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%
\title{Indestructibility and
Level by Level Equivalence and Inequivalence
% and Level by Level Inequivalence
\thanks{2000 Mathematics Subject Classifications:
03E35, 03E55.}
\thanks{Keywords: Supercompact cardinal, strongly
compact cardinal,
indestructibility,
non-reflecting stationary set of ordinals,
level by level equivalence between strong
compactness and supercompactness.}}
\author{Arthur W.~Apter\thanks{The
author's research was partially
supported by PSC-CUNY Grants
and CUNY
Collaborative Incentive Grants.}
\thanks{The author wishes to thank
the referee for helpful suggestions,
comments, and corrections which
considerably improved the presentation
and have been incorporated into the
current version of the paper.}\\
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{August 1, 2006\\
(revised December 5, 2006)}
\begin{document}
\maketitle
\begin{abstract}
If $\gk < \gl$
are such that $\gk$ is
indestructibly supercompact
and $\gl$ is $2^\gl$
supercompact, it is known
from \cite{AH4} that
$\{\gd < \gk \mid \gd$
is a measurable cardinal
which is not a limit of
measurable cardinals and
$\gd$ violates level by
level equivalence between
strong compactness and
supercompactness$\}$
must be unbounded in $\gk$.
On the other hand, using a variant
of the argument used to
establish this fact, it is possible
to prove that
if $\gk < \gl$ are such that
$\gk$ is indestructibly
supercompact and $\gl$ is measurable, then
$\{\gd < \gk \mid \gd$
is a measurable cardinal
which is not a limit of
measurable cardinals and
$\gd$ satisfies level by
level equivalence between
strong compactness and
supercompactness$\}$
must be unbounded in $\gk$.
The two aforementioned
phenomena, however, need not
occur in a universe with
an indestructibly supercompact
cardinal and sufficiently
few large cardinals. In particular, we
show how to construct a model
with an indestructibly
supercompact cardinal $\gk$ in which
if $\gd < \gk$ is a measurable
cardinal which is not a limit of
measurable cardinals, then
$\gd$ must satisfy
level by level equivalence between
strong compactness and supercompactness.
We also, however,
show how to construct a model
with an indestructibly
supercompact cardinal $\gk$ in which
if $\gd < \gk$ is a measurable
cardinal which is not a limit of
measurable cardinals, then
$\gd$ must violate
level by level equivalence between
strong compactness and supercompactness.
\end{abstract}
\baselineskip=24pt
\section{Introduction and Preliminaries}\label{s1}
Say that a model of ZFC satisfies
{\em level by level equivalence
between strong compactness and
supercompactness} if for every
pair of regular cardinals
$\gk < \gl$, $\gk$ is $\gl$
strongly compact iff $\gk$
is $\gl$ supercompact,
except possibly if $\gk$ is
a measurable limit of cardinals $\gd$
which are $\gl$ supercompact.
(The exception comes from a result
of Menas \cite{Me}, who showed that
if $\gk$ is a measurable limit of
cardinals $\gd$ which are $\gl$
strongly compact, then $\gk$ is
$\gl$ strongly compact but need
not be $\gl$ supercompact.)
Say also that the measurable
cardinal $\gk$ satisfies
{\em level by level equivalence between
strong compactness and supercompactness}
iff for every regular cardinal
$\gl > \gk$, $\gk$ is $\gl$
strongly compact iff $\gk$ is
$\gl$ supercompact.
Models in which level by level
equivalence between strong
compactness and supercompactness
holds nontrivially were first
constructed in \cite{AS97a}.
In \cite{AH4}, it was shown
(see Theorem 5) that
%if $\gk$ is indestructibly supercompact
%and level by level equivalence between
%strong compactness and supercompactness
%holds, then no cardinal $\gl > \gk$ is
%$2^\gl$ supercompact.
if $\gk < \gl$ are such that
$\gk$ is indestructibly supercompact and
$\gl$ is $2^\gl$ supercompact, then
level by level equivalence between strong
compactness and supercompactness must
fail unboundedly often below $\gk$.
