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\def\dom{{\rm dom}}
\def\name#1{\mathrin\defg{#1}}
\def\concatx{{\hat{\phantom{a}}}}
\def\dd{{\mathcal D}}
\def\pp{{\hat{p}}}
\def\ga{\gamma}
\def\name#1{\mathring{#1}}
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\def\elemsub{\preceq}
\def\forces{{\;\Vdash}}
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\def\poset{{\mathbb P}}
\def\proof{\par\noindent Proof:\par}
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\def\qed{\par\noindent QED \par}
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\def\rmiff{{\mbox{ iff }}}
\def\rmor{{\mbox{ or }}}
\def\su{\subseteq}
\def\split{{\rm split}}
\def\supp{{\rm supp}}

\newtheorem{theorem}{Theorem}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{define}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}


\begin{document}

\begin{flushright}
Arnold W. Miller \\
March 2016
\end{flushright}

\begin{center}
Nyikos's Joke
\end{center}

In 1988 Peter Nyikos asked me if there
were any Q-sets in Laver's model.
After I solved it (Theorem \ref{mainthm}),  I asked Peter why he wanted to know.
He said he was just joking, ``no Q-points / no Q-sets'', referring
to my paper \cite{laver}.

The file ``nyikos.jok'' \cite{nyjoke} dated Nov 1, 1998 was part of a
letter or email I sent to Jaime Ihoda  (Haim Judah).   At the time Judah
was moving from Berkeley to Tel Aviv and we were writing the
joint paper Judah-Miller-Shelah \cite{sacks}.
What follows may be some version of the proof I had in 1998, but
I haven't been able to locate any other notes on the proof.
At the time I was more interested in
Theorem \ref{ramsey}.   I put this result for one Laver real
(for any Borel $B\su\om^\om$ there is a Laver tree
$p$ with $[p]\su B$ or $[p]\cap B=\emptyset$) 
in the first version of our paper  \cite{sacks-old} (last section).  
A referee for the Journal of Symbolic Logic noted that
it is an immediate consequence
 of Galvin-Prikry Theorem that Borel subsets of $[\om]^\om$
are Ramsey.   Mathias trees are special Laver trees,  this can be seen via the map from  $[\omega]^\omega$
 to $\omega^\omega$ taking a set $X\su\om$ to the increasing enumeration
of $X$.\footnote{Similarly Superperfect trees are Laver trees, so Theorem \ref{ramsey} (for one-step)
for Laver trees implies it for Superperfect trees.}  
 It was rejected and we resubmitted it to the Archive for Mathematical Logic
with the last section deleted.


 Paul Larson recently asked me (May 2015) if I knew how
to prove there are no Q-sets in the iterated perfect set model.
Theorem \ref{ramsey} and  \ref{mainthm} are true for iterated Sacks forcing, as well as,
many other tree forcings, e.g. Superperfect forcing, Mathias forcing, Silver forcing,  or mixed versions.
Theorem \ref{ramsey} could probably be generalized to
the analogues of Silver's Theorem (analytic sets are Ramsay) and Ellentuck's topological
characterization of completely Ramsay sets \cite{ellen}.

I don't know  if Theorem \ref{ramsey} for Laver, Superperfect, and Sacks forcing follows
from Mathias forcing  by a simple argument as in the case of one-step forcing. 
 For the no Q-set result (Theorem \ref{mainthm}),  there are probably simpler arguments,
perhaps based on homogeneity or definability.  

Let $\poset$ be Laver forcing (\cite{laver}).  So $p\in\poset$ iff
$p\su \om^{<\om}$ is a nonempty subtree such that
 there exists $s\in p$ (called the root of $p$)
which is comparable to all $t\in p$ and has the property that every $t\in p$
with $s\su t$ splits, i.e. there are infinitely many $n<\om$ with
$t\concat(n)\in p$.   We use $t\concat(n)$ to denote the sequence
$(t_0,t_1,\ldots,t_k,n)$ where $t=(t_0,t_1,\ldots,t_k)$.
Define $p\leq q$ iff $p\su q$.

