inputs/lecture_05.tex

@@ -103,9 +103,7 @@
     Let $X$ be a completely metrizable space.
     Then every comeager set of $X$ is dense in $X$.
 \end{theorem}
-\gist{%
 \todo{Proof (copy from some other lecture)}
-}{Not proved in the lecture.}
 \begin{theoremdef}
     Let $X$ be a topological space.
     The following are equivalent:
@@ -120,21 +118,7 @@
     \footnote{cf.~\yaref{s5e1}}
 \end{theoremdef}
 \begin{proof}
-    (i) $\implies$ (ii)
-    \gist{%
-        Consider a comeager set $A$.
-        Let $U\neq \emptyset$ be any open set. Since $U$ is
-        non-meager, we have $A \cap U \neq \emptyset$.
-    }{The intersection of a comeager and a non-meager set is nonempty.}
-
-    (ii) $\implies$ (iii)
-    The complement of an open dense set is nwd.
-    \gist{%
-        Hence the intersection of countable
-        many open dense sets is comeager.
-    }{}
-
-
+    \todo{Proof (short)}
 
     (iii) $\implies$ (i)
     Let us first show that $X$ is non-meager.

inputs/lecture_07.tex

@@ -77,7 +77,7 @@
     sets that are not clopen)}{}.
 \end{example}
 
-\subsection{Turning Borel Sets into Clopens}
+\subsection{Turning Borels Sets into Clopens}
 
 \begin{theorem}%
     \gist{%
@@ -109,7 +109,6 @@
     into $B$.
 \end{corollary}
 \begin{proof}
-    \gist{%
     Pick $\cT_B \supset \cT$
     such that $(X, \cT_B)$ is Polish,
     $B$ is clopen in $\cT_B$ and
@@ -122,17 +121,10 @@
     Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
     This is still continuous as $\cT \subseteq \cT_B$.
     Since $2^{\omega}$ is compact, $f$ is an embedding.
-    }{%
-        Clopenize $B$.
-        We can embed $2^{ \omega}$ into Polish spaces.
-        Clopenization makes the topology finer,
-        so this is still continuous wrt.~the original topology.
-        $2^{\omega}$ is compact, so this is an embedding.
-    }
+    %\todo{Think about this}
 \end{proof}
 
 \begin{refproof}{thm:clopenize}
-    \gist{%
     We show that
     \begin{IEEEeqnarray*}{rCl}
         A \coloneqq  \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
@@ -142,26 +134,18 @@
         \}
     \end{IEEEeqnarray*}
     is equal to the set of Borel sets.
-    }{%
-        Let $A$ be the set of clopenizable sets.
-        We show that $A = \cB(X)$.
-    }
-    \gist{The proof rests on two lemmata:}{}
+
+    The proof rests on two lemmata:
     \begin{lemma}
         \label{thm:clopenize:l1}
-        \gist{%
         Let $(X,\cT)$ be a Polish space.
         Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
         there is $\cT_F \supseteq \cT$
         such that $\cT_F$ is Polish,
         $\cB(\cT) = \cB(\cT_F)$
         and  $F$ is clopen in $\cT_F$.
-        }{%
-            Closed sets can be clopenized.
-        }
     \end{lemma}
     \begin{proof}
-        \gist{%
         Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
         Both are Polish spaces.
         Take the coproduct%
@@ -170,13 +154,10 @@
         This space is Polish,
         and the topology is generated by $\cT \cup \{F\}$,
         hence we do not get any new Borel sets.
-        }{Consider $(F, \cT\defon{F}) \oplus (X \setminus F, \cT\defon{X \setminus F})$.}
     \end{proof}
-    \gist{%
     So all closed sets are in $A$.
     Furthermore $A$ is closed under complements,
     since complements of clopen sets are clopen.
-    }{So $\Sigma^0_1(X), \Pi^0_1(X) \subseteq A$.}
 
     \begin{lemma}
         \label{thm:clopenize:l2}
@@ -189,11 +170,11 @@
         is Polish
         and $\cB(\cT_\infty) = \cB(T)$.
     \end{lemma}
-    \begin{refproof}{thm:clopenize:l2}
-        \gist{%
+    \begin{proof}
         We have that $\cT_\infty$ is the smallest
         topology containing all $\cT_n$.
-            To get $\cT_\infty$ consider
+        To get $\cT_\infty$
+        consider
         \[
         \cF \coloneqq  \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
         \]
@@ -205,7 +186,6 @@
         since we may assume that the $A_1, \ldots, A_n$ in the
         definition of $\cF$ belong to
         a countable basis of the respective $\cT_n$).
-        }{}
 
