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\title{Can A Large Cardinal Be Forced From A
Condition Implying Its Negation?
\thanks{Both authors wish to thank the CUNY Research
Foundation for having provided partial support
for this research via the first author's
PSC-CUNY Grant 64455-00-33, under which
the second author was a research assistant.
In addition, we wish to thank the referee
and Carl Jockusch, the corresponding editor,
for helpful comments, corrections, and
suggestions which have been incorporated
into this version of the paper and which
have considerably improved the
presentation of the material contained herein.}
\thanks{2000 Mathematics Subject Classifications:
03E02, 03E35, 03E55}
\thanks{Keywords: Jonsson cardinal, Rowbottom cardinal,
strongly compact cardinal.}}
\author{Arthur W.~Apter\\
Department of Mathematics\\
Baruch College of CUNY\\
New York, New York 10010 USA\\
http://math.baruch.cuny.edu/$\sim$apter\\
awabb@cunyvm.cuny.edu\\
\\
Grigor Sargsyan\\
Group in Logic and the Methodology of Science\\
University of California\\
Berkeley, California 94720 USA\\
grigor@math.berkeley.edu}
\date{August 30, 2003\\
(revised February 8, 2004 and June 11, 2004)}
\begin{document}
\maketitle
\begin{abstract}
In this note, we provide an
affirmative answer to the
title question by giving
two examples of cardinals
satisfying conditions
implying they are non-Rowbottom
which can be
%forced to be Rowbottom cardinals.
turned into Rowbottom cardinals
via forcing.
In our second example,
our cardinal is also
non-Jonsson.
\end{abstract}
\newpage
\baselineskip=24pt
%\section{Introduction and Preliminaries}\label{s1}
A well-known phenomenon is that
under certain circumstances,
it is possible to force over
a given model $V$ of ZFC containing
a cardinal $\gk$ satisfying a
large cardinal property $\varphi(\gk)$ to
create a universe $\ov V$ in which
$\varphi(\gk)$ no longer holds,
yet over which $\varphi(\gk)$
can be resurrected via forcing.
A folklore example of this is provided
by supposing that
$V \models ``$ZFC + GCH + $\gk$ is
measurable''.
If $\FP$ is defined as the
%Easton support iteration of
reverse Easton iteration of
length $\gk$ which adds
a Cohen subset to each
inaccessible cardinal below $\gk$,
then in $V^\FP$, $\gk$
is no longer measurable,
by, e.g., a simpler version of
the argument given in
Lemma 2.4 of \cite{AHa}.
However, if one forces over $V^\FP$
by adding a Cohen subset to $\gk$,
then the standard
reverse Easton arguments show that
$\gk$'s measurability has
been resurrected, since the
entire forcing can now be
viewed as the length $\gk + 1$
reverse Easton iteration over $V$
that adds a Cohen subset to
each inaccessible cardinal less
than or equal to $\gk$.
%For additional details, we
%refer readers to \cite{Ha}.
In this note, we
consider a different,
stronger phenomenon.
Specifically, we examine the
following
\bigskip
\setlength{\parindent}{0pt}
Question: Suppose $\varphi$
is a given large cardinal property.
Is it possible to find a formula
$\psi$ in one free
variable in the language of set theory and
a cardinal $\gk$ such that $\psi(\gk)$
holds,
${\rm ZFC} \proves ``\forall \gl[\psi(\gl)
\implies \neg\varphi(\gl)]$'', yet there is a
partial ordering $\FP$ such that
$\forces_{\FP} \varphi(\gk)$?
\bigskip
\setlength{\parindent}{1.5em}
We show that the answer to
our Question is yes
for $\varphi$ the
%formula in one free variable stating that $x$ is Rowbottom by
large cardinal property of Rowbottomness by
proving the following theorem.
\QuietTheorem Theorem.
\begin{enumerate}
\item
There is a formula
$\psi_1(x)$ in one free variable
such that
%if $\gk$ is a cardinal, then
${\rm ZFC} \proves ``\psi_1(\gk)
\implies [\gk$ is not Rowbottom,
yet for some partial ordering
$\FP$,
$\forces_\FP ``\gk$ is a
Rowbottom cardinal carrying
a Rowbottom filter''] for all
cardinals $\gk$''.
