some small changes, gist

inputs/lecture_04.tex

@@ -2,7 +2,7 @@
 
 % Model-theoretic concepts and ultraproducts
 
-\section{$\ZFC$}
+\section{\texorpdfstring{$\ZFC$}{ZFC}}
 
 % 1900, Russel's paradox
 \todo{Russel's Paradox}

inputs/lecture_05.tex

@@ -104,12 +104,12 @@
 
     We write $f\colon d \to  b$.
     The set of all function from $d$ to $b$
-    is denoted by ${}^d b$ or $b^d$.
+    is denoted by $\leftindex^d b$ or $b^d$.
 \end{definition}
 
 \begin{fact}
     Given sets $d, b$ then
-    ${}^d b$ exists.
+    $\leftindex^d b$ exists.
 \end{fact}
 \begin{proof}
     Apply again \AxAus over $\cP(d \times b)$.

inputs/lecture_08.tex

@@ -103,7 +103,7 @@ Furthermore $F\,''a \coloneqq  \{y : \exists  x \in  a .~(x,y) \in F\}$.
     and $\phi$ does not have quantifiers
     ranging over classes.%
     \footnote{If one removes the restriction regarding
-        quantifiers another theory, called
+        quantifiers, another theory, called
         \vocab{Morse-Kelly} set theory, is obtained.}
 \end{axiom}
 
@@ -113,7 +113,8 @@ Furthermore $F\,''a \coloneqq  \{y : \exists  x \in  a .~(x,y) \in F\}$.
 
 \todo{the following was actually done in lecture 9}
 
-$\BGC$ (in German often NBG) is defined to be  $\BG$
+$\BGC$ (in German often NBG\footnote{\vocab{Neumann-Bernays-Gödel}})
+is defined to be  $\BG$
 together with the additional axiom:
 \begin{axiom}[Choice]
     \[
@@ -212,7 +213,7 @@ $\ZFC \vdash \phi$ then $\BGC \vdash \phi$.
           &&~ ~\overline{g}(0) = x \land \forall m \in  \omega.~(m+1 \in \dom(\overline{g}) \implies \overline{g}(m+1) = f(\overline{g}(m)))\}.
     \end{IEEEeqnarray*}
     $G$ exists as it can be obtained by \AxAus
-    from ${}^{< \omega}M$.
+    from $\leftindex^{< \omega}M$.
     By induction,
     for every $n \in  \omega$,
     there is a $\overline{g} \in G$

inputs/lecture_09.tex

@@ -48,6 +48,7 @@ An alternative way of formulating this is
 \end{definition}
 
 \begin{theorem}
+    \yalabel{Recursion Theorem}{recursion}{thm:recursion}
     Let $R$ be a well-founded
     and set-like relation on $A$ (i.e.~$R \subseteq  A \times A$).
 
@@ -59,11 +60,12 @@ An alternative way of formulating this is
     Then there is a unique function $f$ on $A$
     such that for all $x \in A$,
     \[
-        (F\defon{ \{y \in A : (y,x) \in R\}}, x, F(x)) \in D.
+        (F\defon{ \{y \in A : (y,x) \in R\}}, x, F(x)) \in D\gist{,}{.}
     \]
-    I.e.~$F(x)$ is computed from $F\defon{\{y \in A: (y,x) \in R\}}$.
+    \gist{i.e.~$F(x)$ is computed from $F\defon{\{y \in A: (y,x) \in R\}}$.}{}
 \end{theorem}
 \begin{proof}
+\gist{%
     Uniqueness:
 
     Let $F, F'$ be two such functions.
@@ -106,7 +108,7 @@ An alternative way of formulating this is
 
     If $x \in \dom(F)$ and $y \in A$ with $(y,x) \in R$,
     then $y \in \dom(F)$
-    and $(F\defon{\{y : (y,x) \in R\}}, x,F(x)) \in D$.
+    and \[(F\defon{\{y : (y,x) \in R\}}, x,F(x)) \in D.\]
 
     We need to show that $\dom(F) = A$.
     This holds by induction:
@@ -121,8 +123,26 @@ An alternative way of formulating this is
     But then $f \cup (x,z)$,
     where $z$ is unique such that $(f\defon{\{y : (y,x) \in R\}}, x, z) \in D$,
     is good $\lightning$.
+}{
+    \begin{itemize}
+        \item Uniqueness:
+
+            Consider $\overline{A} \coloneqq \{x \in A : F(x) \neq  F'(x)\}$
+            for two such functions. If $\overline{A} \neq \emptyset$,
+            there is a minimal element $\lightning$.
+        \item Existence:
+
+            \begin{itemize}
+                \item call set-function $f$ \emph{good} iff:
+                    \begin{itemize}
+                        \item $\dom(f) \subseteq A$,
+                        \item $x \in \dom(f), y <_R x \implies y \in \dom(f)$,
+                        \item $\forall x \in \dom(f).~(f\defon{\{y \in A : y <_R x\}}, x, f(x)) \in D$.
+                    \end{itemize}
+                \item $F \coloneqq \bigcup \{f : f \text{ good}\}$ (comprehension)
+                \item $\dom F = A$ (induction).
+            \end{itemize}
+
+    \end{itemize}
+}
 \end{proof}
-
-
-
-

