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\newcommand{\dom}{\operatorname{dom}}
\newcommand{\ran}{\operatorname{ran}}
\newcommand{\cof}{\operatorname{cof}}
\newcommand{\ot}{\text{o.t.}}
\newcommand{\on}{\text{ON}}
\newcommand{\card}{\text{CARD}}
\newcommand{\ac}{\text{AC}}
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\newcommand{\sP}{\mathcal{P}}
\newcommand{\sI}{\mathcal{I}}
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\newcommand{\ch}{\text{CH}}
\newcommand{\gch}{\text{GCH}}
\newcommand{\cub}{\text{c.u.b.}}
\newcommand{\Cub}{\text{Cub}}
\newcommand{\ns}{\text{NS}}
\newcommand{\fin}{\text{FIN}}
\newcommand{\sB}{\mathcal{B}}
\newcommand{\lub}{\text{l.u.b.}}
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\newcommand{\mbar}[1]{\overline{#1}}
\newcommand{\sat}{\text{sat}}
\begin{document}
\begin{center}
\bf The c.u.b.\ Filter and Silver's Theorem
\end{center}
\bigskip
\section{Ideals and Filters}
We first recall the standard notions of ideal and filter.
\begin{defn}
An \emph{ideal} on a set $X$ is a collection $\sI \subseteq \sP(X)$
of subsets of $X$ satisfying:
\begin{enumerate}
\item
If $A \in \sI$ and $B \subseteq A$, then $B \in \sI$.
\item
If $A,B \in \sI$, then $A \cup B \in \sI$.
\end{enumerate}
We say the ideal $\sI$ is proper if $X \notin \sI$ (equivalently
$\sI \neq \sP(X)$).
\end{defn}
We think of an ideal as a notion of smallness for the subsets of $X$;
those subsets of $X$ which are in $\sI$ are the small ones.
The ``dual'' notion is the concept of a filter:
\begin{defn}
A \emph{filter} on a set $X$ is a collection $\sF \subseteq \sP(X)$
of subsets of $X$ satisfying:
\begin{enumerate}
\item
If $A \in \sF$ and $B \supseteq A$, then $B \in \sF$.
\item
If $A,B \in \sF$, then $A \cap B \in \sF$.
\end{enumerate}
We say the filter $\sF$ is proper if $\emptyset \notin \sF$ (equivalently
$\sF \neq \sP(X)$).
\end{defn}
Recall than an \emph{ultrafilter} on a set $X$ is a maximal filter.
Equivalently, an ultrafilter is a filter $\sF$ with the
property that for every $A \in \sP(X)$, either
$A \in \sF$ or $X-A \in \sF$. It is a standard fact that from the
axiom of choice one may extend any filter on a set $X$ to an
ultrafilter.
\begin{exer}
Show that $\sI \subseteq \sP(X)$ is an ideal iff
$\sF=\{ A \colon X-A \in \sI \}$ is a filter.
\end{exer}
We say an ideal $\sI$ is $\kappa$-additive if whenever
$\alpha < \kappa$ and $\{ A_\beta\}_{\beta < \alpha}$ is an
$\alpha$ sequence of members of $\sI$, then $\bigcup_{\beta< \alpha}
A_\beta \in \sI$. The dual notion would be: a filter $\sF$
is $\kappa$-additive if whenever
$\alpha < \kappa$ and $\{ A_\beta\}_{\beta < \alpha}$ is an
$\alpha$ sequence of members of $\sF$, then $\bigcap_{\beta< \alpha}
A_\beta \in \sI$. Note that $\kappa$-additive refers to closure under
\emph{less than } $\kappa$ unions (or intersections).
The notions of ideal and filter are thus interchangeable, and we
will pass back and forth between the two. For $\sI$ an ideal
(or $\sF$ a filter), we sometimes call the sets $A \in \sI$
(or sets $A$ such that $X-A \in \sF$) ``measure zero.''
We call the $A$ such that $X-A \in \sI$ (or $A \in \sF$)
``measure one.'' If neither $A \in \sI$ nor $X-A \in \sI$, we
say $A$ is ``positive.''
\begin{exer}
Show that for any ideal (or filter) there is a largest $\lambda
\in \card$ such that $\sI$ is $\lambda$-additive. We call
this the \emph{additivity} of the ideal (or filter).
