% June 2, 2004 LaTeX2e Burke, Miller, Models in which every nonmeager ..
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\title{Models in which every nonmeager set is nonmeager in a nowhere
dense Cantor set}
\author{
{\sc Maxim R.~Burke} \thanks{Research supported by NSERC. The
author thanks the Department of Mathematics at the University of
Wisconsin for its hospitality during the academic year 1996/1997
when the earlier result mentioned in the introduction was produced
and F.D.~Tall and the Department of Mathematics at the University
of Toronto for their hospitality during the academic year
2003/2004 when the present paper was completed.}
\\
{\footnotesize Department of Mathematics and Statistics} \\
{\footnotesize University of Prince Edward Island} \\
{\footnotesize Charlottetown PE, Canada C1A 4P3} \\
{\footnotesize burke@upei.ca}
%
\and {\sc Arnold W. Miller} \thanks{Thanks to the Fields Institute
for Research in Mathematical Sciences for their support during
part of the time this paper was written and to Juris Steprans who
directed the special program in set theory and analysis.
\endgraf
AMS Subject Classification: Primary 03E35; Secondary 03E17 03E50.
\endgraf Key words and phrases: Property of Baire, Lebesgue measure,
Cantor set, oracle forcing
}
\\
{\footnotesize University of Wisconsin-Madison} \\
{\footnotesize Department of Mathematics, Van Vleck Hall} \\
{\footnotesize 480 Lincoln Drive} \\
{\footnotesize Madison, Wisconsin 53706-1388} \\
{\footnotesize miller@math.wisc.edu}\\
{\footnotesize http://www.math.wisc.edu/$\sim$miller}
}
\date{}
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\begin{document}
\maketitle
%\begin{center}
%Abstract
%\end{center}
%\begin{quote}
\begin{abstract}
We prove that it is relatively consistent with ZFC that in any
perfect Polish space, for every nonmeager set $A$ there exists a
nowhere dense Cantor set $C$ such that $A\cap C$ is nonmeager in
$C$. We also examine variants of this result and establish a
measure theoretic analog.
\end{abstract}
%\end{quote}
\section{Introduction}
Our starting point is the following question of Laczkovich:
%
\begin{quote}
Does there exist (in ZFC) a nonmeager set that is relatively
meager in every nowhere dense perfect set?
\end{quote}
%
Note that the continuum hypothesis implies the existence of
a Luzin set, i.e., an uncountable set of reals which meets
every nowhere dense set in a countable set. Hence, we can think
of Laczkovich's question as asking whether one can construct a
particular weak version of a Luzin set without any extra
set theoretic assumptions.
Recall that a space $X$ is Polish iff it is completely metrizable
and separable. A subset of $X$ is nowhere dense iff its closure
has no interior and it is meager iff it is the countable union of
nowhere dense sets. A subset of $X$ is residual iff it is the
complement of a meager set. A perfect set in a Polish space is a
closed nonempty set without isolated points, and a Polish space
is said to be perfect if it is nonempty and has no isolated
points. As we shall see, the underlying space in the question of
Laczkovich can be taken to be any perfect Polish space. If we
ask, as is quite natural, for the nowhere dense perfect sets in
the statement to be Cantor sets (i.e., sets homeomorphic to the
Cantor middle third set), then we do not know whether the nature
of the Polish space matters. Even for various standard
incarnations of the reals (the real line, the Baire space, and so
on), we have only partial results on their equivalence in this
context. We answered Laczkovich's question for the Cantor set in
1997 by building a model where the answer is negative. (And of
course the perfect nowhere dense sets in this case are
necessarily Cantor sets.) Very shortly afterwards, we noticed the
more elegant solution presented here which uses a slightly
stronger variant of a statement proven consistent by Shelah in
\cite{Sh1980}. We show in Section~\ref{Consistency results} that
the stronger conclusion in which, for any perfect Polish space,
the perfect nowhere dense sets can be taken to be Cantor sets
follows from yet another variant on the same statement. The proof
of the consistency of the variants in question is similar to the
proof of Shelah. Unfortunately, the proof is quite technical and
the argument in \cite{Sh1980} is only a brief sketch, so we give
the argument in some detail in Section~\ref{Order-isomorphisms of
everywhere nonmeager sets} in order to be clear. An alternative
model for the negative answer to Laczkovich's question for the
Cantor set is provided by a paper of Ciesielski and Shelah
\cite{CS}. See Remark~\ref{(b) is enough for perf(R) + model from
CS}. In the final section of the paper, we show how a measure
theoretic version of our results can be deduced from results in
Roslanowski and Shelah \cite{RS}. The authors thank Ilijas Farah
for helpful discussions concerning the models constructed in
\cite{RS}.
