\documentclass[12pt]{article} \usepackage{amssymb,latexsym,hyperref} %\pagestyle{myheadings} %\def\header{{\hfill ny.tex \hfill \today \hfill}} %\markboth\header\header \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}} \def\rmand{\mbox{ and }} \def\si{\sigma} \def\al{\alpha} \def\be{\beta} \def\concat(#1){{\hat{\phantom{a}}\la #1\ra}} \def\elemsub{\preceq} \def\forces{{\;\Vdash}} \def\ka{\kappa} \def\la{\langle} \def\om{\omega} \def\poset{{\mathbb P}} \def\proof{\par\noindent Proof:\par} \def\pr{\prime} \def\qed{\par\noindent QED \par} \def\rank{\mbox{rank}} \def\ra{\rangle} \def\res{\upharpoonright} \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}