         % Proof was finished in lecture 8
         Let $Y = \prod_{n \in \N} (X, \cT_n)$.
@@ -215,7 +195,6 @@
         \begin{claim}
             $\delta$ is a homeomorphism.
         \end{claim}
-        \gist{%
         \begin{subproof}
             Clearly $\delta$ is a bijection.
             We need to show that it is continuous and open.
@@ -240,26 +219,21 @@
             \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}
-        }{}
 
-        \begin{claim}
-                $D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y.$
-        \end{claim}
-        \gist{%
-            \begin{subproof}
-                Let  $(x_n) \in Y \setminus D$.
+        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.
-            \end{subproof}
         It follows that $D$ is Polish.
-        }{}
-    \end{refproof}
+    \end{proof}
 
-    \gist{%
     We need to show that $A$ is closed under countable unions.
     By \yaref{thm:clopenize:l2} there exists a topology
     $\cT_\infty$ such that $A = \bigcup_{n < \omega}  A_n$ is open in $\cT_\infty$
@@ -269,5 +243,4 @@
     $(X, \cT_\infty')$ is Polish,
     $\cB(\cT_\infty') = \cB(\cT)$
     and $A $ is clopen in $\cT_{\infty}'$.
-    }{}
 \end{refproof}

inputs/lecture_08.tex

@@ -1,9 +1,8 @@
-\lecture{08}{2023-11-10}{}%
-\gist{\footnote{%
+\lecture{08}{2023-11-10}{}\footnote{%
     In the beginning of the lecture, we finished
     the proof of \yaref{thm:clopenize:l2}.
     This has been moved to the notes on lecture 7.%
-}}{}
+}
 
 \subsection{Parametrizations}
 %\todo{choose better title}
@@ -23,14 +22,13 @@ where $X$ is a metrizable, usually second countable space.
         \item $\{U_y : y \in Y\} = \Gamma(X)$.
     \end{itemize}
 \end{definition}
-\gist{%
+
 \begin{example}
     Let $X = \omega^\omega$, $Y = 2^{\omega}$
     and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$.
     We will show that there is a $2^{\omega}$-universal
     set for $\Gamma$.
 \end{example}
-}{}
 
 \begin{theorem}
     \label{thm:cantoruniversal}

inputs/lecture_09.tex

@@ -6,7 +6,6 @@
     we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.
 \end{theorem}
 \begin{proof}
-    \gist{%
     Fix $\xi < \omega_1$.
     Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$.
     By \autoref{thm:cantoruniversal},
@@ -20,10 +19,6 @@
     \[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]
     But by the definition of $A$,
     we have $z \in A \iff (z,z) \not\in \cU \lightning$.
-    }{%
-        Let $\cU$ be $X$-universal for $\Sigma^0_\xi(X)$.
-        Consider $\{y \in X : (y,y) \not\in \cU\} \in \Pi^0_\xi(X) \setminus \Sigma^0_\xi(X)$.
-    }
 \end{proof}
 
 
@@ -38,10 +33,8 @@
     f(B) = A.
     \] 
 \end{definition}
-\gist{%
 Trivially, every Borel set is analytic.
 We will see that  not every analytic set is Borel.
-}{}
 \begin{remark}
     In the definition we can replace the assertion that 
     $f$ is continuous
@@ -71,7 +64,6 @@ We will see that not every analytic set is Borel.
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    \gist{%
     To show (i) $\implies$ (ii):
     take $B \in \cB(Y')$
     and $f\colon Y' \to X$
@@ -81,16 +73,11 @@ We will see that not every analytic set is Borel.
     such that $B$ is clopen with respect to the new topology.
     Then let $g = f\defon{B}$
     and $Y = (B, \cT\defon{B})$.
-    }{(i) $\implies$ (ii):
-        Clopenize the Borel set, then restrict.
-    }
 
     (ii) $\implies$ (iii):
     Any Polish space is the continuous image of $\cN$.
-    \gist{%
     Let $g_1: \cN \to  Y$
     and $h \coloneqq g \circ g_1$.
-    }{}
 
     (iii) $\implies$ (iv):
     Let $h\colon \cN \to X$ with $h(\cN) = A$.

inputs/lecture_11.tex

@@ -99,10 +99,9 @@
     (continuous wrt.~to the topology of $X$)
     On the other hand
     \[
-    X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC
+    X \hookrightarrow\cN \overset{\text{continuous embedding}}{\hookrightarrow}\cC
     \]
-    \footnotetext{cf.~\yaref{s2e4}}
-
+    \todo{second inclusion was on a homework sheet}
     For the first inclusion,
     recall that there is a continuous bijection $b\colon  D \to X$,
     where $D \overset{\text{closed}}{\subseteq} \cN$.