%A sketch of a proof of this
%theorem is as follows.
One of
the proofs of this theorem will
be very useful for our purposes,
so we present it now.
Suppose $\gk$ is indestructibly
supercompact and $\gl > \gk$ is
$2^\gl$ supercompact.
%If $2^\gl > \gl^+$, add a
%Cohen subset of $\gl^+$, thereby
%forcing $2^\gl = \gl^+$ and
%preserving both the supercompactness
%of $\gk$ and the $2^\gl = \gl^+$
%supercompactness of $\gl^+$.
%If $2^\gl = \gl^+$, do nothing.
Without loss of generality,
by doing a $\gk$-directed closed
forcing as in Lemma 5.1 of
\cite{AH4} if necessary,
we may assume that
$2^\gl = \gl^+$ and $2^{\gl^+} = \gl^{++}$.
Now, perform a reverse Easton iteration
of length $\gl$ which adds a
non-reflecting stationary set
of ordinals of cofinality $\gk$
to each measurable cardinal in
the open interval $(\gk, \gl)$.
After this forcing, as in Lemma
3 of \cite{AC1}, $\gl$ has
become the least measurable cardinal
above $\gk$, so since $2^\gl = \gl^+$,
$\gl$ is not $\gl^+$ supercompact.
By Lemma 4 of \cite{AC1},
since $2^{\gl^+} = \gl^{++}$,
the iteration preserves
the $\gl^+$ strong compactness of $\gl$.
It also preserves the supercompactness
of $\gk$, since it is $\gk$-directed closed.
Consequently, in the resulting generic
extension, $\gk < \gl$ are such that
$\gk$ is indestructibly supercompact
and $\gl$ is $\gl^+$ strongly compact
but not $\gl^+$ supercompact.
By reflection,
$A = \{\gd < \gk \mid \gd$ is
$\gd^+$ strongly compact but not
$\gd^+$ supercompact$\}$ is
unbounded in $\gk$.
Since the entire forcing is
$\gk$-directed closed, $A$
is unbounded in $\gk$ in our
ground model as well.
We note that in the preceding proof,
$\gl$ becomes a measurable cardinal
which is not a limit of measurable
cardinals. Thus, the above actually
also shows the following.
\begin{theorem}\label{t1}
If $\gk < \gl$ are such that
$\gk$ is indestructibly supercompact
and $\gl$ is $2^\gl$ supercompact, then
$A = \{\gd < \gk \mid \gd$ is both a measurable
cardinal which is not a limit of measurable
cardinals and $\gd$ violates level by
level equivalence between strong
compactness and supercompactness$\}$
is unbounded in $\gk$.
\end{theorem}
In fact, the above method of proof
is quite flexible and also allows
us to establish the following.
\begin{theorem}\label{t2}
If $\gk < \gl$ are such that
$\gk$ is indestructibly supercompact
and $\gl$ is measurable, then
$B = \{\gd < \gk \mid \gd$ is both a measurable
cardinal which is not a limit of measurable
cardinals and $\gd$ satisfies level by
level equivalence between strong
compactness and supercompactness$\}$
is unbounded in $\gk$.
\end{theorem}
\begin{proof}
Suppose $\gk$ is indestructibly
supercompact and $\gl > \gk$ is measurable.
Assume without loss of generality that
$\gl$ is the least measurable
cardinal above $\gk$.
Force to add a non-reflecting
stationary set of ordinals of cofinality
$\gk$ to $\gl^+$.
After this forcing, which is both
$\gk$-directed closed and
$\gl$-strategically closed,
$\gl$ remains the least measurable
cardinal above $\gk$.
(See \cite{A01} and \cite{AS97a}
for a more complete discussion of
the partial ordering which adds
a non-reflecting stationary
set of ordinals of cofinality
$\gg$ to a regular cardinal
$\gr > \gg$.)
By Theorem 4.8 of \cite{SRK},
after the forcing,
because stationary reflection
for a set of ordinals of cofinality $\gk < \gl$
is violated at $\gl^+$,
$\gl$ is not $\gl^+$ strongly compact.
