Merge branch 'main' of https://git.abstractnonsen.se/josia-notes/w23-logic-3

inputs/intro.tex

@@ -1,4 +1,6 @@
-These are my notes on the lecture Probability Theory,
+\gist{%
+These are my notes on the lecture
+Logic 3: Abstract Topological Dynamics and Descriptive Set Theory
 taught by \textsc{Aleksandra Kwiatkowska}
 in the summer term 2023 at the University Münster.
 
@@ -18,3 +20,15 @@ I could not attend!
 This notes follow the way the material was presented in the lecture rather
 closely. Additions (e.g.~from exercise sheets)
 and slight modifications have been marked with $\dagger$.
+}{
+    This document aims to give a very brief summary of
+    my \href{https://josia-notes.users.abstractnonsen.se/w23-logic-3/logic3.pdf}{notes on the course Logic 3}.
+    I try to omit most technical details and only summarize the most important
+    ideas.
+
+    Note that this is currently work in progress.
+    Currently the differences to the original document are only
+    minor (this is still mostly a technical test),
+    but this document will get shorter as I work through
+    and summarize it.
+}

inputs/lecture_01.tex

@@ -76,7 +76,7 @@ However the converse of this does not hold.
     \end{itemize}
 \end{fact}
 \begin{fact}
-    Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4)
+    Compact Hausdorff spaces are \vocab{normal} (T4)
     i.e.~two disjoint closed subsets can be separated
     by open sets.
 \end{fact}
@@ -114,7 +114,7 @@ However the converse of this does not hold.
 \end{absolutelynopagebreak}
 
 \subsection{Some facts about polish spaces}
-
+\gist{%
 \begin{fact}
 Let $(X, \tau)$ be a topological space.
 Let $d$ be a metric on $X$.
@@ -130,9 +130,10 @@ To show that $\tau_d = \tau_{d'}$
 for two metrics  $d, d'$,
 suffices to show that open balls in one metric are unions of open balls in the other.
 \end{fact}
+}{}
 
 \begin{notation}
-    We sometimes denote $\min(a,b)$ by $a \wedge b$.
+    We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$.
 \end{notation}
 
 \begin{proposition}
@@ -142,6 +143,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
 
         Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
 \end{proposition}
+\gist{%
 \begin{proof}
     To check the triangle inequality:
     \begin{IEEEeqnarray*}{rCl}
@@ -154,6 +156,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
 
     Since $d$ is complete, we have that $d'$ is complete.
 \end{proof}
+}{}
 \begin{proposition}
     Let $A$ be a Polish space.
     Then $A^{\omega}$ Polish.
@@ -252,15 +255,15 @@ suffices to show that open balls in one metric are unions of open balls in the o
 \begin{proposition}
     Closed subspaces of Polish spaces are Polish.
 \end{proposition}
-\gist{}{
+\gist{%
 \begin{proof}
     Let $X$ be Polish and $V \subseteq  X$ closed.
     Let $d$ be a complete metric on $X$.
     Then $d\defon{V}$ is complete.
     Subspaces of second countable spaces
     are second countable.
-\end{proof}
+\end{proof}%
-}
+}{}
 
 \begin{definition}
     Let $X$ be a topological space.

inputs/lecture_02.tex

@@ -72,41 +72,51 @@
         \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
         metric is complete.
 
-        $f_U$ is an embedding of $U$ into $X \times \R$\gist{:
+        $f_U$ is an embedding of $U$ into $X \times \R$%
-        \begin{itemize}
+        \gist{:
-            \item It is injective because of the first coordinate.
+            \begin{itemize}
-            \item It is continuous since $d(x, U^c)$ is continuous
+                \item It is injective because of the first coordinate.
-                and only takes strictly positive values. % TODO
+                \item It is continuous since $d(x, U^c)$ is continuous
-            \item The inverse is continuous because projections
+                    and only takes strictly positive values. % TODO
-                are continuous.
+                \item The inverse is continuous because projections
-        \end{itemize}
+                    are continuous.
-    }{.}
+            \end{itemize}
+        }{.}
 
-        So we have shown that $U$ and
+        \gist{%
-        the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
+            So we have shown that $U$ and
-        are homeomorphic.
+            the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
-        The graph is closed \gist{in $U \times  \R$,
+            are homeomorphic.
-        because $\tilde{f_U}$ is continuous.
+            The graph is closed \gist{in $U \times  \R$,
-        It is closed}{} in $X \times \R$ \gist{because
+            because $\tilde{f_U}$ is continuous.
-        $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
+            It is closed}{} in $X \times \R$ \gist{because
-        \todo{Make this precise}
+            $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
+            \todo{Make this precise}
+
+            Therefore we identified $U$ with a closed subspace of
+            the Polish space $(X \times  \R, d_1)$.
+        }{%
+            So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$
+            and the RHS is a close subspace of the Polish space
+            $(X \times \R, d_1)$.
+        }
 
-        Therefore we identified $U$ with a closed subspace of
-        the Polish space $(X \times  \R, d_1)$.
     \end{refproof}
 
     Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
-    Take
+    Consider
     \begin{IEEEeqnarray*}{rCl}
         f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
              x &\longmapsto &
                  \left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
     \end{IEEEeqnarray*}
 
-    As for an open $U$, $f_Y$ is an embedding.
+    \gist{
-    Since $X \times \R^{\N}$
+        As for an open $U$, $f_Y$ is an embedding.
-    is completely metrizable,
+        Since $X \times \R^{\N}$
-    so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
+        is completely metrizable,
+        so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
+    }{}
 
     \begin{claim}
         \label{psubspacegdelta:c2}
@@ -123,36 +133,35 @@
             \item $\diam_d(U) \le \frac{1}{n}$,
             \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
         \end{enumerate}
-        \gist{
+        \gist{%
-        We want to show that $Y = \bigcap_{n \in \N} V_n$.
+            We want to show that $Y = \bigcap_{n \in \N} V_n$.
-        For $x \in Y$, $n \in \N$ we have $x \in V_n$,
+            For $x \in Y$, $n \in \N$ we have $x \in V_n$,
-        as we can choose two neighbourhoods
+            as we can choose two neighbourhoods
-        $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
+            $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
-        such that $\diam_{d_Y}(U) < \frac{1}{n}$
+            such that $\diam_{d_Y}(U) < \frac{1}{n}$
-        and $U_2 \cap Y = U_1$.
+            and $U_2 \cap Y = U_1$.
-        Additionally choose $x \in U_3$ open in $X$
+            Additionally choose $x \in U_3$ open in $X$
-        with $\diam_{d}(U_3) < \frac{1}{n}$.
+            with $\diam_{d}(U_3) < \frac{1}{n}$.
-        Then consider $U_2 \cap U_3 \subseteq V_n$.
+            Then consider $U_2 \cap U_3 \subseteq V_n$.
-        Hence $Y \subseteq  \bigcap_{n \in \N}  V_n$.
+            Hence $Y \subseteq  \bigcap_{n \in \N}  V_n$.
 