\end{exer}
\begin{exer}
Let $\kappa$ be a cardinal and let $\sI$ be the ideal of
subsets of $\kappa$ which have size $< \kappa$. Identify the
additivity of this ideal.
\end{exer}
If $\sI$ is an ideal (or filter) on a set $X$, an \emph{antichain} is a
collection $\{ A_\alpha \}$ of $\sI$-positive subsets of
$X$ such that $A_\alpha \cap A_\beta \in \sI$ for all $\alpha \neq \beta$.
We say the ideal is $\lambda$-saturated if all anti-chains have size
$< \lambda$. The saturation of the ideal, $\sat(\sI)$ is the
largest $\lambda$ such that $\sI$ is $\lambda$-saturated (which
is easily well-defined).
If $\sI$ is an ideal (or $\sF$ a filter) on a set $X$, and $S \subseteq
X$ is positive, then define the notion of the ideal (or filter)
restricted to $S$, which we denote by $\sI_{| S}$ (or $\sF_{| S}$),
and defined by $\sI_{| S}= (\sI \cap \sP(S)) \cup (X-S)$ (that is we
declare complement of $S$ to be in the restricted ideal, i.e, the
restricted ideal ``lives'' on $S$). Equivalently,
$\sF_{| S}= \{ A \cap S \colon A \in \sF\}$.
\section{Boolean Algebras}
\begin{defn}
A Boolean algebra is a set $\sB$ with two distinguished
elements $0$ and $1$ and two binary operations
$+$, $\cdot$, and one unary operations $A \mapsto \mbar{A}$.
The axioms are:
(commutative laws) $a+b=b+a$, $a\cdot b = b \cdot a$.
(associative laws) $a+(b+c)=(a+b)+c$,
$a \cdot (b \cdot c)=(a \cdot b) \cdot c$.
(distributive laws) $a \cdot(b+c)=a \cdot b + a \cdot c$,
$a+(b \cdot c)= (a+b) \cdot (a+c)$.
(identity laws) $a+a=a$, $a \cdot a=a$.
(de Morgan's laws) $\mbar{a+b}=\mbar{a} \cdot \mbar{b}$,
$\mbar{a \cdot b}= \mbar{a} + \mbar{b}$.
(negation laws) $a+\mbar{a}=1$, $a \cdot \mbar{a}=0$.
($0$, $1$ laws) $0+a=a$, $0 \cdot a=0$, $1+a=1$, $1 \cdot a=a$.
\end{defn}
In analogy with set operations, we sometimes write $\vee$ for $+$
and $\wedge$ for $\cdot$ in a Boolean algebra. We also sometimes
write $a^c$ for $\mbar{a}$.
The axioms imply all of the usual set identities.
\begin{exer}
Show that in any Boolean algebra $a= \mbar{\mbar{a}}$.
Show that $a+a \cdot b=a$ and $a \cdot(a+b)=a$.
Show that $a \cdot b= a $ iff $a+ b = b$ iff $a \cdot (\mbar{b})=0$.
\end{exer}
We write $a \leq b$ in a Boolean algebra to denote $a \cdot b =a$,
or equivalently, $a + b=b$.
We also write $a-b$ for $a \cdot(\mbar{b})$.
We have $a \leq b $ iff $\mbar{b} \leq
\mbar{a}$.
The concepts of ideal, filter, ultrafilter generalize naturally
from $\sP(X)$ to any Boolean algebra.
\begin{defn}
An ideal on the Boolean algebra $\sB$ is a collection
$\sI \subseteq \sB$ satisfying:
\begin{enumerate}
\item
If $a \in \sI$, and $b \leq a$ then $b \in sI$.
\item
If $a$, $b$ are in $\sI$, then $a+b \in \sI$.
\end{enumerate}
The ideal $\sI$ is proper if $1 \notin \sI$.
A filter on the Boolean algebra $\sB$ is a collection
$\sF \subseteq \sB$ satisfying:
\begin{enumerate}
\item
If $a \in \sF$, and $a \leq b$ then $b \in sF$.
\item
If $a$, $b$ are in $\sF$, then $a\cdot b \in \sF$.
\end{enumerate}
The filter $\sF$ is proper if $0 \notin \sF$.
An ultrafilter on $\sB$ is a maximal filter.