Write $\perf(X)$ for a Polish space $X$ to mean that for every
nonmeager set $A\sq X$ there is a nowhere dense perfect set $P\sq
X$ such that $A\cap P$ is nonmeager relative to $P$. Write
$\cantor(X)$ if moreover $P$ can be taken to be a Cantor set. Note
that $\perf(X)$ and $\cantor(X)$ are trivially equivalent in
spaces in which nowhere dense perfect sets are necessarily Cantor
sets, e.g., $2^\o$ and $\R$.
We recall for emphasis the following well-known elementary fact of
which we will make frequent use without mention.
%
\prop{rel nwd in dense subspace}{If $X$ is a topological space and
$Y$ is a dense subspace of $X$, then for any $A\sq Y$, $A$ is
nowhere dense in $Y$ if and only if $A$ is nowhere dense in $X$.
Similarly, $A$ is meager in $Y$ if and only if $A$ is meager in
$X$. \qed}
\section{Relationships between various Polish spaces}
\label{Relationships between various Polish spaces}
We begin by showing that for any two perfect Polish spaces $X$ and
$Y$, $\perf(X)$ and $\perf(Y)$ are equivalent statements.
\prop{equivalence of perf(irr) and perf(Polish)}{
We have the following implications.
\begin{enumerate}
\item[{\rm(a)}]
Suppose $X$ is a perfect Polish space and $\perf(X)$ holds. Then
$\perf(\o^\o)$ holds.
\item[{\rm(b)}]
$\perf(\o^\o)$ implies $\perf(X)$ for every perfect Polish space
$X$.
\end{enumerate}
}
\proof We will use the well-known fact that every perfect Polish
space $X$ has a dense $G_\d$ subset $Y$ homeomorphic to $\o^\o$.
(To get $Y$, first remove the boundaries of the elements of a
countable base for $X$. What remains is a zero-dimensional dense
$G_\d$. Remove a countable dense subset of this dense $G_\d$ and
call the result $Y$. Then $Y$ is a perfect Polish space which is
zero-dimensional and has no compact open sets and hence is
homeomorphic to $\o^\o$.)
(a) Let $Y$ be a residual subspace of $X$ homeomorphic to $\o^\o$.
Let $A$ be a nonmeager set in $Y$. In $X$, $A$ is nonmeager so
there is a nowhere dense perfect set $C$ so that $A$ is nonmeager
in $C$. By replacing $C$ by the closure of one of its nonempty
open subsets, we may assume that $A$ is everywhere nonmeager in
$C$. In particular, $A\cap C$ is dense in $C$. Note that $F =
Y\cap C$ is closed relative to $Y$, is nonempty and has no
isolated points (because it contains $A\cap C$ which is dense in
$C$). Since $F$ is dense in $C$, $A\cap C = (A\cap Y)\cap C =
A\cap F$ is not meager in $F$. Also, because $Y$ is dense in $X$
and $F$ is nowhere dense in $X$, $F$ is also nowhere dense in $Y$.
(b) Let $X$ be a perfect Polish space. Let $Y$ be a residual
subspace of $X$ homeomorphic to $\o^\o$. Let $A\sq X$ be
nonmeager. Then $A\cap Y$ is nonmeager in $X$ and hence in $Y$ as
well since $Y$ is dense. By $\perf(\o^\o)$, there is a nowhere
dense perfect set $C$ in $Y$ such that $A\cap C$ is nonmeager in
$C$. If $P$ denotes the closure of $C$ in $X$, then, since $C$ is
dense in $P$, $A\cap C$ is nonmeager in $P$ and hence $A\cap P$ is
also nonmeager in $P$. $P$ is perfect since it is the closure of a
nonempty set without isolated points. $P$ is nowhere dense since
it is the closure of a set which is nowhere dense in $Y$ and hence
in $X$ as well.
