lecture 14

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@ -146,21 +146,21 @@ Relevant concepts to prove this theorem:
iff for all $\beta < \alpha$,
there is some $\gamma \in \alpha$
such that $\beta < \gamma$.
\item We say that $A \subseteq \alpha$
\item We say that $A \subseteq \alpha$
is \vocab{closed},
iff it is closed with respect to the order topology on $\alpha$,
i.e.~for all $\beta < \alpha$,
\sup(A \cap \beta) \in A.
\item $A$ is \vocab{club} (closed unbounded)
\item $A$ is \vocab{club} (\emph{cl}osed \emph{un}bounded)
iff it is closed and unbounded.
The interesting case is that $\alpha$ is a regular uncountable cardinal.
$A \subseteq \alpha$ being unbounded
is equivalent to $f\colon \beta\to \alpha$ being cofinal,
$(\beta, \in ) \overset{f}{\cong} (A, \in )$.

inputs/lecture_14.tex Normal file
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@ -0,0 +1,236 @@
Sometimes we say club
instead of club in $\kappa$.
Let $\kappa$ be a regular uncountable cardinal.
\item $\kappa$ is a club in $\kappa$.
\item $\{\xi + 1 : \xi < \kappa\}$ is unbounded in $\kappa$,
but not closed.
\item For each $\alpha < \kappa$,
the set $\alpha + 1 = \{\xi : \xi \le \alpha\}$
is closed but not unbounded in $\kappa$.
\item $\{\xi < \kappa : \xi \text{ is a limit ordinal}\} $
is club in $\kappa$.
Let $\kappa$ be regular and uncountable.
Let $\alpha < \kappa$
and let $\langle C_{\beta} : \beta < \alpha \rangle$
be a sequence of subsets of $\kappa$ which are all club in $\kappa$.
\bigcap_{\beta < \alpha} C_{\beta}
is club in $\kappa$.
This is false for $\alpha = \kappa$:
Let $C_{\beta} \coloneqq \{\xi : \xi > \beta\}$.
Clearly this is club
but $\bigcap_{\beta < \kappa} C_\beta = \emptyset$.
First let $\alpha = 2$.
Let $C, D \subseteq \kappa$
be a club.
$C \cap D$ is trivially closed:
Let $\beta < \kappa$. Suppose that $(C \cap D) \cap \beta$
is unbounded in $\beta$, so $C \cap \beta$ and $D \cap \beta$
are both unbounded in $\beta$,
so $\beta \in C \cap D$.
$C \cap D$ is unbounded:
Take some $\gamma < \kappa$.
Let $\gamma_0 = \gamma$
and inductively define $\gamma_n$ :
If $n$ is even, let $\gamma_n \coloneqq \min C \setminus (\gamma_{n-1}+1)$,
otherwise $\gamma_n \coloneqq \min D \setminus (\gamma_{n-1}+1)$.
Let $\xi = \sup \{\gamma_n : n < \omega\}$.
Then $\xi = \sup \{\gamma_{2n + 2} : n < \omega\} \in D$
and $\xi \in C$ by the same argument,
so $\xi \in C \cap D$
(here it is important, that $\cf(\kappa) > \omega$)
and $\xi > \gamma$.
The case $\alpha > 2$ is similar:
The intersection is closed by exactly the same argument.%
\footnote{``It is even more closed.''}
Let's prove that $\bigcap \{C_{\beta} : \beta < \alpha\}$
is unbounded in $\kappa$.
We will define a sequence $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$%
\footnote{Ordinal multiplication, i.e.~$\alpha \cdot \omega = \sup_{n < \omega} \underbrace{\alpha + \ldots + \alpha}_{n \text{ times}}$.}
as follows:
Let $\gamma_0 \coloneqq \gamma$.
\[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\]
and at limits choose the supremum.
Let $\xi = \sup_{i < \alpha \cdot \omega} \gamma_i
= \sup_{i < \omega} \gamma_{\alpha \cdot n + \beta + 1} \in \bigcap_{\beta < \alpha} C_\beta$,
where we have used that.
$\cf(\kappa) > \alpha \cdot \omega$.
$F \subseteq \cP(a)$ is a \vocab{filter}
\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 $\emptyset \not\in F$, $\kappa \in F$.
Let $\alpha \le \kappa$.
We call $F$ \vocab{$< \alpha$-closed}
iff for all $\gamma < \alpha$ and $\{X_\beta : \beta < \gamma\} \subseteq F$
then $\bigcap \{X_\beta : \beta < \gamma\} \in F$.
Intuitively, a filter is a collection of ``big'' subsets of $a$.
Let $\kappa$ be regular and uncountable.
The \vocab{club filter} is defined as
\cF_{\kappa} \coloneqq \{X \subseteq \kappa : \exists \text{ club } C \subseteq \kappa .~ C \subseteq X\}.
