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% Carlson collapse is minimal under MA
% A. Miller  Sept 2012

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\def\dd{{\mathcal D}}
\def\force{\forces}
\def\forces{{\;\Vdash}}
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\def\om{\omega}
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\def\rmor{\mbox{ or }}
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\begin{document}

\begin{center}
{\large Carlson Collapse is minimal under MA}
\end{center}

\begin{flushright}
Arnold W. Miller\\
Sept 2012
\end{flushright}

\def\address{\begin{flushleft}
Arnold W. Miller \\
miller@math.wisc.edu \\
http://www.math.wisc.edu/$\sim$miller\\
University of Wisconsin-Madison \\
Department of Mathematics, Van Vleck Hall \\
480 Lincoln Drive \\
Madison, Wisconsin 53706-1388 \\
\end{flushleft}}
 
Namba forcing \cite{namba} may be regarded as a generalization of
Laver forcing \cite{laver} to $\om_2$.  The analogous forcing
for $\om_1$ we call the Carlson collapse.  We first encountered it
when writing our joint paper: Carlson, Kunen, and Miller \cite{min}.
In that paper we
used the Prikry collapse of $\om_1$, which is analogous to superperfect tree
forcing (e.g., Miller \cite{rat}) 
but with subtrees of $\om_1^{<\om}$.
We proved in \cite{min} the analogue of Theorem \ref{one}
for the Prikry collapse, namely that assuming Martin's axiom the generic
extension is minimal.  Our paper \cite{min} 
did not include Theorem \ref{carlson}, although
Carlson had already proved it with an easier proof than
is given here.  
Lemma \ref{main} was obtained 
while giving a topics course \cite{forcing} on forcing.

One of my colleagues many years ago liked to joke about the
referee report that said; ``This paper fills a
much needed gap in the literature.''



\begin{define}
(1) A subtree $p\su \om_1^{<\om}$ is Carlson iff there exists $s\in p$ called
the root of $p$ such that for all $t\in p$ either $s\su t $ or $t\su s$ and
for every $t\in p$ with $s\su t$ there are
uncountably many $\al<\om_1$ with $s\cat(\al)\in p$.

(2) Let $\poset$ be the partial order of Carlson trees under inclusion.

(3) We write $p\leq_0 q$ iff $p\leq q$ and $\root(p)=\root(q)$.

(4) For $s\in p$ define $p_s=\{t\in p \st s\su t \rmor t\su s\}$.

(5) $B(p)=\{t\in p\st \root(p)\su t\}$ $\;\;$ (nodes
beyond the root).
\end{define}


\begin{lemma} \label{main}
Suppose ${\rm MA}_{\om_1}$ and we are given $(p_\al\in\poset:\al<\om_1)$ and
$\tau$ a $\poset$-name such that for each $\al<\om_1$
$$p_\al\forces \tau\in 2^\om\sm V.$$
Then there exists
$(q_\al\leq_0 p_\al:\al<\om_1)$ and $(C_\al:\al<\om_1)$ pairwise
disjoint closed subsets of $2^\om$ such that for every $\al<\om_1$
$$q_\al\forces \tau\in C_\al.$$

\end{lemma}

\begin{theorem}\label{one}
If $M\models {\rm MA}_{\om_1}$ and $G$ is $\poset$-generic over $M$,
then for every $x\in 2^\om\cap M[G]$ either $x\in M$ or $G\in M[x]$.
\end{theorem}
\proof
Given any $p$ such that
$p\forces\tau\in 2^\om\sm M$ construct $q\leq_0p$ and
closed sets $(C_s:s\in B(q))$ such that 
\begin{enumerate}
\item $q_s\forces \tau\in C_s$ for each $s\in B(q)$,
\item $C_s\su C_t$ if $t\su s$, and
\item $C_{s\cat(\al)}\cap C_{s\cat(\be)}=\emptyset$ if $\al\neq\be$.
\end{enumerate}
This is an easy fusion argument combined with
Lemma \ref{main}.  We claim that $$q\forces G\in M[\tau^G].$$
This is because $G$ is determined by the generic collapse
map $g\in \om_1^\om$ defined by $g=\bigcap G$.  Then
$$G=\{p\in\poset\st g\in [p]\}
\rmand g=\bigcup\{s\st \tau^G\in C_s\}.$$
\qed

\begin{theorem}\label{carlson}
(Carlson 1979) Suppose $M\models {\rm MA}_{\om_1}$ and $G$ is
$\poset$-generic over $M$. Then for every $f\in\om^\om\cap M[G]$
there exists $g\in M\cap \om^\om$ such that $\forall n \;\;f(n)<g(n)$.
\end{theorem}

