Added universality of the zero-dimensional spaces.

src/topology/dst/analytic.tex

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+\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}
+
+