\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.


