Josia Pietsch
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11 KiB
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291 lines
11 KiB
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\lecture{16}{2023-12-11}{}
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% TODO ANKI-MARKER
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Recall \yaref{thm:fodor}.
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\begin{question}
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What happens if $S$ is nonstationary?
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\end{question}
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Let $S \subseteq \kappa$ be nonstationary,
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$\kappa$ uncounable and regular.
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Then there is a club $C \subseteq \kappa$
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with $C \cap S = \emptyset$.
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\gist{%
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Let us define $f\colon S \to \kappa$ in the following way:
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If $\alpha \in S$ and $C \cap \alpha \neq \emptyset$,
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then $\max(C \cap \alpha) < \alpha$.
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Define
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\[
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f(\alpha) \coloneqq \begin{cases}
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0 &: C \cap \alpha = \emptyset,\\
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\max(C \cap \alpha) &:C\cap \alpha \neq \emptyset.
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\end{cases}
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\]
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For all $\alpha > 0$,
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we have that $f(\alpha) < \alpha$.
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If $\gamma \in \ran(f)$ then
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$f(\alpha) = \gamma$ implies either $\gamma = 0$ and $\alpha < \min(C)$
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or $\gamma \in C$ and $\gamma < \alpha < \gamma'$
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where $\gamma' = \min(C \setminus (\gamma + 1))$.
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Thus for all $\gamma$,
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there is only an interval of ordinals $\alpha \in S$
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where $f(\alpha) = \gamma$.
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}{%
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Consider $S\setminus \{0\} \ni \alpha \mapsto \underbrace{\max(C \cap \alpha)}_{< \alpha}$
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(here $\max (\emptyset) \coloneqq 0$).
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}
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\gist{%
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Recall that $F \subseteq \cP(\kappa)$ is a filter if
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$X,Y \in F \implies X \cap Y \in F$,
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$X \in F, X \subseteq Y \subseteq \kappa \implies Y \in F$
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and $\emptyset \not\in F, \kappa \in F$.
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}{}
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\begin{definition}
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A filter $F$ is an \vocab{ultrafilter}
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iff for all $X \subseteq \kappa$
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either $X \in F$ or $\kappa \setminus X \in F$.
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\end{definition}
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\gist{%
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\begin{example}
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Examples of filters:
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\begin{enumerate}[(a)]
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\item Let $\kappa \ge \aleph_0$
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and let $F = \{X \subseteq \kappa: \kappa \setminus X \text{ is finite}\}$.
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This is called the \vocab{Fr\'echet filter}
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or \vocab{cofinal filter}.
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It is not an ultrafilter
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(consider for example the even and odd numbers\footnote{we consider limit ordinals to be even}).
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\item Let $\kappa$ be uncountable and regular.
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Then $\cF_\kappa \coloneqq \{X \subseteq \kappa: \exists C \subseteq \kappa \text{ club in $\kappa$}. C \subseteq X\}$.
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\end{enumerate}
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\end{example}
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}{Let $\cF_\kappa \coloneqq \{X \subseteq \kappa : \exists C \subseteq \kappa \text{ club in } \kappa\}$.}
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\begin{question}
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Is $\cF_\kappa$ an ultrafilter?
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\end{question}
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This is certainly not the case if $\kappa \ge \aleph_2$,
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because then $S_0 \coloneqq \{\alpha < \kappa : \cf(\alpha) = \omega\}$
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and $S_1 \coloneqq \{\alpha < \kappa : \cf(\alpha) = \omega_1\} $
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are both stationary
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and clearly disjoint.
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So neither $S_0$ nor $S_1 \subseteq \kappa \setminus S_0$ contains a club.
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For $\kappa < \aleph_1$
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this argument does not work, since there is only
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one cofinality.
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\begin{theorem}[Solovay]
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\yalabel{Solovay's Theorem}{Solovay}{thm:solovay}
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Let $\kappa$ be regular and uncountable.
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If $S \subseteq \kappa$
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is stationary,
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there is a sequence $\langle S_i : i < \kappa \rangle$
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of pairwise disjoint stationary subsets of $\kappa$
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such that $S = \bigcup S_i$.
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\end{theorem}
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\begin{corollary}
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$\cF_{\aleph_1}$ is not an ultrafilter.
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\end{corollary}
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\begin{proof}
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Apply \yaref{thm:solovay} to $S = \aleph_1$.
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Let $\aleph_1 = A \cup B$
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where $A$ and $B$ are both stationary
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and disjoint.
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Then use the argument from above.
