10.1 Polish Spaces

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 Proposition 6.7.3, $A$ 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

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

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 Proposition 5.5.7, Proposition 6.7.2, and Theorem 6.3.10.$\square$

Proof, [Proposition 8.1.5, Coh13]. ($\Rightarrow$): 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 $U \in \mathcal{U}_{n}$,

1. (i)

   $U$ is a non-empty open subset of $X$.

2. (ii)

   $\sup_{x, y \in U}d_{X}(x, y) \le 1/n$.

3. (iii)

   $\sup_{x, y \in U \cap Y}d_{Y}(x, y) \le 1/n$.

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

For each $n \in \natp$, $U_{n}$ is Polish by (2) of Proposition 10.1.2. As a closed subspace of a product of Polish spaces, $\Delta$ is Polish by (1) and (3) of Proposition 10.1.2. Therefore $Y$ is also Polish.$\square$