\qed
\m
Part (b) holds for $\cantor(\cdot)$ by an easier argument.
\prop{(b) for perf^*}{$\cantor(\o^\o)$ implies $\cantor(X)$ for
every perfect Polish space $X$.}
\proof Similar to the proof of Proposition~\ref{equivalence of
perf(irr) and perf(Polish)}(b), except that this time the proof
yields a nowhere dense Cantor set $C\sq Y$ such that $A\cap C$ is
nonmeager in $C$ and then we are done.
\qed
\m
We do not know whether (a) holds for $\cantor(\cdot)$.
\pr{Irrationals vs Cantor set}{Does $\cantor(2^\o)$ imply
$\cantor(\o^\o)$?}
\pr{Unit square vs Cantor set}{Does $\cantor([0,1])$ imply
$\cantor([0,1]\times[0,1])$?}
%
Of course, $\cantor([0,1])\equiv\perf([0,1]) \equiv\perf(2^\o)
\equiv\cantor(2^\o)$, so these two questions have equivalent
hypotheses.
\m
We introduce one more version of $\perf(X)$ based on the following
observation. Suppose that $\perf(\o^\o)$ holds. Then for any
nonmeager set $A$, we have a nowhere dense perfect set $P$ such
that $A\cap P$ is nonmeager in $P$. Replacing $P$ by the closure
of one of its open sets, we may assume that $A\cap P$ is
everywhere nonmeager in $P$. Then if $P$ has a compact open
subset $U$, then $U$ is a Cantor set and $A\cap U$ is nonmeager
in $U$. Otherwise, $P$ itself is homeomorphic to $\o^\o$. Hence,
the perfect set $P$ in the conclusion of $\perf(\o^\o)$ can
always be taken to be either a closed nowhere dense copy of
$\o^\o$ or a Cantor set. Let $\irr(X)$ be the strengthening of
$\perf(X)$ in which we require that the perfect nowhere dense
sets in the definition be homeomorphic to the Baire space
$\o^\o$. Of course a Polish space need not contain any closed
copies of $\o^\o$, so $\irr(X)$ can fail. However, when $X=\o^\o$
it would seem reasonable that $\irr(X)$ might hold, and we will
show in the next section that $\irr(\o^\o)$ is indeed consistent.
Its relationship to $\cantor(\o^\o)$ is unclear to us.
\pr{question on irr} {(a) Does $\perf(\o^\o)$ imply that one of
$\irr(\o^\o)$ or $\cantor(\o^\o)$ must hold? (b) Does either of
$\irr(\o^\o)$ or $\cantor(\o^\o)$ imply the other? }
\section{Consistency results}
\label{Consistency results}
We now turn to the proof of the consistency of
$\cantor(\o^\o)$ and $\irr(\o^\o)$. We need a
variation on the following result which forms part of
the proof of \cite[Theorem~4.7]{Sh1980} which states
that if ZFC is consistent, then so is ZFC +
$2^\o=\o_2$ + ``There is a universal (linear) order of
power $\o_1$.''
\thm{universal order 1}{If ZFC is consistent, then so is ZFC +
both of the following statements.
\begin{enumerate}
\item[{\rm(a)}]
There is a nonmeager set in $\R$ of cardinality $\o_1$.
\item[{\rm(b)}]
Let $A$ and $B$ be everywhere nonmeager subsets of $\R$ of
cardinality $\o_1$. Then $A$ and $B$ are order-isomorphic.
\end{enumerate}
}
We shall need the following variant of this result.
\thm{universal order 2}{If ZFC is consistent, then so is ZFC +
both of the following statements.
\begin{enumerate}
\item[{\rm(a)$'$}]
Every nonmeager set in $\R$ has a nonmeager subset of cardinality
$\o_1$.
