\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}