diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 29c8cf2..b9d0fec 100644 --- a/inputs/lecture_12.tex +++ b/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 $\alpha + 1 = \alpha \cup \{\alpha\}$ diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index a277354..d5c4feb 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -77,7 +77,7 @@ Then \[ \sum_{i \in I} \kappa_i < \prod_{i \in I} \lambda_i. - \] + \] \end{theorem} \begin{proof} \gist{% @@ -88,7 +88,7 @@ such that for all $\eta < \kappa_i$ \[ \underbrace{F((\eta, i))(i)}_{\in \lambda_i} \neq \xi. - \] + \] Such $\xi$ exists, since $\kappa_i < \lambda_i$. Let $f \in \bigtimes_{i \in I} \lambda_i$ @@ -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 $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$. diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 84abc52..0e3f3c5 100644 --- a/inputs/lecture_14.tex +++ b/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} diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 1335f07..5c651c2 100644 --- a/inputs/lecture_15.tex +++ b/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.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \neq$. - \item $\diagi_{\nu < \alpha} C_\nu$ is club. + i.e.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \emptyset$. + \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 $\zeta < \xi$. @@ -207,12 +206,12 @@ Let's do a second proof of \yaref{thm:fodor}. \[ X_\alpha \models D \text{ is club in $\kappa$} \land D \cap T = \emptyset, \] - hence + hence \[ V_\theta \models D \text{ is club in $\kappa$} \land D \cap T = \emptyset. \] In other words, - there is some club + there is some club $D \in X_\alpha$ with $D \cap T = \emptyset$. @@ -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} - - diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index e6104e3..8891ee3 100644 --- a/inputs/lecture_16.tex +++ b/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. +}{} diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index 95da71b..3cc3123 100644 --- a/inputs/lecture_17.tex +++ b/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 -such that $2^{\lambda} = \lambda^+$ for all $\lambda < \kappa$, -then $2^{\kappa} = \kappa^+$. +% 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} The hypothesis of \yaref{thm:silver} is consistent with $\ZFC$. \end{remark} +}{} -We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$. -The general proof differs only in notation. - +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 - \[ - \{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\} = \cP(\aleph_{\omega_1 + 1}) - \] - because if $X \subseteq \aleph_{\omega_1 + 1}$ + On the other hand + $P = \cP(\aleph_{\omega_1})$ + because if $X \subseteq \aleph_{\omega_1}$ 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}