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%
\title{Level by Level Inequivalence beyond Measurability
\thanks{2010 Mathematics Subject Classifications:
03E35, 03E55.}
\thanks{Keywords: Supercompact cardinal, strongly
compact cardinal,
%level by level equivalence between strong
%compactness and supercompactness,
level by level inequivalence between strong
compactness and supercompactness, non-reflecting
stationary set of ordinals, Magidor iteration
of Prikry forcing.}}
\author{Arthur W.~Apter\thanks{The
author's research was partially
supported by PSC-CUNY grants.}\\
Department of Mathematics\\
Baruch College of CUNY\\
New York, New York 10010 USA\\
and\\
The CUNY Graduate Center, Mathematics\\
365 Fifth Avenue\\
New York, New York 10016 USA\\
http://faculty.baruch.cuny.edu/apter\\
awapter@alum.mit.edu}
\date{April 25, 2010\\
(revised May 14, 2011)}
\begin{document}
\maketitle
\begin{abstract}
We construct models
containing exactly one supercompact cardinal in which
%$\gk$ in which
level by level inequivalence between
strong compactness and supercompactness holds.
In each model, above the supercompact cardinal, there are finitely many
%non-supercompact
strongly compact cardinals, and the strongly
compact and measurable cardinals precisely coincide.
\end{abstract}
\baselineskip=24pt
Say that a model containing supercompact cardinals
satisfies {\em level by level inequivalence between
strong compactness and supercompactness} if
for every non-supercompact measurable cardinal
$\gk$, there is some $\gl > \gk$ such that
$\gk$ is $\gl$ strongly compact yet $\gk$
is not $\gl$ supercompact.
Models containing exactly one supercompact
cardinal in which level by level inequivalence
between strong compactness and supercompactness
holds have been constructed in
\cite[Theorem 2]{A02} and \cite[Theorem 2]{A10}.\footnote{Note
that the dual notion of {\em level by level equivalence
between strong compactness and supercompactness}
was first studied by the author and Shelah in
\cite{AS97a}, to which we refer readers for
additional details.}
(See also \cite{AGH}.)
A key feature of each of these models, however, is
a rather restricted large cardinal structure.
In particular, there do not exist in any
of these models cardinals $\gk < \gl$ such that
$\gk$ is $\gl$ supercompact and $\gl$ is measurable,
nor do there exist in any of these models
non-supercompact strongly compact cardinals.
This prompts us to ask
\bigskip\noindent {\bf Question 1:} Is it possible to construct models
containing at least one supercompact cardinal in which
level by level inequivalence between strong compactness
and supercompactness holds, and in which there are
measurable cardinals which are supercompact up to
(and even beyond) a measurable cardinal?
\bigskip\noindent {\bf Question 2:} Is it possible to construct models
containing at least one supercompact cardinal in which
level by level inequivalence between strong compactness
and supercompactness holds, and in which there are
non-supercompact strongly compact cardinals?
\bigskip The purpose of this paper is to answer the
above questions in the affirmative.
Specifically, we will prove the following theorem.
\begin{theorem}\label{t1}
Suppose $V \models ``$ZFC + $\la \gk_i \mid i < \go \ra$
are the first $\go$ supercompact cardinals''.
Let $n \in \go$, $n \ge 1$ be fixed but arbitrary. There
is then a partial ordering $\FP \subseteq V$, a
model $\ov V \subseteq V^\FP$, and a
sequence of cardinals $\gl_0 < \cdots < \gl_n$
such that $\ov V \models ``\gl_0$ is supercompact +
Level by level inequivalence between strong compactness
and supercompactness holds''. In $\ov V$,
$\gl_1, \ldots, \gl_n$ are the first $n$
measurable cardinals above $\gl_0$, each $\gl_i$
for $i = 1, \ldots, n$ is strongly compact,
and there are no measurable cardinals above $\gl_n$.
\end{theorem}
In $\ov V$, the supercompact cardinal $\gl_0$
is clearly an example of a measurable cardinal
which is supercompact beyond a finite number
of measurable cardinals. However, by reflection,
there are unboundedly many in $\gl_0$ measurable
cardinals which are also supercompact beyond
a finite number of measurable cardinals.