-        Now let $x \in \bigcap_{n \in \N} V_n$.
+            Now let $x \in \bigcap_{n \in \N} V_n$.
-        For each $n$ pick $x \in U_n \subseteq X$ open
+            For each $n$ pick $x \in U_n \subseteq X$ open
-        satisfying (i), (ii), (iii).
+            satisfying (i), (ii), (iii).
-        From (i) and (ii) it follows that $x \in \overline{Y}$,
+            From (i) and (ii) it follows that $x \in \overline{Y}$,
-        since we can consider a sequence of points $y_n \in U_n \cap Y$
+            since we can consider a sequence of points $y_n \in U_n \cap Y$
-        and get $y_n \xrightarrow{d} x$.
+            and get $y_n \xrightarrow{d} x$.
-        For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
+            For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
-        is an open set containing $x$,
+            is an open set containing $x$,
-        hence $U_n' \cap Y \neq \emptyset$.
+            hence $U_n' \cap Y \neq \emptyset$.
-        Thus we may assume that the $U_i$ form a decreasing sequence.
+            Thus we may assume that the $U_i$ form a decreasing sequence.
-        We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
+            We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
-        If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
+            If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
-        since $\diam(U_n \cap Y) \xrightarrow{d_Y}  0$
+            since $\diam(U_n \cap Y) \xrightarrow{d_Y}  0$
-        and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y}  0$.
+            and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y}  0$.
-        The sequence $y_n$ converges to the unique point in
+            The sequence $y_n$ converges to the unique point in
-        $\bigcap_{n} \overline{U_n \cap Y}$.
+            $\bigcap_{n} \overline{U_n \cap Y}$.
-        Since the topologies agree, this point is $x$.
+            Since the topologies agree, this point is $x$.
-    }{Then $Y = \bigcap_n U_n$.}
+        }{Then $Y = \bigcap_n U_n$.}
     \end{refproof}
 \end{refproof}
-

inputs/lecture_03.tex

@@ -2,7 +2,7 @@
 
 \subsection{Trees}
 
-
+\gist{%
 \begin{notation}
     Let $A \neq  \emptyset$, $n \in \N$.
     Then
@@ -59,6 +59,7 @@
     define extension, initial segments
     and concatenation of a finite sequence with an infinite one.
 \end{notation}
+}{}
 
 \begin{definition}
     A \vocab{tree}
@@ -127,16 +128,19 @@
 
     We define $U_s$ inductively on the length of $s$.
 
-    For $U_{\emptyset}$ take any non-empty open set
+    \gist{%
-    with small enough diameter.
+        For $U_{\emptyset}$ take any non-empty open set
+        with small enough diameter.
 
-    Given $U_s$, pick $x \neq y \in U_s$
+        Given $U_s$, pick $x \neq y \in U_s$
-    and let $U_{s \concat 0} \ni x$,
+        and let $U_{s \concat 0} \ni x$,
-    $U_{s \concat 1} \ni y$ 
+        $U_{s \concat 1} \ni y$
-    be disjoint, open,
+        be disjoint, open,
-    of diameter $\le \frac{1}{2^{|s| +1}}$
+        of diameter $\le \frac{1}{2^{|s| +1}}$
-    and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq  U_s$.
+        and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq  U_s$.
+    }{}
 
+    \gist{%
     Let $x \in  2^{\N}$.
     Then let $f(x)$ be the unique point in $X$
     such that
@@ -147,19 +151,23 @@
     It is clear that $f$ is injective and continuous.
     % TODO: more details
     $2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
+    }{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\}  = \bigcap_n U_{x\defon n}$.
+    By compactness of $2^{\N}$, we get that  $f^{-1}$ is continuous.}
 \end{proof}
 
 \begin{corollary}
     \label{cor:perfectpolishcard}
     Every nonempty perfect Polish
     space $X$ has cardinality $\fc = 2^{\aleph_0}$
-    % TODO: eulerscript C ?
 \end{corollary}
 \begin{proof}
-    Since the cantor space embeds into $X$,
+    \gist{%
-    we get the lower bound.
+        Since the cantor space embeds into $X$,
-    Since $X$ is second countable and Hausdorff,
+        we get the lower bound.
-    we get the upper bound.
+        Since $X$ is second countable and Hausdorff,
+        we get the upper bound.%
+    }{Lower bound:  $2^{\N} \hookrightarrow X$,
+        upper bound: \nth{2} countable and Hausdorff.}
 \end{proof}
 
 \begin{theorem}
@@ -203,12 +211,14 @@
     countable union of closed sets,
     i.e.~the complement of a $G_\delta$ set.
 \end{definition}
+\gist{%
 \begin{observe}
     \begin{itemize}
         \item Any open set is $F {\sigma}$.
         \item In metric spaces the intersection of an open and closed set is $F_\sigma$.
     \end{itemize}
 \end{observe}
+}{}
 \begin{refproof}{thm:bairetopolish}
     Let $d$ be a complete metric on $X$.
     W.l.o.g.~$\diam(X) \le 1$.
@@ -220,7 +230,7 @@
         \item $F_\emptyset = X$,
         \item $F_s$ is $F_\sigma$ for all  $s$.
         \item The $F_{s \concat i}$ partition $F_s$,
-            i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. % TODO change notation?
+            i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.
 
             Furthermore we want that
             $\overline{F_{s \concat i}} \subseteq F_s$
@@ -228,6 +238,7 @@
         \item $\diam(F_s) \le 2^{-|s|}$.
     \end{enumerate}
 
+    \gist{%
     Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
     We need to construct a partition $(F_i)_{i \in \N}$
     of $F$ with $\overline{F_i} \subseteq F$
@@ -252,6 +263,7 @@
     The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
     are $F_\sigma$, disjoint
     and $F_i^0 = \bigcup_{j} D_j$.
+}{Induction.}
 
 
 

inputs/lecture_04.tex

@@ -5,6 +5,7 @@
 \end{remark}
 
 \begin{refproof}{thm:bairetopolish}
+    \gist{%
     Take
     \[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
 
@@ -13,15 +14,18 @@
     \[
     \bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
     \]
+    }{}
 
     $f\colon D \to  X$ is determined by
     \[
     \{f(x)\} = \bigcap_{n} F_{x\defon{n}}
     \]
 
