\magnification=1100
\voffset=0.00in
\vsize=8.00in
\tolerance = 10000
\def\forces{\hbox{$\|\hskip -2pt\hbox{--}$\hskip 2pt}}
\def\hb{\hfil\break}
\def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
\hbox{\vrule width.#2pt height#1pt \kern#1pt
\vrule width.#2pt}
\hrule height.#2pt}}}}
\def\square{\mathchoice\sqr34\sqr34\sqr{2.1}3\sqr{1.5}3}
\vskip 2in
\centerline{``On Box, Weak Box, and Strong Compactness''}
\vskip .5in
\centerline{by}
\vskip .5in
\centerline{Arthur W. Apter*}
\centerline{Department of Mathematics}
\centerline{Baruch College of CUNY}
\centerline{New York, New York 10010}
\vskip .25in
\centerline{and}
\vskip .25in
\centerline{James. M. Henle*}
\centerline{Department of Mathematics}
\centerline{Smith College}
\centerline{Northampton, Mass. 01060}
\vskip 1.75in
\noindent *The research for this paper was partially supported
by NSF Grant DMS-8616774. In addition, the first author wishes
to thank Menachem Magidor for many explanations of the methods
of his original paper with Ben-David not explicitly stated in
that paper. Without those explanations, this paper would have
been impossible.
\vfill\eject
\vskip 1in
\centerline{``On Box, Weak Box, and Strong Compactness''}
\vskip .5in
\centerline{by}
\vskip .5in
\centerline{Arthur W. Apter}
\vskip .25in
\centerline{and}
\vskip .25in
\centerline{James. M. Henle}
\vskip .75in
\indent One of the most important goals of set theorists over
the last few years has been to reprove old results which
previously had used very strong assumptions from hypotheses
which, at least prima facia, are weaker. Examples of these
abound, including, but certainly not limited to, the work of
Woodin and Cummings (see [C]) on the Singular Cardinals
Problem, in which results previously obtained by Magidor
[Ma1], [Ma2] using supercompactness and hugeness were
reproven using hypermeasurability. \hb
\indent This paper continues the work of [AH] along these
lines, reproving a result of Ben-David and Magidor [BDM]
using strong compactness instead of supercompactness. In
[BDM], the authors show that $Con(ZFC+GCH+
\exists\kappa[\kappa$ is $ \kappa^+$ supercompact])$
\Longrightarrow Con(ZFC+\square^*_{\aleph_\omega}+
\neg\square_{\aleph_\omega})$, where $\square_\kappa$
and $\square^*_\kappa$ are, respectively, the combinatorial
principles box and weak box at the cardinal $\kappa$.
(See [BDM] for the appropriate definitions.) We prove the
following \hfil \break
\noindent Theorem: $Con
(ZFC+GCH+\exists\kappa[\kappa$ is $\kappa^+
$ strongly compact]) $\Longrightarrow Con(ZFC+
\square^*_{\aleph_\omega}+\neg\square_{\aleph_\omega})$.
\hfil\break
\noindent The proof recasts the ideas of [BDM] and [Ma1] in
terms of
strong compactness. As such, we are going to assume complete
familiarity with the ideas and techniques of [BDM] and [Ma1]. \hb
\indent Our notation and terminology will generally be that
of [AH], [BDM], and [Ma1]. $V$ will be our ground model.
We recall that a cardinal $\kappa$ is $\kappa^+$ strongly
compact iff $P_\kappa(\kappa^+)$ carries a $\kappa$-additive
fine ultrafilter {\bf U}. Although {\bf U} can't be assumed
to be normal, as mentioned in [AH], {\bf U} can be chosen so
as to have some weak normality properties. Specifically,
let $k:P_\kappa(\kappa^+) \to \kappa$ by $k(p)=p \cap
\kappa$. Let ${\bf U}_\kappa = k_*({\bf U})$ be the Rudin
~-Keisler projection to an ultrafilter on $\kappa$, i.e.,
$x \in {\bf U}_\kappa$ iff $ k^{-1}(x) \in {\bf U}$. In [AH],
it is shown that {\bf U} can be chosen so that
${\bf U}_\kappa$ is a normal measure on $\kappa$ and so that
{\bf U} is weakly normal in the sense that if
$f:P_\kappa(\kappa^+) \to \kappa$ is a choice function
($f(p) \in p$) then f is constant on a set in {\bf U}. This
means that $D=\{p \in P_\kappa(\kappa^+):p \cap \kappa$ is an
inaccessible cardinal$\} \in {\bf U}$. \hb
\indent We are now in a position to define the partial ordering
$P$ used in the proof of our Theorem. The definition is an
amalgamation of the ideas of [AH] with the corresponding
definition of [BDM] and [Ma1]. Specifically, $P$ is the set
of all finite tuples of the form
$\langle p_1, \ldots , p_n, f_0, \ldots , f_n, A, F \rangle$
satisfying the following properties. \hfil \break
\noindent 1. For $1 \le i \le n, p_i \in D$ and $p_i\cap\kappa
0$. \hfil \break
\indent For each $q \in A(\langle p_1,\ldots, p_n \rangle)$,
$p_n \cap \kappa < q \cap \kappa$, consider $\pi_q =
\langle p_1,\ldots, p_n, q, f_0,\ldots, f_n, \break
F(\langle q \rangle), A, F^{\langle q \rangle} \rangle$.