Since for any cardinal $\gd$,
$\gd$ is measurable iff $\gd$ is
$\gd$ strongly compact iff $\gd$
is $\gd$ supercompact,
$\gl$ satisfies level by level equivalence
between strong compactness and supercompactness.
Since $\gk$ remains indestructibly
supercompact, as before, by reflection,
$B = \{\gd < \gk \mid \gd$ is a measurable
cardinal which is not a limit of
measurable cardinals and $\gd$
satisfies level by level equivalence
between strong compactness
and supercompactness$\}$ is
unbounded in $\gk$ after the
forcing has been performed.
Once more, we infer by the
fact adding a non-reflecting stationary
set of ordinals of cofinality
$\gk$ is $\gk$-directed closed that
$B$ is unbounded in $\gk$ in the ground model.
\end{proof}
We remark that supercompactness is
unnecessary in the two proofs given above.
In fact, as the preceding arguments show,
Theorems \ref{t1} and \ref{t2}
remain true if $\gk$ is an
indestructibly strong cardinal
in the sense of \cite{GS}.
Theorem \ref{t1}
shows that if $\gk$ is
indestructibly supercompact and
there are large enough cardinals
in the universe above $\gk$, then
there are always unboundedly many
measurable cardinals below $\gk$
which are not limits of measurable
cardinals which violate level by
level equivalence between strong
compactness and supercompactness.
On the other hand, Theorem \ref{t2}
shows that if $\gk$ is indestructibly
supercompact and there are large
enough cardinals in the universe
above $\gk$, then
there are always unboundedly many
measurable cardinals below $\gk$
which are not limits of measurable
cardinals which satisfy level by
level equivalence between strong
compactness and supercompactness.
This prompts us to ask the following
\bigskip
\noindent Question: Suppose $\gk$ is
indestructibly supercompact and
there are sufficiently few
large cardinals in the universe.
Must it then always be the case that
the sets $A$ and $B$ of Theorems
\ref{t1} and \ref{t2} are
unbounded in $\gk$?
\bigskip
The answer to the above question is
``no'' and forms the basis for this paper.
Specifically, we have the
following two theorems.
\begin{theorem}\label{t3}
Suppose $V \models ``$ZFC +
$\gk$ is supercompact +
No cardinal is supercompact up
to a measurable cardinal''.
There is then a partial ordering
$\FP \subseteq V$ such that
$V^\FP \models ``$ZFC +
$\gk$ is indestructibly supercompact +
No cardinal is supercompact up to a
measurable cardinal'', and
in $V^\FP$,
if $\gd$ is a measurable
cardinal which is not a limit of
measurable cardinals, then $\gd$
satisfies level by level equivalence
between strong compactness and supercompactness.
\end{theorem}
\begin{theorem}\label{t4}
Suppose $V \models ``$ZFC +
$\gk$ is supercompact +
No cardinal is supercompact up
to a measurable cardinal''.
There is then a partial ordering
$\FP \subseteq V$ such that
$V^\FP \models ``$ZFC +
$\gk$ is indestructibly supercompact +
No cardinal is supercompact up
to a measurable cardinal'', and
in $V^\FP$,
if $\gd$ is a measurable
cardinal which is not a limit of
measurable cardinals, then $\gd$
violates level by level equivalence
between strong compactness and supercompactness.
\end{theorem}
In fact,
as our methods of proof for Theorem \ref{t3} will show,
it is actually possible
to establish somewhat stronger results
than Theorem \ref{t3}.
Namely, when there are
both sufficiently few large cardinals
in the universe and an indestructibly
supercompact cardinal,
level by level equivalence between
strong compactness and supercompactness
holds on a larger subset of the
measurable cardinals below $\gk$.
We will return to this shortly.
We note that the related
question of whether,
in a universe containing sufficiently
few large cardinals, it is
possible to have both an indestructibly
supercompact cardinal and full
level by level equivalence between
strong compactness and supercompactness
is one which has been addressed, in part,
in \cite{A06} and \cite{AH4}.
The question, unfortunately, remains open.
The methods we develop here, however,
will allow us to delve into what
we believe are some interesting variants.