\item
There is a formula
$\psi_2(x)$ in one free variable
such that
%if $\gk$ is a cardinal, then
${\rm ZFC} \proves ``\psi_2(\gk)
\implies [\gk$ is not Jonsson,
yet for some partial ordering
$\FP$,
$\forces_\FP ``\gk$ is a
Rowbottom cardinal carrying
a Rowbottom filter''] for all
cardinals $\gk$''.
\end{enumerate}
\noindent Note that $\cal F$ is
a Rowbottom filter for $\gk$
if for any $\gl < \gk$ and
$f : {[\gk]}^{< \go} \to \gl$,
there is some $A \in {\cal F}$
such that $\card{f''{[A]}^{< \go}} \le \go$
(where as usual, ${[\gk]}^{< \go}$
is the set of
both empty and non-empty
finite sequences of
elements of $\gk$,
where each non-empty finite sequence
is written in
increasing order).
In other words, there is
a cardinal $\gk_1$ which
satisfies a condition
making it automatically non-Rowbottom,
yet which can be turned into
a Rowbottom cardinal carrying
a Rowbottom filter via forcing.
Further, there is another
cardinal $\gk_2$ which satisfies
a condition making it
automatically non-Jonsson, yet
which can be turned into
a Rowbottom cardinal
carrying a Rowbottom filter
via forcing.
This is in sharp contrast
to the example given in
the first paragraph of this note
and the examples to be given
in the next to last paragraph
of this note.
There, one simply has a model
$V$, a large cardinal property
$\varphi$, a cardinal $\gk$,
and a partial ordering $\FP$
such that $V \models \neg \varphi(\gk)$
and $V^\FP \models \varphi(\gk)$.
In the Theorem, however, one has that
for {\it every} model $V$, for the
appropriate $\varphi$ and $\psi$, if
$V \models \psi(\gk)$, then both
$V \models \neg \varphi(\gk)$ and
there is a partial ordering
$\FP \in V$ with
$V^\FP \models \varphi(\gk)$.
This provides a strong positive
answer to our Question.
Before beginning the proof of
the Theorem, we briefly
mention that we assume throughout
a basic understanding of large
cardinals and forcing.
We refer readers to \cite{Ka}
for anything left unexplained.
We turn now to the proof
of clause (1) of the Theorem.
\begin{proof}
To prove clause (1) of the Theorem,
let $\gk_1$ be a singular limit of
$\gd$ measurable cardinals, where
$\gd < \gk_1$ is a regular uncountable cardinal.
%It is a theorem of ZFC (see \cite{Ka})
%that $\gk_1$ is not a Rowbottom cardinal.
Take $\psi_1(x)$ as the formula
in one free variable asserting this fact.
Let $\FP$ be the partial ordering
which collapses $\gd$ to $\go$.
After forcing with $\FP$,
by the L\'evy-Solovay results \cite{LS},
a final segment of measurable cardinals
below $\gk_1$ remains measurable.
Further, $\gk_1$ can now be written
as a limit of $\go$ measurable cardinals.
Therefore, by a theorem of Prikry
(see \cite{Ka}, Theorem 8.7,
page 90), $\gk_1$ has become
a Rowbottom cardinal carrying a
Rowbottom filter.\footnote{Although
Theorem 8.7 of \cite{Ka} does not
explicitly state that a limit of
$\go$ measurable cardinals carries
a Rowbottom filter, this is clear
from the proof presented in \cite{Ka}.}
However, it is a theorem of ZFC (see \cite{Ka},
Exercise 8.6, page 90)
that originally,
since $\gk_1$ had uncountable cofinality,
$\gk_1$ was not a
Rowbottom cardinal.
Thus, we have turned a cardinal
satisfying a condition making it non-Rowbottom
via forcing into a Rowbottom cardinal,
thereby proving clause (1) of
the Theorem.
\end{proof}
We note that since the partial
ordering $\FP$ defined above has
cardinality less than $\gk_1$,
our proof of clause (1) of the Theorem
illustrates the interesting
occurrence that it is
possible for small forcing
to turn a provably non-Rowbottom
cardinal into a Rowbottom cardinal.
We turn now to the proof of
clause (2) of the Theorem.
\begin{proof}
To prove clause (2) of
the Theorem, let $\gk_2$
be the least cardinal which
is both regular and a
limit of strongly compact cardinals.
Take $\psi_2(x)$ as the formula
in one free variable
asserting this fact.
We immediately have the
following lemma.
\begin{lemma}\label{l1}
$\gk_2$ is not a Jonsson cardinal.