inputs/lecture_10.tex

@@ -1,6 +1,5 @@
 \lecture{10}{}{} % Mirko
-
+\subsubsection{Applications of induction and recursion}
-Applications of induction and recursion:
 \begin{fact}
     For every set $x$ there is a transitive set $t$
     such that $x \in t$.
@@ -14,7 +13,7 @@ Applications of induction and recursion:
     Once we have such a function,
      $\{x\} \cup \bigcup \ran(F)$
      is a set as desired.
-     To get this $F$ using the recursion theorem,
+     To get this $F$ using the \yaref{thm:recursion},
      pick $D$ such that
      \[
          (\emptyset, 0, \{x\}) \in D
@@ -23,7 +22,7 @@ Applications of induction and recursion:
      \[
          (f, n+1, \bigcup\bigcup \ran(f)) \in D.
      \]
-     The recursion theorem then gives a function
+     The \yaref{thm:recursion} then gives a function
      such that
      \begin{IEEEeqnarray*}{rCl}
          F(0) &=& \{x\},\\
@@ -51,21 +50,30 @@ Applications of induction and recursion:
 }{}
 \begin{lemma}
     There is a function $F\colon \OR \to  V$
-    such that $F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.
+    such that
+    \gist{$F(\alpha) = \bigcup \{\cP(F(\beta)): \beta < \alpha\}$.}{%
+        $F(\alpha) = \bigcup_{\beta < \alpha} \cP(F(\beta))$.
+    }
+    \gist{}{%
+        $F(\alpha)$ is denoted by $V_\alpha$.
+        These are called the \vocab{rank initial segments} of $V$.
+    }
 \end{lemma}
 \begin{proof}
-    Use the recursion theorem with $R = \in $
+    Use the \yaref{thm:recursion} with $R = \in $
     and $(w,x,y) \in D$ iff
     \[
     y = \bigcup \{\cP(\overline{y}) : \overline{y} \in \ran(w)\}.
     \]
-    This function has the following properties:
+    \gist{%
-    \begin{IEEEeqnarray*}{rCl}
+        This function has the following properties:
-        F(0) &=& \bigcup \emptyset = \emptyset,\\
+        \begin{IEEEeqnarray*}{rCl}
-        F(1) &=& \bigcup \{\cP(\emptyset)\} = \bigcup \{\{\emptyset\} \} = \{\emptyset\},\\
+            F(0) &=& \bigcup \emptyset = \emptyset,\\
-        F(2) &=& \bigcup \{\cP(\emptyset), \cP(\{\emptyset\})\} = \bigcup \{\{\emptyset\}, \{\emptyset, \{\emptyset\} \} \} = \{\emptyset, \{\emptyset\} \},\\
+            F(1) &=& \bigcup \{\cP(\emptyset)\} = \bigcup \{\{\emptyset\} \} = \{\emptyset\},\\
-        \ldots
+            F(2) &=& \bigcup \{\cP(\emptyset), \cP(\{\emptyset\})\} = \bigcup \{\{\emptyset\}, \{\emptyset, \{\emptyset\} \} \} = \{\emptyset, \{\emptyset\} \},\\
-    \end{IEEEeqnarray*}
+            \ldots
+        \end{IEEEeqnarray*}
+    }{}
 
     It is easy to prove by induction:
     \begin{enumerate}[(a)]
@@ -76,16 +84,23 @@ Applications of induction and recursion:
             for $\lambda \in \OR$ a limit.
     \end{enumerate}
 \end{proof}
+\gist{%
 \begin{notation}
-Usually, one writes $V_\alpha$ for $F(\alpha)$.
+    Usually, one writes $V_\alpha$ for $F(\alpha)$.
     They are called the \vocab{rank initial segments}  of $V$.
 \end{notation}
+}{}
 \begin{lemma}
+\gist{%
     If $x$ is any set, then there is some $\alpha \in \OR$
     such that $x \in V_\alpha$,
     i.e.~$V = \bigcup \{V_{\alpha} : \alpha \in \OR\}$.
+}{
+    $V = \bigcup_{\alpha \in \OR} V_\alpha$.
+}
 \end{lemma}
 \begin{proof}
+\gist{%
     We use induction on the well-founded $\in$-relation.
     Let $A = \bigcup \{V_\alpha : \alpha \in \OR\}$.
     We need to show that $A = V$.
@@ -103,6 +118,7 @@ Usually, one writes $V_\alpha$ for $F(\alpha)$.
 
     In other words $x \subseteq V_\alpha$,
     hence $x \in V_{\alpha+1}$.
+}{Induction on $\in$.}
 \end{proof}
 
 \begin{lemma}[\vocab{Transitive collapse}/\vocab{Mostowski collapse}]
@@ -118,29 +134,43 @@ Usually, one writes $V_\alpha$ for $F(\alpha)$.
     with $x \in y \iff (F(x),F(y)) \in R$ for $x,y \in B$.
 \end{lemma}
 \begin{proof}
-    ``$\impliedby$'' Suppose that $R$ is ill-founded
+    ``$\impliedby$''
-    (i.e.~not well-founded).
+    \gist{%
-    Then there is some $(y_n : n < \omega)$ such that $y_n  \in A$
+        Suppose that $R$ is ill-founded
-    and $(y_{n+1}, y_n) \in R$ for all $n < \omega$.
+        (i.e.~not well-founded).
-    But then if $F$ is an isomorphism as above,
+        Then there is some $(y_n : n < \omega)$ such that $y_n  \in A$
-    \[
+        and $(y_{n+1}, y_n) \in R$ for all $n < \omega$.
-    F^{-1}(Y_{n+1}) \in F^{-1}(Y_n)
+        But then if $F$ is an isomorphism as above,
-    \]
+        \[
-    for all $n < \omega$ $\lightning$
+        F^{-1}(Y_{n+1}) \in F^{-1}(Y_n)
+        \]
+        for all $n < \omega$ $\lightning$
+    }{%
+        Trivial, since $\in$ is well-founded.
+    }
 
-    ``$\implies$ '' Suppose that $R$ is well-founded.
+    ``$\implies$ ''
-    We want a transitive class $B$ and a function $F\colon  B \leftrightarrow A$
+    \gist{%
-    such that
+        Suppose that $R$ is well-founded.
-    \[
+        We want a transitive class $B$ and a function $F\colon  B \leftrightarrow A$
-    x \in y \iff (F(x), F(y)) \in R.
+        such that
-    \]
+        \[
-    Equivalently $G\colon  A \leftrightarrow B$
+        x \in y \iff (F(x), F(y)) \in R.
-    with $(x,y) \in R$ iff $G(x) \in G(y)$ for all $x,y \in A$.
+        \]
+        Equivalently $G\colon  A \leftrightarrow B$
+        with $(x,y) \in R$ iff $G(x) \in G(y)$ for all $x,y \in A$.
 
