\lecture{08}{2023-11-10}{}

\todo{put this lemma in the right place}
\begin{lemma}[Lemma 2]
    Let $(X, \cT)$ be a Polish space.
    Let $\cT_n \supseteq \cT$ be Polish
    with $\cB(X, \cT_n) = \cB(X, \cT)$.
    Let  $\cT_\infty$ be the topology generated
    by $\bigcup_n \cT_n$.
    Then $(X, \cT_\infty)$ is Polish
    and $\cB(X, \cT_\infty) = \cB(X, \cT)$.
\end{lemma}
\begin{proof}
    Let $Y = \prod_{n \in \N} (X, \cT_n)$.
    Then $Y$ is Polish.
    Let $\delta\colon (X, \cT_\infty) \to Y$
    defined by $\delta(x) = (x,x,x,\ldots)$.
    \begin{claim}
        $\delta$ is a homeomorphism.
    \end{claim}
    \begin{subproof}
        Clearly $\delta$ is a bijection.
        We need to show that it is continuous and open.

        Let $U \in \cT_i$.
        Then
        \[
        \delta^{-1}(D \cap \left( X \times  X \times \ldots\times  U \times  \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
        \]
        hence $\delta$ is continuous.
        Let $U \in \cT_\infty$.
        Then $U$ is the union of sets of the form
        \[
        V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
        \]
        for some $n_1 < n_2 < \ldots < n_u$
        and $U_{n_i} \in \cT_i$.

        Thus is suffices to consider sets of this form.
        We have that
        \[
        \delta(V) = D \cap (X \times  X \times  \ldots \times  U_{n_1} \times  \ldots \times  U_{n_2} \times  \ldots \times  U_{n_u} \times  X \times  \ldots) \overset{\text{open}}{\subseteq} D.
        \]
    \end{subproof}

    This will finish the proof since
    \[
    D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
    \]
    Why? Let  $(x_n) \in Y \setminus D$.
    Then there are $i < j$ such that $x_i \neq x_j$.
    Take disjoint open $x_i \in U$, $x_j \in V$.
    Then
    \[(x_n) \in  X \times  X \times  \ldots \times  U \times  \ldots \times  X \times  \ldots \times V \times  X \times  \ldots\]
    is open in $Y\setminus D$.
    Hence $Y \setminus D$ is open, thus $D$ is closed.
    It follows that $D$ is Polish.
\end{proof}

\subsection{Parametrizations}
%\todo{choose better title}


Let $\Gamma$ denote a collection of sets in some space.
For us $\Gamma$ will be one of $\Sigma^0_\xi(X), \Pi^0_\xi(X), \Delta^0_\xi(X), \cB(X)$,
where $X$ is a metrizable, usually second countable space.

\begin{definition}
    We say that $\cU \subseteq Y \times X$
    is \vocab{$Y$-universal} for $\Gamma(X)$ /
    $\cU$ \vocab{parametrizes}  $\Gamma(X)$
    iff:
    \begin{itemize}
        \item $\cU \in \Gamma$,
        \item $\{U_y : y \in Y\} = \Gamma(X)$.
    \end{itemize}
\end{definition}

\begin{example}
    Let $X = \omega^\omega$, $Y = 2^{\omega}$
    and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$.
    We will show that there is a $2^{\omega}$-universal
    set for $\Gamma$.
\end{example}

\begin{theorem}
    \label{thm:cantoruniversal}
    Let $X$ be a separable, metrizable space.
    Then for every $\xi \ge 1$,
    there is a $2^{\omega}$-universal
    set for $\Sigma^0_\xi(X)$ and
    similarly for $\Pi^0_\xi(X)$.
\end{theorem}
\begin{proof}
    Note that if $\cU$ is $2^{\omega}$ universal for
    $\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$
    is $2^{\omega}$-universal for $\Pi^0_\xi(X)$.
    Thus it suffices to consider $\Sigma^0_\xi(X)$.

    First let $\xi = 1$.
    We construct $\cU \overset{\text{open}}{\subseteq} 2^{\omega} \times X$
    such that
    \[
    \{U_y : y \in 2^\omega\}  = \Sigma^0_1(X).
    \]

    Let $(V_n)$ be a basis of open sets of $X$.
    For all $y \in 2^\omega$ and $x \in X$
    put $(y,x) \in \cU$ iff
    $x \in \bigcup \{V_n : y_n = 1\}$.
    $\cU$ is open.
    For any $V \overset{\text{open}}{\subseteq} X$,
    define $y \in 2^\omega$
    by $y_n = 1$ iff $V_n \subseteq  V$.
    Then $\cU_y = V$.


    Now suppose that there exists a
    $2^{\omega}$-universal set for $\Sigma^0_{\eta}(X)$
    for all $\eta < \xi$.
    Fix  $\xi_0 \le  \xi_1 \le \ldots < \xi$
    such that $\xi_n \to \xi$ if $\xi$ is a limit,
    or $\xi_n = \xi'$ if $\xi = \xi' +1$ is a successor.

    Recall that $\eta_1 \le  \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.

    Note that if $A = \bigcup_n A_n$,  with $A_n \in \Pi^0_{\eta_n}(X)$
    for some $\eta_n < \xi$,
    we also have
    $A = \bigcup_n A_n'$ with  $A'_n \in \Pi^0_{\xi_n}(X)$.

    We construct a $(2^\omega)^\omega \cong 2^\omega$-universal set
    for $\Sigma^0_\xi(X)$.
    For $(y_n) \in (2^\omega)^\omega$
    and $x \in X$
    we set $((y_n), x) \in \cU$
    iff $\exists n.~(y_n, x) \in U_{\xi_n}$,
    i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$.

    Let $A \in \Sigma^0_\xi(X)$.
    Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
    % TODO
    Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$.
\end{proof}
\begin{remark}
    Since $2^{\omega}$ embeds
    into any uncountable polish space $Y$
    such that the image is closed,
    we can replace $2^{\omega}$ by $Y$
    in the statement of the theorem.%
    \footnote{By definition of the subspace topology
    and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
\end{remark}