\end{lemma}
\begin{proof}
If $\gk_2$ were a Jonsson cardinal,
then since $\gk_2$ is strongly
inaccessible, it is a theorem
of Shelah (see \cite{Sh},
Chapters 3 and 4) that $\gk_2$
must also be a Mahlo cardinal.
Since the set $C$ of strongly
compact cardinals below $\gk_2$
is unbounded in $\gk_2$,
by the Mahloness of $\gk_2$,
the collection $C'$ of limit points
of $C$ must contain a strongly
inaccessible cardinal.
This means that $\gk_2$ is not
the least cardinal which is
both regular and a limit
of strongly compact cardinals,
a contradiction which completes
the proof of Lemma \ref{l1}.
\end{proof}
Now that we know that $\gk_2$
is not a Jonsson cardinal,
we define the partial ordering
$\FP$ which turns $\gk_2$
into a Rowbottom cardinal
carrying a Rowbottom filter.
%The partial ordering $\FP$ used
%in the proof of the Theorem
$\FP$
is a version of the ``modified
Prikry forcing'' given in
\cite{He} and \cite{AH} but
defined using more than one ultrafilter.
More specifically, since
$\gk_2$ is both regular and
a limit of strongly compact cardinals, let
$\la \gk^*_\ga : \ga < \gk_2 \ra$
be a sequence of strongly compact
cardinals whose limit is $\gk_2$, and let
$\la \U_\ga : \ga < \gk_2 \ra$ be a
sequence of ultrafilters such that each
$\U_\ga$ is a $\gk^*_\ga$-additive
uniform ultrafilter over $\gk_2$.
$\FP$ may now be defined as the
set of all finite sequences of
the form
$\la \ga_1, \ldots, \ga_n, f \ra$
satisfying the following properties.
\begin{enumerate}
\item $\la \ga_1, \ldots, \ga_n \ra \in
{[\gk_2]}^{< \go}$.
\item $f$ is a function having domain
$T_{\ga_1, \ldots, \ga_n} =
\{\la \gb_1, \ldots, \gb_m \ra
\in {[\gk_2]}^{< \go} :
\la \ga_1, \ldots, \ga_n \ra$
is an initial segment of
$\la \gb_1, \ldots, \gb_m \ra\}$
such that
$f(\la \gb_1, \ldots, \gb_m \ra)
\in \U_{\gb_m}$.
\end{enumerate}
The ordering on $\FP$ is given by
$\la \gb_1, \ldots, \gb_m, g \ra \ge
\la \ga_1, \ldots, \ga_n, f \ra$
($\la \gb_1, \ldots, \gb_m, g \ra$
is stronger than
$\la \ga_1, \ldots, \ga_n, f \ra$)
iff the following criteria are met.
\begin{enumerate}
\item $\la \ga_1, \ldots, \ga_n \ra$
is an initial segment of
$\la \gb_1, \ldots, \gb_m \ra$.
\item For $i = n + 1, \ldots, m$,
$\gb_i \in f(\la \ga_1, \ldots, \ga_n,
%\gb_1,
\ldots, \gb_{i - 1} \ra)$.
\item For every
%element $s$ of their
%common domain,
$s \in \dom(g)$
(which must be a subset of
$\dom(f)$),
$g(s) \subseteq f(s)$.
\end{enumerate}
\begin{lemma}\label{l2}
Given any formula $\varphi$
in the forcing language with
respect to $\FP$ and any condition
$\la \ga_1, \ldots, \ga_n, f \ra
\in \FP$, there is a condition
$\la \ga_1, \ldots, \ga_n, f' \ra
%\in \FP$
\ge \la \ga_1, \ldots, \ga_n, f \ra$
deciding $\varphi$.
\end{lemma}
\begin{proof}
The proof of Lemma \ref{l2}
is essentially the same as the
proof of Lemma 4.1 of \cite{He}
or Lemma 1.1 of \cite{AH},
taking into account that
different ultrafilters are
used in the definition of $\FP$.
We follow the proofs
of these lemmas almost
verbatim, making the
necessary minor changes where warranted.
Specifically, let
$s = \la \ga_1, \ldots, \ga_n \ra$.
For any $t \in T_s$,
%$t \in T_{\ga_1, \ldots, \ga_n}$,
call $t$ sufficient if, for some $g$,
$\la t, g \ra \decides \varphi$.
For $t$ sufficient, let $g_t$ be a witness,
with $g_t(r) = \gk_2$
for all $r \in \dom(g_t)$
if $t$ is not sufficient.