\proof
Without loss we may suppose that 
$$p\forces \name{f}\in\om^\om\sm M.$$
Let $E\su 2^\om$ be the eventually zero reals.   Let
$\Phi:2^\om\sm E\to \om^\om$ be the natural homeomorphism
and let $\tau$ be a name for $\Phi^{-1}(f)$.  Letting
$p_\al=p$ for all $\al<\om_1$ we obtain $(q_\al\leq_0 p:\al<\om_1)$
and closed pairwise disjoint $(C_\al\su 2^\om:\al<\om_1)$ such that
$q_\al\forces \tau\in C_\al$ for all $\al<\om_1$.  Since the
$C_\al$ are pairwise disjoint there must be $\al$ with
$C_\al\cap E=\emptyset$.  This implies that $\Phi(C_\al)=K_\al\su\om^\om$
is a compact set and so we may find $g\in\om^\om$ such that
$g$ dominates every element of $K_\al$.  But then
$$q_\al\forces \forall n \;\;\name{f}(n)<\check{g}(n).$$
\qed

\begin{center}
Proof of Lemma \ref{main}.
\end{center}

\begin{claim}\label{laverlemma}
Given any sentence $\theta$ and condition $p$ there exists
$q\leq_0p$ such that $$q\forces\theta \rmor q\forces\neg\theta.$$
\end{claim}
\proof
This is the Laver Lemma. It is also true for Namba forcing and
many others.
\qed

\begin{claim}\label{claim1}
Suppose $p_0\forces \tau\in 2^\om\sm M$.
Then there exists $q_1,q_2\leq_0 p_0$ and pairwise
disjoint clopen sets $C_1,C_2$ such that $q_1\forces \tau \in C_1$ and
$q_2\forces \tau \in C_2$.
\end{claim}
\proof
For $p\leq p_0$ define $p$ is good iff there are $q_1,q_2\leq_0 p$ and 
pairwise
disjoint clopen sets $C_1,C_2$ such that $q_i\forces \tau \in C_i$
for $i=1,2$.   Note that if $p$ is bad and $s=\root(p)$ then for
all but countably many $\al<\om_1$ with $s\cat(\al)\in p$ the condition
$p_{s\cat(\al)}$ is bad.  This because there are only countably
many pairs of disjoint clopen sets $C_1,C_2$.

By this observation if the Claim fails then we may construct $q\leq_0 p_0$
such that for every $t\in B(q)=\{t\in q : \root(q)\su t\}$ the condition
$q_t$ is bad.
Using Claim \ref{laverlemma},
it follows that for every $s\in B(q)$  there
exists a unique $x_s\in 2^\om$ such that for every $n<\om$ there exists
$p\leq_0 q_s$ such that $p\forces \tau \res n = x_s\res n$. 
If there were two $x_s$ with this property, we could easily get
a contradiction to the badness of $q_s$.

For every $s\in B(q)$ it must be that $x_s=x_{s\cat(\al)}$ for all 
but countably many $s\cat(\al)\in q$.  To see this suppose not and
let $Q(s)=\{\al<\om_1\st s\cat(\al)\in q\}$. 
Then we would be able
to find $t\in 2^{<\om}$ with the property that uncountably many
$\al\in Q(s)$ had $t\su x_{s\cat(\al)}$ but $t\neq t^\pr = x_s\res |t|$.
But this
means we can find $p\leq_0 q_s$ such that $p\forces t\su \tau$.
We can also find $p^\pr\leq_0q_s$
such that $p^\pr\forces t^pr\su \tau$ by the definition of $x_s$.
This contradicts the badness of $q_s$.

By the above arguments we can find $x\in 2^\om$ and $q\leq_0 p_0$ such
that $x_s=x$ for every $s\in B(q)$.  This contradicts the assumption
that $p_0\forces \tau\neq \check{x}$.

\qed

\begin{claim}\label{claim2}
Suppose $p_i\forces \tau\in 2^\om\sm M$
for $i=1,2$. Then
there exists $q_1\leq_0 p_1$, $q_2\leq_0 p_2$ and pairwise
disjoint clopen sets $C_1,C_2$ such that $q_i\forces \tau \in C_i$
for $i=1,2$.
\end{claim}

\proof
Apply Claim \ref{claim1} to $p_1$ and obtain $q_{1,i}\leq_0 p_1$ and disjoint
clopen $C_1,C_2$ such that $q_{1,i}\forces\tau\in C_i$ for $i=1,2$.
We may as well assume $C_1,C_2$ are complementary.  Apply the Laver
Lemma to $p_2$ and get $q_2\leq_0 p_2$ such that either 
$q_2\forces \tau \in C_1$ or $q_2\forces\tau\in C_2$.  
If $q_2\forces\tau\in C_2$, take $q_1=q_{1,1}$, otherwise
take $q_1=q_{1,2}$.
\qed

\begin{claim}\label{claim3}
Suppose $n<\om$ and
$p_i\forces \tau\in 2^\om\sm M$ for $i<n$.
Then
there exists $(q_i\leq_0 p_i:i<n)$ 
and pairwise
disjoint clopen sets $(C_i :i<n)$
such that $q_i\forces \tau \in C_i$
for $i<n$.
\end{claim}
\proof
Iteratively apply Claim \ref{claim2} to all pairs $i<j<n$.
\qed

\begin{define}
For $T$ a finite subtree of $B(q)$ define 

(1) $p\leq_T q$ iff
$p\leq_0 q$ and $T\su p$.  

