tutorial

inputs/lecture_14.tex

@@ -122,6 +122,13 @@ We have shown (assuming \AxC to choose contained clubs):
     If $\kappa$ is regular and uncountable.
     Then $\cF_\kappa$ is a $< \kappa$-closed filter.
 \end{theorem}
+\begin{proof}
+    Clearly $\emptyset \not\in \cF_\kappa$,
+    $\kappa \in \cF_\kappa$,
+    and $A \in \cF_{\kappa}, A \subseteq B \in \kappa \implies B \in \cF_\kappa$.
+    In \autoref{lem:clubintersection} we are going to show that the intersection
+    of $< \kappa$ many clubs is club.
+\end{proof}
 
 \begin{definition}
     Let $\langle A_\beta : \beta < \alpha \rangle$
@@ -208,6 +215,18 @@ We have shown (assuming \AxC to choose contained clubs):
         in $\gamma$, hence $\delta \in  D_{\overline{\gamma}}$.
     \end{subproof}
 \end{proof}
+\begin{remark}+
+    $\diagi_{\beta < \kappa} D_{\beta}$ actually
+    \emph{is} a club:
+    It suffices to show that $\diagi_{\beta < \kappa} D_\beta$ is closed.
+    Let $\lambda < \kappa$ be a limit ordinal.
+    Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
+    Then there exists $\alpha < \lambda$ such that
+    $\lambda \not\in D_\alpha$.
+    Since $D_\alpha$ is closed,
+    we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
+    In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \max (\alpha ,\sup(D_\alpha \cap \lambda) < \lambda$.
+\end{remark}
 
 \begin{definition}
     Let $\kappa$ be regular and uncountable.

inputs/lecture_15.tex

@@ -35,7 +35,7 @@
     Then for every $\nu$ there exists a club $C_\nu$
     such that  $S_\nu \cap C_\nu = \emptyset$.\footnote{Here we use \AxC to choose the $C_\nu$ uniformly.}
     Let $C = \diagi_{\nu < \alpha} C_\nu$.
-    By \yaref{lem:diagiclub} $C$ is a club.\todo{Show that it \emph{is} a club not just contains one}
+    By \yaref{lem:diagiclub} $C$ is a club.
     So we may pick some $\alpha \in C \cap S$.
     In particular $\alpha \in C_\nu$ for all $\nu < \alpha$.
     Hence $f(\alpha) \neq  \nu$ for all $\nu < \alpha$,

inputs/lecture_16.tex

@@ -216,7 +216,7 @@ 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}
+    \yalabel{Silver's Theorem}{Silver}{thm:silver}
 
     Let $\kappa$ be a singular cardinal of uncountable cofinality.
     Assume that $2^{\lambda} = \lambda^+$ for all (infinite)

inputs/tutorial_2024-01-17.tex

@@ -0,0 +1,34 @@
+\tutorial{}{2024-01-17}{}
+
+\subsection{Sheet 9}
+
+\nr 1
+
+Let $\kappa$ be strongly inaccessible.
+Then $(V_{\kappa}, \in \defon_{V_\kappa}) \models\ZFC$:
+
+Most axioms are trivial.
+
+\begin{itemize}
+    \item \AxUnion: Let $A \in V_{\kappa}$.
+        Then $\rank(x) < \kappa$ for all $x \in A$.
+        Since $\kappa$ is regular, we get
+        $\bigcup A \in V_{\kappa}$.
+    \item \AxPower: This holds since $\kappa$ is strongly inaccessible.
+    \item \AxRep: If  $A \in V_{\kappa}$ and $f\colon A \to V_\kappa$
+        is definable over $V_\kappa$,
+        then $f'' A = \{f(a) : a \in A\}$ has bounded rank below $\kappa$.
+\end{itemize}
+
+
+
+\subsection{Exercise during tutorial}
+
+
+Let $\kappa$ be uncountable and regular
+Then the club filter $\cF_{\kappa}$ is $< \kappa$-closed.
+
+
+
+
+