By induction, let ${\pi_q}' = \langle p_1,\ldots, p_n, q,
f_0,\ldots, f_{j-1}, f^q_j,\ldots, f^q_n, f^q, B^q, G^q
\rangle$ extending $\pi_q$ be so that ${\pi_q}'$
satisfies the conclusions of the lemma for $\pi_q$ and any
appropriate condition extending ${\pi_q}'$ of length
$(n+1)+l$. Then, using the same arguments as in [Ma1], let
$C \in {\bf U}$ be so that for all $q \in C$, $f^q_j,\ldots,
f^q_n$ are constant; we note only that the argument of [Ma1]
producing $f^q_n$ uses that any $f:P_\kappa(\kappa^+) \to
\kappa$ which is a choice function has a {\bf U} measure 1
set, a fact we have already observed can be assumed to be
true about {\bf U}.
\hfil\break
\indent We can now define the condition $\pi'$ which
witnesses the conclusions of Lemma 1 by $\pi' = \langle p_1,
\ldots,p_n, f_0,\ldots, f_{j-1}, f^q_j,\ldots, f^q_n, B,
G \rangle$ where: \hfil \break
\noindent 1. $f^q_j,\ldots, f^q_n$ are the constant values
just mentioned. \hfil \break
\noindent 2. $B(\langle p_1,\ldots, p_n \rangle) = C \cap
A(\langle p_1,\ldots, p_n \rangle)$. \hfil \break
\noindent 3. For $\langle q_1,\ldots, q_m \rangle \in
[D_{p_n}]^{< \omega}$, $B(\langle p_1,\ldots, p_n, q_1,
\ldots,q_m \rangle) = B^{q_1}(\langle p_1,\ldots, p_n, q_1
,\ldots, q_m \rangle)$. If $\langle q_1,\ldots, q_m \rangle
\in [D]^{< \omega} - [D_{p_n}]^{< \omega}$, let $B(\langle
q_1,\ldots, q_m \rangle) = A(\langle q_1,\ldots, q_m \rangle)$
. \hfil \break
\noindent 4. For $q \in B(\langle p_1,\ldots, p_n \rangle)$,
$p_n \cap \kappa < q \cap \kappa$, $G(\langle q \rangle) =
f^q$. \hfil \break
\noindent 5. For $\langle q_1,\ldots, q_m \rangle \in
[D_{p_n}]^{< \omega}$, $G(\langle q_1,\ldots, q_m \rangle) =
G^{q_1}(\langle q_2,\ldots, q_m \rangle)$. \hfil \break
\noindent By the above definition, it is clear that $\pi'$
is a j-length preserving extension of $\pi$. \hfil \break
\indent Let now $\pi''$ extend $\pi'$, $\pi''$ decides
$\varphi$, $\pi'' = \langle p_1,\ldots, p_n, q, q_1,\ldots,
q_l, g_0,\ldots, g_{j-1}, h_j,\ldots, h_l,\break E, H \rangle$.