The splitting nodes of $p$ can be put naturally into one-to-one
correspondence with $\om^{<\om}$ preserving inclusion as
well as the lexicographical order.  Effectively list $\om^{<\om}$  
so that  for any $s\in \om^{<\om}$ and $n<\om$:
\begin{itemize}
\item $s$ is listed before $s\concat(n)$ and
\item $s\concat(n)$ is listed before $s\concat(n+1)$.
\end{itemize}

We let $\split_n(p)$ be the first $n+1$ split nodes of $p$ in
the induced order.   We define $p\leq_nq$ iff $p\leq q$ and
$\split_n(p)=\split_n(q)$.

For $p\in\poset$ and $s\in p$  let
$p_s=\{t\in p\;:\; s\su t \rmor t\su s\}$ and  let 
$$[p]=\{f\in\om^\om\::\; \forall n<\om \; f\res n\in p\}.$$

\begin{lemma}\label{jslemma}(See Judah-Shelah \cite{js})
For any $p\in\poset$ and $D\su \poset$
dense and downward closed there exists $q\leq_0 p$ such that 
for every $f\in [q]$ there is $k<\om$ such that $q_{f\res k}\in D$.
\end{lemma}

\proof
Let
$$R_0=\{s\in p\;:\;\exists q\leq_0p_s \;\;q\in D\}$$
and put
$$R_{\al}=\{s\in p \;:\;\exists^\infty n\;\;s\concat(n)\in 
\bigcup_{\be<\al}R_\be\}.$$
Let $\rank(s)$ be the least $\al$ with $s\in R_\al$ if there is
one, otherwise $\rank(s)=\infty$. 

\bigskip\noindent
Claim. If $r$ is the root of $p$, then $\rank(r)\neq\infty$.

\bigskip
If $\rank(s)=\infty$ then there are infinitely many $n$ such that
$s\concat(n)\in p$ and 
$\rank(s\concat(n))=\infty$.  But then we could build $q\leq_0p$ such
that all $s$ in $q$ have rank $\infty$.  This contradicts the density of
$D$.

Now we can prove the lemma by induction on the rank of the root node.
Let $L$ be the nodes immediately below the root of $p$ and having
smaller rank than the root.
For each $s\in L$ there is $q^s\leq_0p_s$ such that
for every $f\in [q^s]$ there is $k$ with $q^s_{f\res k}\in D$
and so $q=\bigcup_{s\in L}q^s$ satisfies 
the lemma.
\qed

Throughout  we suppose that $\ka$ is any sufficiently large regular cardinal.
$H_\ka$ is the set of all sets whose transitive closure has cardinality less
than $\ka$.   

\begin{lemma}\label{generic}
Suppose $M\elemsub H_\ka$ is countable, $p\in\poset\cap M$.
Then there exists $q\leq_0p$ such that every $f\in [q]$
is Laver generic over $M$, in fact, for all
downward closed dense $D\su\poset$ in $M$ and $f\in [q]$ there
exist $k$ with $q_{f\res k}\in D$.
\end{lemma}
\proof
This is an easy fusion argument.
Let $D_n$ for $n<\om$ list all dense downward closed  subsets of $\poset$ in $M$.
Put $p_0=p$ and build a sequence $p_{n+1}\leq_n p_n$ of conditions
in $\poset\cap M$ such that for every $n$ and $f\in [p_{n+1}]
$ there is $k$ with $(p_{n+1})_{f\res k}\in D_n$.
Let $q$ be the fusion.
\qed



