silver (unfinished gist)

inputs/lecture_12.tex

@@ -9,7 +9,7 @@ Fix an ordinal $\beta$.
 We recursively define
 \begin{IEEEeqnarray*}{rClClr}
     \beta &+& 0 &\coloneqq & \beta\\
-    \beta &+& (\alpha + 1)&\coloneqq &(\beta + \alpha),\\
+    \beta &+& (\alpha + 1)&\coloneqq &(\beta + \alpha) + 1,\\
     \beta &+& \lambda &\coloneqq & \sup_{\alpha < \lambda} \beta + \alpha &~ ~\text{for limit ordinals $\lambda$}
 \end{IEEEeqnarray*}
 (Recall that $\alpha + 1 = \alpha \cup \{\alpha\}$

inputs/lecture_13.tex

@@ -100,12 +100,13 @@
         \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 $(i \mapsto \eta_i) \not\in \ran(F)$.
+        \item $(i \mapsto \xi_i) \not\in \ran(F)$.
     \end{itemize}
 }
 \end{proof}
 
 \begin{corollary}
+    \label{cor:cfpow}
     For infinite cardinals $\kappa$,
     it is $\cf(2^{\kappa}) > \kappa$.
 \end{corollary}
@@ -155,7 +156,7 @@
 The next goal is to show the following:
 (However the method might be more interesting than the result)
 \begin{theorem}[Silver]
-    \yalabel{Silver's Theorem}{Silver}{thm:silver1}
+    \yalabel{Silver's Theorem (case of $\aleph_{\omega_1}$)}{Silver ($\aleph_{\omega_{1}}$)}{thm:silver1}
     If $2^{\aleph_\alpha} = \aleph_{\alpha + 1}$
     for all $\alpha < \omega_1$,
     then $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$.

inputs/lecture_14.tex

@@ -151,7 +151,8 @@ We have shown (assuming \AxC to choose contained clubs):
     is defined to be
     \[
     \diagi_{\beta < \alpha} A_{\beta} \coloneqq 
-    \{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\}  \}.
+    \{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\}  \}
+    = \bigcap_{\beta < \alpha} ([0,\beta] \cup A_\beta)
     \]
 \end{definition}
 \begin{lemma}

inputs/lecture_15.tex

@@ -35,7 +35,7 @@
     Suppose otherwise.
     Then for every $\nu$ there exists a club $C_\nu$
     such that  $S_\nu \cap C_\nu = \emptyset$.\footnote{Here we use \AxC to choose the $C_\nu$ uniformly.}
-    Let $C = \diagi_{\nu < \alpha} C_\nu$.
+    Let $C = \diagi_{\nu < \kappa} C_\nu$.
     By \yaref{lem:diagiclub} $C$ is a club.
     So we may pick some $\alpha \in C \cap S$.
     In particular $\alpha \in C_\nu$ for all $\nu < \alpha$.
@@ -46,8 +46,8 @@
     \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.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \neq$.
+            i.e.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \emptyset$.
-        \item $\diagi_{\nu < \alpha} C_\nu$ is club.
+        \item $\diagi_{\nu < \kappa} C_\nu$ 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}
@@ -104,10 +104,9 @@ to be an elementary substructure of $V_\theta$.
     then $X \prec V_{\theta}$.
 \end{lemma}
 
-% TODO ANKI-MARKER
-
 Let's do a second proof of \yaref{thm:fodor}.
 \begin{refproof}{thm:fodor}
+\gist{%
     Fix $\theta > \kappa$ and look at $V_{\theta}$.
 
