This paper constructs models of ZFC where all inner models with the Axiom of Choice are well-ordered by inclusion, with the order's height being arbitrarily large, using iterated κ-Sacks forcing.
Contribution
It introduces a method to create models of ZFC with a well-ordered hierarchy of inner models satisfying AC, extending the possible height of this hierarchy to any ordinal.
Findings
01
Inner models satisfying AC are well-ordered by inclusion
02
The height of the well-ordering can be arbitrarily large
03
Iterated κ-Sacks forcing does not produce unexpected intermediate models
Abstract
In this paper we show how to build a model of ZFC such that all its inner models satisfying the Axiom of Choice are well-ordered with respect to inclusion, and that said ordering is of arbitrary height (including possibly Ord high). We do this by iterating κ-Sacks forcing for ever-increasing κ, while showing that such forcings do not add any unexpected intermediate inner models.
Equations20
∃M∃I (M⊆I×N∧(M⊊N is an inner model ↔∃!a∈I(M={x∣(a,x)∈M}))∧I is well-ordered ∧a<Ib→((a,x)∈M→(b,x)∈M))
∃M∃I (M⊆I×N∧(M⊊N is an inner model ↔∃!a∈I(M={x∣(a,x)∈M}))∧I is well-ordered ∧a<Ib→((a,x)∈M→(b,x)∈M))
L0[A]
L0[A]
Lα+1[A]
Lδ[A]
L[A]
L0(A)
L0(A)
Lα+1(A)
Lδ(A)
L(A)
p(β)≤β⊩Pβ+1⟨p˙ωβ+1β+1,p˙ωβ+1β+2,...,p˙ωβ+1η,...∣η<δ⟩ is as we constructed it.
p(β)≤β⊩Pβ+1⟨p˙ωβ+1β+1,p˙ωβ+1β+2,...,p˙ωβ+1η,...∣η<δ⟩ is as we constructed it.
g˙=⟨g˙β∣β<δ⟩∧∀β<δ(β is not a limit→(g˙β⊊ℵβ∧g˙β∈M˙β+1∖M˙β))
g˙=⟨g˙β∣β<δ⟩∧∀β<δ(β is not a limit→(g˙β⊊ℵβ∧g˙β∈M˙β+1∖M˙β))
∃M∃I (M⊆I×N∧(M⊊N is an inner ZF model ↔∃!a∈I(M={x∣(a,x)∈M}))∧I is well-ordered ∧a<Ib→((a,x)∈M→(b,x)∈M))
∃M∃I (M⊆I×N∧(M⊊N is an inner ZF model ↔∃!a∈I(M={x∣(a,x)∈M}))∧I is well-ordered ∧a<Ib→((a,x)∈M→(b,x)∈M))
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TopicsAdvanced Topology and Set Theory · Computability, Logic, AI Algorithms · Mathematical and Theoretical Analysis
Full text
Well-Ordered Model Universes
Alon Navon
Tel Aviv University
Abstract
In this paper we show how to build a model of ZFC such
that all its inner models satisfying the Axiom of Choice are well-ordered
with respect to inclusion, and that said ordering is of arbitrary
height (including possibly Ord high). We do this by iterating
κ-Sacks forcing for ever-increasing κ, while showing
that such forcings do not add any unexpected intermediate inner models.
1 Introduction
In this section we aim to present and formalize the concept of a well-ordered
model universe, and establish the proper axiomatic framework to work
with it. We start with an informal presentation to explain the motivation
behind this idea.
Definition 1.1**.**
Let M be a model of ZFC (ZF).
We call a substructure N of M an *inner model of ZFC
*(ZF) if:
N is a model of ZFC (ZF);
2. 2.
The interpretation of ∈N is ∈M∩N2;
3. 3.
The domain of N is a transitive class of M;
4. 4.
N has the same ordinals as M.
A model that satisfies only conditions 1 and 2
is called a standard model. Hence an inner model is a standard
transitive model that has the same ordinals as the base model.
Unless otherwise stated, we reserve the term *inner model *to
refer exclusively to inner models of ZFC. If we want to
discuss an inner model of ZF we shall refer to it explicitly
as such.
It was Kurt Gᅵdel that proved in [8] that any model of
ZF has a least inner model L, called the constructible
universe, which is also a model of ZFC+GCH. Thus, if V=L,
there are non-trivial inner models of V, and we can then partially
order them with respect to inclusion. This partial order has a unique
least element L and a unique greatest element V, and we may
naturally enquire about its other order-theoretic properties. The
aim of this paper is to explore the consistency and implications of
the well-ordering property for this partial order, our main theorem
being the construction of a model of ZFC where the inner
models are not only well-ordered with respect to inclusion, but said
ordering is in fact order-isomorphic to all ordinals.
In practice, this means we construct a model M of ZFC
and a “sequence” of inner models ⟨Mα∣α∈Ord⟩
such that
α<β if and only if Mα⊊Mβ;
2. 2.
For every inner model N of M there is an α∈Ord
such that N=Mα.
This is a sort of tower of inner models, and to construct it we will
need to gradually extend the tower from its base, passing through
the successor case, singular limits, regular limits and finally the
class case. That will be the general path we follow, but first we
need to address a tricky part of our definition - in ZF
we cannot formally talk about a sequence of classes, and each inner
model of V is by definition a proper class. Normally, it is enough
to use the class notation as a shorthand for formulas and ignore the
particulars, but in this paper we’ll also be interested in the interplay
between these classes. So in order to deal with these explicitly,
we turn to Bernays-Gᅵdel set theory, in short BG (or BGC
if we add Global Choice), to serve as our axiomatic framework.
