# A model of second-order arithmetic satisfying AC but not DC

Sy-David Friedman  Kurt Research Center, University of Vienna, Währinger
Strasse 25 A-1090 Vienna, Austria<EMAIL_ADDRESS>http://www.logic.univie.ac.at/ sdf ,  Victoria Gitman  The City University of
New York, CUNY Graduate Center, Mathematics Program, 365 Fifth Avenue, New
York, NY 10016<EMAIL_ADDRESS>https://victoriagitman.github.io and
Vladimir Kanovei  Institute for the Information Transmission Problems (IITP),
19, build.1, Bolshoy Karetny per. Moscow 127051, Russia and Russian University
of Transport (RUT-MIIT), Moscow, Russia<EMAIL_ADDRESS>
###### Abstract.

We show that there is a $\beta$-model of second-order arithmetic in which the
choice scheme holds, but the dependent choice scheme fails for a
$\Pi^{1}_{2}$-assertion, confirming a conjecture of Stephen Simpson. We obtain
as a corollary that the Reflection Principle, stating that every formula
reflects to a transitive set, can fail in models of ${\rm ZFC}^{-}$. This work
is a rediscovery by the first two authors of a result obtained by the third
author in [Kan79].

The first author wishes to acknowledge the support of the Austrian Science
Fund (FWF) through Research Project I 1238.

The third author wishes to acknowledge the support of the RFBR grant
17-01-00705.

##  1\. Introduction

Models of arithmetic are two-sorted structures, having two types of objects,
which we think of as numbers and sets of numbers. Their properties are
formalized using a two-sorted logic with separate variables and quantifiers
for numbers and sets. By convention, we will denote number variables by lower-
case letters and sets variables by upper-case letters. The language of second-
order arithmetic is the language of first-order arithmetic
$\mathcal{L}_{A}=\\{+,\cdot,<,0,1\\}$ together with a membership relation
$\in$ between numbers and sets. A multitude of second-order arithmetic
theories, as well as the relationships between them, have been extensively
studied (see [Sim09]).

An example of a weak second-order arithmetic theory is ${\rm ACA_{0}}$, whose
axioms consist of the modified Peano axioms, where instead of the induction
scheme we have the single second-order induction axiom

$\forall X[(0\in X\wedge\forall n(n\in X\rightarrow n+1\in
X))\rightarrow\forall n(n\in X)],$

and the _comprehension scheme_ for first-order formulas. The latter is a
scheme of assertions stating for every first-order formula, possibly with set
parameters, that there is a set whose elements are exactly the numbers
satisfying the formula. One of the strongest second-order arithmetic theories
is ${\rm Z}_{2}$, often referred to as _full second-order arithmetic_ , which
strengthens comprehension for first-order formulas in ${\rm ACA}_{0}$ to full
comprehension for all second-order assertions. This means that for a formula
with any number of second-order quantifiers, there is a set whose elements are
exactly the numbers satisfying the formula. For example, the reals of any
model of ${\rm ZF}$ give a model of ${\rm Z}_{2}$. We can further strengthen
the theory ${\rm Z}_{2}$ by adding choice principles for sets: the choice
scheme and the dependent choice scheme.

The _choice scheme_ is a scheme of assertions, which states for every second-
order formula $\varphi(n,X,A)$ with a set parameter $A$ that if for every
number $n$, there is a set $X$ witnessing $\varphi(n,X,A)$, then there is a
single set $Y$ collecting witnesses for every $n$, in the sense that
$\varphi(n,Y_{n},A)$ holds, where $Y_{n}=\\{m\mid\langle n,m\rangle\in Y\\}$
and $\langle n,m\rangle$ is any standard coding of pairs. More precisely, an
instance of the choice scheme for the formula $\varphi(n,X,A)$ is

$\forall n\exists X\,\varphi(n,X,A)\rightarrow\exists Y\forall
n\,\varphi(n,Y_{n},A).$

We will denote by $\Sigma^{1}_{n}$-${\rm AC}$ the fragment of the choice
scheme for $\Sigma^{1}_{n}$-assertions, making an analogous definition for
$\Pi^{1}_{n}$, and we will denote the full choice scheme by
$\Sigma^{1}_{\infty}$-${\rm AC}$. The reals of any model of ${\rm ZF}+{\rm
AC}_{\omega}$ (countable choice) satisfy ${\rm
Z}_{2}+\Sigma^{1}_{\infty}$-${\rm AC}$. It is a folklore result, going back
possibly to Mostowski, that the theory ${\rm Z}_{2}+\Sigma^{1}_{\infty}$-${\rm
AC}$ is bi-interpretable with the theory ${\rm ZFC}^{-}$ (${\rm ZFC}$ without
the powerset axiom, with Collection instead of Replacement) together with the
statement that every set is countable.

The _dependent choice scheme_ is a scheme of assertions, which states for
every second-order formula $\varphi(X,Y,A)$ with set parameter $A$ that if for
every set $X$, there is a set $Y$ witnessing $\varphi(X,Y,A)$, then there is a
single set $Z$ making infinitely many dependent choices according to
$\varphi$. More precisely, an instance of the dependent choice scheme for the
formula $\varphi(X,Y,A)$ is

$\forall X\exists Y\,\varphi(X,Y,A)\rightarrow\exists Z\forall
n\,\varphi(Z_{n},Z_{n+1},A).$

We will denote by $\Sigma^{1}_{n}$-${\rm DC}$ the dependent choice scheme for
$\Sigma^{1}_{n}$-assertions, with an analogous definition for $\Pi^{1}_{n}$,
and we will denote the full dependent choice scheme by
$\Sigma^{1}_{\infty}$-${\rm DC}$. The reals of a model of ${\rm ZF}+{\rm DC}$
(dependent choice) satisfy ${\rm Z}_{2}+\Sigma^{1}_{\infty}$-${\rm DC}$.

It is not difficult to see that the theory ${\rm Z}_{2}$ implies
$\Sigma^{1}_{2}$-${\rm AC}$, the choice scheme for
$\Sigma^{1}_{2}$-assertions. Models of ${\rm Z}_{2}$ can build their own
version of ’s constructible universe $L$. If a model of ${\rm Z}_{2}$ believes
that a set $\Gamma$ is a well-order, then it has a set coding a set-theoretic
structure constructed like $L$ along the well-order $\Gamma$. It turns out
that models of ${\rm Z}_{2}$ satisfy a version of Shoenfield’s absoluteness
with respect to their constructible universes. For every
$\Sigma^{1}_{2}$-assertion $\varphi$, a model of ${\rm Z}_{2}$ satisfies
$\varphi$ if and only its constructible universe satisfies $\varphi$ with set
quantifiers naturally interpreted as ranging over the reals. All of the above
generalizes to constructible universes $L[A]$ relativized to a set parameter
$A$.111Indeed, the much weaker theory ${\rm ATR}_{0}$ suffices for everything
we have said so far. See [Sim09] (Section VII.3 and VII.4) for details. Thus,
given a $\Sigma^{1}_{2}$-assertion $\varphi(n,X,A)$ for which the model
satisfies $\forall n\exists X\,\varphi(n,X,A)$, the model can go to its
constructible universe $L[A]$ to pick the least witness $X$ for
$\varphi(n,X,A)$ for every $n$, because $L[A]$ agrees when $\varphi$ is
satisfied, and then put the witnesses together into a single set using
comprehension. So long as the unique witnessing set can be obtained for each
$n$, comprehension suffices to obtain a single set of witnesses. How much more
of the choice scheme follows from ${\rm Z}_{2}$? The reals of the classical
Feferman-Lévy model of ${\rm ZF}$ (see [Lév70], Theorem 8), in which
$\aleph_{1}$ is a countable union of countable sets, is a $\beta$-model of
${\rm Z}_{2}$ in which $\Pi^{1}_{2}$-${\rm AC}$ fails. This is a particulary
strong failure of the choice scheme because, as we explain below,
$\beta$-models are meant to strongly resemble the full standard model given by
$P(\omega)$.

There are two ways in which a model of second-order arithmetic can resemble
the full standard model given by $P(\omega)$. A model of second-order
arithmetic is called an $\omega$-_model_ if its first-order part is $\omega$,
and it follows that its second-order part is some subset of $P(\omega)$. But
even an $\omega$-model can poorly resemble $P(\omega)$ because it may be wrong
about well-foundedness by missing $\omega$-sequences. An $\omega$-model of
second-order arithmetic which is correct about well-foundedness is called a
$\beta$-_model_. The reals of any transitive ${\rm ZF}$-model is a
$\beta$-model of ${\rm Z}_{2}$. One advantage to having a $\beta$-model of
${\rm Z}_{2}$ is that the constructible universe it builds internally is
isomorphic to an initial segment $L_{\alpha}$ of the actual constructible
universe $L$.

The theory ${\rm Z}_{2}$ also implies $\Sigma^{1}_{2}$-${\rm DC}$ (see
[Sim09], Theorem VII.9.2), the dependent choice scheme for
$\Sigma^{1}_{2}$-assertions. In this article, we construct a symmetric
submodel of a forcing extension of $L$ whose reals form a model of second-
order arithmetic in which ${\rm Z}_{2}$ together with
$\Sigma^{1}_{\infty}$-${\rm AC}$ holds, but $\Pi^{1}_{2}$-${\rm DC}$ fails.
The forcing notion we use is a tree iteration of Jensen’s forcing for adding a
unique generic real.

Jensen’s forcing, which we will call here $\mathbb{P}^{J}$, introduced by
Jensen in [Jen70], is a subposet of Sacks forcing constructed in $L$ using the
$\diamondsuit$ principle. The poset $\mathbb{P}^{J}$ has the ccc and adds a
unique generic real over $L$. The collection of all $L$-generic reals for
$\mathbb{P}^{J}$ in any model is $\Pi^{1}_{2}$-definable. Jensen used his
forcing to show that it is consistent with ${\rm ZFC}$ that there is a
$\Pi^{1}_{2}$-definable non-constructible real singleton [Jen70]. Recently
Lyubetsky and the third author extended the “uniqueness of generic filters”
property of Jensen’s forcing to finite-support products of $\mathbb{P}^{J}$
[KL17]. They showed that in a forcing extension $L[G]$ by the $\omega$-length
finite support-product of $\mathbb{P}^{J}$, the only $L$-generic reals for
$\mathbb{P}^{J}$ are the slices of the generic filter $G$. The result easily
extends to $\omega_{1}$-length finite support-products as well.

We in turn extend the “uniqueness of generic filters” property to tree
iterations of Jensen’s forcing. We first define finite iterations
$\mathbb{P}^{J}_{n}$ of Jensen’s forcing $\mathbb{P}^{J}$, and then define an
iteration of $\mathbb{P}^{J}$ along a tree $\mathcal{T}$ to be a forcing whose
conditions are functions from a finite subtree of $\mathcal{T}$ into
$\bigcup_{n<\omega}\mathbb{P}_{n}^{J}$ such that nodes on level $n$ get mapped
to elements of the $n$-length iteration $\mathbb{P}_{n}^{J}$ and conditions on
higher nodes extend conditions on lower nodes. The functions are ordered by
extension of domain and strengthening on each coordinate. We show that in a
forcing extension $L[G]$ by the tree iteration of $\mathbb{P}^{J}$ along the
tree ${}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega_{1}$
(or the tree ${}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega$) the
only $L$-generic filters for $\mathbb{P}_{n}^{J}$ are the restrictions of $G$
to level $n$ nodes of tree. We proceed to construct a symmetric submodel of
$L[G]$ which has the tree of $\mathbb{P}_{n}^{J}$-generic filters added by $G$
but no branch through it. The symmetric model we construct satisfies ${\rm
AC}_{\omega}$ and the tree of $\mathbb{P}_{n}^{J}$-generic filters is
$\Pi^{1}_{2}$-definable in it. The reals of this model thus provide the
desired $\beta$-model of ${\rm Z}_{2}$ in which $\Sigma^{1}_{\infty}$-${\rm
AC}$ holds, but $\Pi^{1}_{2}$-${\rm DC}$ fails.

######  Theorem 1.1.

There is a $\beta$-model of second-order arithmetic ${\rm Z}_{2}$ together
with $\Sigma^{1}_{\infty}$-${\rm AC}$ in which $\Pi^{1}_{2}$-${\rm DC}$ fails.

It should be noted that in our model the instance of $\Pi^{1}_{2}$-${\rm DC}$
failure is parameter-free.

Theorem 1.1 has a long and complicated history. In 1973, Simpson submitted an
abstract claiming a proof of the result [Sim73], but he didn’t follow it up
with a publication because, as he kindly shared with the first author, he
never worked out the details of the argument, which involved Jensen’s forcing,
but did not have all the parts needed to solve the problem. In 1979, the third
author published, in Russian, a technical report with the result [Kan79].
Finally, after the third author’s (joint with Lyubetsky) recent publications
on uniqueness properties of Jensen’s forcing, the first and second author
independently rediscovered the third author’s result, leading to this joint
publication.

Our results also answer a long-standing open question of Zarach from [Zar96]
about whether the Reflection Principle holds in models of ${\rm ZFC}^{-}$. The
_Reflection Principle_ states that every formula can be reflected to a
transitive set, and holds in ${\rm ZFC}$ by the Lévy-Montague reflection
because every formula is reflected by some $V_{\alpha}$. In the absence of the
von Neumann hierarchy, it is not clear how to realize reflection, and indeed
we show that it fails in $H_{\omega_{1}}\models{\rm ZFC}^{-}$ of the symmetric
model we construct.

######  Theorem 1.2.

The theory ${\rm ZFC}^{-}$ does not imply the Reflection Principle.

##  2\. Jensen’s forcing

Jensen’s poset $\mathbb{P}^{J}$, introduced in [Jen70], is a subposet of Sacks
forcing in $L$ with the countable chain condition and the property that it
adds a unique generic real. The poset $\mathbb{P}^{J}$ is constructed in $L$
as the union of a continuous chain
$\langle\mathbb{P}_{\alpha}\mid\alpha<\omega_{1}\rangle$ of length
$\omega_{1}$ of countable perfect posets. At successor stages $\alpha+1$, the
principle $\diamondsuit$ is used to seal a certain countable collection of
maximal antichains of $\mathbb{P}_{\alpha}$, so that by the end of the
construction every maximal antichain of $\mathbb{P}^{J}$ is sealed and
therefore countable.

Recall that a tree $T\subseteq{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ is
_perfect_ if every node in $T$ has a splitting node above it. Given two
perfect trees $T$ and $S$, let’s define their _meet_ $T\wedge S$ as follows.
Let $U_{0}=T\cap S$, and, assuming $U_{\alpha}$ has been defined, let
$U_{\alpha+1}$ be the set of all nodes in $U_{\alpha}$ which have a splitting
node in $U_{\alpha}$ above them. At limit stages, take intersections. Since
$U_{0}$ is countable, there must be a countable stage $\alpha$ in the
construction for which $U_{\alpha}=U_{\alpha+1}$. We define $T\wedge
S=U_{\alpha}$. It is not difficult to see that either $T\wedge S=\emptyset$ or
$T\wedge S$ is a perfect tree. In the latter case, it is the maximal perfect
tree contained in $T\cap S$, so that every perfect tree $U\subseteq T\cap S$
is contained in $T\wedge S$. Recall the standard terminology that if
$T\subseteq{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ is a tree
and $s$ is a node in $T$, then $T_{s}$ denotes the subtree of $T$ consisting
of all nodes compatible with $s$. The following proposition is straightforward
to check.

######  Proposition 2.1.

If $T,S$, and $R$ are perfect trees, then

  1. (1)

$(T\wedge S)_{s}=T_{s}\wedge S_{s}$ for every node $s\in T\wedge S$,

  2. (2)

$(T\wedge S)\wedge R=T\wedge(S\wedge R)$ ($\wedge$ is associative),

  3. (3)

$(T\cup S)\wedge R=(T\wedge R)\cup(S\wedge R)$ ($\wedge$ distributes over
$\cup$).

Let’s recall now some facts about perfect posets, which are subposets of Sacks
forcing closed under certain basic operations.

######  Definition 2.2.

We say that a collection $\mathbb{P}$ of perfect trees ordered by inclusion is
a _perfect poset_ 222Note that this is not a standard definition. In the
literature, a _perfect poset_ is usually defined to be a collection
$\mathbb{P}$ of perfect trees such that ${}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2\in\mathbb{P}$
and whenever $T\in\mathbb{P}$ and $s\in T$, then $T_{s}\in\mathbb{P}$. if

  1. (1)

$({}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{s}\in\mathbb{P}$
for every $s\in{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$, and
for every $T,S\in\mathbb{P}$,

  2. (2)

$T\cup S\in\mathbb{P}$ (closed under unions),

  3. (3)

$T\wedge S\in\mathbb{P}$, if $T\wedge S\neq\emptyset$ (closed under meets).

The smallest perfect poset, which we will denote by $\mathbb{P}^{\text{min}}$,
is the closure under finite unions of the collection
$\\{({}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{s}\mid
s\in{}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2\\}$. Note
that two perfect trees $T$ and $S$ are compatible under the inclusion ordering
precisely when $T\wedge S\neq\emptyset$. So if $\mathbb{P}\subseteq\mathbb{Q}$
are perfect posets, then $T,S\in\mathbb{P}$ are compatible in $\mathbb{Q}$ if
and only if they are already compatible in $\mathbb{P}$. Standard arguments
show that a generic filter $G$ for a perfect poset $\mathbb{P}$ is determined
by the _generic real_ $r$ that is a branch through $T\in\mathbb{P}$ if and
only if $T\in G$.

