\lecture{14}{2023-12-04}{}
\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}
    First let $\alpha = 2$.
    Let $C, D \subseteq \kappa$
    be a 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.
    $\cf(\kappa) > \alpha \cdot \omega$.
    
\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 $\emptyset \not\in F$, $\kappa \in F$.
    \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}
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{definition}
    Let $\langle A_\beta : \beta < \alpha \rangle$
    be a sequence of sets.
    The \vocab{diagonal intersection},
    is defined to be
    \[
    \diagi_{\beta < \alpha} A_{\beta} \coloneqq 
    \{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\}  \}.
    \] 
\end{definition}
\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. % TODO: contains or is?
\end{lemma}
\begin{proof}
    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}
        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 $\gamma \in D_{\beta_0}$.
        For each $\beta'$ with $\beta < \beta' < \gamma$,
        there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
        with $\beta'' \ge  \beta', \beta'' < \gamma$.
        In particular $\beta'' \in D_{\beta_0}$.

        We showed that $D_{\beta_0} \cap \gamma$
        is unbounded in $\gamma$,
        so $\gamma \in D_{\beta_0}$.

        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 $\gamma \in \diagi_{\beta < \kappa} D_\beta$.

        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 $\gamma$, hence $\delta \in  D_{\overline{\gamma}}$.
    \end{subproof}
\end{proof}

\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{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.