-    $f$ is injective and continuous.
+    \gist{%
-    The proof of this is exactly the same as in 
+        $f$ is injective and continuous.
-    \yaref{thm:cantortopolish}.
+        The proof of this is exactly the same as in
+        \yaref{thm:cantortopolish}.
+    }{}
 
     \begin{claim}
         \label{thm:bairetopolish:c1}
@@ -60,7 +64,7 @@
 
     Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
     Clearly $S$ is a pruned tree.
-    Moreover, since $D$ is closed, we have that\todo{Proof this (homework?)}
+    Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})
     \[
     D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
     \]
@@ -76,21 +80,27 @@
         \item $|s| = \phi(|s|)$,
         \item if $s \in S$, then $\phi(s) = s$.
     \end{itemize}
-    Let $\phi(\emptyset) = \emptyset$.
+    \gist{%
-    Suppose that $\phi(t)$ is defined.
+        Let $\phi(\emptyset) = \emptyset$.
-    If $t\concat a \in S$, then set
+        Suppose that $\phi(t)$ is defined.
-     $\phi(t\concat a) \coloneqq t\concat a$.
+        If $t\concat a \in S$, then set
-    Otherwise take some $b$ such that
+         $\phi(t\concat a) \coloneqq t\concat a$.
-    $t\concat b \in S$ and define
+        Otherwise take some $b$ such that
-    $\phi(t\concat a) \coloneqq \phi(t)\concat b$.
+        $t\concat b \in S$ and define
+        $\phi(t\concat a) \coloneqq \phi(t)\concat b$.%
+    }{}%
     This is possible since $S$ is pruned.
 
-    Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
+    \gist{%
-    be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
+        Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
+        be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
+    }{}
 
     $r$ is continuous, since
     $d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
-    It is immediate that $r$ is a retraction.
+    \gist{%
+        It is immediate that $r$ is a retraction.
+    }{}
 \end{refproof}
 
 \section{Meager and Comeager Sets}
@@ -117,9 +127,11 @@
     The complement of a meager set is called
     \vocab{comeager}.
 \end{definition}
+\gist{%
 \begin{example}
     $\Q \subseteq \R$ is meager.
 \end{example}
+}{}
 \begin{notation}
     Let $A, B \subseteq  X$.
     We write $A =^\ast B$
@@ -127,25 +139,29 @@
     $A \symdif B \coloneqq  (A\setminus B) \cup (B \setminus A)$,
     is meager.
 \end{notation}
+\gist{%
 \begin{remark}
     $=^\ast$ is an equivalence relation.
 \end{remark}
+}{}
 \begin{definition}
     A set $A \subseteq  X$
     has the \vocab{Baire property} (\vocab{BP})
     if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
 \end{definition}
+\gist{%
 Note that open sets and meager sets have the Baire property.
+}{}
 
-
+\gist{%
 \begin{example}
     \begin{itemize}
         \item $\Q \subseteq  \R$ is $F_\sigma$.
         \item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
-        \item $\Q \subseteq \R$ is not $G_{\delta}$.
+        \item $\Q \subseteq \R$ is not $G_{\delta}$:
-            (It is dense and meager,
+            It is dense and meager,
             hence it can not be $G_\delta$
-            by the Baire category theorem).
+            by the \yaref{thm:bct}.
     \end{itemize}
-
 \end{example}
+}{}

inputs/lecture_05.tex

@@ -6,10 +6,9 @@
         \item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense.
         \item Any meager set $B$ is contained in a meager $F_{\sigma}$-set.
     \end{itemize}
-
-
 \end{fact}
-\begin{proof} % remove?
+\gist{%
+\begin{proof}
     \begin{itemize}
         \item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
         \item Trivial.
@@ -17,7 +16,9 @@
             Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
     \end{itemize}
 \end{proof}
+}{}
 
+\gist{%
 \begin{definition}
     A  \vocab{$\sigma$-algebra} on a set $X$
     is a collection of subsets of $X$
@@ -32,14 +33,15 @@
     Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
     we have that $\sigma$-algebras are closed under countable intersections.
 \end{fact}
+}{}
 
 \begin{theorem}
     \label{thm:bairesigma}
     Let $X$ be a topological space.
     Then the collection of sets with the Baire property
-    is a $\sigma$-algebra on $X$.
+    is \gist{a $\sigma$-algebra on $X$.
 
-    It is the smallest $\sigma$-algebra
+    It is}{} the smallest $\sigma$-algebra
     containing all meager and open sets.
 \end{theorem}
 \begin{refproof}{thm:bairesigma}
@@ -97,6 +99,7 @@
 %\end{example}
 
 \begin{theorem}[Baire Category theorem]
+    \yalabel{Baire Category Theorem}{Baire Category Theorem}{thm:bct}
     Let $X$ be a completely metrizable space.
     Then every comeager set of $X$ is dense in $X$.
 \end{theorem}
@@ -111,8 +114,8 @@
         \item The intersection of countable many
             open dense sets is dense.
     \end{enumerate}
-    In this case $X$ is called a \vocab{Baire space}.
+    In this case $X$ is called a \vocab{Baire space}.%
-    \footnote{see \yaref{s5e1}}
+    \footnote{cf.~\yaref{s5e1}}
 \end{theoremdef}
 \begin{proof}
     \todo{Proof (short)}
@@ -129,7 +132,6 @@
     This proof can be adapted to other open sets $X$.
 \end{proof}
 
-
 \begin{notation}
     Let $X ,Y$ be topological spaces,
     $A \subseteq  X \times Y$
@@ -273,9 +275,11 @@ but for meager sets:
 % \end{refproof}
 % TODO fix claim numbers
 
+\gist{%
 \begin{remark}
     Suppose that $A$ has the BP.
     Then there is an open $U$ such that
     $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
     Then $A = U \symdif M$.
 \end{remark}
+}{}

inputs/lecture_06.tex

@@ -101,7 +101,6 @@ Then define
 \Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq
 \{X \setminus A | A \in \Sigma^0_\alpha(X)\},
 \]
-% \todo{Define $\lnot$ (element-wise complement)}
 and for $\alpha > 1$
 \[
 \Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.

inputs/lecture_07.tex

@@ -37,17 +37,18 @@
         \item \begin{itemize}
                 \item $\Sigma^0_\xi(X)$ is closed under countable unions.
                 \item  $\Pi^0_\xi(X)$ is closed under countable intersections.
-                \item  $\Delta^0_\xi(X)$ is closed under complements,
+                \item  $\Delta^0_\xi(X)$ is closed under complements.
-                    countable unions and
-                    countable intersections.
             \end{itemize}
         \item \begin{itemize}
                 \item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
                 \item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
+                \item $\Delta^0_\xi(X)$ is closed under finite unions and
+                    finite intersections.
             \end{itemize}
 
     \end{enumerate}
 \end{proposition}
+\gist{%
 \begin{proof}
     \begin{enumerate}[(a)]
         \item This follows directly from the definition.
@@ -67,24 +68,27 @@
 