Given a perfect poset $\mathbb{P}$, we will denote by
$\mathbb{P}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ the
$\omega$-length finite-support product of $\mathbb{P}$. Conditions in
$\mathbb{P}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ are
functions $p:\omega\to\mathbb{P}$ such that for all but finitely many $n$,
$p(n)={}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$. We will
sometimes abuse notation by writing $p=\langle T_{0},\ldots,T_{n-1}\rangle$
for conditions in $\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, meaning
that all remaining coordinates are trivial.

Following [Abr84], we associate to a perfect poset $\mathbb{P}$, the poset
$\mathbb{Q}(\mathbb{P})$ whose conditions are pairs $(T,n)$ with
$T\in\mathbb{P}$ and $n\in\omega$ ordered so that
$(T_{2},n_{2})\leq(T_{1},n_{1})$ whenever $T_{2}\subseteq T_{1}$, $n_{2}\geq
n_{1}$, and $T_{1}\cap{}^{n_{1}}2=T_{2}\cap{}^{n_{1}}2$. We will refer to
$\mathbb{Q}(\mathbb{P})$ as the _fusion poset_ for $\mathbb{P}$ because fusion
arguments involving $\mathbb{P}$ amount to producing a filter for
$\mathbb{Q}(\mathbb{P})$ meeting sufficiently many dense sets. If
$G\subseteq\mathbb{Q}(\mathbb{P})$ is $V$-generic, then the union of
$T\cap{}^{n}2$ for $(T,n)\in G$ is the generic perfect tree $\mathcal{T}$
added by $\mathbb{Q}(\mathbb{P})$. Note that $\mathcal{T}\leq T$ for every $T$
that appears in some condition in $G$.

Let $\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ denote the
$\omega$-length finite-support product of the $\mathbb{Q}(\mathbb{P})$.
Conditions in $\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ are function
$q:\omega\to\mathbb{Q}(\mathbb{P})$ such that for all but finitely many $n$,
$q(n)=({}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2,0)$, but we
will sometimes abuse notation by writing the conditions as finite tuples.

Suppose that $G\subseteq\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is
$V$-generic. Let $\langle\mathcal{T}_{n}\mid n<\omega\rangle\in V[G]$ be the
$\omega$-length sequence of generic perfect trees derived from $G$ and let
$\dot{\mathcal{T}}_{n}$ be the canonical
$\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$-names for
the trees $\mathcal{T}_{n}$. In $V[G]$, let

$\mathbb{U}=\\{\mathcal{T}_{n}\wedge S\mid
S\in\mathbb{P},\,n\in\omega,\,\mathcal{T}_{n}\wedge S\neq\emptyset\\}.$

We will denote by $\mathbb{P}^{*}$ the collection of perfect trees that is the
closure under finite unions of $\mathbb{P}$ and $\mathbb{U}$ and argue that
$\mathbb{P}^{*}$ is a perfect poset extending $\mathbb{P}$. This will be the
case even when $G$ is only generic for a countable transitive model $M$
containing $\mathbb{P}$. At successor stages $\alpha+1$ in the construction of
Jensen’s poset $\mathbb{P}^{J}$, if $\diamondsuit$ codes a certain countable
transitive model $M$ with $\mathbb{P}_{\alpha}\in M$, we will extend
$\mathbb{P}_{\alpha}$ to $\mathbb{P}_{\alpha+1}=\mathbb{P}_{\alpha}^{*}$
constructed in a generic extension $M[G]$ by
$\mathbb{Q}(\mathbb{P}_{\alpha})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$.

Suppose that $M$ is a countable transitive model of ${\rm
ZFC}^{-}+``\mathcal{P}(\omega)$ exists” and $\mathbb{P}\in M$ is a perfect
poset. Clearly we have $\mathbb{Q}(\mathbb{P})\in M$. We will argue that if
$G\subseteq\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is
$M$-generic, then the poset $\mathbb{P}^{*}$ constructed in $M[G]$ is a
perfect poset with the property that every maximal antichain of $\mathbb{P}$
that is an element of $M$ remains maximal in $\mathbb{P}^{*}$. First, we need
the following easy proposition.

######  Proposition 2.3.

Suppose that $\mathbb{P}$ is a perfect poset and a condition
$q\in\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ forces that
$\dot{\mathcal{T}}_{n}\wedge\check{S}\neq\emptyset$ with $S\in\mathbb{P}$.
Then there is a condition $\bar{q}\leq q$ with $\bar{q}(n)=(\bar{T},m)$ and a
node $s\in\bar{T}\cap{}^{m}2$ for which $(\bar{T})_{s}\leq S$.

###### Proof.

Let $q(n)=(T,m)$. Since
$q\Vdash\dot{\mathcal{T}_{n}}\wedge\check{S}\neq\emptyset$, there must be some
$s\in T\cap{}^{m}2$ such that $U=T_{s}\wedge S\neq\emptyset$. Let $\bar{T}$ be
the perfect tree we get by replacing $T_{s}$ with $U$ in $T$ and let
$\bar{q}\leq q$ be such that $\bar{q}(n)=(\bar{T},m)$ and $\bar{q}(i)=q(i)$
for all $i\neq n$. Note that the closure under unions property of perfect
posets is needed to conclude that $\bar{T}\in\mathbb{P}$. ∎

######  Proposition 2.4.

Suppose $M$ is a countable transitive model of ${\rm
ZFC}^{-}+``\mathcal{P}(\omega)$ exists” and $\mathbb{P}\in M$ is a perfect
poset. If $G\subseteq\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is
$M$-generic and $\mathbb{P}^{*}$ is constructed in $M[G]$ as above, then:

  1. (1)

$\mathbb{P}^{*}$ is a perfect poset.

  2. (2)

$\mathbb{U}$ is dense in $\mathbb{P}^{*}$ and $\\{\mathcal{T}_{n}\mid
n<\omega\\}$ is a maximal antichain of $\mathbb{P}^{*}$.

  3. (3)

Every maximal antichain of $\mathbb{P}$ from $M$ remains maximal in
$\mathbb{P}^{*}$.

###### Proof.

First, let’s prove (2). We argue that every $T\in\mathbb{P}$ has some generic
tree $\mathcal{T}_{n}$ below it. Fix $T\in\mathbb{P}$. Let
$q\in\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Since $q$
has finite support, we can choose some $n$ such that
$q(n)=({}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2,0)$ and
strengthen $q$ to a condition $\bar{q}$ such that $q(m)=\bar{q}(m)$ for all
$m\neq n$ and $\bar{q}(n)=(T,0)$. Clearly
$\bar{q}\Vdash\dot{\mathcal{T}}_{n}\leq T$. So by density, there must be some
such $\bar{q}\in G$.

Next, let’s show that for $i\neq j$,
$\mathcal{T}_{i}\wedge\mathcal{T}_{j}=\emptyset$, and so the generic trees
$\mathcal{T}_{n}$ form an antichain. Fix any condition
$p\in\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ with
$p(n)=(T_{n},m_{n})$. By strengthening further, we can assume that
$m_{i}=m_{j}=m$. For each node $s$ that is on level $m$ of both $T_{i}$ and
$T_{j}$, we can choose disjoint perfect trees $U^{s,i}\subseteq(T_{i})_{s}$
and $U^{s,j}\subseteq(T_{j})_{s}$. Let $S_{i}$ be the perfect tree which we
get by replacing, for each $s$ on level $m$ of both $T_{i}$ and $T_{j}$,
$(T_{i})_{s}$ with $U^{s,i}$ in $T_{i}$, and let $S_{j}$ be obtained
similarly. Now let $\bar{p}\leq p$ be the condition where on coordinate $i$,
we put $S_{i}$ instead of $T_{i}$ and on coordinate $j$, we put $S_{j}$
instead of $T_{j}$. Clearly $\bar{p}$ forces that
$\mathcal{T}_{i}\wedge\mathcal{T}_{j}=\emptyset$. It remains to show that
every tree $T$ in $\mathbb{P}^{*}$ is compatible with some $\mathcal{T}_{n}$.
If $T$ is a finite union of trees one of which is in $U$, then this is clear,
and otherwise $T\in\mathbb{P}$, and we just argued that then some
$\mathcal{T}_{n}\leq T$.

To prove (1), it suffices to show that $\mathbb{P}^{*}$ is closed under meets.
So suppose that $R=\bigcup_{i<n}R_{i}$ and
$R^{\prime}=\bigcup_{j<m}R^{\prime}_{j}$, where
$R_{i},R^{\prime}_{j}\in\mathbb{P}\cup\mathbb{U}$, and consider $R\wedge
R^{\prime}=\bigcup_{i<n,j<m}R_{i}\wedge R^{\prime}_{j}$. We will argue that
each $R_{i}\wedge R^{\prime}_{j}$ is either empty or in
$\mathbb{P}\cup\mathbb{U}$. If both $R_{i}$ and $R^{\prime}_{j}$ are in
$\mathbb{P}$, then the conclusion follows since $\mathbb{P}$ is a perfect
poset. If one of the $R_{i}$ or $R^{\prime}_{j}$ is $\mathcal{T}_{k}\wedge S$
with $S\in\mathbb{P}$ and the other is $S^{\prime}\in\mathbb{P}$, then
$R_{i}\wedge R^{\prime}_{j}=\mathcal{T}_{k}\wedge(S\wedge S^{\prime})$ and the
conclusion follows. Finally if $R_{i}=\mathcal{T}_{k}\wedge S$ and
$R^{\prime}_{j}=\mathcal{T}_{k^{\prime}}\wedge S^{\prime}$ with
$S,S^{\prime}\in\mathbb{P}$, then $R_{i}\wedge
R^{\prime}_{j}=(\mathcal{T}_{k}\wedge\mathcal{T}_{k^{\prime}})\wedge(S\wedge
S^{\prime})$. If $k=k^{\prime}$, then the conclusion follows and if $k\neq
k^{\prime}$, we just argued above that
$\mathcal{T}_{k}\wedge\mathcal{T}_{k^{\prime}}=\emptyset$.

Let’s now prove (3). Fix a maximal antichain $\mathcal{A}\in M$ of
$\mathbb{P}$. It suffices to show that every tree in $\mathbb{U}$ is
compatible with some element of $\mathcal{A}$. So suppose that
$\mathcal{T}_{n}\wedge S\in\mathbb{U}$ with $S\in\mathbb{P}$, and fix a
condition $q\in G$ forcing that
$\dot{\mathcal{T}}_{n}\wedge\check{S}\neq\emptyset$. We will argue that there
is a a condition $q^{\prime}\leq q$ with $q^{\prime}(n)=(T^{\prime},m)$ such
that there is a node $s\in T^{\prime}\cap{}^{m}2$ and $A\in\mathcal{A}$ for
which $T^{\prime}_{s}\leq S,A$. By Proposition 2.3, there is $\bar{q}\leq q$
with $\bar{q}(n)=(\bar{T},m)$ such that there is $s\in{}^{m}2$ with
$\bar{T}_{s}\leq S$. Since $\mathcal{A}$ is maximal in $\mathbb{P}$, there is
$A\in\mathcal{A}$ and $U\in\mathbb{P}$ such that $U\leq\bar{T}_{s},A$. Let
$T^{\prime}$ be the tree we get by replacing $\bar{T}_{s}$ with $U$ in
$\bar{T}$ and let $q^{\prime}\leq q$ be the condition where
$q^{\prime}(k)=\bar{q}(k)$ for all $k\neq n$ and
$q^{\prime}(n)=(T^{\prime},m)$. By density, there is some such condition
$q^{\prime}\in G$. But this means that
$(\mathcal{T}_{n})_{s}\leq\mathcal{T}_{n}\wedge S$ and
$(\mathcal{T}_{n})_{s}\leq A$, so that $\mathcal{T}_{n}\wedge S$ is compatible
with $A\in\mathcal{A}$. ∎

We let $\mathbb{U}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ denote the
collection of elements $p\in\mathbb{P}^{*{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ such that
for all $n$, $p(n)\in\mathbb{U}$ or $p(n)={}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$. Note that
it follows from Proposition 2.4 that $\mathbb{U}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is dense in
$\mathbb{P}^{*{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$.

######  Proposition 2.5.

Suppose that $M$ is a countable transitive model of ${\rm
ZFC}^{-}+``\mathcal{P}(\omega)$ exists” and $\mathbb{P}\in M$ is a perfect
poset. If $G\subseteq\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is
$M$-generic and $\mathbb{P}^{*}$ is constructed in $M[G]$ as above, then every
maximal antichain of $\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ from $M$
remains maximal in $\mathbb{P}^{*{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$.

###### Proof.

Fix a maximal antichain $\mathcal{A}\in M$ of
$\mathbb{P}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. It suffices
to show that every condition $p\in\mathbb{U}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is
compatible with some element of $\mathcal{A}$. Let

$p=\langle\mathcal{T}_{i_{0}}\wedge S_{0},\mathcal{T}_{i_{1}}\wedge
S_{1},\ldots,\mathcal{T}_{i_{m}}\wedge S_{m}\rangle.$

Note that the values $i_{n}$ do not have to be distinct and we need to
carefully address this possibility. Fix a condition $q\in G$ forcing that
$\dot{\mathcal{T}}_{i_{n}}\wedge\check{S}_{n}\neq\emptyset$ for every $n\leq
m$. Repeatedly using the construction in the proof of Proposition 2.3 and
going to a large enough level, we can find a condition $\bar{q}\leq q$ with
$\bar{q}(n)=(\bar{T}_{n},k_{n})$ such that for every $n\leq m$, there is a
node $s_{n}$ on level $k_{i_{n}}$ of $\bar{T}_{i_{n}}$ such that
$(\bar{T}_{i_{n}})_{s_{n}}\leq S_{n}$, and for $n\neq n^{\prime}$, $s_{n}\neq
s_{n^{\prime}}$. Let
$\bar{p}=\langle(\bar{T}_{i_{0}})_{s_{0}},(\bar{T}_{i_{1}})_{s_{1}},\ldots,(\bar{T}_{i_{m}})_{s_{m}}\rangle$.
Since $\mathcal{A}$ is maximal in $\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, the
condition $\bar{p}$ is compatible with some $a\in\mathcal{A}$, and so for
every $n\leq m$, we can let $U_{n}=\bar{p}(n)\wedge a(n)\in\mathbb{P}$. Now,
for every $i\in\\{i_{n}\mid n\leq m\\}$, let $R_{i}$ be the tree we get by
replacing $(\bar{T}_{i})_{s_{n}}$ with $U_{n}$ in $\bar{T}_{i}$ whenever
$i_{n}=i$. Let $r$ be the condition such that $r(i)=(R_{i},k_{i})$ for
$i\in\\{i_{n}\mid n\leq m\\}$ and $r(i)=\bar{q}(i)$ otherwise. By density,
there is some such $r\in G$. But this means that
$(\mathcal{T}_{i_{n}})_{s_{n}}\leq S_{n}\wedge a(n)$, so that $p$ is
compatible with $a$. ∎

All our constructions will take place inside the constructible universe $L$.
For reasons that will become obvious during the course of the constructions,
we will restrict ourselves to countable models of ${\rm
ZFC}^{-}+``\mathcal{P}(\omega)$ exists” which happen to be initial segments
$L_{\alpha}$ of $L$. So let’s call countable $L_{\alpha}$ satisfying ${\rm
ZFC}^{-}+``\mathcal{P}(\omega)$ exists” _suitable_ models. Relevant examples
of suitable models for us will be transitive collapses of countable $M\prec
L_{\omega_{2}}$.

We are now ready to review the construction of Jensen’s perfect poset
$\mathbb{P}^{J}$, which will use the $\diamondsuit$-principle to anticipate
and seal maximal antichains. So let’s start by fixing a canonically defined
$\diamondsuit$-sequence $\langle S_{\alpha}\mid\alpha<\omega_{1}\rangle$. Note
that if $M$ is a suitable model and $\delta=\omega_{1}^{M}$, then $\langle
S_{\alpha}\mid\alpha<\delta\rangle$ is an element of $M$.

Jensen’s poset $\mathbb{P}^{J}$ will be the union of the following increasing
sequence $\langle\mathbb{P}_{\alpha}\mid\alpha<\omega_{1}\rangle$ of perfect
posets. Let $\mathbb{P}_{0}=\mathbb{P}^{\text{min}}$. At limit stages, we will
take unions. Suppose $\mathbb{P}_{\alpha}$ has been defined. We let
$\mathbb{P}_{\alpha+1}=\mathbb{P}_{\alpha}$, unless the following happens.
Suppose $S_{\alpha}$ codes a well-founded and extensional binary relation
$E\subseteq\alpha\times\alpha$ such that the collapse of $E$ is a suitable
model $M_{\alpha}$ with $\mathbb{P}_{\alpha}\in M_{\alpha}$ and
$\alpha=\omega_{1}^{M_{\alpha}}$. In this case, we take the $L$-least
$M_{\alpha}$-generic filter
$G\subseteq\mathbb{Q}(\mathbb{P}_{\alpha})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ and let
$\mathbb{P}_{\alpha+1}=\mathbb{P}^{*}_{\alpha}$ as constructed in
$M_{\alpha}[G]$.

As we observed in Proposition 2.4, $\mathbb{P}_{\alpha+1}$ is a perfect poset
with the property that every maximal antichain of $\mathbb{P}_{\alpha}$ in
$M_{\alpha}$ remains maximal in $\mathbb{P}_{\alpha+1}$. Also, by Proposition
2.5, every maximal antichain of
$\mathbb{P}_{\alpha}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ in
$M_{\alpha}$ remains maximal in
$\mathbb{P}_{\alpha+1}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Now let’s
argue that every maximal antichain of
$\mathbb{P}_{\alpha}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ in
$M_{\alpha}$ remains maximal in the final poset
$\mathbb{P}^{J{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. It suffices
to argue that the models $M_{\alpha}$ form an increasing sequence, which
follows because if $\beta>\alpha$, then $\beta=\omega_{1}^{M_{\beta}}$, and
therefore $M_{\beta}$ has $S_{\alpha}$ as an element and can collapse it to
obtain $M_{\alpha}$. This shows that every maximal antichain of
$\mathbb{P}_{\alpha}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ that is an
element of $M_{\alpha}$ is sealed.

