Added universality of the zero-dimensional spaces.

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 +1,113 @@
+\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{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{proposition}
+\label{proposition:cantor-universality}
+    Let $X$ be a non-empty compact metrisable space, then there exists a surjective mapping $f \in C(2^{\natp}; X)$. 
+\end{proposition}
+\begin{proof}
+    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}
+