-    In other words, $G(y) = \{G(x) : (x,y) \in R\}$.
+        In other words, $G(y) = \{G(x) : (x,y) \in R\}$.
-    Such a function $G$ and class $B$ exist by the recursion theorem.
+        Such a function $G$ and class $B$ exist by the \yaref{thm:recursion}.
+    }{Use recursion to define $F(y) \coloneqq \{F(x) : (x,y) \in R\}$.}
 \end{proof}
 
+As a consequence of the \yaref{lem:mostowski},
+we get that if $<$ is a well-order on a set $a$
+then there is some transitive set $b$
+with $(b, \in\defon{b}) \cong (a, <)$.
+
+
 \begin{lemma}[\vocab{Rank function}]
     Let $R$ be a well-founded and set-like binary relation
     on a class $A$.
@@ -149,7 +179,7 @@ Usually, one writes $V_\alpha$ for $F(\alpha)$.
     \[(x,y) \in R \implies F(x) < F(y).\]
 \end{lemma}
 \begin{proof}
-    By the recursion theorem,
+    By the \yaref{thm:recursion},
     there is $F$ such that
     \[
     F(y) = \sup \{F(x) + 1 : (x,y) \in R\}.

inputs/lecture_11.tex

@@ -2,12 +2,6 @@
 
 \subsection{Cardinals}
 
-Consequence of the Mostowski collapse:
-
-If $<$ is a well-order on a set $a$
-then there is some transitive $b$
-with $(b, \in\defon{b}) \cong (a, <)$.
-
 \begin{definition}
     Let $a$ be any set.
     The \vocab{cardinality} of $a$
@@ -16,8 +10,10 @@ with $(b, \in\defon{b}) \cong (a, <)$.
     such that there is some bijection $f\colon  \alpha \to a$.
 
     An ordinal $\alpha$ is called a \vocab{cardinal},
-    iff there is some set $a$ with $|a| = \alpha$
+    iff \gist{%
-    (equivalently, $|\alpha| = \alpha$).
+        there is some set $a$ with $|a| = \alpha$
+        (equivalently, $|\alpha| = \alpha$).%
+    }{$|\alpha| = \alpha$.}
 \end{definition}
 
 We often write $\kappa, \lambda, \ldots$ for cardinals.
@@ -27,23 +23,28 @@ We often write $\kappa, \lambda, \ldots$ for cardinals.
     there is come cardinal $\lambda > \kappa$.
 \end{lemma}
 \begin{proof}
-    Consider the powerset of  $\kappa$.
+    \gist{%
-    We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$.
+        Consider the powerset of  $\kappa$.
-    Hence $\kappa < |2^{\kappa}|$.
+        We know that there is no surjection $\kappa \twoheadrightarrow \cP(\kappa)$.
+        Hence $\kappa < |2^{\kappa}|$.
+    }{$\kappa < |2^{\kappa}|$.}
 \end{proof}
 
 \begin{definition}
     For each cardinal $\kappa$,
     $\kappa^+$ denotes the least cardinal $\lambda > \kappa$.
 \end{definition}
+\gist{%
 \begin{warning}
     This has nothing to do with the ordinal successor of $\kappa$.
 \end{warning}
+}{}
 \begin{lemma}
     Let $X$ be any set of cardinals.
     Then $\sup X$ is a cardinal.
 \end{lemma}
 \begin{proof}
+\gist{%
     If there is some $\kappa \in X$ with $\lambda \le \kappa$
     for all $\lambda \in X$,
     then $\kappa = \sup(X)$ is a cardinal.
@@ -62,8 +63,19 @@ We often write $\kappa, \lambda, \ldots$ for cardinals.
     However, there exists $\mu \twoheadrightarrow \sup(X)$,
     hence also $\mu \twoheadrightarrow \lambda$
     (which is in contradiction to $\lambda$ being a cardinal).
+}{%
+    \begin{itemize}
+        \item If $\sup(X) \in X$ this is trivial.
+        \item Let $\sup(X) \not\in X$, $\mu \coloneqq  |\sup(X)|$.
+          \begin{itemize}
+              \item Suppose that $\sup(X)$ is not a cardinal.
+                  Then $\mu \in \sup(X)$, since $\sup(X) \in \OR$.
+              \item $\exists  \lambda \in X.~\lambda > \mu$ $\lightning$.
+          \end{itemize}
+    \end{itemize}
+}
 \end{proof}
-We may now use the recursion theorem
+We may now use the \yaref{lem:recursion}
 to define a sequence $\langle \aleph_\alpha : \alpha \in \OR \rangle$
 with the following properties:
 \begin{IEEEeqnarray*}{rCl}
@@ -72,7 +84,7 @@ with the following properties:
     \aleph_{\lambda} &=& \sup \{\aleph_\alpha : \alpha < \lambda\}.
 \end{IEEEeqnarray*}
 Each $\aleph_\alpha$ is a cardinal.
-Also, a trivial induction show that $\alpha \le \aleph_\alpha$.
+Also, a trivial induction shows that $\alpha \le \aleph_\alpha$.
 In particular $|\alpha| \le \aleph_{\alpha}$.
 Therefore the $\aleph_\alpha$ are all the infinite cardinals:
 If $a$ is any infinite set, then $|a| \le  \aleph_{|a|}$,
@@ -84,7 +96,7 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
 \end{notation}
 
 \begin{notation}
-    Let ${}^a b \coloneqq  \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
+    Let $\leftindex^a b \coloneqq  \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
 \end{notation}
 
 \begin{definition}[Cardinal arithmetic]
@@ -93,40 +105,44 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
     \begin{IEEEeqnarray*}{rCl}
         \kappa + \lambda &\coloneqq & |\{0\} \times \kappa \cup \{1\} \times \lambda|,\\
         \kappa \cdot \lambda &\coloneqq & |\kappa \times  \lambda|,\\
-        \kappa^{\lambda} &\coloneqq & |{}^{\lambda}\kappa|.
+        \kappa^{\lambda} &\coloneqq & |\leftindex^{\lambda}\kappa|.
     \end{IEEEeqnarray*}
 \end{definition}
+\gist{%
 \begin{warning}
     This is very different from ordinal arithmetic!
 \end{warning}
+}{}
 