######  Theorem 2.6.

The finite-support product $\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ of Jensen’s
poset $\mathbb{P}^{J}$ has the ccc.

###### Proof.

Fix a maximal antichain $\mathcal{A}$ of
$\mathbb{P}^{J{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Choose some
transitive $M\prec L_{\omega_{2}}$ of size $\omega_{1}$ with $\mathcal{A}\in
M$. We can decompose $M$ as the union of a continuous elementary chain of
countable substructures

$X_{0}\prec X_{1}\prec\cdots\prec X_{\alpha}\prec\cdots\prec M$

with $\mathcal{A}\in X_{0}$. By properties of $\diamondsuit$, there is some
$\alpha$ such that $\alpha=\omega_{1}\cap{X_{\alpha}}$,
$\mathbb{P}_{\alpha}=\mathbb{P}^{J}\cap X_{\alpha}$, and $S_{\alpha}$ codes
$X_{\alpha}$. Let $M_{\alpha}$ be the transitive collapse of $X_{\alpha}$.
Then $\mathbb{P}_{\alpha}$ is the image of $\mathbb{P}^{J}$ under the collapse
and $\alpha$ is the image of $\omega_{1}$. Let
$\bar{\mathcal{A}}=\mathcal{A}\cap X_{\alpha}$ be the image of $\mathcal{A}$
under the collapse. So at stage $\alpha$ in the construction of
$\mathbb{P}^{J}$, we chose a forcing extension $M_{\alpha}[G]$ of $M_{\alpha}$
by $\mathbb{Q}(\mathbb{P}_{\alpha})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ and let
$\mathbb{P}_{\alpha+1}=\mathbb{P}^{*}_{\alpha}$ as constructed in
$M_{\alpha}[G]$. Thus, by our observation above, $\bar{\mathcal{A}}$ remains
maximal in $\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, and hence
$\bar{\mathcal{A}}=\mathcal{A}$ is countable. ∎

Finally, we would like to observe that every maximal antichain of
$\mathbb{P}_{\alpha+1}$ from $M_{\alpha}[G]$, in particular the antichain
$\langle\mathcal{T}_{n}\mid n<\omega\rangle$ of generic perfect trees, remains
maximal in $\mathbb{P}^{J}$.

######  Proposition 2.7.

Suppose that $\mathcal{A}\in M_{\alpha}[G]$ is a maximal antichain of
$\mathbb{P}_{\alpha+1}$ (or
$\mathbb{P}_{\alpha+1}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega})$. Then
$\mathcal{A}$ remains maximal in $\mathbb{P}^{J}$ (or
$\mathbb{P}^{J{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega})$.

###### Proof.

Fix $\beta>\alpha$. We already argued that $M_{\alpha}\in M_{\beta}$. Since
$M_{\beta}$ sees that $M_{\alpha}$ is countable, it has some $M$-generic
filter for $\mathbb{Q}(\mathbb{P}_{\alpha})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. But since
$G_{\alpha}$ was chosen to be the $L$-least such filter, then by suitability,
$M_{\beta}$ must contain $G_{\alpha}$. So $M_{\alpha}[G_{\alpha}]\subseteq
M_{\beta}$. ∎

##  3\. The Kanovei-Lyubetsky Theorem

We will reprove here the Kanovei-Lyubetsky theorem from [KL17] showing that
Jensen’s poset $\mathbb{P}^{J}$ from Section 2, has the property that in a
forcing extension $L[G]$ of $L$ by the finite support $\omega$-length product
$\mathbb{P}^{J{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, the only
$L$-generic reals for $\mathbb{P}$ are the $\omega$-many slices of $G$. This
is a generalization of Jensen’s uniqueness of generic filters property to
products.

Suppose that $\mathbb{P}$ is a perfect poset and $H$ is a generic filter for
the product $\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. We will
call $x_{n}$ the real on the $n$-th coordinate of $H$ and let $\dot{x}_{n}$ be
its canonical name.

For the next lemma, we suppose that $\mathbb{P}$ is a countable perfect poset
that is an element of a suitable model $M$. We should think of $\mathbb{P}$ as
one of the perfect posets $\mathbb{P}_{\alpha}$ arising at stage $\alpha$ in
the construction of Jensen’s poset $\mathbb{P}^{J}$ and of $M$ as the model
$M_{\alpha}$ from that stage.

######  Theorem 3.1 (Kanovei-Lyubetsky, [KL17]).

In $M$, suppose that $\dot{r}$ is a $\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$-name for a
real such that for all $n\in\omega$, $\mathop{1\hskip-2.5pt{\rm
l}}\Vdash_{\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\dot{r}\neq\dot{x}_{n}$.
Then in a forcing extension $M[G]$ by
$\mathbb{Q}(\mathbb{P})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, for every
generic perfect tree $\mathcal{T}_{n}$, conditions forcing that
$\dot{r}\notin[\mathcal{T}_{n}]$ are dense in
$\mathbb{P}^{*{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$.

###### Proof.

Fix a condition $p\in\mathbb{P}^{*{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ and
$d\in\omega$. We need to find $p^{\prime}\leq p$ such that
$p^{\prime}\Vdash\dot{r}\not\in[\mathcal{T}_{d}]$. Since
$\mathbb{U}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is dense in
$\mathbb{P}^{*{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, we can
assume without loss that $p\in\mathbb{U}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. So let

$p=\langle\mathcal{T}_{i_{0}}\wedge S_{0},\mathcal{T}_{i_{1}}\wedge
S_{1},\ldots,\mathcal{T}_{i_{m}}\wedge S_{m}\rangle.$

By strengthening if necessary, we can assume that for some $n\leq m$,
$i_{n}=d$. As we argued in the proof of Proposition 2.5, there is a condition
$q\in G$ with $q(n)=(T_{n},k_{n})$ such that for every $n\leq m$, there is a
node $s_{n}$ on level $k_{i_{n}}$ of $T_{i_{n}}$ such that
$(T_{i_{n}})_{s_{n}}\leq S_{n}$ and for $n\neq n^{\prime}$, $s_{n}\neq
s_{n^{\prime}}$.

Now we are going to construct a condition $a_{q}=\langle
W_{0},\ldots,W_{l^{\prime}},\ldots,W_{l}\rangle\in\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ associated
to $q$ and satisfying the following properties.

  1. (1)

For every $n\leq m\leq l^{\prime}$, $W_{n}\leq(T_{i_{n}})_{s_{n}}$.

  2. (2)

For every node $s$ on level $k_{d}$ of $T_{d}$, there is $i\leq l^{\prime}$
such that $W_{i}\leq(T_{d})_{s}$.

  3. (3)

For every $n\leq l^{\prime}$,
$a_{q}\Vdash_{\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\dot{r}\notin[W_{n}]$.

Let
$a=\langle(T_{i_{0}})_{s_{0}},(T_{i_{1}})_{s_{1}},\ldots,(T_{i_{m}})_{s_{m}},(T_{d})_{r_{0}},\ldots,(T_{d})_{r_{k}}\rangle$,
where $r_{0},\ldots,r_{k}$ are nodes on level $k_{d}$ of $T_{d}$ excluding
those amongst the $s_{n}$. Since $\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\Vdash\dot{r}\neq\dot{x}_{n}$
for all $n\in\omega$, we can strengthen $a$ to a condition $a_{q}=\langle
W_{0},W_{1},\ldots,W_{m+k+2},\ldots,W_{l}\rangle$ forcing that
$\dot{r}\notin[W_{n}]$ for all $n\leq m+k+2$.

Using $a_{q}$, we construct the following condition $\bar{q}\leq q$. For
$i\in\\{i_{n}\mid n\leq m\\}$, let $R_{i}$ be the tree we get by replacing
$(T_{i})_{s_{n}}$ with $W_{n}$ in $T_{i}$ whenever $i_{n}=i$ and if $i=d$,
then we also replace all $T_{r_{n}}$ with the appropriate $W_{j}$ as well. Let
$\bar{q}(i)=(R_{i},k_{i})$ for $i\in\\{i_{n}\mid n\leq m\\}$ and
$\bar{q}(i)=q(i)$ otherwise. By density, some such condition $\bar{q}$,
constructed from $a_{q}$, is in $G$. It follows that each
$(\mathcal{T}_{i_{n}})_{s_{n}}\leq W_{n}\wedge S_{n}$ and $\mathcal{T}_{d}$
has a level $k_{d}$ such that for every node $s$ on level $k_{d}$,
$\mathcal{T}_{d}\leq W_{j}$ for some $j$. The first part gives us that
$p=\langle\mathcal{T}_{i_{0}}\wedge S_{0},\mathcal{T}_{i_{1}}\wedge
S_{1},\ldots,\mathcal{T}_{i_{m}}\wedge S_{m}\rangle$ is compatible with
$a_{q}$. Finally, we would like to argue that
$a_{q}\Vdash_{\mathbb{P}^{*{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\dot{r}\notin[\mathcal{T}_{d}]$.

Let $H^{*}$ be any $V$-generic filter for
$\mathbb{P}^{*{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ containing
$a_{q}$ ($V=M[G]$). Since every maximal antichain of
$\mathbb{P}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ in $M$
remains maximal in $\mathbb{P}^{*{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, it follows
that $H^{*}$ restricts to an $M$-generic filter $H$ for
$\mathbb{P}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ and
$a_{q}\in H$. Thus in $M[H]$, $\dot{r}_{H}$ is not a branch through any
$W_{n}$. But this is absolute, and so $\dot{r}_{H}$ is not a branch through
$\mathcal{T}_{d}$ in $V[H^{*}]$. ∎

######  Theorem 3.2 (Kanovei-Lyubetsky, [KL17]).

Suppose $H\subseteq\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ is
$L$-generic. If $r\in L[H]$ is $L$-generic for $\mathbb{P}^{J}$, then
$r=x_{n}$ for some $n<\omega$.

###### Proof.

Let’s suppose that $r\in L[H]$ is a real which is not one of the $x_{n}$. Let
$\dot{r}$ be a nice $\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$-name for $r$
such that for all $n\in\omega$, $\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}^{j{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\Vdash\dot{r}\neq\dot{x}_{n}$.

Choose some transitive model $M\prec L_{\omega_{2}}$ of size $\omega_{1}$ with
$\dot{r}\in N$. We can decompose $M$ as the union of a continuous elementary
chain of countable substructures

$X_{0}\prec X_{1}\prec\cdots\prec X_{\alpha}\prec\cdots\prec M$

with $\dot{r}\in X_{0}$. By properties of $\diamondsuit$, there is some
$\alpha$ such that $\alpha=\omega_{1}\cap X_{\alpha}$,
$\mathbb{P}_{\alpha}=\mathbb{P}^{J}\cap X_{\alpha}$, and $S_{\alpha}$ codes
$X_{\alpha}$. Let $M_{\alpha}$ be the collapse of $X_{\alpha}$. Then
$\mathbb{P}_{\alpha}$ is the image of $\mathbb{P}^{J}$ under the collapse and
$\alpha$ is the collapse of $\omega_{1}$. Clearly $\dot{r}$ is fixed by the
collapse because it is a nice name and all antichains of
$\mathbb{P}^{J{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ are
countable (by Theorem 2.6). So at stage $\alpha$ in the construction of
$\mathbb{P}^{J}$, we chose a forcing extension $M_{\alpha}[G]$ by
$\mathbb{Q}(\mathbb{P}_{\alpha})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ and let
$\mathbb{P}_{\alpha+1}=\mathbb{P}_{\alpha}^{*}$ as constructed in
$M_{\alpha}[G]$. By elementarity, $M_{\alpha}$ satisfies that
$\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}_{\alpha}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\Vdash\dot{r}\neq\dot{x}_{n}$
for all $n\in\omega$. Thus, by Theorem 3.1, for every $n<\omega$,
$\mathbb{P}_{\alpha+1}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ has a
maximal antichain $\mathcal{A}_{n}\in M_{\alpha}[G]$ consisting of conditions
$q$ such that $q\Vdash_{\mathbb{P}_{\alpha+1}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\dot{r}\notin[\mathcal{T}_{n}]$.
It follows, using Proposition 2.7, that all the antichains $\mathcal{A}_{n}$
remain maximal in $\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Also, by
Proposition 2.7, $\langle\mathcal{T}_{n}\mid n<\omega\rangle$ remains maximal
in $\mathbb{P}^{J}$.

So let’s argue that if $q\in\mathcal{A}_{n}$, then
$q\Vdash_{\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}}\dot{r}\notin[\mathcal{T}_{n}]$.
Let $\bar{H}^{*}\subseteq\mathbb{P}^{J{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ be an
$L$-generic filter containing $q$ and let $\bar{H}$ be the restriction of
$\bar{H}^{*}$ to $\mathbb{P}_{\alpha+1}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Since
$q\in\bar{H}$, it follows that $\dot{r}_{\bar{H}}\notin[\mathcal{T}_{n}]$
holds in $M_{\alpha}[G][\bar{H}]$, but this statement is absolute and so also
holds in $L[\bar{H}^{*}]$. Since $H$ must meet every $\mathcal{A}_{n}$, it
holds in $L[H]$ that $\dot{r}_{H}=r$ is not a branch through any
$\mathcal{T}_{n}$. So $r$ cannot be $L$-generic for $\mathbb{P}^{J}$. ∎

##  4\. Finite iterations of perfect posets

An iteration of perfect posets is an iteration of forcing notions in which
every initial segment forces that the next poset is perfect. Here we will only
be dealing with finite iterations of perfect posets, so we are not concerned
with issues of support.

######  Definition 4.1.

A _finite iteration of perfect posets_ is a finite iteration

$\mathbb{P}_{n}=\mathbb{Q}_{0}*\dot{\mathbb{Q}}_{1}*\cdots*\dot{\mathbb{Q}}_{n-1}$

such that $\mathbb{Q}_{0}$ is a perfect poset and for $1\leq i<n$,

$\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{P}_{i}}\Vdash``\dot{\mathbb{Q}}_{i}$
is a perfect poset”.

Suppose $G\subseteq\mathbb{P}_{n}$ is $V$-generic. For $1\leq i<n$, let
$G_{i}$ be the restriction of $G$ to $\mathbb{P}_{i}$. Let $G(0)=G_{1}$ and
for $1\leq i<n$, let $G(i)=\\{p(i)_{G_{i}}\mid p\in G\\}$. Let $r_{i}$ be the
unique real determined by $G(i)$. It is not difficult to see that the sequence
of reals $\vec{r}=\langle r_{0},\ldots,r_{n-1}\rangle$ determines $G$.
Elements of $G_{1}$ are trees with $r_{0}$ as a branch, and inductively,
elements of $G_{i+1}$ are conditions $p\in\mathbb{P}_{i}$ such that
$p\upharpoonright i\in G_{i}$ and $r_{i}\in p(i)_{G_{i}}$.

The analogue of the fusion poset $\mathbb{Q}(\mathbb{P})$ for a finite
iteration $\mathbb{P}_{n}$ of perfect posets is the fusion poset
$\mathbb{Q}(\mathbb{P}_{n})$ whose conditions are pairs $(p,F)$ with
$p\in\mathbb{P}_{n}$ and $F:n\to\omega$, ordered so that
$(p_{2},F_{2})\leq(p_{1},F_{1})$ whenever $p_{2}\leq p_{1}$ and for every
$i<n$, we have $F_{2}(i)\geq F_{1}(i)$ and

$p_{2}\upharpoonright i\Vdash
p_{1}(i)\cap{}^{F_{1}(i)}2=p_{2}(i)\cap{}^{F_{1}(i)}2.$

Fusion arguments with names for perfect trees require that we have some
information about a fixed level $n$ of the tree. We will now argue that there
are densely many conditions in $\mathbb{Q}(\mathbb{P}_{n})$ where this is the
case.

Suppose $p\in\mathbb{P}_{n}$ and $\sigma:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$. Following
[Abr84], let’s define, by induction on $n$, what it means for $\sigma$ to
_lie_ on $p$.333Abraham [Abr84] uses the terminology $\sigma$ is _consistent_
with $p$. For $n=1$, we shall say that $\sigma$ _lies_ on $p$ whenever
$\sigma(0)\in p(0)$. If $\sigma$ lies on $p$, we shall denote by $p\mid\sigma$
the condition $p(0)_{\sigma(0)}$. Note that $p\mid\sigma\leq p$. So suppose
that we have defined when $\sigma$ lies on $p$ for $p\in\mathbb{P}_{n}$, and
for $\sigma$ which lies on $p$, we have defined $p\mid\sigma$ so that
$p\mid\sigma\leq p$. Let $p\in\mathbb{P}_{n+1}$. We define that $\sigma$ lies
on $p$ if $\sigma\upharpoonright n$ lies on $p\upharpoonright n$ and
$(p\upharpoonright n)\mid(\sigma\upharpoonright n)\Vdash\sigma(n)\in p(n)$. If
$\sigma$ lies $p$, we shall denote by $p\mid\sigma$ the condition $\bar{p}$
such that $\bar{p}\upharpoonright n=(p\upharpoonright
n)\mid(\sigma\upharpoonright n)$ and $\bar{p}(n)=\dot{T}$, where $\dot{T}$ is
a $\mathbb{P}_{n}$-name that is interpreted as $p(n)_{\sigma(n)}$ by any
$\mathbb{P}_{n}$-generic filter containing $(p\upharpoonright
n)\mid(\sigma\upharpoonright n)$ and as $p(n)$ otherwise. Clearly this gives
that $p\mid\sigma\leq p$.

