241 lines
8 KiB
TeX
241 lines
8 KiB
TeX
|
\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.
|
||
|
|
|