w23-logic-3/inputs/lecture_08.tex

\lecture{08}{2023-11-10}{}%
\gist{\footnote{%
    In the beginning of the lecture, we finished
    the proof of \yaref{thm:clopenize:l2}.
    This has been moved to the notes on lecture 7.%
}}{}

\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}
\gist{%
    \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}