some changes

inputs/lecture_02.tex

@@ -1,4 +1,5 @@
 \lecture{02}{2023-10-19}{Topology on $\R$}
+\gist{%
 \begin{definition}
     A set $O \subseteq \R$ is called \vocab{open} in $\R$ iff it is the union
     of a set of open intervals.
@@ -20,6 +21,7 @@
 \begin{remark}+
     $\{O \subseteq \R\} \sim 2^{\aleph_0} < \cP(\R)$.
 \end{remark}
+}{}
 
 \begin{definition}
     We call $x \in \R$
@@ -28,9 +30,11 @@
     there is some $y \in A$, $y \in (a,b)$, $y \neq x$.
     We write \vocab{$A'$} for the set of all accumulation points of $A$.
 \end{definition}
+\gist{%
 \begin{example}
     $\{\frac{1}{n+1} | n \in \N\}' = \{0\}$.
 \end{example}
+}{}
 
 \begin{lemma}
     \label{lem:closedaccumulation}
@@ -38,9 +42,11 @@
 \end{lemma}
 \begin{refproof}{lem:closedaccumulation}
     ``$\implies$''
+    \gist{%
     Let $A$ be closed. Suppose that $x \in A' \setminus A$.
     Then there exists $(a,b) \ni x$
     disjoint from $A$. Hence $x \not\in A' \lightning$
+    }{trivial.}
 
     ``$\impliedby$''
     Suppose $A' \subseteq  A$.
@@ -48,6 +54,7 @@
         $A \subseteq  \R$ is closed iff all Cauchy sequences
         in $A$ converge in $A$.
     \end{claim}
+    \gist{%
     \begin{subproof}
         Let $A$ be closed and $\langle x_n : n \in \omega \rangle$
         a Cauchy sequence in $A$.
@@ -62,6 +69,7 @@
         we may pick $x_n \in (x - \frac{1}{n+1}, x + \frac{1}{n+1}) \cap A$
         for all $n < \omega$.
     \end{subproof}
+    }{}
 
     Now if $A' \subseteq  A$ and $A$ were not closed,
     there would be some Cauchy-sequence $(x_n)$
@@ -92,6 +100,7 @@ We want to prove two things:
     Then $P \sim \R$.
 \end{lemma}
 \begin{proof}
+\gist{%
     It suffices to find an injection $f\colon \R \hookrightarrow P$.
     We have $\underbrace{\{0,1\}^{\omega}}_{\text{infinite 0-1-sequences}} \sim \R$,
     hence it suffices to construct $f\colon \{0,1\}^\omega\hookrightarrow P$.
@@ -132,4 +141,5 @@ We want to prove two things:
     and $f(t') \in [a_{t'\defon{n}}, b_{t'\defon{n}}]$
     which are disjoint.
     Thus $f(t) \neq f(t')$, i.e.~$f$ is injective.
+}{Cantor scheme.}
 \end{proof}

inputs/lecture_03.tex

@@ -77,7 +77,7 @@ all condensation points are accumulation points.
     \begin{subproof}
         $P \neq \emptyset$: $\checkmark$ 
 
-        $P \subseteq  P'$ (i.e. $P$ is closed):
+        $P \subseteq  P'$:
         % \begin{IEEEeqnarray*}{rCl}
         %     P &=& \{x \in A | \text{every open neighbourhood of $x$ is uncountable}\}\\
         %     &\subseteq & \{x \in A | \text{every open neighbourhood of $x$ is at least countable}\} = P'.
@@ -97,7 +97,7 @@ all condensation points are accumulation points.
         But then $(a,b) \cap A$ is at most countable
         contradicting $ x \in P$.
 
-        $P' \subseteq P$ :
+        $P' \subseteq P$ (i.e.~$P$ is closed):
         Let $x \in P'$.
         Then for $a < x < b$ the set
         $(a,b) \cap P$