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\end{proof}
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\begin{yarefproof}{thm:solovay}%
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\gist{%
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\footnote{``This is one of the arguments where it is certainly worth it to look at it again.''}
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We will only prove this for $\aleph_1$.
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Fix $S \subseteq \aleph_1$ stationary.
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For each $0 < \alpha < \omega_1$,
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either $\alpha$ is a successor ordinal
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or $\alpha$ is a limit ordinal and $\cf(\alpha) = \omega$.
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Let $S^\ast \coloneqq \{\alpha \in S \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $.
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$S^\ast$ is still stationary:
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Let $C \subseteq \omega_1$
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be a club,
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then $D = \{\alpha \in C \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $
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is still a club,
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so
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\[S^\ast \cap C = S^\ast \cap D = S \cap D \neq \emptyset.\]
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Let
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\[
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\langle \langle \gamma_n^{\alpha} : n < \omega \rangle : \alpha \in S^\ast\rangle
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\]
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be such that $ \langle \gamma_n^\alpha : n < \omega \rangle$
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is cofinal in $ \alpha$.
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\begin{claim}
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\label{thm:solovay:p:c1}
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There exists $n < \omega$
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such that for all $\delta < \omega_1$
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the set
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\[
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\{\alpha \in S^\ast : \gamma_n^\alpha > \delta\}
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\]
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is stationary.
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\end{claim}
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\begin{subproof}
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Otherwise for all $n < \omega$,
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there is a $\delta$ such that
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$\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$
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is nonstationary.
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Let $\delta_n$ be the least such $\delta$.
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Let $C_n$ be a club disjoint from
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\[
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\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta_n\},
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\]
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i.e.~if $\alpha \in S^\ast \cap C_n$, then $\gamma_n^{\alpha} \le \delta_n$.
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Let $\delta^\ast \coloneqq \sup_{n< \omega}\delta_n$.
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Let $C = \bigcap_{n < \omega} C_n$.
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Then $C$ is a club.
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We must have that if $\alpha \in S^\ast \cap C$
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then $\gamma_n^{\alpha} \le \delta^\ast$ for all $n$.
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% But now things get a bit fishy:
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Let $C' \coloneqq C \setminus (\delta^\ast + 1)$.
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$C'$ is still club.
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As $S^\ast$ is stationary,
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we may pick some $\alpha \in S^\ast \cap C'$.
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But then $\gamma_n^{\alpha} > \delta^\ast$
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for $n$ large enough
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as $\langle \gamma_n^{\alpha} : n < \omega \rangle$
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is cofinal in $\alpha$ $\lightning$.
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\end{subproof}
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Let $n < \omega$ be as in \yaref{thm:solovay:p:c1}.
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Consider
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\begin{IEEEeqnarray*}{rCl}
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f\colon S^\ast&\longrightarrow & \omega_1\\
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\alpha&\longmapsto & \gamma^{\alpha}_n.
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\end{IEEEeqnarray*}
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Clearly this is regressive.
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We will now define a strictly increasing sequence
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$\langle \delta_i : i < \omega_1 \rangle$
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as follows:
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Let $\delta_0 = 0$.
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For $0 < i < \omega_1$ suppose that $\delta_j, j < i$ have been defined.
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Let $\delta \coloneqq (\sup_{j < i} \delta_j) + 1$.
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By \yaref{thm:solovay:p:c1} (rather, by the choice of $n$),
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we have that $\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$
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is stationary.
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Hence by Fodor there is some stationary $T \subseteq S^\ast$
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and some $\delta'$ such that for all $\alpha \in T$
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we have
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$\gamma_n^{\alpha} = \delta'$.
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Write $\delta_i = \delta'$ and $T_i = T$.
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By construction, all the $T_i$ are stationary.
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Since the $\delta_i$ are strictly increasing
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and since $\gamma_n^{\alpha} = \delta_i$ for all $\alpha \in T_i$,
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we have that the $T_i$ are disjoint.
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% \begin{claim}
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% \label{thm:solovay:p:c2}
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% Each $T_i$ is stationary
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% and if $i \neq j$, then $T_i \cap T_j = \emptyset$.
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% \end{claim}
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% \begin{subproof}
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% The first part is true by construction.
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% Let $j < i$.
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% Then if $\alpha \in T_i$, $\alpha' \in T_j$,
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% we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$
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% hence $\alpha \neq \alpha'$.
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% \end{subproof}
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Now let
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\[
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S_i \coloneqq \begin{cases}
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T_i &: i > 0,\\
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T_0 \cup (S \setminus \bigcup_{j > 0} T_j) &: i = 0.
