+

@@ -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.

inputs/lecture_03.tex

@@ -150,9 +150,8 @@
 \end{proof}
 
 \begin{corollary}
-    \label{cor:perfectpolishcard}
     Every nonempty perfect Polish
-    space $X$ has cardinality $\fc = 2^{\aleph_0}$ 
+    space $X$ has cardinality $C = 2^{\aleph_0}$ 
     % TODO: eulerscript C ?
 \end{corollary}
 \begin{proof}
@@ -164,11 +163,9 @@
 
 \begin{theorem}
     Any Polish space is countable
-    or it has cardinality $\fc$. % TODO C
+    or it has cardinality $C$. % TODO C
 \end{theorem}
-\begin{proof}
+\todo{Homework 3}
-    See \autoref{cor:polishcard}.
-\end{proof}
 
 
 \begin{definition}

inputs/lecture_07.tex

@@ -1,5 +1,3 @@
-\subsection{Sheet 6}
-
 \lecture{07}{2023-11-07}{}
 
 \begin{proposition}
@@ -190,3 +188,5 @@
     $\cB(\cT_\infty') = \cB(\cT)$
     and $A $ is clopen in $\cT_{\infty}'$.
 \end{refproof}
+
+

inputs/lecture_09.tex

@@ -119,7 +119,7 @@ We will see that  not every analytic set is Borel.
             Take $f^+\colon X \times  Z \to Y \times Z, f^+ = f \times \id$.
             Then
             \[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
-        \item See \yaref{ex:7.2}.
+        \item \todo{Exercise}
     \end{enumerate}
 \end{proof}
 

inputs/lecture_15.tex

@@ -1,3 +0,0 @@
-\lecture{15}{2023-12-05}{}
-
-\todo{Lecture 15 is missing}

inputs/tutorial_01.tex

@@ -1,5 +1,6 @@
-\tutorial{01}{2023-10-17}{}
+\tutorial{01}{202-10-17}{}
 
+% TODO MAIL
 
 \begin{fact}
     A countable product of separable spaces $(X_n)_{n \in \N}$ is separable.
@@ -65,4 +66,5 @@
 
     In arbitrary topological spaces,
     Lindelöf is the strongest of these notions.
+    
 \end{remark}

inputs/tutorial_02.tex

@@ -1,181 +1,11 @@
-\subsection{Sheet 1}
 \tutorial{02}{2023-10-24}{}
+
 % Points: 15 / 16
 
-\nr 1
+\subsubsection{Exercise 4}
-\todo{handwritten solution}
-
-\nr 2
-
-\begin{enumerate}[(a)]
-    \item $d$ is an ultrametric:
-
-        Let $f,g,h \in X^{\N}$.
-
-        We need to show that $d(f,g) \le \max(d(f,h), d(g,h))$.
-
-        If $f = g$ this is trivial.
-        Otherwise let $n$ be minimal such that $f(n) \neq g(n)$.
-        Then $h(n) \neq f(n)$ or $ h(n) \neq g(n)$
-        must be the case.
-        W.l.o.g.~$h(n) \neq f(n)$.
-        Then $d(f,g) = \frac{1}{1+n} \le d(f,h)$.
-
-    \item $d$ induces the product topology on $X^{\N}$:
-
-        It suffices to show that the $\epsilon$-balls with respect to $d$
-        are exactly the basic open set of the product topology,
-        i.e.~the sets of the form
-        \[
-        \{x_1\} \times \ldots \times  \{x_n\}  \times X^{\N}
-        \]
-        for some $n \in \N$, $x_1,\ldots,x _n \in X$.
-
-        Let $\epsilon > 0$. Let $n$ be minimal such that $\frac{1}{1+n} \ge \epsilon$.
-        Then $B_{\epsilon}((x_i)_{i \in \N}) = \{x_1\} \times  \{x_n\} \times X^{\N}$.
-        Since $\N \ni n \mapsto \frac{1}{1+n}$
-        is injective, every basic open set of the product topology
-        can be written in this way.
-
-    \item $d$ is complete:
-
-        Let $(f_n)_{n \in \N}$ be a Cauchy sequence with respect to $d$.
-        For $n \in \N$ take $N_n \in \N$
-        such that $d(f_i, f_j) < \frac{1}{1 + n}$.
-        Clearly $f_i(n) = f_j(n)$ for all  $n > N_n$.
-
-        Define $f \in  X^\N$ by $f(n) \coloneqq f_{N_n}(n)$.
-        Then $ (f_n)_{n \in \N}$
-        converges to $f$,
-        since for all $n > N_n$ $f_n$
-
-
-    \item If $X$ is countable, then $X^{\N}$ with the product topology
-        is a Polish space:
-
-        (We assume that $X$ is non-empty, as otherwise the claim is wrong)
-
-        We need to show that there exists a countable dense subset.
-        To this end, pick some $x_0 \in X$ and
-        consider the set $D \coloneqq \bigcup_{n\in \N} (X^n \times \{x_{0}\}^{\N})$.
-        Since $X$ is countable, so is $D$.
-        Take some $(a_n)_{n \in \N} \in X^{\N}$
-        and consider $B \coloneqq  B_{\epsilon}((a_n)_{n \in \N})$.
-        Let $m$ be such that $\frac{1}{1+m} < \epsilon$.
-        Then $(b_{n})_{n \in \N} \in B \cap D$,
-        where $b_n \coloneqq  a_n$ for $n \le  m$ and $b_n \coloneqq x_0$
-        otherwise.
-        Hence $D$ is dense.
-\end{enumerate}
-
-\nr 3
-
-\begin{enumerate}[(a)]
-    \item $S_{\infty}$ is a Polish space:
-
-        From (2) we know that $\N^{\N}$ is Polish.
-        Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
-        with respect to $\N^\N$.
-
-        Consider the sets $I \coloneqq \bigcap_{(i,j) \in \N^2, i < j} \{f \in \N^{\N} | f(i) \neq f(j)\}$
-        and $S \coloneqq \bigcap_{n \in \N} \{f \in \N^\N | n \in \im f\}$.
-
-        We have that $\{f \in \N^\N | f(i) \neq f(j)\} = \bigcup_{n \in \N}  \N^{i-1} \times \{n\}  \times \N^{i - j -1 } \times (\N \setminus \{n\} ) \times \N^\N$
-        is open.
-        Hence $I$ is $G_{\delta}$.
-
-        Furthermore $\{f \in \N^{\N} | n \in \im f\} = \bigcup_{k \in \N} \N^k \times \{n\} \times \N^\N$j
-        is open,
-        thus $S$ is $G_\delta$ as well.
-        In particular $S \cap G$ is $G_\delta$.
-        Since $I$ is the subset of injective functions
-        and $S$ is the subset of surjective functions,
-        we have that $S_{\infty} = I \cap S$.
-
-    \item $S_{\infty}$ is not locally compact:
-
-        Consider the point $x = (i)_{i \in \N} \in S_{\infty}$.
-        Let $x \in B$ be open. We need to show that there
-        is no closed compact set $C \supseteq B$
-        W.l.o.g.~let $B = (\{0\} \times \ldots \times  \{n\}  \times \N^\N) \cap S_\infty$
-        for some $n \in \N$.
-        Let $C \supseteq B$ be some closed set.
-        Consider the open covering
-        \[
-        \{S_{\infty} \setminus B\} \cup \{ B_j | j > n\}.
-        \]
-        where
-        \[
-            B_j \coloneqq (\{0\} \times  \ldots \times \{n\} \times \{j\} \times \N^{\N}) \cap S_\infty.
-        \]
-        Clearly there cannot exist a finite subcover
-        as $B$ is the disjoint union of the $B_j$.
-
-         % TODO Think about this
-\end{enumerate}
-
-\nr 4
-
-% (uniform metric)
-% 
-% \begin{enumerate}[(a)]
-%     \item $d_u$ is a metric on $\cC(X,Y)$:
-% 
-%         It is clear that $d_u(f,f) = 0$.
-% 
-%         Let  $f \neq g$. Then there exists $x \in X$ with
-%         $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$.
-% 
-%         Since $d$ is symmetric, so is $d_u$.
-% 
-%         Let $f,g,h \in \cC(X,Y)$.
-%         Take some $\epsilon > 0$
-%         choose $x_1, x_2 \in X$
-%         with $d_u(f,g) \le   d(f(x_1), g(x_1)) + \epsilon$,
-%         $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$.
-% 
-%         Then for all $x \in X$
-%         \begin{IEEEeqnarray*}{rCl}
-%             d(f(x), h(x)) &\le &
-%             d(f(x), g(x)) + d(g(x), h(x))\\
-%                           &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\
-%                           &\le & d_u(f,g) + d_u(g,h) - 2\epsilon.
-%         \end{IEEEeqnarray*}
-%         Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$.
-%         Taking $\epsilon \to 0$ yields the triangle inequality.
-% 
-%     \item $\cC(X,Y)$ is a Polish space:
-%         \todo{handwritten solution}
-% 
-%         \begin{itemize}
-%             \item $d_u$ is a complete metric:
-% 
-%                 Let $(f_n)_n$ be a Cauchy series with respect to $d_u$.
-% 
-%                 Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect
-%                 to  $d$ for every $x$.
-%                 Hence there exists a pointwise limit $f$ of the $f_n$.
-%                 We need to show that $f$ is continuous.
-% 
-%                 %\todo{something something uniform convergence theorem}
-% 
-%             \item $(\cC(X,Y), d_u)$ is separable:
-% 
-%                 Since $Y$ is separable, there exists a countable
-%                 dense subset $S \subseteq Y$.
-% 
-%                 Consider $\cC(X,S) \subseteq  \cC(X,Y)$.
-%                 Take some $f \in \cC(X,Y)$.
-%                 Since $X$ is compact,
-% 
-% 
-%                 % TODO
-% 
-%         \end{itemize}
-% \end{enumerate}
 