In fact, as our proof will show, {\em every}
measurable cardinal $\gk < \gl_0$ will be
strongly compact beyond a finite number of
measurable cardinals (and much more).
We will comment further on this towards the
end of the paper.
Before beginning the proof of Theorem \ref{t1},
we elaborate briefly on our terminology.
Suppose $\gk < \gl$ are cardinals.
The partial ordering $\FP$ is {\em $\gk$-directed closed}
if every directed subset of $\FP$ of cardinality
less than $\gk$ has a common extension.
$\gk$ is {\em ${<} \gl$ strongly compact} if
$\gk$ is $\gd$ strongly compact for every $\gd < \gl$.
\begin{pf}
Turning to the proof of Theorem \ref{t1},
let $V$ be as in the hypotheses of this theorem.
Without loss of generality, by first forcing
GCH and then forcing with a partial ordering
such as the ones described in \cite{A83}
and \cite{A98}, we may assume in addition that
$V \models ``$For each $i < \go$, $\gk_i$
has its supercompactness indestructible
under $\gk_i$-directed closed forcing \cite{L} and
$2^{\gk_i} = \gk^+_i$''.
We now describe the first partial ordering $\FQ$
used in the proof of Theorem \ref{t1}.
For any $i < \go$, let $\FQ_i$ be the
reverse Easton iteration of length $\gk_{i + 1}$
which adds a non-reflecting stationary set of
ordinals of cofinality $\gk_i$ to each
measurable cardinal in the open interval
$(\gk_i, \gk_{i + 1})$.
(See, e.g., \cite{A98}, \cite{AC1},
or \cite{AS97a} for a
more complete description of this partial ordering.)
Next, take $\FQ = \prod_{i < \go} \FQ_i$ as the full
support product.
By its definition, $\FQ$ is $\gk_0$-directed closed.
Consequently, by our hypotheses on $V$,
$V^\FQ \models ``\gk_0$ is supercompact''.
However, the following is also true.
\begin{lemma}\label{l1}
For each $i < \go$,
$V^\FQ \models ``$No cardinal $\gd \in (\gk_i, \gk_{i + 1})$
is measurable, and $\gk_{i + 1}$ is ${<}\gk_{i + 2}$ strongly compact''.
\end{lemma}
\begin{proof}
Let $i < \go$ be fixed but arbitrary.
Write $\FQ = \FQ^i \times \FQ_i \times \FQ_{< i}$, where
$\FQ^i = \prod_{j > i} \FQ_j$ and
$\FQ_{< i} = \prod_{j < i} \FQ_j$. Because by
its definition, $\FQ^i$ is in fact
$\gk^+_{i + 1}$-directed closed,
$V^{\FQ^i} \models ``\gk_{i + 1}$ is supercompact, and
$2^{\gk_{i + 1}} = \gk^+_{i + 1}$''.
Therefore, the same argument as mentioned in
\cite[page 1908, last paragraph]{AC1} literally
unchanged now shows that
$V^\FQ \models ``$No cardinal $\gd \in (\gk_i, \gk_{i + 1})$
is measurable, and $\gk_{i + 1}$ is ${<}\gk_{i + 2}$ strongly compact''.
This completes the proof of Lemma \ref{l1}.
\end{proof}
Let ${V_*} = V^\FQ$. By Lemma \ref{l1} and the
sentence immediately preceding its statement,
${V_*} \models ``\la \gk_i \mid i < \go \ra$
is a sequence of measurable cardinals such that
for each $i < \go$, $\gk_{i + 1}$ is the least
measurable cardinal greater than $\gk_i$, and
$\gk_i$ is ${<}\gk_{i + 1}$ strongly compact''.
Therefore, since ${V_*} \models ``\gk_0$ is
supercompact'', let $j : {V_*} \to M$ be a
$\gl$ supercompactness embedding for
$\gl$ sufficiently large with the property that
$M \models ``\la \gk_i \mid i < \go \ra$
is a sequence of measurable cardinals such that
for each $i < \go$, $\gk_{i + 1}$ is the least
measurable cardinal greater than $\gk_i$, and
$\gk_i$ is ${<}\gk_{i + 1}$ strongly compact''.