Bernays-Gᅵdel set theory, sometimes known as Von Neumann–Bernays–Gᅵdel
set theory (NBG for short), has its origins in a 1925
paper by John von Neumann [13], which formally introduced
classes into set theory for the first time. Von Neumann’s theory employed
functions and arguments as its primitive notions, and used them to
define sets and classes. However, in the 1930s Paul Bernays reformulated
the theory by taking classes and sets as the primitive notions [2].
Later, while working on his proof for the relative consistency of
the Axiom of Choice [8], Gᅵdel significantly simplified
Bernay’s theory, leading to what is now known as Bernays-Gᅵdel set
theory.
Unlike Zermelo-Fraenkel, Bernays-Gᅵdel set theory allows for two types
of objects: classes and sets. Every set is also considered a class,
and if a class is a member of another class then it is also a set.
Thus, a model ⟨V,V,∈⟩ of BG
consists of a collection V of classes together with a
subcollection V⊊V of sets and a relation ∈⊊V×V.
The axioms of this system are mostly very similar to those of ZF,
and a formal exposition of them and their application in class forcing
can be found in [15]. To avoid any ambiguity, we always
denote a model of BG in the form of a ⟨V,V,∈⟩
triplet, and a model of ZF as plain V.
In order to justify our use of BG, we quote the following
facts about the link between it and ZF.
Fact 1.2**.**
BG* (BGC) is a conservative extension of ZF
(ZFC).*
This fact was proven by Paul Cohen in [3]. Furthermore,
Mostowski [12] showed that every set-theoretical statement
provable in ZF (ZFC) is provable in BG
(BGC), and that if a sentence involving only set variables
is provable in BG (BGC) then it is provable
in ZF (ZFC) as well.
Fact 1.3**.**
Let ⟨V,V,∈⟩
be a model of BG (BGC), where V
is the collection of classes and V⊆V is the collection
of sets. Then by taking V and ∈∩(V×V)
we get a model of ZF (ZFC).
So if we have a model of BG, by “throwing away” the
classes, we are left with a model of ZF. What about the
other way round?
Fact 1.4**.**
Let V be a model of ZF. Take V
to be the collection of all classes definable in V with set parameters,
and take ∈ to be the obvious extension of the membership relation
to V. Then ⟨V,V,∈⟩
is a model of BG.
Note that even if V is a model of ZFC then ⟨V,V,∈⟩
as defined above might only satisfy BG, not BGC.
But using class forcing one can add a uniform choice function that
is κ-closed for each κ, and so adds no new sets to
the universe. The resulting model ⟨V,V′,∈⟩
extends ⟨V,V,∈⟩, satisfies
BGC and has the same sets and the same restriction of
∈ to sets as ⟨V,V,∈⟩
(see A.1 in [15]).
Facts 1.3 and 1.4 establish a useful
correspondance between models of ZF and BG.
Corollary 1.5**.**
M* is a definable proper inner model of V if
and only if M=V and ⟨M,M,∈⟩⊊⟨V,V,∈⟩
(as defined in fact 1.4).*
Proof.
(⇒)M is proper and so M=V. M
is a definable class of V, therefore M∈V, and also
every class definable in M with set parameters is similarly definable
in V. Therefore M⊊V and ⟨M,M,∈⟩⊊⟨V,V,∈⟩.
(⇐)M∈M and so M∈V,
so M is a definable class in V. M⊊V and according
to fact 1.3 is a model of ZF.
∎
But moving to BG doesn’t quite work out all the kinks.
In particular, BG doesn’t allow for class membership within
another class. Therefore in order to speak of a sequence of classes
we need to abandon our standard definition of a sequence, and instead
use an alternative definition that is also suitable for classes.
Definition 1.6**.**
Let I be a class. Then we call S⊆I×V
an I-indexed family of classes, and for each i∈I we denote
Si={x∣(i,x)∈S}. If I is well-ordered,
then we say S is a sequence, and I is its underlying
order.
Thus, instead of having a ’family of classes’ which each class is
a member of, each enumerated class is generated using straightforward
class comprehension. This perspective allows us to speak of sequences
of classes within BG.
Now that we’ve set up this notion, we finally come to the primary
definition of this article:
Definition 1.7**.**
Let ⟨V,V,∈⟩
be a model of BG. We call a model N⊆V of ZFC
where all its inner models are well-ordered with respect to inclusion
a well-ordered model universe. Formally, we postulate the existence
of a class M in ⟨V,V,∈⟩,
which is the sequence of all proper inner models of N ordered by
inclusion. This means:
M⊆I×N;
2. 2.
M is a proper inner model of N if and
only if there exists a unique a∈I such that M=Ma={x∣(a,x)∈M};
3. 3.
I is a well-ordered class;
4. 4.
If a<Ib then (a,x)∈M→(b,x)∈M.
In summary, applying the convention that lower-case letters indicate
sets and upper-case letters indicates classes, we demand the following
be true:
[TABLE]
Where there’s any ambiguity about the base model we denote class M
as M(N).
A few important remarks are in order.
Remark 1.8*.*
Class M(N) belongs
by definition to ⟨V,V,∈⟩.
It need not be definable in N, nor even in V. However in the
example that we build later in the article, M(V)
actually will be definable in V, and then instead of working
with some background model of BG we will take V
to be the collection of all classes definable in V with set parameters,
exactly as we did in fact 1.4.
Remark 1.9*.*
Our demand that M be a proper inner model is superfluous,
and only used to simplify discussion of the case I=Ord,
when all the proper inner models are in a bijection with the ordinals.
This convention allows us to prove general theorems about all models
Mαα∈Ord, without having to constantly
special-case “MOrd”.
Remark 1.10*.*
It is natural to ask why when defining the sequence we only demand
I be well-ordered, instead of being equal to some ordinal or Ord.