\end{defn}
It is straightforward to check that a filter $\sF$ on a Boolean
algebra $\sB$ is an ultrafilter iff for ever $a \in \sB$
either $a \in \sF$ or $\mbar{a} \in \sF$. With $\ac$, every filter
on a Boolean algebra can be extended to an ultrafilter (the proof
is the same as that for filters on $\sP(X)$.
If $X$ is any set, then all $\sB \subseteq \sP(X)$ which
contains $\emptyset$, $X$, and is closed under finite unions,
finite intersections, and complements is a Boolean algebra
under the operations of union, intersection, and complement. We
call such a $\sB$ an algebra of subsets of $X$.
Conversely, Stone's theorem says
any boolean algebra is isomorphic to an algebra of subsets of some
set $X$:
\begin{thm} ($\zfc$)
Every Boolean algebra is isomorphic to an algebra of subsets of
some set $X$.
\end{thm}
\begin{proof}
Let $\sB$ be a Boolean algebra. Let $X=\{ u \colon u$ is an ultrafilter
on $\sB \}$.
Define $\pi \colon \sB \to \sP(X)$ by
$\pi(a)= \{ u \in X \colon a \in u \}$. Let $S=\ran(\pi)$.
We claim that $\pi$ is a Boolean algebra isomorphism between
$\sB$ and the algebra of sets $S$. Clearly $\pi(0)=\emptyset$
and $\pi(1)=X$.
That $S$ is a field of sets
and $\pi$ is a homomorphism (i.e., preserves the Boolean operations)
follows from the equations: $\pi(a \cdot b)= \pi(a) \cap \pi(b)$,
$\pi(a+b)=\pi(a) \cup \pi(b)$, $\pi(\mbar{a})=X-\pi(a)$.
For example, the first equation says $a \cdot b$ is in an ultrafilter
iff $a$ and $b$ are. It is immediate from the definition that this
in fact holds for all filters. For the second equation, note that
$\pi(a), \pi(b) \subseteq \pi(a+b)$ as $a, b \leq a+b$.
If $u \in \pi(a+b)$ but $u \notin \pi(a)$ and $u \notin \pi(b)$, then
since $u$ is an ultrafilter, $u \in \pi(\mbar{a})$ and
$u \in \pi(\mbar{b})$. Since $u$ is a filter,
$u \in \pi(\mbar{a} \cdot \mbar{b})$, and so
$u \in \pi((\mbar{a} \cdot \mbar{b} ) \cdot (a+b))= \pi(0)=\emptyset$.
The third equation follows from the fact that any ultrafilter
must contain either $a$ or $\mbar{a}$. It remains to show that
$\pi$ is one-to-one. Suppose $a \neq b$. Without loss of
generality $a \nleq b$ (since if $a \leq b$ and $b \leq a$ then
$a=a\cdot b=b$). So, $a-b \neq 0$. Let $u$ be an ultrafilter on
$\sB$ with $a-b \in u$. Then $u \in \pi(a)$ but $u \notin \pi(b)$.
\end{proof}
\begin{defn}
A Boolean algebra $\sB$ is \emph{complete} is for every
$A \subseteq \sB$, a least upper bound, $\lub(A)$ for
the elements of $A$ under $\leq$ exists. Equivalently,
for every $A$ the greatest lower bound $\glb(A)$ exists. We
also write $\Sigma(A)$, $\sup(A)$ for $\lub(A)$ and
$\Pi(A)$, $\inf(A)$ for $\glb(A)$. A Boolean algebra is
said to be $\kappa$-complete if $\Sigma(A)$, $\Pi(A)$
exists for all $A$ of size $< \kappa$.
\end{defn}
For example, $\sP(X)$ is a complete Boolean algebra.
On the other hand, $\sP(\omega)/\fin$ is not complete, where
$\fin$ denotes the ideal of finite subsets of $\omega$.
To see this, let $A_n$ be disjoint infinite subsets of $\omega$
whose union is $\omega$. Then $\{ [A_n] \}_{n \in \omega}$ does
not have a least upper bound.
The notions of $\kappa$-additive and $\kappa$-saturated generalize
from ideals on a set $X$ (i.e., the Boolean algebra $\sP(X)$)
to arbitrary Boolean algebras:
\begin{defn}
An ideal $\sI$ on a $\kappa$-complete
Boolean algebra $\sB$ is said to be $\kappa$-additive if
$\Sigma(A) \in \sI$ whenever $A \subseteq \sI$ and $|A| < \kappa$.