inputs/lecture_04.tex

@@ -12,6 +12,7 @@ $\ZFC$ stands for
     \item \textsc{Fraenkel}'s axioms,
     \item the \yaref{ax:c}.
 \end{itemize}
+\gist{
 \begin{notation}
     We write $x \subseteq y$ as a shorthand
     for $\forall z.~(z \in x \implies z \in y)$.
@@ -45,6 +46,7 @@ $\ZFC$ stands for
         \forall u.~((u \in z) \iff (u \in x \land u \not\in y)).
     \]
 \end{notation}
+}{Trivial, boring notation.}
 $\ZFC$ consists of the following axioms:
 \begin{axiom}[\vocab{Extensionality}]
     \yalabel{Axiom of Extensionality}{(Ext)}{ax:ext} % (AoE)
@@ -148,13 +150,15 @@ $\ZFC$ consists of the following axioms:
 \begin{axiomschema}[\vocab{Replacement} (Fraenkel)]
     \yalabel{Axiom of Replacement}{(Rep)}{ax:rep}
     Let $\phi$ be some $\cL_{\in }$ formula
-    with free variables $x, y$.\todo{Allow more variables}
+    with free variables $x, y$.
     Then
-    \[
+    \begin{IEEEeqnarray*}{l}
-        \forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
+        \forall v_1 \ldots \forall v_p.~\\
+        \left[\left( \forall x \exists! y.~\phi(x,y,\overline{v})\right) \to \forall a .~\exists b .~\forall y.~(y \in b \leftrightarrow \exists x (x \in a \land \phi(x,y, \overline{v}))\right]
+    \end{IEEEeqnarray*}
+        %\forall x \in a.~\exists !y \phi(x,y) \implies \exists b.~\forall x \in a.~\exists y \in b.~\phi(x,y).
         % \forall v_1 \ldots \forall v_p .~\forall x.~ \exists y'.~\forall y.~(y = y' \iff \phi(x))
         % \implies \forall a.~\exists b.~\forall y.~(y \in b \iff \exists x.~(x \in a \land \phi(x))
-    \]
 \end{axiomschema}
 
 \begin{axiom}[\vocab{Choice}]

inputs/lecture_05.tex

@@ -23,6 +23,7 @@
     \]
 \end{definition}
 
+\gist{%
 \begin{definition}
     For sets $x, y$ we write
     $(x,y)$ for $\{\{x\}, \{x,y\}\}$.
@@ -130,7 +131,7 @@
    (In other mathematical fields, this is sometimes
    denoted as $f(a)$. We don't do that here.)
 \end{notation}
-
+}{[Some boring definitions omitted.]}
 \begin{definition}
     A binary relation $\le $ on a set $a$
     is a \vocab{partial order}
@@ -168,7 +169,6 @@
     x \in b \land \forall y \in b.~y \le x.
     \]
 
-
     In a similar way we define \vocab[Minimal element]{minimal elements} 
     and the \vocab{minimum} of $b$.
     We say that $x $ is an \vocab{upper bound} 
@@ -181,6 +181,7 @@
     (This does not necessarily exist.)
     Similarly $\text{\vocab{$\inf$}}(b)$ is defined.
 \end{definition}
+\gist{
 \begin{remark}+
     Note that in a partial order,
     a maximal element is not necessarily a maximum.
@@ -200,6 +201,7 @@
     We write $(a,\le_a) \cong (b, \le_b)$
     if they are isomorphic.
 \end{definition}
+}{}
 \begin{definition}
     Let $(a,\le)$ be a partial order.
     Then $(a,\le)$ is 

inputs/lecture_06.tex

@@ -8,6 +8,7 @@
     Then $a$ has a maximal element.
 \end{theorem}
 \begin{refproof}{thm:zorn}
+    \gist{%
     Fix $(a, \le )$ as in the hypothesis.
     Let $A \coloneqq  \{ \{(b,x) : x \in b\}  : b \subseteq a, b \neq \emptyset\}$.
     Note that $A$ is a set (use separation on  $\cP(\cP(a) \times  \bigcup \cP(a))$).
@@ -65,6 +66,36 @@
     Then $B = B_{u_0}^{\le^{\ast\ast}}$.
     So $\le^{\ast\ast} \in W$, but now $u_0 \in b$.
     So $b$ must have a maximum.
+    \todo{Why does this prove the lemma?}
+}{
+    \begin{itemize}
+        \item $A  \coloneqq  \{\{(b,x) : x \in b\}, b \subseteq a, b \neq \emptyset\}$.
+        \item \AxC $\leadsto$ choice function on $A$,
+            $f\colon  \cP(a) \setminus \{\emptyset\}  \to  a$, $f(b) \in b$.
+        \item $\le^\ast$ on $a$:
+            \begin{itemize}
+            \item $W \coloneqq  \{\le' \text{wo on} b \subseteq a : \forall u,b \in b.~u \le' v \implies u \le v, B_u^{\le '} \neq \emptyset, u = f(B^{\le '}_u)\}$ where
+                \item $B_u^{\le'} = \{w \in  a : w \text{ $\le $-upper bound of } \{v \in  b : v <' u\} \}$.
+            \end{itemize}
+        \item $\le', \le '' \in W \implies \le' \substack{\subseteq\\\supseteq} \le''$:
+            \begin{itemize}
+                \item $(b, \le') \overset{g}{\cong} (c, \le'') (\defon{v})$.
+                \item $g = \id_b$:
+                    \begin{itemize}
+                        \item $u_0$ $\le'$ minimal with $g(u_0) \neq u_0$.
+                        \item $\{w \in b : w <' u_0\} \overset{g \defon{\ldots}}{=} \{w \in c : w <'' g(u_0)\}$.
+                        \item $B^{\le '}_{u_0}  = B_{g(u_0)}^{\le ''} \neq \emptyset$,
+                        so $u_0 = f(B^{\le '}_{u_0}) = f(B^{\le ''}_{g(u_0)}) = g(u_0) \lightning$
+                    \end{itemize}
+            \end{itemize}
+        \item  $\le^\ast \coloneqq  \bigcup W$ is wo on $b \subseteq a$.
+        \item Suppose $b$ has no maximum. Then  $B \cap b = \emptyset$.
+        \item $u_0 \coloneqq f(B)$, $\le^{\ast\ast} = \le^\ast \cup \{(u,u_0) | u \in b\}  \cup \{(u_0,u_0)\}$.
+        \item $B = B_{u_0}^{\le^{\ast\ast}}$, so $\le^{\ast\ast} \in W$,
+            but $u_0 \in b \lightning$. ?
+    \end{itemize}
+}
+
 \end{refproof}
 