\item[{\rm(b)$'$}]
Let $A$ and $B$ be everywhere nonmeager subsets of $\R$ of
cardinality $\o_1$. Suppose we are given countable dense subsets $A_0\sq A$
and $B_0\sq B$. Then $A$ and $B$ are order-isomorphic by an order
isomorphism taking $A_0$ isomorphically to $B_0$.
\end{enumerate}
}
\pr{Is added part automatic}{In the presence of (a), does (b)
imply (b)$'$?}
We shall in fact verify in Theorem~\ref{universal order 3} that in
(b)$'$ we can even ask that given pairwise disjoint countable
dense subsets $A_i$, $i<\o$, of $A$ and pairwise disjoint
countable dense subsets $B_i$, $i<\o$, of $B$, the
order-isomorphism of $A$ and $B$ takes $A_i$ isomorphically to
$B_i$ for each $i<\o$. As explained in the introduction, the proof
is similar to the one in \cite{Sh1980}, but as the proof is quite
technical and the argument in \cite{Sh1980} is only a brief
sketch, we need to give the argument in some detail in order to be
clear. We do that in the next section. Here, we derive the
consequences of interest to us for this paper. The definition of
$\irr(X)$ is given at the end of Section~\ref{Relationships
between various Polish spaces}.
\thm{consequences of interest}{Assume {\rm(a)$'$} and {\rm(b)$'$}.
Then $\cantor(\o^\o)$ and $\irr(\o^\o)$ both hold.}
\proof We will use the following elementary fact.
\faact{extending order isomorphisms}{If $K,L\subseteq\R$ are dense
and $h\colon K\to L$ is an order isomorphism, then $h$ extends to
an order isomorphism of $\R$. \qed}
%
Suppose that $A\subseteq\R\sm\Q$ is not meager. We wish to find a
Cantor set $C\sq\R\sm\Q$ such that $A\cap C$ is nonmeager relative
to $C$. By (a)$'$, we may assume that $A$ has cardinality exactly
$\o_1$. $A$ is everywhere nonmeager in some open interval
$(a,b)$. Let $C\subseteq (a,b)\sm\Q$ be a Cantor set, and, by
(a)$'$, let $B\subseteq C$ be a set of cardinality $\o_1$ which is
nonmeager relative to $C$. Then $(A\cup\Q)\cap (a,b)$ and $(A\cup
B\cup\Q)\cap (a,b)$ are both everywhere nonmeager in $(a,b)$ and
both have cardinality $\o_1$. By (b)$'$, there is an
order-isomorphism $h\colon (A\cup\Q)\cap (a,b)\to (A\cup
B\cup\Q)\cap (a,b)$ such that $h[\Q\cap (a,b)]=\Q\cap (a,b)$.
Extend $h$ to $(a,b)$ and denote the extension also by $h$. Since
$h$ is a homeomorphism, $h^{-1}[C]$ is a Cantor set and
$h^{-1}[B]$ is non meager relative to $h^{-1}[C]$. Since
$h^{-1}[C]\sq\R\sm\Q$ and $h^{-1}[B]\subseteq A$, we are done.
To get $\irr(\o^\o)$, we make a different choice of $C$ in the
preceding argument. This time, choose $C$ to be any Cantor set so
that $C\cap\Q$ is dense in $C$. Then $h^{-1}[C]$ will have the
same property, so $h^{-1}[C\sm\Q]=h^{-1}[C]\sm\Q$ is closed
nowhere dense in $\R\sm\Q$ and homeomorphic to $\o^\o$. \qed
\rem{(b) is enough for perf(R) + model from CS}{The reader can
easily verify that a similar but simpler argument yields that
(a)$'$ and (b) imply $\perf(\R)$. An alternative proof of the
consistency of $\perf(\R)$ can by had by using Theorem 2 of
\cite{CS} which states that the following statement is consistent
relative to ZFC:
%
\begin{quote}
For every $A\subseteq 2^\omega\times 2^\omega$ for which the sets
$A$ and $A^c=(2^\omega\times 2^\omega\setminus A)$ are nowhere
meager in $2^\omega\times 2^\omega$ there is a homeomorphism
$f:2^\omega\to 2^\omega$ such that the set $\{x\in 2^\omega:
(x,f(x))\in A\}$ does not have the Baire property in $2^\omega$.