 \begin{theorem}[Hessenberg]
     \label{thm:hessenberg}
     For all $\alpha$ we have
     \[
     \aleph_\alpha \cdot \aleph_\alpha = \aleph_\alpha.
-    \] 
+    \]
-
 \end{theorem}
 \begin{corollary}
     For all $\alpha, \beta$ it is
     \[
     \aleph_\alpha + \aleph_\beta = \aleph_\alpha \cdot  \aleph_\beta = \max \{\aleph_\alpha, \aleph_\beta\}.
-    \] 
+    \]
 \end{corollary}
 \begin{proof}
     Wlog.~$\alpha \le \beta$.
-    Trivially $\aleph_\alpha \le \aleph_\beta$.
+    \gist{%
-    It is also clear that
+        Trivially $\aleph_\alpha \le \aleph_\beta$.
-    \[
+        It is also clear that
-    \aleph_{\beta} \le  \aleph_\alpha + \aleph_{\beta} \le  \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta.
+        \[
-    \] 
+            \aleph_{\beta} \le  \aleph_\alpha + \aleph_{\beta} \le  \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta.
+        \]
+    }{Then $\aleph_{\beta} \le  \aleph_\alpha + \aleph_{\beta} \le  \aleph_\alpha \cdot \aleph_\beta \le \aleph_\beta \cdot \aleph_\beta = \aleph_\beta$.}
 \end{proof}
 \begin{refproof}{thm:hessenberg}
+\gist{%
     Define a well-order $<^\ast$ on $\OR \times \OR$ by setting
     \[
     (\alpha,\beta) <^\ast (\gamma,\delta)
-    \] 
+    \]
     iff
     \begin{itemize}
         \item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
@@ -139,15 +155,15 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
     There is an isomorphism
     \[
         (\OR, <) \cong^{\Gamma^{-1}} (\OR \times  \OR, <^\ast).
-    \] 
+    \]
     $\Gamma$ is called the \vocab{Gödel pairing function}.
     \begin{claim}
         For all $\alpha$ it is
-        $\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})) = \aleph_\alpha$,
+        $\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) = \aleph_\alpha$,
         i.e.
         \[
         \aleph_\alpha = \{\xi: \exists  \eta, \eta' < \aleph_\alpha.~\xi = \Gamma((\eta,\eta'))\}.
-        \] 
+        \]
     \end{claim}
     \begin{subproof}
         We use induction of $\alpha$.
@@ -173,7 +189,7 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
         If $(\gamma,\delta) <^\ast (\eta, \eta')$,
         then $\max \{\gamma,\delta\} \le \max \{\eta, \eta'\}$.
         Say $\eta \le  \eta' < \aleph_\alpha$
-        and let $\aleph_\alpha = |\eta'|$.
+        and let $\aleph_\beta = |\eta'|$.
         There is a surjection
         \[f\colon  \underbrace{(\eta +1)}_{ \le \aleph_\beta} \times \underbrace{(\eta' + 1)}_{\sim  \aleph_\beta} \twoheadrightarrow \aleph_\alpha.\]
 
@@ -181,15 +197,38 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
         The inductive hypothesis then produces a surjection
         $f^\ast\colon  \aleph_\beta \to  \aleph_\alpha \lightning$.
     \end{subproof}
+}{
+    \begin{itemize}
+        \item Define wellorder $<^\ast \subseteq \OR \times \OR$,
+            $(\alpha,\beta) <^\ast (\gamma,\delta)$ iff
+            \begin{itemize}
+                \item $\max(\alpha,\beta) < \max(\gamma,\delta)$ or
+                \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha < \gamma$ or
+                \item $\max(\alpha,\beta) = \max(\gamma,\delta)$ and $\alpha = \gamma$ and $\beta < \delta$.
+            \end{itemize}
+        \item Gödel pairing function $\Gamma : (\OR \times \OR, <^\ast) \xrightarrow{\cong} (\OR, <)$.
+        \item $\forall \alpha.~\ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha}) = \aleph_\alpha$.
+            \begin{itemize}
+                \item Induction on $\alpha$.
+                \item Clearly $\aleph_\alpha \subseteq \ran(\Gamma\defon{\aleph_\alpha \times \aleph_\alpha})$
+                    ($\aleph_\alpha$ is a cardinal).
+                \item Suppose $\aleph_\alpha \subsetneq \ran(\Gamma\defon{\aleph_\alpha \times  \aleph_\alpha})$,
+                    We get a surjection $\aleph_\beta \xrightarrow{\text{induction}} \aleph_\beta \times  \aleph_\beta \to \aleph_\alpha$
+                    for some $\beta < \alpha$.
+            \end{itemize}
+    \end{itemize}
+}
 \end{refproof}
+\gist{%
 However, exponentiation of cardinals is far from trivial:
 \begin{observe}
     $2^{\kappa} = |\cP(\kappa)|$,
-    since ${}^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$.
+    since $\leftindex^{\kappa} \{0,1\} \leftrightarrow\cP(\kappa)$.
 
     Hence by Cantor $2^{\kappa}\ge \kappa^+$.
 \end{observe}
 This is basically all we can say.
+}{}
 
 The \vocab{continuum hypothesis} states that $2^{\aleph_0} = \aleph_1$.
 

inputs/lecture_12.tex

@@ -26,17 +26,20 @@ and
     \beta^{\lambda} &\coloneqq & \sup_{\alpha < \lambda} \beta^{\alpha} &~ ~\text{for limit ordinals $\lambda$}.
 \end{IEEEeqnarray*}
 
+\gist{%
 \begin{example}
     \leavevmode
     \begin{itemize}
         \item $2+ 2 =4$,
+        \item $196883 + 1 = 196884$,
         \item $1 + \omega = \sup_{n < \omega} 1 + n = \omega \neq \omega +1$,
         \item $2 \cdot \omega = \sup_{n < \omega} 2\cdot n = \omega$,
-        \item $\omega \cdot 2 =\omega \cdot  1 + \omega = \omega + \omega$..
+        \item $\omega \cdot 2 =\omega \cdot  1 + \omega = \omega + \omega$.
     \end{itemize}
 \end{example}
+}{}
 
-
+\gist{%
 \begin{warning}
     Cardinal arithmetic and ordinal arithmetic are very different!
     The symbols are the same, but usually we will distinguish
@@ -46,6 +49,7 @@ and
 \end{warning}
 