(2) For each $t\in T$ define
$$q_{t,T}=\{s\in q\st s\su t \rmor 
(t \subset s \rmand s\res (|t|+1)\notin T\}$$
\end{define}
Note that $\{q_{t,T}\st t\in T\}$ is a finite maximal antichain
beneath $q$.  This is analogous to Laver's $q\leq_n p$ except
there are uncountably many $T$.

\begin{claim}\label{claim9}
Suppose $p\forces \tau\in 2^\om\sm M$ and 
$p^\pr\forces \tau\in 2^\om\sm M$
and $T\su B(p)$ and $T^\pr\su B(p^\pr)$ are finite subtrees.
Then there are $q\leq_{T} p$ and $q^\pr\leq_{T^\pr} p^\pr$ 
and pairwise disjoint
clopen sets $C$ and $C^\pr$ such that $q\forces\tau\in C$
and $q^\pr\forces\tau\in C^\pr$.
\end{claim}
\proof
Let $p_i$ for $i<m$ list all
$p_{T,t}$ for $t\in T$ and let $p_i$ for $m\leq i<n$ list all $p^\pr_{T^\pr,t^\pr}$
for $t^\pr\in T^\pr$.
Apply Claim \ref{claim3} to obtain $q_i$ and $C_i$.
Let $q=\bigcup\{q_i:i<m\}$ and $C=\bigcup\{C_i:i<m\}$.
Similarly put $q^\pr=\bigcup\{q_i:m\leq i<n\}$
and $C^\pr=\bigcup\{C_i:m\leq i <n\}$.
\qed

\begin{claim}\label{laverfussion}
Suppose $p\forces \tau\in 2^\om$, then there exists $q\leq_0p$ such
and $\pi:q\to 2^{<\om}$ such that for every $s\in q$
$\;\;\;|\pi(s)|=|s|$ and $q_s\forces \pi(s)\su\tau$.
\end{claim}
\proof
Use the Laver lemma (Claim \ref{laverlemma})
and fusion to get the result by considering the sequence of
sentences ``$\tau(n)=0$''.
\qed

\begin{define}
For $q$ as in Claim \ref{laverfussion} define the poset
$\qq(q)$ as follows:
 
$(T,C)\in\qq(q)$ iff $T$ is a finite
subtree of $B(q)$, $C$ is clopen subset of $2^\om$, and
there exists $p\leq_T q$ such that $p\forces \tau\in C$.

Define $(T_1,C_1)\leq (T_2,C_2)$ iff $T_1\supseteq T_2$ and $C_1\su C_2$.
\end{define}

From now on for the poset $\qq(q)$ the condition $q$ will always
have the property of Claim \ref{laverfussion} and hence for
any $p\leq q$ and clopen set $C$ we have that $p\forces\tau\in C$
iff the range of the induced continuous map $\pi:[p]\to 2^\om$ is
a subset of $C$.

\begin{claim}
$\qq(q)$ has the ccc.
\end{claim}
\proof
Since there are only countably many clopen sets it is enough
to see that any pair with the same clopen set,
$(T_1,C)$ and $(T_2,C)$ is compatible.  We claim
that $(T_1\cup T_2,C)\in\qq(q)$.  Note that for $p\leq q$ that $p\forces \tau \in C$ iff 
$[\pi(s)]\cap C\neq\emptyset$ for every $s\in p$.
It follows that if $p_1\leq_{T_1} q$
and $p_2\leq_{T_2} q$ and each force $\tau\in C$, 
then $(p_1\cup p_2)\leq_{T_1\cup T_2}q$ and 
$p_1\cup p_2\forces\tau\in C$.
\qed

\begin{define}
For $G$ and $\qq(q)$-filter define
$$q^G=\bigcup\{T\st \exists C \; (T,C)\in G\}$$
and
$$C^G=\bigcap\{C\st \exists T \; (T,C)\in G\}.$$
\end{define}