\end{quote}
%
(A set has the Baire property if it has the form $U\triangle M$
where $U$ is open and $M$ is meager.) Note that the map $2^\o\to
f$ given by $x\mapsto(x,f(x))$ is a homeomorphism. Hence the
conclusion could be stated as ``$f\cap A$ does not have the Baire
property in $f$''. Since $2^\omega\times 2^\omega$ is
homeomorphic to $2^\omega$ and the graph of a homeomorphism of
$2^\o$ is a perfect nowhere dense set in $2^\omega\times
2^\omega$, the statement above implies the following special case
of $\perf(2^\o)$ (which is equivalent to $\perf(\R)$).
%
\begin{quote}
For every $A\subseteq 2^\omega$ for which the sets $A$ and $A^c$
are both nowhere meager in $2^\omega$, there is a perfect nowhere
dense set $P$ such that the set $A\cap P$ is not meager in $P$.
\end{quote}
%
To reduce $\perf(2^\o)$ to this special case, consider a nonmeager
set $A\subseteq 2^\omega$. $A$ is everywhere nonmeager in some
clopen set $U$. If $A$ is comeager in some clopen set, then it
contains a nowhere dense perfect set and we are done. Hence we may
assume that, relative to $U$, $A$ and $A^c$ are both everywhere
nonmeager. The clopen set $U$ is homeomorphic to $2^\omega$, so we
now find ourselves in the special case described above.
}
\section{Order-isomorphisms of everywhere nonmeager sets}
\label{Order-isomorphisms of everywhere nonmeager sets}
We now turn to the proof of the consistency of (a)$'$
and (b)$'$. We begin by recalling the basic properties
of oracle-cc forcing. See \cite[Chapter IV]{Sh1998}
for the details. A version of this material is also
explained in \cite[Sections 4--6]{Bu}.
\defi{oracle}{A sequence
\[
\barM=\la M_\d:\d\ \mbox{is a limit ordinal}\ <\o_1\ra
\]
is called an {\em oracle} if each $M_\d$ is a countable transitive
model of a sufficiently large fragment of ZFC, $\d\in M_\d$ and
for each $A\sq\o_1$, $\{\d:A\cap\d\in M_\d\}$ is stationary in
$\o_1$.}
%
The meaning of ``sufficiently large'' depends on the context. In a
particular proof, some fragment of ZFC for which models can be
produced in ZFC must suffice for all the oracles in the proof. The
existence of an oracle is equivalent to $\diamondsuit$, (see
\cite[Theorem II 7.14]{Ku}) and hence implies CH. We limit the
definition of the $\barM$-chain condition to partial orders of
cardinality $\o_1$. This covers our present needs.
\m
Associated with an oracle $\barM$, there is a filter $\Trap$
generated by the sets
\[
\{\d<\o_1:\d\ \mbox{is a limit ordinal and}\ A\cap\d\in
M_\d\},\quad A\sq\o_1.
\]
This is a proper normal filter containing all closed unbounded
sets.
\defi{completely embedded modulo D}{
If $P$ is any partial order, $P'\sq P$, and ${\mathfrak D}$ is any
class of sets, then we write $P'<_{\mathfrak D}P$ to mean that
every predense subset of $P'$ which belongs to ${\mathfrak D}$ is
predense in $P$.}
\defi{oracle-cc}{A partial order $P$ {\it satisfies the $\barM$-chain
condition}, or simply {\it is $\barM$-cc}, if there is a
one-to-one function $f\colon P\to\o_1$ such that
\[
\{\d<\o_1:\d\ \mbox{is a limit ordinal and}\
f^{-1}(\d)<_{M_{\d,f}}P\}
\]
belongs to $\Trap$, where $M_{\d,f}=\{f^{-1}(A):A\sq\d,\,A\in
M_\d\}$.}
It is not hard to verify that if $P$ is $\barM$-cc, then $P$ is
ccc. Also, any one-to-one function $g\colon P\to\o_1$ can replace
$f$ in the definition.