 \begin{fact}
-    Let $X $ be a compact Hausdorff space.
+    Let $X $ be a compact Hausdorffspace.
     Then the following are equivalent:
     \begin{enumerate}[(i)]
         \item $X$ is Polish,
@@ -202,7 +32,7 @@
     Hausdorff spaces are normal
 \end{proof}
 
-Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
+Let $X$ be compact Polish (compact metrisable $\implies$ compact Polish)
 and $Y $ Polish.
 Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
 Consider the metric $d_u(f,g) \coloneqq  \sup_{x \in X} |d(f(x), g(x))|$.
@@ -229,4 +59,3 @@ Clearly $d_u$ is a metric.
 \begin{claim}
     There exists a countable dense subset.
 \end{claim}
-\todo{handwritten solution}

inputs/tutorial_03.tex

@@ -1,61 +1,15 @@
-\subsection{Sheet 2}
+\ctutorial{03}{2023-10-31}{}
-
-\tutorial{03}{2023-10-31}{}
-
 
 % 15 / 16
 
 \begin{remark}
-    $F_\sigma$ stands for \vocab{ferm\'e sum denumerable}.
+    $F_\sigma$
+    stands for ferm\'e sum denumerable.
 \end{remark}
 
 
-\nr 1
+\subsection{Exercise 2}
 
-Let $(U_i)_{i < \omega}$ be a countable base of $X$.
-Define
-\begin{IEEEeqnarray*}{rCl}
-    f\colon X &\longrightarrow & 2^{\omega} \\
-    x &\longmapsto & (x_i)_{i < \omega}
-\end{IEEEeqnarray*}
-where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
-Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
-is open.
-We have that $f$ is injective since $X$ is T1.
-
-Let $f\colon X \hookrightarrow 2^\omega$ be such that
-$f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\})$.
-
-Let $V \subseteq 2^{\omega}$ be closed.
-Then $2^{\omega} \setminus V$ is open, i.e.~has the form
-$\bigcup_{i \in I} ((\prod_{j<n_j} X_{i,j}) \times 2^{\omega})$
-for some $X_{i,j} \subseteq 2$.
-As $2^{\omega}$ is second countable, we may assume $I$ to be countable.
-
-Then $V = \bigcap_{i \in  I} \left(2^{\omega} \setminus  ((\prod_{i <n_j} X_{i,j}) \times 2^{\omega})\right)$.
-Since  $f$ is injective, we have $f^{-1}(\bigcap_{a \in A} a) = \bigcap_{a \in A} f^{-1}(a)$.
-Thus it suffices to show that $f^{-1}(2^{\omega} \setminus  ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
-is $G_{\delta}$, as a countable intersection of $G_{\delta}$-sets
-is $G_{\delta}$.
-
-We have that $U_k \coloneqq f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 1\})$
-is open. Since $f$ is injective
-$f^{-1}(\{y = (y_i) \in 2^{\omega} : y_k = 0\}) = X \setminus U_k$
-is closed, in particular it is $G_\delta$.
-Let $x = (x_1,\ldots, x_n) \in 2^{n} \setminus (\prod_{i < n} X_i)$.
-Then $f^{-1}(\{x\} \times 2^\omega) = \bigcap_{i < n}\bigcap U'_n$
-is $G_{\delta}$, were $U'_i = U_i$ if $x_k = i$
-and $U'_i = X \setminus U_i$ otherwise.
-
-Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
-is finite, we get that
-$f^{-1}(2^{\omega} \setminus  ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
-is $G_\delta$ as a finite union of $G_{\delta}$ sets.
-
-
-
-\nr 2
-\todo{handwritten solution}
 (b)
 Let $f(x^{(i)})$ be a sequence in $f(X)$.
 Suppose that $f(x^{(i)}) \to y$.
@@ -64,7 +18,8 @@ Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
 Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
 
 
-\nr 3
+\subsection{Exercise 3}
+
 \begin{example}
     Consider
     \begin{IEEEeqnarray*}{rCl}
@@ -76,6 +31,7 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
     and $\osc_f(x) = 0$ for $x \not\in \Q$.
 \end{example}
 
+
 \begin{definition}
     We say that $f\colon X \to Y$ is continuous
     at $a \in X$,
@@ -84,11 +40,13 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
     then $f^{-1}(N)$ is a neighbourhood of $a$.
 \end{definition}
 
+
+
 \begin{theorem}[Kuratowski]
     Let $X$ be metrizable, $Y$ completely metrizable,
-    $S \subseteq X$ and $f\colon S \to Y$ continuous.
+    $f\colon S \to Y$ continuous.
-    Then $f$ can be extended to a continuous function $\tilde{f}$
+    Then $f$ can be extended to a continuous fnuction $f_n$
-    on a $G_\delta$ set  $G$ with $S \subseteq  G \subseteq \overline{S}$.
+    on a $G_\delta$ set  $G$ with $S \subseteq  G \subseteq \overline{G}$.
 \end{theorem}
 \begin{proof}
     Let $G \coloneqq  \overline{S} \cap \{x \in X | \osc_f(x) = 0\}$.
@@ -107,12 +65,15 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
     is an intersection of open sets.
 