The reason for this is that we also can also consider sequences that
are longer than the ordinals, and we don’t want to unnecessarily
exclude them from the definition. We revisit this issue in the last
section of this article, but in the meanwhile we define what it means
for a well-ordered model universe to be nice.
Definition 1.11**.**
We call a well-ordered model universe nice
if the underlying order of M is equal to some ordinal
or to Ord.
In essence, a well-ordered model universe is nice if its model tower
isn’t ’too tall’. This restriction is not superficial. We shall later
see an interesting property that fails if the well-ordered model universe
isn’t nice.
Definition 1.12**.**
Let N be a nice well-ordered model universe. We define its height
to be the order-type of the underlying order of M, so
ht(N)=otp(I). In case I=Ord,
we instead define ht(N)=∞. Note that for convenience,
we designate Mht(N)=N, even though it is not
formally part of the sequence of proper inner models.
To summarize our notational conventions, throughout this article:
Models of BG are always denoted as a triplet ⟨V,V,∈⟩,
whereas models of ZF are denoted using plain letters V.
2. 2.
M refers exclusively to the sequence of proper inner models
as defined in 1.7.
3. 3.
Mα shall refer to the αth inner model of sequence
M.
4. 4.
The height of a well-ordered model universe, denoted ht(V),
is the order-type of the underlying order of M.
5. 5.
A well-ordered model universe is considered nice if the underlying
order of M isn’t longer than Ord.
2 Implications
Now, it is time to explore some of the implications of the inner model
well-ordering property. For the rest of this section we assume ⟨V,V,∈⟩⊨BG,
V is a well-ordered model universe, and M⊆I×V
is its sequence of proper inner models ordered by inclusion.
Lemma 2.1**.**
M0=L.
Proof.
We know from Gᅵdel [8] that L is the least inner model
of V. Therefore, in order to be included in our hierarchy, we must
have M0=L.
∎
This begs the question of how ’close’ are V and L, assuming
our well-ordered model universe exists. One quick observation, resulting
directly from the well-ordering of the inner models, is that V
cannot contain a measurable cardinal.
Theorem 2.2**.**
V⊨There is no measurable cardinal.
Proof.
Suppose to the contrary, that there is a measurable cardinal κ∈V.
Then there exists an elementary embedding j:V→M, where
M is an inner model of V [17]. Therefore M=Mα
for some α∈I. But because the embedding is elementary,
j(κ) is measurable in Mα, so there exists
an additional elementary embedding j(j):Mα→N.
But this means there is an elementary embedding V→N,
and so N is itself an inner model of V as well, so N=Mβ
for some β∈I. However, N⊊Mα, and therefore
β<Iα.
By induction, we can repeat this process and construct an infinite
descending chain of inner models. But V is a well-ordered model
universe, so the inner models are well-ordered and this is impossible.
Therefore, there is no measurable cardinal in V.
∎
However, not only are there no measurable cardinals in a well-ordered
model universe, but we can further show it has no 0♯ as
well, although this is a bit less straightforward.
Theorem 2.3**.**
V⊨0♯ does not exist.
Proof.
Suppose to the contary, that 0♯ does exist. Then every
uncountable cardinal in V is inaccessible in L (see corollary
18.3 in [9]).
Now remember Cohen forcing [3], where we use finite partial
functions, and define P=Fin(ω,2)L.
In L we have ∣P∣=ℵ0, and therefore
∣P(P)∣=2ℵ0=ℵ1,
which is obviously smaller than the first inaccessible cardinal. Hence
in V we have ∣P(P)∣=ℵ0,
meaning there are at most countable many dense subsets of P,
and therefore by the Rasiowa-Sikorski lemma [14] there
exists a generic set G∈V∖L that intersects them all.
Hence V⊨∃G(L[G]⊋L∧L[G]⊨ZFC).
Therefore L[G] is an inner model of V.
However each Cohen forcing is isomorphic to the product of two separate
Cohen forcings. Namely, for each I0⊊I we have Fin(I,2)≅Fin(I0,2)×Fin(I∖I0,2)
(see Kunen [11] ch. VIII 2.1). So take I=ω and
I0 the set of even natural numbers. According to the theorem
G0=G∩Fin(I0,2) is Fin(I0,2)-generic
over L, G1=G∩Fin(ω∖I0,2)
is Fin(ω∖I0,2)-generic over
L[G0], and L[G]=L[G0][G1].
So G1∈/L[G0], and for the same reasoning
we have G0∈/L[G1]. But as both L[G0]
and L[G1] are inner models of L[G]
and so of V, either L[G0]⊊L[G1]
or L[G1]⊊L[G0]. Either way
we arrive at a contradiction. Therefore 0♯ does not exist.
∎
Now that we know V cannot be too far off L, we may wonder if
there is perhaps a deeper connection between the two. For this we
turn to the notion of relative constructibility.
Note that there is a lot of confusion regarding its notation, so we
shall now present the notation used by Jech in [9] and
which we adhere to.
Constructibility can be generalized in two different ways. One way
is to consider sets constructive relative to a given set A, resulting
in the inner model L[A].
This is done by defining defA(M)={X⊆M∣X is definable over (M,∈,A∩M)},
where A∩M is a unary predicate, and then defining a cumulative
hierarchy:
[TABLE]
The resulting model L[A] is a model of ZFC
(see ch. 13 in [9]).
Another way, yields for every set A the smallest inner model of
ZF that contains it. However, in general, this model need
not satisfy the Axiom of Choice.
Let T=TC({A}) be the transitive closure
of A, and define the following cumulative hierarchy:
[TABLE]
The resulting model L(A) is an inner model of ZF,
contains A, and is the smallest such model.
Theorem 2.4**.**
Let V be a nice well-ordered model
universe. For all α∈IMα=L[A] for
some A∈V.
Proof.
