+

@@ -0,0 +1,223 @@
+\subsection{Sheet 7}
+
+\tutorial{08}{2023-12-05}{}
+% 17 / 20
+\nr 1
+\begin{itemize}
+    \item For $\xi = 1$ this holds by the definition of the
+        subspace topology.
+
+        We now use transfinite induction, to show that
+        the statement holds for all $\xi$.
+        Suppose that $\Sigma^0_{\zeta}(Y)$ and $\Pi^0_{\zeta}(Y)$
+        are as claimed for all $\zeta < \xi$.
+
+        Then
+        \begin{IEEEeqnarray*}{rCl}
+            \Sigma^0_\xi(Y) &=& \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(Y), \alpha_n < \xi\}\\
+                            &=& \{\bigcup_{n < \omega} (A_n \cap Y) : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
+                            &=& \{Y \cap \bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
+                            &=& \{Y \cap A : A \in \Sigma^0_{\xi}(X)\}.
+        \end{IEEEeqnarray*}
+        and
+        \begin{IEEEeqnarray*}{rCl}
+            \Pi^0_\xi(Y) &=& \lnot \Sigma^0_\xi(Y)\\
+                         &=& \{Y \setminus A : A \in \Sigma^0_\xi(Y)\}\\
+                         &=& \{Y \setminus (A \cap Y) : A \in \Sigma^0_\xi(X)\}\\
+                         &=& \{Y \cap (X \setminus A) : A \in \Sigma^0_\xi(X)\}\\
+                         &=& \{Y \cap A : A \in \Pi^0_\xi(X)\}.
+        \end{IEEEeqnarray*}
+    \item Let $V \in \cB(Y)$.
+
+        We show that $f^{-1}(V) \in \cB(Y)$,
+        by induction on the minimal $\xi$ such that $V \in \Sigma_\xi^0$.
+        For $\xi = 0$ this is clear.
+        Suppose that we have already shown $f^{-1}(V') \in \cB(Y)$
+        for all $V' \in \Sigma^0_\zeta$, $\zeta < \xi$.
+        Then $f^{-1}(Y \setminus V')  = X \setminus f^{-1}(V') \in \cB(V)$,
+        since complements of Borel sets are Borel.
+        In particular, this also holds for $\Pi^0_\zeta$ sets
+        and $\zeta < \xi$.
+        Let $V \in \Sigma^0_\xi$.
+        Then $V = \bigcap_{n} V_n$ for some $V_n \in \Pi^{0}_{\alpha_n}$,
+        $\alpha_n < \xi$.
+        In particular $f^{-1}(V) = \bigcup_n f^{-1}(V_n) \in \cB(X)$.
+\end{itemize}
+
+\nr 2
+
+
+Recall \autoref{thm:analytic}.
+
+Let $(A_i)_{i<\omega}$ be analytic subsets of a Polish space $X$.
+Then there exists Polish spaces $Y_i$ and $f_i\colon Y_i \to X$
+continuous such that $f_i(B_i) = A_i$
+for some $B_i \in \cB(Y_i)$.
+
+ \begin{itemize}
+    \item $\bigcup_i A_i$ is analytic:
+        Consider the Polish space $Y \coloneqq  \coprod_{i < \omega} Y_i$
+        and $f \coloneqq \coprod_i f_i$, i.e.~
+        $Y_i \ni y \mapsto f_i(y)$.
+        $f$ is continuous,
+        $\coprod_{i < \omega} B_i \in \cB(Y)$
+        and
+        \[f(\coprod_{i < \omega} B_i) = \bigcup_i A_i.\]
+    \item $\bigcap_i A_i$ 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 $\bigcap A_i = \proj_X(C)$.
+\end{itemize}
+
+\nr 3
+\todo{Wait for mail}
+\todo{Find a countable clopen base}
+
+\begin{itemize}
+    \item We use the same construction as in exercise 2 (a)
+        on sheet 6.
+        Let $A \subseteq X$ be analytic,
+        i.e.~there exists a Polish space $Y$ and $f\colon Y \to X$ Borel
+        with $f(Y) = X$.
+        Then $f$ is still Borel with respect to the
+        new topology, since Borel sets are preserved
+        and by exercise 1 (b).
+        % Let $(B_i)_{i < \omega}$ be a countable basis of $(X,\tau)$.
+        % By a theorem from the lecture, there exists Polish
+        % topologies $\cT_i$ such that $B_i$ is clopen wrt.~$\cT_i$
+        % and $\cB(\cT_i) = \cB(\tau)$.
+        % By a lemma from the lecture,
+        % $\tau' \coloneqq \bigcup_i  \cT_i$
+        % is Polish as well and $\cB(\tau') = \cB(\tau)$.
+        % \todo{TODO: Basis}
+    \item Suppose that there exist no disjoint clopen sets $U_0,U_1$,
+        such that $W \cap U_0$ and $W \cap U_1$ are uncountable.
+
+        Let $W_0 \coloneqq  W$
+        Then there exist disjoint clopen sets $C_i^{(0)}$
+        such that $W_0 \subseteq \bigcup_{i < \omega} C_i^{(0)}$
+        and $\diam(C_i) < 1$,
+        since $X$ is zero-dimensional.
+