\begin{define}\label{iteration}
We define the countable support iteration of Laver forcing $\poset_\al$ for $\al\leq\om_2$:
\begin{enumerate}
\item $\al=1$.  $\poset_1=\poset$ (Laver forcing). 
\item $\al$ limit.   $p\in\poset_\al$ iff $p\res\be\in\poset_\be$ for all $\be<\al$
and its support \par $\supp(p)=^{def}\{\be\;:\; p(\be)\neq 1\}$ is countable.
\item $\al=\be+1$.  $p\in\poset_{\be+1}$ iff $p\res\be\in\poset_\be$,
 $p\res\be\forces p(\be)\in\name{\poset}$, and $p(\be)$ is
a nice $\poset_\be$-name for a nonempty subtree of $\om^{<\om}$, which means:
\begin{enumerate}
  \item $p(\be)\su \{\la q,\check{s}\ra\;:\; s\in\om^{<\om} \rmand q\in\poset_\be\}$
  \item  if $\la q,\check{s}\ra \in p(\be)$, $t\su s$,  then  $\la q,\check{t}\ra\in p(\be)$
  \item $\la1,\check{\la\ra}\ra\in p(\be)$
\end{enumerate}
\end{enumerate}
$p\leq q$ iff $\forall \be<\al\;\; p\res\be\forces p(\be)\leq q(\be)$
\end{define}

\begin{define}
For  $p\in\poset_\al$, and we define
$[p]\su  (\om^\om)^\al$
by induction on $\al\leq\om_2$. 

$\al=1$. $$[p]=\{f\in\om^\om\;:\; \forall n<\om\;\; f\res n\in p\}.$$

$\al$ limit.  $$[p]=\{f\in(\om^\om)^\al\;:\; \forall \be<\al\;\; f\res\be\in [p\res\be].$$

$\al=\be+1$. $$[p]=\{f\in(\om^\om)^{\be+1}\;:\;  f\res\be\in [p\res\be] \rmand
f(\be)\in [p(\be)^{G_{f\res\be}}]\}.$$

where as usual
$\;\;p(\be)^{G_{f\res\be}}=\{s\;:\; \exists q\in G_{f\res\be}\;\; \la q,\check{s}\ra \in p(\be)\}.$

\noindent In all cases for $f\in (\om^\om)^\al$
 we define $G_f=\{p\in\poset_\al\;:\; f\in [p]\}$.
\end{define}

In general, $[p]$ could be the empty set since without a little genericity
$[p(\be)^{G_{f\res\be}}]$ may be empty.   Note that by using nice names,  it will
always be a nonempty subtree of $\om^{<\om}$.

\begin{define}
$\si:\al\to\om$ is finitely supported iff $$\supp(\si)=^{def}\{\be\;:\;\si(\be)\neq 0\}$$ is finite.
\end{define}

\begin{define}
For $q\in\poset_\al$, $f\in [q]$, and $\si:\al\to\om$ finitely supported, we say
that $q_{f,\si}\in\poset_\al$ is defined iff $\supp(\si)\su\supp(q)$ and for all $\be<\al$ 
$\;\;(q_{f,\si})\res\be$ is defined and if $n=\si(\be)$ and $s=f(\be)\res n$ then
$$(q_{f,\si})\res\be\forces \check{s}\in q(\be).$$
We then put $q_{f,\si}(\be)=q(\be)_s=^{def}\{(p,\hat{t})\in q(\be)\;: t\su s \rmor s\su t\}$.
\end{define}

\noindent Since $\la1,\check{\la\ra}\ra\in q(\be)$ if $\si(\be)=0,\;$ then
$(q_{f,\si})\res\be\forces \check{s}\in q(\be)$ is trivially true.


\begin{define}
 $p\in\poset_\al$ is determined is defined by induction on $\al$.

$\al=1$.  Every $p\in\poset$ is determined.

$\al$ limit.  $p\in\poset_\al$ is determined iff $p\res \be$ is determined for every $\be<\al$.
 