     Fix $S \subseteq \kappa$ stationary
@@ -158,13 +157,13 @@ Let's do a second proof of \yaref{thm:fodor}.
         Let us show that $C$ is unbounded in $\kappa$.
         Let $\zeta < \kappa$.
         Let us define a strictly increasing sequence
-        $ \langle \xi_n  n < \omega \rangle$
+        $ \langle \xi_n  : n < \omega \rangle$
         a follows.
         Set $\xi_0 \coloneqq \zeta$.
         Suppose $\xi_n$ has been chosen.
         Look at $X_{\xi_n} \cap \kappa$.
         Since $|X_{\xi_n} \cap \kappa| < \kappa$,
-        $\sup (X_{\xi_n \cap \kappa}) < \kappa$.
+        $\sup (X_{\xi_n} \cap \kappa) < \kappa$.
         Set $\xi_{n+1} \coloneqq  \sup(X_{\xi_n} \cap \kappa) + 1$.
         Set $\xi \coloneqq \sup_{n<\omega} \xi_n$.
         Clearly $\zeta < \xi$.
@@ -234,8 +233,51 @@ Let's do a second proof of \yaref{thm:fodor}.
         \]
         Hence there is some $\eta \in X_\alpha$ with $\eta \in D$.
         This means that
-        $\xi < \underbrace{\eta}_{\in D} < \alpha$..
+        $\xi < \underbrace{\eta}_{\in D} < \alpha$.
     \end{subproof}
+}{%
+    \begin{itemize}
+        \item Fix $S \subseteq  \kappa$ stationary, $f\colon  S \to \kappa$ regressive.
+        \item Fix $\theta > \kappa$ look at $V_\theta$,
+            fix Skolem functions $f_\phi$ over  $V_{\theta}$ for all $\phi$.
+        \item Define sequence of elementary substructures $\langle X_\xi, \xi < \kappa \rangle$:
+             \begin{itemize}
+                \item $X_0$ minimal st.~closed under $f_\phi$ and $S,f \in X$ (note: countable).
+                \item $X_{\xi+1}$ minimal st.~$X_{\xi} \subseteq X_{\xi+1}$, closed under $f_\phi$ and
+                    $\min(\kappa \setminus X_\xi) \in X$. (note: $|X_\xi| = |X_{\xi + 1}|$)
+                \item $X_{\lambda} \coloneqq  \bigcup_{\xi < \lambda} X_\xi$
+                    (note: cardinality may increase)
+            \end{itemize}
+        \item Note: strictly increasing, $|X_\xi| < \kappa$, $\xi \subseteq X_\xi$.
+        \item $C\coloneqq  \{\xi < \kappa: X_\xi \cap \kappa = \xi\}$ is club:
+            \begin{itemize}
+                \item clearly closed.
+                \item unbounded in $\kappa$:
+                    \begin{itemize}
+                        \item Take  $\zeta < \kappa$, Define $\langle \xi_n : n < \omega \rangle$ by
+                            $\xi_0 \coloneqq  \zeta$, $\xi_{n+1} \coloneqq  \sup(X_{\xi_n} \cap \kappa)$
+                            ($ < \kappa$), $\xi \coloneqq  \sup \xi_n < \kappa$.
+                        \item $X_\xi \cap \kappa = \xi$, i.e.~$\xi \in C$:
+                            \begin{itemize}
+                                \item $\eta < \xi \implies \exists n.~\eta < \xi_n \implies \eta \in  \xi_n \subseteq X_{\xi_n} \subseteq X_\xi$ $\implies \xi \subseteq X_\xi \cap \kappa$.
+                                \item $\eta \in X_\xi \cap \kappa \implies \exists n.~\eta \in X_{\xi_n} \implies
+                                    \eta < \xi_{n+1} < \xi \implies X_\xi \cap \kappa \subseteq \xi$.
+                            \end{itemize}
+                    \end{itemize}
+            \end{itemize}
+        \item Take $\alpha \in S \cap C$, $\nu \coloneqq  f(\alpha) < \alpha$,
+            $T \coloneqq  \{\xi \in S : f(\xi) = \nu\}$.
+            ($T \in X_\alpha$, since it can be defined from $S, f, \nu$)
+        \item  $T$ is stationary:
+            Suppose $D \cap T = \emptyset$ for a club $D$.
+             $X_\alpha \prec V_\theta \implies$ find such $D \in X_\alpha$.
+        \item Clearly $\alpha \in T$. Want ($\lightning$) $\alpha \in D$:
+            \begin{itemize}
+                \item Suffices $\sup (D \cap \alpha) = \alpha$. Let $\xi < \alpha$.
+                \item $D$ unbounded, so $V_\theta \models \exists  \eta > \xi : \eta \in D$
+                \item $\implies X_\alpha \models \exists \eta > \xi .  ~\eta \in D$
+                \item $\implies \exists \eta > \xi. ~\eta \in D \land \underbrace{\eta < \alpha}_{\in X_\alpha}$.
+            \end{itemize}
+    \end{itemize}
+}
 \end{refproof}
-
-