Working in ${V_*}$,
by reflection, for each $\gg < \gk_0$, there is
a sequence $\la \gd_i \mid i < \go \ra$ of
measurable cardinals with $\gd_0 > \gg$ such
that for each $i < \go$,
$\gd_{i + 1}$ is the least measurable cardinal
greater than $\gd_i$, and $\gd_i$ is ${<}\gd_{i + 1}$
strongly compact. This now allows us to define
$\FR \in {V_*}$ as the Magidor iteration of
Prikry forcing \cite{Ma} of length $\gk_0$
which adds a Prikry sequence to each measurable cardinal
$\gd < \gk_0$ for which for $\la \gd_i \mid i < \go \ra$
the first $\go$ many measurable cardinals greater than
or equal to $\gd$ (with $\gd = \gd_0$), there is some
$i < \go$ such that $\gd_i$ is not ${<}\gd_{i + 1}$ strongly compact.
In other words, $\FR$ iteratively destroys
via a Magidor iteration of Prikry forcing of
length $\gk_0$ any measurable cardinal $\gd < \gk_0$
which cannot be the first member of a sequence of measurable
cardinals reflecting the aformentioned properties of
$\la \gk_i \mid i < \go \ra$.
\begin{lemma}\label{l2}
Let $\gl = \sup_{i < \go} \gk_i$. Then
$V_*^\FR \models ``\gk_0$ is $\gl$ supercompact''.
\end{lemma}
\begin{proof}
Let $j : {V_*} \to M$ be an elementary embedding
witnessing the $\gl$ supercompactness of $\gk_0$
generated by a supercompact ultrafilter over $P_{\gk_0}(\gl)$.
%such that $M \models ``\gk_0$ is not $\gl$ supercompact''.
Since $M^\gl \subseteq M$, $M \models ``\la \gk_i \mid
i < \go \ra$ is a sequence of measurable cardinals
such that for each $i < \go$, $\gk_{i + 1}$ is
the least measurable cardinal greater than $\gk_i$,
and $\gk_i$ is ${<}\gk_{i + 1}$ strongly compact''.
Thus, since $j$ is generated by a supercompact
ultrafilter over $P_{\gk_0}(\gl)$,
by the definition of $\FR$, $j(\FR) =
\FR \ast \dot \FR'$, where the first ordinal $\gg$
at which $\dot \FR'$ is forced to do nontrivial
forcing is well above $\gl$.
We follow now the proof of the Lemma of \cite{A00}.
Let $|\ \ \ |$ be the distance function of \cite{Ma}.
Define a term $\dot {\cal U}$ in ${V_*}$ by
$p \forces ``\dot B \in \dot {\cal U}$''
iff $p \forces ``\dot B \subseteq
{(P_{\gk_0}(\gl))}^{V_*^\FR}
$''
and there is $q \in j(\FR)$ such that
$q \ge j(p)$ ($q$ extends $j(p)$),
$|j(p) - q| = 0$,
$j(p) \rest {\gg} = q \rest {\gg} =
j(p) \rest {\gk_0} = q \rest \gk_0 = p$, and
$q \forces ``\la j(\b) \mid \b < \gl \ra \in j(\dot B)$''.
By \cite[Theorem 3.4]{Ma}, $\dot{\cal U}$ is a
well-defined term for a strongly compact measure over $
{(P_{\gk_0}(\gl))}^{V_*^\FR}
$ in $V_*^\FR$.
To see that $\forces_\FR ``\dot {\cal U}$ is normal'', let
$p \forces ``\dot f :
{(P_{\gk_0}(\gl))}^{V_*^\FR}
\to \gl$ is a function such that $\dot f(s) \in s$
for all $s \in \dot B$ where $\dot B \in \dot {\cal U}$''.
%Let $\la \varphi_\a \mid \a < \gl \ra$ be the sequence of
%statements in the forcing language with respect to
%$j(\FR)$, where $\varphi_\a$ is the statement
%$``j(\dot f)(\la j(\b) \mid \b < \gl \ra) = j(\a)$''.