     For $x \in G$, as $x \in  \overline{S}$,
-    there exists $(x_n)_{x_n < \omega}$, $x_n \in S$
+    there exists $(x_n)_{x_n < \omega}$, $x_n \in S_$
     such that $x_n \to x$.
     We have that $(f(x_n))_n$ is Cauchy,
     as $\osc_f(x) = 0$.
+    % TODO
 
     \todo{Something is missing here}
+
+
 \end{proof}
 
 \begin{corollary}
@@ -120,74 +81,20 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
     Then $Y$ is $G_{\delta}$.
 \end{corollary}
 \begin{proof}
-    \todo{TODO}
+    Consider the identity $f\colon Y \hookrightarrow X$.
-    % Consider the identity $f\colon Y \hookrightarrow X$.
+    Then $f$ can be extended to a $G_{\delta}$ set.
-    % Then $f$ can be extended to a $G_{\delta}$ set $G \subseteq X$
+    $f$ and $\id_G$ agree on $Y$.
-    % with $Y \subseteq G \subseteq  \overline{Y}$.
+    Hence $Y \subseteq G \subseteq  \overline{Y}$.
-    % $\tilde{f}$ and $\id_G$ agree on $Y$.
+    $Y$ is dense in $G$ and $\cod(f)$ is ltd.\todo{????}
-    % $Y$ is dense in $G$ and the codomain of $f$ is ltd.\todo{????}
+    So $f = \id_G$, i.e.~$G = Y$.
-    % So $f = \id_G$, i.e.~$G = Y$.
 \end{proof}
 
-\nr 4
+\subsection{Exercise 4}
-\begin{enumerate}[(1)]
-    \item $f$ is a topological embedding:
-        Consider a basic open set
-        $B = \prod_{i < n} X_i \times \omega^{\omega}$
-        for some $X_i \subseteq \omega$.
 
-        Then $f(B) = \left(\bigcup_{x \in \prod_{i<n} X_i} B_x \right) \cap f(\omega^{\omega})$
+Let $C$ be the subspace of $2^{\omega}$ consisting
-        is open in $f(\omega^{\omega})$,
+of sequences with finitely many $1$s.
-        where $B_x \coloneqq \{0^{x_0}10^{x_1}1\ldots 10^{x_n-1}1\} \times 2^{\omega}$.
+We want to show that $C \cong \Q$.
 
-        On the other hand let $C = \{x_0x_1x_2x_3x_4 \ldots x_{n-1}\}\times 2^{\omega}$
+Go to the right in the even digits, go to the left for the odd digits,
-        be some basic open set of $2^{\omega}$.
+i.e.~let $C = (1,-1,1,-1, \ldots)$
-        W.l.o.g.~$x_0x_1\ldots x_{n-1}$
+and set $x < y$ iff $C \cdot  x <_{\text{lex}} C \cdot y$.
-        has the form $0^{a_0}10^{a_1}1\ldots 10^{a_k}x_{n-1}$.
-        If $x_{n-1} = 1$,
-        we get
-        \[
-            f^{-1}(C \cap f(\omega^{\omega})) = \{(a_0,a_1,\ldots,a_k)\} \times \omega^{\omega}.
-        \]
-        In the case of $x_{n-1} = 0$,
-        it is
-        \[
-        f^{-1}(C \cap f(\omega^\omega)) = \bigcup_{b > a_k} \{(a_0,a_1,\ldots,a_{k-1}, b)\} \times \omega^{\omega}.
-        \]
-        In both cases the preimage is open.
-
-
-    \item  $C \coloneqq 2^{\omega} \setminus f(\omega^\omega)$
-        is countable and dense in $2^{\omega}$.
-
-        We have $C = \{x \in 2^{\omega} | x_i = 0 \text{ for all but finitely many $i$}\} = \bigcup_{i < \omega} (2^{i} \times 1^{\omega})$.
-        Clearly this is countable.
-
-        For denseness take some $x \in 2^\omega$.
-        Let $x^{(n)}$ be defined by $x^{(n)}_i = x_i$ for $i < n$
-        and $x^{(n)}_i = 0$ for $i \ge n$.
-        Then $x^{(n)} \in C$ for all $n$,
-        and $x^{(n)}$ converges to $x$.
-
-    \item $f(\omega^\omega)$ is $G_\delta$:
-
-        We have
-        \begin{IEEEeqnarray*}{rCl}
-            f(\omega^\omega) &=& 2^\omega \setminus \left(\bigcup_{i < \omega} (2^{i} \times 1^{\omega})\right)\\
-                             &=& \bigcap_{i < \omega} \left(2^{\omega} \setminus(2^{i} \times 1^{\omega})\right).
-        \end{IEEEeqnarray*}
-    \item $C$ as in (2) is homeomorphic to $\Q$.
-
-        Go to the right in the even digits, go to the left for the odd digits,
-        i.e.~let $C = (1,-1,1,-1, \ldots)$
-        and set $x < y$ iff $C \cdot  x <_{\text{lex}} C \cdot y$,
-        where $<_{\text{lex}}$ denotes the lexicographical ordering.
-        Note that the order topology of $<$ on $C$ 
-        agrees with the subspace topology from $2^\omega$.
-
-        By Cantor's theorem for countable, unbounded, dense linear
-        linear orders,
-        we get an order isomorphism $C \leftrightarrow \Q$.
-        This is also a homeomorphism, as the topologies
-        on $C$ and $\Q$ are the respective order topologies.
-\end{enumerate}

inputs/tutorial_04.tex

@@ -1,208 +1,70 @@
-\subsection{Sheet 3}
+\tutorial{04}{2023-11-14}{}
 
-\tutorial{04}{}{}
+\subsection{Sheet 4}
 
-\nr 1
+% 14 / 20
+
+\subsubsection{Exercise 1}
 
-Let $A \neq \emptyset$ be discrete.
-For $D \subseteq  A^{\omega}$,
-let
-\[
-    T_D \coloneqq  \{x\defon{n} \in  A^{<\omega} | x \in D, n \in \N\}..
-\]
 \begin{enumerate}[(a)]
-    \item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:
+    \item $\langle  X_\alpha : \alpha\rangle$
+    is a descending chain of closed sets (transfinite induction).
 
-        Clearly $T_D$ is a tree.
+    Since $X$ is second countable, there cannot exist
-        Let $x \in T_D$.
+    uncountable strictly decreasing chains of closed sets:
-        Then there exists $d \in D$ such that $x = d\defon{n}$.
-        Hence  $x \subseteq d\defon{n+1} \in T_D$.
-        Thus $x$ is not a leaf, i.e.~$T_D$ is pruned.
-    \item For any $T \subseteq  A^{<\omega}$, $[T]$ is a closed subset
-        of $A^{\omega}$:
 
-        Let $a \in A^{\omega} \setminus [T]$.
+    Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$
-        Then there exists some $n$ such that $a\defon{n} \not\in T$.
+    was such a sequence,
-        Hence $\{a_0\} \times \ldots \times \{a_{n+1}\} \times A^{\omega}$
+    then $X \setminus X_{\alpha}$ is open for every $\alpha$,
-        is an open neighbourhood of $a$ disjoint from $[T]$.
+    Let $\{U_n : n < \omega\}$ be a countable basis.
+    Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
+    is a strictly ascending chain in $\omega$.
 
-    \item $T \mapsto [T]$ is a bijection between the
+    \item We need to show that $X_{\alpha_0}$ is perfect and closed.
-        pruned trees on $A$ and the closed subsets of $A^{\omega}$.
+        It is closed since all $X_{\alpha}$ are,
+        and perfect, as a closed set $F$ is perfect
+        iff it coincides $F'$.
 