A Boolean algebra is $\kappa$-saturated if every antichain in
$\sB$ has size $< \kappa$. $\sat(\sB)$ is the largest $\kappa$
such that $\sB$ is $\kappa$-saturated.
\end{defn}
Thus, an ideall $\sI$ on $\kappa$ is $\lambda$-saturated iff
the Boolean algebra $\sP(\kappa)/ \sI$ is $\lambda$-saturated.
We will be mainly interested in complete Boolean algebras.
For complete Boolean algebras it is a theorem that
$\sat(\sB)$ is a regular cardinal.
\section{The \cub\ Filter}
We introduce now a specific filter of basic importance called the
\cub\ filter (the corresponding ideal is called the
non-stationary ideal).
\begin{defn}
If $A \subseteq \on$, then the \emph{closure} of $A$, $\mbar{A}$,
is the set of
all $\alpha \in \on$ such that $\forall \beta < \alpha\
\exists \gamma \ (\beta < \gamma \leq \alpha \wedge \gamma \in A)$.
We cay $A$ is \emph{closed} if $A= \mbar{A}$.
\end{defn}
It is easy to see that $\mbar{A}$ consists of $A$ together with
the ordinals $\alpha$ which are limit points of $A$, that is,
$\alpha$ is a limit ordinal and $A$ is unbounded in $\alpha$.
Thus, $A$ is closed iff it contains all its limit points.
The topological terminology is justified. We can put a
topology on the ordinals (order topology) by defining the basic open
sets to be of the form $(\alpha, \beta)= \{ \gamma \colon
\alpha < \gamma < \beta\}$ (together with $\{ 0\}$ since $0$
is the least element in the ordering of ordinals). It is easily
checked that is a base for a topology (in fact, this is true for
any linear ordering on any set). A neighborhood base at $\alpha$
consists of sets of the form $(\beta, \alpha]$, where $\beta < \alpha$.
In this topology, the closure operation defined
above is just topological closure in the order topology.
\begin{defn}
Let $\alpha$ be a limit ordinal. We say $C \subseteq \alpha$ is
\cub\ is $C$ is closed and unbounded in $\alpha$.
If $\cof(\alpha)> \omega$, then the \cub\ filter on $\alpha$,
$\Cub(\alpha)$ is defined to be the collection of subsets of $\alpha$
which contain a \cub\ set. The corresponding ideal is denoted $\ns(\alpha)$;
the ideal of non-stationary subsets of $\alpha$ (terminology
explained below).
\end{defn}
\begin{exer}
Show that if $\cof(\alpha)=\omega$ then $\Cub(\alpha)$ is not
a filter.
\end{exer}
We let $\sP(\kappa)/\ns$ denote the set of equivalence classes
$[A]$, for $A \in \sP(\kappa)$, under the equivalence relation
$A \sim B$ iff $A \triangle B \in \ns(\kappa)$. In fact, for
any ideal $\sI$ on $\kappa$ we may consider
$\sP(\kappa)/\sI$. This forms a Boolean algebra.
\begin{lem}
Suppose $\cof(\alpha) > \omega$. Then $\Cub(\alpha)$
is a filter, and is $\cof(\alpha)$-additive. In fact, the
intersection of $< \cof(\alpha)$ many \cub\ subsets of $\alpha$ is
\cub.
\end{lem}
\begin{proof}
By definition if $A \in \Cub(\alpha)$ and $B \supseteq A$ then
$B \in \Cub(\alpha)$. Suppose $\delta < \cof(\alpha)$
and $\{ A_ \beta \}_{\beta < \delta}$ is a sequence of
sets in $\Cub(\alpha)$. Using $\ac$, we may assume that all of the
$A_\beta$ are actually \cub\ subsets of $\alpha$, and
show their intersection is \cub.
Clearly $\bigcap_{\beta< \delta} A_\beta$ is closed. We must show it
is unbounded in $\alpha$. For each $\beta < \delta$, let
$f_\beta \colon \alpha \to \alpha$ be given by $f_\beta(\gamma)=
$ the least element of $A_\beta$ which is $> \gamma$.