######  Definition 4.2.

Let $F:n\to\omega$ and $\sigma:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$. We shall
say that $\sigma$ _lies on levels_ $F$ if $\sigma(i)\in{}^{F(i)}2$ for all
$i<n$.444Abraham [Abr84] uses the terminology $\sigma$ is _bounded_ by $F$. We
shall say that a pair $(p,F)$ with $p\in\mathbb{P}_{n}$ is _determined_ if for
every $\sigma$ lying on levels $F$, either $\sigma$ lies on $p$ or there is
some $i<n$ such that $\sigma\upharpoonright i$ lies on $p\upharpoonright i$
and $(p\upharpoonright i)\mid(\sigma\upharpoonright i)\Vdash\sigma(i)\notin
p(i)$. If $\sigma$ lies on levels $F$ and lies on $p$, we shall say that
$\sigma$ _lies_ on $(p,F)$.

Observe that whenever a pair $(p,F)$ is determined, for every $\sigma$ which
lies on $p$, $(p\mid\sigma)\upharpoonright i$ decides the $F(i)$-th level of
$p(i)$, and indeed $(p\mid\sigma)\upharpoonright i$ forces that $p(i)$ is the
union of the $p\mid\sigma^{\prime}(i)$ for $\sigma^{\prime}\upharpoonright
i=\sigma\upharpoonright i$ that lie on $(p,F)$.

######  Proposition 4.3.

Suppose $\mathbb{P}_{n}$ is a finite iteration of perfect posets.

  1. (1)

Determined conditions $(p,F)$ are dense in $\mathbb{Q}(\mathbb{P}_{n})$.

  2. (2)

Given a determined condition $(p,F)$, the finite set

$\\{p\mid\sigma\mid\sigma\text{ lies on }(p,F)\\}$

is a maximal antichain below $p$ in $\mathbb{P}_{n}$.

######  Proposition 4.4.

Suppose $\mathbb{P}_{n}$ is a finite iteration of perfect posets and
$D\subseteq\mathbb{P}_{n}$ is open dense. For any determined
$(p,F)\in\mathbb{Q}(\mathbb{P}_{n})$, there is a determined $(q,F)\leq(p,F)$
satisfying

$q\mid\sigma\in D\text{ whenever }\sigma\text{ lies on }(q,F).$

For details of proofs, see [Abr84].

Since we will be working mainly with determined conditions, we will now
introduce a kind of normal form for them.

######  Definition 4.5.

Suppose $\mathbb{P}_{n}$ is a finite iteration of perfect posets and
$\sigma:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$. Let’s call
a condition $p\in\mathbb{P}_{n}$ a $\sigma$-_condition_ if

  1. (1)

$p(0)\leq({}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{\sigma(0)}$,

  2. (2)

for all $1\leq i<n$, $p\upharpoonright i\Vdash
p(i)\leq({}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{\sigma(i)}$.

Suppose that $X_{F}$ is a collection of
$\sigma:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ lying on
levels $F:n\to\omega$. An $X_{F}$-_assignment_ is a function
$\varphi:X_{F}\to\mathbb{P}_{n}$ such that each $\varphi(\sigma)$ is a
$\sigma$-condition and $\varphi(\sigma)\upharpoonright
i=\varphi(\sigma^{\prime})\upharpoonright i$ whenever $\sigma\upharpoonright
i=\sigma^{\prime}\upharpoonright i$.

To motivate these definitions, consider a determined condition
$(p,F)\in\mathbb{Q}(\mathbb{P}_{n})$ and let $X_{F}^{p}$ be the set of all
$\sigma$ that lie on it. In this case, the map $\varphi_{p}$ defined by
$\varphi_{p}(\sigma)=p\mid\sigma$ for all $\sigma\in X_{F}^{p}$ is clearly an
$X_{F}^{p}$-assignment. Thus, a determined condition $(p,F)$ gives us a
natural $X_{F}$-assignment, and now conversely we would like to argue that any
$X_{F}$-assignment has a naturally associated determined condition $(q,F)$.

Suppose $X_{F}$ and $\varphi$ are as in Definition 4.5. Observe that given any
$\sigma,\sigma^{\prime}\in X_{F}$, either $\sigma(0)=\sigma^{\prime}(0)$ and
so $\varphi(\sigma)(0)=\varphi(\sigma^{\prime})(0)$, or $\sigma(0)=s\neq
t=\sigma^{\prime}(0)$ are two nodes on level $F(0)$ such that
$\varphi(\sigma)(0)\leq({}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{s}$ and
$\varphi(\sigma^{\prime})(0)\leq({}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{t}$. More
generally, if $\sigma\neq\sigma^{\prime}$, then there is some least $i$ such
that $\sigma\upharpoonright i=\sigma^{\prime}\upharpoonright i$ and there are
nodes $\sigma(i)=s\neq t=\sigma^{\prime}(i)$ on level $F(i)$ such that
$\varphi(\sigma)\upharpoonright
i\Vdash\varphi(\sigma)(i)\leq({}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{s}$ and
$\varphi(\sigma^{\prime})\upharpoonright
i\Vdash\varphi(\sigma^{\prime})(i)\leq({}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2)_{t}$. It
follows, in particular, that for any $i<n$, the conditions
$\varphi(\sigma)\upharpoonright i$ for $\sigma\in X_{F}$ form an antichain.

######  Proposition 4.6.

Suppose $X_{F}$ is a collection of
$\sigma:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ on level
some $F:n\to\omega$ and $\varphi$ is an $X_{F}$-assignment. Then there is a
determined condition $(q,F)\in\mathbb{Q}(\mathbb{P}_{n})$ such that

  1. (1)

$\sigma$ lies on $(q,F)$ if and only if $\sigma\in X_{F}$,

  2. (2)

for every $\sigma\in X_{F}$,

$q\mid\sigma(0)=\varphi(\sigma)(0),$

and for all $i<n$, $(q\mid\sigma)\upharpoonright i$ and
$\varphi(\sigma)\upharpoonright i$ extend each other and force

$q\mid\sigma(i)=\varphi(\sigma)(i).$

###### Proof.

Let $q(0)$ be the union of the $\varphi(\sigma)(0)$ for $\sigma\in X_{F}$,
which is in the perfect poset $\mathbb{Q}_{0}$. Let $q(1)$ be the (canonical)
$\mathbb{P}_{1}$-name for the tree which is the union of the collection of
trees given by the interpretation of the name

$\\{(\varphi(\sigma)(1),\varphi(\sigma)(0))\mid\sigma\in X_{F}\\}.$

Since each
$\varphi(\sigma)(0)\Vdash\varphi(\sigma)(1)\in\dot{\mathbb{Q}}_{1}$, and
$\dot{\mathbb{Q}}_{1}$ is forced to be a perfect poset, it follows that
$q(0)\Vdash q(1)\in\dot{\mathbb{Q}}_{1}$. Let’s see what $q(1)_{G}$ looks like
in a forcing extension $V[G]$ by $\mathbb{P}_{1}$. If $\varphi(\sigma)(0)$ and
$\varphi(\sigma^{\prime})(0)$ are in $G$, then
$\sigma(0)=\sigma^{\prime}(0)=s$ for some $s$, and so
$\varphi(\sigma)(0)=\varphi(\sigma^{\prime})(0)=p$ for some $p$. Thus, the
interpretation $q(1)_{G}$ is the tree which is the union of the
$\varphi(\sigma)(1)_{G}$ for $\sigma(0)=s$. Similarly, let $q(i)$ be the
$\mathbb{P}_{i}$-name for the tree which is the union of the collection of
trees given by the interpretation of the name
$\\{(\varphi(\sigma)(i),\varphi(\sigma)\upharpoonright i)\mid\sigma\in
X_{F}\\}$. Again, we have $q\upharpoonright i\Vdash
q(i)\in\dot{\mathbb{Q}}_{i}$. Let’s see now what $q(i)_{G}$ looks like in a
forcing extension $V[G]$ by $\mathbb{P}_{i}$. Since
$\varphi(\sigma)\upharpoonright i$ for $\sigma\in X_{F}$ form an antichain, if
$\varphi(\sigma)\upharpoonright i$ and
$\varphi(\sigma^{\prime})\upharpoonright i$ are both in $G$, then
$\sigma^{\prime}\upharpoonright i=\sigma\upharpoonright i=\tau$ for some
$\tau$. So the interpretation $q(i)_{G}$ is the union of the
$\varphi(\sigma)(i)_{G}$ for $\sigma\upharpoonright i=\tau$.

First, we argue that every $\sigma\in X_{F}$ lies on $(q,F)$ and
simultaneously show (2). So fix some $\sigma\in X_{F}$. By construction
$\sigma(0)\in q(0)$ and $q(0)_{\sigma(0)}=\varphi(\sigma)(0)$. So assume
inductively that for some $i<n$,

  1. (1)

$\sigma\upharpoonright i$ lies on $q\upharpoonright i$,

  2. (2)

$(q\upharpoonright i)\mid(\sigma\upharpoonright
i)\leq\varphi(\sigma)\upharpoonright i$,

  3. (3)

$\varphi(\sigma)\upharpoonright i\leq(q\upharpoonright
i)\mid(\sigma\upharpoonright i)$.

Suppose that $G\subseteq\mathbb{P}_{i}$ is a $V$-generic filter containing
$(q\upharpoonright i)\mid(\sigma\upharpoonright i)$. Then also, by assumption,
$\varphi(\sigma)\upharpoonright i\in G$. By definition of $q$, we have

$(q(i)_{G})_{\sigma(i)}=\varphi(\sigma)(i)_{G}.$

So $\sigma\upharpoonright i+1$ lies on $q\upharpoonright i+1$ and we have

$(q\upharpoonright i+1)\mid(\sigma\upharpoonright
i+1)\leq\varphi(\sigma)\upharpoonright i+1\text{ and
}\varphi(\sigma)\upharpoonright i+1\leq(q\upharpoonright
i+1)\mid(\sigma\upharpoonright i+1).$

Now suppose that $\tau:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ lies on
$(q,F)$. By definition of $q$, it follows that $\tau(0)=\sigma(0)$ for some
$\sigma\in X_{F}$. So suppose inductively that for some $i<n$, there is
$\sigma\in X_{F}$ such that $\tau\upharpoonright i=\sigma\upharpoonright i$.
Since $\tau$ lies on $q$, it follows that $(q\upharpoonright
i)\mid(\sigma\upharpoonright i)\Vdash\tau(i)\in q(i)$. But then there is some
$\sigma^{\prime}$ such that $\sigma\upharpoonright
i=\sigma^{\prime}\upharpoonright i$ and $\tau(i)=\sigma^{\prime}(i)$. So in
the last step, we will obtain $\sigma\in X_{F}$ such that $\sigma=\tau$. ∎

We shall call the condition $q$ constructed in Proposition 4.6 the
_amalgamation_ of $\varphi$.

######  Proposition 4.7.

Suppose $(p,F)$ is determined and $q$ is the amalgamation of the
$X_{F}^{p}$-assignment $\varphi_{p}$. Then $q\leq p$ and $p\leq q$.

###### Proof.

Clearly $p(0)=q(0)$. So let’s suppose inductively that for some $i<n$, we have
$q\upharpoonright i\leq p\upharpoonright i$ and $p\upharpoonright i\leq
q\upharpoonright i$. We will argue that both $p\upharpoonright i$ and
$q\upharpoonright i$ force $p(i)=q(i)$. So suppose $G\subseteq\mathbb{P}_{i}$
is a $V$-generic filter containing $p\upharpoonright i$, and hence also
$q\upharpoonright i$. Since the conditions $(p\mid\sigma)\upharpoonright i$
for $\sigma\in X_{F}^{p}$ form a maximal antichain below $p\upharpoonright i$,
it follows that one such $(p\mid\sigma)\upharpoonright i$ is in $G$. Thus,
$p(i)_{G}$ is the union of the $(p\mid\sigma^{\prime}(i))_{G}$ for
$\sigma^{\prime}\in X_{F}^{p}$ with $\sigma^{\prime}\upharpoonright
i=\sigma\upharpoonright i$. Similarly, there is some $\tau\in X_{F}^{p}$ such
that $(q\mid\tau)\upharpoonright i\in G$, so we must have $\tau\upharpoonright
i=\sigma\upharpoonright i$. Also, $q(i)_{G}$ is the union of the
$(q\mid\sigma^{\prime}(i))_{G}$ for $\sigma^{\prime}\in X_{F}^{p}$ with
$\sigma^{\prime}\upharpoonright i=\sigma\upharpoonright i$. Now, by
Proposition 4.6, $(q\mid\sigma)\upharpoonright i\Vdash
q\mid\sigma^{\prime}(i)=p\mid\sigma^{\prime}(i)$ for all $\sigma^{\prime}\in
X_{F}^{p}$ with $\sigma^{\prime}\upharpoonright i=\sigma\upharpoonright i$. So
$p(i)_{G}=q(i)_{G}$. ∎

Let us say that a map $\tau:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ _extends_
another map $\sigma:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ if
$\tau(i)$ extends $\sigma(i)$ for every $i<n$. Now we would like to argue that
given a determined condition $(q,F)$ and strengthenings $q_{\sigma}\leq
q\mid\sigma$ for every $\sigma$ lying on $(q,F)$, we can obtain a determined
condition $(\bar{q},\bar{F})\leq(q,F)$ such that for every $\sigma$ lying on
$(q,F)$, there is some $\tau$ lying on $(\bar{q},\bar{F})$ extending $\sigma$
with $\bar{q}\mid\tau\leq q_{\sigma}$. Indeed, we will get the following
stronger statement.

######  Proposition 4.8.

Suppose $(q,F)\in\mathbb{Q}(\mathbb{P}_{n})$ is determined and for every
$\sigma$ which lies on $(q,F)$, there is a finite set $\mathscr{X}_{\sigma}$
of conditions $p\leq q\mid\sigma$. Then there is a condition
$(\bar{q},\bar{F})\leq(q,F)$, also determined, such that for every $\sigma$
which lies on $(q,F)$ and $p\in\mathscr{X}_{\sigma}$, there is $\tau$ which
lies on $(\bar{q},\bar{F})$ and extends $\sigma$ having $\bar{q}\mid\tau\leq
p$. Moreover, for every $\tau$ which lies on $(\bar{q},\bar{F})$,
$\bar{q}\mid\tau\leq p$ for some $p\in\mathscr{X}_{\sigma}$ with $\tau$
extending $\sigma$.

###### Proof.

First, suppose $n=1$. Let $(T,n)$ be a condition in $\mathbb{Q}(\mathbb{P})$
and suppose that for every node $t$ on level $n$ of $T$, we have a finite set
$\mathscr{X}_{t}$ of trees $S\leq T_{t}$. Let $T^{\prime}$ be the tree we get
by replacing $T_{t}$ with the union of $S$ in $\mathscr{X}_{t}$ for every $t$
on level $n$ of $T$. Fix $t$ on level $n$ of $T^{\prime}$. Find a large enough
level $n_{t}\geq n$ such that for every $s$ on level $n_{t}$ of $T^{\prime}$,
we can thin out $T^{\prime}_{s}$ to $P^{s}$ with the property that $P^{s}\leq
S$ for some $S\in\mathscr{X}_{t}$, and moreover for every
$S\in\mathscr{X}_{t}$, there is some node $s$ such that $P^{s}\leq S$. Let
$\bar{T}$ be the tree we obtain by replacing $T^{\prime}_{s}$ with $P^{s}$ in
$T^{\prime}$. Let $\bar{n}\geq n_{t}$ for every $t$ on level $n$ of
$T^{\prime}$. Clearly the condition $(\bar{T},\bar{n})$ has all the desired
properties.