We prove this theorem by induction on I. It is trivially true for
M0=L=L[∅].
For the successor stage, note that Mα⊨AC.
By a theorem of Vopěnka [19], this means there exists
a set of ordinals A∈Mα+1∖Mα. Thus Mα+1⊇L(A)⊋Mα.
However, note that for sets of ordinals L(A)=L[A],
because A⊊L. Therefore Mα+1⊇L[A]⊋Mα.
However we know there is no model of ZFC strictly between
Mα and Mα+1. Therefore Mα+1=L[A].
We turn to the limit stage. Let δ be a limit ordinal, and
assume the theorem was proven for all β<δ. We want to
show that even working in Mδ we can enumerate all the inner
models preceding it on the model tower. Still working in V, for
every β<δ there exists a set of ordinals Cβ
such that Mβ=L[Cβ] (see ex. 13.27 in [9]).
Cβ∈Mβ and so Cβ∈Mδ, meaning
L[Cβ]Mδ=L[Cβ]=Mβ
and therefore Mβ is a definable with set parameters in Mδ.
So, working in ⟨Mδ,Mδ,∈⟩
as defined using fact 1.4, for each β<δMβ∈Mδ.
Also, note that any inner model of the form L[A] where
A∈Mδ is definable using set parameters in V as well,
and therefore is equal to Mβ for some β<δ.
Despite having each individual model definable with set parameters
in Mδ, we still can’t be sure we can actually enumerate
all the inner models preceding Mδ within ⟨Mδ,Mδ,∈⟩.
So next we define an eqivalence relation on sets of ordinals: A∼B⇔L[A]=L[B].
For each equivalence class [A] let r[A]
be the sets of minimal rank in [A]. Obviously for each
Ar[A] is a set, and by the induction hypothesis
and the note above there are at most δ different models of
the form L[A]. So {r[A]∣A∈V}
is a set of sets, and using the Axiom of Choice we can choose a representative
from each r[A].
Next, because all of said models are equal to some Mβ on
the chain, we can sort the representatives according to the binary
relation A≤B⇔A∈L[B]. The models
are well-ordered because as noted above they all belong on the tower.
We also already established that each one is definable with set parameters
in Mδ, so what we get is a sequence of representatives
⟨Aβ∣β<δ⟩, which completely
enumerates the Mβ’s for all β<δ, and which is
defined using set parameters within Mδ.
We define inductively two sequences ⟨Bβ∣β<δ⟩
and ⟨γβ∣β<δ⟩.
Let B0=∅, γ0=0. For each β<δ
define γβ=sup(α<β⋃Bα)
and Bβ={γβ+ϵ∣ϵ∈Aβ}.
It is clear by the definitions that γ is strictly monotonously
increasing, and that all the B’s are mutually pairwise disjoint.
Let Bδ=β<δ⋃Bβ. Clearly
Bδ is a set of ordinals. We claim L[Bδ]=Mδ.
First note that for each β<δBβ=(Bδ∩γβ+1)∖γβ
and Aβ={ϵ∣γβ+ϵ∈Bβ}.
Therefore Aβ∈L[Bδ] and so Mβ=L[Aβ]⊆L[Bδ].
Therefore L[Bδ]⊇Mδ.
On the other hand, we’ve already shown that ⟨Aβ∣β<δ⟩∈Mδ.
So ⟨Bβ∣β<δ⟩∈Mδ
and therefore Bδ∈Mδ, implying L[Bδ]=L(Bδ)⊆Mδ.
We conclude that L[Bδ]=Mδ, and so the
induction is complete.
∎
It is instructive to note that we used the niceness property exactly
once, to justify how we could simultaneously choose a representative
from each r[A]. To do this for class-many sets would
have required the Axiom of Global Choice (see [4]), which
as noted could be false in ⟨Mδ,Mδ,∈⟩.
Moreover, if the underlying order was longer than Ord,
this proof would fail because “γOrd” would
be undefinable, as γ is a strictly increasing sequence of
ordinals.
Corollary 2.5**.**
If ht(V)<∞ then V=L[A] for some A∈V.
Proof.
Use the proof above, only substitute Mht(V) for V.
∎
Corollary 2.6**.**
If ⟨V,V,∈⟩⊨BGC
and I=Ord then V=L[A] for some class A⊆Ord.
Proof.
Using Global Choice, we can choose in the limit stage class-many representatives
from all the r[A]’s simultaneously. Then we take Bδ=α∈Ord⋃Bα.
By the same arguments as in the theorem, for all α∈OrdL[Bδ]⊇L[Bα]. But
this means L[Bδ] contains all the proper inner
models, hence L[Bδ]=V.
∎
It now emerges that the models in our tower are not arbitrary at all.
They are in fact the very familiar models of the form L[A].
We thus conclude that a nice well-ordered model universe V is inherently
quite ’small’ and ’close’ to L, especially if ht(V)<∞.
Before proceeding to the next section, it is worth noting what would
happen if instead of basing our model tower on L, we would base
it on some arbitrary inner model M. Obviously M wouldn’t
be well-ordered anymore, so we would have to relax our definition.
We will only require that all inner models containing M be on a
well-ordered chain, and that all other inner models be contained in
M. So below M everything could be completely chaotic, but above
M we would have a well-ordered tower. Now let’s consider the implications.
First of all, as for theorem 2.2, this alteration
potentially allows for an infinite descending chain of models. So
let’s assume that V does have a measurable cardinal and j:V→N
is the corresponding elementary embedding. Then N must also contain
a measurable cardinal, and repeating this process, due to the well-ordering
we arrive at a model N0⊊M after a finite number of
steps. Thus there is an elementary embedding k:V→N0,
and so k↾M is an elementary embedding of M into
some smaller inner model. Therefore M must also include a measurable
cardinal.