Fix $\eta < \alpha$. Let $\eta_0=\eta$, and let
$\eta_{n+1}= \sup_{\beta< \delta} f_\beta(\eta_n)$. Note that
$\eta_{n+1} < \alpha$ as $\cof(\alpha)> \delta$. Since also
$\cof(\alpha)> \omega$, $\eta_\omega= \sup_n(\eta_n) < \alpha$.
Each $A_\beta$ is unbounded in $\eta_\omega$ (as there is a point of
$A_\beta$ between $\eta_n$ and $\eta_{n+1}$ for any $n$), and thus
$\eta_\omega \in A_\beta$ for all $\beta< \delta$.
\end{proof}
In discussing the \cub\ filter $\Cub(\alpha)$, there is actually no loss
of generality is assuming $\alpha$ is a regular cardinal.
For assume $\cof(\alpha)=\kappa$. Let $\{ \gamma_\eta \}_{\eta < \kappa}$
be a continuous, increasing, cofinal sequence in $\alpha$.
By continuous we mean that for $\eta$ limit that
$\gamma_\eta=\sup_{\eta' < \eta} \gamma_{\eta'}$. Let $C=\{ \gamma_\eta
\colon \eta < \kappa\}$. Then $C$ is \cub\ in $\alpha$.
The map $A \mapsto A'=\{ \gamma_\eta \colon \eta \in A\}$ is a bijection
between $\sP(\kappa)$ and $\sP(C)$ which preserves the notion of
\cub\ since the $\gamma_\eta$ are continuous. Thus we have an isomorphism
between $\sP(\kappa)/\Cub$ and $\sP(C)/\Cub$. Finally, the map
$A \mapsto A \cap C$ is a Boolean algebra isomprphism between
$\sP(\alpha)/\Cub$ and $\sP(C)/\Cub$ (the map is one-to-one since
$C$ is \cub). Thus, $\sP(\kappa)/\Cub \cong \sP(\alpha)/\Cub$ as
Boolean algebras. Thus, as far as discussions concerning the \cub\
filter are concerned, we may replace $\alpha$ by the set $C$ of
size $\kappa=\cof(\alpha)$.
\begin{defn}
Let $\kappa$ be a cardinal and $A_\alpha \subseteq \kappa$
for $\alpha < \kappa$. The \emph{diagonal intersection} of the
$A_\alpha$ is defined by $\triangledown A_\alpha=
\{ \beta < \kappa \colon \forall \alpha < \beta\ (\beta \in A_\alpha \}$.
The \emph{diagonal union} is defined by
$\triangle A_\alpha = \{ \beta < \kappa \colon \exists \alpha < \beta\
(\beta \in A_\alpha)\}$.
\end{defn}
\begin{defn}
A filter $\sF$ (or ideal $\sI$) is said to be \emph{normal}
if whenever $A_\alpha$, $\alpha< \kappa$, are in $\sF$ (or $\sI$), then
$\triangledown A_\alpha \in \sF$ (resp.\ $\triangle A_\alpha \in \sI$).
\end{defn}
\begin{lem} \label{normal}
For every regular cardinal $\kappa$, the filter $\Cub(\kappa)$
is normal.
\end{lem}
\begin{proof}
Assume $A_\alpha \in \cub(\kappa)$ for all $\alpha < \kappa$. Using
$\ac$, let $C_\alpha \subseteq A_\alpha$ be \cub. It suffices to show that
$\triangledown C_\alpha$ is \cub\ in $\kappa$. The
diagonal intersection is easily closed, we show it is also unbounded.
For each $\alpha < \kappa$, let $f_\alpha \colon \kappa \to \kappa$
be given by $f_\alpha(\eta)=$ least element of $C_\alpha$ greater than
$\eta$. Let $\eta_0< \kappa$. Define $\eta_{n+1}=
\sup_{\alpha < \eta_n} f_\alpha(\eta_n)$. Let $\eta_\omega=
\sup_n \eta_n$. Note that if $\alpha < \eta_\omega$, then for
all $n$ such that $\eta_n > \alpha$, there is a point of $C_\alpha$
between $\eta_n$ and $\eta_{n+1}$, and hence $\eta_\omega \in C_\alpha$.
Thus, $\eta_\omega \in \triangledown A_\alpha$.