Let $\chi$= j-int($\pi'',\pi'$)=$\langle p_1,\ldots, p_n,
q, q_1,\ldots, q_l, g_0,\ldots, g_{j-1}, f^q_j,\ldots,
f^q_n,\nobreak G(\langle q \rangle), G(\langle q, q_1 \rangle),
\ldots,\break G(\langle q, q_1,\ldots, q_l \rangle), B,
G^{\langle q, q_1,\ldots, q_l \rangle} \rangle$. Since
the $j^{th}$ through $n^{th}$ function coordinates
$f^q_j,\ldots, f^q_n$ are the same in both ${\pi_q}'$ and
$\pi''$, and since $G(\langle q \rangle) = f^q$, the
definitions of B and G ensure that $\chi$ extends
${\pi_q}'$. We thus have that $\pi''$ extends $\chi$ and
$\chi$ extends ${\pi_q}'$, so j-int($\pi'', \chi) = \chi$
extends j-int($\pi'', {\pi_q}')$, and by construction,
j-int($\pi'', {\pi_q}'$) decides $\varphi$. Thus, $\chi$
decides $\varphi$, proving Lemma 1. \hfil \break
\hfill$\square$
Lemma 1 \hfil \break
\indent With the same proof as in [Ma1], the following
corollary to Lemma 1, stated as Lemma 2, is also true.
\hfil \break
\noindent Lemma 2: Let $\pi$ be a condition of length $n$,
$j \le n$, and $\mu \le \kappa^+_j$. If $\pi \forces
$ ``$\dot f: \mu \to \alpha$ is a function into the
ordinal $\alpha$'', then there is a j-length preserving
extension $\pi'$ of $\pi$ so that if $\pi''$ extends
$\pi'$ and $\pi'' \forces$ ``$\dot f(\lambda) = \beta$''
where $\lambda < \mu$ is arbitrary, then j-int($\pi'',
\pi'$) $\forces$ ``$\dot f(\lambda) = \beta$''.
\hfil \break
\indent Let now $G$ be $V$-generic over $P$. As in [Ma1],
a certain submodel $V'$ of $V[G]$, to be described later,
will be the model witnessing the conclusions of our Theorem.
Before describing $V'$, as in [Ma1] we first let, for $j<
\omega$, $G \vert j = \{ \langle f_0,\ldots, f_{j-1} \rangle
$ : Some $\pi\in G$ has the form $\langle p_1,\ldots,p_n,f_0,
\ldots,f_n, A, F \rangle$ where $j \le n \}$. As in [Ma1],
$G \vert j$ is $V$-generic over $Col(\omega_2, \kappa_1)\times
Col(\kappa^+_1, \kappa_2) \times \cdots \times Col(\kappa^+_{j-1},
\kappa_j)$ where $\kappa_i= p_i \cap \kappa$ for $i \le j$.
The genericity of $G$ ensures that $\kappa_i$ is well
defined. \hfil \break
\noindent Lemma 3 (Theorem 3.2 of [Ma1]): Let $\pi \in P$
have length at least $j$, and let $\mu \le \kappa^+_j$ be
an ordinal and $\dot f$ be a term so that $\pi \forces$ ``$\dot
f \subseteq \mu$''. Then for some condition $\pi_0$
extending
$\pi$, $\pi_0 \forces$ ``$\dot f \in V[G \vert j]$''.
\hfil \break
\indent Proof of Lemma 3: By Lemma 2, we can let $\pi'=
\langle p_1,\ldots, p_n, f_0,\ldots, f_n, A, F \rangle$
be a j-length preserving extension of $\pi$ so that if
$\lambda < \mu$ is arbitrary and $\pi''$ extending
$\pi'$ decides ``$\lambda \in \dot f$'', then
j-int($\pi'', \pi'$) decides ``$\lambda \in \dot f$'' in
the same way. Fix now $\lambda < \mu$, and let $\langle
g_0,\ldots, g_{j-1} \rangle$ be the appropriate portion of
the condition $\pi'' = \langle p_1,\ldots, p_m, g_0,\ldots,
g_{j-1}, g_j,\ldots, g_m, B, H \rangle$ extending $\pi'$
so that $\pi''$ decides ``$\lambda \in \dot f$''. If we
form the condition $\chi = \langle p_1,\ldots, p_n, g_0,
\ldots,g_{j-1}, f_j,\ldots, f_n, A, F \rangle$, then Lemma
2 shows that $\chi$ behaves as a modified Prikry condition
of [AH] or [H] does with respect to deciding the statement
``$\lambda \in \dot f$'' in the sense that if $\chi' =
\langle p_1,\ldots, p_n, p_{n+1},\ldots, p_l, h_0,\ldots,
h_{j-1}, h_j,\ldots, h_l, C, I \rangle$ extends $\chi$ and
$\chi'$ decides ``$\lambda \in \dot f$'' then $\chi'' =
\langle p_1,\ldots, p_n, p_{n+1},\ldots, p_l, g_0,\ldots,
g_{j-1}, f_j,\ldots, f_n, F(\langle p_{n+1} \rangle),\ldots,
F(\langle p_{n+1},\ldots, p_l \rangle), A, F^{\langle p_{n+1},\ldots,
p_l \rangle} \rangle $ decides ``$\lambda \in \dot f$'' in the same
way, i.e., the additional information necessary to decide
``$\lambda \in \dot f$'' is essentially added by $\langle
p_{n+1},\ldots, p_l \rangle$ with the role of $F$ (and, in
this case, $A$) largely irrelevant. Thus, we can mimic the
proof of Lemma 1.2 of [AH] or Lemma 4.3 of [H] and obtain a
function $A_{\langle g_0,\ldots, g_{j-1} \rangle}
\subseteq^* A$ so that $\psi = \langle p_1,\ldots, p_n, g_0,
\ldots,g_{j-1}, f_j,\ldots, f_n, A_{\langle g_0,\ldots,
g_{j-1} \rangle}, F \rangle$ decides ``$\lambda \in \dot
f$''. \hfil \break
\indent For any arbitrary $\langle h_0,\ldots, h_{j-1}
\rangle$ in a condition $\pi''$ as above, let $A_{\langle
h_0,\ldots, h_{j-1} \rangle}$ have the property just stated.