-        \begin{claim}
+        $X \setminus X_{\alpha_0}$
-            $[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$.
+        is countable:
-        \end{claim}
+        $X_{\alpha} \setminus X_{\alpha + 1}$ is
-        \begin{subproof}
+        countable as for every $x$ there exists a basic open set $U$,
-            Clearly $D \subseteq [T_D]$.
+        such that $U \cap X_{\alpha} = \{x\}$,
-            Let $x \in [T_D]$.
+        and the space is second countable.
-            Then for every $n < \omega$,
+        Hence $X \setminus X_{\alpha_0}$
-            there exists some $d_n \in D$ such that
+        is countable as a countable union of countable sets.
-            $d_n\defon{n} = x\defon{n}$.
-            Clearly the $d_n$ converge to $x$.
-            Since $D$ is closed, we get $x \in D$.
-        \end{subproof}
-
-        This shows that $T \mapsto [T]$ is surjective.
-
-        Now let $T \neq T'$ be pruned trees.
-        Then there exists $x \in T \mathop{\triangle} T'$,
-        wlog.~$x \in T \setminus T'$.
-        Since $T$ is pruned
-        by applying the axiom of countable choice
-        we get an infinite branch $x' \in  [T] \setminus [T']$.
-        Hence the map is injective.
-
-    \item Let $N_s \coloneqq  \{x \in A^{\omega} | s \subseteq x\}$.
-        Show that every open $U \subseteq A^{\omega}$
-        can be written as $U = \bigcup_{s \in S}  N_s$
-        for some set of pairwise incompatible $S \subseteq A^{<\omega}$.
-
-
-        Let $U$ be open.
-        Then $U$ has the form
-        \[
-        U = \bigcup_{i \in I} X_i \times  A^{\omega}
-        \]
-        for some $X_i \subseteq A^{n_i}$, $n_i < \omega$.
-        Clearly
-        $U = \bigcup_{s \in S'} N_s$
-        for
-        $S' \coloneqq  \bigcup_{i \in I} X_i$.
-        Define
-        \[
-        S \coloneqq \{s \in S' | \lnot\exists t \in S'.~t\subseteq s \land |t| < |s|\}.
-        \]
-        Then the elements of $S$ are pairwise incompatible
-        and $U = \bigcup_{s \in S}  N_s$.
-
-    \item Let $T \subseteq A^{<\omega}$ be an infinite tree which is finitely
-        splitting.
-        Then $[T]$ is nonempty:
-
-        Let us recursively construct a sequence of compatible $s_n \in T$
-        with $|s_n| = n$
-        such that  $\{s_n\} \times A^{<\omega} \cap T$ is infinite.
-        Let  $s_0$ be the empty sequence;
-        by assumption $T$ is infinite.
-        Suppose that $s_n$ has been chosen.
-        Since $T$ is finitely splitting, there are only finitely
-        many $a \in A$ with $s_n\concat a \in T$.
-        Since $ \{s_n\} \times A^{<\omega} \cap T$
-        is infinite,
-        there must exist at least on $a \in A$
-        such that $\{s_n\concat a\} \times A^{<\omega} \cap T$
-        is infinite.
-        Define $s_{n+1} \coloneqq  s_n \concat a$.
-
-        Then the union of the $s_n$ is an infinite branch of $T$,
-        i.e.~$[T]$ is nonempty.
-
-
-    \item Then $[T]$ is compact:
-        \todo{TODO}
-
-        % Let $\langle s_n, n <\omega \rangle$
-        % be a Cauchy sequence in $[T]$.
-
-        % Then for every $m < \omega$
-        % there exists an $N < \omega$ such that
-        % $s_n\defon{m} = s_{n'}\defon{m}$
-        % for all  $n, n' > N$.
-        % Thus there exists a pointwise limit $s$ of the $s_n$.
-
-        % Since for all $m$ we have $s\defon{m} = s_n\defon{m} \in [T]$
-        % for $m$ large enough,
-        % we get $s \in [T]$.
-
-        % Hence $[T]$ is sequentially compact.
 \end{enumerate}
 
-\nr 2
-\todo{handwritten}
-\nr 3
-\todo{handwritten}
 
-\nr 4
+\subsection{Exercise 3}
 
-\begin{notation}
+\begin{itemize}
-    For $A \subseteq X$ let $A'$ denote the set of
+    \item Let $Y \subseteq \R$ be $G_\delta$
-    accumulation points of $A$.
+        such that $Y$ and $\R \setminus Y$ are dense in $\R$.
-\end{notation}
+        Then $Y \cong \cN$.
 
-\begin{theorem}
+        $Y$ is Polish, since it is $G_\delta$.
-    Let $X$ be a Polish space.
-    Then there exists a unique partition $X = P \sqcup U$
-    of $X$ into a perfect closed subset $P$ and a countable open subset $U$.
-\end{theorem}
-\begin{proof}
 
-    Let $P$ be the set of condensation points of $X$
+        $Y$ is 0-dimensional,
-    and $U \coloneqq X \setminus P$.
+        since the sets $(a,b) \cap Y$ for  $a, b \in \R \setminus Y$
+        form a clopen basis.
 
-    \begin{claim}
+        Each compact subset of $Y$ has empty interior:
-        $U$ is open and countable.
+        Let $K \subseteq Y$ be compact
-    \end{claim}
+        and $U \subseteq K$ be open in $Y$.
-    \begin{subproof}
+        Then we can find cover of $U$ that has no finite subcover $\lightning$.
-        Let $S$ be a countable dense subset.
-        For each $x \in U$,
-        there is an $\epsilon_x > 0$, $s_x \in S$ such that $x \in B_{\epsilon_x}(s_x)$
-        is at most countable.
-        Clearly $B_{\epsilon_x}(s_x) \subseteq U$,
-        as for every $y \in B_{\epsilon_x}(s_x)$,
-        $B_{\epsilon_x}(s_x)$ witnesses that $y \not\in P$.
-        Thus $U = \bigcup_{x \in U} B_{\epsilon_x}(s_x)$
-        is open.
-        Wlog.~$\epsilon_x \in \Q$ for all $x$.
-        Then the RHS is the union of at most
-        countably many countable sets, as $S \times \Q$
-        is countable.
-    \end{subproof}
 
-    \begin{claim}
+    \item Let $Y \subseteq \R$ be $G_\delta$ and dense
-        $P$ is perfect.
+        such that $\R \setminus Y$ is dense as well.
-    \end{claim}
+        Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\}   \subseteq  \R^2$.
-    \begin{subproof}
+        Then $Z$ is dense in $\R^2$
-        Let $x \in P'$ and $x \in U$ an open neighbourhood.
+        and $\R^2 \setminus \Z$ is dense in $\R^2$.
-        Then there exists $y \in P \cap U$.
-        In particular, $U$ is an open neighbourhood of $ y$,
-        hence $U$ is uncountable.
-        It follows that $x \in P$.
 
