
+    \[
+    \Delta := \bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracs{f_m \circ \pi_m = f_n \circ \pi_n}
+    \]
+
+    is closed, and Polish by \autoref{proposition:polish-space-extension}. Hence $\bigcap_{n \in \natp}A_n = f_1 \circ \pi_1(\Delta)$ is also analytic.
+
+    (3): By \autoref{proposition:polish-space-extension}, every open and closed subset of $X$ is analytic. Thus (1), (2), and \autoref{lemma:monotone-borel-characterisation} imply that every element of $\cb_X$ is analytic.
+\end{proof}
+
+

src/topology/dst/index.tex

@@ -2,3 +2,5 @@
 \label{chap:polish-spaces}
 
 \input{./polish.tex}
+\input{./analytic.tex}
+\input{./zero.tex}

src/topology/dst/zero.tex

@@ -0,0 +1,124 @@
+\section{Zero Dimensional Spaces}
+\label{section:zero-dimensional}
+
+\begin{definition}[Zero-Dimensional]
+\label{definition:zero-dimentional}
+    Let $X$ be a topological space, then $X$ is \textbf{zero-dimensional} if $X$ admits a base consisting of clopen sets. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:zero-dimensional-extension}
+    Let $\seqi{X}$ and $X$ be zero-dimensional spaces, then:
+    \begin{enumerate}
+        \item For any $A \subset X$, $A$ is zero-dimensional. 
+        \item $\prod_{i \in I}X_i$ is zero-dimensional.
+        \item $\bigsqcup_{i \in I}X_i$ is zero-dimensional. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted. 
+
+\begin{definition}[Cantor Space]
+\label{definition:cantor-space}
+    Let $2 = \bracs{0, 1}$ be equipped with the discrete topology, then $2^{\natp}$ is the \textbf{Cantor space}. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:cantor-space-embedding}
+    The mapping
+    \[
+    2^{\natp} \to [0, 1] \quad \seq{x_n} \mapsto 2\sum_{n \in \natp} \frac{x_n}{3^n}
+    \]
+
+    is an embedding. 
+\end{proposition}
+
+
+\begin{definition}[Baire Space]
+\label{definition:the-baire-space}
+    Let $\natp$ be equipped with the discrete topology, then $\mathscr{N} = (\natp)^{\natp}$ is the \textbf{Baire space}. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:baire-universality}
+    Let $X$ be a non-empty Polish space, then there exists a surjective mapping $f \in C(\mathscr{N}; X)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 8.2.7]{CohnMeasure}}}. ]
+    Let $d: X^2 \to [0, \infty)$ be a complete metric on $X$. To construct the desired map, it is sufficient to construct $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $N \in \natp$, $\bracsn{n_k}_1^N \subset \natp$, $C(n_1, \cdots, n_N) \subset X$ is closed and non-empty. 
+        \item For each $N \in \natp$, $\bracsn{n_k}_1^N \subset \natp$, $\text{diam}(C(n_1, \cdots, n_N)) \le 1/N$. 
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^{N} \subset \natp$,
+        \[
+        C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_1, \cdots, n_{N+1})
+        \]
+        \item $X = \bigcup_{n_1 \in \natp}C(n_1)$. 
+    \end{enumerate}
+
+    Since $X$ is Polish, there exists a countable dense subset $\seq{x_k} \subset X$. For each $n_1 \in \natp$, let $C(n_1) = \ol{B(x_{n_1}, 1/2)}$, then $\bracsn{C(n_1)|n_1 \in \natp}$ satisfies (i) and (ii) by definition. As $\seq{x_k}$ is dense in $X$, $X = \bigcup_{n_1 \in \natp}C(n_1)$, and (iv) is also satisfied. 
+
+    Let $N \in \natp$ and suppose inductively that $\bracsn{C(n_1, \cdots, n_k)| \bracsn{n_j}_1^k \subset \natp,1 \le k \le N}$ has been constructed to satisfy (i)-(iv) for each $1 \le k \le N$. 
+    
+    Fix $\bracsn{n_k}_1^N \subset \natp$. By \autoref{proposition:separable-metric-space}, there exists a countable dense subset $\seq{y_k} \subset C(n_1, \cdots, n_N)$. For each $n_{N+1} \in \natp$, let $C(n_1, \cdots, n_{N+1}) = \ol{B(y_{n_{N+1}}, 1/[2(N+1)])} \cap C(n_1, \cdots, n_N)$, then $\bracsn{C(n_1, \cdots, n_{N+1})|n_{N+1} \in \natp}$ satisfies (i) and (ii) by definition. Since $\seq{y_k}$ is dense in $C(n_1, \cdots, n_N)$,
+    \[
+     C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_1, \cdots, n_{N+1})
+    \]
+
+    so (iii) is also satisfied. 
+