\end{proof}
An immediate but important consequence of this lemma is Fodor's
theorem. To state it, we introduce the important notion of
stationarity.
\begin{defn}
Let $\kappa$ be a regular cardinal. Then $S \subseteq \kappa$ is
\emph{stationary} if $S \cap C \neq \emptyset$ for every \cub\
$C \subseteq \kappa$.
\end{defn}
Note that $S$ being stationary is just saying that $S$ is positive
with respect to the $\Cub$ filter on $\kappa$. That is, $S$ is not
in the corresponding ideal (which is why we called this
ideal the non-statioary ideal).
\begin{thm} (Fodor's Theorem) \label{fodor}
Let $\kappa$ be a regular cardinal, $S \subseteq \kappa$
be stationary, and $f \colon S \to \kappa$
be \emph{pressing down}, that is, $f(\alpha) < \alpha$ for all
$\alpha \in S$. Then there is a stationary set $S' \subseteq S$ on which $f$
is constant.
\end{thm}
\begin{proof}
If not, then for all $\alpha \in S$ there is a set $A_\alpha \in
\Cub(\kappa)$ such that $f(\beta) \neq \alpha$ for all
$\beta \in A_\alpha \cap S$. From lemma~\ref{normal},
$\triangledown A_\alpha \in \Cub(\kappa)$ (for $\alpha \notin S$ we may
take $A_\alpha=\kappa$), and thus there is some
$\beta \in (\triangledown A_\alpha) \cap S$ as $S$ is stationary.
Then $f(\alpha)<\alpha$ and so $\alpha \in A_{f(\alpha)} \cap S$,
a contradiction to the definition of $A_{f(\alpha)}$.
\end{proof}
\begin{exer}
Let $\kappa$ be regular and $f_\alpha \colon \kappa \to \kappa$ for all
$\alpha < \kappa$. Show that $C=\{ \beta< \kappa \colon
\forall \alpha < \beta\ (\beta$ is closed under $f_\alpha) \}$ is
\cub\ in $\kappa$.
\end{exer}
If $\lambda < \kappa$ are regular cardinals, then $S^\kappa_\lambda=
\{ \alpha < \kappa \colon \cof(\alpha)=\lambda \}$ is stationary in
$\kappa$. For example, for $\kappa=\aleph_2$ this gives two disjoint
stationary subsets of $\aleph_2$, namely $S_{\omega}$ and $S_{\omega_1}$.
We will show now more generally that any stationary subset
$S \subseteq \kappa$ of a regular cardinal $\kappa$ can
be split into $\kappa$ many disjoint stationary subsets. For
successor $\kappa$ this is due to Ulam, and for limit $\kappa$
to Solovay.
We consider first the successor case and prove a slightly
more general result.
\begin{thm} (Ulam) \label{ulam}
Let $\kappa$ be a successor cardinal and $\sI$ a $\kappa$-additive
ideal on $\kappa$ containing all the singletons. Then there is a $\kappa$
size family of pairwise disjoint $\sI$-positive subsets of $\kappa$.
\end{thm}
\begin{proof}
Let $\kappa=\lambda^+$. For each $\rho < \kappa$ let $f_\rho
\colon \lambda \to \kappa$ be a bijection. For each $\alpha < \lambda$
and $\beta < \kappa$ let $X^\alpha_\beta=
\{ \rho > \beta \colon f_\rho(\alpha)=\beta \}$.
For each $\beta < \kappa$ there is an $\alpha(\beta)< \lambda$
such that $X^{\alpha(\beta)}_{\beta} \notin \sI$ since $\sI$ is
$\kappa$-additive and $\bigcup_{\alpha< \lambda} X^\alpha_\beta
=\kappa- (\beta+1)$, which is not in $\sI$. For some $\alpha_0 < \lambda$
we must have $|\{ \beta \colon \alpha(\beta)=\alpha_0 \}|=\kappa$.
If $S=\{ \beta \colon \alpha(\beta)=\alpha_0 \}$, then for
$\beta_1 \neq \beta_2 \in S$ we have $X^{\alpha_0}_{\beta_1} \cap
X^{\alpha_0}_{\beta_2}=\emptyset$.
\end{proof}
\begin{cor} \label{corulam}
If $\kappa$ is a successor cardinal and $S \subseteq \kappa$
is stationary, then $S$ can be split into $\kappa$ many
paiwise disjoint stationary subsets.