-        On the other hand let $x \in P$
-        and let $U$ be an open neighbourhood.
-        We need to show that $U \cap P \setminus \{x\}$
-        is not empty.
-        Suppose that for all $y \in U \cap P \setminus \{x\}$,
-        there is an open neighbourhood $U_y$
-        such that $U_y$ is at most countable.
-        Wlog.~$U_y = B_{\epsilon_y}(s_y)$ for some $s_y \in S$, $\epsilon_y > 0$,
-        where $S$ is again a countable dense subset.
-        Wlog.~$\epsilon_y \in \Q$.
-        But then
-        \[
-            U = \{x\} \cup \bigcup_{y \in U} B_{\epsilon_y}(s_y)
-        \]
-        is at most countable as a countable union of countable sets,
-        contradiction $x \in P$.
-    \end{subproof}
 
-    \begin{claim}
+        We have that for every $y \in Y$
-        Let $P,U$ be defined as above
+        $\partial B_y(0) \subseteq Z$.
-        and let $P_2 \subseteq  X$, $U_2 \subseteq  X$
-        be such that $P_2$ is perfect and closed,
-        $U_2$ is countable and open
-        and $X = P_2 \sqcup U_2$.
-        Then $P_2 = P$ and $U_2 = U$.
-    \end{claim}
-    \todo{TODO}
-\end{proof}
 
-\begin{corollary}\label{cor:polishcard}
+
-    Any Polish space is either countable or has cardinality equal to $\fc$.
+        Other example:
-\end{corollary}
+        Consider $\R^2 \setminus \Q^2$.
-\begin{subproof}
+\end{itemize}
-    Let $X = P \sqcup U$
-    where  $P$ is perfect and $U$ is countable.
-    If $P \neq \emptyset$, we have $|P| = \fc$
-    by \yaref{cor:perfectpolishcard}.
-\end{subproof}

inputs/tutorial_05.tex

@@ -1,123 +1,35 @@
-\subsection{Sheet 4}
+\tutorial{05}{}{}
-\tutorial{05}{2023-11-14}{}
 
-% 14 / 20
+% Sheet 5 - 18.5 / 20
 
-\nr 1
+\subsection{Exercise 1}
+
+Let $B \subseteq C$ be comeager.
+Then $B = B_1 \cup B_2$,
+where $B_1$ is dense $G_\delta$
+and $B_2$ is meager.
 
 
-\begin{enumerate}[(a)]
+\begin{fact}
-    \item $\langle  X_\alpha : \alpha\rangle$
+     $X$ is Baire iff every non-empty open set is non-meager.
-    is a descending chain of closed sets (transfinite induction).
 
-    Since $X$ is second countable, there cannot exist
+     In particular, let $X$ be Baire,
-    uncountable strictly decreasing chains of closed sets:
+     then $U \overset{\text{open}}{\subseteq} X$
+     is Baire.
+\end{fact}
 
-    Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$
+\subsection{Exercise 4}
-    was such a sequence,
-    then $X \setminus X_{\alpha}$ is open for every $\alpha$,
-    Let $\{U_n : n < \omega\}$ be a countable basis.
-    Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$,
-    is a strictly ascending chain in $\omega$.
 
-    \item We need to show that $X_{\alpha_0}$ is perfect and closed.
+\begin{enumerate}[(i)]
-        It is closed since all $X_{\alpha}$ are,
+    \item $|B| = \fc$, since $B$ contains a comeager
-        and perfect, as a closed set $F$ is perfect
+        $G_\delta$ set, $B'$:
-        iff it coincides $F'$.
+        $B'$ is Polish,
-
+        hence $B' =  P \cup C$
-        $X \setminus X_{\alpha_0}$
+        for $P$ perfect and $C$ countable,
-        is countable:
+        and $|P| \in \{\fc, 0\}$.
-        $X_{\alpha} \setminus X_{\alpha + 1}$ is
+        But $B'$ can't contain isolated point$.
-        countable as for every $x$ there exists a basic open set $U$,
-        such that $U \cap X_{\alpha} = \{x\}$,
-        and the space is second countable.
-        Hence $X \setminus X_{\alpha_0}$
-        is countable as a countable union of countable sets.
-\end{enumerate}
-
-\nr 2
-\todo{handwritten}
-
-
-\nr 3
-\begin{itemize}
-    \item Let $Y \subseteq \R$ be $G_\delta$
-        such that $Y$ and $\R \setminus Y$ are dense in $\R$.
-        Then $Y \cong \cN$.
-
-        $Y$ is Polish, since it is $G_\delta$.
-
-        $Y$ is 0-dimensional,
-        since the sets $(a,b) \cap Y$ for  $a, b \in \R \setminus Y$
-        form a clopen basis.
-
-        Each compact subset of $Y$ has empty interior:
-        Let $K \subseteq Y$ be compact
-        and $U \subseteq K$ be open in $Y$.
-        Then we can find cover of $U$ that has no finite subcover $\lightning$.
-
-    \item Let $Y \subseteq \R$ be $G_\delta$ and dense
-        such that $\R \setminus Y$ is dense as well.
-        Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\}   \subseteq  \R^2$.
-        Then $Z$ is dense in $\R^2$
-        and $\R^2 \setminus \Z$ is dense in $\R^2$.
-
-
-        We have that for every $y \in Y$
-        $\partial B_y(0) \subseteq Z$.
-
-
-        Other example:
-        Consider $\R^2 \setminus \Q^2$.
-\end{itemize}
-
-\nr 4
-
-\begin{enumerate}[(a)]
-    \item Let $d$ be a compatible, complete metric on $X$, wlog.~$d \le 1$.
-        Set $ U_{\emptyset} \coloneqq X$.
-        Suppose that $U_{s}$ has already been chosen.
-        Then $D_s \coloneqq X \setminus U_s$ is closed.
-        Hence $U_s^{(n)} \coloneqq  \{x \in X | \dist(x,D_s) > \frac{1}{n}\}$ 
-        is open.
-        Let $m$ be such that $D_s^{(m)} \neq \emptyset$.
-        Clearly $\overline{U_s^{(n)}} \subseteq U_s$.
-        Let $(B_k)_{k < \omega}$ be a countable cover of $X$
-        consisting of balls of diameter  $2^{-|s|-2}$.
-        Take some bijection  $\phi\colon \omega \to \omega \times  (\omega \setminus m)$
-        and set $U_{s\concat i} \coloneqq  U_s^{(\pi_1\left( \phi(i) \right))} \cap B_{\pi_2(\phi(i))}$,
-        where there $\pi_i$ denote the projections
-        (if this is empty, set $U_{s \concat i} \coloneqq U_s^{\pi_1(\phi(j))} \cap B_{\pi_2(\phi(j))}$
-    for some arbitarily chosen $j < \omega$ such that it is not empty).
-        Then $\overline{U_{s \concat i}} \subseteq  \overline{U_s^{(n)}} \subseteq U_s$,
-        \[
-        \bigcup_{i < \omega} U_{s \concat i} = \bigcup_{n < \omega} U_{s}^{\left( n \right) } = U_s
-        \]
-        and $\diam(U_{s \concat i}) \le \diam(B_{\pi_2(\phi(i))})$.
-    \item Let $s \in \omega^\omega$.
-        Then
-        \[
-        \bigcap_{n < \omega} \overline{U_{s\defon{n}}}
-        \]
-        contains exactly one point.
-        Let $f$ be the function that maps an $s \in \omega^{\omega}$
-        to the unique point in the intersection of the $\overline{U_{s\defon{n}}}$.
-        Let $x \in X$ be some point.
-        Then by induction we can construct a sequence $s \in \omega^{\omega}$
-        such that $x \in U_{s\defon{n}}$ for all $n$,
-        hence $x = f(s)$, i.e.~$f$ is surjective.
-
-        Let $B \overset{\text{open}}{\subseteq} X$.
-        Then $B = \bigcup_{i \in  I} U_i$
-        for some $i \subseteq \omega^{<\omega}$,
-        as every basic open set can be recovered as a union of $U_i$
-        and $f^{-1}(B) = \bigcup_{i \in  I} \left( \{i_0\}  \times  \ldots \{i_{|i|-1}\}  \right) \times \omega^{\omega}$ is open,
-        hence $f$ is continuous.
-
-
-        On the other hand, consider an open ball  $B \coloneqq  \{\prod_{i < n} \{x_i\}\} \times \omega^{\omega} \subseteq \omega^{\omega}$.
-        Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
-        hence $f$ is open.
 \end{enumerate}
 
 
+
+

inputs/tutorial_06.tex

@@ -1,91 +1,180 @@
-\subsection{Sheet 5}
+\tutorial{06}{2023-11-28}{}
 
-\tutorial{06}{}{}
+% 5 / 20
 
-% Sheet 5 - 18.5 / 20
+\subsection{Exercise 1}
 
-\nr 1
+\begin{warning}
-\todo{handwritten}
+    Note that not every set has a density!
-Let $B \subseteq C$ be comeager.
+\end{warning}
-Then $B = B_1 \cup B_2$,
-where $B_1$ is dense $G_\delta$
-and $B_2$ is meager.
 
