# Descriptive set theory

### 1 Polish space

Definition A topological space is called polish space if it is separable and completely metrizable (i.e. admits a complete compatible metric).

We work with Polish topological spaces as opposed to Polish metric spaces becasue we don’t want to fix a particular complete metric, we may change it to serve different purposes; all we care about is that such a complete compatible metric exists.

Lemma If $X$ is a topological space with a compatible metric $d$, then  the following metric is also compatible: for $x,y\in X$, $\overline{d}(x,y):=\min(d(x,y),1)$.

Propositioin

1. Completion of any separable metric space is Polish.
2. A closed subset of a Polish space is Polish (with respect to relative topology).
3. A countable disjoint union of Polish spaces is Polish.
4. A countable product of Polish spaces is Polish (with respect to the product topology).

Examples

1. $\mathbf{R}^{\mathbf{N}}$, $\mathbf{C}^{\mathbf{N}}$;
2. The cantor space $\mathcal{C}=2^{\mathbf{N}}$ with the discrete topology in $2$;
3. The Baire space $\mathcal{N}=\mathbf{N}^{\mathbf{N}}$ with the discrete topology on $\mathbf{N}$;
4. The Hilbert cube $I^{\mathbf{N}}$, where $I=[0,1]$.

Lemma If $X$ is a metric space, then closed sets are $G_\delta$; equivalently, open sets are $F_\sigma$.

Proposition A subset of a Polish space is Polish if and only if it is $G_\delta$.

Proof Let $X$ be a Polish space and let $d_X$ be a complete compatible metric on $X$.

$\Leftarrow$ Considering first an open set $U\subset X$, we exploit the fact that it does not contain its boundary point to define a compatible metric topology of $U$ that makes the boundary of $U$ “look like infinite” in order to prevent sequences that converge to the boundary from being Cauchy. In fact, instead of defining a metric  explicitly, we define a homeomorphism of $U$ with a closed subset of $X\times \mathbf{R}$ by

$\displaystyle x\mapsto (x,\frac{1}{d_X(x,\partial U)}),$

where $d_X$ is a complete compatible metric for $X$.

$\Rightarrow$ Let $Y\subset X$ be  completely metrizable and let $d_Y$ be a complete compatible metric for $Y$. Define an open set  $V_n\subset X$ as the union of all open sets $U\subset X$ that satisfy

1. $U\cap Y=\emptyset$,
2. $\text{diam}_{d_X}(U)<1/n$,
3. $\text{diam}_{d_Y}(U\cap Y)<1/n$.

We show that $Y=\bigcap_{n\in\mathbf{N}}V_n$.

### 2 Trees

For a nonempty set $A$, we denote by $A^{<\mathbf{N}}$  the set of  finite tuples of elements of $A$, i.e.

$\displaystyle A^{<\mathbf{N}}=\bigcup_{n\in\mathbf{N}}A^n,$

where $A^{0}=\{\emptyset\}$. For $s\in A^{<\mathbf{N}}$, we denote by $|s|$ the length of $s$.

Definition  For a set $A$, a subset $T$ of $A^{<\mathbf{N}}$ is called a (set theoretic) tree if it is cloded downward under $\subset$.