Theorem 2.4 would still work as well, using
M as the base for the induction. Accordingly, corollary 2.6
would still hold up as well.
After analyzing the structure of well-ordered model universes, we
turn to the problem of constructing one of arbitrary height.
3 Perfect set forcing
In lemma 2.1 we proved the base of our model tower
is L. In this section we show how to build the first step in our
tower. Unlike the previous section, from here on we only assume that
we’re working within a model of ZFC, not BG.
Also note that throughout this paper we follow the Israeli convention
for forcing, i.e if p>q are forcing conditions, then p is the
stronger condition.
To construct the first floor in the tower, we call upon the notion
of Sacks forcing [16], first invented by Gerald
Sacks, which is useful for creating minimal generic extensions. In
this section we present the original Sacks forcing and some of its
most important properties.
We assume the reader has a basic understanding of forcing. For a general
introduction to the technique of forcing, the reader may refer to
ch. VII of Kunen’s book [11]. For a more thorough exposition
and analysis of Sacks forcing, the reader may consult Geschke and
Quickert [7].
Definition 3.1**.**
Let Seq denote the set of all finite binary sequences.
A tree is a set p⊆Seq, such that for each s∈p
if s↾n∈p then for all m<ns↾m∈p.
2. 2.
If p⊆Seq and s∈p, we say that ssplits
in p if s⌢0∈p and s⌢1∈p.
3. 3.
If p⊆Seq and s splits in p then we say s is an
order n splitting node if ∣{t⊊s∣t splits in p}∣=n.
4. 4.
If p⊆Seq, we say s is a stem of p if s
is a splitting node and for all t⊊st is not a splitting
node.
Definition 3.2**.**
We say p⊆Seq is a perfect tree if:
p is a tree;
2. 2.
And for every s∈p there exists a splitting node t∈p such
that t⊇s.
Definition 3.3**.**
If p is a perfect tree and s∈p we denote p↾s={t∈p∣s⊆t∨t⊆s}.
Plainly p↾s is perfect as well.
Definition 3.4**.**
We call P={p⊆Seq∣p is a perfect tree},
where P is ordered by reverse inclusion: p≤q⟺p⊇q,
Sacks forcing. Later, after we present the generalized form,
we shall refer to it as ℵ0-Sacks forcing.
We can identify the generic set G with a function f:ω→2.
First, note that the set of perfect trees with a stem of height at
least n is a dense set in P. Thus there are trees of
arbitrarily long finite stems in G. Also, if two trees p,q both
have stems of height greater or equal than n, but the restrictions
of the stems on n differ, then ht(p∩q)<n, and so there is
no r∈P such that r>p,q. Hence all trees belonging
to the generic set G must agree on their stems. Thus we can define
f(n)=s(n), where s is part of the stem of any p∈G. Due
to their agreement, the function s is well defined, and due to
the arbitrary finite length of the stems f is defined on ω.
For the other direction, we may define G={p∈P∣∀n∈ω(f↾n∈p)}
[7]. So in essence G is equivalent to a new real number,
called a Sacks real.
Lemma 3.5**.**
CH* implies that ∣P∣=ℵ1
and so P satisfies the ℵ2-antichain condition.*
Proof.
We simply count the number of possible conditions. There are at most
ℵ0 finite binary sequences, and therefore at most 2ℵ0
possible trees. Assuming CH2ℵ0=ℵ1,
and so there are at most ℵ1 conditions and no antichains
of cardinality ℵ2.
∎
We note that P does not offer much in way of closure.
It is plainly not ℵ1-closed, as one may take any perfect
tree p and build the following sequence: ⟨pn∣n∈ω⟩
where p0=p and pn+1=pn↾s⌢0 where
s is the single order [math] splitting node of pn. This is
obviously a sequence of perfect trees such that for all npn+1>pn,
however n∈ω⋂pn has no splitting nodes
at all, and therfore is not a perfect tree.
Luckily, perfect trees offer a slightly weaker form of closure, using
the technique of fusion.
Definition 3.6**.**
Suppose p,q∈P. We say p≥nq if:
p≥q;
2. 2.
And s∈p is an order n splitting in p node if and only
if s∈q is an order n splitting node in q.
Lemma 3.7**.**
Fusion: Let ⟨pn∈P∣n∈ω⟩
be a sequence of conditions such that for all npn+1≥npn.
Then n∈ω⋂pn∈P.
Proof.
Define n∈ω⋂pn=pω. We claim pω∈P,
meaning it’s a perfect tree. Take s∈pω. Let ∣{t⊊s∣t splits in pω}∣=m.
Take pm+1. By definition s∈pm+1. However pm+1
is a perfect tree, and so has a splitting node of order m+1 above
s, which we denote t⊇s. But because it is an order
m+1 splitting node and pm+2≥m+1pm+1 we have t∈pm+2.
By induction we get t∈pω, but s⊆t so we
found a splitting node in pω above our arbitrary s.
Hence pω is indeed perfect.
∎
Note that it is obvious from the chain condition that all cardinals
greater than or equal to ℵ2 are preserved, as is of course
ℵ0. We now complete the picture with showing ℵ1
is preserved.
Lemma 3.8**.**
ℵ1* is preserved under
ℵ0-Sacks forcing.*
Proof.
Assume X is a countable set of ordinals in V[G].
We show the existence of a set A∈V countable in V such that
X⊆A˙. Let F˙ be a name and let p be a condition
such that p⊩F˙ witnesses that X˙ is countable,
that is p⊩F˙:ω→X˙ is surjective.