If $s$ is sufficient, then we are done.
If not, then for any $t \in T_s$,
sufficient or otherwise, one of the sets
$$X_t = \{\ga < \gk_2 : \exists g
[\la t^\frown \ga, g \ra \forces \varphi]\},$$
$$Y_t = \{\ga < \gk_2 : \exists g
[\la t^\frown \ga, g \ra \forces \neg
\varphi]\}, {\hbox{ \rm or}}$$
$$Z_t = \{\ga < \gk_2 : \forall g
[\la t^\frown \ga, g \ra\
{\hbox{\rm does not decide}} \ \varphi]\}$$
is an element of $\U_{\max(t)}$.
Let $A_t$ be that set, and for
$i \le {\rm length}(t)$, let
$t \rest i$ be the first
$i$ members of $t$.
For $t \in T_s$,
define $f'$ by
$$f'(t) = f(t) \cap \bigcap_{n \le i \le
{\rm length}(t)} g_{t \rest i}(t) \cap A_t.$$
Note that by the definition of $\FP$,
$f'(t) \in \U_{\max(t)}$, which means
that $\la s, f' \ra$ is a well-defined
member of $\FP$ extending $\la s, f \ra$.
Now, let $t$ be sufficient and of
minimal length $m + 1 > n$, with
$\la t, f'' \ra \ge \la s, f' \ra$ and
$f'' = f' \rest T_t$.
Let $t'$ be the sequence $t$
without its last element.
It then follows that
$A_{t'}$ must be either
$X_{t'}$ or $Y_{t'}$, so
we suppose without loss of
generality that $A_{t'} = X_{t'}$.
It must be the case that
$\la t', f' \rest T_{t'} \ra \forces \varphi$,
since if some extension
$\la t'', g' \ra \forces \neg \varphi$,
such a condition must add elements
to $t'$, since $t'$ isn't sufficient.
The first element added to $t'$,
$\ga$, must come from $X_{t'}$,
yielding a condition
$\la t'^\frown \{\ga\} ^\frown u, g' \ra
\forces \neg \varphi$.
However, by construction,
$$\la t'^\frown \{\ga\} ^\frown u, g' \ra \ge
\la t'^\frown \{\ga\}, f' \rest
T_{t'^\frown \{\ga\}} \ra \ge
\la t'^\frown \{\ga\}, g_{t'^\frown \{\ga\}} \ra
\forces \varphi,$$
which is a contradiction. Thus,
$\la t', f' \rest T_{t'} \ra \decides \varphi$,
which contradicts the minimality
of the length of $t$ for
sufficiency.
This completes the proof of
Lemma \ref{l2}.
\end{proof}
\begin{lemma}\label{l3}
Forcing with $\FP$ adds
no new subsets to
any $\gd < \gk_2$.
\end{lemma}
\begin{proof}
Given $\gd < \gk_2$, suppose that
$p = \la \ga_1, \ldots, \ga_n, f \ra \forces
``\tau \subseteq \gd$''.
Without loss of generality, by
extending $p$
%$\la \ga_1, \ldots, \ga_n, f \ra$
if necessary, we also assume that
$\gk^*_{\ga_n} > \gd$.
Further, by Lemma \ref{l2},
for each $\gb < \tau$, we let
$\la \ga_1, \ldots, \ga_n, f_\gb \ra$
be such that
$\la \ga_1, \ldots, \ga_n, f_\gb \ra
\decides ``\gb \in \tau$''.
Note that the domains of
all of the $f_\gb$ s for
$\gb < \gd$ and $f$ are the same,
namely $T_{\ga_1, \ldots, \ga_n}$.
Therefore, by the choice of $p$ and the
definition of $\FP$, for
each $s \in T_{\ga_1, \ldots, \ga_n}$,
%in the domain
%of all of the $f_\gb$ s and $f$,
$f_\gb(s)$ and $f(s)$ lie in
an ultrafilter $\U_{\max(s)}$ that is
$\gk^*_{\ga_n}$-additive.
This means that
$g(s) = \bigcap_{\gb < \gd} f_\gb(s)
\cap f(s)$ is such that
$g(s) \in \U_{\max(s)}$, and
$q = \la \ga_1, \ldots, \ga_n, g \ra$
is a well-defined element of $\FP$
such that $q \ge p$ and $q$ decides
the statement ``$\gb \in \tau$''
for every $\gb < \gd$.
