removed junk

+

@@ -1,223 +0,0 @@
-\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.