     \end{enumerate}
 \end{proof}
+}{}
 \begin{example}
     Consider the cantor space $2^{\omega}$.
     We have that $\Delta^0_1(2^{\omega})$
-    is not closed under countable unions
+    is not closed under countable unions%
-    (countable unions yield all open sets, but there are open
+    \gist{ (countable unions yield all open sets, but there are open
-    sets that are not clopen).
+    sets that are not clopen)}{}.
 \end{example}
 
 \subsection{Turning Borels Sets into Clopens}
 
 \begin{theorem}%
+    \gist{%
     \footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
             unfortunately seems to be non-standard vocabulary.
             Our tutor repeatedly advised against using it in the final exam.
             Contrary to popular belief
             the very same tutor was \textit{not} the one first to introduce it,
             as it would certainly be spelled ``to clopenise'' if that were the case.
-        }
+        }%
+    }{}%
     \label{thm:clopenize}
     Let $(X, \cT)$ be a Polish space.
     For any  Borel set $A \subseteq  X$,
@@ -163,7 +167,7 @@
         such that $\cT_n \supseteq \cT$
         and $\cB(\cT_n) = \cB(\cT)$.
         Then the topology $\cT_\infty$ generated by  $\bigcup_{n} \cT_n$
-        is still Polish
+        is Polish
         and $\cB(\cT_\infty) = \cB(T)$.
     \end{lemma}
     \begin{proof}
@@ -183,7 +187,51 @@
         definition of $\cF$ belong to
         a countable basis of the respective $\cT_n$).
 
-        \todo{This proof will be finished in the next lecture}
+        % Proof was finished in lecture 8
+        Let $Y = \prod_{n \in \N} (X, \cT_n)$.
+        Then $Y$ is Polish.
+        Let $\delta\colon (X, \cT_\infty) \to Y$
+        defined by $\delta(x) = (x,x,x,\ldots)$.
+        \begin{claim}
+            $\delta$ is a homeomorphism.
+        \end{claim}
+        \begin{subproof}
+            Clearly $\delta$ is a bijection.
+            We need to show that it is continuous and open.
+
+            Let $U \in \cT_i$.
+            Then
+            \[
+            \delta^{-1}(D \cap \left( X \times  X \times \ldots\times  U \times  \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
+            \]
+            hence $\delta$ is continuous.
+            Let $U \in \cT_\infty$.
+            Then $U$ is the union of sets of the form
+            \[
+            V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
+            \]
+            for some $n_1 < n_2 < \ldots < n_u$
+            and $U_{n_i} \in \cT_i$.
+
+            Thus is suffices to consider sets of this form.
+            We have that
+            \[
+            \delta(V) = D \cap (X \times  X \times  \ldots \times  U_{n_1} \times  \ldots \times  U_{n_2} \times  \ldots \times  U_{n_u} \times  X \times  \ldots) \overset{\text{open}}{\subseteq} D.
+            \]
+        \end{subproof}
+
+        This will finish the proof since
+        \[
+        D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
+        \]
+        Why? Let  $(x_n) \in Y \setminus D$.
+        Then there are $i < j$ such that $x_i \neq x_j$.
+        Take disjoint open $x_i \in U$, $x_j \in V$.
+        Then
+        \[(x_n) \in  X \times  X \times  \ldots \times  U \times  \ldots \times  X \times  \ldots \times V \times  X \times  \ldots\]
+        is open in $Y\setminus D$.
+        Hence $Y \setminus D$ is open, thus $D$ is closed.
+        It follows that $D$ is Polish.
     \end{proof}
 
     We need to show that $A$ is closed under countable unions.

inputs/lecture_08.tex

@@ -1,61 +1,8 @@
-\lecture{08}{2023-11-10}{}
+\lecture{08}{2023-11-10}{}\footnote{%
-
+    In the beginning of the lecture, we finished
-\todo{put this lemma in the right place}
+    the proof of \yaref{thm:clopenize:l2}.
-\begin{lemma}[Lemma 2]
+    This has been moved to the notes on lecture 7.%
-    Let $(X, \cT)$ be a Polish space.
+}
-    Let $\cT_n \supseteq \cT$ be Polish
-    with $\cB(X, \cT_n) = \cB(X, \cT)$.
-    Let  $\cT_\infty$ be the topology generated
-    by $\bigcup_n \cT_n$.
-    Then $(X, \cT_\infty)$ is Polish
-    and $\cB(X, \cT_\infty) = \cB(X, \cT)$.
-\end{lemma}
-\begin{proof}
-    Let $Y = \prod_{n \in \N} (X, \cT_n)$.
-    Then $Y$ is Polish.
-    Let $\delta\colon (X, \cT_\infty) \to Y$
-    defined by $\delta(x) = (x,x,x,\ldots)$.
-    \begin{claim}
-        $\delta$ is a homeomorphism.
-    \end{claim}
-    \begin{subproof}
-        Clearly $\delta$ is a bijection.
-        We need to show that it is continuous and open.
-
-        Let $U \in \cT_i$.
-        Then
-        \[
-        \delta^{-1}(D \cap \left( X \times  X \times \ldots\times  U \times  \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
-        \]
-        hence $\delta$ is continuous.
-        Let $U \in \cT_\infty$.
-        Then $U$ is the union of sets of the form
-        \[
-        V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
-        \]
-        for some $n_1 < n_2 < \ldots < n_u$
-        and $U_{n_i} \in \cT_i$.
-
-        Thus is suffices to consider sets of this form.
-        We have that
-        \[
-        \delta(V) = D \cap (X \times  X \times  \ldots \times  U_{n_1} \times  \ldots \times  U_{n_2} \times  \ldots \times  U_{n_u} \times  X \times  \ldots) \overset{\text{open}}{\subseteq} D.
-        \]
-    \end{subproof}
-
-    This will finish the proof since
-    \[
-    D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
-    \]
-    Why? Let  $(x_n) \in Y \setminus D$.
-    Then there are $i < j$ such that $x_i \neq x_j$.
-    Take disjoint open $x_i \in U$, $x_j \in V$.
-    Then
-    \[(x_n) \in  X \times  X \times  \ldots \times  U \times  \ldots \times  X \times  \ldots \times V \times  X \times  \ldots\]
-    is open in $Y\setminus D$.
-    Hence $Y \setminus D$ is open, thus $D$ is closed.
-    It follows that $D$ is Polish.
-\end{proof}
 
 \subsection{Parametrizations}
 %\todo{choose better title}

inputs/lecture_09.tex

@@ -148,6 +148,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
 
 
 \begin{theorem}
+    \label{thm:universals11}
     Let $X,Y$ be uncountable Polish spaces.
     There exists a $Y$-universal $\Sigma^1_1(X)$ set.
 \end{theorem}

inputs/lecture_11.tex

@@ -153,13 +153,14 @@ We will not proof this in this lecture.
 