Next, let’s consider the case $n=2$. Let $(q,F)$ be a condition in
$\mathbb{Q}(\mathbb{P}_{2})$ and suppose that for every $\sigma$ which lies on
$(q,F)$, we have a finite set $\mathscr{X}_{\sigma}$ of conditions $p\leq
q\mid\sigma$. For every node $t$ on level $F(0)$ of $q(0)$, let
$\mathscr{X}_{t}$ be the set of all trees $p(0)$ with
$p\in\mathscr{X}_{\sigma}$ for $\sigma(0)=t$. Using the case $n=1$, let
$(\bar{T},\bar{n})$ be the condition for the sets $\mathscr{X}_{t}$. Fix a
node $s$ on level $\bar{n}$ of $\bar{T}$ and let $t_{s}$ be the node on level
$F(0)$ of $\bar{T}$ which $s$ extends. Since $(q,F)$ was determined,
$\bar{T}_{s}$ decides the $F(1)$-th level of $q(1)$. So let $\dot{T}^{s}$ be a
$\mathbb{P}_{1}$-name for the tree we get by replacing, for every node $t$ on
level $F(1)$ of $q(1)$, each $q(1)_{t}$ with the union of $p(1)$ for
$p\in\mathscr{X}_{\sigma}$ with $\sigma=\\{(0,t_{s}),(1,t)\\}$ and
$\bar{T}_{s}\leq p(0)$. Strengthen $\bar{T}_{s}$ to $P^{s}$ deciding a level
$m_{s}\geq F(1)$ such that for every node $u$ on level $m_{s}$ of
$\dot{T}^{s}$ above a node $t$ on level $F(1)$, we can thin out
$\dot{T}^{s}_{u}$ to $\dot{P}^{s,u}$ such that $\dot{P}^{s,u}\leq p(1)$ for
some $p\in\mathscr{X}_{\sigma}$ with $\sigma=\\{(0,t_{s}),(1,t)\\}$ and
$\bar{T}_{s}\leq p(0)$, and moreover for every such $p(1)$, there is some node
$u$ such that $\dot{P}^{s,u}\leq p(1)$. We can assume, with some more thinning
out, that all $m_{s}=m$ for a fixed $m$. Let $\bar{T}^{\prime}$ be the tree we
get by replacing each $\bar{T}_{s}$ with $P^{s}$ in $\bar{T}$. Let
$\dot{T}^{{}^{\prime}s}$ be a $\mathbb{P}_{1}$-name for the tree we get by
replacing, for every node $u$ on level $m$ of $\dot{T}^{s}$, each
$\dot{T}^{s}_{u}$ with $\dot{P}^{s,u}$. Let $\bar{F}=\\{(0,\bar{n}),(1,m)\\}$.
Let $X_{\bar{F}}$ consist of $\sigma:2\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ such that
$\sigma(0)=s$ on level $\bar{n}$ of $\bar{T}^{\prime}$ and $\sigma(1)=u$ on
level $m$ of $\dot{T}^{{}^{\prime}s}$. Let $\varphi$ be the
$X_{\bar{F}}$-assignment such that
$\varphi(\sigma)=\\{(0,\bar{T}^{\prime}_{\sigma(0)}),(1,\dot{T}^{{}^{\prime}\sigma(0)}_{\sigma(1)})\\}$.
Let $\bar{q}$ be the amalgamation of $\varphi$. Clearly the condition
$(\bar{q},\bar{F})$ has all the desired properties.

Finally, assuming that the statement holds for $n$, let’s argue that it holds
for $n+1$ by mimicking how the argument moves from $n=1$ to $n=2$. Let $(q,F)$
be a condition in $\mathbb{Q}(\mathbb{P}_{n+1})$ and suppose that for every
$\sigma$ which lies on $(q,F)$, we have a finite set $\mathscr{X}_{\sigma}$ of
conditions $p\leq q\mid\sigma$. For every $\tau$ which lies on
$(q\upharpoonright n,F\upharpoonright n)$, let $\mathscr{X}_{\tau}$ be the set
of all conditions $p\upharpoonright n$ with $p\in\mathscr{X}_{\sigma}$ for
$\sigma\upharpoonright n=\tau$. Using the inductive hypothesis for $n$, let
$(r,J)$ be the condition for the sets $\mathscr{X}_{\tau}$. Fix $\rho$ lying
on $(r,J)$, and find $\tau_{\rho}$ which lies on $(q\upharpoonright
n,F\upharpoonright n)$ such that $\rho$ extends $\tau_{\rho}$. Since $(q,F)$
was determined, $r\mid\rho$ decides the $F(n)$-th level of $q(n)$. So let
$\dot{T}^{\rho}$ be a $\mathbb{P}_{n}$-name for the tree we get by replacing,
for every node $t$ on level $F(n)$ of $q(n)$, each $q(n)_{t}$ with the union
of $p(n)$ for $p\in\mathscr{X}_{\sigma}$ with
$\sigma=\tau_{\rho}\cup\\{(n,t)\\}$ and $r\mid\rho\leq p\upharpoonright n$.
Strengthen $r\mid\rho$ to $r^{\prime}_{\rho}$ deciding a level $m_{\rho}\geq
F(n)$ such that for every node $u$ on level $m_{\rho}$ of $\dot{T}^{\rho}$
above a node $t$ on level $F(n)$, we can thin out $\dot{T}^{\rho}_{u}$ to
$\dot{P}^{\rho,u}$ such that $\dot{P}^{\rho,u}\leq p(n)$ for some
$p\in\mathscr{X}_{\sigma}$ with $\sigma=\tau_{\rho}\cup\\{(n,t)\\}$ and
$r\mid\rho\leq p\upharpoonright n$, and moreover for every such $p(n)$, there
is some node $u$ with $\dot{P}^{\rho,u}\leq p(n)$. We can assume, with some
more thinning out, that all $m_{\rho}=m$ for a fixed $m$. Let
$(r^{\prime},J^{\prime})$ be the condition we get by applying the inductive
hypothesis to $(r,J)$ and $\mathscr{Y}_{\rho}=\\{r^{\prime}_{\rho}\\}$. Let
$\dot{T}^{{}^{\prime}\rho}$ a $\mathbb{P}_{n}$-name for the tree we get by
replacing, for every node $u$ on level $m$ of $\dot{T}^{\rho}$, each
$\dot{T}^{\rho}_{u}$ with $\dot{P}^{\rho,u}$. Let
$\bar{F}=J^{\prime}\cup\\{(n,m)\\}$. Let $X_{\bar{F}}$ consist of
$\sigma:n+1\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ such that
$\sigma=\tau\cup\\{(n,u)\\}$ for some $\tau$ which lies on
$(r^{\prime},J^{\prime})$ with $\tau$ extending $\rho$ lying on $(r,J)$ and
$u$ on level $m$ of $\dot{T}^{{}^{\prime}\rho}$. Let $\varphi$ be the
$X_{\bar{F}}$-assignment such that
$\varphi(\sigma)=r^{\prime}\mid(\sigma\upharpoonright
n)\cup\\{(n,T^{{}^{\prime}\rho}_{\sigma(n)})\\}$, where $\sigma\upharpoonright
n$ extends $\rho$ lying on $(r,J)$. Let $\bar{q}$ be the amalgamation of
$\varphi$. Clearly the condition $(\bar{q},\bar{F})$ has all the desired
properties. ∎

##  5\. Growing finite iterations of perfect posets

In the construction of Jensen’s poset $\mathbb{P}^{J}$, at nontrivial stages
$\alpha$, we used the $\omega$-many perfect trees obtained from a partially
generic filter for
$\mathbb{Q}(\mathbb{P}_{\alpha})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ to grow the
perfect poset $\mathbb{P}_{\alpha}$ to $\mathbb{P}_{\alpha+1}$. What we would
like to do now is to find an appropriate generalization of this construction
for growing a finite iteration $\mathbb{P}_{n}$ of perfect posets using
partially generic filters for the associated fusion poset
$\mathbb{Q}(\mathbb{P}_{n})$. More precisely, we would like the following.

Given a finite iteration
$\mathbb{P}_{n}=\mathbb{Q}_{0}*\dot{\mathbb{Q}}_{1}*\cdots*\dot{\mathbb{Q}}_{n-1}$
of perfect posets, we would like to be able to extend it to a finite iteration
$\mathbb{P}_{n}^{*}=\mathbb{Q}_{0}^{*}*\dot{\mathbb{Q}}_{1}^{*}*\cdots*\dot{\mathbb{Q}}_{n-1}^{*}$
of perfect posets, constructed in a generic extension of some suitable model
$M$, with the following properties:

  1. (1)

$\mathbb{Q}_{0}\subseteq\mathbb{Q}_{0}^{*}$,

  2. (2)

For all $1\leq i<n$, $\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{P}_{i}^{*}}$
forces that $\dot{\mathbb{Q}}_{i}$ is a perfect poset and
$\dot{\mathbb{Q}}_{i}^{*}$ extends it.

  3. (3)

$\mathbb{P}_{n}\subseteq\mathbb{P}_{n}^{*}$.

  4. (4)

Every maximal antichain $\mathcal{A}\in M$ of $\mathbb{P}_{n}$ remains maximal
in $\mathbb{P}_{n}^{*}$.

The next theorem from [Abr84] holds the main idea for constructing
$\mathbb{P}_{n}^{*}$. The set-up for the theorem is left intentionally vague
with the details forthcoming in the next section.

Fix a suitable model $M$ with $\mathbb{P}_{n}\in M$. We carry out the
construction of $\mathbb{P}_{n}^{*}$ in $n$-steps, at each step $i$,
constructing a $\mathbb{P}_{i}^{*}$-name $\dot{\mathbb{Q}}_{i}^{*}$ for a
perfect poset extending $\dot{\mathbb{Q}}_{i}$. We extend $\mathbb{Q}_{0}$ to
$\mathbb{Q}_{0}^{*}$ as before in a (carefully chosen) forcing extension of
$M$ by $\mathbb{Q}(\mathbb{Q}_{0})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. So suppose
inductively that we already extended $\mathbb{P}_{i}$ to $\mathbb{P}_{i}^{*}$
satisfying requirements (1)-(4). Thus, in particular, every maximal antichain
$\mathcal{A}\in M$ of $\mathbb{P}_{i}$ remains maximal in
$\mathbb{P}_{i}^{*}$, and so every $V$-generic filter $H^{*}$ for
$\mathbb{P}_{i}^{*}$ restricts to an $M$-generic filter $H$ for
$\mathbb{P}_{i}$. From this it follows that $\dot{\mathbb{Q}}_{i}$ is forced
to be a perfect poset by $\mathbb{P}_{i}^{*}$ because $M[H]$ satisfies that
$(\dot{\mathbb{Q}}_{i})_{H}$ is a perfect poset, and this statement is clearly
absolute between $M[H]$ and $V[H^{*}]$. Fix a carefully chosen $M$-generic
filter $G\subseteq\mathbb{Q}(\mathbb{P}_{i+1})$. Let $\tau(G)$ be a
$\mathbb{P}_{i}^{*}$-name for a subset of $\mathbb{Q}(\dot{\mathbb{Q}}_{i})$
such that in any forcing extension $V[H^{*}]$ by $\mathbb{P}_{i}^{*}$
$\tau(G)$ gets interpreted as the collection of all pairs
$(p(i)_{H^{*}},F(i))$ for $(p,F)\in G$. Provided that the poset
$\mathbb{P}_{i}^{*}$ contains a kind of master condition for $G$, we will be
able to conclude that $\tau(G)$ is $M[H]$-generic for
$\mathbb{Q}((\dot{\mathbb{Q}}_{i})_{H})$, where $H$ is the restriction of the
$V$-generic filter $H^{*}\subseteq\mathbb{P}_{i}^{*}$.

######  Theorem 5.1.

Suppose $\bar{p}\in\mathbb{P}^{*}_{i}$ is such that for every $(p,F)\in G$,
$\bar{p}\leq p\upharpoonright i$. Then

$\bar{p}\Vdash``\tau(G)$ is an $M[\dot{H}]$-generic filter for
$\mathbb{Q}(\dot{\mathbb{Q}}_{i})$”,

where $\dot{H}$ is the canonical name for the restriction of the generic
filter to $\mathbb{P}_{i}$.

###### Proof.

Suppose $H^{*}\subseteq\mathbb{P}^{*}_{i}$ is $V$-generic with $\bar{p}\in
H^{*}$. Let $H$ be the restriction of $H^{*}$ to $\mathbb{P}_{i}$. Let

$K=\tau(G)_{H^{*}}=\\{(p(i)_{H^{*}},F(i))\mid p\in G\\}$

and $\mathbb{Q}_{i}=(\dot{\mathbb{Q}}_{i})_{H}$. First, we argue that $K$ is a
filter on $\mathbb{Q}(\mathbb{Q}_{i})$. Suppose for $(p,F)\in G$,
$(p(i)_{H},F(i))\leq(T,b)$ in $\mathbb{Q}(\mathbb{Q}_{i})$, so that $b\leq
F(i)$ and $p(i)_{H}\cap{}^{b}2=T\cap{}^{b}2$. It follows that there is a
$\mathbb{P}_{i}$-name $\dot{T}$ for $T$ such that

$\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{P}_{i}}\Vdash``p(i)\leq\dot{T}\text{
and }p(i)\cap{}^{b}2=\dot{T}\cap{}^{b}2"$.

Let $p^{\prime}=p\upharpoonright i\cup\\{(i,\dot{T})\\}$ and
$F^{\prime}=F\cup\\{(i,b)\\}$. Clearly $(p,F)\leq(p^{\prime},F^{\prime})$,
which means that $(p^{\prime},F^{\prime})\in G$. It follows that
$p^{\prime}(i)_{H}=T$, and so $(T,b)\in K$.

Next, we fix $(p,F)$ and $(p^{\prime},F^{\prime})$ both in $G$ and argue that
$(p(i)_{H},F(i))$ and $(p^{\prime}(i)_{H},F^{\prime}(i))$ are compatible in
$K$. There is $(q,J)\in G$ below both $(p,F)$ and $(p^{\prime},F^{\prime})$.
It follows that $J(i)\geq F(i),F^{\prime}(i)$, $q\upharpoonright
i\Vdash_{\mathbb{P}_{i}}q(i)\leq p(i),p^{\prime}(i)$,

$q\upharpoonright
i\Vdash_{\mathbb{P}_{i}}``q(i)\cap{}^{F(i)}2=p(i)\cap{}^{F(i)}2\text{ and
}q(i)\cap{}^{F^{\prime}(i)}2=p^{\prime}(i)\cap{}^{F^{\prime}(i)}2"$,

and $(q(i)_{H},J(i))\in K$. Since $\bar{p}\leq q\upharpoonright i$ and
$\bar{p}\in H^{*}$, we have $q\upharpoonright i\in H^{*}$. Now observe that
$q\upharpoonright i$ must force the statements mentioned above over
$\mathbb{P}^{*}_{i}$ as well because any $V$-generic filter for
$\mathbb{P}_{i}^{*}$ restricts to an $M$-generic filter for $\mathbb{P}_{i}$
and the statements in question are absolute. Thus,

$q(i)_{H}\leq p(i)_{H},p^{\prime}(i)_{H}$

and

$q(i)_{H}\cap{}^{F(i)}2=p(i)_{H}\cap{}^{F(i)}2\text{ and
}q(i)_{H}\cap{}^{F^{\prime}(i)}2=p^{\prime}(i)_{H}\cap{}^{F^{\prime}(i)}2.$

So $(q(i)_{H},J(i))\leq(p(i)_{H},F(i)),\,(p^{\prime}(i)_{H},F^{\prime}(i))$.

Finally, we have to see that $K$ is $M[H]$-generic. So suppose $D\in M[H]$ is
dense open in $\mathbb{Q}(\mathbb{Q}_{i})$. Let $\dot{D}\in M$ be a
$\mathbb{P}_{i}$-name for $D$ such that

$\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}_{i}}\Vdash_{\mathbb{P}_{i}}``\dot{D}$ is dense open in
$\mathbb{Q}(\dot{\mathbb{Q}}_{i})"$.

In $M$, define

$E=\\{(p,F)\in\mathbb{Q}(\mathbb{P}_{i+1})\mid p\upharpoonright
i\Vdash(p(i),F(i))\in\dot{D}\\}.$

We claim that $E$ is dense open in $\mathbb{Q}(\mathbb{P}_{i+1})$. It is easy
to see that $E$ is open, so let’s argue that it is dense. Fix some
$(q,J)\in\mathbb{Q}(\mathbb{P}_{i+1})$ and assume without loss that $(q,J)$ is
determined. There must be some pair $(\dot{T},\dot{k})$ such that $\dot{T}$ is
a $\mathbb{P}_{i}$-name for an element of $\dot{\mathbb{Q}}_{i}$, $\dot{k}$ is
a $\mathbb{P}_{i}$-name for a natural number, and

$q\upharpoonright i\Vdash``(\dot{T},\dot{k})\in\dot{D}\text{ and
}(\dot{T},\dot{k})\leq(q(i),J(i))".$

The set of conditions which decide the value of $\dot{k}$ is dense open below
$q\upharpoonright i$ in $\mathbb{P}_{i}$. So, by Proposition 4.4, there a
determined condition $(p^{\prime},F^{\prime})\leq(q\upharpoonright
i,J\upharpoonright i)$ in $\mathbb{Q}(\mathbb{P}_{i})$ such that for every
$\sigma$ which lies on it, $p^{\prime}\mid\sigma$ decides that
$\dot{k}=k(\sigma)$. Let $k\in\omega$ be above all the $k(\sigma)$. Then

$p^{\prime}\Vdash``(\dot{T},k)\leq(\dot{T},\dot{k})\leq(q(i),J(i))\text{ and
}(\dot{T},k)\in\dot{D}".$

Let $p=p^{\prime}\cup\\{(i,\dot{T})\\}$ and $F=F^{\prime}\cup\\{(i,k)\\}$.
Clearly $(p,F)\in E$. Let’s argue that $(p,F)\leq(q,J)$. By construction
$p\leq q$ and each $k(\sigma)\geq J(i)$, so $k\geq J(i)$. Finally,
$p^{\prime}=p\upharpoonright i\Vdash(\dot{T},k)\leq(q(i),J(i))$ and so
$p\upharpoonright i\Vdash\dot{T}\cap{}^{J(i)}2=q(i)\cap{}^{J(i)}2$.