$\al=\be+1$. $p\in\poset_\al$ is determined iff $p\res \be$ is determined and
 for every $f\in [p\res\be]$ 
for every $n<\om$ there is a finitely supported $\si:\be\to\om$ such that
$(p\res\be)_{f,\si}$ is defined and for some $T_n\su\om^{<\om}$
$$(p\res\be)_{f,\si}\forces \split_n(p(\be))=\check{T_n}.$$
\end{define}
Note that if $p$ is determined, then for any $f\in[p]$ and $\be<\al$ that
$p(\be)^{G_{f\res\be}}$ is a Laver tree.

\begin{lemma}\label{closed}
If $p\in\poset_\al$ is determined, then
$[p]\su(\om^\om)^\al$ is closed.
\end{lemma}
\proof
Suppose $f\notin [p]$.   Choose $\be<\al$ minimal so that
$f\res\be\in [p\res\be]$ but $f\res(\be+1)\notin [p\res(\be+1)]$.
There must be some $n,N<\om$ and $T_n\su\om^{<\om}$ such that $\split_n(p(\be)^{G_f})=T_n$
and $s=f(\be)\res N$ is a node which cannot be on any Laver tree whose
$n$-split nodes are $T_n$.    Let $\si:\be\to\om$ be finitely
supported so that $(p\res\be)_{f,\si}$ is defined and forces
``$\split_n(p(\be))=\check{T_n}$''.    It follows that
$$\{g\in(\om^\om)^\al\;:\; g(\be)\res N = f(\be)\res N\rmand \forall \ga<\be \;\;g\res \si(\ga)= f\res \si(\ga)\}$$
is a basic open set containing $f$ and disjoint from $[p]$.
\qed

\begin{lemma}\label{extension}
If $p\in\poset_\al$ is determined, $f\in[p]$, and $\si:\al\to\om$ is
finitely supported, then there exists finitely supported $\rho:\al\to\om$
with $\rho(\be)\geq \si(\be)$ all $\be$ and
$p_{f,\rho}$ is defined.
\end{lemma}
\proof
By induction on the maximum of the support of $\si$.   If $\si$ is identically
zero, then $\rho=\si$ works.   Otherwise, let $\be<\al$ be the maximum such
that $\si(\be)\neq 0$.  By inductive hypothesis there is $\si_1:\be\to\om$ which
bounds $\si\res\be$ and $(p\res\be)_{f\res\be,\si_1}$ is defined.   By the definition
of determined there is $\si_2:\be\to\om$ such that $(p\res\be)_{f\res\be,\si_2}$ is defined
and forces $f(\be)\res\si(\be)\in p(\be)$.   (It is enough to decide $\split_n(p)$ for $n=\si(\be)$.)
But note that it is easy to show that for $\rho\res\be$ the maximum of $\si_1$ and $\si_2$ that
$(p\res\be)_{f\res\be,\rho\res\be}$ is defined.   So now extend $\rho\res\be$ to $\al$ by letting $\rho(\be)=\si(\be)$.
\qed

\begin{define} For $\al\leq\om_2$, $n<\om$, $F\in [\al]^{<\om}$ and $p,q\in\poset_\al$

$q\leq^{F}_{n} p$ iff $p\leq q$ and for all $\be\in F\;\;$  $q\res\be \forces q(\be)\leq_ n p(\be)$.
\end{define}

\begin{lemma}\label{main}
Suppose $\al\leq\om_2$, $n<\om$, $F\in [\al]^{<\om}$ and $p\in\poset_\al$.
\begin{enumerate}
\item[[det]] There is a determined $q\leq^F_n p$.
\item[[one]] If $D\su\poset_\al$ dense and downward closed, then 
there is a determined $q\leq^F_n p$ such that for every
$f\in [q]$ there is a finite partial $\si:\al\to\om$
such that $q_{f,\si}$ is defined and $q_{f,\si}\in D$.
\item[[ctlb]] If $\dd$ is a countable collection of dense and downward closed subsets
of $\poset_\al$, then \label{one}
there is a determined $q\leq^F_n p$ such that for every
$f\in [q]$ and $D\in\dd$ there is a finite partial $\si:\al\to\om$
such that $q_{f,\si}$ is defined and $q_{f,\si}\in D$.
\item[[gen]] For any countable $M\elemsub H_\ka$ with $p$ and $\al$ in $M$ there\label{two}
is a determined $q\leq^F_n p$ such that for every 
$f\in [q]$,  $G_f^M$ is $\poset_\al$-generic over $M$ where
$$G_f^M=^{def}\{r\in\poset_\al^M\;:\; \exists \si\in\Sigma\;\; q_{f,\si}\leq r\}$$
and 
$\Sigma$ is the set of finite partial maps  $\si:\al\to\om$ such that
$q_{f,\si}$ is defined.
\end{enumerate}
\end{lemma}
\proof
Induction on $\al$ followed by the size of $F$.  The case $\al=1$ is done
by Lemmas \ref{jslemma} and \ref{generic}.

