inputs/lecture_14.tex

@@ -93,11 +93,7 @@
     \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
+        \item $\emptyset \not\in F$, $\kappa \in F$.
-                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}
 
 

inputs/lecture_16.tex

@@ -1,246 +1,3 @@
-<<<<<<< HEAD
-\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}
-
-\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}
-\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]
-    \yaref{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.
-
-||||||| ede97ee
-=======
 \lecture{16}{2023-12-14}{Silver's Theorem}
 
 We now want to prove \yaref{thm:silver}.
@@ -461,4 +218,3 @@ The general proof differs only in notation.
     then $X_i \le X$ for all $i < \aleph_{\omega_1 + 1}$,
     so such a set $X$ does not exist by \yaref{thm:silver:p:c3}.
 \end{refproof}
->>>>>>> 5ee970194987e24c54bfc9c787cb219e15e71520