 \subsection{Ill-Founded Trees}
 
-
+\gist{%
 Recall that a \vocab{tree} on $\N$ is a subset of
 $\N^{<\N}$
 closed under taking initial segments.
 
 We now identify trees with their characteristic functions,
-i.e.~we want to associate a tree $T \subseteq  \N^{<\N}$
+i.e.~we want to associate a tree $T \subseteq  \N^{<\N}$}%
+{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:}
 \begin{IEEEeqnarray*}{rCl}
     \One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\}  \\
      x &\longmapsto &  \begin{cases}
@@ -167,16 +168,15 @@ i.e.~we want to associate a tree $T \subseteq  \N^{<\N}$
          0 &: x \not\in T.
      \end{cases}
 \end{IEEEeqnarray*}
-Note that $\One_T \in  {\{0,1\}^\N}^{< \N}$.
+\gist{Note that $\One_T \in  {\{0,1\}^\N}^{< \N}$.}{}
 
-Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
+Let \vocab{$\Tr$} $ \coloneqq  \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
 
 \begin{observe}
-    \[
+    $\Tr \subseteq  {2^{\N}}^{<\N}$ is closed
-    \Tr \subseteq  {2^{\N}}^{<\N}
+    (where we take the topology of the Cantor space).
-    \]
-    is closed (where we take the topology of the Cantor space).
 \end{observe}
+\gist{%
 Indeed, for any $ s \in \N^{<\N}$
 we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
 and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
@@ -185,6 +185,4 @@ In particular for $s$ fixed,
 we have that
 \[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
 is clopen in ${2^{\N}}^{<\N}$.
-
+}{}
-
-

inputs/lecture_12.tex

@@ -21,102 +21,135 @@
         T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
     \end{IEEEeqnarray*}
 
-    Consider $D \coloneqq \{(T, \beta) \in \Tr \times  \cN : \forall n.~ T(\beta\defon{n}) = 1\}$.
+    Consider
-    Note that this set is closed in $\Tr \times \cN$,
+    \[D \coloneqq \{(T, \beta) \in \Tr \times  \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
-    since it is a countable intersection of clopen sets.
+    \gist{%
-    % TODO Why clopen?
+        Note that this set is closed in $\Tr \times \cN$,
-    Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
+        since it is a countable intersection of clopen sets.
+        Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
+    }{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.}
 \end{proof}
 
 \begin{definition}
     An analytic set $B$ in some Polish space $Y$
     is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
+    \gist{%
     iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$,
     there exists a Borel function $f\colon X\to Y$
-    such that $x \in A \iff f(x) \in B$.
+    such that $x \in A \iff f(x) \in B$,
+    i.e.~$f^{-1}(B) = A$.
+    }{%
+    iff for any $A \in \Sigma^1_1(X)$, $X$ Polish
+    there exists $f\colon  X \to Y$ Borel such that $f^{-1}(B) = A$.}
 
-    Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}).
+    \gist{%
+    Similarly, a conalytic set $B$ is called
+    \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete})
+    iff for any $A \subseteq \Pi^1_1(X)$
+    there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.
+    }{Similarly we define \vocab{complete coanalytic} / \vocab{$\Pi_1^1$-complete}.}
 \end{definition}
 \begin{observe}
     \leavevmode
     \begin{itemize}
         \item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
-        \item $\Sigma^1_1$-complete sets are never Borel:
+        \item $\Sigma^1_1$-complete sets are never Borel%
-            Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
+            \gist{:
-            Take $A \in \Sigma^1_1(X) \setminus \cB(X)$
+                Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
-            and $f\colon X \to  Y$ Borel.
+                Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
-            But then $f^{-1}(B)$ is Borel.
+                \footnote{e.g.~\yaref{thm:universals11}}
+                and $f\colon X \to  Y$ Borel.
+                But then we get that $f^{-1}(B)$ is Borel $\lightning$.
+            }{.}
     \end{itemize}
 \end{observe}
 
 \begin{theorem}
     \label{thm:lec12:1}
     Suppose that $A \subseteq \cN$ is analytic.
-    Then there is $f\colon \cN \to \Tr$\todo{Borel?}
+    \gist{%
-    such that $x \in A \iff f(x)$ is ill-founded.
+        Then there is a continuous function $f\colon \cN \to \Tr$
+        such that $x \in A \iff f(x)$ is ill-founded,
+        i.e.~$A = f^{-1}(\IF)$.
+    }{%
+        Then there exists $f\colon \cN \to \Tr$ continuous
+        such that $A = f^{-1}(\IF)$.
+    }
 \end{theorem}
 For the proof we need some prerequisites:
-\begin{enumerate}[1.]
+
-    \item Recall that for $S$ countable,
+\gist{%
-        the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees
+    Recall that for $S$ countable,
-        $T \subseteq S^{<\N}$ on $S$ correspond
+    the pruned%
-        to closed subsets of $S^{\N}$:
+    \footnote{no maximal elements,
-        \begin{IEEEeqnarray*}{rCl}
+        in particular this implies ill-founded if the tree is non empty.
-            T &\longmapsto & [T]\\
+    } trees $T \subseteq S^{<\N}$ on $S$ correspond
-            \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
+    to closed subsets of $S^{\N}$:%
-        \end{IEEEeqnarray*}
+    \footnote{cf.~\yaref{s3e1} (c)}
-        \todo{Copy from exercises}
+    \begin{IEEEeqnarray*}{rCl}
-    \item \leavevmode\begin{definition}
+        T &\longmapsto & [T]\\
-        If $T$  is a tree on $\N \times \N$
+        \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
-        and $x \in \cN$,
+    \end{IEEEeqnarray*}
-        then the \vocab{section at $x$}
+}{%
-        %denoted $T(x)$,
+    For $S$ countable,
-        is the following tree on $\N$ :
+    pruned trees on $S$ correspond to closed subsets of $S^{\N}$
-        \[
+    via $T \mapsto [T]$.
-        T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
+}
-        \]
+\begin{definition}
-    \end{definition}
+    If $T$  is a tree on $\N \times \N$
-    \item \leavevmode
+    and $x \in \cN$,
-        \begin{proposition}
+    then the \vocab{section at $x$}
-            \label{prop:lec12:2}
+    denoted $T(x)$,
-            Let $A \subseteq  \cN$.
+    is the following tree on $\N$ :
-            The following are equivalent:
+    \[
-            \begin{itemize}
+    T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
-                \item   $A$ is analytic.
+    \]
-                \item There is a pruned tree on $\N \times \N$
+\end{definition}
-                    such that
+\begin{proposition}
-                    \[A = \proj_1 ([T]) = \{x \in \cN : \exists  y \in \cN.~ (x,y) \in [T]\}.\]
+    \label{prop:lec12:2}
-            \end{itemize}
+    Let $A \subseteq  \cN$.
-        \end{proposition}
+    The following are equivalent:
-        \begin{proof}
+    \begin{itemize}
-            $A$ is analytic iff
+        \item   $A$ is analytic.
-            there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$
+        \item There is a pruned tree on $\N \times \N$
-            such that $A = \proj_1(F)$.
+            such that
-            But closed sets of $\N \times \N$ correspond to pruned trees,
+            \[A = \proj_1 ([T]) = \{x \in \cN : \exists  y \in \cN.~ (x,y) \in [T]\}.\]
-            by the first point.
+    \end{itemize}
-        \end{proof}
+\end{proposition}
-\end{enumerate}
+\begin{proof}
+    \gist{%
+        $A$ is analytic iff
+        there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
+        such that $A = \proj_1(F)$.
+        But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
+        by the first point.
+    }{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
+\end{proof}
 \begin{refproof}{thm:lec12:1}
-    Take a tree $T$ on $\N \times \N$
+    \gist{%
-    as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
+        Take a tree $T$ on $\N \times \N$
+        as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
+    }{Write $A = \proj_1([T])$ for a pruned tree $T$ on $\N \times \N$.}
     Consider
     \begin{IEEEeqnarray*}{rCl}
         f\colon \cN &\longrightarrow & \Tr \\
         x &\longmapsto & T(x).
     \end{IEEEeqnarray*}
-    Clearly $x \in A \iff f(x)$ is ill-founded.
+    \gist{%
-    $f$ is continuous:
+        Clearly $x \in A \iff f(x) \in \IF$.
-    Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
+        $f$ is continuous:
-    Then for all $m \le n, s,t \in \N^{<\N}$
+        Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
-    such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
+        Then for all $m \le n, s,t \in \N^{<\N}$
-    we have
+        such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
-    \begin{itemize}
+        we have
-        \item $t \in T(x) \iff (s,t) \in T$,
+        \begin{itemize}
-        \item $t \in T(y) \iff (s,t) \in T$.
+            \item $t \in T(x) \iff (s,t) \in T$,
-    \end{itemize}
+            \item $t \in T(y) \iff (s,t) \in T$.
+        \end{itemize}
 