\bigskip\noindent  {\bf [det]:}  

To simplify notation assume $n=0$. Put $F_0=F$ and $q_0=p$ and
construct a fusion sequence
$q_n\in\poset_\al$, $F_n\in[\al]^{<\om}$,   and
$\be_n<\al$,  for $n<\om$
 so that the following is true:
\begin {itemize}
\item $q_{n+1}\leq^{F_n}_nq_n$,
\item $\bigcup_n F_n = \bigcup_n \supp(q_n)$,  $F_n\su F_{n+1}$,
\item $\be_n\in F_n$, 
\item  for every  $\be\in\cup_n F_n$ there are infinitely many $n$ with $\be_n=\be$,
\item $q_n\res\be_n$ is determined, and 
\item for every $f\in [q_n\res\be_n]$ there is a finitely supported
$\si:\be_n\to\om$ such that $(q_n)_{f,\si}\res\be_n $ is defined and for some 
$T_n\su\om^{<\om}$ 
$$(q_n)_{f,\si}\res\be_n\forces \split_n(q_n(\be_n))=\check{T_n}.$$
\end{itemize}
Let $q$ be the fusion.   We prove that $q$ is determined.  Fix $\be<\al$ and assume
we have shown that $q\res\be$ is determined  and let $f\in[q\res\be]$. 
Fix $n$ so that $\be_n=\be$.   By construction 
there is  $T_n\su\om^{<\om}$ 
finitely supported $\si_n:\be\to\om$ so that
$$(q_n)_{f,\si_n}\res\be\forces \split_n(q_n(\be))=\check{T_n}.$$
Since $\be=\be_n\in F_n$ and $q\leq^{F_n}_nq_n$
$$q\res\be \forces \split_n(q(\be))= \split_n(q_n(\be)).$$
By Lemma \ref{extension}  there exists $\rho_n:\be\to\om$ dominating $\si_n$ so
that $(q)_{f,\rho_n}\res\be$ is determined.   Since $(q)_{f,\rho_n}\res\be\leq (q_n)_{f,\si_n}\res\be$
$$(q)_{f,\rho_n}\res\be\forces \split_n(q(\be))=\check{T_n}.$$
Since there are infinitely many $n$ such that $\be=\be_n$, hence $q\res(\be+1)$ is determined.

\bigskip\noindent  {\bf [one]:}  

Case 1. $F$ is empty.   
First extend to be in $D$, then extend to a determined condition.
For any $f$  take $\si$ to be the constant zero function.  

Case 2.  $0\in F$.   Let $F=F_0\cup\{0\}\su\al$ where $0\notin F_0$. 
By the maximality principle there is a name $\name{q}$ so that
$$p(0)\forces \name{q}\leq^{F_0}_n p\res[1,\al) \mbox{ is determined and }
\forall f\in [\name{q}]\;\exists \name{\si}:\al\to\om \;\; \name{q}_{f,\si}^{def}\in \name{E}
$$
where in $V[G]$
$$E=\{(r\res[1,\al))^{G}\;\;:\;\; r(0)\in G \rmand r\in D\}. $$

Put all relevant sets in
a countable $M\elemsub H_\ka$.
By Lemma \ref{generic} there exists $q(0)\leq_n p(0)$ such that
 $G_f$ is $\poset$-generic over $M$ for every $f\in [q(0)]$.
Consider the concatenation of $q(0)$ with $\name{q}$.   More precisely,
for $\be>0$ define
$$q(\be)=\{\la r,\check{s}\ra\;:\; r\leq q\res\be \rmand r\forces \check{s}\in \name{q}(\be)\}.$$

