diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index ab8970f..c91f05f 100644 --- a/inputs/lecture_14.tex +++ b/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. diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 23acc5c..878c124 100644 --- a/inputs/lecture_15.tex +++ b/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 $\nu < \alpha$, diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index 54da5b0..b62940e 100644 --- a/inputs/lecture_16.tex +++ b/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) diff --git a/inputs/tutorial_2024-01-17.tex b/inputs/tutorial_2024-01-17.tex new file mode 100644 index 0000000..1a733a3 --- /dev/null +++ b/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. + + + + +