\end{cor}
\begin{proof}
Consider $\sI_{| S}$,
the non-stationary ideal
restricted to $S$. This is a $\kappa$-additive, proper ideal containing
all the singletons. From theorem~\ref{ulam}, let
$A_\alpha$, $\alpha < \kappa$, be a $\kappa$ sequence of pairwise
disjoint $\sI$-positive subsets of $\kappa$. Then $A'_\alpha=
A_\alpha \cap S$ form a $\kappa$-sequence of pairwise disjoint $\sI$-positive
subsets of $S$. We can enlarge one, if necessary, so they
union to $S$.
\end{proof}
If $\kappa$ is a regular cardinal, and $S \subseteq \kappa$ is
stationary, we define the set of \emph{thin points} $\tilde{s} \subseteq
S$ by $\alpha \in \tilde{S}$ iff $S \cap \alpha$ is not stationary
in $\alpha$.
\begin{lem}
Let $\kappa$ be regular and $S \subseteq \kappa$ be stationary and
consist of limit ordinals. Then $\tilde{S}$ is stationary.
\end{lem}
\begin{proof}
Let $C \subseteq \kappa$ be \cub. Let $\alpha$ be the least limit point of
$C$ which is in $S$ (which exists as $C'$ is also \cub). If $\cof(\alpha)
> \omega$, then $C' \cap \alpha$ is \cub\ in $\alpha$ and is
disjoint from $S$, and so $\alpha \in S'$. If $\cof(\alpha)=
\omega$, then there is an $\omega$ sequence of successor ordinals
cofinal in $\alpha$, which also gives a \cub\ subset of
$\alpha$ missing $S$.
\end{proof}
We now prove the limit case of theorem~\ref{corulam}. Actually, the
proof (due to Solovay) works for both limit and successor cardinals,
and provides a different proof of theorem~\ref{ulam}.
\begin{thm}
Let $\kappa$ be a regular cardinal. Then every stationary $S \subseteq \kappa$
can be split into $\kappa$ many pairwise disjoint stationary subsets.
\end{thm}
\begin{proof}
Let $\kappa$ be regular, and $S \subseteq \kappa$ be stationary. Without
loss of generality we may assume $S$ consists of limit ordinals.
Let $\tilde{S}$ be the thin points of $S$. For each $\alpha \in \tilde{S}$,
let $\eta^\alpha_\xi$, $\xi < \cof(\alpha)$, be an increasing
continuous sequence with supremum $\alpha$ which misses $S$.
We claim that there is $\xi$ such that for all $\delta < \kappa$ the
set $\{ \alpha \in \tilde{S} \colon \eta^\alpha_\xi > \delta\}$ is stationary.
If not, then for each $\xi$ there is a $\rho(\xi) < \kappa$ and a
\cub\ set $C_\xi$ such that for all $\alpha \in \tilde{S} \cap C_\xi$
we have $\eta^\alpha_\xi < \rho(\xi)$ (if $\eta^\alpha_\xi$ is defined).
Let $C = \triangledown C_\xi$, and
let $D \subseteq C$ be \cub\ and closed under the function
$\xi \mapsto \rho(\xi)$. Since $\tilde{S}$ is stationary, let
$\alpha < \beta$ be two elements of $\tilde{S} \cap D$.
Then for each $\xi \in \cof(\beta) \cap \alpha$ we have
$\eta^\beta_\xi < \alpha$. This shows that $\cof(\beta) \geq \alpha$
and that $\eta^\beta_\alpha$ is defined and equal to $\alpha$ (since
the sequence $\eta^\beta_\xi$ is continuous). By definition of the
$\eta^\beta_\xi$, this shows $\alpha \notin \tilde{S}$, a contradiction.
Fix now $\xi$ as in the claim. Note that $\alpha \mapsto
\eta^\alpha_\xi$ is pressing down. For each $\gamma< \kappa$, by the claim
and Fodor's theorem there is a $\tau(\gamma)> \gamma$ and a stationary
set $S_\gamma \subseteq \tilde{S}$ such that $\eta^\alpha_\xi=\tau(\gamma)$
for all $\alpha \in S_\gamma$. Since $\kappa$ is regular,
there is a $\kappa$ size set $A \subseteq \kappa$ such that
$\tau(\alpha) \neq \tau(\beta)$ for $\alpha\neq \beta \in A$.