-\begin{fact}
-     $X$ is Baire iff every non-empty open set is non-meager.
-
-     In particular, let $X$ be Baire,
-     then $U \overset{\text{open}}{\subseteq} X$
-     is Baire.
-\end{fact}
-
-\nr 2
-
-Let $(U_i)_{i < \omega}$ be a countable base of $Y$.
-We want to find a $G_\delta$ set $A \subseteq X$
-such that $f\defon{A}$ is continuous.
-It suffices make sure that $f^{-1}\defon{A}(U_i)$ is open for all  $i < \omega$.
-Take some  $i < \omega$.
-Then  $V_i \setminus M_i \subseteq f^{-1}(U_i) \subseteq V_i \cup M_i$,
-where $V_i$ is open and $M_i$ is meager.
-Let $M'_i \supseteq M_i$ be a meager $F_\sigma$-set.
-Now let $A \coloneqq X \setminus \bigcup_{i <\omega} M_i'$.
-We have that $A$ is a countable intersection of open dense sets,
-hence it is dense and $G_\delta$.
-For any $i < \omega$,
-$V_i \cap A \subseteq  f\defon{A}^{-1}(U_i) \subseteq  (V_i \cup M_i) \cap A = V_i \cap A$,
-so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
-
-\nr 3
-\todo{handwritten}
-
-
-\nr 4
 \begin{enumerate}[(a)]
-    \item $|B| = \fc$, since $B$ contains a comeager
+    \item Let $X = \bI^{\omega}$.
-        $G_\delta$ set, $B'$:
+        Let $C_0 = \{(x_n) : x_n \to 0\}$.
-        $B'$ is Polish,
+        Claim: $C_0 \in \Pi^0_3(X)$ (intersections of $F_\sigma$ sets).
-        hence $B' =  P \cup C$
-        for $P$ perfect and $C$ countable,
-        and $|P| \in \{\fc, 0\}$.
-        But $B'$ can't contain isolated point.
-    \item To ensure that (a) holds, it suffices to chose
-        $a_i \not\in F_i$.
-        Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$,
-        there exists some $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
-        where $\pi$ denotes the projection.
-        Choose one such $x$.
-        We need to find $y \in \R$,
-        such that $(x,y) \not\in F_i$
-        and $\{a_j | j < i\} \cup \{(x,y)\}$
-        does not contain three collinear points.
 
-        Since $(F_i)_x$ is meager, we have that
+        We have
-        $|\{x\} \times \R \setminus F_i| = |\R \setminus (F_i)_x| = \fc$.
+        \[
-        Let $L \coloneqq \{y \in \R | \exists j < k < i. ~a_j, a_k,(x,y) \text{are collinear}\}$.
+        x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q,
-        Since every pair $a_j \neq a_k, j < k < i$,
+        \]
-        adds at most one point to $L$,
+        i.e.
-        we get $|L| \le |i|^2 < \fc$.
+        \[
-        Hence $|\R \setminus (F_i)_x \setminus L| = \fc$.
+        C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}.
-        In particular, the set is non empty, and we find $y$ as desired
+        \]
-        and can set $a_i \coloneqq (x,y)$.
+        Clearly this is a $\Pi^0_3$ set.
-
-        % We have not chosen too many points so far.
-        % So there are not too many lines,
-        % we can not choose a point from,
-        % but there are many points in $B$.
-    \item $A$ is by construction not a subset of any $F_\sigma$ meager set.
-        Hence it is not meager, since any meager set is contained
-        in an $F_\sigma$ meager set.
-    \item For every $x \in \R$ we have that $A_x$ contains at most
-        two points, hence it is meager.
-        In particular $\{x \in \R | A_x \text{ is meager}\} = \R$
-        is comeager.
-        However $A$ is not meager.
-        Hence $A$ can not be a set with the Baire property
-        by the theorem.
-        In particular, the assumption of the set having
-        the BP is necessary.
 
+    \item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$.
+        Claim: $Z \in  \Pi^0_3(2^{\N})$.
+        It is
+        \[
+        Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega}
+        \bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}.
+        \]
+        Clearly this is a  $\Pi^0_3$-set.
 \end{enumerate}
 