Fix some $(p,F)\in E\cap G$. Since $\bar{p}\leq p\upharpoonright i$, it
follows that $p\upharpoonright i\in H^{*}$. Thus, $(p(i)_{H},F(i))\in D\cap
K$, completing the argument that $K$ is an $M[H]$-generic filter for
$\mathbb{Q}(\mathbb{Q}_{i})$. ∎

Next, we are going to obtain a stronger version of Theorem 5.1 that tells us
how to get an $M[H]$-generic filter for
$\mathbb{Q}(\mathbb{Q}_{i})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, which is
really what we need to extend $\dot{\mathbb{Q}}_{i}$ to
$\dot{\mathbb{Q}}_{i}^{*}$. For this, we will need to enlarge our fusion poset
$\mathbb{Q}(\mathbb{P}_{i+1})$. Let $\bar{\mathbb{Q}}_{i}(\mathbb{P}_{i+1})$
be the following modification of $\mathbb{Q}(\mathbb{P}_{i+1})$. Conditions in
$\bar{\mathbb{Q}}(\mathbb{P}_{i+1})$ are pairs $(p,F)$ such that
$(p\upharpoonright i,F\upharpoonright i)\in\mathbb{Q}(\mathbb{P}_{i})$, $p(i)$
is some finite tuple
$\langle\dot{T}_{0},\dot{T}_{1},\ldots,\dot{T}_{k-1}\rangle$ with
$p\upharpoonright i\Vdash\dot{T}_{j}\in\dot{\mathbb{Q}}_{i}$ for all $j<k$,
and $F(i)=f:k\to\omega$. The ordering is $(p^{\prime},F^{\prime})\leq(p,F)$
whenever

  1. (1)

$(p^{\prime}\upharpoonright i,F^{\prime}\upharpoonright
i)\leq(p\upharpoonright i,F\upharpoonright i)$, and
for $j<k$,

  2. (2)

$F^{\prime}(i)(j)\geq F(i)(j)$,

  3. (3)

$p^{\prime}\upharpoonright i\Vdash
p^{\prime}(i)(j)\cap{}^{F(i)(j)}2=p(i)(j)\cap{}^{F(i)(j)}2$.

The point is that if $H^{*}\subseteq\mathbb{P}_{i}^{*}$ is $V$-generic, then

$\langle((\dot{T}_{0})_{H^{*}},f(0)),\ldots,((\dot{T}_{k-1})_{H^{*}},f(k-1))\rangle$

is a condition in
$\mathbb{Q}((\dot{\mathbb{Q}}_{i})_{H})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Suppose now
that $G\subseteq\bar{\mathbb{Q}}(\mathbb{P}_{i+1})$ is $M$-generic. Let
$\tau(G)$ be a $\mathbb{P}_{i}^{*}$-name for a subset of
$\mathbb{Q}(\mathbb{Q}_{i})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ such that in
any forcing extension $V[H^{*}]$ by $\mathbb{P}_{i}^{*}$, $\tau(G)$ gets
interpreted as the collection of all

$\langle((\dot{T}_{0})_{H^{*}},f(0)),\ldots,((\dot{T}_{k-1})_{H^{*}},f(k-1))\rangle$

for $(p,F)\in G$. If it so happens that $\mathbb{P}_{i}^{*}$ has a master
condition $\bar{p}$ for $G$, then $\tau(G)$ will be $M[H]$-generic for
$\mathbb{Q}((\mathbb{Q}_{i})_{H})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, where $H$
is the restriction of the $V$-generic filter
$H^{*}\subseteq\mathbb{P}_{i}^{*}$.

######  Theorem 5.2.

Suppose $\bar{p}\in\mathbb{P}^{*}_{i}$ is such that for every $(p,F)\in G$,
$\bar{p}\leq p\upharpoonright i$. Then

$\bar{p}\Vdash``\tau(G)$ is an $M[\dot{H}]$-generic filter for
$\mathbb{Q}(\dot{\mathbb{Q}}_{i})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$”,

where $\dot{H}$ is the canonical name for the restriction of the generic
filter to $\mathbb{P}_{i}$.

The proof is essentially the same as of Theorem 5.1. Using Theorem 5.2, we can
let $\dot{\mathbb{Q}}_{i}^{*}$ be the canonical $\mathbb{P}_{i}^{*}$-name for
the extension of $\dot{\mathbb{Q}}_{i}$ formed in $M[\dot{H}][\tau(G)]$, where
$\dot{H}$ is the restriction of the generic filter to $\mathbb{P}_{i}$. In the
next section, we will show how to obtain the required $M$-generic filters $G$
so that the inductive assumptions hold for $\mathbb{P}_{i}^{*}$.

##  6\. Tree iterations of perfect posets

Let’s define that an _$\omega$ -iteration of perfect posets_ is a sequence

$\vec{P}=\langle\mathbb{P}_{n}\mid n<\omega\rangle,$

where $\mathbb{P}_{0}=\\{\emptyset\\}$ is a trivial poset and each
$\mathbb{P}_{n}$, for $n\geq 1$, is a finite iteration of perfect posets with
the coherence requirement that for $0<m<n$, $\mathbb{P}_{n}\upharpoonright
m=\mathbb{P}_{m}$. The initial poset $\mathbb{P}_{0}$ is included in the
sequence to make the subsequent definitions more uniform. For this reason, we
will also make the ad hoc definition that
$\mathbb{Q}(\mathbb{P}_{0})=\\{\emptyset\\}$. Note that an $\omega$-iteration
of perfect posets is not itself an iteration, rather it is a coherent sequence
of finite iterations.

A _tree iteration_ is a non-linear forcing iteration along some tree. Given a
tree of height $\omega$, the tree iteration of perfect posets will use an
$\omega$-iteration $\vec{P}=\langle\mathbb{P}_{n}\mid n<\omega\rangle$ of
perfect posets with conditions assigned to level $n$ nodes of the tree coming
from the poset $\mathbb{P}_{n}$. Conditions will be assigned to the nodes
coherently so that if a node $s$ on level $n$ extends a node $t$ on level $m$,
then the condition $p$ on node $s$ will be such that $p\upharpoonright m$ is
on node $t$.

######  Definition 6.1.

Let $\vec{P}=\langle\mathbb{P}_{n}\mid n<\omega\rangle$ be an
$\omega$-iteration of perfect posets and let $\mathscr{T}$ be a tree of height
$\omega$. A _$\mathscr{T}$ -iteration of perfect posets_ is the following
partial order $\mathbb{P}(\vec{P},\mathscr{T})$. Conditions in
$\mathbb{P}(\vec{P},\mathscr{T})$ are functions $f_{X}$ with domain some
finite subtree $X$ of $\mathscr{T}$ such that:

  1. (1)

For every node $s$ on level $n$ of $X$, $f_{X}(s)\in\mathbb{P}_{n}$.

  2. (2)

Whenever $s\leq t$ are two nodes in $X$, then $f_{X}(t)\upharpoonright{\rm
lev}(s)=f_{X}(s)$.

The ordering is $f_{Y}\leq f_{X}$ whenever $Y$ extends $X$ and for every node
$s\in X$, $f_{Y}(s)\leq f_{X}(s)$.

The analogue of the fusion posets $\mathbb{Q}(\mathbb{P})$ and
$\mathbb{Q}(\mathbb{P}_{n})$ for $\mathbb{P}(\vec{P},\mathscr{T})$ will be the
fusion poset $\mathbb{Q}(\vec{P},\mathscr{T})$ whose conditions are functions
$f_{X}$ with domain some finite subtree $X$ of $\mathscr{T}$ such that:

  1. (1)

For every node $s$ on level $n$ of $X$,
$f_{X}(s)\in\mathbb{Q}(\mathbb{P}_{n})$.

  2. (2)

Whenever $s\leq t$ are two nodes in $X$, with $f_{X}(s)=(p_{s},F_{s})$ and
$f_{X}(t)=(p_{t},F_{t})$, then $p_{t}\upharpoonright{\rm lev}(s)=p_{s}$ and
$F_{t}\upharpoonright{\rm lev}(s)=F_{s}$.

The ordering is $f_{Y}\leq f_{X}$ whenever $Y$ extends $X$ and for every node
$s\in X$, $f_{Y}(s)\leq f_{X}(s)$.

######  Proposition 6.2.

  1. (1)

Suppose $f_{X}\in\mathbb{P}(\vec{P},\mathscr{T})$, $f_{X}(s)=p$, and $q\leq
p$. Then there is a condition $g_{X}\leq f_{X}$ in
$\mathbb{P}(\vec{P},\mathscr{T})$ with $g_{X}(s)=q$.

  2. (2)

Suppose $f_{X}\in\mathbb{Q}(\vec{P},\mathscr{T})$, $f_{X}(s)=(p,F)$, and
$(q,F^{\prime})\leq(p,F)$. Then there is $g_{X}\leq p_{X}$ such that
$g_{X}(s)=(q,F^{\prime})$.

###### Proof.

We will only prove (1) because the proof of (2) is analogous. Define $g_{X}$
as follows. Fix a node $t\in X$. If $t\leq s$, then let
$g_{X}(t)=q\upharpoonright{\rm lev}(t)$. If $s\leq t$, then let $g_{X}(t)$ be
$q$ concatenated with the tail of $f_{X}(t)$. Otherwise, let $t^{\prime}$ be
the largest node that is compatible with both $t$ and $s$. Let $g_{X}(t)$ be
$q\upharpoonright{\rm lev}(t^{\prime})$ concatenated with the tail of
$f_{X}(t)$. ∎

Let us call a condition $f_{X}\in\mathbb{Q}(\vec{P},\mathscr{T})$ _determined_
if every $f_{X}(s)$ is determined. Clearly a condition $f_{X}$ is determined
if and only if conditions on the terminal nodes are determined.

######  Proposition 6.3.

The set of all determined conditions is dense in
$\mathbb{Q}(\vec{P},\mathscr{T})$.

###### Proof.

Fix $f_{X}\in\mathbb{Q}(\vec{P},\mathscr{T})$. Let $\langle s_{i}\mid i\leq
m\rangle$ be an enumeration of the terminal nodes of $X$. Using the
construction in the proof of Proposition 6.2, strengthen $f_{X}$ to
$f^{0}_{X}$ such that $f^{0}_{X}(s_{0})$ is determined. Inductively, given
$f^{i}_{X}$, we let $f^{i+1}_{X}$ be the strengthening of $f^{i}_{X}$
constructed as in the proof of Proposition 6.2 such that
$f^{i+1}_{X}(s_{i+1})$ is determined. Let’s argue that $f^{m}_{X}$ is
determined. We can assume inductively that the conditions $f^{m-1}_{X}(s_{i})$
for $i<m$ are determined. By construction $f^{m}_{X}(s_{m})$ is determined. So
fix $i<m$. Let $s$ be the node where the branch of $s_{i}$ and the branch of
$s_{m}$ split. By construction, $f^{m}_{X}(s_{i})$ is
$f^{m}_{X}(s_{m})\upharpoonright{\rm lev}(s)$ concatenated with the tail of
$f^{m-1}_{X}(s_{i})$. Clearly $f^{m}_{X}(s_{m})\upharpoonright{\rm lev}(s)$ is
determined and $f^{m}_{X}(s_{m})\upharpoonright{\rm lev}(s)\leq
f^{m-1}_{X}(s_{i})\upharpoonright{\rm lev}(s)$. Now using the definition of
what it means to be determined, it is easy to see that the whole condition is
determined. ∎

We will initially consider tree iterations along the countable tree
${}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega$, and
then later extend our results to tree iterations along the uncountable tree
${}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega_{1}$.

It is easy to see that the poset
$\mathbb{Q}(\mathbb{Q}_{0})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ completely
embeds into $\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ via
the map sending a condition to the corresponding tree of height $\leq 2$.

######  Proposition 6.4.

The poset $\mathbb{Q}(\mathbb{Q}_{0})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ completely
embeds into $\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ via
the following map $\varphi$:

  1. (1)

$\varphi(\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{Q}(\mathbb{Q}_{0})})=f_{X}$,
where $X$ consists of the root node $s$ and $f_{X}(s)=\emptyset$.
For $p$ with some non-trivial $p(i)$,

  2. (2)

$\varphi(p)=f_{X}$, where $X$ consists of the root node together with nodes
$\langle i\rangle$ for non-trivial $p(i)$, such that $f_{X}(\langle
i\rangle)=p(i)$.

More generally, for each node $s$ on level $n$ of
${}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega$,
$\bar{\mathbb{Q}}(\mathbb{P}_{n+1})$ completely embeds into
$\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ via
the map sending a condition to the corresponding tree of height $\leq n+2$
whose stem stretches up to $s$.

######  Proposition 6.5.

Fix a node $s$ on level $n$ of ${}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}{\omega}$. The
poset $\bar{\mathbb{Q}}(\mathbb{P}_{n+1})$ completely embeds into
$\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ via
the following map $\varphi_{s}$:

  1. (1)

$\varphi_{s}(\mathop{1\hskip-2.5pt{\rm
l}}_{\bar{\mathbb{Q}}(\mathbb{P}_{n+1})})=f_{X}$, where $X$ is the branch
ending in $s$, such that $f_{X}(s)=\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{Q}(\mathbb{P}_{n})}$.
For $(p,F)$ with $\text{dom}(p(n))=k$,

  2. (2)

$\varphi_{s}((p,F))=f_{X}$, where $X$ consists of the branch ending in $s$
together with nodes $\langle s\mathbin{{}^{\smallfrown}}i\rangle$ for $i<k$,
such that $f_{X}(s)=(p\upharpoonright n,F\upharpoonright n)$ and

$f_{X}(s\mathbin{{}^{\smallfrown}}i)=(p\upharpoonright
n\mathbin{{}^{\smallfrown}}p(n)(i),F\upharpoonright
n\mathbin{{}^{\smallfrown}}F(n)(i))$.

Suppose $G\subseteq\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ is
$V$-generic and fix some node $s$ on level $n$ of
${}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}{\omega}$. We
will use the notation $G_{s}$ for the $V$-generic filter for
$\bar{\mathbb{Q}}(\mathbb{P}_{n+1})$ added by $G$ via the embedding
$\varphi_{s}$ and we will use the notation $G_{\emptyset}$ for the $V$-generic
filter for $\mathbb{Q}(\mathbb{Q}_{0})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$ added by $G$
via the embedding $\varphi$.

Suppose now that $\vec{P}=\langle\mathbb{P}_{n}\mid n<\omega\rangle$ is an
$\omega$-iteration of perfect posets that is an element of a suitable model
$M$. Let $G\subseteq\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ be
$M$-generic. We will argue that in $M[G]$, we can grow each iteration
$\mathbb{P}_{n}$ to an iteration $\mathbb{P}_{n}^{*}$ satisfying requirements
(1)-(4) from Section 5:

  1. (1)

$\mathbb{Q}_{0}\subseteq\mathbb{Q}_{0}^{*}$,

  2. (2)

For all $1\leq i<n$, $\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{P}_{i}^{*}}$
forces that $\dot{\mathbb{Q}}_{i}$ is a perfect poset and
$\dot{\mathbb{Q}}_{i}^{*}$ extends it.

  3. (3)

$\mathbb{P}_{n}\subseteq\mathbb{P}_{n}^{*}$.

  4. (4)

Every maximal antichain $\mathcal{A}\in M$ of $\mathbb{P}_{n}$ remains maximal
in $\mathbb{P}_{n}^{*}$.

It is straightforward to extend $\mathbb{Q}_{0}$ to $\mathbb{Q}^{*}_{0}$. By
Proposition 6.4, $G$ adds an $M$-generic filter $G_{\emptyset}$ for
$\mathbb{Q}(\mathbb{Q}_{0})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$. Let
$\mathcal{T}_{i}^{0}$ for $i<\omega$ be the generic perfect trees added by
$G_{\emptyset}$ and construct $\mathbb{Q}_{0}^{*}$ as before. Recall that
$\langle\mathcal{T}_{i}^{0}\mid i<\omega\rangle$ is a maximal antichain in
$\mathbb{Q}_{0}^{*}$ and every maximal antichain $\mathcal{A}\in M$ of
$\mathbb{Q}_{0}$ remains maximal in $\mathbb{Q}_{0}^{*}$.

Now let’s show how to extend $\dot{\mathbb{Q}}_{1}$ to a
$\mathbb{P}_{1}^{*}$-name $\dot{\mathbb{Q}}^{*}_{1}$ for a perfect poset. By
Proposition 6.5, for each node $s$ on level 1 of the tree
${}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega$, $G$
adds an $M$-generic filter $G_{s}$ for $\bar{\mathbb{Q}}(\mathbb{P}_{2})$.
Observe that each $\mathcal{T}_{i}^{0}\leq p\upharpoonright 1$ for all $p$
with $(p,F)\in G_{\langle i\rangle}$. Thus, by Theorem 5.2, whenever
$H^{*}\subseteq\mathbb{P}_{1}^{*}$ is a $V$-generic filter containing
$\mathcal{T}_{i}^{0}$, then the interpretation $\tau(G_{\langle
i\rangle})_{H^{*}}$ is an $M[H]$-generic filter for
$\mathbb{Q}((\dot{\mathbb{Q}}_{1})_{H})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$, where $H$
is the restriction of $H^{*}$ to $\mathbb{P}_{1}$. Let $\tau$ be a mixed
$\mathbb{P}_{1}^{*}$-name that is interpreted as $\tau(G_{\langle
i\rangle})_{H^{*}}$, whenever $\mathcal{T}_{i}^{0}\in H^{*}$. Since the
conditions $\mathcal{T}_{i}^{0}$ form a maximal antichain in
$\mathbb{P}_{1}^{*}$,

$\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{P}_{1}^{*}}\Vdash``\tau$ is an
$M[\dot{H}]$-generic filter for
$\mathbb{Q}(\dot{\mathbb{Q}}_{1})^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}$,”

where $\dot{H}$ is the canonical name for the restriction of the generic
filter to $\mathbb{P}_{1}$. So let $\dot{\mathbb{Q}}_{1}^{*}$ be a
$\mathbb{P}_{1}^{*}$-name for the perfect poset constructed as usual from
$\dot{\mathbb{Q}}_{1}$ and $\tau$.

