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} (\emph{cl}osed \emph{un}bounded)
+        \item $A$ is \vocab{club} (closed unbounded)
             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

@@ -1,236 +0,0 @@
-\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

@@ -142,6 +142,4 @@
 \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

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