Let $\varphi_\ga$ for $\ga < \gl$ be the statement
$``j(\dot f)(\la j(\b) \mid \b < \gl \ra) = j(\a)$''
in the forcing language with respect to $j(\FR)$, and
consider the sequence $\la \varphi_\a \mid \a < \gl \ra$.
Since $M^\gl \subseteq M$,
$\la \varphi_\a \mid \a < \gl \ra \in M$. Thus, since
$\gg$ is the least $M$ cardinal
in the half-open interval $[{\gk_0}, j({\gk_0}))$ at which
$j(\FR)$ is forced to do nontrivial forcing and
$\gg > \gl$, we can apply \cite[Lemma 2.4]{Ma} in
$M$ to $\la \varphi_\a \mid \a < \gl \ra$ and obtain a condition
$q \ge j(p)$, $q \in j(\FR)$ such that
$|j(p) - q| = 0$,
$j(p) \rest {\gg} = q \rest {\gg} =
j(p) \rest {\gk_0} = q \rest \gk_0 = p$, and
%$j(p) \rest \gk_0 = q \rest \gk_0 =
%j(p) \rest {\gl} = q \rest {\gl} = p$,
if $q' \ge q$, $q'$ decides $\varphi_\a$ for some
$\a < \gl$, then
$q' \rest {\gk_0} \cup (q - p)$ decides $\varphi_\a$
in the same way. Hence, since
$p \forces ``\dot B \in \dot {\cal U}$'' implies we can assume
(by extending $q$ if necessary) that
$q \forces ``\la j(\b) \mid \b < \gl \ra \in j(\dot B)$'',
there must be some $\a < \gl$ such that for some
$q' \ge q$, $q' \forces \varphi_\a$, i.e., such that
$q' \forces ``j(\dot f)(\la j(\b) \mid \b < \gl \ra) = j(\a)$''.
By choice of $q$, $q' \rest {\gk_0} \cup (q - p) \forces \varphi_\a$,
i.e., $q' \rest {\gk_0} \ge p$ is such that for some
$r \in j(\FR)$ ($r$ can be taken as
$q' \rest {\gk_0} \cup (q - p))$,
$|j(q' \rest {\gk_0}) - r| = 0$,
$j(q' \rest {\gk_0}) \rest {\gk_0} = r \rest {\gk_0} =
q' \rest {\gk_0}$, and
$r \forces \varphi_\a$. Since $r \forces \varphi_\a$,
$r \forces
``\la j(\b) \mid \b < \gl \ra \in
j(\{ s \in \dot B \mid \dot f(s) = \a \})$'', so
$q' \rest {\gk_0} \ge p$ is such that $q' \rest {\gk_0} \forces
``\{ s \in \dot B \mid \dot f(s) = \a \} \in \dot {\cal U}$''.
This completes the proof of Lemma \ref{l2}.
\end{proof}
Let $V_{**} = V_*^\FR$.
We observe that the proof of Lemma \ref{l2}
shows that if ${V_*} \models ``$There are no
measurable cardinals above $\gl$'', then
$V_{**} \models ``\gk_0$ is supercompact''.
\begin{lemma}\label{l3}
$V_{**} \models ``$If $\gk < \gk_0$ is measurable, then
$\gk$ is ${<}\gk'$ strongly compact for $\gk'$ the least
measurable cardinal greater than $\gk$''.
\end{lemma}
\begin{proof}
Suppose $V_{**} \models ``\gk < \gk_0$ is measurable''.
%In particular, $V_{**} \models ``\gk$ is a regular cardinal''.
By \cite[Theorem 3.1]{Ma}, $V_{*} \models ``\gk$ is
measurable'' as well.
Therefore, by its definition,
we may write $\FR = \FR_0 \ast \dot \FR_1$, where
$\card{\FR_0} \le \gk$, and
the first ordinal at which $\dot \FR_1$ is forced
to do nontrivial forcing is greater than $\gr$,
the supremum of the first $\go$ many $V_*$-measurable
cardinals greater than $\gk$.