Since $\vert Col(\omega_2, \kappa_1) x Col(\kappa^+_1,
\kappa_2) x \cdots x Col(\kappa^+_{j-1}, \kappa_j) \vert =
\kappa_j < \kappa$, the additivity of {\bf U} ensures that
there is some $A_\lambda$ so that $A_\lambda(\langle q_1,
\ldots, q_i \rangle) \subseteq A_{\langle h_0, \ldots,
h_{j-1} \rangle}(\langle q_1, \ldots, q_i \rangle)$ for
all $A_{\langle h_0, \ldots, h_{j-1} \rangle}$. (Simply
let $A_\lambda(\langle q_1, \ldots, q_i, \rangle) =
\cap_{\langle h_0, \ldots, h_{j-1} \rangle}
A_{\langle h_0, \ldots, h_{j-1} \rangle}(\langle q_1,
\ldots, q_i \rangle)$.) Since $\mu < \kappa$ and $\lambda <
\mu$ was arbitrary, we can use the just mentioned argument
to form a function $E$ so that $E(\langle q_1, \ldots, q_i
\rangle) \subseteq A_\lambda(\langle q_1, \ldots, q_i
\rangle)$ for all $\lambda < \mu$. We now claim that the
condition $\pi_0 = \langle p_1, \ldots, p_n, f_0, \ldots,
f_n, E, F \rangle$ extending $\pi'$ is so that $\pi_0
\forces$ ``$\dot f \in V[G \vert j]$''. \hfil \break
\indent Define $\dot g$ to be the realization of the term
$\{\exists \lambda < \mu \exists \langle g_0, \ldots,
g_{j-1} \rangle \in G \vert j$ : For $0 \le i \le j-1$, $
g_i \supseteq f_i$ and $\langle p_1, \ldots, p_n, g_0,
\ldots, g_{j-1}, f_j, \ldots, f_n, E, F \rangle \forces$
``$\lambda \in \dot f$''$\}$. Clearly, $\dot g$ denotes
a subset of $\mu$ present in $V[G \vert j]$. We show
that $\pi_0 \forces$ ``$\dot f = \dot g$''. If for some
$\langle g_0, \ldots, g_{j-1} \rangle \in G \vert j$,
$g_i \supseteq f_i$ for $ 0 \le i \le j-1$ and $\langle p_1,
\ldots, p_n, g_0, \ldots, g_{j-1}, f_j, \ldots, f_n,
E, F \rangle \forces$ ``$\lambda \in \dot f$'', then
since $\pi_0$ can be assumed to be an element of $G$,
$\lambda \in i_G(\dot f)$, i.e., $\pi_0 \forces$ ``$
\dot g \subseteq \dot f$''. Let now $\pi_1 = \langle
p_1, \ldots, p_m, g_0, \ldots, g_m, E', F' \rangle
\in G$ and $\lambda < \mu$ be so that $\pi_1 \forces$ ``
$\lambda \in \dot f$''. By the genericity of $G$, we
can assume that $\pi_1$ extends $\pi_0$. The definition
of $\pi_0$ ensures, since $\pi_2 = \langle p_0, \ldots,
p_n, g_0, \ldots, g_{j-1}, f_j, \ldots, f_n, E, F \rangle$
must decide ``$\lambda \in \dot f$'', that $\pi_2
\forces$ ``$\lambda \in \dot f$''. Thus, $\lambda \in
i_{G \vert j}(\dot g)$, so $\pi_0 \forces$ ``$\dot f
\subseteq \dot g$'', i.e., $\pi_0 \forces$ ``$\dot f =
\dot g$''. This proves Lemma 3. \hfil \break
\hfill$\square$
Lemma 3 \hfil \break
\indent As in [Ma1], $r = \{\langle p_1, \ldots, p_n
\rangle : \exists \pi \in G[\langle p_1, \ldots, p_n
\rangle \in \pi]\}$ generates a cofinal $\omega$
sequence through $\kappa$ and $\kappa^+$ which collapses
