\tutorial{07}{2023-12-05}{}
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\subsection{Exercise 2}

Recall \autoref{thm:analytic}.

Let $(A_i)_{i < \omega}$ be analytic subsets of a Polish space $X$.

$\bigcap_i A_i$ is $\Sigma^1_1$:

% Let $Y_i$ be Polish such that $f_i(Y_i) = A_i$.
% Let $Y \coloneqq \coprod Y_i$, $f = \coprod f_i$ and $Z = \prod Y_i$.
% Note that $Y$ and $Z$ are Polish.
% We can embed $Z$ into $Y^{\N}$.
% 
% Define a tree $T$ on $Y$ as follows:
% $(y_0, \ldots, y_n) \in T$ iff
% \begin{itemize}
%     \item $\forall 0 \le i \le n.~ y_i \in Y_i$ and
%     \item $\forall  i,j .~ f(y_i) = f(y_j)$.
% \end{itemize}
% 
% Then $[T]$ consists of sequences $y = (y_n)$
% such that $\forall j \in \N.~f(y) \in  \im (f_j)$,
% so $\forall y \in [T].~f(y) \in \bigcap_{i \in \N} \im(f_i) = \bigcap_{i \in \N} A_i$.
% $[T] \subseteq  i(Z) \subseteq Y^{\N}$,
% and $[T]$ is closed.
% 
% 
% Other solution:

Let $Z = \prod Y_i$
and let $D \subseteq  Z$
be defined by
\[
D \coloneqq \{(y_n) : f_i(y_i) = f_j(y_j) ~ \forall i,j\}.
\]
$D$ is closed,
at it is the preimage of the diagonal
under $Z \xrightarrow{(f_0,f_1,\ldots)} X^{\N}$.
Then $\bigcap A_i$ is the image of  $D$
under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.

\paragraph{Other solution}
Let $F_n \subseteq X \times  \cN$ be closed,
and $C \subseteq  X \times \cN^{\N}$ defined by
\[
C \coloneqq  \{(x,(y^{(n)}) ) : \forall n.~(x, y^{(n)}) \in  F_n\}.
\]
$C$ is closed
and $\bigcap A_i = \proj_X(C)$.

\subsection{Exercise 3}

\begin{itemize}
    \item Make $X$ zero dimensional preserving the Borel structure.
    \item \todo{Find a countable clopen base}
    \item 
\end{itemize}


\subsection{Exercise 4}


Proof of Schröder-Bernstein:

Let $X_0 \coloneqq  X$, $Y_0 \coloneqq Y$
and define $X_{i+1} \coloneqq  g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
We have $X_{i+1} \subseteq X_i$ and similarly for $Y_i$.
$f$ and $g$ are bijections between
$X_\omega \coloneqq \bigcap X_i$ and $Y_\omega \coloneqq \bigcap Y_i$.

% https://q.uiver.app/#q=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
\[\begin{tikzcd}
	{X \setminus X_\omega} & {=} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
	{Y\setminus Y_\omega} & {=} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
	\arrow["f"'{pos=0.7}, from=1-3, to=2-5]
	\arrow["g"{pos=0.1}, from=2-3, to=1-5]
	\arrow["f"{pos=0.8}, from=1-7, to=2-9]
	\arrow["g"{pos=0.1}, from=2-7, to=1-9]
\end{tikzcd}\]


By \autoref{thm:lusinsouslin}
the injective image via a Borel set of a Borel set is Borel.

\autoref{thm:lusinsouslin} also gives that the inverse
of a bijective Borel map is Borel.