Given $h\in [q(0]$ since $G_h$ is $\poset$-generic over $M$ we have that
$$M[G_h]\models q^{G_h}\in \poset_{[1,\al)} \mbox{ is determined.}$$
Also in $M[G_h]$ for any $f\in  [q^{G_h}]$ there is a finitely supported
$\si:[1,\al)\to\om$ such that $q^{G_h}_{f,\si}$ is defined and an element of
$E$.  It follows that there is an $N<\om$ such that 
$$(q(0),q\res[1,\al))_{h\concatx f,N\concatx \si}\in D.$$

Thus the conclusion of [one] holds for all $f\in [q]$ such that
$f(0)\in [q(0)]$ and $f\res[1,\al)\in M[G_{f(0)}]$.   
To show that it holds for all $f\res[1,\al)\in V$
note that $[q\res [1,\al)^{G_{f(0)}}]$ is a closed set 
(Lemma \ref{closed}) and so by $\bf\Pi^1_1$ absoluteness
[one] holds for all $f$ in $V$.

\bigskip
Case 3. $0<\ga=\min(F)\su\al$.
This is similar to case 2.  Let 
$$p\res\ga\forces \name{q}\leq^F_n p\res[\ga,\al) \mbox{ is determined and }
\forall f\in [\name{q}]\;\exists \name{\si}:[\ga,\al)\to\om \;\; \name{q}_{f,\si}^{def}\in \name{E}
$$
where 
$$E=\{(r\res[\ga,\al))^{G}\;\;:\;\; r\res\ga\in G \rmand r\in D\}. $$
Taking $M$ as before,
by inductive hypotheses we can find determined $q\res\ga\leq p\leq\ga$ such that
$G_h$ is $\poset_\ga$-generic over $M$ for  every $h\in [q]$.  
Combining $q\res\ga$ and $\name{q}\res[\ga,\al)$ gives us the
required condition $q$.

\bigskip\noindent  {\bf [ctble]:}  

This will follow from [one] by the usual fusion arguments:
  We may assume
without loss that $\dd$ includes the countably many dense sets need
in the proof of [det] to show that the fusion is determined.
Let $D_n$ for $n<\om$ list $\dd$.   
To simplify notation assume $n=0$.  Put $F_0=F$ and $p_0=p$.
Let $F_n$ be an increasing sequence of finite subsets of
$\al$ constructed inductively so that
$$\bigcup_n F_n = \bigcup_n \supp(p_n),$$
where the $p_n$ are 
 determined conditions with $p_n\in M$,   
$p_{n+1}\leq^{F_n}_np_n$, and for every $f\in [p_{n+1}]$
there is a finitely supported $\si$ such that  
$(p_{n+1})_{f,\si}$ is defined and an element of $D_n$.
Let $q$ be the fusion.  

Given $f\in [q]$ and $n$ since $f\in [p_{n+1}]$ there is 
a finitely supported $\si:\al\to\om$ such that 
$(p_{n+1})_{f,\si}$ is defined and an element of $D_n$.
By Lemma \ref{extension} there is a $\rho:\al\to\om$ bounding $\si$ such that
$q_{f,\rho}$ is defined.  But $q_{f,\rho}\leq (p_{n+1})_{f,\si}$ and since
$D_n$ is downward closed we are done.


\bigskip\noindent  {\bf [gen]:}  

This is immediate from [ctlbe] just taking $\dd$ to be all downward closed subsets
of $\poset_\al$ which are in $M$.

