some small changes

inputs/lecture_08.tex

@@ -124,9 +124,12 @@ together with the additional axiom:
 \begin{fact}
     $\BGC$ is conservative over $\ZFC$,
     i.e.~for all formulae $\phi$ in the language
-    of set theory (only set variables):
+    of set theory (only set variables)
+    we have that
     if $\BGC \vdash \phi$ then $\ZFC \vdash \phi$.
 \end{fact}
+
+
 We cannot prove this fact at this point,
 as the proof requires forcing.
 The converse is easy however, i.e.~if
@@ -172,9 +175,9 @@ $\ZFC \vdash \phi$ then $\BGC \vdash \phi$.
 \end{lemma}
 \begin{proof}
     Suppose such sequence exists.
-    Then $\{x_n : n < \omega\}$
-    (this exists as by definition sequence of the $x_n$ is a function
-    and this set is the range of that function)
+    Then $\{x_n : n < \omega\}$%
+    \footnote{This exists as by definition the sequence $(x_n)$ is a function
+    $f\colon \omega \to V$ and this set is the image of $f$.}
     violates \AxFund.
 
     For the other direction let $M \neq \emptyset$ be some set.

inputs/lecture_09.tex

@@ -4,12 +4,12 @@
 \begin{definition}
     Let $R$  be a binary relation.
     $R$ is called \vocab{well-founded}
-    if for all classes $X$,
+    iff for all classes $X$,
     there is an $R$-least $y$ such that
     there is no $z \in X$ with $(z,y) \in R$.
 \end{definition}
 
-\begin{theorem}[Induction (again, but now with classes)]
+\begin{theorem}[Induction (again, but now for classes)]
     Suppose that $R$ is a well-founded relation.
     Let $X$ be a class such that for all sets $x$,
     \[
@@ -22,7 +22,7 @@
     Consider $Y = \{x : x \not\in X\} \neq \emptyset$.
     By hypothesis, there is some
     $x \in Y$ such that $(y,x) \not\in R$ for all $y  \in Y$.
-    In other words, if $(y,x) \not\in R$,
+    In other words, if $(y,x) \in R$,
     then $x \not\in Y$, i.e.~$x \in X$.
     Thus $\{y: (y,x) \in R\} \subseteq X$.
     Hence $x \in X \lightning$.

inputs/lecture_11.tex

@@ -52,7 +52,7 @@ We often write $\kappa, \lambda, \ldots$ for cardinals.
     Let us now assume that for all $\kappa \in X$
     there is some $\lambda \in X$
     with $\lambda > \kappa$.
-    Suppose that $\sup(X)$ is no a cardinal
+    Suppose that $\sup(X)$ is not a cardinal
     and write $\mu = |\sup(X)|$.
     Then $\mu \in \sup(X)$,
     since $\sup(X)$ is an ordinal.
@@ -96,7 +96,7 @@ so $|a| = \aleph_\beta$ for some $\beta \le |\alpha|$.
 \end{notation}
 
 \begin{notation}
-    Let $\leftindex^a b \coloneqq  \{f : f \text{ is a function}, \dom(f) = \alpha, \ran(f) \subseteq b\}$.
+    Let $\leftindex^a b \coloneqq  \{f : f \text{ is a function}, \dom(f) = a, \ran(f) \subseteq b\}$.
 \end{notation}
 
 \begin{definition}[Cardinal arithmetic]

inputs/lecture_12.tex

@@ -215,15 +215,17 @@ cf.~\yaref{def:inaccessible}.
     \end{IEEEeqnarray*}
     \begin{itemize}
         \item First case: $\beta \ge \alpha+1$.
-            Then
+            Note that for all $\gamma \le \beta$
+            we have
             \[
-            \aleph_{a+1}^{\aleph_\beta} \le  \aleph_{\beta}^{\aleph_\beta}
-            \le \left( 2^{\aleph_{\beta}} \right)^{\aleph_\beta}
-            = 2^{\aleph_{\beta} \cdot \aleph_\beta} = 2^{\aleph_\beta} \le  \aleph_{\alpha+1}^{\aleph_\beta}.
+            \aleph_{\gamma}^{\aleph_\beta} \le \aleph_\beta^{\aleph_\beta}
+            \le \left( 2^{\aleph_\beta} \right)^{\aleph_\beta}
+            = 2^{\aleph_\beta \cdot \aleph_\beta} = 2^{\aleph_{\beta}}
+            \le  \aleph_\gamma^{\aleph_\beta}.
             \]
-            Also $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$
-            in this case (by the same argument),
-            so
+            So in this case $\aleph_\alpha^{\aleph_{\beta}} = 2^{\aleph_\beta}$
+            and $\aleph_{\alpha+1}^{\aleph_\beta} = 2^{\aleph_{\beta}}$.
+            Thus
             \[
             \aleph_{\alpha+1}^{\aleph_{\beta}} = 2^{\aleph_{\beta}}
             = \aleph_{\alpha}^{\aleph_\beta} = \aleph_{\alpha}^{\aleph_\beta} \cdot  \aleph_{\alpha+1}.

