\lecture{16}{2023-12-11}{} % TODO ANKI-MARKER Recall \yaref{thm:fodor}. \begin{question} What happens if $S$ is nonstationary? \end{question} 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 $C \cap \alpha \neq \emptyset$, then $\max(C \cap \alpha) < \alpha$. Define \[ f(\alpha) \coloneqq \begin{cases} 0 &: C \cap \alpha = \emptyset,\\ \max(C \cap \alpha) &:C\cap \alpha \neq \emptyset. \end{cases} \] For all $\alpha > 0$, we have that $f(\alpha) < \alpha$. If $\gamma \in \ran(f)$ then $f(\alpha) = \gamma$ implies either $\gamma = 0$ and $\alpha < \min(C)$ or $\gamma \in C$ and $\gamma < \alpha < \gamma'$ 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 $\max (\emptyset) \coloneqq 0$). } \gist{% 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 $\emptyset \not\in F, \kappa \in F$. }{} \begin{definition} A filter $F$ is an \vocab{ultrafilter} iff for all $X \subseteq \kappa$ either $X \in F$ or $\kappa \setminus X \in F$. \end{definition} \gist{% \begin{example} Examples of filters: \begin{enumerate}[(a)] \item Let $\kappa \ge \aleph_0$ and let $F = \{X \subseteq \kappa: \kappa \setminus X \text{ is finite}\}$. This is called the \vocab{Fr\'echet filter} or \vocab{cofinal filter}. It is not an ultrafilter (consider for example the even and odd numbers\footnote{we consider limit ordinals to be even}). \item Let $\kappa$ be uncountable and regular. 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} This is certainly not the case if $\kappa \ge \aleph_2$, because then $S_0 \coloneqq \{\alpha < \kappa : \cf(\alpha) = \omega\}$ and $S_1 \coloneqq \{\alpha < \kappa : \cf(\alpha) = \omega_1\} $ are both stationary and clearly disjoint. 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 one cofinality. \begin{theorem}[Solovay] \yalabel{Solovay's Theorem}{Solovay}{thm:solovay} Let $\kappa$ be regular and uncountable. If $S \subseteq \kappa$ is stationary, there is a sequence $\langle S_i : i < \kappa \rangle$ of pairwise disjoint stationary subsets of $\kappa$ such that $S = \bigcup S_i$. \end{theorem} \begin{corollary} $\cF_{\aleph_1}$ is not an ultrafilter. \end{corollary} \begin{proof} Apply \yaref{thm:solovay} to $S = \aleph_1$. Let $\aleph_1 = A \cup B$ where $A$ and $B$ are both stationary and disjoint. Then use the argument from above. \end{proof} \begin{yarefproof}{thm:solovay}% \gist{% \footnote{``This is one of the arguments where it is certainly worth it to look at it again.''} 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 $\cf(\alpha) = \omega$. Let $S^\ast \coloneqq \{\alpha \in S \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $. $S^\ast$ is still stationary: Let $C \subseteq \omega_1$ be a club, then $D = \{\alpha \in C \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $ is still a club, so \[S^\ast \cap C = S^\ast \cap D = S \cap D \neq \emptyset.\] Let \[ \langle \langle \gamma_n^{\alpha} : n < \omega \rangle : \alpha \in S^\ast\rangle \] be such that $ \langle \gamma_n^\alpha : n < \omega \rangle$ is cofinal in $ \alpha$. \begin{claim} \label{thm:solovay:p:c1} There exists $n < \omega$ such that for all $\delta < \omega_1$ the set \[ \{\alpha \in S^\ast : \gamma_n^\alpha > \delta\} \] is stationary. \end{claim} \begin{subproof} Otherwise for all $n < \omega$, there is a $\delta$ such that $\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$ is nonstationary. Let $\delta_n$ be the least such $\delta$. Let $C_n$ be a club disjoint from \[ \{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta_n\}, \] i.e.~if $\alpha \in S^\ast \cap C_n$, then $\gamma_n^{\alpha} \le \delta_n$. Let $\delta^\ast \coloneqq \sup_{n< \omega}\delta_n$. Let $C = \bigcap_{n < \omega} C_n$. Then $C$ is a club. We must have that if $\alpha \in S^\ast \cap C$ then $\gamma_n^{\alpha} \le \delta^\ast$ for all $n$. % But now things get a bit fishy: Let $C' \coloneqq C \setminus (\delta^\ast + 1)$. $C'$ is still club. 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 as $\langle \gamma_n^{\alpha} : n < \omega \rangle$ is cofinal in $\alpha$ $\lightning$. \end{subproof} Let $n < \omega$ be as in \yaref{thm:solovay:p:c1}. Consider \begin{IEEEeqnarray*}{rCl} f\colon S^\ast&\longrightarrow & \omega_1\\ \alpha&\longmapsto & \gamma^{\alpha}_n. \end{IEEEeqnarray*} Clearly this is regressive. We will now define a strictly increasing sequence $\langle \delta_i : i < \omega_1 \rangle$ as follows: Let $\delta_0 = 0$. For $0 < i < \omega_1$ suppose that $\delta_j, j < i$ have been defined. Let $\delta \coloneqq (\sup_{j < i} \delta_j) + 1$. By \yaref{thm:solovay:p:c1} (rather, by the choice of $n$), we have that $\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$ is stationary. Hence by Fodor there is some stationary $T \subseteq S^\ast$ and some $\delta'$ such that for all $\alpha \in T$ we have $\gamma_n^{\alpha} = \delta'$. Write $\delta_i = \delta'$ and $T_i = T$. By construction, all the $T_i$ are stationary. Since the $\delta_i$ are strictly increasing and since $\gamma_n^{\alpha} = \delta_i$ for all $\alpha \in T_i$, we have that the $T_i$ are disjoint. % \begin{claim} % \label{thm:solovay:p:c2} % Each $T_i$ is stationary % and if $i \neq j$, then $T_i \cap T_j = \emptyset$. % \end{claim} % \begin{subproof} % The first part is true by construction. % Let $j < i$. % Then if $\alpha \in T_i$, $\alpha' \in T_j$, % we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$ % hence $\alpha \neq \alpha'$. % \end{subproof} Now let \[ S_i \coloneqq \begin{cases} T_i &: i > 0,\\ T_0 \cup (S \setminus \bigcup_{j > 0} T_j) &: i = 0. \end{cases} \] 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} \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{yarefproof} \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} (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} Let $\kappa$ be a singular cardinal of uncountable cofinality. Assume that $2^{\lambda} = \lambda^+$ for all (infinite) cardinals $\lambda < \kappa$. Then $2^{\kappa} = \kappa^+$. \end{theorem} \begin{definition} $\GCH$, the \vocab{generalized continuum hypothesis} 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 $\GCH \implies \CH$. \yaref{thm:silver} says that if $\GCH$ is true below $\kappa$, then it is true at $\kappa$. The proof of \yaref{thm:silver} is quite elementary, so we will do it now, but the statement can only be fully appreciated later. }{}