Because $\card{\FR_0} \le \gk$, by the
L\'evy-Solovay results \cite{LS}, the first
$\go$ many measurable cardinals greater than $\gk$
are the same in both $V_*$ and $V^{\FR_0}_*$.
By \cite[Lemma 2.1]{Ma},
$\forces_{\FR_0} ``$Forcing with $\dot \FR_1$ adds no
subsets of $\gr$''. Consequently, it is the case that
$V_{**} = V_*^\FR \models ``\gk$ is ${<}\gk'$ strongly compact''
iff
$V^{\FR_0}_* \models ``\gk$ is ${<}\gk'$ strongly compact''.
For this last fact, we consider the following two cases.
\bigskip
\noindent Case 1: $\card{\FR_0} < \gk$. In this situation,
by the results of \cite{LS}, since
$V_* \models ``\gk$ is ${<}\gk'$ strongly compact'',
$V^{\FR_0}_* \models ``\gk$ is ${<}\gk'$ strongly compact''.
\bigskip
\noindent Case 2: $\card{\FR_0} = \gk$. In this situation,
by \cite[Theorem 3.4]{Ma}, forcing with $\FR_0$ preserves
any degree of strong compactness $\gk$ exhibits in $V_*$.
Hence, as in Case 1,
$V^{\FR_0}_* \models ``\gk$ is ${<}\gk'$ strongly compact''.
\bigskip
Cases 1 and 2 complete the proof of Lemma \ref{l3}.
\end{proof}
Fix now an arbitrary $n \in \go$, $n \ge 1$.
Because $\card{\FR} = \gk_0$, again by the results
of \cite{LS}, the first $\go$ many measurable
cardinals greater than $\gk_0$ are the same
in both $V_*$ and $V_{**} = V_*^\FR$. Consequently,
by Lemma \ref{l2} and reflection, in $V_{**}$, we can
find $\gl_0 < \gk_0$ such that
$\gl_0$ is the least cardinal which is
$\gl_{n + 1}$ supercompact for $\gl_{n + 1}$ the
$(n + 1)^{\rm st}$ measurable cardinal greater than $\gl_0$.
Denote by
$\gl_0 < \gl_1 < \cdots < \gl_n < \gl_{n + 1} < \gk_0$
the sequence composed of the first $n + 2$
many consecutive measurable cardinals starting with $\gl_0$.
%Let $\gs$ be the least inaccessible cardinal greater than $\gl_n$, and let
Let $\ov V = (V_{\gl_{n + 1}})^{V_{**}}$.
Clearly, $\ov V \models ``\gl_0$ is supercompact +
$\gl_1, \ldots, \gl_n$ are the first $n$ measurable
cardinals above $\gl_0$ + There are no measurable
cardinals above $\gl_n$''.
\begin{lemma}\label{l4}
$\ov V \models ``$Each $\gl_i$ for $i = 1, \ldots, n$
is strongly compact''.
\end{lemma}
\begin{proof}
By Lemma \ref{l3}, $V_{**} \models ``\gl_i$ is
${<}\gl_{i + 1}$ strongly compact for $i = 1, \ldots, n$''.
Therefore, $\ov V =
{(V_{\gl_{n + 1}})}^{V_{**}} \models ``\gl_n$ is strongly compact''.
By Ketonen's characterization of strong
compactness \cite{Ke}\footnote{Ketonen
characterized strong compactness in
\cite{Ke} by showing that for
$\gk \le \gl$ regular cardinals, $\gk$
is $\gl$ strongly compact iff for every
regular cardinal $\gd$ such that
$\gk \le \gd \le \gl$, there is
a $\gk$-additive, uniform ultrafilter
over $\gd$.}, if $\ga$ is
${<}\gb$ strongly compact, $\gb$ is $\gg$ strongly
compact, and $\gg$ is regular,
then $\ga$ is $\gg$ strongly compact.
Applying this theorem finitely often and doing a
``downwards induction'' going from $\gl_{n - 1}$
to $\gl_1$ then tells us that
$\ov V \models ``$Each $\gl_i$ for $i = n - 1, \ldots, 1$
is strongly compact''.
This completes the proof of Lemma \ref{l4}.