We now build a fusion sequence ⟨pα∣α∈ω⟩
starting with p0=p. Assume we defined pn. Let Sn
be the set of all order n splitting nodes of pn. For each
s∈Sn let qs⌢0,qs⌢1 and as⌢0,as⌢1
be such that qs⌢i≥pn↾s⌢i
and qs⌢i⊩F˙(n)=as⌢i. Let pn+1=s∈Sn,i=0,1⋃qs⌢i.
Note that the union of perfect trees is a perfect tree, and that all
splitting nodes of order ≤n are preserved: if t is an order
m<n splitting node in pn then it is also a splitting node
in qs⌢0 for the s∈Sn that is s⊋t,
and so is in pn+1; whereas if t is an order n splitting
node in pn then t is a splitting node in qt⌢0∪qt⌢1
and so is in pn+1. Thus all splitting nodes of order n are
preserved in pn+1, and so pn+1≥npn. Using lemma
3.7 we get q=n∈ω⋂pn∈P.
Now define A=n∈ω⋃{as⌢i∣s∈Sn∧i=0,1}.
Note that A is a countable union of finite sets, hence A is
countable in V. Now observe that q⊩ran(F˙)⊆A.
As F˙ is the name of the function that witnesses the countability
of X this means q⊩X˙⊆A.
In this process we built a specific q≥p, so q is not guaranteed
to be in the generic set G. However, as we found a q⊩X˙⊆A
above any condition stronger or equal to p, due to density, there
is somer≥p in G such that r⊩X˙⊆A.
Therefore V[G]⊨X⊆A, where A is countable
in V, which implies that ℵ1 is preserved.
∎
Theorem 3.9**.**
Sacks forcing produces a minimal extension
of V, meaning that for every model W of ZFC if V⊆W⊆V[G],
then either W=V or W=V[G].
Proof.
According to theorem 15.43 of [9], every intermediate
model of ZFC is equal to V[A], where A
is a set of ordinals. Hence it is sufficient to show that for any
set of ordinals A in V[G], either V[A]=V
or V[A]=V[G].
Let A˙ be the name of a set of ordinals in V[G].
There is an ordinal α such that 0⊩A˙⊆α,
and let z˙ be the name of the characteristic function of A,
z:α→2. If A∈V then obviously V[A]=V.
Assume then p∈G is a condition that forces A˙∈/V.
For a condition q∈P let z˙q be the longest
initial segment of z˙ that is decided by q, and γq
be the first ordinal for which z˙ is undecided. For q≥p
they must be well-defined, because if q decides all of z˙,
it decides all of A˙, and then A∈V, in contradiction
to p⊩A˙∈/V. Plainly γq<α.
Mark p0=p. Assume we’ve already chosen pn. For every splitting
node s∈Sn, where Sn is defined as in lemma 3.8,
let’s look at γpn↾s and conditions pn↾s⌢i.
Suppose that for both i=0,1 we have pn↾s⌢i⊩z˙(γpn↾s)=j.
γpn↾s is undecided, so take q≥pn↾s
such that q⊩z˙(γpn↾s)=1−j.
Either q∩pn↾s⌢0∈P, or
q∩pn↾s⌢1∈P. But q and
pn↾s⌢i are incompatible for i=0,1,
hence our supposition is impossible.
Thus, if for a certain i there is a j such that pn↾s⌢i⊩z˙(γpn↾s)=j,
then there is some qs⌢(1−i)≥pn↾s⌢(1−i)
such that qs⌢(1−i)⊩z˙(γpn↾s)=1−j,
and we take qs⌢i=pn↾s⌢i so
that qs⌢i⊩z˙(γpn↾s)=j.
If there is no such i, then we are free to take for both i=0,1qs⌢i≥pn↾s⌢i such that
qs⌢i⊩z˙(γpn↾s)=i.
The point is that in both cases we found qs⌢i that decide
z˙(γpn↾s) in conflicting
ways for i=0,1.
We now take pn+1=s∈Sn,i=0,1⋃qs⌢i.
Again, exactly as in lemma 3.8, we recognize
⟨pn∣n∈ω⟩ is a fusion sequence.
Thus we can take condition q=n∈ω⋂pn.
Let f={s∈q∣z˙q↾s⊆z˙G}.
This is a branch of q, because if s is a splitting node of q,
then either for i=0 or i=1, but not both, z˙q↾s⌢i(γq↾s)=z˙q(γq↾s),
hence for only one i we have s⌢i∈f. Thus f is
a completely definable branch in V[A], and so f∈V[A].
We now note that given p we created a stronger condition q and
so from density we can assume q∈G. We claim that f is our
Sacks real f. Mark the Sacks real as g. If f disagrees with
g, then because both are branches in q, there must be a splitting
node s of q where they diverge. But that would imply z˙f(γq↾s)=z˙G(γq↾s)
in contradiction to the definition of f. Thus f=g, f is our
Sacks real, and we get G∈V[A].
Therefore V[G]⊆V[A]⊆V[G]
and we conclude V[A]=V[G].
∎
Corollary 3.10**.**
V⊨CH⇒V[G]⊨CH.
Proof.
In the proof of theorem 3.9 let’s assume A
is a ’new’ subset of ℵ0, meaning we have V[G]⊨A⊊ℵ0∧A∈/V.
Using fusion, we generate a perfect tree q∈P in the
ground model that is used to interpret A according to the Sacks
real G.
Viewed another way, and taking [q] to signify the branches
of q, q is in fact a continuous map q:[q]→P(ℵ0)
such that V[G]⊨q(G)=A. Therefore given
G, there can be no two subsets of ℵ0A1=A2
that produce the same q.
But according to lemma 3.5 there are at most ℵ1
conditions in P, so there are at most ℵ1 new
subsets of ℵ0 in V[G]. Hence, (2ℵ0=ℵ1)V[G].
∎
4 κ-Sacks forcing
In this section we show how we can extend our model tower through
the successor steps.