For each $j<\omega$, we can choose a $\mathbb{P}_{1}^{*}$-name
$\dot{\mathcal{T}}_{j}^{1}$ for the perfect tree on coordinate $j$ of $\tau$.
The pairs $(\mathcal{T}_{i}^{0},\dot{\mathcal{T}}_{j}^{1})$ form a maximal
antichain in $\mathbb{P}_{2}^{*}$ because the collection of
$\mathcal{T}_{i}^{0}$ is a maximal antichain and each $\mathcal{T}_{i}^{0}$
forces that the trees $\dot{T}_{j}^{1}$ form a maximal antichain in
$\dot{\mathbb{Q}}_{1}^{*}$. Also, clearly for every $(p,F)\in G_{\langle
i,j\rangle}$, we have $(\mathcal{T}_{i}^{0},\dot{\mathcal{T}}_{j}^{1})\leq
p\upharpoonright 2$. This will allow us to apply Theorem 5.2 to grow
$\dot{\mathbb{Q}}_{2}$ to $\dot{\mathbb{Q}}_{2}^{*}$. Finally, let’s argue
that $\mathbb{P}_{2}$ is actually a subset of $\mathbb{P}_{2}^{*}$. Suppose
$p$ is a condition in $\mathbb{P}_{2}$. Then $p(0)\in\mathbb{P}_{1}$ and hence
$p(0)\in\mathbb{P}_{1}^{*}$ as well. Also, clearly $p(1)$ is a
$\mathbb{P}_{1}^{*}$-name. So we need to argue that
$p(0)\Vdash_{\mathbb{P}_{1}^{*}}p(1)\in\dot{\mathbb{Q}}_{1}^{*}$. Fix a
$V$-generic filter $H^{*}\subseteq\mathbb{P}_{1}^{*}$ with $p(0)\in H^{*}$ and
consider $M[H]$ where $H$ is the restriction of $H^{*}$ to $\mathbb{P}_{1}$.
Since $p(0)\Vdash_{\mathbb{P}_{1}}p(1)\in\dot{\mathbb{Q}}_{1}$, in $M[H]$, we
have
$p(1)_{H}\in(\dot{\mathbb{Q}}_{1})_{H}\subseteq(\dot{\mathbb{Q}}_{1}^{*})_{H^{*}}$.
It remains to show (4), that every maximal antichain of $\mathbb{P}_{2}$ from
$M$ remains maximal in $\mathbb{P}_{2}^{*}$. We will provide an inductive
proof of this later in Lemma 6.8.

For now to finish the construction, we assume that properties (1)-(4) hold for
$\mathbb{P}_{n}^{*}$. We will additionally assume that:

  1. (1)

For each node $s$ on level $n$ of ${}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega$ there
is a condition

$(\mathcal{T}_{s(0)}^{0},\dot{\mathcal{T}}_{s(1)}^{1},\ldots,\dot{\mathcal{T}}_{s(n-1)}^{n-1})\in\mathbb{P}_{n}^{*}$

which is below all $p\upharpoonright n$ with $(p,F)\in G_{s}$.

  2. (2)

The collection of all such conditions
$(\mathcal{T}_{s(0)}^{0},\dot{\mathcal{T}}_{s(1)}^{1},\ldots,\dot{\mathcal{T}}_{s(n-1)}^{n-1})$
is a maximal antichain in $\mathbb{P}_{n}^{*}$.

With this set-up, we extend $\dot{\mathbb{Q}}_{n}$ to
$\dot{\mathbb{Q}}_{n}^{*}$ identically to the case $n=1$ above, using Theorem
5.2. It is also easy to see inductively that $\mathbb{P}_{n}$ is a subset of
$\mathbb{P}_{n}^{*}$.

To get some intuition for the construction, let’s fix an $L$-generic filter
$H^{*}\subseteq\mathbb{P}_{1}^{*}$ and see what a condition
$\mathcal{T}^{1}_{j}=(\dot{\mathcal{T}}^{1}_{j})_{H^{*}}$ looks like. Since
$\langle\mathcal{T}^{0}_{i}\mid i<\omega\rangle$ is a maximal antichain of
$\mathbb{P}_{1}^{*}$, there is a unique $i$ such that $\mathcal{T}^{0}_{i}\in
H^{*}$. Let

$K=\\{(p(1)_{H^{*}},F(1))\mid f_{X}\in G\text{ and }f_{X}(\langle
i,j\rangle)=(p,F)\\}.$

Then $\mathcal{T}_{j}^{1}$ is the union of $T\cap{}^{n}2$ for $(T,n)\in K$.

To give the promised argument that every maximal antichain of $\mathbb{P}_{n}$
from $M$ remains maximal in $\mathbb{P}_{n}^{*}$, we first need to define the
analogue of $\mathbb{U}\subseteq\mathbb{P}^{*}$ from Section 2 for
$\mathbb{P}_{n}^{*}$.

Let $\mathbb{U}_{n}$ be the subset of $\mathbb{P}_{n}^{*}$ consisting of
conditions $p=(p_{0},\dot{p}_{1},\ldots,\dot{p}_{n-1})$ such that
$p_{0}=\mathcal{T}^{0}_{j_{0}}\wedge S_{0}$ for some $j_{0}<\omega$ and
$S_{0}\in\mathbb{Q}_{0}$, and for $i<n$, $p\upharpoonright
i\Vdash\dot{p}_{i}=\dot{\mathcal{T}}^{i}_{j_{i}}\wedge\dot{S}_{i}$ for some
$j_{i}<\omega$ and a $\mathbb{P}_{i}$-name $\dot{S}_{i}$ such that
$\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}_{i}}\Vdash\dot{S}_{i}\in\dot{\mathbb{Q}}_{i}$.

######  Proposition 6.6.

$\mathbb{U}_{n}$ is dense in $\mathbb{P}^{*}_{n}$.

###### Proof.

We argue by induction on $n$. By Proposition 2.4, $\mathbb{U}_{1}$ is dense in
$\mathbb{P}_{1}^{*}$ and every maximal antichain of $\mathbb{P}_{1}$ from $M$
remains maximal in $\mathbb{P}_{1}^{*}$. So let’s suppose inductively that
$\mathbb{U}_{n}$ is dense in $\mathbb{P}_{n}^{*}$ and every maximal antichain
of $\mathbb{P}_{n}$ from $M$ remains maximal in $\mathbb{P}_{n}^{*}$. This
argument is meant to take place simultaneously with the inductive proof of
Lemma 6.8, where we use the density of $\mathbb{U}_{n}$ to argue that maximal
antichains stay maximal.

Fix a condition
$(q_{0}^{\prime},\dot{q}_{1}^{\prime},\ldots,\dot{q}_{n}^{\prime})$ in
$\mathbb{P}_{n+1}^{*}$. In particular, we have
$(q_{0}^{\prime},\dot{q}_{1}^{\prime},\ldots,\dot{q}_{n-1}^{\prime})\Vdash\dot{q}_{n}^{\prime}\in\mathbb{Q}_{n}^{*}$.
Consider any forcing extension $V[H^{*}]$ by $\mathbb{P}_{n}^{*}$ with
$(q_{0}^{\prime},\dot{q}_{1}^{\prime},\ldots,\dot{q}_{n-1}^{\prime})\in
H^{*}$. In $V[H^{*}]$, there is $j_{n}<\omega$ and
$S\in(\dot{\mathbb{Q}}_{n})_{H^{*}}$ such that
$(\dot{\mathcal{T}}_{j_{n}}^{n})_{H^{*}}\wedge S\leq(q_{n}^{\prime})_{H^{*}}$.
So $S\in M[H]$, where $H$ is the restriction of $H^{*}$ to $\mathbb{P}_{n}$.
Let $\dot{S}$ be a $\mathbb{P}_{n}$-name in $M$ such that $(\dot{S})_{H}=S$
and let $p\in H$ force that $\dot{S}\in\dot{\mathbb{Q}}_{n}$. Let
$\mathcal{A}\in M$ be any maximal antichain of $\mathbb{P}_{n}$ extending
$\\{p\\}$ and let $\dot{S}_{n}$ be the mixed name such that
$p\Vdash\dot{S}=\dot{S}_{n}$ and every other $q\in\mathcal{A}$ forces that
$\dot{S}_{n}={}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$. So
$\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}_{n}}\Vdash\dot{S}_{n}\in\dot{\mathbb{Q}}_{n}$. Now let
$(q_{0},\dot{q}_{1},\ldots,\dot{q}_{n-1})\leq(q_{0}^{\prime},\dot{q}_{1}^{\prime},\ldots,\dot{q}_{n-1}^{\prime})$
be a condition in $H^{*}$ forcing that
$\dot{\mathcal{T}}^{n}_{j_{n}}\wedge\dot{S}_{n}\leq\dot{q}_{n}^{\prime}$. We
have just argued that there is a condition
$(q_{0},\dot{q}_{1},\ldots,\dot{q}_{n-1})\leq(q_{0}^{\prime},\dot{q}_{1}^{\prime},\ldots,\dot{q}_{n-1}^{\prime})$
and a $\mathbb{P}_{n}$-name $\dot{S}_{n}$ such that $\mathop{1\hskip-2.5pt{\rm
l}}_{\mathbb{P}_{n}}\Vdash\dot{S}_{n}\in\dot{\mathbb{Q}}_{n}$ and

$(q_{0},\dot{q}_{1},\ldots,\dot{q}_{n-1})\Vdash\dot{\mathcal{T}}^{n}_{j_{n}}\wedge\dot{S}_{n}\leq\dot{q}_{n}^{\prime}$

for some $j_{n}<\omega$.

Now, by the inductive assumption, there is a condition

$(p_{0},\dot{p}_{1},\ldots,\dot{p}_{n-1})\leq(q_{0},\dot{q}_{1},\ldots,\dot{q}_{n-1})$

in $\mathbb{U}_{n}$. So clearly the condition
$(p_{0},\dot{p}_{1},\ldots,\dot{p}_{n-1},\dot{\mathcal{T}}^{n}_{j_{n}}\wedge\dot{S}_{n})\in\mathbb{U}_{n+1}$.
∎

We will usually abuse notation by writing conditions in $\mathbb{U}_{n}$ in
the form

$(\mathcal{T}^{0}_{j_{0}}\wedge
S_{0},\dot{\mathcal{T}}^{1}_{j_{1}}\wedge\dot{S}_{1},\ldots,\dot{\mathcal{T}}^{n-1}_{j_{n-1}}\wedge\dot{S}_{n-1}).$

The next lemma is a generalization of Proposition 2.3.

######  Lemma 6.7.

Suppose $f_{X}\in\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$
forces that

$(\mathcal{T}^{0}_{j_{0}}\wedge
S_{0},\dot{\mathcal{T}}_{j_{1}}^{1}\wedge\dot{S}_{1},\ldots,\dot{\mathcal{T}}_{j_{n-1}}^{n-1}\wedge\dot{S}_{n-1})\in\mathbb{U}_{n}.$

Then there is a condition $g_{Y}\leq f_{X}$ such that $g_{Y}(\langle
j_{0},j_{1},\ldots,j_{n-1}\rangle)=(q,F)$ is determined and
$\tau:n\to{}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ lies on
$(q,F)$ such that

$q(0)_{\tau(0)}\leq S_{0}$, and for all $1\leq i<n$, $(q\upharpoonright
i)\mid(\tau\upharpoonright i)\Vdash q(i)_{\tau(i)}\leq\dot{S}_{i}$.

###### Proof.

The case $n=1$ follows by Proposition 2.3. So suppose $n=2$. Recall that every
maximal antichain of $\mathbb{P}_{1}$ from $M$ remains maximal in
$\mathbb{P}_{1}^{*}$. By strengthening $f_{X}$, we can assume that $\langle
j_{0},j_{1}\rangle\in X$. Let $f_{X}(\langle
j_{0},j_{1}\rangle)=(\bar{q},\bar{F})$. By strengthening further, using the
case $n=1$, we can assume that there is a node $s_{0}$ on level $\bar{F}(0)$
such that $\bar{q}(0)_{s_{0}}\leq S_{0}$. By strengthening some more, we can
assume that $(\bar{q},\bar{F})$ is determined.

Let’s argue that $\bar{q}(0)_{s_{0}}$ forces that for some node $s_{1}$ on
level $\bar{F}(1)$ of $\bar{q}(1)$,
$\bar{q}(1)_{s_{1}}\wedge\dot{S}_{1}\neq\emptyset$. If this is not the case,
then there is some $T\leq\bar{q}(0)_{s_{0}}$ which forces that there is no
such node. Let $T^{\prime}$ be the tree we get by replacing
$\bar{q}(0)_{s_{0}}$ with $T$ in $\bar{q}(0)$. Let
$\bar{q}^{\prime}=(T^{\prime},\bar{q}(1))$ and let $f^{\prime}_{X}$ be a
condition strengthening $f_{X}$ so that $f^{\prime}_{X}(\langle
j_{0},j_{1}\rangle)=(\bar{q}^{\prime},\bar{F})$. Let $G$ be an $M$-generic
filter for $\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$
containing $f^{\prime}_{X}$ and consider $M[G]$. The tree
$R=\mathcal{T}_{j_{0}}^{0}\wedge S_{0}\wedge\bar{q}^{\prime}(0)_{s_{0}}(=T)$
is a condition in $\mathbb{P}_{1}^{*}$, and so we can fix some $V$-generic
filter $H^{*}\subseteq\mathbb{P}^{*}_{1}$ containing $R$. In particular,
$H^{*}$ contains $\mathcal{T}_{j_{0}}^{0}\wedge S_{0}$ and $T$. Thus, in
$M[H]$ (where $H$ is the restriction of $H^{*}$ to $\mathbb{P}_{1}$),
$\bar{q}(1)_{H}$ does not have any subtree on level $F(1)$ which has a common
subtree with $(\dot{S}_{1})_{H}$. But then
$(\dot{\mathcal{T}}_{j_{1}}^{1})_{H^{*}}\wedge(\dot{S}_{1})_{H}=\emptyset$
contradicting our assumption that $f_{X}$ forced
$(\mathcal{T}_{j_{0}}^{0}\wedge
S_{0},\dot{\mathcal{T}}_{j_{1}}^{1}\wedge\dot{S}_{1})\in\mathbb{U}_{2}$.

Let’s strengthen $\bar{q}(0)_{s_{0}}$ to some $T_{0}$, so that there is a node
$s_{1}$ on level $\bar{F}(1)$ such that
$T_{0}\Vdash\bar{q}(1)_{s_{1}}\wedge\dot{S}_{1}\neq\emptyset$. Let
$\dot{T}_{1}$ be a $\mathbb{P}_{1}$-name such that

$T_{0}\Vdash\dot{T}_{1}=\bar{q}(1)_{s_{1}}\wedge\dot{S}_{1}.$

Since $(\bar{q},\bar{F})$ is determined, it follows that
$\bar{q}(0)_{s_{0}}\Vdash s_{1}\in\bar{q}(1)$. Thus,
$\tau=\\{(0,s_{0}),(1,s_{1})\\}$ lies on $(\bar{q},\bar{F})$ and
$(T_{0},\dot{T}_{1})\leq\bar{q}\mid\tau$. By Proposition 4.8, there is a
determined condition $(q,F)\leq(\bar{q},\bar{F})$ with $\rho$ lying on
$(\bar{q},\bar{F})$ and extending $\tau$ such that
$q\mid\rho\leq(T_{0},\dot{T}_{1})$. So finally, we strengthen $f_{X}$ to a
condition $g_{X}$ with $g_{X}(\langle j_{0},j_{1}\rangle)=(q,F)$ using
Proposition 6.2 (2).

Suppose inductively that the statement holds for $n$ and every maximal
antichain of $\mathbb{P}_{n}$ from $M$ remains maximal in
$\mathbb{P}_{n}^{*}$. Let’s argue that the statement holds for $n+1$. This
will basically be a generalization of the argument passing from $n=1$ to
$n=2$. By strengthening $f_{X}$, we can assume that $\langle
j_{0},\ldots,j_{n}\rangle\in X$ and that $f_{X}$ is determined. Let
$f_{X}(\langle j_{0},\ldots,j_{n}\rangle)=(\bar{q},\bar{F})$. By using our
inductive assumption for $n$ and strengthening further, we can assume that
$\tau:n\to{}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ lies on
$(\bar{q}\upharpoonright n,\bar{F}\upharpoonright n)$ such that

$q(0)_{\tau(0)}\leq S_{0}$, and for all $1\leq i<n$, $(q\upharpoonright
i)\mid(\tau\upharpoonright i)\Vdash q(i)_{\tau(i)}\leq\dot{S}_{i}$.

Let’s argue that $(\bar{q}\upharpoonright n)\mid\tau$ forces that for some
node $s_{n}$ on level $\bar{F}(n)$ of $\bar{q}(n)$,
$\bar{q}(n)_{s_{n}}\wedge\dot{S}_{n}\neq\emptyset$. If this is not the case,
then there is a condition $p\leq(\bar{q}\upharpoonright n)\mid\tau$ which
forces that there is no such node. By Proposition 4.8, there is a determined
condition $(p^{\prime},F^{\prime})\leq(\bar{q}\upharpoonright
n,F\upharpoonright n)$ with $\rho$ lying on $(p^{\prime},F^{\prime})$ and
extending $\tau$ such that $p^{\prime}\mid\rho\leq p$. Let
$\bar{q}^{\prime}=p^{\prime}\cup\\{(n,\bar{q}(n))\\}$ and
$\bar{F}^{\prime}=F^{\prime}\cup\\{(n,\bar{F}(n))\\}$. Let $f^{\prime}_{X}$ be
the strengthening of $f_{X}$ in which $f^{\prime}_{X}(\langle
j_{0},\ldots,j_{n}\rangle)=(\bar{q}^{\prime},\bar{F}^{\prime})$. Let $G$ be an
$M$-generic filter for $\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$
containing $f^{\prime}_{X}$ and consider $M[G]$. The condition

$R=(\mathcal{T}_{j_{0}}^{0}\wedge
S_{0}\wedge\bar{q}^{\prime}(0)_{\rho(0)},\ldots,\dot{T}_{j_{n-1}}^{n-1}\wedge\dot{S}_{n-1}\wedge\bar{q}^{\prime}(n-1)_{\rho(n-1)})$

is in $\mathbb{P}^{*}_{n}$ (where
$(\bar{q}^{\prime}(0)_{\rho(0)},\ldots,\bar{q}^{\prime}(n-1)_{\rho(n-1)})\leq
p$ by construction), and so we can fix a $V$-generic filter $H^{*}$ containing
$R$, and argue as in the case $n=2$ towards a contradiction.

