\lecture{16}{2023-12-11}{}

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 $C \cap S = \emptyset$.
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 $f(\alpha) = \gamma$.

\todo{Move this to the definition of filter}
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 $\emptyset \not\in F, \kappa \in F$.

\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
on 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 sets 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{refproof}{thm:solovay}%
    %\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.

    We will only proof 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 $\cf(\alpha) = \omega_1$.

    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 $\delta^\ast$ 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 $T_i = T$.
    
    \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.
\end{refproof}

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}.

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}
Recall that $\CH$ says that $2^{\aleph_0} = \aleph_1$.
So $\GCH \implies \CH$.

\yalabel{thm:silver} says that if $\GCH$ is true below $\kappa$,
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.