%Hence, $q \forces ``\tau \in V$''.
Hence, forcing with $\FP$
adds no new subsets to $\gd$.
This completes the proof of
Lemma \ref{l3}.
\end{proof}
The proof of clause (2) of
the Theorem now easily follows.
The usual density arguments show
that after forcing with $\FP$,
$\cof(\gk_2) = \go$.
Further, by Lemma \ref{l3},
since forcing with $\FP$
adds no new bounded subsets
to $\gk_2$,
%in $V^\FP$
after forcing with $\FP$,
$\gk_2$ remains a limit of
measurable cardinals.
Therefore, once again by
Prikry's theorem used in
the proof of clause (1)
of the Theorem,
$\gk_2$ is
%in $V^\FP$
after forcing with $\FP$
a Rowbottom cardinal
carrying a Rowbottom filter.
This completes the proof of
clause (2) of the Theorem,
and hence also completes the
proof of the Theorem.
\end{proof}
It is a theorem of ZFC (see \cite{Ka},
Theorem 8.7, page 90)
that a $\gk_1$ as in clause (1)
of the Theorem
is Jonsson (or more specifically,
is $\gd^+$ Rowbottom).
Thus, in clause (1) of the Theorem,
we actually have that a cardinal
which is both Jonsson and partially
Rowbottom is turned into a
fully Rowbottom cardinal carrying a
Rowbottom filter via small forcing.
This contrasts with what occurs
in clause (2) of the Theorem,
where a cardinal satisfying
a condition making it
automatically non-Jonsson
is transformed into a Rowbottom cardinal
carrying a Rowbottom filter.
Further, as readers may verify for
themselves, our methods of proof
yield that the least cardinal $\gl$
which is both regular and a limit
of cardinals which are $\gl$
strongly compact suffices as the
$\gk_2$ of clause (2) of the Theorem.
We take this opportunity to make some
remarks concerning relative
consistency results related to
the Theorem.
First, as Devlin has shown in \cite{De},
it is consistent, relative to the
existence of a Ramsey cardinal,
for the least Jonsson cardinal $\gk$
not to be Rowbottom.
Since Kleinberg has shown
(see \cite{Kl1}, \cite{Kl2}, or
\cite{Ka}, Propositions 8.15(c)
and 10.18, pages 95 - 96 and
128 - 129 respectively)
%(see \cite{Kl})
that under these circumstances,
$\gk$ is $\gd$ Rowbottom
for some $\gd < \gk$ and
can be transformed into
a Rowbottom cardinal via a
forcing having size
less than $\gk$, we have that
%the conclusions of
%clause (1) of the Theorem
the existence of a non-Rowbottom
Jonsson cardinal which can be changed
into a Rowbottom cardinal
%carrying a Rowbottom filter
via small forcing is
consistent relative to
a Ramsey cardinal.
Further, Woodin has shown
(see page 57 of \cite{La})
that the stationary tower
forcing can be used,
assuming either a proper
class of measurable cardinals
(for class forcing),
or an inaccessible limit of
measurable cardinals with
a Woodin cardinal above it
(for set forcing),
to change the cofinality of
the least inaccessible limit $\gk$
of measurable cardinals to
$\go$ (or indeed, to any
regular cardinal $\gd < \gk$)
without adding bounded subsets
to $\gk$.
This thereby gives that
%the conclusions of
%clause (2) of the Theorem
the existence of a non-Jonsson
cardinal which can be turned into
a Rowbottom cardinal carrying a
Rowbottom filter via forcing
is consistent relative to
assumptions much weaker than
even the existence of a
cardinal $\gl$ which is
$\gl^+$ strongly compact.
However, we emphasize again
that these results,
like the one given in the
first paragraph of this note,
are only relative
consistency results.
This is in sharp contrast
to the Theorem, which is
an outright ZFC theorem, and
thus provides a much stronger
phenomenon.
%We conclude this note with some
In conclusion, we ask
%some additional questions which
%are raised by the results and methods of this note.
%In particular,
if there are
other large cardinal or
combinatorial properties
that provide answers to
our main Question for
a particular cardinal $\gk$.
If so, can these properties
be added by a forcing
which doesn't add bounded subsets
to $\gk$ and doesn't collapse cardinals,
or by a forcing which is small
relative to $\gk$?
Is there such a property implying that
$\gk$ is a regular cardinal?
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\end{document}