w23-logic-3/inputs/lecture_08.tex

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\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}
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%\todo{choose better title}
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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}
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\label{thm:cantoruniversal}
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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.
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For any $V \overset{\text{open}}{\subseteq} X$,
define $y \in 2^\omega$
by $y_n = 1$ iff $V_n \subseteq V$.
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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 $A_n \in \Pi^0_{\eta_n}(X)$
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for some $\eta_n < \xi$,
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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}
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\begin{remark}
Since $2^{\omega}$ embeds
into any uncountable polish space $Y$
such that the image is closed,
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we can replace $2^{\omega}$ by $Y$
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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}