Thus, there is a condition

$(T_{0},\dot{T}_{1},\ldots,\dot{T}_{n-1})\leq(\bar{q}\upharpoonright
n)\mid\tau,$

a $\mathbb{P}_{n}$-name $\dot{T}_{n}$ for a perfect tree, and a node $s_{n}$
on level $\bar{F}(n)$ such that

$(T_{0},\dot{T}_{1},\ldots,\dot{T}_{n-1})\Vdash\dot{T}_{n}=\bar{q}(n)_{s_{n}}\wedge\dot{S}_{n}.$

Let $\sigma=\tau\cup\\{(n,s_{n})\\}$. So
$(T_{0},\dot{T}_{1},\ldots,\dot{T}_{n-1},\dot{T}_{n})\leq\bar{q}\mid\sigma$,
and it remains to construct the required condition using Proposition 4.8. ∎

We are now ready to prove that every maximal antichain of $\mathbb{P}_{n}$
from $M$ remains maximal in $\mathbb{P}_{n}^{*}$.

######  Lemma 6.8.

Every maximal antichain of $\mathbb{P}_{n}$ from $M$ remains maximal in
$\mathbb{P}_{n}^{*}$.

###### Proof.

The statement is true for $n=1$. So we can assume inductively that
$\mathbb{U}_{n}$ is dense in $\mathbb{P}_{n}^{*}$. Let $\mathcal{A}\in M$ be a
maximal antichain in $\mathbb{P}_{n}$. It suffices to show that every
condition in $\mathbb{U}_{n}$ is compatible with an element of $\mathcal{A}$.
So let $f_{X}$ be some condition which forces that

$(\mathcal{T}_{j_{0}}^{0}\wedge
S_{0},\dot{T}^{1}_{j_{1}}\wedge\dot{S}_{1},\ldots,\dot{T}_{j_{n-1}}^{n-1}\wedge\dot{S}_{n-1})\in\mathbb{U}_{n}.$

By Proposition 6.7, we can strengthen $f_{X}$ to a condition $g_{Y}$ such that
$g_{Y}(\langle j_{0},\ldots,j_{n-1}\rangle)=(q,F)$ is determined and
$\tau:n\to{}^{{\mathrel{\mathchoice{\raise 2.0pt\hbox{$\scriptstyle<$}}{\raise
1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ lies on
$(q,F)$ such that $q(0)_{\tau(0)}\leq S_{0}$, and for all $1\leq i<n$,
$(q\upharpoonright i)\mid(\tau\upharpoonright i)\Vdash
q(i)_{\tau(i)}\leq\dot{S}_{i}$. Since $\mathcal{A}$ is maximal in
$\mathbb{P}_{n}$, $q\mid\tau$ is compatible with some $p\in\mathcal{A}$. So
let $q^{\prime}\leq q\mid\tau,p$. By Proposition 4.8, there is a determined
condition $(\bar{q},\bar{F})\leq(q,F)$ with $\rho$ lying on
$(\bar{q},\bar{F})$ and extending $\tau$ such that $\bar{q}\mid\rho\leq
q^{\prime}$. Strengthen $g_{Y}$ to $\bar{g}_{Y}$ with $(q,F)$ strengthened to
$(\bar{q},\bar{F})$ as in Proposition 6.2 (2). The condition $\bar{g}_{Y}$
forces that $(\mathcal{T}_{j_{0}}^{0}\wedge
S_{0},\dot{T}^{1}_{j_{1}}\wedge\dot{S}_{1},\ldots,\dot{T}_{j_{n-1}}^{n-1}\wedge\dot{S}_{n-1})$
is compatible with $p\in\mathcal{A}$. ∎

Let $\vec{P}^{*}=\langle\mathbb{P}_{n}^{*}\mid n<\omega\rangle$ be the
$\omega$-iteration made up of the extended iterations $\mathbb{P}_{n}^{*}$.
Let $\mathbb{U}$ be the subset of
$\mathbb{P}(\vec{P}^{*},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$
consisting of conditions $f_{X}$ such that for all $s\in X$ on level $n$,
$f_{X}(s)\in\mathbb{U}_{n}$.

######  Proposition 6.9.

$\mathbb{U}$ is dense in
$\mathbb{P}(\vec{P}^{*},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$.

Next, we will prove the analogue of Proposition 2.5.

######  Lemma 6.10.

Every maximal antichain of
$\mathbb{P}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ from
$M$ remains maximal in
$\mathbb{P}(\vec{P}^{*},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$.

###### Proof.

Fix a maximal antichain $\mathcal{A}\in M$ of
$\mathbb{P}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$. By
Proposition 6.9, it suffices to show that every condition $f_{X}\in\mathbb{U}$
is compatible with some element of $\mathcal{A}$. So fix $f_{X}\in\mathbb{U}$.
For a node $s\in X$ on level $n$ of ${}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega$, let

$f_{X}(s)=(\mathcal{T}_{j_{0,s}}^{0}\wedge
S_{0,s},\ldots,\dot{\mathcal{T}}_{j_{n-1,s}}^{n-1}\wedge\dot{S}_{n-1,s}).$

Let $\langle j_{0,s},\ldots,j_{n-1,s}\rangle=\vec{j}_{s}$, and note that we
can have $\vec{j}_{s}=\vec{j}_{s^{\prime}}$ for $s^{\prime}\neq s$. Fix a
condition $g_{Y}\in G$ forcing that $f_{X}\in\mathbb{U}$. Repeatedly using the
construction in the proof of Lemma 6.7 on terminal nodes, we can find a
determined condition $\bar{g}_{Y}\leq g_{Y}$ with
$\bar{g}_{Y}(\vec{j}_{s})=(\bar{q}_{\vec{j}_{s}},\bar{F}_{\vec{j}_{s}})$ such
that for every node $s\in X$, some
$\tau_{s}:n\to{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}2$ lies on
$(\bar{q}_{\vec{j}_{s}},\bar{F}_{\vec{j}_{s}})$ with

$\bar{q}_{\vec{j}_{s}}(0)_{\tau_{s}(0)}\leq S_{0,s}$, and for all $1\leq i<n$,
$(\bar{q}_{\vec{j}_{s}}\upharpoonright i)\mid(\tau_{s}\upharpoonright
i)\Vdash\bar{q}_{\vec{j}_{s}}(i)_{\tau_{s}(i)}\leq\dot{S}_{i,s}$.

The construction gives $\tau_{s}$ satisfying that if $s^{\prime}$ extends $s$,
then $\tau_{s^{\prime}}$ extends $\tau_{s}$.

Let $\bar{f}_{X}$ be a condition in
$\mathbb{P}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ such
that for every node $s\in X$, we have
$\bar{f}_{X}(s)=\bar{q}_{\vec{j}_{s}}\mid\tau_{s}$. Since $\mathcal{A}$ is
maximal in $\mathbb{P}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$,
there is $a_{Z}\in\mathcal{A}$ compatible with $\bar{f}_{X}$. So let
$r_{W}\leq a_{Z},\bar{f}_{X}$.

We then carry out the construction from the proof of Proposition 4.8, working
our way up the tree instead of up the iteration, to obtain a condition
$h_{Y}\leq\bar{g}_{Y}$ with
$h_{Y}(\vec{j}_{s})=(q_{\vec{j}_{s}},F_{\vec{j}_{s}})$ such that for every
$s\in X$, some $\sigma$ lies on $(q_{\vec{j}_{s}},F_{\vec{j}_{s}})$ having
$q_{\vec{j}_{s}}\mid\sigma\leq r_{W}(s)$. The condition $h_{Y}$ forces that
$f_{X}$ and $a_{Z}$ are compatible. By density, some such $h_{Y}$ must be in
$G$. ∎

We now summarize our results in the following lemma, which will serve as the
analogue of Proposition 2.4.

######  Lemma 6.11.

Suppose $M$ is a suitable model and $\vec{P}=\langle\mathbb{P}_{n}\mid
n<\omega\rangle\in M$ is an $\omega$-iteration of perfect posets. If
$G\subseteq\mathbb{Q}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ is
$M$-generic and $\vec{P}^{*}=\langle\mathbb{P}_{n}^{*}\mid n<\omega\rangle$ is
constructed in $M[G]$ as above, then

  1. (1)

$\vec{P}^{*}$ is an $\omega$-iteration of perfect posets,

  2. (2)

$\mathbb{Q}_{0}\subseteq\mathbb{Q}_{0}^{*}$ and for all $n<\omega$,
$\mathop{1\hskip-2.5pt{\rm l}}_{\mathbb{P}_{n}^{*}}$ forces that
$\dot{\mathbb{Q}}_{n}$ is a perfect poset and $\dot{\mathbb{Q}}_{n}^{*}$
extends it,

  3. (3)

$\mathbb{P}_{n}\subseteq\mathbb{P}_{n}^{*}$,

  4. (4)

every maximal antichain of $\mathbb{P}_{n}$ from $M$ remains maximal in
$\mathbb{P}_{n}^{*}$,

  5. (5)

every maximal antichain of
$\mathbb{P}(\vec{P},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ from
$M$ remains maximal in
$\mathbb{P}(\vec{P}^{*},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$.

Now we will define an $\omega$-iteration
$\vec{P}^{J}=\langle\mathbb{P}_{n}^{J}\mid n<\omega\rangle$ of perfect posets
such that the tree iteration
$\mathbb{P}(\vec{P}^{J},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ is
going to have the property that in a forcing extension $L[G]$ of $L$ by
$\mathbb{P}(\vec{P}^{J},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$, the
only $L$-generic filters for the iteration $\mathbb{P}_{n}^{J}$ are the
restrictions of $G$ to a level $n$ node. This will be the generalization of
Jensen’s uniqueness of generic filters property to tree iterations.

Let $\langle S_{\alpha}\mid\alpha<\omega_{1}\rangle$ be a canonically defined
$\diamondsuit$-sequence. We will construct $\vec{P}^{J}$ in $\omega_{1}$-many
steps using $\diamondsuit$ to seal maximal antichains along the way. Let
$\vec{P}_{0}$ be the $\omega$-iteration of perfect posets where
$\mathbb{Q}_{0}=\mathbb{P}_{\text{min}}$ and each
$\dot{\mathbb{Q}}_{n}=\check{\mathbb{P}}_{\text{min}}$. Suppose the
$\omega$-iteration $\vec{P}_{\alpha}=\langle\mathbb{P}_{n}^{\alpha}\mid
n<\omega\rangle$ of perfect poset has been defined. We let
$\vec{P}_{\alpha}=\vec{P}_{\alpha+1}$, unless the following happens. Suppose
$S_{\alpha}$ codes a well-founded binary relation
$E\subseteq\alpha\times\alpha$ such that the collapse of $E$ is a suitable
model $M_{\alpha}$ with $\vec{P}_{\alpha}\in M_{\alpha}$ and
$\alpha=\omega_{1}^{M_{\alpha}}$. In this case, we take the $L$-least
$M_{\alpha}$-generic filter
$G\subseteq\mathbb{Q}(\vec{P}_{\alpha},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ and
let $\vec{P}_{\alpha+1}=\vec{P}_{\alpha}^{*}$ as constructed in
$M_{\alpha}[G]$. At limit stages $\lambda$, to obtain the $\omega$-iteration
$\vec{P}_{\lambda}$, we let $\mathbb{Q}^{\lambda}_{0}$ be the union of the
$\mathbb{Q}^{\xi}_{0}$ for $\xi<\lambda$, and given that we have defined
$\mathbb{P}_{n}^{\lambda}$, we let $\dot{\mathbb{Q}}_{n}^{\lambda}$ be a
$\mathbb{P}^{\lambda}_{n}$-name for the poset that is the union of the
$\dot{\mathbb{Q}}_{n}^{\xi}$ for $\xi<\lambda$. In order for this limit
definition to make sense, we need to verify that each
$\dot{\mathbb{Q}}^{\xi}_{n}$ is a $\mathbb{P}^{\lambda}_{n}$-name for a
perfect poset. So let’s argue that this is indeed the case.

Clearly $\mathbb{P}_{1}^{\lambda}$ makes sense,
$\mathbb{P}_{1}^{\xi}\subseteq\mathbb{P}_{1}^{\lambda}$ for every
$\xi<\lambda$, and every maximal antichain of $\mathbb{P}_{1}^{\xi}$ from
$M_{\xi}$ remains maximal in $\mathbb{P}_{1}^{\lambda}$. So we can assume
inductively that we have defined $\mathbb{P}_{n}^{\lambda}$ so that
$\mathbb{P}_{n}^{\xi}\subseteq\mathbb{P}_{n}^{\lambda}$ for every
$\xi<\lambda$, and every maximal antichain of $\mathbb{P}_{n}^{\xi}$ in
$M_{\xi}$ remains maximal in $\mathbb{P}_{n}^{\lambda}$. Let
$H^{*}\subseteq\mathbb{P}_{n}^{\lambda}$ be $V$-generic. By our assumption,
the filter $H^{*}$ restricts to an $M_{\xi}$-generic filter $H$ for
$\mathbb{P}^{\xi}_{n}$. The model $M_{\xi}[H]$ satisfies that
$(\dot{\mathbb{Q}}^{\xi}_{n})_{H}$ is a perfect poset and so this must be the
case in $V[H^{*}]$ as well since this is absolute. So we can extend the
definition to $\mathbb{P}_{n+1}^{\lambda}$. It is easy to see that
$\mathbb{P}_{n+1}^{\xi}\subseteq\mathbb{P}_{n+1}^{\lambda}$. It remains to
argue that every maximal antichain $\mathcal{A}\in M_{\xi}$ of
$\mathbb{P}_{n+1}^{\xi}$ remains maximal in $\mathbb{P}_{n+1}^{\lambda}$. Fix
$p\in\mathbb{P}_{n+1}^{\lambda}$. We can assume inductively that we have
already showed for every $\mu<\nu<\lambda$, that every maximal antichain of
$\mathbb{P}_{n+1}^{\mu}$ in $M_{\mu}$ remains maximal in
$\mathbb{P}_{n+1}^{\nu}$. Then, by definition of $\mathbb{P}^{\lambda}_{n+1}$,
there is some $\xi<\eta<\lambda$ such that $p(0)\in\mathbb{Q}_{0}^{\eta}$, and
$p\upharpoonright i\Vdash p(i)\in\dot{\mathbb{Q}}^{\eta}_{i}$ for $1\leq i<n$.
Since $\mathcal{A}$ is maximal in $\mathbb{P}^{\eta}_{n+1}$, it follows that
there is some $q\in\mathcal{A}$ that is compatible with
$p\in\mathbb{P}^{\eta}_{n+1}\subseteq\mathbb{P}^{\lambda}_{n+1}$. In
particular, we have just shown the following.

######  Lemma 6.12.

If $\lambda$ is a limit ordinal and $\xi<\lambda$, then
$\mathbb{P}_{n}^{\xi}\subseteq\mathbb{P}_{n}^{\lambda}$ and every maximal
antichain of $\mathbb{P}_{n}^{\xi}$ from $M_{\xi}$ remains maximal in
$\mathbb{P}_{n}^{\lambda}$.

######  Lemma 6.13.

If $\alpha<\beta<\omega_{1}$, then every maximal antichain of
$\mathbb{P}(\vec{P}_{\alpha},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ from
$M_{\alpha}$ remains maximal in
$\mathbb{P}(\vec{P}_{\beta},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$.

The proof uses Lemma 6.8 for successor stages and follows easily for limit
stages.

For $n<\omega$, let $\mathbb{P}_{n}^{J}=\mathbb{P}_{n}^{\omega_{1}}$ be
constructed as all other limit stages and let
$\vec{\mathbb{P}}^{J}=\langle\mathbb{P}_{n}^{J}\mid n<\omega\rangle$ be the
corresponding $\omega$-iteration.

It will be useful for future arguments to assume that conditions in
$\mathbb{P}_{n}^{J}$ are coded by reals.

######  Theorem 6.14.

The poset $\mathbb{P}(\vec{P}^{J},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$ has
the ccc.

###### Proof.

Fix a maximal antichain $\mathcal{A}$ of
$\mathbb{P}(\vec{P}^{J},{}^{{\mathrel{\mathchoice{\raise
2.0pt\hbox{$\scriptstyle<$}}{\raise 1.0pt\hbox{$\scriptstyle<$}}{\raise
0.0pt\hbox{$\scriptscriptstyle<$}}{\scriptscriptstyle<}}}\omega}\omega)$.
Choose some transitive $M\prec L_{\omega_{2}}$ of size $\omega_{1}$ with
$\mathcal{A}\in M$. We can decompose $M$ as the union of an elementary chain
of countable substructures

$X_{0}\prec X_{1}\prec\cdots\prec X_{\alpha}\prec\cdots\prec M$

with $\mathcal{A}\in X_{0}$. By the properties of $\diamondsuit$, there is
some $\alpha$ such that $\alpha=\omega_{1}\cap{X_{\alpha}}$,