w23-logic-2/inputs/lecture_16.tex

294 lines
11 KiB
TeX
Raw Normal View History

2023-12-11 15:45:36 +01:00
\lecture{16}{2023-12-11}{}
2024-02-14 02:06:55 +01:00
% TODO ANKI-MARKER
2023-12-11 15:45:36 +01:00
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 $C \cap S = \emptyset$.
2024-02-14 02:06:55 +01:00
\gist{%
2023-12-11 15:45:36 +01:00
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 $f(\alpha) = \gamma$.
2024-02-14 02:06:55 +01:00
}{%
Consider $S\setminus \{0\} \ni \alpha \mapsto \underbrace{\max(C \cap \alpha)}_{< \alpha}$
(here $\max (\emptyset) \coloneqq 0$).
}
2023-12-11 15:45:36 +01:00
2024-02-14 02:06:55 +01:00
\gist{%
2023-12-11 15:45:36 +01:00
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 $\emptyset \not\in F, \kappa \in F$.
2024-02-14 02:06:55 +01:00
\todo{Move this to the definition of filter?}
}{}
2023-12-11 15:45:36 +01:00
\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}
2024-02-13 02:00:58 +01:00
\gist{%
2023-12-11 15:45:36 +01:00
\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}
2024-02-13 02:00:58 +01:00
}{Let $\cF_\kappa \coloneqq \{X \subseteq \kappa : \exists C \subseteq \kappa \text{ club in } \kappa\}$.}
2023-12-11 15:45:36 +01:00
\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
2024-02-14 02:06:55 +01:00
one cofinality.
2023-12-11 15:45:36 +01:00
\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}%
2024-02-14 02:06:55 +01:00
\gist{%
2023-12-11 15:45:36 +01:00
%\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.
2024-02-14 02:06:55 +01:00
% TODO TODO TODO
2023-12-11 15:45:36 +01:00
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.
2024-02-14 02:06:55 +01:00
}{%
(Proof for $\aleph_1$)
Fix $S \subseteq \aleph_1$ stationary.
$\alpha \in (0, \omega_1)$ is successor ordinal or limit with $\cf(\alpha) = \omega_1$.
\begin{itemize}
\item $S^\ast \coloneqq \{\alpha \in S \setminus \{0\} : \alpha \text{ limit ordinal}\}$.
Still stationary:
$C$ club $\implies D = \{\alpha \in C \setminus \{0\} : \alpha \text{ limit}\}$ club.
\item For all $\alpha \in S^\ast$ choose $\langle \gamma^{\alpha}_n : n < \omega \rangle$ cofinal.
\item $\exists n < \omega.~\forall \delta < \omega_1: \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\}$
stationary:
\begin{itemize}
% TODO THINK!
% TODO TODO TODO
\item Otherwise $\forall n < \omega.~\exists \delta.~\{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta\} $ nonstationary.
\item $\delta_n\coloneqq $ least such $\delta$,
$C_n$ club s.t.~$C_n \cap \{\alpha \in S^\ast : \gamma^{\alpha}_n > \delta_n\} = \emptyset$.
(i.e.~$\alpha \in S^\ast \cap C_n \implies \gamma_n^{\alpha} \le \delta_n$).
\item $\delta^\ast \coloneqq \sup_{n < \omega} \delta_n$, $C \coloneqq \bigcap_{n < \omega} C_n$
is club. $\alpha \in S^\ast \cap C \implies \forall n.~\gamma_n^\alpha \le \delta^\ast$.
\item $C' \coloneqq C \setminus (\delta^\ast + 1)$ still club.
\item Pick $\alpha \in S^\ast \cap C'$.
But $\gamma_n^{\alpha} < \delta^\ast$ is cofinal $\lightning$.
\end{itemize}
\item Consider $f\colon S^\ast \to \omega_1, \alpha \mapsto \gamma_n^{\alpha}$ (note: regressive)
\item Define $\langle \delta_i : i < \omega_1 \rangle$ strictly increasing:
\begin{itemize}
\item $\delta_0 \coloneqq 0$,
\item $\delta'_{i} \coloneqq (\sup_{j < i} \delta_i) + 1$,
$\{\alpha \in S^\ast : \gamma_n^\alpha > \delta'_i\}$
is stationary $\overset{\yaref{thm:fodor}}{\implies} $ $\exists$ stationary $T_i \subseteq S^\ast$
s.t.~$f''T_i = \{\delta_i\} $ constant.
\end{itemize}
\item $T_i$ are disjoint: $i \neq j \implies \underbrace{f''T_i}_{\{\delta_i\} } \cap \underbrace{f''T_j}_{\{\delta_j\}} = \emptyset$.
\item $S_0 \coloneqq T_0 \cup \left( S \setminus \bigcup_{j > 0} T_j \right)$, $S_i \coloneqq T_i$.
\end{itemize}
}
2023-12-11 15:45:36 +01:00
\end{refproof}
2024-02-14 02:06:55 +01:00
\gist{%
2023-12-11 15:45:36 +01:00
We now want to do another application of \yaref{thm:fodor}.
Recall that $2^{\kappa} > \kappa$, in fact $\cf(2^{\kappa}) > \kappa$
2024-02-14 02:06:55 +01:00
by \yaref{thm:koenig}
(cf.~\yaref{cor:cfpow}).
2023-12-11 15:45:36 +01:00
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:
2024-02-14 02:06:55 +01:00
}{}
2023-12-11 15:45:36 +01:00
\begin{theorem}[Silver]
2024-01-17 11:51:59 +01:00
\yalabel{Silver's Theorem}{Silver}{thm:silver}
2023-12-11 15:45:36 +01:00
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}
2024-02-14 02:06:55 +01:00
\gist{%
2023-12-11 15:45:36 +01:00
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.
2024-02-14 02:06:55 +01:00
}{}