$\kappa^+$ to $\kappa$; in particular, $\langle \kappa_n :
n < \omega \rangle = \langle p_n \cap \kappa : n < \omega
\rangle$ is cofinal in $\kappa$, and $\langle \lambda_n :
n < \omega \rangle = \langle sup(p_n) : n < \omega
\rangle$ is cofinal in $\kappa^+$. Lemma 3 shows that
$\kappa$ isn't collapsed, and in fact, that $V[G] \models$
``$\kappa = \aleph_\omega$''. We thus need to define a
certain submodel $V'$ of $V[G]$ which will satisfy the
conclusions of our Theorem. \hfil \break
\indent In order to define $V'$, we need to define a
certain group {\bf G} of automorphisms of $P$.
Specifically, let $\langle p_1, \ldots, p_n \rangle,
\langle p_1', \ldots, p_n' \rangle \in [D]^{< \omega}$
be so that for $1 \le i \le n$, $p_i \cap \kappa = p_i'
\cap \kappa$. For $1 \le i \le n$, let $\psi$ be the
permutation of $[D]^{< \omega}$ given by $\psi(p_i) =
p_i'$, $\psi(p_i') = p_i$, and $\psi$ is the identity
otherwise. $\psi$ naturally generates an automorphism
$\rho$ of $P$ by, for $\pi = \langle q_1, \ldots, q_m,
f_0, \ldots, f_m, A, F \rangle \in P$, $\rho(\pi) =
\langle \psi(q_1), \ldots, \psi(q_m), f_0, \ldots, f_m,
\psi(A), \psi(F) \rangle$, where the action of $\psi$ on
$A$ and $F$ is defined as follows: \hfil \break
\noindent 1. If $A(\langle u_1, \ldots, u_l \rangle) =
A' \in {\bf U}$, then $\psi(A(\langle u_1, \ldots, u_l
\rangle)) = A(\psi(u_1), \ldots, \psi(u_l) \rangle)$ has
value $\{\psi(u) : u \in A' \}$ which, since $\psi$
permutes only finitely many elements, is a {\bf U}
measure 1 set. \hfil \break
\noindent 2. If $F(\langle u_1, \ldots, u_l \rangle) =
h$, then $\psi(F(\langle u_1, \ldots, u_l \rangle)) =
F(\langle \psi(u_1), \ldots, \psi(u_l) \rangle)$ has value
$h$ as well. \hfil \break
\noindent For {\bf G} the collection of all such $\rho$
and any $\rho \in {\bf G}$, the reader may verify that
$\rho$ is an automorphism of $P$. Further, for $\pi =
\langle p_1, \ldots, p_n, f_0, \ldots, f_n, A, F
\rangle$ and $\pi' = \langle p_1', \ldots, p_n', f_0,
\ldots, f_n, A', F \rangle$ so that for $1 \le i \le n$,
$p_i \cap \kappa = p_i' \cap \kappa$, there is an
automorphism
$\rho \in {\bf G}$ so that $\rho(\pi)$ and
$\pi'$ are compatible. \hfil \break
\indent Define now, for any functions $F$ and $F'$ so
that there are conditions $\pi, \pi' \in P$ with $F \in
\pi$, $F' \in \pi'$, and $F$ and $F'$ are the constraint
portions of their respective conditions which tell how to
extend the generic sequence of collapses of $\kappa$ to
$\aleph_\omega$, a relationship $\sim$ by $F \sim F'$ iff
$\exists A \in {\bf U} \forall \langle p_1, \ldots, p_n
\rangle \in [A]^{< \omega}[F(\langle p_1, \ldots, p_n
\rangle) = F'(\langle p_1, \ldots, p_n \rangle)]$.