 We will very rarely use ordinal arithmetic.
+}{}
 
 \subsection{Cofinality}
 
@@ -100,7 +104,7 @@ We will very rarely use ordinal arithmetic.
     \begin{itemize}
         \item $\cf(\aleph_\omega) = \omega$.
             In fact $\cf(\aleph_{\lambda}) \le \lambda$
-            for limit ordinals $\lambda$
+            for limit ordinals $\lambda \neq 0$
             (consider $\alpha \mapsto \aleph_\alpha$).
         \item $\cf(\aleph_{\omega + \omega}) = \omega$.
     \end{itemize}
@@ -110,20 +114,24 @@ We will very rarely use ordinal arithmetic.
     $\cf(\beta)$ is a cardinal.
 \end{lemma}
 \begin{proof}
+\gist{%
     Let $f\colon \alpha \to \beta$ be cofinal.
     Then $\tilde{f}\colon |\alpha| \to \beta$,
     the composition with $\alpha \leftrightarrow|\alpha|$
     is cofinal as well and $|\alpha| \le \alpha$.
+}{Trivial.}
 \end{proof}
 
+\gist{%
 \begin{question}
     How does one imagine ordinals with
     cofinality $> \omega$?
 \end{question}
 No idea.
+}{}
 
 \begin{definition}
-    An ordinal $\beta$ is \vocab{regular} 
+    An ordinal $\beta$ is \vocab{regular}
     iff $\cf(\beta) = \beta$.
     Otherwise $\beta$ is called \vocab{singular}.
 \end{definition}
@@ -172,6 +180,7 @@ In particular, a regular ordinal is always a cardinal.
     by \yaref{thm:hessenberg}.
 \end{proof}
 
+\gist{%
 \begin{itemize}
     \item $\aleph_0, \aleph_1, \aleph_2, \ldots$ are regular,
     \item $\aleph_\omega$ is singular,
@@ -184,11 +193,13 @@ In particular, a regular ordinal is always a cardinal.
     \item $\aleph_{\omega_1 + 1}, \ldots$ is regular,
     \item $\aleph_{\omega_2}$ is singular.
 \end{itemize}
+}{}
 
 \begin{question}[Hausdorff]
     Is there a regular limit cardinal?
 \end{question}
-Maybe. This is independent of $\ZFC$.
+Maybe. This is independent of $\ZFC$,
+cf.~\yaref{def:inaccessible}.
 
 
 \begin{theorem}[Hausdorff]
@@ -197,9 +208,10 @@ Maybe. This is independent of $\ZFC$.
     \]
 \end{theorem}
 \begin{proof}
+\gist{%
     Recall that
     \begin{IEEEeqnarray*}{rCl}
-        \aleph_{\alpha+1}^{\aleph_\beta} &=& |{}^{\aleph_\beta} \aleph_{\alpha+1}|.
+        \aleph_{\alpha+1}^{\aleph_\beta} &=& |\leftindex^{\aleph_\beta} \aleph_{\alpha+1}|.
     \end{IEEEeqnarray*}
     \begin{itemize}
         \item First case: $\beta \ge \alpha+1$.
@@ -209,8 +221,8 @@ Maybe. This is independent of $\ZFC$.
             \le \left( 2^{\aleph_{\beta}} \right)^{\aleph_\beta}
             = 2^{\aleph_{\beta} \cdot \aleph_\beta} = 2^{\aleph_\beta} \le  \aleph_{\alpha+1}^{\aleph_\beta}.
             \]
-            Also $\aleph_\alpha^{\aleph_{\beta} = 2^{\aleph_\beta}}$
+            Also $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$
-            in this case,
+            in this case (by the same argument),
             so
             \[
             \aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}}
@@ -224,14 +236,35 @@ Maybe. This is independent of $\ZFC$.
             is unbounded.
             So
             \[
-            {}^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} {}^{\aleph_\beta} \xi
+            \leftindex^{\aleph_{\beta}}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} \leftindex^{\aleph_\beta} \xi
             \]
             for each $\xi < \aleph_{\alpha+1}$, $|\xi| \le \aleph_\alpha$,
             hence
             \[
-            |{}^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
+            |\leftindex^{\aleph_{\beta}}\xi| \le \aleph_\alpha^{\aleph_\beta}
             \]
             for each $\xi < \aleph_{\alpha+1}$.
             Therefore, \[\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \cdot \aleph_{\alpha}^{\aleph_{\beta}} \le \aleph_{\alpha+1}^{\aleph_\beta}.\]
     \end{itemize}
+}{%
+    $\ge$ is trivial.
+    \begin{itemize}
+        \item First case: $\beta \ge \alpha + 1$.
+            Note that $\gamma \le \beta \implies \aleph_{\gamma}^{\aleph_\beta} = 2^{\aleph_\beta}$,
+            so
+            \[
+                \aleph_{\alpha+1}^{\aleph_\beta} \overset{\alpha+1 \le \beta}{=}
+                2^{\aleph_\beta}
+                \overset{\alpha < \beta}{=} \aleph_\alpha^{\aleph_\beta}
+                \le \aleph_\alpha^{\aleph_\beta} \cdot \aleph_{\alpha+1}.
+            \]
+        \item Second case: $\beta < \alpha+1$:
+            \begin{itemize}
+                \item $\aleph_{\alpha+1}$ is regular, so all $f\colon \aleph_\beta \to  \aleph_{\alpha+1}$ are bounded.
+                \item Thus $\leftindex^{\aleph_\beta}\aleph_{\alpha+1} = \bigcup_{\xi < \aleph_{\alpha+1}} \leftindex^{\aleph_\beta} \xi$ for all $\xi < \aleph_{\alpha+1}$.
+                \item So $\aleph_{\alpha+1}^{\aleph_\beta} \le \aleph_{\alpha+1} \aleph_\alpha^{\aleph_\beta}$.
+            \end{itemize}
+            
+    \end{itemize}
+}
 \end{proof}

inputs/lecture_13.tex

@@ -1,5 +1,5 @@
 \lecture{13}{2023-11-30}{}
-
+\gist{%
 \begin{remark}[``Constructive'' approach to $\omega_1$ ]
     There are many well-orders on $\omega$.
     Let $W$ be the set of all such well-orders.
@@ -41,7 +41,9 @@
     Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
     \todo{move this}
 \end{remark}
+}{}
 
