inputs/lecture_13.tex

@@ -39,6 +39,7 @@
     is not a well-order on a countable set.
 
     Thus $\otp(\faktor{W}{\sim}, <) = \omega_1$.
+    \todo{move this}
 \end{remark}
 }{}
 

inputs/lecture_14.tex

@@ -150,20 +150,11 @@ We have shown (assuming \AxC to choose contained clubs):
     The \vocab{diagonal intersection},
     is defined to be
     \[
-    \diagi_{\beta < \alpha} A_{\beta} \coloneqq
+    \diagi_{\beta < \alpha} A_{\beta} \coloneqq 
     \{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\}  \}
     = \bigcap_{\beta < \alpha} ([0,\beta] \cup A_\beta)
     \]
 \end{definition}
-\begin{remark}+
-    \label{rem:diagiclosed}
-    Note that if $A$ is closed,
-    so is $[0,\alpha] \cup A$.
-    Since the intersection of arbitrarily many
-    closed sets is closed,
-    we get that the diagonal intersection
-    of closed sets is closed.
-\end{remark}
 \begin{lemma}
     \label{lem:diagiclub}
     Let $\kappa$ be a regular, uncountable cardinal.
@@ -190,23 +181,22 @@ We have shown (assuming \AxC to choose contained clubs):
         $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$.
     \end{claim}
     \begin{subproof}
-        Cf.~\yaref{rem:diagiclosed}.
+        Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$
-        % Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$
+        is unbounded in $\gamma$.
-        % is unbounded in $\gamma$.
+        We aim to show that $\gamma \in  \diagi_{\beta < \kappa} D_{\beta}$.
-        % We aim to show that $\gamma \in  \diagi_{\beta < \kappa} D_{\beta}$.
+        Let $\beta_0 < \gamma$.
-        % Let $\beta_0 < \gamma$.
+        We need to see that $\gamma \in D_{\beta_0}$.
-        % We need to see that $\gamma \in D_{\beta_0}$.
 
-        % For each $\beta_0 \le \beta' < \gamma$
+        For each $\beta_0 \le \beta' < \gamma$
-        % there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
+        there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
-        % such that $\beta' \le \beta'' < \gamma$,
+        such that $\beta' \le \beta'' < \gamma$,
-        % since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$.
+        since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$.
-        % In particular $\beta'' \in D_{\beta_0}$.
+        In particular $\beta'' \in D_{\beta_0}$.
 
-        % So $D_{\beta_0} \cap \gamma$
+        So $D_{\beta_0} \cap \gamma$
-        % is unbounded in $\gamma$.
+        is unbounded in $\gamma$.
-        % Since $D_{\beta_0}$ is closed
+        Since $D_{\beta_0}$ is closed
-        % it follows that $\gamma \in D_{\beta_0}$.
+        it follows that $\gamma \in D_{\beta_0}$.
 
         %As $\beta_0 < \gamma$ was arbitrary,
         %this shows that $\gamma \in \diagi_{\beta < n} D_\beta$.
@@ -280,9 +270,9 @@ We have shown (assuming \AxC to choose contained clubs):
 \end{refproof}
 \begin{remark}+
     $\diagi_{\beta < \kappa} C_{\beta}$ actually
-    \emph{is} a club,
+    \emph{is} a club:
-    since $\diagi_{\beta < \kappa} C_\beta$ is closed,
+    It suffices to show that $\diagi_{\beta < \kappa} C_\beta$ is closed.
-    again cf.~\yaref{rem:diagiclosed}.
+    This can be shown in the same way as for $\diagi_{\beta < \kappa} D_\beta$.
 %     Let $\lambda < \kappa$ be a limit ordinal.
 %     Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
 %     Then there exists $\alpha < \lambda$ such that
@@ -298,20 +288,6 @@ We have shown (assuming \AxC to choose contained clubs):
     iff $C \cap S \neq \emptyset$
     for every club $C \subseteq \kappa$.
 \end{definition}
-\begin{remark}+[\url{https://mathoverflow.net/q/37503}]
-    Informally, club sets and stationary sets
-    can be viewed as large sets of a measure space
-    of measure $1$.
-    Clubs behave similarly to sets of measure $1$
-    and stationary sets are analogous to
-    sets of positive measure:
-    \begin{itemize}
-        \item Every club is stationary,
-        \item the intersection of two clubs is a club,
-        \item the intersection of a club and a stationary set is stationary,
-        \item there exist disjoint stationary sets.
-    \end{itemize}
-\end{remark}
 \begin{example}
     \begin{itemize}
         \item Every $D \subseteq \kappa$ which is club in $\kappa$

inputs/lecture_16.tex

@@ -43,6 +43,7 @@ Recall that $F \subseteq \cP(\kappa)$ is a filter if
 $X,Y \in F \implies X \cap Y \in F$,
 $X \in F, X \subseteq Y \subseteq \kappa \implies Y \in F$
 and $\emptyset \not\in F, \kappa \in F$.
+\todo{Move this to the definition of filter?}
 }{}
 \begin{definition}
     A filter $F$ is an \vocab{ultrafilter} 
@@ -103,7 +104,11 @@ one cofinality.
 
 \begin{refproof}{thm:solovay}%
 \gist{%
-    \footnote{``This is one of the arguments where it is certainly worth it to look at it again.''}
+    %\footnote{``This is one of the arguments where it is certainly
+    %    worth it to look at it again''}
+    % TODO: Look at this again and think about it.
+    % TODO TODO TODO
+
     We will only prove this for $\aleph_1$.
     Fix $S \subseteq \aleph_1$ stationary.
 