We conclude Section \ref{s1}
with a discussion of
some preliminary material.
We presume a basic knowledge
of large cardinals and forcing.
Good references in this
regard are \cite{J} and \cite{SRK}.
We also mention that the partial
ordering $\FP$ is {\em $\gk$-directed
closed} if for every directed set $D$ of
conditions of size less than $\gk$,
there is a condition in $\FP$
extending each member of $D$.
$\FP$ is {\em $\gk$-strategically closed}
if in the two person game in
which the players construct a sequence
$\seq{p_\ga \mid \ga \le \gk}$
such that if $\ga < \gb \le \gk$,
$p_\gb$ is a stronger condition
than $p_\ga$, player II has
a strategy ensuring the game
can always be continued.
We abuse notation slightly and
take $V^\FP$ as being the generic
extension of $V$ by $\FP$.
An {\em indestructibly supercompact
cardinal} is one as first given
by Laver in \cite{L}, i.e.,
$\gk$ is indestructibly supercompact
if $\gk$'s supercompactness is
preserved in any generic extension
via a $\gk$-directed closed
partial ordering.
The cardinal $\gk$ is {\em supercompact
up to the cardinal $\gl$} if
$\gk$ is $\gd$ supercompact
for every $\gd < \gl$.
A corollary of Hamkins' work on
gap forcing found in
\cite{H2} and \cite{H3}
will be employed in the
proofs of Theorems \ref{t3}
and \ref{t4}.
We therefore state as a
separate theorem
what is relevant for this paper, along
with some associated terminology,
quoting from \cite{H2} and \cite{H3}
when appropriate.
Suppose $\FP$ is a partial ordering
which can be written as
$\FQ \ast \dot \FR$, where
$\card{\FQ} < \gd$,
$\FQ$ is nontrivial, and
$\forces_\FQ ``\dot \FR$ is
$\gd^+$-directed closed''.
In Hamkins' terminology of
\cite{H2} and \cite{H3},
$\FP$ {\em admits a gap at $\gd$}.
In Hamkins' terminology of \cite{H2}
and \cite{H3},
$\FP$ is {\em mild}
with respect to a cardinal $\gk$
iff every set of ordinals $x$ in
$V^\FP$ of size below $\gk$ has
a ``nice'' name $\tau$
in $V$ of size below $\gk$,
i.e., there is a set $y$ in $V$,
$|y| <\gk$, such that any ordinal
forced by a condition in $\FP$
to be in $\tau$ is an element of $y$.
Also, as in the terminology of
\cite{H2}, \cite{H3}, and elsewhere,
an embedding
$j : \ov V \to \ov M$ is
{\em amenable to $\ov V$} when
$j \rest A \in \ov V$ for any
$A \in \ov V$.
The specific corollary of
Hamkins' work from
\cite{H2} and \cite{H3}
we will be using
is then the following.
\begin{theorem}\label{t5}
%(Hamkins' Gap Forcing Theorem)
{\bf(Hamkins)}
Suppose that $V[G]$ is a generic
extension obtained by forcing that
admits a gap
at some regular $\gd < \gk$.
Suppose further that
$j: V[G] \to M[j(G)]$ is an embedding
with critical point $\gk$ for which
$M[j(G)] \subseteq V[G]$ and
${M[j(G)]}^\gd \subseteq M[j(G)]$ in $V[G]$.
Then $M \subseteq V$; indeed,
$M = V \cap M[j(G)]$. If the full embedding
$j$ is amenable to $V[G]$, then the
restricted embedding
$j \rest V : V \to M$ is amenable to $V$.
If $j$ is definable from parameters
(such as a measure or extender) in $V[G]$,
then the restricted embedding
$j \rest V$ is definable from the names
of those parameters in $V$.
Finally, if $\FP$ is mild with
respect to $\gk$ and $\gk$ is
$\gl$ strongly compact in $V[G]$
for any $\gl \ge \gk$, then
$\gk$ is $\gl$ strongly compact in $V$.