+        By assumption, exactly one of the $C_i^{(0)}$ has
+        uncountable intersection with $W_0$.
+        Let $i_0$  be such that  $W_0 \cap C_{i_0}^{(0)}$ is uncountable
+        and set $W_1 \coloneqq W_0 \cap C_{i_0}^{(0)}$.
+        Note that $W_0 \setminus W_1 = \bigcup_{i \neq i_0} C_i^{(0)}$ is countable.
+
+        Let us recursively continue this construction:
+        Suppose that $W_n$ uncountable has been chosen.
+        Then choose $C_{i}^{(n)}$ clopen,
+        disjoint with diameter $\le \frac{1}{n}$
+        such that $W_n \subseteq \bigcup_{i} C_i^{(n)}$
+        and let $i_n$ be the unique index
+        such that $W_n \cap C_{i_n}^{(n)}$ is uncountable.
+
+        Since $\diam(C_{i_n}^{(n)}) \xrightarrow{n \to \infty} 0$
+        and the $C_{i_n}^{(n)}$ are closed,
+        we get that $\bigcap_n C_{i_n}^{(n)}$
+        contains exactly one point. Let that point be  $x$.
+
+        However then
+        \[
+        W = \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right)
+        \cup \bigcap_{n} (W \cap C_{i_n}^{(n)})
+        = \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right) \cup \{x\}
+        \]
+        is countable as a countable union of countable sets $\lightning$.
+
+    \item Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
+        as in the first part.
+        Clearly $f$ is also continuous with respect to the new topology,
+        so we may assume that $X$ is zero dimensional.
+
+        Let $W \subseteq X$ be such that $f\defon{W}$ is injective
+        and $f(W) = f(X)$ (this exists by the axiom of choice).
+        Since $f(X)$ is uncountable, so is $W$.
+        By the second point, there exist disjoint clopen sets
+        $U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
+        are uncountable.
+        Inductively construct $U_s$ for $s \in 2^{<\omega}$
+        as follows:
+        Suppose that $U_{s}$ has already been chosen.
+        Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
+        be disjoint clopen such that $U_{s\concat 1} \cap W$
+        and $U_{s\concat  0} \cap W$ are uncountable.
+        Such sets exist, since $ U_s \cap W$ is uncountable
+        and $U_s$ is a zero dimensional space with the subspace topology.
+        And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
+        in $U_s$ iff it is clopen in $X$.
+
+        Clearly this defines a Cantor scheme.
+    \item \todo{TODO}
+
+\end{itemize}
+ 
+
+\nr 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 $Y_\omega \coloneqq \bigcap Y_i$.
+
+
+%\resizebox{\textwidth}{!}{%
+% https://q.uiver.app/#q=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
+\[\begin{tikzcd}
+	{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
+	{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
+	\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
+	\arrow["g"{pos=0.1}, from=2-2, to=1-4]
+	\arrow["f"{pos=0.8}, from=1-6, to=2-8]
+	\arrow["g"{pos=0.1}, from=2-6, to=1-8]
+\end{tikzcd}\]
+%}
+
+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.
+So we can just do the same proof and every set will be Borel.

inputs/lecture_15.tex

@@ -1,31 +1,5 @@
 \lecture{15}{2023-12-05}{}
 
-Recall:
-\begin{definition}+
-    Let $X$ be a set.
-    A \vocab{group action} of a group $G$ on $X$
-    is a function
-    $\alpha\colon G \times X \to X$
-    such that
-    \begin{itemize}
-        \item $\forall x \in X.~\alpha(1_G,x) = x$,
-        \item $\forall g,h \in G, x \in X.~\alpha(gh,x) = \alpha(g,\alpha(h,x))$.
-    \end{itemize}
-
-    Often we will abbreviate $\alpha(g,x)$ as $g\cdot x$.
-\end{definition}
-\begin{remark}+
-    Group actions of a group $G$ on a set $X$
-    correspond to group-homomorphisms
-    $G \to \Sym(X)$.
-    Indeed for a group action $\alpha\colon G \times X \to X$
-    consider
-    \begin{IEEEeqnarray*}{rCl}
-        G&\longrightarrow & \Sym(X) \\
-        g&\longmapsto & (x \mapsto g \cdot x).
-    \end{IEEEeqnarray*}
-\end{remark}
-
 \begin{theorem}[The Boundedness Theorem]
     \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}