+\gist{%
 \begin{notation}
     Let $I \neq \emptyset$
     and let $\{\kappa_i : i \in I\}$
@@ -57,13 +59,13 @@
     \] 
     where 
     \[
-    \bigtimes_{i \in I} A_i \coloneqq \{f : f \text{is a function with domain  $I$ and $f(i) \in A_i$}\}.
+    \bigtimes_{i \in I} A_i \coloneqq \{f : f \text{ is a function},  \dom(f) = I, \forall i.~f(i) \in A_i\}.
     \] 
 \end{notation}
 \begin{remark}
     \AxC is equivalent to $\forall i \in I.~A_i \neq \emptyset \implies \bigtimes_{i \in I} A_i \neq \emptyset$.
 \end{remark}
-
+}{}
 \begin{theorem}[K\H{o}nig]
     \yalabel{K\H{o}nig's Theorem}{K\H{o}nig}{thm:koenig}
     Let $I \neq \emptyset$.
@@ -78,6 +80,7 @@
     \] 
 \end{theorem}
 \begin{proof}
+\gist{%
     Consider a function $F\colon \bigcup_{i \in I} (\kappa_i \times \{i\}) \to \bigtimes_{i \in I}\lambda_i$.
     We want to show that $F$ is not surjective.
 
@@ -92,6 +95,14 @@
     be defined by $i \mapsto \xi_i$.
 
     Then $f \not\in \ran(F)$.
+}{%
+    \begin{itemize}
+        \item Consider $F\colon \bigcup_{i \in I}(\kappa_i \times \{i\}) \to \bigtimes_{i \in I} \lambda_i$. Want: $F$ is not surjective.
+        \item Let $\xi_i$ minimal such that $\underbrace{F((\eta,i))(i)}_{\in \lambda_i} \neq \xi$
+            for all $\eta < \kappa_i$.
+        \item $(i \mapsto \eta_i) \not\in \ran(F)$.
+    \end{itemize}
+}
 \end{proof}
 
 \begin{corollary}
@@ -99,6 +110,7 @@
     it is $\cf(2^{\kappa}) > \kappa$.
 \end{corollary}
 \begin{proof}
+\gist{%
     If $2^{\kappa}$ is a successor cardinal,
     then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$,
     since successor cardinals are regular.
@@ -115,6 +127,19 @@
     \sup \{\kappa_i : i  < \kappa\} \le \sum_{i \in \kappa} \kappa_i < \prod_{i \in \kappa} \lambda_i = \left( 2^{\kappa} \right)^{\kappa} = 2^{\kappa \cdot \kappa} = 2^{\kappa}
     \]
     and $f$ is not cofinal.
+}{%
+    \begin{itemize}
+        \item Trivial for successors ($\cf(2^{\kappa}) \overset{\text{regular}}{=} 2^{\kappa} > \kappa$).
+        \item Suppose $2^\kappa$ is a limit, $\exists  f\colon \kappa \to 2^{\kappa}$ cofinal.
+            Wlog.~$f(i) \in \Card$.
+            But
+            \[
+            \sup \{f(i)) : i < \kappa\} \le \sum_{i \in \kappa} f(i)
+            \overset{\yaref{thm:koenig}}{<} \prod_{i \in \kappa} 2^{\kappa}
+            = (2^{\kappa})^{\kappa} = 2^{\kappa}. ~ ~\qedhere
+            \]
+    \end{itemize}
+}
 \end{proof}
 
 \begin{fact}

inputs/lecture_14.tex

@@ -1,4 +1,5 @@
 \lecture{14}{2023-12-04}{}
+\gist{%
 \begin{abuse}
     Sometimes we say club
     instead of club in  $\kappa$.
@@ -17,6 +18,7 @@
              is club in $\kappa$.
     \end{itemize}
 \end{example}
+}{}
 \begin{lemma}
     \label{lem:clubintersection}
     Let $\kappa$ be regular and uncountable.
@@ -34,12 +36,12 @@
     Let $C_{\beta} \coloneqq  \{\xi : \xi > \beta\}$.
     Clearly this is club
     but $\bigcap_{\beta < \kappa} C_\beta = \emptyset$.
-    
 \end{warning}
 \begin{refproof}{lem:clubintersection}
+\gist{%
     First let $\alpha = 2$.
     Let $C, D \subseteq \kappa$
-    be a club.
+    be club.
     $C \cap D$ is trivially closed:
 
     Let $\beta < \kappa$. Suppose that $(C \cap D) \cap \beta$
@@ -84,11 +86,21 @@
     = \sup_{i < \omega} \gamma_{\alpha \cdot n + \beta + 1} \in \bigcap_{\beta < \alpha} C_\beta$,
     where we have used that.
     $\cf(\kappa) > \alpha \cdot \omega$.
-    
+}{%
+    \begin{itemize}
+        \item Trivially closed.
+        \item Recursively define $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$, by
+            $\gamma_0 \coloneqq \gamma$,
+    \[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\]
+    and $\sup$ at limits.
+    \item Then $\sup_{i < \alpha\cdot \omega} \gamma_i \in \bigcap_{\beta < \alpha} C_\beta$
+        (we used $\cf(\kappa) > \alpha\cdot \omega$).
+    \end{itemize}
+}
 \end{refproof}
 
 \begin{definition}
-    $F \subseteq  \cP(a)$ is a \vocab{filter} 
+    $F \subseteq  \cP(a)$ is a \vocab{filter}
     iff
     \begin{enumerate}[(a)]
         \item $X,Y \in F \implies X \cap Y \in F$,
@@ -106,14 +118,16 @@
     iff for all $\gamma < \alpha$ and $\{X_\beta : \beta < \gamma\} \subseteq F$
     then $\bigcap \{X_\beta : \beta < \gamma\} \in F$.
 \end{definition}
+\gist{%
 Intuitively, a filter is a collection of ``big'' subsets of  $a$.
+}{}
 
 \begin{definition}
     Let $\kappa$ be regular and uncountable.
     The \vocab{club filter} is defined as
     \[
     \cF_{\kappa} \coloneqq  \{X \subseteq  \kappa : \exists \text{ club } C \subseteq  \kappa .~ C \subseteq X\}.
-    \] 
+    \]
 \end{definition}
 Clearly this is a filter.
 