\prop{oracle-cc properties}{The $\barM$-cc satisfies the following
properties.
\begin{enumerate}
\item[{\rm(1)}]
If $\a<\o_2$ is a limit ordinal, $\la\la
P_\beta\ra_{\beta\leq\a},\la \dot Q_\beta\ra_{\beta<\a}\ra$ is a
finite-support $\a$-stage iteration of partial orders, and for
each $\beta<\a$, $P_\beta$ is $\barM$-cc, then $P_\a$ is
$\barM$-cc.
\item[{\rm(2)}]
If $P$ is $\barM$-cc, then there is a $P$-name $\barM^*$ for an
oracle such that for each $P$-name $\dot Q$ for a partial order,
if $\Vdash_P$~``$\dot Q$ is $\barM^*$-cc'' then $P*\dot Q$ is
$\barM$-cc.
\item[{\rm(3)}]
If $\barM_\a$, $\a<\o_1$, are oracles, then there is an oracle
$\barM$ such that for any partial order $P$, if $P$ is $\barM$-cc,
then $P$ is $\barM_\a$-cc for all $\a<\o_1$.
\end{enumerate}
}
We will need the following lemmas.
\lem{Main Lemma 1}{Let $\barM=\la M_\d:\d<\o_1\ra$ be an oracle
and let $A$ and $B$ be everywhere nonmeager subsets of $\R$.
Suppose we are given pairwise disjoint countable dense subsets
$A_i$, $i<\o$, of $A$ and pairwise disjoint countable dense
subsets $B_i$, $i<\o$, of $B$. Then there is a forcing notion $P$
satisfying the $\barM$-cc such that for every $G\sq P$ generic
over $V$, $V[G]\models$ $A$ and $B$ are order-isomorphic by an
order isomorphism taking $A_i$ isomorphically to $B_i$ for each
$i<\o$.
}
\proof Fix well-orderings of $A$ and $B$ in type $\o_1$. (CH holds
because there is an oracle.) We will inductively define one-to-one
enumerations $\la a_\a:\a<\o_1\ra$ of $A$ and $\la
b_\a:\a<\o_1\ra$ of $B$ and functions $f_\d$, $\d<\o_1$. We let
$A_\d=\{a_\a:\o\d\leq\a<\o(\d+1)\}$ and
$B_\d=\{b_\a:\o\d\leq\a<\o(\d+1)\}$ for $\d<\o_1$. For $A'\sq A$
and $B'\sq B$, let $P(A',B')$ denote the set of finite partial
order-preserving maps $p\colon A'\to B'$ such that $p[A_\d]\sq
B_\d$ for all $\d<\o_1$. We also use the notation
\[
A\restr\a=\{a_\beta:\beta<\a\},\ B\restr\a=\{b_\beta:\beta<\a\}.
\]
We will arrange that the following conditions hold.
\begin{enumerate}
\item[(1)] The sets $A_\d$ and $B_\d$ are dense in $\R$.
\item[(2)]
For $\d<\o$, the sets $A_\d$ and $B_\d$ are as in the hypothesis.
\item[(3)]
For each $\d<\o_1$, $f_\d$ is a bijective map of
$P(A\restr\o\d,B\restr\o\d)$ onto $\o\d$.
\item[(4)] For each $\d<\d'<\o_1$, $f_\d\sq f_{\d'}$.
\item[(5)]
For each infinite $\d<\o_1$, the predense subsets of
$P(A\restr\o\d,B\restr\o\d)$ which have the form $f_\d^{-1}[S]$
for some $S\in \bigcup_{\eta\leq\d}M_{\eta}$ remain predense in
$P(A\restr\o(\d+1),B\restr\o(\d+1))$.