inputs/lecture_16.tex

@@ -1,5 +1,8 @@
 \lecture{16}{2023-12-11}{}
 
+% TODO ANKI-MARKER
+
+
 Recall \yaref{thm:fodor}.
 \begin{question}
     What happens if $S$ is nonstationary?
@@ -8,6 +11,7 @@ Let $S \subseteq \kappa$ be nonstationary,
 $\kappa$ uncounable and regular.
 Then there is a club $C \subseteq  \kappa$
 with $C \cap S = \emptyset$.
+\gist{%
 Let us define $f\colon S \to  \kappa$ in the following way:
 
 If $\alpha \in S$ and $C \cap \alpha \neq \emptyset$,
@@ -29,13 +33,18 @@ where $\gamma' = \min(C \setminus (\gamma + 1))$.
 Thus for all $\gamma$,
 there is only an interval of ordinals $\alpha \in S$
 where $f(\alpha) = \gamma$.
+}{%
+    Consider $S\setminus \{0\}  \ni \alpha \mapsto \underbrace{\max(C \cap \alpha)}_{< \alpha}$
+    (here  $\max (\emptyset) \coloneqq 0$).
+}
 
-\todo{Move this to the definition of filter}
+\gist{%
 Recall that $F \subseteq \cP(\kappa)$ is a filter if
 $X,Y \in F \implies X \cap Y \in F$,
 $X \in X, X \subseteq Y \subseteq \kappa \implies Y \in F$
 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} 
     iff for all $X \subseteq \kappa$
@@ -69,7 +78,7 @@ So neither $S_0$ nor $S_1 \subseteq \kappa \setminus S_0$ contains a club.
 
 For $\kappa < \aleph_1$
 this argument does not work, since there is only
-on cofinality.
+one cofinality.
 
 \begin{theorem}[Solovay]
     \yalabel{Solovay's Theorem}{Solovay}{thm:solovay}
@@ -94,9 +103,11 @@ on cofinality.
 
 
 \begin{refproof}{thm:solovay}%
+\gist{%
     %\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 proof this for $\aleph_1$.
     Fix $S \subseteq \aleph_1$ stationary.
@@ -208,15 +219,54 @@ on cofinality.
     \]
     Then $\langle S_i : i < \omega_1 \rangle$ is
     as desired.
+}{%
+    (Proof for $\aleph_1$)
+    Fix $S \subseteq  \aleph_1$ stationary.
+    $\alpha \in (0, \omega_1)$ is successor ordinal or limit with $\cf(\alpha) = \omega_1$.
+    \begin{itemize}
+        \item $S^\ast \coloneqq  \{\alpha \in S \setminus \{0\}  : \alpha \text{ limit ordinal}\}$.
+            Still stationary:
+            $C$ club $\implies D = \{\alpha \in C \setminus \{0\} : \alpha \text{ limit}\}$ club.
+        \item For all $\alpha \in S^\ast$ choose $\langle \gamma^{\alpha}_n : n < \omega \rangle$ cofinal.
+        \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.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$.
+                    (i.e.~$\alpha \in S^\ast \cap C_n \implies \gamma_n^{\alpha} \le  \delta_n$).
+                \item $\delta^\ast \coloneqq \sup_{n < \omega} \delta_n$, $C \coloneqq \bigcap_{n < \omega} C_n$
+                    is club. $\alpha \in S^\ast \cap C \implies \forall n.~\gamma_n^\alpha \le \delta^\ast$.
+                \item $C' \coloneqq  C \setminus (\delta^\ast + 1)$ still club.
+                \item Pick $\alpha \in S^\ast \cap C'$.
+                    But $\gamma_n^{\alpha} < \delta^\ast$ is cofinal $\lightning$.
+            \end{itemize}
+        \item Consider $f\colon S^\ast \to \omega_1, \alpha \mapsto \gamma_n^{\alpha}$ (note: regressive)
+        \item Define $\langle \delta_i : i < \omega_1 \rangle$ strictly increasing:
+            \begin{itemize}
+                \item $\delta_0 \coloneqq  0$,
+                \item $\delta'_{i} \coloneqq  (\sup_{j < i} \delta_i) + 1$,
+                    $\{\alpha \in S^\ast : \gamma_n^\alpha > \delta'_i\}$
+                    is stationary $\overset{\yaref{thm:fodor}}{\implies} $ $\exists$ stationary $T_i \subseteq S^\ast$
+                    s.t.~$f''T_i = \{\delta_i\} $ constant.
+            \end{itemize}
+        \item $T_i$ are disjoint: $i \neq j \implies \underbrace{f''T_i}_{\{\delta_i\} } \cap \underbrace{f''T_j}_{\{\delta_j\}} = \emptyset$.
+        \item $S_0 \coloneqq T_0 \cup \left( S \setminus \bigcup_{j > 0} T_j \right)$, $S_i \coloneqq  T_i$.
+    \end{itemize}
+}
 \end{refproof}
-
+\gist{%
 We now want to do another application of \yaref{thm:fodor}.
 Recall that $2^{\kappa} > \kappa$, in fact $\cf(2^{\kappa}) > \kappa$
-by \yaref{thm:koenig}.
+by \yaref{thm:koenig}
+(cf.~\yaref{cor:cfpow}).
 
