\lecture{14}{2023-12-04}{} \gist{% \begin{abuse} Sometimes we say club instead of club in $\kappa$. \end{abuse} \begin{example} Let $\kappa$ be a regular uncountable cardinal. \begin{itemize} \item $\kappa$ is a club in $\kappa$. \item $\{\xi + 1 : \xi < \kappa\}$ is unbounded in $\kappa$, but not closed. \item For each $\alpha < \kappa$, the set $\alpha + 1 = \{\xi : \xi \le \alpha\}$ is closed but not unbounded in $\kappa$. \item $\{\xi < \kappa : \xi \text{ is a limit ordinal}\} $ is club in $\kappa$. \end{itemize} \end{example} }{} \begin{lemma} \label{lem:clubintersection} Let $\kappa$ be regular and uncountable. Let $\alpha < \kappa$ and let $\langle C_{\beta} : \beta < \alpha \rangle$ be a sequence of subsets of $\kappa$ which are all club in $\kappa$. Then \[ \bigcap_{\beta < \alpha} C_{\beta} \] is club in $\kappa$. \end{lemma} \begin{warning} This is false for $\alpha = \kappa$: Let $C_{\beta} \coloneqq \{\xi : \xi > \beta\}$. Clearly this is club but $\bigcap_{\beta < \kappa} C_\beta = \emptyset$. \end{warning} \begin{refproof}{lem:clubintersection} \gist{% First let $\alpha = 2$. Let $C, D \subseteq \kappa$ be club. $C \cap D$ is trivially closed: Let $\beta < \kappa$. Suppose that $(C \cap D) \cap \beta$ is unbounded in $\beta$, so $C \cap \beta$ and $D \cap \beta$ are both unbounded in $\beta$, so $\beta \in C \cap D$. $C \cap D$ is unbounded: Take some $\gamma < \kappa$. Let $\gamma_0 = \gamma$ and inductively define $\gamma_n$ : If $n$ is even, let $\gamma_n \coloneqq \min C \setminus (\gamma_{n-1}+1)$, otherwise $\gamma_n \coloneqq \min D \setminus (\gamma_{n-1}+1)$. Let $\xi = \sup \{\gamma_n : n < \omega\}$. Then $\xi = \sup \{\gamma_{2n + 2} : n < \omega\} \in D$ and $\xi \in C$ by the same argument, so $\xi \in C \cap D$ (here it is important, that $\cf(\kappa) > \omega$) and $\xi > \gamma$. The case $\alpha > 2$ is similar: The intersection is closed by exactly the same argument.% \footnote{``It is even more closed.''} Let's prove that $\bigcap \{C_{\beta} : \beta < \alpha\}$ is unbounded in $\kappa$. We will define a sequence $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$% \footnote{Ordinal multiplication, i.e.~$\alpha \cdot \omega = \sup_{n < \omega} \underbrace{\alpha + \ldots + \alpha}_{n \text{ times}}$.} as follows: Let $\gamma_0 \coloneqq \gamma$. Choose \[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\] and at limits choose the supremum. Let $\xi = \sup_{i < \alpha \cdot \omega} \gamma_i = \sup_{i < \omega} \gamma_{\alpha \cdot n + \beta + 1} \in \bigcap_{\beta < \alpha} C_\beta$, where we have used that. $\cf(\kappa) > \alpha \cdot \omega$. }{% \begin{itemize} \item Trivially closed. \item Recursively define $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$, by $\gamma_0 \coloneqq \gamma$, \[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\] and $\sup$ at limits. \item Then $\sup_{i < \alpha\cdot \omega} \gamma_i \in \bigcap_{\beta < \alpha} C_\beta$ (we used $\cf(\kappa) > \alpha\cdot \omega$). \end{itemize} } \end{refproof} \begin{definition} $F \subseteq \cP(a)$ is a \vocab{filter} iff \begin{enumerate}[(a)] \item $X,Y \in F \implies X \cap Y \in F$, \item $X \in F \land X \subseteq Y \subseteq \kappa \implies Y \in F$, \item $\emptyset \not\in F$,\footnote{Some authors don't require $\emptyset \not\in F$, but that is a degenerate case anyway, since $\emptyset \in F \iff F = \cP(a)$.