lecture 08
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\lecture{07}{2023-11-07}{}
\begin{proposition}
Let $X$ be second countable.
Then $|\cB(X)| \le \fc$.
% $\fc := 2^{\aleph_0}$
\end{proposition}
\begin{proof}
We use strong induction on $\xi < \omega_1$.
We have $\Sigma^0_1(X) \le \fc$
(for every element of the basis, we can decide
whether to use it in the union or not).
Suppose that $\forall \xi' < \xi.~|\Sigma^0_{\xi'}(X)| \le \fc$.
Then $|\Pi^0_{\xi'}(X)| \le \fc$.
We have that
\[
\Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
\]
Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.
We have
\[
\cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X).
\]
Hence
\[
|\cB(X)| \le \omega_1 \cdot \fc = \fc.
\]
\end{proof}
\begin{proposition}[Closure properties]
Suppose that $X$ is metrizable.
Let $1 \le \xi < \omega_1$.
Then
\begin{enumerate}[(a)]
\item \begin{itemize}
\item $\Sigma^0_\xi(X)$ is closed under countable unions.
\item $\Pi^0_\xi(X)$ is closed under countable intersections.
\item $\Delta^0_\xi(X)$ is closed under complements,
countable unions and
countable intersections.
\end{itemize}
\item \begin{itemize}
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
\end{itemize}
\end{enumerate}
\end{proposition}
\begin{proof}
\begin{enumerate}[(a)]
\item This follows directly from the definition.
Note that a countable intersection can be written
as a complement of the countable union of complements:
\[
\bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
\]
\item If suffices to check this for $\Sigma^0_{\xi}(X)$.
Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
Then
\[
A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right)
\]
and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
\end{enumerate}
\end{proof}
\begin{example}
Consider the cantor space $2^{\omega}$.
We have that $\Delta^0_1(2^{\omega})$
is not closed under countable unions
(countable unions yield all open sets, but there are open
sets that are not clopen).
\end{example}
\subsection{Turning Borels Sets into Clopens}
\begin{theorem}
\label{thm:clopenize}
Let $(X, \cT)$ be a Polish space.
For any Borel set $A \subseteq X$,
there is a finer Polish topology,
\footnote{i.e.~$\cT_A \supseteq \cT$ and $(X, \cT_A)$ is Polish}
such that
\begin{itemize}
\item $A$ is clopen in $\cT_A$,
\item the Borel sets do not change,
i.e.~$\cB(X, \cT) = \cB(X, \cT_A)$.
\end{itemize}
\end{theorem}
\begin{corollary}[Perfect set property]
Let $(X, \cT)$ be Polish,
and let $B \subseteq X$ be Borel and uncountable.
Then there is an embedding
of the cantor space $2^{\omega}$
into $B$.
\end{corollary}
\begin{proof}
Pick $\cT_B \supset \cT$
such that $(X, \cT_B)$ is Polish,
$B$ is clopen in $\cT_B$ and
$\cB(X,\cT) = \cB(X, \cT_B)$.
Therefore $(\cB, \cT_B\defon{B})$ is Polish.
We know that there is an embedding
$f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$.
Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
This is still continuous as $\cT \subseteq \cT_B$.
Since $2^{\omega}$ is compact, $f$ is an embedding.
%\todo{Think about this}
\end{proof}
\begin{refproof}{thm:clopenize}
We show that
\begin{IEEEeqnarray*}{rCl}
A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
&& (X, \cT_B) \text{ is Polish},\\
&& \cB(X, \cT) = \cB(X, \cT_B)\\
&& B \text{ is clopen in $\cT_B$}\\
\}
\end{IEEEeqnarray*}
is equal to the set of Borel sets.
The proof rests on two lemmata:
\begin{lemma}
\label{thm:clopenize:l1}
Let $(X,\cT)$ be a Polish space.
Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
there is $\cT_F \supseteq \cT$
such that $\cT_F$ is Polish,
$\cB(\cT) = \cB(\cT_F)$
and $F$ is clopen in $\cT_F$.
\end{lemma}
\begin{proof}
Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
Both are Polish spaces.
Take the coproduct\footnote{topological sum} $F \oplus (X \setminus F)$
of these spaces.
This space is Polish,
and the topology is generated by $\cT \cup \{F\}$,
hence we do not get any new Borel sets.
\end{proof}
So all closed sets are in $A$.
Furthermore $A$ is closed under complements,
since complements of clopen sets are clopen.
\begin{lemma}
\label{thm:clopenize:l2}
Let $(X, \cT)$ be Polish.
Let $\{\cT_n\}_{n < \omega}$
be Polish topologies
such that $\cT_n \supseteq \cT$
and $\cB(\cT_n) = \cB(\cT)$.
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
is still Polish
and $\cB(\cT_\infty) = \cB(T)$.
\end{lemma}
\begin{proof}
We have that $\cT_\infty$ is the smallest
topology containing all $\cT_n$.
To get $\cT_\infty$
consider
\[
\cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
\]
Then
\[
\cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}.
\]
(It suffices to take countable unions,
since we may assume that the $A_1, \ldots, A_n$ in the
definition of $\cF$ belong to
a countable basis of the respective $\cT_n$).
\todo{This proof will be finished in the next lecture}
\end{proof}
We need to show that $A$ is closed under countable unions.
By \yaref{thm:clopenize:l2} there exists a topology
$\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$
and $\cB(\cT_\infty) = \cB(\cT)$.
Applying \yaref{thm:clopenize:l1}
yields a topology $\cT_\infty'$ such that
$(X, \cT_\infty')$ is Polish,
$\cB(\cT_\infty') = \cB(\cT)$
and $A $ is clopen in $\cT_{\infty}'$.
\end{refproof}

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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}
\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}
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.
Let $V = \bigcup \{V_n : V_n \subseteq V\}$.
Pick $y \in 2^\omega$
and let $y_n = 1$ 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 $A_n \in \Pi^0_{\eta_n}(X)$
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}

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