 Trivially, if $\kappa \le  \lambda$ then $2^{\kappa} \le  2^{\lambda}$.
 This is in some sense the only thing we can prove about successor cardinals.
 However we can say something about singular cardinals:
+}{}
 \begin{theorem}[Silver]
     \yalabel{Silver's Theorem}{Silver}{thm:silver}
 
@@ -231,6 +281,7 @@ However we can say something about singular cardinals:
     is the statement
     that $2^{\lambda} = \lambda^+$ holds for all infinite cardinals $\lambda$,
 \end{definition}
+\gist{%
 Recall that $\CH$ says that $2^{\aleph_0} = \aleph_1$.
 So $\GCH \implies \CH$.
 
@@ -239,3 +290,4 @@ then it is true at $\kappa$.
 
 The proof of \yalabel{thm:silver} is quite elementary,
 so we will do it now, but the statement can only be fully appreciated later.
+}{}

inputs/lecture_17.tex

@@ -1,18 +1,24 @@
 \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
+% More generally, if $\kappa$ is a singular cardinal of uncountable cofinality
-such that $2^{\lambda} = \lambda^+$ for all $\lambda < \kappa$,
+% such that $2^{\lambda} = \lambda^+$ for all $\lambda < \kappa$,
-then $2^{\kappa} = \kappa^+$.
+% then $2^{\kappa} = \kappa^+$.
 
+\gist{%
 \begin{remark}
     The hypothesis of \yaref{thm:silver}
     is consistent with $\ZFC$.
 \end{remark}
+}{}
 
-We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$.
+We will only proof
+\gist{%
+    \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$
+    (see \yaref{thm:silver1}).
     The general proof differs only in notation.
-
+}{\yaref{thm:silver1}.}
+\gist{%
 \begin{remark}
     It is important that the cofinality is uncountable.
     For example it is consistent
@@ -20,8 +26,10 @@ The general proof differs only in notation.
     $2^{\aleph_n} = \aleph_{n+1}$ for all $n < \omega$
     but at the same time $2^{\aleph_{\omega}} = \aleph_{\omega + 2}$.
 \end{remark}
+}{}
 
-\begin{refproof}{thm:silver}
+\begin{refproof}{thm:silver1}
+\gist{%
     We need to count the number of $X \subseteq \aleph_{\omega_1}.$
     Let us fix $\langle f_\lambda : \lambda < \kappa \text{ an infinite cardinal} \rangle$
     such that $f_{\lambda}\colon  \cP(\lambda) \to \lambda^+$
@@ -203,18 +211,58 @@ The general proof differs only in notation.
 
     The set
     \[
-    \{Y \subseteq \aleph_{\omega_1} : \exists  i < \aleph_{\omega_1 + 1} .~Y \le X_i\}
+    P \coloneqq \{Y \subseteq \aleph_{\omega_1} : \exists  i < \aleph_{\omega_1 + 1} .~Y \le X_i\}
     = \bigcup_{i < \aleph_{\omega_1 + 1}} \{Y \subseteq  \aleph_{\omega_1} : Y \le X_i\}
     \]
     has size $\le  \aleph_{\omega_1 + 1}$
     (in fact the size is exactly $\aleph_{\omega_1 + 1}$).
 