\end{theorem}
\section{The Proofs of Theorems
\ref{t3} and \ref{t4}}\label{s2}
We turn now to the proof of
Theorem \ref{t3}.
\begin{proof}
Suppose $V \models ``$ZFC +
$\gk$ is supercompact +
No cardinal is supercompact up
to a measurable cardinal''.
If necessary, by first doing
the appropriate form of the forcing
found in \cite{AS97a},
we may assume
in addition that
$V \models ``$GCH + Level by
level equivalence between strong
compactness and supercompactness holds''.
Let $f$ be a Laver function
\cite{L} for $\gk$, i.e.,
$f : \gk \to V_\gk$ is such that
for every $x \in V$ and every
$\gl \ge \card{{\rm TC}(x)}$, there is
an elementary embedding
$j : V \to M$ generated by a
supercompact ultrafilter over
$P_\gk(\gl)$ such that
$j(f)(\gk) = x$.
The partial ordering $\FP$
which is used to establish
Theorem \ref{t3} is the
reverse Easton iteration of
length $\gk$ which begins by
adding a Cohen subset of
$\go$ and then does nontrivial
forcing only at those
cardinals $\gd < \gk$ which
are at least $\gd^+$ supercompact in $V$.
At such a stage $\gd$, if
$f(\gd)$ is a term always forced
to denote a $\gd$-directed
closed partial ordering $\FQ$ having
rank below the least $V$-measurable
cardinal above $\gd$, then we
force with $\FQ$.
If this is not the case, then
we perform trivial forcing.
%In other words, speaking loosely,
%we only do nontrivial forcing
%exactly when $\gd$ is at least
%$\gd^+$ supercompact and the
%Laver function hands us a
%$\gd$-directed closed partial
%ordering having rank below
%the least $V$-measurable cardinal
%above $\gd$.
\begin{lemma}\label{l1}
$V^\FP \models ``\gk$ is
indestructibly supercompact''.
\end{lemma}
\begin{proof}
Let $\FQ \in V^\FP$ be such that
$V^\FP \models ``\FQ$ is
$\gk$-directed closed''.
Take $\dot \FQ$ as a term for
$\FQ$ such that
$\forces_{\FP} ``\dot \FQ$ is
$\gk$-directed closed''.
Suppose $\gl \ge
\max(\gk^+, \card{{\rm TC}(\dot \FQ)})$
is an arbitrary cardinal, and let
$\gg = 2^{\card{[\gl]^{< \gk}}}$. Take
$j : V \to M$ as an elementary
embedding witnessing the $\gg$
supercompactness of $\gk$
generated by a supercompact ultrafilter
over $P_\gk(\gg)$ such that
$j(f)(\gk) = \dot \FQ$. Since
$V \models ``$No cardinal above
$\gk$ is measurable'', $\gg \ge 2^\gk$,
and $M^\gg \subseteq M$,
$M \models ``\gk$ is measurable and
no cardinal in the half-open interval
$(\gk, \gg]$ is measurable''. Hence,
the definition of $\FP$ implies that
$j(\FP \ast \dot \FQ) = \FP \ast
\dot \FQ \ast \dot \FR \ast
j(\dot \FQ)$, where
the first nontrivial stage of
forcing in $\dot \FR$ takes
place well above $\gg$.
%$2^{[\gg]^{< \gk}}$.
Laver's original argument from \cite{L} now applies
and shows
$V^{\FP \ast \dot \FQ} \models
``\gk$ is $\gl$ supercompact''.
(Simply let $G_0 \ast G_1
\ast G_2$ be $V$-generic over
$\FP \ast \dot \FQ \ast \dot \FR$,
lift $j$ in $V[G_0][G_1][G_2]$ to
$j : V[G_0] \to M[G_0][G_1][G_2]$,
take a master condition $p$
for $j '' G_1$ and a
$V[G_0][G_1][G_2]$-generic object
$G_3$ over $j(\FQ)$ containing $p$, lift $j$
again in $V[G_0][G_1][G_2][G_3]$ to
$j : V[G_0][G_1] \to M[G_0][G_1][G_2][G_3]$,
and show by the $\gg^+$-directed closure of
$\FR \ast j(\dot \FQ)$ that the supercompactness
measure over ${(P_\gk(\gl))}^{V[G_0][G_1]}$
generated by $j$ is actually a member of
$V[G_0][G_1]$.)