\qed

I don't know if using Shelah's notion of $q$-generic condition (as in his definition
of proper forcing)  would give an easier proof of Lemma \ref{main}.

\begin{theorem}\label{ramsey}
For any $\al<\om_1$,  Borel set $B\su(\om^\om)^\al$,
and
$p\in \poset_{\al}$, there exists a determined condition
$q\leq p$ such that either $[q]\su B$ or
$[q]\cap B=\emptyset$.
\end{theorem}
\proof
Choose countable
$M\elemsub H_\ka$ containing  $p,\poset_\al$, and a Borel code for $B$. 
Extend $p$ so that
we may assume that $p\forces \name{f}\in B$ or
$p\forces \name{f}\notin B$.  Where $f$ is the generic Laver sequence. 
Assume the $p\forces \name{f}\in B$.
By  Lemma \ref{main} 
we have a determined $q\leq p$ such that all $f \in [q]$ are
generic over $M$.  Hence 
$$M[G_f]\models {f}\in B$$
By absoluteness of Borel predicates $[q]\su B$.
Similarly, if $p\forces \name{f}\notin B$, then 
$[q]\cap B=\emptyset$.
\qed

\begin{define}
$p\in\poset_\al$ is canonically determined (c.d.) iff it
is determined and for every $\be<\al$ and $s\in\om^{<\om}$ if
$\la q,\hat{s}\ra\in p(\be)$, then $q=(p\res\be)_{f,\si}$ for some
$f\in [p\res\be]$ and finitely supported $\si:\be\to\om$ such that
 $q=(p\res\be)_{f,\si}$ is defined.
\end{define}


\begin{lemma}\label{cannon}
If $p\in\poset_\al$ is determined, then there exists $p^\prime\in\poset_\al$ canonically
determined with the same support, $p\equiv p^\prime$, and $[p]=[p^\prime]$.
\end{lemma}
\proof
Note that if $q$ is canonically determined, then  $q_{f,\si}$ is canonically determined whenever
it is defined.
Define $p^\prime(\be)$ by induction. 
Let $p(0)=p^\prime(0)$.  Define
$$\Gamma_\be=\{ (p^\prime\res\be)_{f,\si}\mbox{ defined }\::\; f\in [p\res\be]\rmand
\si:\be\to\om \mbox{ finitely supported}\}$$
$$p^\prime(\be)=\{\la q,\hat{s}\ra \;:\; q\in \Gamma_\be \rmand
q\forces \hat{s}\in p(\be)\}.
$$
Assume $p\res\be\equiv p^\prime\res\be$.  Then for any $G$ $\poset_\be$-generic over $V$,
by determinateness $p(\be)^G=(p^\prime(\be))^G$ and so $p\res(\be+1)\equiv p^\prime\res(\be+1)$.
\qed

\begin{lemma}\label{canisom}
Let $\Sigma\in [\al]^\om$ and define $\poset_{\Sigma}^{c.d.}$ to be the
canonically determined conditions of $\poset_\al$ with support $\Sigma$.
Let $\be$ be the order type of $\Sigma$ and  $\poset_{\be}^{c.d.}$ the
canonically determined conditions of $\poset_\be$.   Then there is
natural isomorphism $j: \poset_{\Sigma}^{c.d.} \to \poset_{\be}^{c.d.}$
\end{lemma}
\proof
Let $j:\Sigma\to\be$ be the order isomorphism.
For $\ga\in\Sigma$ define 
$$j(p)(\ga)(j(\ga))=\{\la j(q),\hat{s}\ra\;:\; \la q,\hat{s}\ra \in q(\ga)\}.$$
\qed

I think that $\poset_\be$ is isomorphic to the set of conditions in $\poset_\al$ which have
hereditary support $\Sigma$,  i.e. $\supp(q)\su \Sigma$ and
if $\la p,\hat{s}\ra\in q(\be)$, then $p$ is hereditarily of support $\Sigma$.

\begin{theorem}\label{mainthm}
In Laver's model for the Borel conjecture \cite{laver} there is no
uncountable $Q$-set.
\end{theorem}
\proof
If an uncountable Q-set does occur by the usual
arguments due to Laver (\cite{laver} page 164) 
we may assume that this Q-set $X$ is in the ground model
$V$.  The continuum
hypothesis is true in $V$, hence by a transfinite argument of
length $\om_1$ and Theorem \ref{ramsey} we may construct $Y\su X$ such that for every $\al<\om_1$,
canonically determined $q\in\poset_\al$, and
Borel set $B\su [q]\times 2^\om$ coded
in $V$  there is a $u\in X$ and a canonically determined $r\leq q$ such that
for all ${f}\in [r]$ 
$$u\in Y\rmiff (f,u)\notin B.$$

\bigskip\noindent{\bf Claim}. In $V[G]$ where $G$ is $\poset_{\om_2}$-generic 
over $V$, the set $Y$ is not relatively Borel in $X$.