Naᅵvely we could try and repeat the Sacks forcing, hoping that no
unexpected models ’pop up’ along the way. However, ultimately we desire
to iterate our forcing class-many times, so we need to be wary of
preserving the Power Set Axiom. Because each application of classical
Sacks forcing adds a real number, were we simply to iterate the forcing
class-many times, 2ℵ0 would ’explode’, and the resultant
model would fail to satisfy ZF. Instead, what we need
to do is find a way to build minimal models where the subsets of each
cardinal eventually stabilize.
For this, we turn to perfect trees of height κ, via Kanamori’s
extension of Sacks forcing to uncountable cardinals [10].
The following definitions are an almost perfect analogue to the definitions
of the previous section, except where noted otherwise.
Definition 4.1**.**
Let Seq=α<κ⋃2α.
A tree is a set p⊆Seq, such that for each s∈p
if s↾α∈p then for all β<αs↾β∈p.
2. 2.
If p⊆Seq and s∈p, we say that ssplits
in p if s⌢0∈p and s⌢1∈p.
3. 3.
If p⊆Seq and s splits in p then we say s is an
order α splitting node if when we order {t⊆s∣t splits in p}
by inclusion it is the αth node.
4. 4.
If p⊆Seq, we say s is a stem of p if s
is a splitting node and for all t⊊st is not a splitting
node.
Definition 4.2**.**
We say p⊆Seq is a perfect tree
if:
p is a tree.
2. 2.
For every s∈p there exists a splitting node t∈p such that
t⊇s.
3. 3.
If δ<κ is a limit ordinal, s∈δ2
and s↾β∈p for every β<δ, then
s∈p. Intuitively ’p is closed’.
4. 4.
If δ<κ is a limit ordinal, s∈δ2
and for arbitrarily large β<αs↾β
splits in p, then s splits in p. Intuitively ’the splitting
nodes of p are closed’.
The last two conditions are new, though it is easy to see that the
original ℵ0-Sacks forcing satisfies them by default. Conditions
3 and 4 are necessary to ensure the
closure property in lemma 4.5. Without condition
3 the limit of ω trees might be empty, and
without condition 4 the limit might consist of just
a branch without any splitting nodes.
Definition 4.3**.**
If p is a perfect tree and s∈p we denote p↾s={t∈p∣s⊆t∨t⊆s}.
Plainly p↾s is perfect as well.
Definition 4.4**.**
We call P={p⊆Seq∣p is a perfect tree},
where P is ordered by reverse inclusion p≤q⟺p⊇q,
κ-Sacks forcing.
As before, we can identify the generic set G with a function f:κ→2.
There are trees with arbitrarily long stems in G, and these stems
must coincide on their mutual domain. Thus we can define f(α)=s(α),
where s is part of a stem for some p∈G. This function is
well-defined on κ. For the other direction, we may define
G={p∈P∣∀α<κ(f↾α∈p)}.
So in essence G defines a new subset of κ.
Next, to achieve maximal closure in P, we require κ
to be regular. So from here on it is assumed κ is a regular
cardinal.
Lemma 4.5**.**
P* is κ-closed.*
Proof.
Let δ<κ, ⟨pα∣α<δ⟩
be a sequence of increasing conditions. We claim p=α<δ⋂pα∈P.
Conditions 3 and 4 of definition 4.2
are trivially true in p. It is left to show that each node in p
has a splitting node above it.
Let S be the splitting nodes of p and for each α<δ
let Sα be the set of splitting nodes of pα.
p is not empty because ∅∈pα for all α.
Assume s∈p. Then for each α<δs∈pα,
and denote tα as the order [math] splitting node of pα↾s.
If α<β<δ then Sα⊇Sβ⊇S.
Therefore either ⟨tα∣α<δ⟩
stabilizes, in which case for some γtγ∈Sα
for all α<δ, and therefore s⊆tγ∈S.
Or for each β<δ⟨tα∣β≤α<δ⟩∈pβ
is an unbounded sequence of splitting nodes under α<δ⋃tα,
and therefore due to definition 4.2 condition 4α<δ⋃tα∈Sβ. But that
means s⊊α<δ⋃tα∈S.
Either way, we found a splitting node above an arbitrary s∈p,
and therefore p is perfect.
∎
Lemma 4.6**.**
If 2<κ=κ and 2κ=κ+
then ∣P∣=κ+ and so P satisfies
the κ++-antichain condition.
Proof.
We simply count the number of possible conditions. There are at most
κ binary sequences of length <κ, and therefore at
most 2κ=κ+ possible trees. So there are at most
κ+ conditions and no antichains of cardinality κ++.
∎
We now extend the technique of fusion to this forcing.
Definition 4.7**.**
Suppose p,q∈P. We say p≥αq
if:
p≥q;
2. 2.
And for all β≤α, s∈p is an order β splitting
in p node if and only if s∈q is an order β splitting
node in q.
Note that because of the closure of the splitting nodes (definition
4.2 condition 4), if δ is a limit
ordinal p≥δq⇔∀α<δ(p≥αq).
Lemma 4.8**.**
Fusion: Let ⟨pα∈P∣α<κ⟩
be a sequence of conditions such that for all αpα+1≥αpα,
and for δ a limit ordinal pδ=α<δ⋂pα.
Then α<κ⋂pα∈P.
Proof.
Define α<κ⋂pα=pκ. We
claim pκ∈P, meaning it is a perfect tree. Conditions
3 and 4 of definition 4.2
are trivially true in pκ. It is left to show that each
node in pκ has a splitting node above it.
Take s∈pκ. Let β be the order type of {t⊊s∣t splits in pκ}
ordered by inclusion. Take pβ+1. By definition s∈pβ+1.
However pβ+1 is a perfect tree, and so has a splitting node
of order β+1 above s, which we denote t⊇s.