+    Finally, suppose that $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \mathscr{N}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \mathscr{N}$ such that $x = f(\seq{n_k})$. As such, $f: \mathscr{N} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \mathscr{N}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 1/N$. Therefore $f \in C(\mathscr{N}; X)$. 
+\end{proof}
+
+\begin{theorem}[Alexandroff-Hausdorff]
+\label{theorem:cantor-universality}
+    Let $X$ be a non-empty compact metrisable space, then there exists a surjective mapping $f \in C(2^{\natp}; X)$. 
+\end{theorem}
+\begin{proof}[Proof, adapted from {{\cite[Proposition 8.2.7]{CohnMeasure}}}. ]
+    Let $d: X \times X \to [0, \infty)$ be a metric on $X$. Since $X$ is compact, for each $N \in \natp$, there exists $K_N \in \natp$ and $\seqf{x_{N, k}|1 \le k \le K_N} \subset X$ such that $X = \bigcup_{k = 1}^{K_N}B(x_{N, k}, 1/N)$.
+
+    To construct the desired map, it is sufficient to construct
+    \[
+    \bracs{C(n_1, \cdots, n_N) \bigg | \bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}, N \in \natp} \subset 2^X
+    \]
+    
+    
+    such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^N$, $C(n_1, \cdots, n_N) \subset X$ is closed and non-empty. 
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^N$, $\text{diam}(C(n_1, \cdots, n_N)) \le 4/N$. 
+        \item For each $N \in \natp$ and $\bracsn{n_k}_1^{N}$,
+        \[
+        C(n_1, \cdots, n_{N}) = \bigcup_{n_{N+1} = 1}^{K_{N+1}}C(n_1, \cdots, n_{N+1})
+        \]
+        \item $X = \bigcup_{n_1 = 1}^{K_1}C(n_1)$. 
+    \end{enumerate}
+
+    For each $1 \le n_1 \le K_1$, let $C(n_1) = \ol{B(x_{1, n_1}, 1)}$, then $\bracsn{C(n_1)|1 \le n_1 \le K_1}$ satisfies (i), (ii), and (iv) by definition. 
+
+    Let $N \in \natp$ and suppose inductively that
+    \[
+    \bracs{C(n_1, \cdots, n_K) \bigg | \bracsn{n_k}_1^K \in \prod_{n = 1}^{K}\bracs{1, \cdots, K_n}, 1 \le K \le N} \subset 2^X
+    \]
+    
+    
+    has been constructed to satisfy (i)-(iv) for each $1 \le K \le N$. Fix $\bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}$. Since $X = \bigcup_{k = 1}^{K_{N+1}}B(x_{N+1, k}, 1/(N+1))$, there exists $\bracsn{y_k}_1^{K_{N+1}} \subset C(n_1, \cdots, n_N)$ such that $C(n_1, \cdots, n_N) \subset \bigcup_{k = 1}^{K_{N+1}}B(y_k, 2/(N+1))$. For each $1 \le n_{N+1} \le K_{N+1}$, let $C(n_1, \cdots, n_{N+1}) = C(n_1, \cdots, n_N) \cap \ol{B(y_{n_{N+1}}, 2/(N+1))}$, then $\bracsn{C(n_1, \cdots, n_{N+1})|1 \le n_{N+1} \le K_{N+1}}$ satisfies (i)-(iii). 
+
+    Now, suppose that
+    \[
+    \bracs{C(n_1, \cdots, n_N) \bigg | \bracsn{n_k}_1^N \in \prod_{n = 1}^{N}\bracs{1, \cdots, K_n}, N \in \natp} \subset 2^X
+    \]
+    
+    has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$ such that $x = f(\seq{n_k})$. As such, $f: \prod_{n \in \natp}\bracs{1, \cdots, K_n} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \prod_{n \in \natp}\bracs{1, \cdots, K_n}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 4/N$. Therefore $f \in C(\prod_{n \in \natp}\bracs{1, \cdots, K_n}; X)$. 
+
+    Finally, for each $n \in \natp$, there exists $L_n \in \natp$ and a surjective mapping $g_n: 2^{L_n} \to \bracs{1, \cdots, K_n}$. Let
+    \[
+    g = \prod_{n \in \natp}g_n: \prod_{n \in \natp}2^{L_n} \to \prod_{n \in \natp}\bracs{1, \cdots, K_n}
+    \]
+    
+    be the product of $\seq{g_n}$, then $g \in C(\prod_{n \in \natp}2^{L_n}; \prod_{n \in \natp}\bracs{1, \cdots, K_n})$, and
+    \[
+    f \circ g: 2^{\natp} \iso \prod_{n \in \natp}2^{L_n} \to X
+    \]
+
+    is the desired continuous surjection. 
+\end{proof}
+

src/topology/index.tex

@@ -6,4 +6,5 @@
 \input{./functions/index.tex}
 \input{./metric/index.tex}
 \input{./groups/index.tex}
+\input{./dst/index.tex}
 \input{./notation.tex}

src/topology/notation.tex

@@ -26,4 +26,7 @@
     $f \prec U$ & $f \in C_c(X; [0,1])$ with $\mathrm{supp}(f) \subset U$. & \autoref{definition:compactly-supported-01} \\
     $C_0(X; E)$ & Continuous functions vanishing at infinity. & \autoref{definition:vanish-at-infinity} \\
     $BC(X; E)$ & Bounded continuous functions $X \to E$. & \autoref{definition:bounded-continuous-function-space} \\
+
+    % DST 
+    $\mathscr{N}$ & The Baire space. & \autoref{definition:the-baire-space} \\ 
 \end{tabular}