-    So if $x\defon{n} = y\defon{n}$,
+        So if $x\defon{n} = y\defon{n}$,
-    then $t \in T(x) \iff t \in  T(y)$ as long as $|t| \le n$..
+        then $t \in T(x) \iff t \in  T(y)$ as long as $|t| \le n$.
+    }{}
 \end{refproof}
 
 \begin{corollary}
@@ -125,7 +158,9 @@ For the proof we need some prerequisites:
 \end{corollary}
 \begin{proof}
     Let $X$ be Polish.
-    Suppose that $A \subseteq X$ is analytic and uncountable.
+    Suppose that $A \subseteq X$ is analytic and uncountable%
+    \gist{}{ (trivial for countable)}.
+
     Then
     % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
 \[\begin{tikzcd}
@@ -138,16 +173,18 @@ For the proof we need some prerequisites:
 \end{tikzcd}\]
     where $f$ is chosen as in \yaref{thm:lec12:1}.
 
-    If $X$ is Polish and countable and $A \subseteq  X$ analytic,
+    \gist{%
-    just consider
+        If $X$ is Polish and countable and $A \subseteq  X$ analytic,
-    \begin{IEEEeqnarray*}{rCl}
+        just consider
-        g \colon X &\longrightarrow & \Tr \\
+        \begin{IEEEeqnarray*}{rCl}
-        x &\longmapsto & \begin{cases}
+            g \colon X &\longrightarrow & \Tr \\
-            a &: x \in A,\\
+            x &\longmapsto & \begin{cases}
-            b &: x \not\in A,\\
+                a &: x \in A,\\
-        \end{cases}
+                b &: x \not\in A,\\
-    \end{IEEEeqnarray*}
+            \end{cases}
-    where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
+        \end{IEEEeqnarray*}
+        where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
+    }{}
 \end{proof}
 
 \subsection{Linear Orders}
@@ -161,18 +198,19 @@ Let
 \[
 \WO \coloneqq  \{x \in \LO: x \text{ is a well ordering}\}.
 \]
+\gist{%
+    Recall that
+    \begin{itemize}
+        \item $(A,<)$ is a well ordering iff there are no infinite descending chains.
+        \item Every well ordering is isomorphic to an ordinal.
+        \item Any two well orderings are comparable,
+            i.e.~they are isomorphic,
+            or one is isomorphic to an initial segment of the other.
 
-Recall that
+            Let $(A, <_A) \prec (B, <_B)$ denote that
-\begin{itemize}
+            $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
-    \item $(A,<)$ is a well ordering iff there are no infinite descending chains.
+    \end{itemize}
-    \item Every well ordering is isomorphic to an ordinal.
+}{}
-    \item Any two well orderings are comparable,
-        i.e.~they are isomorphic,
-        or one is isomorphic to an initial segment of the other.
-
-        Let $(A, <_A) \prec (B, <_B)$ denote that
-        $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
-\end{itemize}
 
 \begin{definition}
     A \vocab{rank} on some set $C$
@@ -181,13 +219,14 @@ Recall that
     \phi\colon  C \to  \Ord.
     \]
 \end{definition}
-\begin{example}
+\gist{%
-    Let $C = \WO$
+    \begin{example}
-    and
+        Let $C = \WO$
-    \begin{IEEEeqnarray*}{rCl}
+        and
-        \phi\colon \WO &\longrightarrow & \Ord \\
+        \begin{IEEEeqnarray*}{rCl}
-    \end{IEEEeqnarray*}
+            \phi\colon \WO &\longrightarrow & \Ord \\
-    where $\phi((A,<_A))$ is the unique ordinal
+        \end{IEEEeqnarray*}
-    isomorphic to $(A, <_A)$.
+        where $\phi((A,<_A))$ is the unique ordinal
-\end{example}
+        isomorphic to $(A, <_A)$.
-
+    \end{example}
+}{}

inputs/lecture_13.tex

@@ -1,11 +1,12 @@
 \lecture{13}{2023-11-08}{}
-
+\gist{%
 % Recap
 $\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
 $\LO \subseteq 2^{\N \times \N}$ is closed
 and $\WO = \{x \in  \LO: x \text{ is a wellordering}\} $
  is coanalytic in $\LO$.
 % End Recap
+}{}
 
 Another way to code linear orders:
 
@@ -32,6 +33,7 @@ with $(f^{-1}(\{1\}), <)$.
     and  $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
 \end{proof}
 
+
 \begin{definition}[\vocab{Kleene-Brouwer ordering}]
     Let $(A,<)$ be a linear order and $A$ countable.
     We define the linear order $<_{KB}$ on $A^{<\N}$
@@ -55,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$.
     $(T, <_{KB}\defon{T})$ is well ordered.
 \end{proposition}
 \begin{proof}
-    If $T$ is ill-founded and $x \in [T]$,
+    \gist{%
-    then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
+        If $T$ is ill-founded and $x \in [T]$,
-    Thus $(T, <_{KB}\defon{T})$ is not well ordered.
+        then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
+        Thus $(T, <_{KB}\defon{T})$ is not well ordered.
 