@@ -126,7 +140,7 @@ We have shown (assuming \AxC to choose contained clubs):
     Clearly $\emptyset \not\in \cF_\kappa$,
     $\kappa \in \cF_\kappa$,
     and $A \in \cF_{\kappa}, A \subseteq B \in \kappa \implies B \in \cF_\kappa$.
-    In \autoref{lem:clubintersection} we are going to show that the intersection
+    In \autoref{lem:clubintersection} showed that the intersection
     of $< \kappa$ many clubs is club.
 \end{proof}
 
@@ -138,7 +152,7 @@ We have shown (assuming \AxC to choose contained clubs):
     \[
     \diagi_{\beta < \alpha} A_{\beta} \coloneqq 
     \{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\}  \}.
-    \] 
+    \]
 \end{definition}
 \begin{lemma}
     \label{lem:diagiclub}
@@ -146,9 +160,11 @@ We have shown (assuming \AxC to choose contained clubs):
     If $\langle C_{\beta} : \beta < \kappa \rangle$
     is a sequence of club subsets of $\kappa$,
     then $\diagi_{\beta < \kappa} C_{\beta}$
-    contains a club. % TODO: contains or is?
+    contains a club.
 \end{lemma}
-\begin{proof}
+\begin{refproof}{lem:diagiclub}
+    % TODO THINK
+\gist{
     Let us fix $\langle C_{\beta} : \beta < \alpha \rangle$.
     Write $D_{\beta} \coloneqq  \bigcap \{C_{\gamma} : \gamma \le  \beta\} $
     for $\beta < \kappa$.
@@ -192,7 +208,7 @@ We have shown (assuming \AxC to choose contained clubs):
     \begin{subproof}
         Fix $\gamma < \kappa$.
         We need to find $\delta > \gamma$
-        with $\gamma \in \diagi_{\beta < \kappa} D_\beta$.
+        with $\delta \in \diagi_{\beta < \kappa} D_\beta$.
 
         Define $\langle \gamma_n : n < \omega \rangle$
         as follows:
@@ -212,20 +228,57 @@ We have shown (assuming \AxC to choose contained clubs):
         for some $n < \omega$.
         For $m \ge  n$, $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_{\overline{\gamma}}$.
         So $D_{\overline{\gamma}} \cap \delta$ is unbounded
-        in $\gamma$, hence $\delta \in  D_{\overline{\gamma}}$.
+        in $\delta$, hence $\delta \in  D_{\overline{\gamma}}$.
     \end{subproof}
-\end{proof}
+}{%
+    \begin{itemize}
+        \item Fix $\langle C_\beta : \beta < \alpha \rangle$.
+            Set $D_{\beta} \coloneqq  \bigcap_{\gamma \le \beta} D_{\gamma}$.
+            It suffices to analyze $D_{\beta}$.
+        \item $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$:
+            \begin{itemize}
+                \item Let $\gamma < \kappa$ such that $\left( \diagi_{\beta < \kappa} D_{\beta}\right) \cap \gamma$ unbounded in $\gamma$.
+                    Want $\gamma \in \diagi_{\beta < \kappa} D_\beta$.
+                \item Let $\beta_0 < \gamma$. Want $\gamma \in D_{\beta_0}$.
+                \item $D_{\beta_0} \cap \gamma$ is unbounded in $\gamma$
+                    ($D_{\beta_0} \setminus \beta_0 \supseteq \diagi_{\beta < \kappa} D_{\beta} \setminus \beta_0$)
+                    $\overset{D_{\beta_0} \text{ closed}}{\implies} \gamma \in D_{\beta_0}$.
+            \end{itemize}
+        \item $\diagi_{\beta < \kappa} D_\beta$ is unbounded in $\kappa$:
+            \begin{itemize}
+                \item Fix $\gamma < \kappa$. We need to find $\diagi_{\beta < \kappa} D_\beta \ni \delta > \gamma$.
+                \item Define $\langle \gamma_n : n < \omega \rangle$
+                    by $\gamma_0 \coloneqq  \gamma$,
+                    $\gamma_{n+1} \coloneqq  \min D_{\gamma_n} \setminus (\gamma_n + 1)$,
+                    $\delta \coloneqq  \sup_n \gamma_n \overset{\cf(\kappa) > \omega}{<} \kappa$
+                \item Want $\delta \in \diagi_{\beta < \kappa} D_\beta$,
+                    i.e.~$\forall \epsilon < \delta.~\delta\in D_\epsilon$.
+
+                    If $\epsilon < \delta$, then $\epsilon \le \gamma_n$
+                    for $n$ large enough,
+                    so $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_\epsilon$
+                    for $m \ge n$.
+                    Thus $\sup(D_\epsilon \cap \delta) = \delta$
+                    $\overset{D_\epsilon \text{ closed}}{\implies} \delta \in D_{\epsilon}$.
+            \end{itemize}
+
+
+    \end{itemize}
+
+}
+\end{refproof}
 \begin{remark}+
-    $\diagi_{\beta < \kappa} D_{\beta}$ actually
+    $\diagi_{\beta < \kappa} C_{\beta}$ actually
     \emph{is} a club:
-    It suffices to show that $\diagi_{\beta < \kappa} D_\beta$ is closed.
+    It suffices to show that $\diagi_{\beta < \kappa} C_\beta$ is closed.
-    Let $\lambda < \kappa$ be a limit ordinal.
+    This can be shown in the same way as for $\diagi_{\beta < \kappa} D_\beta$.
-    Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
+%     Let $\lambda < \kappa$ be a limit ordinal.
-    Then there exists $\alpha < \lambda$ such that
+%     Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
-    $\lambda \not\in D_\alpha$.
+%     Then there exists $\alpha < \lambda$ such that
-    Since $D_\alpha$ is closed,
+%     $\lambda \not\in D_\alpha$.
-    we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
+%     Since $D_\alpha$ is closed,
-    In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$.
+%     we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
+%     In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$.
 \end{remark}
 
 \begin{definition}

inputs/lecture_15.tex

@@ -28,6 +28,7 @@
     $f(\alpha) = \nu$ for all $\alpha \in T$.
 \end{theorem}
 \begin{proof}
+\gist{%
     Let $S, f$ be given.
     For $\nu < \kappa$ set $S_\nu \coloneqq  \{\alpha \in S : f(\alpha) = \nu\}$. % f^{-1}(\nu)$.
     We aim to show that one of the $S_\nu$ is stationary.
@@ -41,6 +42,16 @@
     Hence $f(\alpha) \neq  \nu$ for all $\nu < \alpha$,
     so $f(\alpha) \ge \alpha$.
     But $f$ is regressive $\lightning$
+}{%
+    \begin{itemize}
+        \item For $\nu < \kappa$ set $S_\nu \coloneqq  \{\alpha \in S : f(\alpha) = \nu\}$.
+        \item Suppose none of the $S_\nu$ is stationary,
+            i.e.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \neq$.
+        \item $\diagi_{\nu < \alpha} C_\nu$ is club.
+        \item Pick $\alpha \in C \cap S$.
+            But then $\forall  \nu < \alpha.~\alpha \in C_\nu$, i.e.~$f(\alpha) \ge \alpha$.
+    \end{itemize}
+}
 \end{proof}
 