 \begin{remark}
@@ -82,6 +113,7 @@
     Then $A$ contains a $\subseteq$-maximal element.
 \end{corollary}
 
+\gist{%
 \begin{remark}[Cultural enrichment]
     Other assertions which are equivalent
     to the \yaref{ax:c}:
@@ -96,12 +128,14 @@
         \item Every set can be well-ordered.%\footnote{This is clearly false.}
     \end{itemize}
 \end{remark}
+}{}
 % \begin{remark}
 %     The axiom of choice is true.
 % \end{remark}
 
 \pagebreak
 \subsection{The Ordinals}
+\gist{
 \begin{goal}
     We want to define nice representatives of the equivalence classes
     of well-orders.
@@ -115,7 +149,7 @@ We can hence form the smallest inductive set
 Note that $\omega$ exists, as it is a subset of the inductive
 set given by \AxInf.
 We call $\omega$ the set of \vocab{natural numbers}.
-
+}{}
 \begin{notation}
     We write $0$ for $\emptyset$,
     and $y + 1$ for $y \cup  \{y\}$.
@@ -147,16 +181,19 @@ We have the following principle of induction:
     and for all $y, z \in x$,
     we have that $y = z$, $y \in z$ or $y \ni z$.
 \end{definition}
+\gist{
 Clearly, the $\in$-relation is a well-order on an ordinal $x$.
 \begin{remark}
     This definition is due to \textsc{John von Neumann}.
 \end{remark}
+}{}
 
 \begin{lemma}
     Each natural number (i.e.~element of $\omega$)
     is an ordinal.
 \end{lemma}
 \begin{proof}
+\gist{
     We use \yaref{lem:induction}.
     Clearly $\emptyset$ is an ordinal.
     Now let $\alpha$ be an ordinal.
@@ -169,6 +206,7 @@ Clearly, the $\in$-relation is a well-order on an ordinal $x$.
     since $\alpha$ is an ordinal.
     Suppose $x = \alpha$.
     Then either $y = x$ or $y \in \alpha = x$.
+}{Induction}
 \end{proof}
 
 \begin{lemma}

inputs/lecture_07.tex

@@ -20,6 +20,7 @@
     \end{enumerate}
 \end{lemma}
 \begin{refproof}{lem:7:ordinalfacts}
+\gist{
     We have already proved (a) before.
 
     (b) Fix $x \in \alpha$. Then $x \subseteq \alpha$.
@@ -113,6 +114,7 @@
         But this violates \AxFund,
         as $\alpha_0 \in  \beta_0 \in \alpha_0$.
     \end{subproof}
+}{Long and tedious, but not many ideas.}
 \end{refproof}
 