\end{enumerate}
To do this, we proceed as follows. The construction of the
functions $f_\d$ is dictated by (4) at limit stages, and
$f_{\d+1}$ is an arbitrary extension of $f_\d$ satisfying (3). The
elements of $A_\d$ and $B_\d$ for $\d<\o$ are given by (2). For
$\d\geq\o$, by induction on $\d$ we choose the elements of $A_\d$
and $B_\d$ by alternately defining $a_{\o\d+n}$ and $b_{\o\d+n}$,
beginning with $a_{\o\d}$ when $\d$ is even and with $b_{\o\d}$
when $\d$ is odd. Let us illustrate the construction with the case
where $\d$ is even. Fix an enumeration $\la I_m:00$, we pick $a_{\o\d+n}$ and $b_{\o\d+n}$ from
$I_n$ to ensure $A_\d$ and $B_\d$ will be dense.
To choose one of these elements, say $b_{\o\d+n}$, let $N$ be a
countable elementary submodel of $H_\t$, for a suitably large
$\t$, such that $A$, $B$, $f_\d$, the sequences $\la
a_\a:\a\leq\o\d+n\ra$ and $\la b_\a:\a<\o\d+n\ra$, and
$\bigcup_{\eta\leq\d}M_\eta$ are all elements of $N$. Choose
$b_{\o\d+n}$ to be a member of $B$ which is a Cohen real over $N$.
We must check that the construction gives (5). Let $D$ be a
predense subset of $P(A\restr\o\d,B\restr\o\d)$ of the appropriate
form. In particular, we have $D\in N$. We will show that $D$
remains predense in $P(A\restr\o\d+n+1,B\restr\o\d+n+1)$.
\rem{why it works}{We are showing by induction on $n$ that $D$
remains predense in $P(A\restr\o\d+n+1,B\restr\o\d+n)$ and then in
$P(A\restr\o\d+n+1,B\restr\o\d+n+1)$. (This establishes (5) since
each member of $P(A\restr\o(\d+1),B\restr\o(\d+1))$ belongs to
$P(A\restr\o\d+n,B\restr\o\d+n)$ for some $n<\o$.) Our current
stage has the second form. Note that at the stage $n=0$, we first
consider the passage from $P(A\restr\o\d,B\restr\o\d)$ to
$P(A\restr\o\d+1,B\restr\o\d)$. But these two partial orders are
equal because there is no legal target value for $a_{\o\d}$ until
$b_{\o\d}$ is chosen. So the preservation of the predense sets
trivially holds at that stage. In particular, it does not matter
that $a_{\d\o}$ is not Cohen generic over the previous
construction.}
%
Let
\[
p\in P(A\restr\o\d+n+1,B\restr\o\d+n+1)\sm
P(A\restr\o\d+n+1,B\restr\o\d+n).
\]
Then $p$ has the form $q\cup\{(a,b_{\o\d+n})\}$ for some $q\in
P(A\restr\o\d+n+1,B\restr\o\d+n)$ and $a\in\{a_{\o\d+m}:m\leq
n\}$. Fix $r\in D$. The set
\[
\{b\in\R: q\cup\{(a,b)\}\ \mbox{is compatible with}\ r\}\in N
\]
(``compatible with'' here means only that $q\cup\{(a,b)\}\cup r$
is a finite order isomorphism) is open and hence its complement
$C_r$ is closed, as is the set $C_D=\bigcap_{r\in D} C_r$ of $b$
for which $q\cup\{(a,b)\}$ is incompatible with every member of
$D$. Since $p$ is an partial order isomorphism, there are open
rational intervals $J_1$ and $J_2$ such that $J_1\cap\dom
p=\{a\}$, $J_2\cap \range p=\{b_{\o\d+n}\}$. Note that whenever
$x\in J_1$ and $b\in J_2$, $q\cup\{(x,b)\}$ is a partial
isomorphism.
\clm{C_D is nowhere dense in J_2}{$C_D$ is nowhere dense in
$J_2$.}
\proof Fix a nonempty open subinterval $J$ of $J_2$. There is an
extension of $q$ by members of $A_0\times B_0$---the point of
using $A_0$ and $B_0$ being simply that they are dense and
contained in $A(\o\d+n+1)$ and $B(\o\d+n)$, respectively---which
adds two points in $J_1\times J$ straddling the line $x=a$. So
this extension has the form
\[
q'=q\cup\{(x_1,y_1),(x_2,y_2)\},\ x_1