$\sim$ can easily be verified to be an equivalence
relation on the collection of all such $F$, so we let
$[F]_\sim$ be the equivalence class of $F$ under $\sim$.
$V'$ can now be defined as $V[\langle \kappa_n : n <
\omega \rangle, \langle G \vert j : j < \omega \rangle,
\langle {\bf H_{\rm n}} : n < \omega \rangle]$ where for
$n < \omega$, ${\bf H}_n = \{[F]_\sim : \exists \pi \in
G[F \in \pi$ and $\pi$ has length $n]\}$. For any $\rho \in
{\bf G}$ and any reasonably chosen term $\tau$ for a set in
$V'$, the definition of {\bf G} ensures that $\tau$ is
invariant under $\rho$. Thus, the exact same argument as in
Theorem 3.3 of [Ma1], using the fact mentioned in the last
sentence of the last paragraph, can be used to show that
$V' \models$ ``$\kappa = \aleph_\omega$ and ${(\kappa^+)}^V
= \aleph_{\omega + 1}$'', i.e., that ${(\kappa^+)}^V$
isn't collapsed in $V'$. \hfil \break
\indent We complete the proof of our Theorem by showing that
$V' \models$ ``$\square^*_{\aleph_\omega} +
\neg \square_{\aleph\omega}$''. To show that $V' \models$ ``
$\square^*_{\aleph_\omega}$'', we note that since $\kappa =
{\aleph_\omega}^{V'}$,$\kappa$ is regular in $V$,
${\aleph_1}^V = {\aleph_1}^{V'}$, ${p(\aleph_1)}^V =
{p(\aleph_1)}^{V'}$, ${(\kappa^+)}^V = {(\kappa^+)}^{V'}$,
and $V \models GCH$, Theorem 3 of [BDM] (which states that if
for some cardinal $\delta$ there is an inner model $M$ of
$ZFC + GCH$ containing all subsets of $\aleph_1$ in which
$\delta$ is regular and
$\delta^+ = {(\delta^+)}^M$, then $V \models$ ``$
\square^*_\delta$'') can be applied to show $V' \models $``
$\square^*_{\aleph_\omega}$''; here, $V'$ plays the role of
the universe, $V$ plays the role of the inner model,
and $\delta = \kappa$. To show that $V' \models$ ``$
\neg \square_{\aleph_\omega}$'',we note that as in [BDM]
the filter ${\bf F} = \{x \subseteq \kappa^+ : x \in
V'$ and $\exists m[\{\lambda_n : m \le n < \omega\}
\subseteq x] \}$ is a uniform filter which can be extended
to an ultrafilter ${\bf U}^*$ over ${p(\kappa^+)}^
{V'}$ which is $\lambda$-indecomposable for any $V'$
regular $\lambda$ with $\omega < \lambda <
{\aleph_\omega}^{V'}$. Assuming ${\bf F} \in V'$, the
extension can be done in $V'$, yielding that ${\bf U}^*
\in V'$ and, for the same reasons as in [BDM], $V'
\models$ ``$\neg \square_{\aleph_\omega}$''. It thus
suffices to show that ${\bf F} \in V'$. \hfil \break
\indent To show ${\bf F} \in V'$, define a filter
${\bf F}' \in V'$ by $x \in {\bf F}'$ iff there is a term
$\tau$ denoting $x$ so that for every $m < \omega$ there is
some $n \ge m$ and a condition $\pi_n$ of length $n$,
$\pi_n = \langle p_0, \ldots, p_n, f_0, \ldots, f_n, A_n,
F_n \rangle$ satisfying: \hfil \break
\noindent 1. $\pi_n \forces$ ``$\tau \subseteq
{(\kappa^+)}^{V'}$, $\tau \in V'$, and $\exists m_0
[\{\lambda_j : m_0 \le j < \omega \} \subseteq \tau]$''.
\hfil \break
\noindent 2. $\langle p_1 \cap \kappa, \ldots, p_n \cap
\kappa \rangle = \langle \kappa_1, \ldots, \kappa_n
\rangle$. \hfil \break
\noindent 3. $\langle f_0, \ldots, f_n \rangle \in
G \vert (n+1)$. \hfil \break
\noindent 4. $F_n \in {[F]}_\sim$ for some ${[F]}_\sim
\in {\bf H}_n$. \hfil \break
\noindent Clearly, the definition of ${\bf F}'$ can be
given in $V'$. We show that ${\bf F}' = {\bf F}$.