@@ -139,7 +144,7 @@ one cofinality.
     \end{claim}
     \begin{subproof}
         Otherwise for all $n < \omega$,
-        there is a  $\delta$ such that
+        there is a  $\delta$ such that 
         $\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$
         is nonstationary.
         Let $\delta_n$ be the least such $\delta$.
@@ -191,23 +196,20 @@ one cofinality.
     $\gamma_n^{\alpha} = \delta'$.
 
     Write $\delta_i = \delta'$ and $T_i = T$.
-
+    
-    By construction, all the $T_i$ are stationary.
+    \begin{claim}
-    Since the $\delta_i$ are strictly increasing
+        \label{thm:solovay:p:c2}
-    and since $\gamma_n^{\alpha} = \delta_i$ for all $\alpha \in T_i$,
+        Each $T_i$ is stationary
-    we have that the $T_i$ are disjoint.
+        and if $i \neq j$, then $T_i \cap T_j = \emptyset$.
-    % \begin{claim}
+        \footnote{maybe this should not be a claim}
-    %     \label{thm:solovay:p:c2}
+    \end{claim}
-    %     Each $T_i$ is stationary
+    \begin{subproof}
-    %     and if $i \neq j$, then $T_i \cap T_j = \emptyset$.
+        The first part is true by construction.
-    % \end{claim}
+        Let $j < i$.
-    % \begin{subproof}
+        Then if $\alpha \in T_i$, $\alpha' \in T_j$,
-    %     The first part is true by construction.
+        we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$
-    %     Let $j < i$.
+        hence $\alpha \neq  \alpha'$.
-    %     Then if $\alpha \in T_i$, $\alpha' \in T_j$,
+    \end{subproof}
-    %     we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$
-    %     hence $\alpha \neq  \alpha'$.
-    % \end{subproof}
 
     Now let
     \[
@@ -230,6 +232,8 @@ one cofinality.
         \item $\exists  n < \omega.~\forall  \delta < \omega_1: \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\}$
             stationary:
             \begin{itemize}
+                % TODO THINK!
+                % TODO TODO TODO
                 \item Otherwise $\forall n < \omega.~\exists \delta.~\{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\} $ nonstationary.
                 \item $\delta_n\coloneqq $ least such $\delta$,
                     $C_n$ club s.t.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$.

inputs/lecture_17.tex

@@ -1,6 +1,9 @@
 \lecture{17}{2023-12-14}{Silver's Theorem}
 
 We now want to prove \yaref{thm:silver}.
+% More generally, if $\kappa$ is a singular cardinal of uncountable cofinality
+% such that $2^{\lambda} = \lambda^+$ for all $\lambda < \kappa$,
+% then $2^{\kappa} = \kappa^+$.
 
 \gist{%
 \begin{remark}
@@ -9,7 +12,7 @@ We now want to prove \yaref{thm:silver}.
 \end{remark}
 }{}
 
-We will only prove
+We will only proof
 \gist{%
     \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$
     (see \yaref{thm:silver1}).
@@ -150,7 +153,6 @@ We will only prove
 
         Let $\beta < \omega_1$ be such that for all $Y \in A_2$
         and for all $\alpha \in T$, $h_Y(\alpha) = \beta$.
-        % TODO WHY DOES THIS WORK?
         Then $\overline{f}_Y(\alpha) < \aleph_\beta$
         for all $Y \in A_2$ and $\alpha \in T$.
 

inputs/lecture_18.tex

@@ -27,6 +27,7 @@
     Since $2^{\lambda} < \kappa$,
     \AxPow works.
     The other axioms are trivial.
+    \todo{Exercise}
 \end{proof}
 \begin{corollary}
     $\ZFC$ does not prove the existence of inaccessible
@@ -39,7 +40,7 @@
 \end{proof}
 
 \begin{definition}[Ulam]
-    A cardinal $\kappa > \aleph_0$ is \vocab{measurable}
+    A cardinal $\kappa > \aleph_0$ is \vocab{measurable} 
     iff there is an ultrafilter $U$ on $\kappa$,
     such that $U$ is not principal\gist{\footnote{%
         i.e.~$\{\xi\}  \not\in U$ for all $\xi < \kappa$%

inputs/lecture_19.tex

@@ -79,7 +79,6 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
 A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas
 and \vocab{downwards absoluteness} for $\Pi_1$-formulas:
 \begin{lemma}
-    \label{lem:pi1downardsabsolute}
     Let $M$ be transitive.
     Let $\phi(x_0,\ldots,x_n) \in \cL_\in$ and $a_0,\ldots,a_n \in M$.
     Then

inputs/lecture_22.tex

@@ -112,7 +112,7 @@ is well founded.
 
             The formula $\forall x.~\forall y .~((\forall  z \in x.~z \in y \land \forall z \in y.~z \in x) \to  x = y)$
             is $\Pi_1$, hence it is true in $M[g]$
-            by \yaref{lem:pi1downardsabsolute}.
+            by %TODO REF downward absolutenes.
         \item \AxFund:
             Again,
              \[
@@ -137,7 +137,7 @@ is well founded.
             so $M[g] \models \text{``$\{x,y\}$ is the pair of $x$ and $y$''}$.
             Hence $M[g] \models \AxPair$.
         \item \AxUnion:
-            Similar to \AxPair.
+            Similar to \AxPair.\gist{\todo{Exercise}}{}
     \end{itemize}
 \end{proof}
 
@@ -194,6 +194,7 @@ Still missing are
             \[
             p  \Vdash^{\mathbb{P}}_M \phi(\tau_1,\ldots, \tau_k).
             \]
+            
     \end{enumerate}
 \end{theorem}
 \begin{proof}