Then the sets $S_\delta=\{ \alpha \in \tilde{S} \colon
\eta^\alpha_\xi= \tau(\delta) \}$, for $\delta \in A$,
are pairwise disjoint and stationary.
\end{proof}
\section{Silver's Theorem}
We prove a theorem of Silver which shows a significant restriction
on the continuum function at singular cardinals of uncountable
cofinality.
\begin{thm}
Let $\kappa$ be a singular cardinal of uncountable cofinality.
If the $\gch$ hols below $\kappa$ (i.e., $\forall \lambda < \kappa\
(2^\lambda= \lambda^+)$), then it holds at $\kappa$ as well.
\end{thm}
\begin{proof}
Let $\kappa_\alpha$, $\alpha< \cof(\kappa)$ be an increasing, continuous
sequence of cardinals cofinal in $\kappa$. For $A \subseteq \kappa$
consider the function $f_A$ with domain $\cof(\kappa)$ where
$F_A(\alpha)=A \cap \kappa_{\alpha}$. Since
$2^{\kappa_\alpha}=\kappa_{\alpha+1}$, we may identify $f_A$ with a function
satisfying $f(\alpha) \in \kappa_{\alpha+1}$. Consider the collection
$F=\{ f_A \colon A \subseteq \kappa \}$ of all such functions.
Note that this forms an \emph{almost disjoint} family of functions, that is,
if $A \neq B$ then $\exists \alpha < \cof(\kappa)\ \forall \beta > \alpha\
(f_A(\beta) \neq f_B(\beta))$.
Let $g \colon \cof(\kappa) \to \kappa$ with $g(\alpha) < \kappa_{\alpha
+1}$ for all $\alpha$. Let $F_g$ denote those $f \in F$ such that
$\{ \alpha < \cof(\kappa) \colon f(\alpha) \leq g(\alpha)\}$ is stationary.
We claim that for any such $g$, $|F_g| \leq \kappa$. To see this,
let $\pi_\alpha \colon g(\alpha)+1 \to \kappa_\alpha$ be a bijection.
If $f \in F_g$
then there is a stationary set $S_f \subseteq \cof(\kappa)$ and an ordinal
$\delta_f < \kappa$ such that for all $\alpha \in S_f$,
$\pi_\alpha(f(\alpha)) < \delta_f$ (by Fodor's theorem).
Let $h_f \colon S_f \to \delta_f$ be the function $h_f(\alpha)=
\pi_\alpha(f(\alpha))$. The map $f \mapsto (S_f, \delta_f, h_f)$
is one-to-one on $F$ since $(S_f, \delta_f, h_f)$ determines $f$ on $S$, which
determines $f \in F$ as $F$ is an almost disjoint family.
There are at most $2^{\cof(\kappa)}< \kappa$ choices for $S_f$,
$\kappa$ many choices for $\delta_f$, and $\sup_{\delta < \kappa} \delta^
{\cof(\kappa)}< \kappa$ many choices for $h_f$. Thus, $|F_g|
\leq \kappa$.
We now show that $|F| \leq \kappa^+$. We define a sequence
$f_\alpha \in F$ recursively so that for $\beta < \alpha$
we have $\{ \xi \colon
f_\beta(\xi) < f_{\alpha}(\xi) \}$ is stationary.
Assume $f_\beta$ has been defined for
$\beta < \alpha$. If for every $f \in F$ there is a $\beta < \alpha$
such that $\{ \xi \colon f(\xi) \leq f_\beta(\xi)\}$ is stationary,
then stop the construction. Otherwise, let $f_{\alpha} \in F$ be such
that $\forall \beta < \alpha\ \{ \xi \colon
f_\beta(\xi) < f_{\alpha}(\xi) \} \in \Cub(\cof(\kappa))$.
In particular, for all $\beta < \alpha$,
$\{ \xi \colon
f_\beta(\xi) < f_{\alpha}(\xi) \}$ is stationary. $f_{\kappa^+}$
cannot be defined by the claim. Thus, we end with a collection
$\{ f_\alpha\}_{\alpha < \lambda}$, where $\lambda \leq \kappa^+$.
Every $f \in F$ is then in some $F_{f_{\alpha}}$, and from the claim
it follows that $|F| \leq \kappa \cdot \lambda \leq \kappa^+$.
\end{proof}
\end{document}