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\end{cases}
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\]
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Then $\langle S_i : i < \omega_1 \rangle$ is
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as desired.
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}{%
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(Proof for $\aleph_1$)
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Fix $S \subseteq \aleph_1$ stationary.
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$\alpha \in (0, \omega_1)$ is successor ordinal or limit with $\cf(\alpha) = \omega_1$.
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\begin{itemize}
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\item $S^\ast \coloneqq \{\alpha \in S \setminus \{0\} : \alpha \text{ limit ordinal}\}$.
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Still stationary:
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$C$ club $\implies D = \{\alpha \in C \setminus \{0\} : \alpha \text{ limit}\}$ club.
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\item For all $\alpha \in S^\ast$ choose $\langle \gamma^{\alpha}_n : n < \omega \rangle$ cofinal.
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\item $\exists n < \omega.~\forall \delta < \omega_1: \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\}$
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stationary:
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\begin{itemize}
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\item Otherwise $\forall n < \omega.~\exists \delta.~\{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\} $ nonstationary.
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\item $\delta_n\coloneqq $ least such $\delta$,
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$C_n$ club s.t.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$.
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(i.e.~$\alpha \in S^\ast \cap C_n \implies \gamma_n^{\alpha} \le \delta_n$).
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\item $\delta^\ast \coloneqq \sup_{n < \omega} \delta_n$, $C \coloneqq \bigcap_{n < \omega} C_n$
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is club. $\alpha \in S^\ast \cap C \implies \forall n.~\gamma_n^\alpha \le \delta^\ast$.
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\item $C' \coloneqq C \setminus (\delta^\ast + 1)$ still club.
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\item Pick $\alpha \in S^\ast \cap C'$.
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But $\gamma_n^{\alpha} < \delta^\ast$ is cofinal $\lightning$.
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\end{itemize}
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\item Consider $f\colon S^\ast \to \omega_1, \alpha \mapsto \gamma_n^{\alpha}$ (note: regressive)
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\item Define $\langle \delta_i : i < \omega_1 \rangle$ strictly increasing:
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\begin{itemize}
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\item $\delta_0 \coloneqq 0$,
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\item $\delta'_{i} \coloneqq (\sup_{j < i} \delta_i) + 1$,
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$\{\alpha \in S^\ast : \gamma_n^\alpha > \delta'_i\}$
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is stationary $\overset{\yaref{thm:fodor}}{\implies} $ $\exists$ stationary $T_i \subseteq S^\ast$
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s.t.~$f''T_i = \{\delta_i\} $ constant.
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\end{itemize}
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\item $T_i$ are disjoint: $i \neq j \implies \underbrace{f''T_i}_{\{\delta_i\} } \cap \underbrace{f''T_j}_{\{\delta_j\}} = \emptyset$.
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\item $S_0 \coloneqq T_0 \cup \left( S \setminus \bigcup_{j > 0} T_j \right)$, $S_i \coloneqq T_i$.
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\end{itemize}
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}
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\end{yarefproof}
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\gist{%
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We now want to do another application of \yaref{thm:fodor}.
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Recall that $2^{\kappa} > \kappa$, in fact $\cf(2^{\kappa}) > \kappa$
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by \yaref{thm:koenig}
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(cf.~\yaref{cor:cfpow}).
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Trivially, if $\kappa \le \lambda$ then $2^{\kappa} \le 2^{\lambda}$.
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This is in some sense the only thing we can prove about successor cardinals.
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However we can say something about singular cardinals:
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}{}
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\begin{theorem}[Silver]
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\yalabel{Silver's Theorem}{Silver}{thm:silver}
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Let $\kappa$ be a singular cardinal of uncountable cofinality.
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Assume that $2^{\lambda} = \lambda^+$ for all (infinite)
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cardinals $\lambda < \kappa$.
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Then $2^{\kappa} = \kappa^+$.
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\end{theorem}
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\begin{definition}
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$\GCH$, the \vocab{generalized continuum hypothesis}
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is the statement
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that $2^{\lambda} = \lambda^+$ holds for all infinite cardinals $\lambda$,
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\end{definition}
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\gist{%
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Recall that $\CH$ says that $2^{\aleph_0} = \aleph_1$.
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So $\GCH \implies \CH$.
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\yaref{thm:silver} says that if $\GCH$ is true below $\kappa$,
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then it is true at $\kappa$.
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The proof of \yaref{thm:silver} is quite elementary,
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so we will do it now, but the statement can only be fully appreciated later.
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}{}
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