inputs/lecture_13.tex

@@ -79,6 +79,8 @@ with $(f^{-1}(\{1\}), <)$.
     }{easy}
 \end{proof}
 
+% TODO ANKI-MARKER
+
 \begin{theorem}[Lusin-Sierpinski]
     The set $\LO \setminus \WO$
     (resp.~$2^{\Q} \setminus \WO$)
@@ -87,23 +89,19 @@ with $(f^{-1}(\{1\}), <)$.
 \begin{proof}
     We will find a continuous function
     $f\colon \Tr \to \LO$ such that
-    \gist{%
     \[
     x \in \WF \iff f(x) \in \WO
     \]
     (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$).
     This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete
-    }{
-         $f^{-1}(\LO \setminus \WO) = \IF$.
-         This suffices
-    }
     (see \yaref{cor:ifs11c}).
 
     Fix a bijection $b\colon \N \to \N^{<\N}$.
 
     \begin{idea}
         For $T \in \Tr$ consider
-        $<_{KB}\defon{T}$.
+        $<_{KB}\defon{T}$
+        % TODO?
     \end{idea}
 
     Let $\alpha \in \Tr$.
@@ -111,7 +109,7 @@ with $(f^{-1}(\{1\}), <)$.
     (i.e.~$m \le_{f(\alpha)} n$)
     iff
     \begin{itemize}
-        \item $\alpha(b(m)) = \alpha(b(n)) = 1$
+        \item $(\alpha(b(m)) = \alpha(b(n)) = 1$
             and $b(m) \le_{KB} b(n)$
             (recall that we identified $\Tr$
             with a subset of ${2^{\N}}^{<\N}$),
@@ -125,7 +123,7 @@ with $(f^{-1}(\{1\}), <)$.
 \end{proof}
 
 % TODO: new section?
-\gist{%
+
 Recall that a \vocab{rank} on a set $C$
 is a map $\phi\colon C \to \Ord$.
 \begin{example}
@@ -134,7 +132,6 @@ with $(f^{-1}(\{1\}), <)$.
          x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}.
     \end{IEEEeqnarray*}
 \end{example}
-}{}
 
 \begin{definition}
     A \vocab{prewellordering} $\preceq$
@@ -144,7 +141,7 @@ with $(f^{-1}(\{1\}), <)$.
         \item reflexive,
         \item transitive,
         \item total (any two $x,y$ are comparable),
-        \item $\prec$ \gist{($x \prec y \iff x \preceq y \land y \not\preceq x$)}{} is well-founded,
+        \item $\prec$ ($x \prec y \iff x \preceq y \land y \not\preceq x$) is well-founded,
             in the sense that there are no descending infinite chains.
     \end{itemize}
 \end{definition}
@@ -152,7 +149,7 @@ with $(f^{-1}(\{1\}), <)$.
     \begin{itemize}
         \item A prewellordering may not be a linear order since
             it is not necessarily antisymmetric.
-        %\item The linearly ordered wellfounded sets are exactly the wellordered sets.
+        \item The linearly ordered wellfounded sets are exactly the wellordered sets.
         \item Modding out $x \sim y :\iff x \preceq y \land y \preceq x$
             turns a prewellordering into a wellordering.
     \end{itemize}
@@ -166,11 +163,11 @@ between downwards-closed ranks and prewellorderings:
                             \phi_{\preceq}&\longmapsfrom& \preceq,
 \end{IEEEeqnarray*}
 where $\phi_\preceq(x)$ is defined as
-\gist{\begin{IEEEeqnarray*}{rCl}
+\begin{IEEEeqnarray*}{rCl}
     \phi_{\preceq}(x) &\coloneqq &0 \text{ if $x$ is minimal},\\
     \phi_{\preceq}(x) &\coloneqq & \sup \{\phi_{\preceq}(y) + 1 : y \prec x\},
 \end{IEEEeqnarray*}
-i.e.}{}
+i.e.
 \[
     \phi_{\preceq}(x) = \otp\left(\faktor{\{y \in C : y \prec x\}}{\sim}\right).
 \]

inputs/lecture_14.tex

@@ -1,6 +1,5 @@
 \lecture{14}{2023-12-01}{}
 
-% TODO ANKI-MARKER
 \begin{theorem}[Moschovakis]
     If $C$ is coanalytic,
     then there exists a $\Pi^1_1$-rank on $C$.