From 42af65da7a7e803cd1bfbf35a26bc3e09c1c4789 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Mon, 11 Dec 2023 15:45:36 +0100 Subject: [PATCH] lecture 16 --- inputs/lecture_14.tex | 6 +- inputs/lecture_16.tex | 240 ++++++++++++++++++++++++++++++++++++++++++ logic.sty | 1 + logic2.tex | 1 + 4 files changed, 247 insertions(+), 1 deletion(-) create mode 100644 inputs/lecture_16.tex diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 250793c..e0118d9 100644 --- a/inputs/lecture_14.tex +++ b/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} diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex new file mode 100644 index 0000000..82bf656 --- /dev/null +++ b/inputs/lecture_16.tex @@ -0,0 +1,240 @@ +\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. + diff --git a/logic.sty b/logic.sty index 5f0f75a..7c3a7af 100644 --- a/logic.sty +++ b/logic.sty @@ -127,6 +127,7 @@ \DeclareSimpleMathOperator{CH} +\DeclareSimpleMathOperator{GCH} \DeclareSimpleMathOperator{DC} \DeclareSimpleMathOperator{Ord} \DeclareSimpleMathOperator{OR} % Ordinals diff --git a/logic2.tex b/logic2.tex index 70c2c44..84f179a 100644 --- a/logic2.tex +++ b/logic2.tex @@ -39,6 +39,7 @@ \input{inputs/lecture_13} \input{inputs/lecture_14} \input{inputs/lecture_15} +\input{inputs/lecture_16} \cleardoublepage