\tutorial{07}{2023-12-05}{} % 17 / 20 \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.