\hfil \break
\indent If $\pi = \langle p_1, \ldots, p_n, f_0, \ldots,
f_n, A, F \rangle \in G$ is so that $\pi \forces$ ``$\tau
\in \dot {\bf F}$'' and $x = i_G(\tau)$, then the definition
of $P$ and $G$ will ensure that properties (1) - (4) above
hold for any condition $\pi_m$ extending $\pi$ of length $m$
where $m \ge n$ and $\pi_m \in G$, meaning that $x \in
{\bf F}'$. If now $x \in {\bf F}'$ and $\tau$ is a term
denoting $x$, let $\pi = \langle p_1, \ldots, p_n, f_0,
\ldots, f_n, A, F \rangle \in G$ be so that $\pi$ decides
``$\tau \in \dot {\bf F}$''. We are done if $\pi \forces
$ ``$\tau \in \dot {\bf F}$'', so assume that $\pi \forces
$ ``$\tau \not \in \dot {\bf F}$'', i.e., that $\pi \forces
$ ``$\forall m_0[\{\lambda_j : m_0 \le j < \omega \} \not
\subseteq \tau]$''. By the definition of ${\bf F}'$, let
$m > n$ be so that $\pi' = \langle p_1', \ldots, p_m',
f_0', \ldots, f_m', A', F' \rangle \forces$ ``$\exists m_0
[\{\lambda_j : m_0 \le j < \omega \} \subseteq \tau]$'' and
$\pi'$ satisfies properties (1) - (4). By the genericity of
$G$, let $\pi'' \in G$ extend $\pi$, $\pi'' = \langle p_1,
\ldots, p_n, p_{n+1}, \ldots, p_m, f_0, \ldots, f_n, f_{n+1},
\ldots, f_m, A'', F'' \rangle$. \hfil \break
\indent Let $\rho \in {\bf G}$ be an automorphism of $P$
generated by the permutation of $[D]^{< \omega}$ given by
$\psi(p_i) = p_i'$ and $\psi(p_i') = p_i$ for $1 \le i \le
m$ and $\psi$ is the identity otherwise. By the definition
of $\psi$ and $F'$, $\rho(\pi') = \langle \psi(p_1'),\ldots,
\psi(p_m'), f_0', \ldots, f_m', \psi(A'), \psi(F') \rangle =
\langle p_1, \ldots, p_m, f_0', \ldots, f_m', B, F' \rangle$
for $B = \psi(A')$, and since $\tau$ can be chosen so as to
be invariant under $\rho$ and each $\lambda_j$ for $j > m$
has the same interpretation for any generic object containing
$\rho(\pi')$ as it does for any generic object containing
$\pi'$, $\rho(\pi') \forces$ ``$\exists m_0[\{\lambda_j :
m_0 \le j < \omega \} \subseteq \tau]$''. For any measure 1
set $C_0$ so that $\forall \langle r_1, \ldots, r_l \rangle
\in {[C_0]}^{< \omega}[F'(\langle r_1, \ldots, r_l \rangle) =
F'''(\langle r_1, \ldots, r_l \rangle)]$ for some $[F''']_\sim
\in
{\bf H}_m$, let $C'(\langle r_1, \ldots, r_l \rangle)$ for
$\langle r_1, \ldots, r_l \rangle \in {[D]}^{< \omega}$ be
defined as $A''(\langle r_1, \ldots, r_l \rangle) \cap
B(\langle r_1, \ldots, r_l \rangle) \cap A'(\langle r_1,
\ldots, r_l \rangle) \cap C_0 \cap C_1(\langle r_1, \ldots,
r_l \rangle)$ where $\pi''' = \langle p_1, \ldots, p_m,
f_0'', \ldots, f_m'', C_1, F''' \rangle \in G$. Since
$F'''$ and $F''$ are both parts of conditions of length
$m$ which are elements of $G$ and $F' \vert {[C_0]}^
{< \omega} = F''' \vert {[C_0]}^{< \omega}$, the
compatibility of $G$ ensures that for any $\langle r_1, \ldots,
r_l \rangle \in {[D_{p_m}]}^{< \omega}$ so that $r_i \in
C'(\langle p_1, \ldots, p_m, r_1, \ldots, r_{i-1} \rangle)$
for $1 \le i \le l$, $\{r: F'(\langle r_1, \ldots, r_l,
r \rangle)$ and $F''(\langle r_1, \ldots, r_l, r \rangle)$
are compatible$\}\in {\bf U}$ and $\{r: F'(\langle r
\rangle)$ and $F''(\langle r \rangle)$ are compatible$\}\in
{\bf U}$. For any such $r$, define $H(\langle r_1, \ldots,
r_l, r \rangle)$ as $F'(\langle r_1, \ldots, r_l, r