-    But
+    On the other hand
-    \[
+    $P = \cP(\aleph_{\omega_1})$
-    \{Y \subseteq \aleph_{\omega_1} : \exists  i < \aleph_{\omega_1 + 1} .~Y \le X_i\} = \cP(\aleph_{\omega_1 + 1})
+    because if $X \subseteq \aleph_{\omega_1}$
-    \]
-    because if $X \subseteq \aleph_{\omega_1 + 1}$
     is such that $X \nleq X_i$ for all $i < \aleph_{\omega_1 + 1}$,
     then $X_i \le X$ for all $i < \aleph_{\omega_1 + 1}$,
     so such a set $X$ does not exist by \yaref{thm:silver:p:c3}.
+}{Need to count $X \subseteq \aleph_{ \omega_{1}}$.
+    \begin{itemize}
+        \item Fix bijections $f_\lambda\colon 2^{\lambda} \to \lambda^+$
+            for all infinite cardinals $\lambda < \kappa$.
+        \item For $X \subseteq \aleph_{ \omega_1}$ define $f_X\colon \omega_1 \to \aleph_{ \omega_1},
+            \alpha \mapsto f_{\aleph_\alpha}(X \cap \aleph_\alpha)$.
+        \item (1) $X \neq Y \iff f_X \neq f_Y$:
+            \begin{itemize}
+            \item $X \neq Y \iff \exists \alpha.~X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha \iff \exists \alpha.~f_X(\alpha) \neq  f_Y(\alpha)$.
+            \end{itemize}
+        \item $X \le Y :\iff \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\}$ stationary.
+        \item (2) $X\le Y \lor Y \le X$:
+            Suppose $X \not\le Y, Y \not\le X$. Choose witnessing clubs $C, D$.
+            $C \cap D$ is club, but then $f_X(\alpha) \le f_Y(\alpha)$ or
+            $f_X(\alpha) \ge f_Y(\alpha)$ for $\alpha \in C \cap D$.
+        \item (3) $X \subseteq \aleph_{ \omega_1}$, then $|\underbrace{\{Y \subseteq \aleph_{ \omega_1} : Y \le X\}}_{A}| \le \aleph_{ \omega_1}$
+            \begin{itemize}
+                \item Suppose $|A| \ge \aleph_{ \omega_1 + 1}$.
+                \item $S_Y \coloneqq  \{\alpha : f_Y(\alpha) \le f_X(\alpha)\}$ stationary for all $Y \in A$.
+                    $2^{\aleph_1} = \aleph_2 \implies$ at most $\aleph_2$ such $S_Y$.
+                \item If $\forall  S \in \cP( \omega_1).~ |\underbrace{\{Y \in A : S_Y = S\}}_{A_S}| < \aleph_{ \omega_1 + 1}$,
+                    then $|A| \le \aleph_2 \cdot  <\aleph_{ \omega_1 + 1}$ $\lightning$  $\aleph_{ \omega_1 + 1}$ regular.
+                \item So $\exists S \in \omega_1.~|A_S| = \aleph_{ \omega_1 + 1 + 1}$.
+                    % TODO TODO TODO hier weiter!
+            \end{itemize}
+        \item Define sequence $\langle X_i : i < \aleph_{ \omega_1 + 1} \rangle$
+            of subsets of $\aleph_{\omega_1}$:
+            \begin{itemize}
+                \item Consider $\{Y \subseteq \aleph_{ \omega_1} : \exists  j < i.~Y \le X_j\}$
+                    (cardinality $ \le \aleph_{ \omega_1}$),
+                    Take $X_i \subseteq \aleph_{ \omega_1}$
+                    such that $X_i \subseteq \aleph_{ \omega_1}$
+                    such that $X_i \not\le X_j$ for all $j < i$.
+                \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 $X \in \cP(\aleph_{ \omega_1}) \setminus M \implies \forall i < \aleph_{ \omega_1 + 1}.~X_i \le X$
+            $\lightning$ (3).
+            Thus $P = \cP(\aleph_{ \omega_1})$.
+    \end{itemize}
+}
+
 \end{refproof}