inputs/lecture_13.tex

@@ -4,7 +4,7 @@
     There are many well-orders on $\omega$.
     Let $W$ be the set of all such well-orders.
     For $R, S \in W$,
-    write $R \le S$ if $R$ is isomorphic to
+    write $R \le S$ iff $R$ is isomorphic to
     an initial segment of $S$.
     Consider $\faktor{W}{\sim}$,
     where $R \sim S :\iff R \le S \land S \le R$.
@@ -15,7 +15,7 @@
     Suppose that $\{R_n : n \in \omega\} \subseteq W$
     is such that $R_{n+1} < R_n$.
     Then there exist $n_i \in \omega$
-    such that $R_i \cong R_0\defon{x <_{R_0} n_i}$
+    such that $R_i \cong R_0\defon{\{x : x <_{R_0} n_i\}}$
     and these form a $<_{R_0}$ strictly decreasing sequence.
 
     So $(\faktor{W}{\sim})$
@@ -116,13 +116,13 @@
     then $\cf(2^{\kappa}) = 2^{\kappa} > \kappa$,
     since successor cardinals are regular.
     
-    Suppose $\cf(2^{k}) \le  \kappa$ is a limit cardinal.
+    Suppose $\cf(2^{\kappa}) \le  \kappa$ is a limit cardinal.
     Then there is some cofinal $f\colon  \kappa\to 2^{\kappa}$.
     Write $\kappa_i = f(i)$
     (replacing $f(i)$ by $|f(i)|^{+}$ we may assume
     that every $\kappa_i$ is a cardinal).
 
-    For $i \in \kappa$, write $\lambda_i = 2^{k}$.
+    For $i \in \kappa$, write $\lambda_i = 2^{\kappa}$.
     By \yaref{thm:koenig},
     \[
     \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}
@@ -171,14 +171,14 @@ Relevant concepts to prove this theorem:
         \item We say that $A \subseteq \alpha$
             is \vocab{unbounded} (in $\alpha$),
             iff for all $\beta < \alpha$,
-            there is some $\gamma \in \alpha$
+            there is some $\gamma \in A$
             such that $\beta < \gamma$.
         \item We say that $A \subseteq \alpha$
             is \vocab{closed},
             iff it is closed with respect to the order topology on $\alpha$,
             i.e.~for all $\beta < \alpha$,
             \[
-            \sup(A \cap \beta) \in A.
+            \sup(A \cap \beta) \in A \cup \{0\} .
             \]
         \item $A$ is \vocab{club} (\emph{cl}osed \emph{un}bounded)
             iff it is closed and unbounded.

inputs/lecture_15.tex

@@ -136,7 +136,7 @@ Let's do a second proof of \yaref{thm:fodor}.
     \]
 
     Note that $|X_{\xi}| = |X_{\xi + 1}|$
-    but the size is increased at limits.
+    but the size may increase at limits.
     It is easy to see inductively that $|X_{\xi}| < \kappa$
     for every $\xi < \kappa$,
     while $X_\xi \subsetneq X_{\xi'}$
@@ -158,7 +158,7 @@ Let's do a second proof of \yaref{thm:fodor}.
         Let $\zeta < \kappa$.
         Let us define a strictly increasing sequence
         $ \langle \xi_n  : n < \omega \rangle$
-        a follows.
+        as follows.
         Set $\xi_0 \coloneqq \zeta$.
         Suppose $\xi_n$ has been chosen.
         Look at $X_{\xi_n} \cap \kappa$.