As $\gl$ and $\FQ$ were arbitrary,
this completes the proof of
Lemma \ref{l1}.
\end{proof}
\begin{lemma}\label{l2}
$V^\FP \models ``$No cardinal
is supercompact up to a
measurable cardinal''.
\end{lemma}
\begin{proof}
Write
$\FP = \FR_0 \ast \dot \FR_1$,
where $\card{\FR_0} = \go$,
$\FR_0$ is nontrivial, and
$\forces_{\FR_0} ``\dot \FR_1$ is
$\ha_2$-directed closed''.
Since $\FP$ admits the factorization
just given,
%by Hamkins' Gap Forcing Theorem of \cite{H2} and \cite{H3},
by Theorem \ref{t5},
for any pair of cardinals
$\gd \le \gl$, if
$V^\FP \models ``\gd$ is $\gl$
supercompact'', then
$V \models ``\gd$ is $\gl$ supercompact''
as well. Consequently, since
$V \models ``$No cardinal is
supercompact up to a measurable cardinal'',
it is also the case that
$V^\FP \models ``$No cardinal is
supercompact up to a measurable cardinal''.
%This completes the proof of Lemma \ref{l2}.
\end{proof}
\begin{lemma}\label{l3}
If $V^\FP \models ``\gd$ is a
measurable cardinal which is not a limit of
measurable cardinals'', then
$V \models ``\gd$ is a
measurable cardinal which is not a limit of
measurable cardinals''.
\end{lemma}
\begin{proof}
Suppose
$V^\FP \models ``\gd$ is
a measurable cardinal which is not
a limit of measurable cardinals''.
Since by Lemmas \ref{l1} and \ref{l2},
$V^\FP \models ``\gk$ is supercompact
and no cardinal is supercompact up to
a measurable cardinal'',
we know that
$V^\FP \models ``$No cardinal above
$\gk$ is measurable''.
We may therefore immediately
infer that $\gd < \gk$. In addition,
by the factorization of $\FP$
given in Lemma \ref{l2} and Theorem \ref{t5},
$V \models ``\gd$ is measurable'' as well.
Thus, to prove Lemma \ref{l3}, it suffices
to show that
$V \models ``\gd$ is not a limit of
measurable cardinals''.
To do this, assume to the contrary that
$V \models ``\gd$ is a limit of
measurable cardinals''. In particular,
$V \models ``\gd$ is a limit of measurable
cardinals which themselves are not limits
of measurable cardinals''.
For any such measurable cardinal
$\gg < \gd$, write
$\FP = \FP_\gg \ast \dot \FP^\gg$. Since
$\FP_\gg$ is forcing equivalent to a
partial ordering having size below $\gg$,
by the L\'evy-Solovay results \cite{LS},
$V^{\FP_\gg} \models ``\gg$ is measurable''.
Since
$\forces_{\FP_\gg} ``$Forcing with
$\dot \FP^\gg$
adds no subsets of the least inaccessible
cardinal above $\gg$'',
$V^{\FP_\gg \ast \dot \FP^\gg} = V^\FP \models
``\gg$ is measurable''. Thus,
$V^\FP \models ``\gd$ is a limit
of measurable cardinals''.
This contradiction completes the
proof of Lemma \ref{l3}.
\end{proof}
\begin{lemma}\label{l4}
In $V^\FP$, if $\gd$
is a measurable cardinal which
is not a limit of measurable cardinals,
then $\gd$ satisfies level by level
equivalence between strong compactness
and supercompactness.
\end{lemma}
\begin{proof}
Suppose
$V^\FP \models ``\gd$ is a measurable
cardinal which is not a limit
of measurable cardinals''.
By Lemma \ref{l3}, it is the case that
$\gd < \gk$ and
$V \models ``\gd$ is a measurable
cardinal which is not a limit
of measurable cardinals''.