-    Conversely, let $<\defon{KB}$ be not a well-ordering
+        Conversely, let $<\defon{KB}$ be not a well-ordering
-    on $T$.
+        on $T$.
-    Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
+        Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
-    be an infinite descending chain.
+        be an infinite descending chain.
-    We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
+        We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
-    stabilizes for $n > n_0$.
+        stabilizes for $n > n_0$.
-    Let $a_0 \coloneqq  s_{n_0}(0)$.
+        Let $a_0 \coloneqq  s_{n_0}(0)$.
-    Now for $n \ge  n_0$ we have that $s_n(0)$ is constant,
+        Now for $n \ge  n_0$ we have that $s_n(0)$ is constant,
-    hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
+        hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
-    Thus there is $n_1 \ge n_0$ such that $s_n(1)$
+        Thus there is $n_1 \ge n_0$ such that $s_n(1)$
-    is constant for all $n \ge n_1$.
+        is constant for all $n \ge n_1$.
-    Let $a_1 \coloneqq s_{n_1}(1)$
+        Let $a_1 \coloneqq s_{n_1}(1)$
-    and so on.
+        and so on.
-    Then $(a_0,a_1,a_2, \ldots) \in [T]$.
+        Then $(a_0,a_1,a_2, \ldots) \in [T]$.
+    }{easy}
 \end{proof}
+
+% TODO ANKI-MARKER
+
 \begin{theorem}[Lusin-Sierpinski]
     The set $\LO \setminus \WO$
     (resp.~$2^{\Q} \setminus \WO$)

inputs/lecture_14.tex

@@ -85,9 +85,8 @@
     \[
     \forall  x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast).
     \]
-    We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.
+    We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
-    \todo{missing picture
+    \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
-        \url{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}}
 
 \end{theorem}
 \begin{proof}

inputs/lecture_16.tex

@@ -175,9 +175,10 @@ By Zorn's lemma, this will follow from
     \end{IEEEeqnarray*}
 
     The topology induced by the metric
-    is given by basic open subsets\footnote{see Exercise Sheet 9% TODO REF
+    is given by basic open subsets\footnote{cf.~\yaref{s9e2}}
-    }
+    of the form
-    $[U_0; U_1,\ldots, U_n]$, $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
+    $[U_0; U_1,\ldots, U_n]$,
+    for $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
     where
     \[
         [U_0; U_1,\ldots,U_n] \coloneqq
@@ -188,8 +189,8 @@ By Zorn's lemma, this will follow from
 We want to view flows as a metric space.
 For a fixed compact metric space $X$,
 we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$.
-Note that $\cC(X,X)$ is Polish.
+Note that $\cC(X,X)$ is Polish.\footnote{cf.~\yaref{s1e4}}
-Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.
+Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.\footnote{Exercise} % TODO
 
 However we do not want to consider only flows on a fixed space $X$,
 but we want to look all flows at the same time.

inputs/tutorial_01.tex

@@ -27,8 +27,10 @@
 
 \begin{definition}
     A topological space is \vocab{Lindelöf}
-    if every open cover has a countable subcover.
+    iff every open cover has a countable subcover.
 \end{definition}
+
+
 \begin{fact}
     Let $X$ be a metric space.
     If $X$ is Lindelöf,
@@ -64,5 +66,12 @@
     and Lindelöf coincide.
 
     In arbitrary topological spaces,
-    Lindelöf is the strongest of these notions.
+    Lindelöf is the weakest of these notions.
 \end{remark}
+
+\begin{definition}+
+    A metric space $X$ is  \vocab{totally bounded}
+    iff for every $\epsilon > 0$ there exists
+    a finite set of points $x_1,\ldots,x_n$
+    such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.
+\end{definition}

inputs/tutorial_09.tex

@@ -140,16 +140,7 @@ amounts to a finite number of conditions on the preimage.
             \end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X).
         \end{IEEEeqnarray*}
         Clearly this has the desired properties.
-    \item We have
+   \item Let $X$ be a compact Polish space.
-        \begin{IEEEeqnarray*}{Cl}
-            & \Z \circlearrowright X \text{ has a dense orbit}\\
-            \iff& \exists  x \in X.~ \overline{\Z\cdot x} = X\\
-            \iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~
-            z \cdot x \in U\\
-            \iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~
-            \exists z \in \Z.~f^z(x) \in U.
-        \end{IEEEeqnarray*}
-    \item Let $X$ be a compact Polish space.
         What is the Borel complexity of $\Homeo(X)$ inside  $\cC(X,X)$?
 
         Recall that $\cC(X,X)$ is a Polish space with the uniform topology.
@@ -160,14 +151,9 @@ amounts to a finite number of conditions on the preimage.
         \end{IEEEeqnarray*}
         by the general fact
         \begin{fact}
-            Let $X$ be comapct and  $Y$ Hausdorff,
+            Let $X$ be compact and  $Y$ Hausdorff,
             $f\colon  X \to  Y$ a continuous bijection.
             Then $f$ is a homeomorphism.
         \end{fact}
-    \item It suffices to check the condition from part (b)
-        for open sets $U$ of a countable basis
-        and points  $x \in X$ belonging to a countable dense subset.
-        Replacing quantifiers by unions resp.~intersections
-        gives that $D$ is Borel.
 \end{itemize}
 