inputs/lecture_16.tex

@@ -41,7 +41,7 @@ where $f(\alpha) = \gamma$.
 \gist{%
 Recall that $F \subseteq \cP(\kappa)$ is a filter if
 $X,Y \in F \implies X \cap Y \in F$,
-$X \in X, X \subseteq Y \subseteq \kappa \implies Y \in F$
+$X \in F, X \subseteq Y \subseteq \kappa \implies Y \in F$
 and $\emptyset \not\in F, \kappa \in F$.
 \todo{Move this to the definition of filter?}
 }{}
@@ -86,7 +86,7 @@ one cofinality.
     If $S \subseteq \kappa$
     is stationary,
     there is a sequence $\langle S_i : i < \kappa \rangle$
-    of pairwise disjoint stationary sets of $\kappa$
+    of pairwise disjoint stationary subsets of $\kappa$
     such that $S = \bigcup S_i$.
 \end{theorem}
 \begin{corollary}
@@ -109,12 +109,12 @@ one cofinality.
     % TODO: Look at this again and think about it.
     % TODO TODO TODO
 
-    We will only proof this for $\aleph_1$.
+    We will only prove this for $\aleph_1$.
     Fix $S \subseteq \aleph_1$ stationary.
 
     For each $0 < \alpha < \omega_1$,
     either $\alpha$ is a successor ordinal
-    or $\alpha$ is a limit ordinal and $\cf(\alpha) = \omega_1$.
+    or $\alpha$ is a limit ordinal and $\cf(\alpha) = \omega$.
 
     Let $S^\ast \coloneqq  \{\alpha \in  S \setminus \{0\}  : \alpha \text{ is a limit ordinal}\} $.
     $S^\ast$ is still stationary:
@@ -163,7 +163,7 @@ one cofinality.
 
         Let $C' \coloneqq  C \setminus (\delta^\ast + 1)$.
         $C'$ is still club.
-        As $\delta^\ast$ is stationary,
+        As $S^\ast$ is stationary,
         we may pick some $\alpha \in S^\ast \cap C'$.
         But then $\gamma_n^{\alpha} > \delta^\ast$
         for $n$ large enough
@@ -201,6 +201,7 @@ one cofinality.
         \label{thm:solovay:p:c2}
         Each $T_i$ is stationary
         and if $i \neq j$, then $T_i \cap T_j = \emptyset$.
+        \footnote{maybe this should not be a claim}
     \end{claim}
     \begin{subproof}
         The first part is true by construction.
@@ -285,9 +286,9 @@ However we can say something about singular cardinals:
 Recall that $\CH$ says that $2^{\aleph_0} = \aleph_1$.
 So $\GCH \implies \CH$.
 
-\yalabel{thm:silver} says that if $\GCH$ is true below $\kappa$,
+\yaref{thm:silver} says that if $\GCH$ is true below $\kappa$,
 then it is true at $\kappa$.
 
-The proof of \yalabel{thm:silver} is quite elementary,
+The proof of \yaref{thm:silver} is quite elementary,
 so we will do it now, but the statement can only be fully appreciated later.
 }{}

inputs/lecture_18.tex

@@ -42,13 +42,14 @@
 \begin{definition}[Ulam]
     A cardinal $\kappa > \aleph_0$ is \vocab{measurable} 
     iff there is an ultrafilter $U$ on $\kappa$,
-    such that $U$ is not principal\footnote{%
+    such that $U$ is not principal\gist{\footnote{%
         i.e.~$\{\xi\}  \not\in U$ for all $\xi < \kappa$%
-    }
-    and
-    if $\theta < \kappa$
+    }}{}
+    and $< \kappa$-closed\gist{,%
+    i.e.~if $\theta < \kappa$
     and $\{X_i : i < \theta\} \subseteq  U$,
-    then $\bigcap_{i < \theta} X_i \in U$
+    then $\bigcap_{i < \theta} X_i \in U$.
+    }{.}
 \end{definition}
 
 \begin{goal}
@@ -77,7 +78,7 @@
 \end{theorem}
 \begin{proof}
     2. $\implies$ 1.:
-    Fox $j\colon  V \to M$.
+    Fix $j\colon  V \to M$.
     Let $U = \{X \subseteq  \kappa : \kappa \in j(X)\}$.
     We need to show that $U$ is an ultrafilter:
     \begin{itemize}