\end{proof}
\begin{lemma}\label{l5}
$\ov V \models ``$Level by level inequivalence
between strong compactness and supercompactness holds''.
\end{lemma}
\begin{proof}
Because $\ov V \models ``$The $\gl_i$ for $i = 1, \ldots n$
are both strongly compact and the first $n$ measurable
cardinals above $\gl_0$'', $\ov V \models ``$No
$\gl_i$ for $i = 1, \ldots, n$ is supercompact''.
Thus, $\ov V \models ``$Level by level inequivalence
between strong compactness and supercompactness
holds above $\gl_0$''. Since $\ov V \models ``\gl_0$
is supercompact'', the proof of Lemma \ref{l5}
will be complete once we have shown that
$\ov V \models ``$Level by level inequivalence
between strong compactness and supercompactness
holds below $\gl_0$''. To do this, let
$\gk < \gl_0 < \gk_0$ be measurable.
By Lemma \ref{l3} and finitely many applications of
Ketonen's characterization of strong compactness
of \cite{Ke},
$\ov V \models ``\gk$ is $\gg$ strongly compact for
$\gg$ the $(n + 1)^{\rm st}$ measurable cardinal above $\gk$''.
By the choice of $\gl_0$, this means that there is a
cardinal $\gg' > \gk$, $\gg' \le \gg < \gl_0$ such that
$\ov V \models ``\gk$ is $\gg'$ strongly compact yet
$\gk$ is not $\gg'$ supercompact''.
This completes the proof of Lemma \ref{l5}.
\end{proof}
With $\FP = \FQ \ast \dot \FR$,
Lemmas \ref{l1} -- \ref{l5} and the intervening remarks
complete the proof of Theorem \ref{t1}.
\end{pf}
We observe that the proof of Lemma \ref{l5} actually shows that
in $\ov V$, for $\gk < \gl_0$ measurable,
$\gk$ is ${<}\gr$ strongly compact, where
$\gr$ is the supremum of the first $\go$ many measurable
cardinals greater than $\gk$.
Thus, $\gk$ exhibits a significant degree of
level by level inequivalence between strong
compactness and supercompactness.
Also, suppose we build $\ov V$ by
taking $n = 0$ in the above construction. We explicitly note
that although there are no measurable cardinals above $\gl_0$
(so $\gl_0$ of course is not supercompact beyond
a measurable cardinal), every measurable cardinal
$\gk < \gl_0$ both exhibits level by level inequivalence
between strong compactness and supercompactness
and is ${<}\gr$ strongly compact.
We finish with two questions.
First, we ask if it is possible to
prove Theorem \ref{t1} using somewhat
weaker hypotheses. Our current methods
of proof seem to require something along the
lines of the existence
of an $\go$ sequence of supercompact cardinals.
Second, we note that the large cardinal structure
of the model witnessing the conclusions of
Theorem \ref{t1} remains somewhat limited.
We conclude by asking if it is possible
to remove the restrictions inherent to our proof,
and obtain results
analogous to those of this paper in which
the large cardinal structure of the universe
can be arbitrary.
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\end{document}
Note that by Lemma \ref{l2} and the
L\'evy-Solovay results \cite{LS},
$V_{**} \models ``\gk_0$ is $\gl$ supercompact +
$\la \gk_i \mid i < \go \ra$ is a sequence of
measurable cardinals such that for each $i < \go$,
$\gk_{i + 1}$ is the least measurable cardinal
greater than $\gk_i$ +
$\gk_i$ is $\gk_{i + 1}$ strongly compact''.
Therefore, by reflection,
%for each $\gg < \gk_0$,
%there is a sequence $\la \gd_i \mid i < \go \ra$
%of measurable cardinals with $\gd_0 > \gg$
%having the properties that for each
%$i < \go$, $\gd_{i + 1}$ is the least measurable
%cardinal greater than $\gd_i$, $\gd_i$ is
%$\gd_{i + 1}$ strongly compact, and $\gd_0$ is
%$\gl'$ supercompact for $\gl' = \sup_{i < \go} \gd_i$.
%This means that
we can let $\gl_0 < \gl_1 < \cdots < \gl_n < \gk_0$
be a sequence of measurable cardinals such that