inputs/tutorial_13.tex

@@ -0,0 +1,146 @@
+\tutorial{13}{2024-01-23}{}
+
+
+Continuation of sheet 8, exercise 4.
+ % Whiteboard https://wbo.ophir.dev/boards/Dphc7eylJJcIA0WbQsn7jzec1domqyx51gXb5qe6rzw-#263,0,0.5
+
+\begin{definition}
+    Let $X$ be a compact metric space.
+    For $K \subseteq X$ compact and $U \overset{\text{open}}{\subseteq} X$
+    let
+    \[
+    S_{K,U} \coloneqq  \{f \in \cC(X,X): f(K) \subseteq U\}.
+    \]
+    The \vocab{compact open topology} on $\cC(X,X)$
+    is the topology that has $S_{K,U}$ as a subbase.
+\end{definition}
+\begin{fact}
+    If $X$ is compact,
+    then the compact open topology
+    is the topology induced by the uniform metric $d_\infty$.
+\end{fact}
+\begin{proof}
+    Take some $S_{K,U}$. We need to show that this can be written
+    as a union of open $d_{\infty}$-balls.
+    Let $f_0 \in S_{K,U}$.
+    Consider the continuous function $d(-, U^c)$.
+    Since $f_0(K)$ is compact,
+    there exists $\epsilon \coloneqq \min d(f_0(K), U^c)$
+    and $B_{\epsilon}(f_0) \subseteq S_{K,U}$.
+
+
+    On the other hand, consider $B_{\epsilon}(f_0)$ for some $\epsilon > 0$
+    and $f_0 \in \cC(X,X)$.
+
+    As $f_0$ is uniformly continuous,
+    there exists $\delta > 0$ such that $d(x,x') < \delta \implies d(f_0(x), f_0(x')) < \frac{\epsilon}{3}$.
+    Cover $X$ with finitely many $\delta$-balls $B_\delta(a_1), \ldots, B_{\delta}(a_k)$.
+    Then
+    \[f_0(\overline{B_{\delta}(a_i)}) \subseteq \overline{f_0(B_{\delta}(a_i)} \subseteq \overline{B_{\frac{\epsilon}{3}}(f_0(a_i))} \subseteq  B_{\frac{\epsilon}{2}}(f_0(a_i)).\]
+
+    For $i \le  k$, let $S_i \coloneqq  S_{\overline{B_{\delta}(a_i)}, B_{\frac{\epsilon}{2}}(f_0(a_i))}$.
+    Take $\bigcap_{i \le k} S_i$. This is open
+    in the compact open topology and
+    $B_{\epsilon}(f_0) \subseteq \bigcap_{i \le k} S_i$.
+\end{proof}
+
+\begin{claim}
+    $f \in \cC(X,X)$ is surjective
+    iff for all basic open $\emptyset\neq U \subseteq  X$
+    there exists a basic open $\emptyset \neq V \subseteq  X$
+    with $f(\overline{V}) \subseteq U$.
+    
+    Note that we can write this as a $G_\delta$-condition.
+
+\end{claim}
+\begin{subproof}
+    Take $B_\epsilon(f(x_0))\subseteq U$.
+    Then there exists $\delta > 0$
+    such that $f(B_{\delta}(x_0)) \subseteq B_{\frac{\epsilon}{2}}(f(x_0))$
+    hence $f(\overline{B_{\delta}(x_0)}) \subseteq B_\epsilon(f(x_0))$.
+
+
+    For the other direction take $y \in X$.
+    We want to find a preimage.
+    For every $B_{\frac{1}{n}}(y)$,
+    there exists a basic open set $V_n$ with $f(\overline{V}) \subseteq B_{\frac{1}{n}}(y)$.
+    Take $x_n \in  V_n$.
+    Since $X$ is compact, it is sequentially compact,
+    so there exists a converging subsequence.
+    Wlog.~$x_n \to x$,
+    so $f(x_n) \to f(x) = y$.
+\end{subproof}
+
+\begin{claim}
+    $f \in \cC(X,X)$ is injective iff
+    for all basic open $U$,$V$
+    with $\overline{U} \cap \overline{V} = \emptyset$
+    we have $f(\overline{U}) \cap f(\overline{V}) = \emptyset$.
+
+    This is a $G_\delta$-condition,
+    since we can write it as
+    \[
+    \bigcap_{U,V} S_{\overline{U}, f(\overline{V})^c}.
+    \]
+\end{claim}
+\begin{subproof}
+    $\implies$ is trivial.
+
+    $\impliedby$ follows since for all pairs $x,y \in X$,
+    we can find $x \in U$, $y \in V$ such that $\overline{U} \cap \overline{V} = \emptyset$.
+\end{subproof}
+
+Hence $\Homeo(X,X)$ is $G_\delta$.
+In particular it is a Polish space.
+
+
+
+Let $D$ be the set of $\Z$-flows with dense orbit.
+\begin{claim}
+    $f \in D$ $\iff$
+    for all basic open $U,V \subseteq X$,
+    there exists $n \in \Z$
+    such that $f^n(U) \cap V \neq \emptyset$.
+
+    \end{claim}
+\begin{subproof}
+    Suppose that the orbit of $x_0 \in X$ is dense.
+    Then there exist $k,l \in \Z$
+    such that $f^k(x_0)\in U$ and $f^l(x_0) \in V$,
+    so $f^{l-k} U \cap V \neq \emptyset$.
+
+
+    For basic open sets $V$
+    let
+    \[
+     A_V \coloneqq   \{ x \in X: \exists n.~ f^n(x) \in V\}.
+    \]
+    By assumption, all the $A_V$ are dense.
+    Hence $\bigcap_{V}A_V$ is dense by the \yaref{thm:bct}.
+
+    $A_V = \bigcup_{n \in \Z} f^n(V)$ is open.
+\end{subproof}
+
+\begin{claim}
+The condition can be written as a $G_\delta$ set.
+\end{claim}
+\begin{subproof}
+
+    For  $f_0(U) \cap V \neq \emptyset$
+    take $u \in U$ such that $f_0(u) \in V$.
+    Then there exists  $\epsilon > 0$ such that $B_{\epsilon}(f_0(u)) \subseteq U$,
+    hence $B_{\epsilon}(f_0)$ is an open neighbourhood contained
+    in $\{f : f(U) \cap V \neq \emptyset \} $.
+
+    For $n = 2$ note that
+    $d(f^2(u), f^2_0(u) \le  d(f(f(u)), f_0(f(u))) + d(f_0(f(u)), f_0(f_0(u)))$.
+    The first part can be bounded by $d(f,f_0)$.
+    For the second part,
+    note that there exists $\delta$ such that
+    \[d(a,b) < \delta \implies d(f_0(a), f_0(b)) < \frac{\epsilon}{2}.\]
+    Let $\eta \coloneqq  \min \{\delta, \frac{\epsilon}{2}\}$
+    and consider $d_\infty(f,f_0) < \epsilon$.
+
+    For other  $n$ it is some more work, which is left as an exercise.
+\end{subproof}
+

jrpie-gist.sty

@@ -1,6 +1,10 @@
 \NeedsTeXFormat{LaTeX2e}
 \ProvidesPackage{jrpie-gist}[2023/01/22 - gist version for lecture notes]
 
+% TODO gist info
+% TODO link to long version (provide link to main document)
+
+% TODO \phantomsection to cross link
 \newcommand{\gist}[2]{%
     \ifcsname EnableGist\endcsname%
         #2%

logic3.tex

@@ -66,6 +66,7 @@
 \input{inputs/tutorial_07}
 \input{inputs/tutorial_08}
 \input{inputs/tutorial_09}
+\input{inputs/tutorial_13} % sic!
 \input{inputs/tutorial_10}
 \input{inputs/tutorial_11}
 \input{inputs/tutorial_12b}