 \begin{lemma}
@@ -146,6 +148,7 @@ for example $\bigcup \omega = \omega$.
     Otherwise $\alpha$ is called
     a \vocab[Ordinal!limit]{limit ordinal}.
 \end{definition}
+\gist{
 \begin{observe}
     Note that $\alpha$ is a limit ordinal iff for all
     $\beta \in \alpha$, $\beta + 1 \in \alpha$:
@@ -158,7 +161,8 @@ for example $\bigcup \omega = \omega$.
     then by definition there is some $\beta \in \alpha$,
     with $\beta + 1 = \alpha$, so $\beta + 1 \not\in \alpha$.
 \end{observe}
-
+}{}
+\gist{
 \begin{notation}
     If $\alpha, \beta$ are ordinals,
     we write $\alpha < \beta$ for $\alpha \in \beta$
@@ -185,3 +189,4 @@ for example $\bigcup \omega = \omega$.
         \item $\omega +1 = \omega \cup \{\omega\} , \omega + 2, \ldots$,
     \end{itemize}
 \end{example}
+}{}

inputs/lecture_08.tex

@@ -1,6 +1,7 @@
 \lecture{08}{2023-11-13}{Induction and recursion}
 
 \subsection{Classes}
+\gist{
 It is often very handy to work in a class theory rather than
 in set theory.
 
@@ -11,10 +12,11 @@ sets (denoted by lower case letters)
 and classes (denoted by capital letters),
 as well as one binary relation symbol $\in$
 for membership.
+}{}
 
 \vocab{Bernays-Gödel class theory} (\vocab{BG})
 has the following axioms:
-
+\gist{
 \begin{axiom}[Extensionality]
     \yalabel{Axiom of Extensionality}{(Ext)}{ax:bg:ext}
     \[
@@ -54,6 +56,7 @@ has the following axioms:
     \exists x .~(\emptyset \in x \land \left( \forall y \in x .~y \cup \{y\}  \in x \right)).
     \]
 \end{axiom}
+}{\AxExt, \AxFund, \AxPair, \AxUnion, \AxPow, \AxInf as from $\ZF$.}
 
 Together with the following axioms for classes:
 

inputs/lecture_14.tex

@@ -298,7 +298,7 @@ We have shown (assuming \AxC to choose contained clubs):
             and $S_1 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega_1\}$.
             Clearly these are disjoint.
             They are both stationary:
-            Let $c \subseteq \kappa$ be a club.
+            Let $C \subseteq \kappa$ be a club.
             Let $(\xi_i : i \le \omega_1)$
             be defined as follows:
             $\xi_0 \coloneqq  \min C$,

inputs/lecture_17.tex

@@ -252,7 +252,7 @@ We will only proof
                 \item $\forall Y \in A$ define $h_Y\colon S^\ast \to  \omega_1, \alpha \mapsto \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_\beta\}$.
                 \item Apply \yaref{thm:fodor} to $h_Y, S^\ast$ to get $T_Y \subseteq S^\ast$ stationary with $h_Y\defon{T_Y}$ constant.
                 \item $\exists T.~|\underbrace{\{Y \in A_S : T_Y = T\}}_{A_{S,T}}| = \aleph_{ \omega_1 + 1}$.
-                \item Let $\{\beta\}  = g_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$.
+                \item Let $\{\beta\}  = h_Y''T$, i.e.~$\overline{f}_Y(\alpha) < \aleph_{\beta}$ for $Y\in A_{S,T}, \alpha \in T$.
                 \item $\leftindex^T \aleph_\beta \le 2^{\aleph_\beta \cdot \aleph_1} = \aleph_{\beta+1} \aleph_2 < \aleph_{ \omega_1}$.
                 \item $\exists \tilde{f}\colon T \to \aleph_\beta .~ |\underbrace{\{Y \in A_{S,T} : \overline{f}_Y\defon{T} = \tilde{f}\} }_{A_{S,T,\tilde{f}}}| = \aleph_{ \omega_1 + 1}$.
                 \item $Y,Y' \in A_{S,T,\tilde{f}} \implies \forall \alpha \in T.~f_Y(\alpha) = f_{Y'}\left( \alpha \right) \overset{T \text{ unbounded}}{\implies} Y = Y'$
@@ -270,7 +270,7 @@ We will only proof
                 \item $P \coloneqq \{Y \subseteq \aleph_{ \omega_1} : \exists  i < \aleph_{ \omega_1 + 1}.~Y \le X_i\}$
             \end{itemize}
         \item $|P| \overset{\text{in fact } =}{\le} \aleph_{ \omega_1 + 1} \aleph_{ \omega_1} = \aleph_{ \omega_1 + 1}$ by (3).
-        \item $X \in \cP(\aleph_{ \omega_1}) \setminus M \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
+        \item $X \in \cP(\aleph_{ \omega_1}) \setminus P \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
             $\lightning$ (3).
             Thus $P = \cP(\aleph_{ \omega_1})$.
     \end{itemize}