tutorial

inputs/lecture_09.tex

@@ -43,6 +43,7 @@ We will see that  not every analytic set is Borel.
 \end{remark}
 
 \begin{theorem}
+    \label{thm:borel}
     Let $X$ be Polish,
     $\emptyset \neq A \subseteq X$.
     Then the following are equivalent:

inputs/lecture_12.tex

@@ -51,7 +51,7 @@
 \begin{theorem}
     \label{thm:lec12:1}
     Suppose that $A \subseteq \cN$ is analytic.
-    Then there is $f\colon \cN \to \Tr$
+    Then there is $f\colon \cN \to \Tr$\todo{Borel?}
     such that $x \in A \iff f(x)$ is ill-founded.
 \end{theorem}
 For the proof we need some prerequisites:

inputs/tutorial_07.tex

@@ -0,0 +1,93 @@
+\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=WzAsMTgsWzAsMCwiWCBcXHNldG1pbnVzIFhfXFxvbWVnYSJdLFswLDEsIllcXHNldG1pbnVzIFlfXFxvbWVnYSJdLFsxLDAsIj0iXSxbMSwxLCI9Il0sWzIsMCwiKFhfMCBcXHNldG1pbnVzIFhfMSkiXSxbMiwxLCIoWV8wIFxcc2V0bWludXMgWV8xKSJdLFs0LDEsIihZXzAgXFxzZXRtaW51cyBZXzEpIl0sWzYsMSwiKFlfMCBcXHNldG1pbnVzIFlfMSkiXSxbNCwwLCIoWF8wIFxcc2V0bWludXMgWF8xKSJdLFs2LDAsIihYXzAgXFxzZXRtaW51cyBYXzEpIl0sWzcsMCwiXFxjZG90cyJdLFs3LDEsIlxcY2RvdHMiXSxbMywwLCJcXGN1cCJdLFszLDEsIlxcY3VwIl0sWzUsMCwiXFxjdXAiXSxbNSwxLCJcXGN1cCJdLFs4LDFdLFs4LDBdLFs0LDYsImYiLDIseyJsYWJlbF9wb3NpdGlvbiI6NzB9XSxbNSw4LCJnIiwwLHsibGFiZWxfcG9zaXRpb24iOjEwfV0sWzksMTYsImYiLDAseyJsYWJlbF9wb3NpdGlvbiI6ODB9XSxbNywxNywiZyIsMCx7ImxhYmVsX3Bvc2l0aW9uIjoxMH1dXQ==
+\[\begin{tikzcd}
+	{X \setminus X_\omega} & {=} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
+	{Y\setminus Y_\omega} & {=} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
+	\arrow["f"'{pos=0.7}, from=1-3, to=2-5]
+	\arrow["g"{pos=0.1}, from=2-3, to=1-5]
+	\arrow["f"{pos=0.8}, from=1-7, to=2-9]
+	\arrow["g"{pos=0.1}, from=2-7, to=1-9]
+\end{tikzcd}\]
+
+
+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.
+
+
+
+