Josia Pietsch bc8b5a8b6c
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2024-02-07 22:48:53 +01:00

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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.%
%\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.
We say that $\cU \subseteq Y \times X$
is \vocab{$Y$-universal} for $\Gamma(X)$ /
$\cU$ \vocab{parametrizes} $\Gamma(X)$
\item $\cU \in \Gamma(Y \times X)$,
\item $\{U_y : y \in Y\} = \Gamma(X)$.
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 $\Gamma$.
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 $\Pi^0_\xi(X)$.
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 $(2^{\omega \times \omega}) \cong 2^\omega$-universal set
for $\Sigma^0_\xi(X)$.
For $(y_{m,n}) \in (2^{\omega \times \omega})$
and $x \in X$
we set $((y_{m,n}), x) \in \cU$
iff $\exists n.~((y_{m,n})_{m < \omega}, x) \in U_{\xi_n}$,
i.e.~iff $\exists n.~x \in (U_{\xi_n})_{(y_{m,n})_{m < \omega}}$.
Let $A \in \Sigma^0_\xi(X)$.
Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
Furthermore $\cU \in \Sigma^0_{\xi}((2^{\omega \times \omega} \times X)$.
\item Suffices for $\Sigma^0_\xi$ (complement is $\Pi^0_\xi$-universal).
\item $2^\omega$-universal set for $\Sigma^0_1(X)$, since
$X$ is second countable ($(y,x) \in \cU \iff x \in \bigcup_n \{V_n : y_n = 1\}$).
\item Induction: Take $\xi_k \to \xi$,
and $\cU_{\xi_k}$ $\Sigma^0_{\xi_k}$-universal.
Construct $2^{ \omega \times \omega}$-universal:
$(y_{m,n}, x) \in \cU :\iff \exists n.~((y_{m,n}), x) \in \cU_{\xi_n}$.
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.%
\gist{\footnote{By definition of the subspace topology
and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}}{}