Thus, as in Lemma \ref{l3},
we may write
$\FP = \FP_{\gd} \ast
\dot \FP^\gd$, where
$\FP_\gd$ is forcing equivalent
to a partial ordering having
size below $\gd$ and
$\forces_{\FP_\gd} ``$Forcing with
$\dot \FP^\gd$ adds no new subsets of
the least inaccessible cardinal above
$\gd$''. From this, it is easily
seen that any subset of $\gd$ in
$V^\FP$ of size below $\gd$ has
a ``nice'' name
(in the same sense as in
the preliminaries)
of size below $\gd$ in $V$.
Consequently, by the factorization
of $\FP$ given in Lemma \ref{l2}
%and the results of \cite{H2} and \cite{H3},
and Theorem \ref{t5},
for any $\gl > \gd$ such that
$V^\FP \models ``\gd$ is $\gl$
strongly compact'',
$V \models ``\gd$ is $\gl$ strongly compact''
as well.
Suppose now
$V^\FP \models ``\gl > \gd$ is a
regular cardinal such that
$\gd$ is $\gl$ strongly compact''.
By the preceding paragraph,
$V \models ``\gd$ is $\gl$ strongly compact'',
so by level by level equivalence between
strong compactness and supercompactness
and the fact that
$V \models ``\gl$ is regular'',
$V \models ``\gd$ is $\gl$ supercompact''.
As $\gl > \gd$, by GCH in $V$,
$\gl \ge \gd^+ = 2^\gd$, from which we
immediately infer that
$V \models ``\gd$ is a measurable
cardinal which is a limit of
measurable cardinals''. However, since
$V \models ``\gd$ is a measurable
cardinal which is not a limit of
measurable cardinals'', this is
an immediate contradiction.
This means that there is no
$V^\FP$-cardinal $\gl > \gd$ such that
$V^\FP \models ``\gd$ is $\gl$
strongly compact'', i.e.,
$V^\FP \models ``$Level by level
equivalence between strong compactness
and supercompactness holds at $\gd$''.
%This completes the proof of Lemma \ref{l4}.
\end{proof}
Lemmas \ref{l1} -- \ref{l4} complete
the proof of Theorem \ref{t3}.
\end{proof}
As we remarked in Section \ref{s1},
the techniques used to prove
Theorem \ref{t3} may also be
employed to establish a result
in which the conclusions of
Theorem \ref{t3} hold on a
larger subset of the measurable
cardinals below $\gk$. For instance,
speaking somewhat loosely,
assume we start with a ground model
$V$ as in Theorem \ref{t3} and
do nontrivial forcing exactly
when $\gd < \gk$ is at
least $\gd^{+ 17}$ supercompact and
the Laver function hands us a
$\gd$-directed closed partial
ordering having rank below the least
$V$-measurable cardinal above $\gd$.
In our generic extension $V^\FP$,
not only do we end up with an
indestructibly supercompact cardinal
$\gk$ and no cardinal being
supercompact up to a measurable cardinal,
but level by level equivalence between
strong compactness and supercompactness
holds at every cardinal
$\gd < \gk$ which is $\gd^{+ n}$
supercompact but is not a limit of
cardinals $\gg$ which are themselves
$\gg^{+ n}$ supercompact, where
$n \le 16$ is a natural number.
%The proof, which we leave to the
%readers of this paper, is exactly
%as given in Lemmas \ref{l1} -- \ref{l4},
%under the proviso that the phrase
%``measurable cardinal which is not a
%limit of measurable cardinals''
%is replaced by the phrase
%``$\gd^{+ n}$ supercompact cardinal
%which is not a limit of cardinals
%$\gg$ which are themselves
%$\gg^{+ n}$ supercompact, where
%$n \le 16$ is a natural number''.
The proof is essentially as given in
Lemmas \ref{l1} -- \ref{l4}, with
suitable modifications which we
leave to the readers of this paper.
Having completed the proof of
Theorem \ref{t3}, we turn now
to the proof of Theorem \ref{t4}.
\begin{proof}
Suppose
$V \models ``$ZFC + $\gk$ is
supercompact + No cardinal is
supercompact up to a measurable cardinal''.