 \subsection{Some model theory and a second proof of Fodor's Theorem}
@@ -70,7 +81,7 @@ Recall the following:
     over $V_\theta$ for $\phi$
     is a function
     \[
-    f\colon {}^k V_\theta \to V_\theta,
+    f\colon \leftindex^k V_\theta \to V_\theta,
     \]
     where $k$ is the number of free variables of
     $\exists v.~\phi$
@@ -91,9 +102,10 @@ to be an elementary substructure of $V_\theta$.
     (where $k$ is the number of free variables of $\exists v.~\phi$)
     $f_\phi(x_1,\ldots,x_k) \in X$,
     then $X \prec V_{\theta}$.
-    
 \end{lemma}
 
+% TODO ANKI-MARKER
+
 Let's do a second proof of \yaref{thm:fodor}.
 \begin{refproof}{thm:fodor}
     Fix $\theta > \kappa$ and look at $V_{\theta}$.

inputs/lecture_16.tex

@@ -42,6 +42,7 @@ and $\emptyset \not\in F, \kappa \in F$.
     either $X \in F$ or $\kappa \setminus X \in F$.
 \end{definition}
 
+\gist{%
 \begin{example}
     Examples of filters:
     \begin{enumerate}[(a)]
@@ -55,6 +56,7 @@ and $\emptyset \not\in F, \kappa \in F$.
             Then $\cF_\kappa \coloneqq  \{X \subseteq  \kappa: \exists  C \subseteq \kappa \text{ club in $\kappa$}. C \subseteq X\}$.
     \end{enumerate}
 \end{example}
+}{Let $\cF_\kappa \coloneqq  \{X \subseteq \kappa : \exists C \subseteq  \kappa \text{ club in } \kappa\}$.}
 \begin{question}
     Is $\cF_\kappa$ an ultrafilter?
 \end{question}

inputs/lecture_18.tex

@@ -1,6 +1,7 @@
 \lecture{18}{2023-12-18}{Large cardinals}
 
 \begin{definition}
+    \label{def:inaccessible}
     \begin{itemize}
         \item A cardinal $\kappa$ is called \vocab{weakly inaccessible}
             iff $\kappa$ is uncountable,\footnote{dropping this we would get that $\aleph_0$ is inaccessible}
@@ -19,7 +20,7 @@
 
 \begin{theorem}
     If $\kappa$ is inaccessible,
-    then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in ) \models\ZFC$.}
+    then $V_\kappa \models \ZFC$.\footnote{More formally $(V_{\kappa}, \in\defon{V_\kappa}) \models\ZFC$.}
 \end{theorem}
 \begin{proof}
     Since $\kappa$ is regular, \AxRep works.
@@ -145,11 +146,11 @@
 
     1. $\implies$ 2.
     Fix $U$.
-    Let ${}^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
+    Let $\leftindex^{\kappa}V$ be the class of all function from $\kappa$ to $V$.
-    For $f,g \in {}^{\kappa}V$ define
+    For $f,g \in \leftindex^{\kappa}V$ define
     $f \sim g :\iff \{\xi < \kappa : f(\xi) = g(\xi)\} \in U$.
     This is an equivalence relation since $U$ is a filter.
-    Write $[f] = \{g \in {}^\kappa V : g \sim ~ f \land g \in V_\alpha \text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
+    Write $[f] = \{g \in \leftindex^\kappa V : g \sim ~ f \land g \in V_\alpha \text{ for the least $\alpha$ such that there is some $h \in V_\alpha$ with $h \sim f$}\}$.%
     \footnote{This is know as \vocab{Scott's Trick}.
         Note that by defining equivalence classes
         in the usual way (i.e.~without this trick),
@@ -164,7 +165,7 @@
 
     This is independent of the choice of the representatives,
     so it is well-defined.
-    Now write $\cF = \{[f] : f \in {}^{\kappa}V\}$
+    Now write $\cF = \{[f] : f \in \leftindex^{\kappa}V\}$
     and look at $(\cF, \tilde{\in})$.
 
     The key to the construction is \yaref{thm:los} (see below).
@@ -188,7 +189,7 @@ Then
 Let us show that $(\cF, \tilde{\in })$
 is well-founded.
 Otherwise there is $\langle f_n : n < \omega \rangle$
-such that $f_n \in {}^\kappa V$
+such that $f_n \in \leftindex^\kappa V$
 and $[f_{n+1}] \tilde{\in } [f_n]$ for all $n < \omega$.
 
 Then $X_n \coloneqq \{\xi < \kappa : f_{n+1}(\xi) \in  f_n(\xi)\} \in U$,
@@ -228,7 +229,7 @@ So $j(\alpha) \le \alpha$.
 
 \begin{theorem}[\L o\'s]
     \yalabel{\L o\'s's Theorem}{\L o\'s}{thm:los}
-    For all formulae $\phi$ and for all $f_1, \ldots, f_k \in {}^{\kappa} V$,
+    For all formulae $\phi$ and for all $f_1, \ldots, f_k \in \leftindex^{\kappa} V$,
     \[
         (\cF, \tilde{\in}) \models \phi([f_1], \ldots, [f_k])
         \iff \{\xi < \kappa : (V, \in ) \models \phi(f_1(\xi), \ldots, f_k(\xi))\} \in U.

logic.sty

@@ -24,6 +24,7 @@
 \usepackage{multirow}
 \usepackage{float}
 \usepackage{scalerel}
+\usepackage{leftindex}
 %\usepackage{algorithmicx}
 
 \newcounter{subsubsubsection}[subsubsection]