Merge branch 'master' of https://git.abstractnonsen.se/josia-notes/w23-logic-2

inputs/lecture_14.tex

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

inputs/lecture_16.tex

@@ -1,3 +1,246 @@
+<<<<<<< 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}.
@@ -218,3 +461,4 @@ 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