If necessary, by first doing a
preliminary forcing such as given in
the proof of Theorem 2 of \cite{A01}
and elaborated upon at the end of that paper,
we may assume in addition that
$V \models ``$GCH + Every measurable cardinal
$\gd$ is $\gd^+$ strongly compact''.
The partial ordering $\FP$ which
we then use to establish Theorem \ref{t4}
is exactly the same as the one
given in the proof of Theorem \ref{t3}.
Also, the proofs of the analogous versions of
Lemmas \ref{l1} and \ref{l2}
go through exactly as earlier and show that
$V^\FP \models ``\gk$ is indestructibly
supercompact + No cardinal is supercompact
up to a measurable cardinal''.
The proof of Theorem \ref{t4}
will therefore be complete once
we have established the following.
\begin{lemma}\label{l5}
In $V^\FP$, if $\gd$ is
a measurable cardinal which
is not a limit of measurable cardinals, then
$\gd$ violates level by level equivalence
between strong compactness and
supercompactness.
\end{lemma}
\begin{proof}
The proof of the
analogous version of Lemma \ref{l3}
goes through exactly as before
and shows that if
$V^\FP \models ``\gd$ is a measurable
cardinal which is not a limit of
measurable cardinals'', then $\gd < \gk$ and
$V \models ``\gd$ is a measurable
cardinal which is not a limit of
measurable cardinals'' as well.
Therefore, the proof of Lemma \ref{l5}
will be complete once we have shown that
any $V$-measurable cardinal $\gd$ which is not
a limit of $V$-measurable cardinals
violates level by level equivalence
between strong compactness
and supercompactness in $V^\FP$.
However, as in the proof of Lemma \ref{l4},
we may write
$\FP = \FP_{\gd} \ast
\dot \FP^\gd$, where
$\FP_\gd$ is forcing equivalent
to a partial ordering having
size below $\gd$ and
$\forces_{\FP_\gd} ``$Forcing with
$\dot \FP^\gd$ adds no new subsets of
the least inaccessible cardinal above
$\gd$''. Therefore, since
$V \models {\rm GCH}$, standard
arguments in conjunction with the
results of \cite{LS} yield that
$V^{\FP_\gd} \models ``2^\gd =
\gd^+$ + $\gd$ is
a measurable cardinal which is not
a limit of measurable cardinals +
$\gd$ is $\gd^+$ strongly compact'',
which means we may now immediately
infer that
$V^{\FP_\gd \ast \dot \FP^\gd} =
V^\FP \models ``2^\gd = \gd^+$ + $\gd$ is
a measurable cardinal which is not
a limit of measurable cardinals +
$\gd$ is $\gd^+$ strongly compact''.
Further,
$V^\FP \models ``\gd$ is not
$\gd^+$ supercompact'', since
otherwise, as $V^\FP \models ``2^\gd
= \gd^+$'',
$V^\FP \models ``\gd$ is a measurable
cardinal which is a limit of
measurable cardinals''. Thus,
$V^\FP \models ``\gd$ is $\gd^+$
strongly compact yet $\gd$ is not
$\gd^+$ supercompact'', i.e.,
$V^\FP \models ``\gd$ violates
level by level equivalence between
strong compactness and
%\break\newpage\noindent
supercompactness''.
This completes both the proof of Lemma \ref{l5}
and the proof of Theorem \ref{t4}.
\end{proof}
\end{proof}
As we mentioned in Section \ref{s1},
the question of whether it is possible
to have both an indestructibly
supercompact cardinal and full
level by level equivalence between
strong compactness and supercompactness
in a universe with sufficiently few
large cardinals is still open.
We conclude by posing an analogous
question for a model in which
level by level equivalence between
strong compactness and supercompactness
fails everywhere. In particular,
is it possible
to have both an indestructibly
supercompact cardinal and full
level by level {\em inequivalence} between
strong compactness and supercompactness
in a universe with sufficiently few
large cardinals?
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\end{document}