+\subsection{Exercise 2}
+
+\begin{fact}
+    Let $(X,\tau)$ be a Polish space and
+    $A \in \cB(X)$.
+    Then there exists $\tau' \supseteq \tau$
+    with the same Borel sets as  $\tau$
+    such that $A$ is clopen.
+
+    (Do it for $A$ closed,
+    then show that the sets which work
+    form a $\sigma$-algebra).
+\end{fact}
+
+\begin{enumerate}[(a)]
+    \item Let $(X, \tau)$ be Polish.
+        We want to expand $\tau$ to a Polish topology
+        $\tau_0$ maintaining the Borel sets,
+        such that $(X, \tau')$ is 0d.
+
+        Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$.
+        Each  $U_n$ is open, hence Borel,
+        so by a theorem from the lecture$^{\text{tm}}$
+        there exists a Polish topology $\tau_n$
+        such that $U_n$ is clopen, preserving Borel sets.
+
+
+        Hence we get $\tau_\infty$
+        such that all the $V_n$ are clopen in $\tau_\infty$.
+        Let $\tau^{1} \coloneqq \tau_\infty$.
+        Do this $\omega$-many times to get $\tau^{\omega}$.
+        $\tau^{\omega}$ has a base consisting
+        of finite intersections $A_1 \cap \ldots \cap A_n$,
+        where $A_i$ is a basis element we chose
+        to construct $\tau_i$,
+        hence clopen.
+    \item Let $(X, \tau_X), Y$ be Polish
+        and $f\colon  X \to Y$ Borel.
+        Show $\exists \tau' \supseteq \tau$ maintaining the Borel structure
+        with $f$ continuous.
+
+        Let $(U_n)_n$ be a countable base of $Y$.
+        Clopenize all the preimages of the $(U_n)_n$.
+
+    \item Let $f\colon  X \to Y$ be a Borel isomorphism.
+        Then there are finer topologies preserving the Borel
+        structure
+        such that $f\colon  X' \to Y'$ is a homeomorphism.
+
+        Repeatedly apply (c).
+        Get $\tau_X^1$ to make $f$ continuous.
+        Then get $\tau_Y^1$ to make $f^{-1}$ continuous
+        (possibly violating continuity of $f$)
+        and so on.
+
+        Let $\tau_X^\omega \coloneqq  \langle \tau_X^n \rangle$
+        and similarly for $\tau_Y^\omega$.
+\end{enumerate}
+\begin{idea}
+    If you do something and it didn't work,
+    try doing it again ($\omega$-many times).
+\end{idea}
+
+\subsection{Exercise 3}
+
+\begin{enumerate}[(a)]
+    \item Show that if  $\Gamma$ is self-dual (closed under complements)
+        and closed under continuous preimages,
+        then for any topological space $X$,
+        there does not exist an $X$-universal set for $\Gamma(X)$.
+
+
+        Suppose there is an $X$-universal set for $\Gamma(X)$,
+        i.e.~$U \subseteq  X \times X$
+        such that  $U \in  \Gamma(X \times X) \land \{U_x :  \in X\}  = \Gamma(X)$.
+
+        Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$.
+
+        Let $V = U^c$.
+        Then $V \in \Gamma(X \times  X)$ and $d^{-1}(V) \in \Gamma(X)$.
+        Then $d^{-1}(V) = U_x$ for some $x$.
+        But then $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$.
+
+
+
+    \item Let $\xi$ be an ordinal
+        and let $X$ be a topological space.
+        Show that neither $\cB(X)$ nor $\Delta^0_\xi(X)$ can have $X$-universal
+        sets.
+
+        Clearly $\cB(X)$ is  self-dual and closed under continuous preimages.
+        Clearly $\Delta^0_\xi(X)$ is self-dual
+        and closed under continuous preimages (by a trivial induction).
+\end{enumerate}
+
+\subsection{Exercise 4}
+
+Recall:
+\begin{fact}[Sheet 5, Exercise 1]
+Let $\emptyset\neq X$ be a Baire space.
+Then $\forall  A \subseteq  X$,
+$A$ is either meager or locally comeager.
+\end{fact}
+
+\begin{theorem}[Kechris 16.1]
+    Let $X, Y$ be Polish.
+
+    Let
+    \[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~
+    A_U \coloneqq  \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\]
+
+    Then $\cA$ contains all Borel sets.
+\end{theorem}
+\begin{proof}
+    \begin{enumerate}[(i)]
+        \item Show for $V \in \cB(X), W \overset{\text{open}}{\subseteq}  Y$
+            that $V \times W \in \cA$.
+
+            Clearly $V \times W$ is Borel
+            and $\{x \in X: W \cap U \text{ is not meager}\} \in  \{\emptyset, V\}$.
+        \item Let $(A_n)_{n < \omega} \in \cA^{\omega}$.
+            Then $\bigcap_n A_n \in \cA$.
+            ($(\bigcup_n A_n)_U = \bigcup_n (A_{n})_U$).
+        \item Let $A \in \cA$ and $B = A^c$.
+            Fix $\emptyset\neq U \subseteq Y$.
+            Then $\{x : A_x \text{is not meager in $U$}\}$ is Borel,
+            i.e.~$\{x : A_x^c \text{ is not meager in $U$}\}$ is Borel.
+
+            Since $A$ is Borel, $A_x$ is Borel as well.
+            Hence by the fact:
+            \begin{IEEEeqnarray*}{rCl}
+                && \{x : A_x^c \text{ is not meager in $U$}\}\\
+                &=& \{x \colon A_x^c \text{ is locally comeager in $U$}\}\\
+                &=& \{x \colon \exists \emptyset\neq V \overset{\text{open}}{\subseteq} V.~ A_x \text{ is meager in $V$}\}\\
+                &=& \bigcup_{\emptyset \neq  V \overset{\text{open}}{\subseteq} U} A_V^c
+            \end{IEEEeqnarray*}
+            (a countable union suffices, since we only need to check this for $V$ of the basis; if  $A \subseteq V$ is nwd, then $A \cap U \subseteq U$ is nwd for all $U \overset{\text{open}}{\subseteq} V$).
+    \end{enumerate}
+\end{proof}
+
+
+
+
+
+

inputs/tutorial_07.tex

@@ -1,174 +1,93 @@
-\subsection{Sheet 6}
+\tutorial{07}{2023-12-05}{}
-\tutorial{07}{2023-11-28}{}
+% 17 / 20
 
-% 5 / 20
+\subsection{Exercise 2}
 
-\nr 1
+Recall \autoref{thm:analytic}.
-\begin{warning}
-    Note that not every set has a density!
-\end{warning}
 
-\begin{enumerate}[(a)]
+Let $(A_i)_{i < \omega}$ be analytic subsets of a Polish space $X$.
-    \item Let $X = \bI^{\omega}$.
-        Let $C_0 = \{(x_n) : x_n \to 0\}$.
-        Claim: $C_0 \in \Pi^0_3(X)$ (intersections of $F_\sigma$ sets).
 
-        We have
+$\bigcap_i A_i$ is $\Sigma^1_1$:
-        \[
-        x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q,
-        \]
-        i.e.
-        \[
-        C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}.
-        \]
-        Clearly this is a $\Pi^0_3$ set.
 
-    \item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$.
+% Let $Y_i$ be Polish such that $f_i(Y_i) = A_i$.
-        Claim: $Z \in  \Pi^0_3(2^{\N})$.
+% Let $Y \coloneqq \coprod Y_i$, $f = \coprod f_i$ and $Z = \prod Y_i$.
-        It is
+% Note that $Y$ and $Z$ are Polish.
-        \[
+% We can embed $Z$ into $Y^{\N}$.
-        Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega}
+% 
-        \bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}.
+% Define a tree $T$ on $Y$ as follows:
-        \]
+% $(y_0, \ldots, y_n) \in T$ iff
-        Clearly this is a  $\Pi^0_3$-set.
+% \begin{itemize}
-\end{enumerate}
+%     \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:
 
-\nr 2
+Let $Z = \prod Y_i$
-\begin{fact}
+and let $D \subseteq  Z$
-    Let $(X,\tau)$ be a Polish space and
+be defined by
-    $A \in \cB(X)$.
+\[
-    Then there exists $\tau' \supseteq \tau$
+D \coloneqq \{(y_n) : f_i(y_i) = f_j(y_j) ~ \forall i,j\}.
-    with the same Borel sets as  $\tau$
+\]
-    such that $A$ is clopen.
+$D$ 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 $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
 
-    (Do it for $A$ closed,
+\paragraph{Other solution}
-    then show that the sets which work
+Let $F_n \subseteq X \times  \cN$ be closed,
-    form a $\sigma$-algebra).
+and $C \subseteq  X \times \cN^{\N}$ defined by
-\end{fact}
+\[
+C \coloneqq  \{(x,(y^{(n)}) ) : \forall n.~(x, y^{(n)}) \in  F_n\}.
+\]
+$C$ is closed
+and $\bigcap A_i = \proj_X(C)$.
 
-\begin{enumerate}[(a)]
+\subsection{Exercise 3}
-    \item Let $(X, \tau)$ be Polish.
-        We want to expand $\tau$ to a Polish topology
-        $\tau_0$ maintaining the Borel sets,
-        such that $(X, \tau')$ is 0d.
 