\proof
Suppose for contradiction that it is $\bf\Sigma^0_\be$.  
Let $U\su 2^\om\times 2^\om$
be a universal $\bf\Sigma^0_\be$ set coded in the ground model $V$. 
 Let $\tau$ be name for
an element of $2^\om$ and $p\in G$ a condition such that
$$p\forces \;\forall u\in X \;\;( u\in Y \rmiff (\tau,u)\in U ).$$

Put $ p,X,Y,\tau$ into a countable $M\elemsub H_\ka$.  Using 
 Lemma \ref{main} and \ref{cannon}  let $q\leq p$ be 
canonically determined with $\supp(q)=\Sigma=M\cap\om_2$.
Let $j: \poset_{\Sigma}^{c.d.} \to \poset_{\be}^{c.d.}$ be the isomorphism
from Lemma \ref{canisom}.  Let $r=j(q)$.

Define $B\su [r]\times 2^\om$ by 
$$(f,u)\in B\rmiff (\tau^{G_{j(f)}},u)\in U.$$
This map is continuous and so $B$ is Borel.  
By our construction there is $u_0\in X$ and a canonically 
determined $r_0\leq r$ in $\poset_\al$ so that for every $h\in [r_0]$
$$u_0\in Y \rmiff (h,u_0)\notin B.$$
Let $j(r_0)=q_0\leq q$,

 Now we get a contradiction.   Suppose
$G$ is $\poset_{\om_2}$-generic over $V$ with $q_0\in G$.
Let $f_G$ be the $\om_2$-sequence of Laver reals.
Let $h\in [r_0]$ be the Laver sequence with $h(j(\be))= f_G(\be)$.
By the choice of $u_0$ and $[r_0]$ we have that
$$ u_0\in Y \rmiff (h,u_0)\notin B.$$
By the definition of $B$ 
$$(h,u_0)\in B \rmiff (\tau^G,u_0)\in U$$
so
$$ u_0\in Y \rmiff (\tau^G,u_0)\notin U$$
which contradicts
$$p\forces  u_0\in Y \rmiff (\tau,u_0)\in U.$$
\qed



\begin{thebibliography}{99}

\bibitem{ellen}
Ellentuck, Erik;
A new proof that analytic sets are Ramsey.
J. Symbolic Logic 39 (1974), 163–165. 

\bibitem{js} % laver tree of laver reals
Judah, Haim; Shelah, Saharon;
The Kunen-Miller chart (Lebesgue measure, the Baire property, Laver reals and preservation theorems for forcing).
J. Symbolic Logic 55 (1990), no. 3, 909-927. 

\bibitem{sacks}
Judah, Haim; Miller, Arnold W.; Shelah, Saharon;
Sacks forcing, Laver forcing, and Martin's axiom
Archive for Mathematical Logic, 31(1992), 145-161.
\par\url{http://www.math.wisc.edu/~miller/res/sacks.pdf}

\bibitem{sacks-old} 
Unpublished rejected version of \cite{sacks} submitted to JSL 
in July 1990.
\par\url{http://www.math.wisc.edu/~miller/res/unpub/perf.pdf}

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

\bibitem{laver}
Miller, Arnold W.;
There are  no Q-Points in Laver's Model for the Borel Conjecture;
Proceedings of the American Mathematical Society, 78(1980), 103-106.
\par\url{http://www.math.wisc.edu/~miller/res/laver.pdf}

\bibitem{nyjoke}
Miller, Arnold W.; nyikos.jok, email to Judah Nov 1 1998. 
\par\url{http://www.math.wisc.edu/~miller/res/unpub/nyjoke.pdf}

\end{thebibliography}

\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}


\end{document}