\begin{claim}\label{claim14}
We can find $\dd$ a family of dense subsets of $\qq(q)$ with 
$|\dd|=\om_1$ such that for every $G$ a $\qq(q)$ filter which meets
each element of $\dd$ we have that
$q^G\leq_0p$ and $q^G\forces \tau\in C^G$.
\end{claim}
\proof
Note that the trivial condition $(\{root(q)\},2^\om)$ is always in
$G$.
For $s\in B(q)$ and $\al<\om_1$ define
$$D_{s,\al}=\{(T,C)\in \qq(q)\st (T,C)\forces s\notin q^G \rmor
\exists \be>\al \;\; s\cat(\be)\in T\}$$
To see that it is dense note that any condition can be extended
to a condition $(T,C)$ such that
either $(T,C)\forces s\notin q^G$ or $(T,C)\forces s\in q^G$.
In the first case $(T,C)\in D_{s,\al}$ and we are done.  
In the second case we must be able to find
$(T^\pr,C^\pr)\leq (T,C)$ with $s\in T^\pr$.
By the definition of $\qq(q)$ there exists $p\leq_{T^\pr} q$
such that $p\forces \tau\in C^\pr$.  Choose any $\be>\al$ with
$s\cat(\be)\in p$.  Then $p$ witnesses that $(T^\pr\cup\{s\cat(\be)\},C^\pr)$
is in $\qq(q)$ and $D_{s,\al}$.

Meeting all the $D_{s,\al}$ guarantees that $q^G\in\poset$
and $q^G\leq_0q$.

To see that $q^G\forces \tau\in C^G$ it is enough to show that
for every $(T,C)\in G$ that
$$q^G\forces \tau\in C.$$
Choose $n$ large enough so that there exists $\Gamma\su 2^n$ such that
we may write $C=\bigcup\{[s]\st s\in\Gamma\}$.  We claim that
$(T,C)$ forces that
for every $s\in q^G\cap \om_1^n$ that $\pi(s)\in \Gamma$.  If this were
not the case, then for some $(T^\pr,C^\pr)\leq (T,C)$ and
$s\in T^\pr\cap \om_1^n$ we would have that $\pi(s)\notin\Gamma$.
But this means that $q_s\forces \tau\in [\pi(s)]$ 
and therefor $q_s\forces \tau\notin C$ contradicting the 
definition of $\qq(q)$ that $C^\pr\su C$ and 
there exists $p\leq_{T^\pr}q$ (so $p_s\leq_0 q_s$)
and $p\forces \tau\in C^\pr$.
\qed

Finally we prove Lemma \ref{main}.  We assume that the
$q_\al\leq_0p_\al$ have the
property as in Claim \ref{laverfussion}.  We let
$$\qq=\sum \{\qq(q_\al)\st\al<\om_1\}$$
be the direct sum which has the ccc by ${\rm MA}_{\om_1}$.  
For any $\al<\be<\om_1$ let
$$D_{\al,\be}=\{p\in \qq\st p_\al=(T_\al,C_\al),\;\; p_\be=(T_\be,C_\be),
\rmand C_\al\cap C_\be=\emptyset\}
$$
By Claim \ref{claim9} this set is dense.  By Claim \ref{claim14}
we may find a family $\dd$ of dense subsets of $\qq$ with $|\dd|=\om_1$
such that if $G$ is a $\qq$-filter meeting each element of $\dd$ 
then each $q_\al^G\leq_0 q_\al$ has the property that $q_\al^G\forces 
\tau \in C_\al^G$.  If $G$ meets all the $D_{\al,\be}$ then
the $C_\al^G$ will be pairwise disjoint.  Applying 
${\rm MA}_{\om_1}$ gives us the sequences
$(q^G_\al\leq_0q_\al\leq_0p_\al:\al<\om_1)$ and $(C^G_\al:\al<\om_1)$
to prove Lemma \ref{main}.

\qed


\begin{thebibliography}{99}

\bibitem{min} Carlson, Tim; Kunen, Kenneth; Miller, Arnold W.; A minimal degree
which collapses $\om_1$. J. Symbolic Logic 49 (1984), no. 1, 298-300.

\url{http://www.math.wisc.edu/~miller/res/min.pdf}

\bibitem{laver} Laver, Richard; On the consistency of Borel's conjecture. Acta
Math. 137 (1976), no. 3-4, 151-169.

\bibitem{rat} Miller, Arnold W.; Rational perfect set forcing. Axiomatic set
theory (Boulder, Colo., 1983), 143\-159, Contemp. Math., 31, Amer. Math.
Soc., Providence, RI, 1984.

\url{http://www.math.wisc.edu/~miller/res/rat.pdf}

\bibitem{forcing} Miller, A.; Math 873 Forcing Fall 2009

\url{http://www.math.wisc.edu/~miller/old/m873-09/index.html}

\bibitem{namba} Namba, Kanji; Independence proof of
$(\om,\om_\al)$-distributive law in complete Boolean algebras. Comment.
Math. Univ. St. Paul. 19 (1971), 1-12.

\end{thebibliography}

\bigskip

\address


\end{document}