-        Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$.
+\begin{itemize}
-        Each  $U_n$ is open, hence Borel,
+    \item Make $X$ zero dimensional preserving the Borel structure.
-        so by a theorem from the lecture$^{\text{tm}}$
+    \item \todo{Find a countable clopen base}
-        there exists a Polish topology $\tau_n$
+    \item 
-        such that $U_n$ is clopen, preserving Borel sets.
+\end{itemize}
 
 
-        Hence we get $\tau_\infty$
+\subsection{Exercise 4}
-        such that all the $V_n$ are clopen in $\tau_\infty$.
-        Let $\tau^{1} \coloneqq \tau_\infty$.
-        Do this $\omega$-many times to get $\tau^{\omega}$.
-        $\tau^{\omega}$ has a base consisting
-        of finite intersections $A_1 \cap \ldots \cap A_n$,
-        where $A_i$ is a basis element we chose
-        to construct $\tau_i$,
-        hence clopen.
-    \item Let $(X, \tau_X), Y$ be Polish
-        and $f\colon  X \to Y$ Borel.
-        Show $\exists \tau' \supseteq \tau$ maintaining the Borel structure
-        with $f$ continuous.
-
-        Let $(U_n)_n$ be a countable base of $Y$.
-        Clopenize all the preimages of the $(U_n)_n$.
-
-    \item Let $f\colon  X \to Y$ be a Borel isomorphism.
-        Then there are finer topologies preserving the Borel
-        structure
-        such that $f\colon  X' \to Y'$ is a homeomorphism.
-
-        Repeatedly apply (c).
-        Get $\tau_X^1$ to make $f$ continuous.
-        Then get $\tau_Y^1$ to make $f^{-1}$ continuous
-        (possibly violating continuity of $f$)
-        and so on.
-
-        Let $\tau_X^\omega \coloneqq  \langle \tau_X^n \rangle$
-        and similarly for $\tau_Y^\omega$.
-\end{enumerate}
-\begin{idea}
-    If you do something and it didn't work,
-    try doing it again ($\omega$-many times).
-\end{idea}
-
-\nr 3
-\begin{enumerate}[(a)]
-    \item Show that if  $\Gamma$ is self-dual (closed under complements)
-        and closed under continuous preimages,
-        then for any topological space $X$,
-        there does not exist an $X$-universal set for $\Gamma(X)$.
 
 
-        Suppose there is an $X$-universal set for $\Gamma(X)$,
+Proof of Schröder-Bernstein:
-        i.e.~$U \subseteq  X \times X$
-        such that  $U \in  \Gamma(X \times X) \land \{U_x :  \in X\}  = \Gamma(X)$.
 
-        Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$.
+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 $Y_\omega \coloneqq \bigcap Y_i$.
 
-        Let $V = U^c$.
+% https://q.uiver.app/#q=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
-        Then $V \in \Gamma(X \times  X)$ and $d^{-1}(V) \in \Gamma(X)$.
+\[\begin{tikzcd}
-        Then $d^{-1}(V) = U_x$ for some $x$.
+	{X \setminus X_\omega} & {=} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
-        But then $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$.
+	{Y\setminus Y_\omega} & {=} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
+	\arrow["f"'{pos=0.7}, from=1-3, to=2-5]
+	\arrow["g"{pos=0.1}, from=2-3, to=1-5]
+	\arrow["f"{pos=0.8}, from=1-7, to=2-9]
+	\arrow["g"{pos=0.1}, from=2-7, to=1-9]
+\end{tikzcd}\]
+
+
+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.
 
 
 
-    \item Let $\xi$ be an ordinal
-        and let $X$ be a topological space.
-        Show that neither $\cB(X)$ nor $\Delta^0_\xi(X)$ can have $X$-universal
-        sets.
 
-        Clearly $\cB(X)$ is  self-dual and closed under continuous preimages.
-        Clearly $\Delta^0_\xi(X)$ is self-dual
-        and closed under continuous preimages (by a trivial induction).
-\end{enumerate}
-
-\nr 4
-Recall:
-\begin{fact}[Sheet 5, Exercise 1]
-Let $\emptyset\neq X$ be a Baire space.
-Then $\forall  A \subseteq  X$,
-$A$ is either meager or locally comeager.
-\end{fact}
-
-\begin{theorem}\footnote{See Kechris 16.1}
-    Let $X, Y$ be Polish.
-
-    For $\emptyset \neq U \overset{\text{open}}{\subseteq} Y$
-    let
-    \[A_U \coloneqq  \{ x \in X : A_x \text{ is not meager in $U$}\}.\]
-
-    Define
-    \[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ A_U \text{ is Borel}\}.\]
-
-    Then $\cA$ contains all Borel sets.
-\end{theorem}
-\begin{proof}
-    \begin{enumerate}[(i)]
-        \item Show for $V \in \cB(X), W \overset{\text{open}}{\subseteq}  Y$
-            that $V \times W \in \cA$.
-
-            Clearly $V \times W$ is Borel
-            and $\{x \in X: W \cap U \text{ is not meager}\} \in  \{\emptyset, V\}$.
-        \item Let $(A_n)_{n < \omega} \in \cA^{\omega}$.
-            Then $\bigcap_n A_n \in \cA$.
-            ($(\bigcup_n A_n)_U = \bigcup_n (A_{n})_U$).
-        \item Let $A \in \cA$ and $B = A^c$.
-            Fix $\emptyset\neq U \subseteq Y$.
-            Then $\{x : A_x \text{is not meager in $U$}\}$ is Borel,
-            i.e.~$\{x : A_x^c \text{ is not meager in $U$}\}$ is Borel.
-
-            Since $A$ is Borel, $A_x$ is Borel as well.
-            Hence by the fact:
-            \begin{IEEEeqnarray*}{rCl}
-                && \{x : A_x^c \text{ is not meager in $U$}\}\\
-                &=& \{x \colon A_x^c \text{ is locally comeager in $U$}\}\\
-                &=& \{x \colon \exists \emptyset\neq V \overset{\text{open}}{\subseteq} V.~ A_x \text{ is meager in $V$}\}\\
-                &=& \bigcup_{\emptyset \neq  V \overset{\text{open}}{\subseteq} U} A_V^c
-            \end{IEEEeqnarray*}
-            (a countable union suffices, since we only need to check this for $V$ of the basis; if  $A \subseteq V$ is nwd, then $A \cap U \subseteq U$ is nwd for all $U \overset{\text{open}}{\subseteq} V$).
-    \end{enumerate}
-\end{proof}

inputs/tutorial_08.tex

@@ -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
-\yalabel{Exercise}{}{ex:7.2}
-
-
-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 analytic:
-        % 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 $Y_\omega \coloneqq \bigcap Y_i$.
-
-
-% https://q.uiver.app/#q=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
-\adjustbox{scale=0.7,center}{%
-\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.

logic.sty

@@ -22,7 +22,6 @@
 \usepackage{listings}
 \usepackage{multirow}
 \usepackage{float}
-\usepackage{adjustbox}
 \usepackage{quiver}
 %\usepackage{algorithmicx}
 
@@ -133,9 +132,6 @@
 \DeclareSimpleMathOperator{LO} % linear orders
 \DeclareSimpleMathOperator{WO} % well orderings
 
-
-\DeclareSimpleMathOperator{osc} % oscillation
-
 \newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}
 
 %https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly
@@ -147,5 +143,3 @@
 \newcommand{\fc}{\mathfrak{c}}
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
-\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
-\newcommand\nr[1]{\subsubsection{Exercise #1}}

logic3.tex

@@ -15,7 +15,6 @@
 
 \cleardoublepage
 
-\setcounter{tocdepth}{2}
 \tableofcontents
 \cleardoublepage
 
@@ -39,9 +38,6 @@
 \input{inputs/lecture_12}
 \input{inputs/lecture_13}
 \input{inputs/lecture_14}
-\input{inputs/lecture_15}
-
-
 
 
 
@@ -49,17 +45,6 @@
 
 \appendix
 
-\section{Tutorial and Exercises}
-\input{inputs/tutorial_01}
-\input{inputs/tutorial_02}
-\input{inputs/tutorial_03}
-\input{inputs/tutorial_04}
-\input{inputs/tutorial_05}
-\input{inputs/tutorial_06}
-\input{inputs/tutorial_07}
-\input{inputs/tutorial_08}
-
-
 \PrintVocabIndex