\rangle) \cup F''(\langle r_1, \ldots, r_l, r \rangle)$
and $H(\langle r \rangle)$ as $F'(\langle r \rangle) \cup
F''(\langle r \rangle)$; for any appropriate sequence for
which $H$ is not defined by the above, take $H$ on that
sequence to be the empty function. Since $f_i$ and
$f_i'$ for $0 \le i \le m$ must be compatible, the condition
$\langle p_1, \ldots, p_m, f_0 \cup f_0', \ldots, f_m \cup
f_m', C', H \rangle$ is now an extension of $\rho(\pi')$
and $\pi''$, contradicting $\rho(\pi') \forces$ ``$\exists
m_0[\{\lambda_j : m_0 \le j < \omega\} \subseteq \tau]$''
and $\pi'' \forces$ ``$\forall m_0[\{\lambda_j : m_0 \le j
< \omega\} \not \subseteq \tau]$''. This contradiction
shows that ${\bf F} \in V'$, so $V' \models$ ``$\neg
\square_{\aleph_\omega}$''. This completes the proof of
our Theorem. \hfil \break
\hfill$\square$
Theorem \hfil \break
\indent We note that the above proof that $V' \models$ ``
$\neg \square_{\aleph_\omega}$'' shows that the definition of
${\bf F}'$ makes sense, i.e., is well defined. Also, for
the same reasons as in [BDM], the model theoretic transfer
principle $\langle \aleph_1,\aleph_0 \rangle \to \langle
\aleph_{\omega + 1}, \aleph_\omega \rangle$, a consequence
of $\square_{\aleph_\omega}$, holds in $V'$.
\hfil \break
\indent In conclusion, we remark that an alternate proof of
the Theorem can be given using Magidor's definition of
[Ma1] or [BDM] of the constraint portion $F$ of a condition,
assuming the ultrafilter {\bf U} on $P_\kappa(\kappa^+)$
is so that ${\bf U}_\kappa$ is a normal ultrafilter over
$\kappa$ and $\{p$ : The order type of $p$ is ${(p \cap
\kappa)}^+\} \in {\bf U}$. This last fact is of course
true if {\bf U} is a normal ultrafilter over
$P_\kappa(\kappa^+)$, but if $\kappa$ is $\kappa^+$ strongly
compact, i.e., if {\bf U} is not assumed to be fully normal,
then it is unknown if such an ultrafilter exists. (Blass
and Zwicker have pointed out that such an ultrafilter
{\bf U} exists assuming $\kappa$ is $\kappa^{++}$ strongly
compact and GCH if we drop the requirement that
${\bf U}_\kappa$ is a normal ultrafilter over $\kappa$.
The normality of ${\bf U}_\kappa$, however, is essential
in the proof of Lemma 1.) This
prompts us to ask if the existence or consistency of such an
ultrafilter can be proven from just $\kappa^+$ strong
compactness, or indeed, from any amount of strong compactness.
\vfill \eject
\vskip 1in
\raggedright\frenchspacing
\centerline{References}
\vskip 1.25in
\item{[AH]} A. Apter, J. Henle, ``Relative
Consistency
Results via Strong Compactness'', to appear in {\it Fundamenta
Mathematicae}. \hfil \break
\item{[BDM]} S. Ben-David, M. Magidor, ``The Weak Box is
Really Weaker than the Full Box'', {\it J. Symbolic Logic
51}, 1986, 1029-1033. \hfil \break
\item{[C]} J. Cummings, ``A Model in which GCH Holds
at Successors but Fails at Limits'', to appear in {\it
Transactions of the American Math. Soc.} \hfil \break
\item{[H]} J. Henle, ``Partition Properties and
Prikry Forcing on Simple Spaces'', {\it J. Symbolic Logic
55}, 1990, 938-947. \hfil \break
\item{[Ma1]} M. Magidor, ``On the Singular Cardinals
Problem I'', {\it Israel J. Math. 28}, 1977, 1-31.
\hfil \break
\item{[Ma2]} M. Magidor, ``On the Singular Cardinals
Problem II'', {\it Annals of Math. 106}, 1977, 517-549.
\hfil \break
\vfill \eject \end