lecture 14

inputs/lecture_13.tex

@@ -146,21 +146,21 @@ Relevant concepts to prove this theorem:
             iff for all $\beta < \alpha$,
             there is some $\gamma \in \alpha$
             such that $\beta < \gamma$.
-        \item We say that $A \subseteq \alpha$ 
+        \item We say that $A \subseteq \alpha$
             is \vocab{closed},
             iff it is closed with respect to the order topology on $\alpha$,
             i.e.~for all $\beta < \alpha$,
             \[
             \sup(A \cap \beta) \in A.
-            \] 
+            \]
-        \item $A$ is \vocab{club} (closed unbounded)
+        \item $A$ is \vocab{club} (\emph{cl}osed \emph{un}bounded)
             iff it is closed and unbounded.
     \end{itemize}
 \end{definition}
+The interesting case is that $\alpha$ is a regular uncountable cardinal.
 \begin{fact}
     $A \subseteq \alpha$ being unbounded
     is equivalent to $f\colon \beta\to \alpha$ being cofinal,
     where
     $(\beta, \in ) \overset{f}{\cong} (A, \in )$.
 \end{fact}
-

inputs/lecture_14.tex

@@ -0,0 +1,236 @@
+\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}
+    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{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.
+
+

logic.sty

@@ -142,4 +142,6 @@
 \DeclareSimpleMathOperator{cf}
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
+%\newcommand\diagi{\mathop{\large \Delta}\limits}
+\newcommand\diagi{\mathop{\large \Delta}}
 

logic2.tex

@@ -37,6 +37,7 @@
 \input{inputs/lecture_11}
 \input{inputs/lecture_12}
 \input{inputs/lecture_13}
+\input{inputs/lecture_14}
 
 
 \cleardoublepage