} $\kappa \in F$. \end{enumerate} Let $\alpha \le \kappa$. We call $F$ \vocab{$< \alpha$-closed} iff for all $\gamma < \alpha$ and $\{X_\beta : \beta < \gamma\} \subseteq F$ then $\bigcap \{X_\beta : \beta < \gamma\} \in F$. \end{definition} \gist{% Intuitively, a filter is a collection of ``big'' subsets of $a$. }{} \begin{definition} Let $\kappa$ be regular and uncountable. The \vocab{club filter} is defined as \[ \cF_{\kappa} \coloneqq \{X \subseteq \kappa : \exists \text{ club } C \subseteq \kappa .~ C \subseteq X\}. \] \end{definition} Clearly this is a filter. We have shown (assuming \AxC to choose contained clubs): \begin{theorem} If $\kappa$ is regular and uncountable. Then $\cF_\kappa$ is a $< \kappa$-closed filter. \end{theorem} \begin{proof} Clearly $\emptyset \not\in \cF_\kappa$, $\kappa \in \cF_\kappa$, and $A \in \cF_{\kappa}, A \subseteq B \in \kappa \implies B \in \cF_\kappa$. In \autoref{lem:clubintersection} showed that the intersection of $< \kappa$ many clubs is club. \end{proof} \begin{definition} Let $\langle A_\beta : \beta < \alpha \rangle$ be a sequence of sets. The \vocab{diagonal intersection}, is defined to be \[ \diagi_{\beta < \alpha} A_{\beta} \coloneqq \{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\} \} = \bigcap_{\beta < \alpha} ([0,\beta] \cup A_\beta) \] \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. If $\langle C_{\beta} : \beta < \kappa \rangle$ is a sequence of club subsets of $\kappa$, then $\diagi_{\beta < \kappa} C_{\beta}$ contains a club. \end{lemma} \begin{refproof}{lem:diagiclub} % TODO THINK \gist{ Let us fix $\langle C_{\beta} : \beta < \alpha \rangle$. Write $D_{\beta} \coloneqq \bigcap \{C_{\gamma} : \gamma \le \beta\} $ for $\beta < \kappa$. Each $D_{\beta}$ is a club, $D_{\beta} \subseteq C_{\beta}$ and $D_{\beta} \supseteq D_{\beta'}$ for $\beta \le \beta' < \kappa$. It suffices to show that $\diagi_{\beta < \kappa} D_{\beta}$ contains a club. \begin{claim} $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$. \end{claim} \begin{subproof} Cf.~\yaref{rem:diagiclosed}. % Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$ % is unbounded in $\gamma$. % We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$. % Let $\beta_0 < \gamma$. % We need to see that $\gamma \in D_{\beta_0}$. % For each $\beta_0 \le \beta' < \gamma$ % there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$ % such that $\beta' \le \beta'' < \gamma$, % since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$. % In particular $\beta'' \in D_{\beta_0}$. % So $D_{\beta_0} \cap \gamma$ % is unbounded in $\gamma$. % Since $D_{\beta_0}$ is closed % it follows that $\gamma \in D_{\beta_0}$. %As $\beta_0 < \gamma$ was arbitrary, %this shows that $\gamma \in \diagi_{\beta < n} D_\beta$. \end{subproof} \begin{claim} $\diagi_{\beta < \kappa} D_{\beta}$ is unbounded in $\kappa$. \end{claim} \begin{subproof} Fix $\gamma < \kappa$. We need to find $\delta > \gamma$ with $\delta \in \diagi_{\beta < \kappa} D_\beta$. Define $\langle \gamma_n : n < \omega \rangle$ as follows: $\gamma_0 \coloneqq \gamma$ and \[ \gamma_{n+1} \coloneqq \min D_{\gamma_n} \setminus (\gamma_n + 1) \] We have $\delta \coloneqq \sup_{n < \omega} \gamma_n \in \kappa$ by cofinality of $\kappa$. We need to show that $\delta \in D_{\overline{\gamma}}$ for all $\overline{\gamma} < \delta$. If $\overline{\gamma} < \delta$, then $\overline{\gamma} \le \gamma_n$ for some $n < \omega$. For $m \ge n$, $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_{\overline{\gamma}}$. So $D_{\overline{\gamma}} \cap \delta$ is unbounded in $\delta$, hence $\delta \in D_{\overline{\gamma}}$. \end{subproof} }{% \begin{itemize} \item Fix $\langle C_\beta : \beta < \alpha \rangle$. Set $D_{\beta} \coloneqq \bigcap_{\gamma \le \beta} D_{\gamma}$. It suffices to analyze $D_{\beta}$. \item $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$: \begin{itemize} \item Let $\gamma < \kappa$ such that $\left( \diagi_{\beta < \kappa} D_{\beta}\right) \cap \gamma$ unbounded in $\gamma$. Want $\gamma \in \diagi_{\beta < \kappa} D_\beta$. \item Let $\beta_0 < \gamma$. Want $\gamma \in D_{\beta_0}$. \item $D_{\beta_0} \cap \gamma$ is unbounded in $\gamma$ ($D_{\beta_0} \setminus \beta_0 \supseteq \diagi_{\beta < \kappa} D_{\beta} \setminus \beta_0$) $\overset{D_{\beta_0} \text{ closed}}{\implies} \gamma \in D_{\beta_0}$. \end{itemize} \item $\diagi_{\beta < \kappa} D_\beta$ is unbounded in $\kappa$: \begin{itemize} \item Fix $\gamma < \kappa$. We need to find $\diagi_{\beta < \kappa} D_\beta \ni \delta > \gamma$. \item Define $\langle \gamma_n : n < \omega \rangle$ by $\gamma_0 \coloneqq \gamma$, $\gamma_{n+1} \coloneqq \min D_{\gamma_n} \setminus (\gamma_n + 1)$, $\delta \coloneqq \sup_n \gamma_n \overset{\cf(\kappa) > \omega}{<} \kappa$ \item Want $\delta \in \diagi_{\beta < \kappa} D_\beta$, i.e.~$\forall \epsilon < \delta.~\delta\in D_\epsilon$. If $\epsilon < \delta$, then $\epsilon \le \gamma_n$ for $n$ large enough, so $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_\epsilon$ for $m \ge n$. Thus $\sup(D_\epsilon \cap \delta) = \delta$ $\overset{D_\epsilon \text{ closed}}{\implies} \delta \in D_{\epsilon}$. \end{itemize} \end{itemize} } \end{refproof} \begin{remark}+ $\diagi_{\beta < \kappa} C_{\beta}$ actually \emph{is} a club: It suffices to show that $\diagi_{\beta < \kappa} C_\beta$ is closed. 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 % $\lambda \not\in D_\alpha$. % Since $D_\alpha$ is closed, % we get $\sup(D_{\alpha} \cap \lambda) < \lambda$. % In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$. \end{remark} \begin{definition} Let $\kappa$ be regular and uncountable. $S \subseteq \kappa$ is called \vocab{stationary} (in $\kappa$) 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 $\kappa$ is stationary in $\kappa$. \item There exist disjoint stationary sets:\footnote{Note that clubs can never be disjoint, since their intersection is a club.} Let $\kappa = \omega_2$. Let $S_0 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega\}$ and $S_1 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega_1\}$. Clearly these are disjoint. They are both stationary: Let $c \subseteq \kappa$ be a club. Let $(\xi_i : i \le \omega_1)$ be defined as follows: $\xi_0 \coloneqq \min C$, $\xi_i \coloneqq \min (C \setminus \sup_{j < i} \xi_j)$. For $i \le \omega_1$ we have that $\xi_i = \sup_{j < i} \xi_j$. In particular $\xi_\omega \in S_0 \cap C$ and $\xi_{\omega_1} \in S_1 \cap C$. \end{itemize} \end{example} We will show later that if $ \kappa$ is a regular uncountable cardinal, then every stationary $S \subseteq \kappa$ can be written as $S = \bigcup_{i < \kappa} S_i$, where the $S_i$ are stationary and pairwise disjoint.