Clearly this is a filter.
We have shown (assuming \AxC to choose contained clubs):
If $\kappa$ is regular and uncountable.
Then $\cF_\kappa$ is a $< \kappa$-closed filter.
Let $\langle A_\beta : \beta < \alpha \rangle$
be a sequence of sets.
The \vocab{diagonal intersection},
is defined to be
\diagi_{\beta < \alpha} A_{\beta} \coloneqq
\{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\} \}.
Let $\kappa$ be a regular, uncountable cardinal.
If $\langle C_{\beta} : \beta < \kappa \rangle$
is a sequence of club subsets of $\kappa$,
then $\diagi_{\beta < \kappa} C_{\beta}$
contains a club.
Let us fix $\langle C_{\beta} : \beta < \alpha \rangle$.
Write $D_{\beta} \coloneqq \bigcap \{C_{\gamma} : \gamma \le \beta\} $
for $\beta < \kappa$.
Each $D_{\beta}$ is a club,
$D_{\beta} \subseteq C_{\beta}$
and $D_{\beta} \supseteq D_{\beta'}$
for $\beta \le \beta' < \kappa$.
It suffices to show that $\diagi_{\beta < \kappa} D_{\beta}$
contains a club.
$\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$.
Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$
is unbounded in $\gamma$.
We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$.
Let $\beta_0 < \gamma$.
We need to see $\gamma \in D_{\beta_0}$.
For each $\beta'$ with $\beta < \beta' < \gamma$,
there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
with $\beta'' \ge \beta', \beta'' < \gamma$.
In particular $\beta'' \in D_{\beta_0}$.
We showed that $D_{\beta_0} \cap \gamma$
is unbounded in $\gamma$,
so $\gamma \in D_{\beta_0}$.
As $\beta_0 < \gamma$ was arbitrary,
this shows that $\gamma \in \diagi_{\beta < n} D_\beta$.
$\diagi_{\beta < \kappa} D_{\beta}$
is unbounded in $\kappa$.
Fix $\gamma < \kappa$.
We need to find $\delta > \gamma$
with $\gamma \in \diagi_{\beta < \kappa} D_\beta$.
Define $\langle \gamma_n : n < \omega \rangle$
as follows:
$\gamma_0 \coloneqq \gamma$
\gamma_{n+1} \coloneqq \min D_{\gamma_n} \setminus (\gamma_n + 1)
We have $\delta \coloneqq \sup_{n < \omega} \gamma_n \in \kappa$
by cofinality of $\kappa$.
We need to show that $\delta \in D_{\overline{\gamma}}$
for all $\overline{\gamma} < \delta$.
If $\overline{\gamma} < \delta$, then $\overline{\gamma} \le \gamma_n$
for some $n < \omega$.
For $m \ge n$, $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_{\overline{\gamma}}$.
So $D_{\overline{\gamma}} \cap \delta$ is unbounded
in $\gamma$, hence $\delta \in D_{\overline{\gamma}}$.
Let $\kappa$ be regular and uncountable.
$S \subseteq \kappa$ is called \vocab{stationary} (in $\kappa$)
iff $C \cap S \neq \emptyset$
for every club $C \subseteq \kappa$.
\item Every $D \subseteq \kappa$ which is club in $\kappa$
is stationary in $\kappa$.
\item There exist disjoint stationary sets:\footnote{Note that clubs can never be disjoint, since their intersection is a club.}
Let $\kappa = \omega_2$.
Let $S_0 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega\}$
and $S_1 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega_1\}$.
Clearly these are disjoint.
They are both stationary:
Let $c \subseteq \kappa$ be a club.
Let $(\xi_i : i \le \omega_1)$
be defined as follows:
$\xi_0 \coloneqq \min C$,
$\xi_i \coloneqq \min (C \setminus \sup_{j < i} \xi_j)$.
For $i \le \omega_1$ we have that $\xi_i = \sup_{j < i} \xi_j$.
In particular $\xi_\omega \in S_0 \cap C$
and $\xi_{\omega_1} \in S_1 \cap C$.
We will show later that if $ \kappa$ is a regular uncountable cardinal,
then every stationary $S \subseteq \kappa$ can be written as
$S = \bigcup_{i < \kappa} S_i$,
where the $S_i$ are stationary and pairwise disjoint.

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@ -142,4 +142,6 @@
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
%\newcommand\diagi{\mathop{\large \Delta}\limits}
\newcommand\diagi{\mathop{\large \Delta}}

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@ -37,6 +37,7 @@