Now we proceed by transfinite induction. Assume t is an order β+1
splitting node in pϵ, where ϵ>β. Then pϵ+1≥ϵpϵ,
so t is also an order β+1 splitting node in pϵ+1.
Let δ<κ be a limit ordinal, such that for all ϵ
with β<ϵ<δt is an order β+1 splitting
node in pϵ. Then by definition 4.2 condition
4t is an order β+1 splitting node in pδ.
Therefore by induction t⊇s is a splitting node in pκ.
We found a splitting node in pκ above our arbitrary s.
Hence pκ is indeed perfect.
∎
Note that it is obvious from the chain condition that all cardinals
greater than or equal to κ++ are preserved. On the other
hand, due to closure all cardinals less than or equal to κ
are preserved. So to complete the picture we must show κ+
is preserved.
Lemma 4.9**.**
If 2<κ=κ then κ+ is preserved
under κ-Sacks forcing.
Proof.
This proof closely mirrors the proof of lemma 3.8.
Assume X is a set of ordinals in V[G], such that
(∣X∣=κ)V[G]. We show the
existence of a set A∈V of cardinality κ in V such
that X⊆A˙. Let F˙ be a name and let p be
a condition such that p⊩F˙:κ→X˙ is surjective.
We now build a fusion sequence ⟨pα∣α<κ⟩
with p0=p. Assume we defined pα. Let Sα
be the set of all order α splitting nodes of pα.
For each s∈Sα let qs⌢0,qs⌢1
and as⌢0,as⌢1 be such that qs⌢i≥pα↾s⌢i
and qs⌢i⊩F˙(α)=as⌢i. Let pα+1=s∈Sn,i=0,1⋃qs⌢i.
All splitting nodes of order ≤α are preserved: if t
is an order β<α splitting node in pα then it
is also a splitting node in qs⌢0 for the s∈Sα
that is s⊋t, and so in pα+1; whereas if t
is an order α splitting node in pα then t is
a splitting node in qt⌢0∪qt⌢1 and so
is in pα+1. Thus all splitting nodes of order α
are preserved in pα+1, and so pα+1≥αpα.
In the limit case we define pδ=α<δ⋂pα.
Thus we have a fusion sequence, and using lemma 4.8
we get q=α<κ⋂pα∈P.
Now define A=α<κ⋃{as⌢i∣s∈Sα∧i=0,1}.
Note that A is a union of κ sets of at most 2<κ=κ
cardinality, so ∣A∣=κ. Now observe that q⊩ran(F˙)⊆A.
As F˙ is the name of the function that witnesses the cardinality
of X, this means q⊩X˙⊆A.
Although we built a specific q≥p, due to density there is somer≥p in G such that r⊩X˙⊆A. Therefore
V[G]⊨X⊆A, where A is of cardinality
κ in V, which implies that κ+ is preserved.
∎
Theorem 4.10**.**
κ-Sacks forcing produces
a minimal extension of V, such that for every model W of ZFC
if V⊆W⊆V[G], then either W=V or
W=V[G].
Proof.
This proof closely mirrors the proof of theorem 3.9,
and so is given here in a more concise form.
We show that for any set of ordinals A∈V[G] either
V[A]=V or V[A]=V[G]. Let A˙
be the name of a set of ordinals in V[G], and let z˙
be the name of its characteristic function. Assume p∈G forces
A˙∈/V.
For a condition q∈P let z˙q be the longest
initial segment of z˙ that is decided by q, and γq
be the first ordinal for which z˙ is undecided.
Mark p0=p. Assume we’ve already chosen pα. For every
splitting node s∈Sα, where Sα is defined
as in lemma 4.9, let’s look at γpα↾s
and conditions pα↾s⌢i. If for a
certain i there is a j such that pα↾s⌢i⊩z˙(γpα↾s)=j
and we take qs⌢i=pα↾s⌢i,
there will be qs⌢(1−i)≥pα↾s⌢(1−i)
such that qs⌢(1−i)⊩z˙(γpα↾s)=1−j.
If there is no such i, then we are free to take for both i=0,1qs⌢i≥pα↾s⌢i such
that qs⌢i⊩z˙(γpα↾s)=i.
Either case we found qs⌢i that decide z˙(γpα↾s)
in conflicting ways for i=0,1.
We now take pα+1=s∈Sn,i=0,1⋃qs⌢i,
and for limit ordinals pδ=α<δ⋂pα.
Exactly as in lemma 4.9, ⟨pα∣α<κ⟩
is a fusion sequence and we can take q=α<κ⋂pα.
Let f={s∈q∣z˙q↾s⊆z˙G}.
f is a branch of q, and using A is completely definable,
so f∈V[A].
Due to density we may assume q∈G, in which case f is actually
our new function κ→2, which we identify with G. Thus
we get G∈V[A], and so V[A]=V[G].
∎
Corollary 4.11**.**
V⊨2κ=κ+∧2<κ=κ⇒V[G]⊨2κ=κ+.
Proof.
Again, in direct analogy to corollary 3.10. In the proof
of theorem 4.10 let’s assume A is
a ’new’ subset of κ, meaning we have V[G]⊨A⊊κ∧A∈/V.
Using fusion, we generate a perfect tree q∈P in the
ground model that is used to interpret A according to G.
Taking [q] to signify the branches of q, we can view
q as a mapping q:[q]→P(κ)
between the branches of q and subsets of κ. Our construction
method for q implies that V[G]⊨q(G)=A.
Therefore for a given G, there cannot be two subsets of κA1=A2 that produce the same q.
But according to lemma 4.6 there are at most κ+
conditions in P, so there can be at most κ+
new subsets of κ in V[G]. Hence, (2κ=κ+)V[G].
∎
After proving minimality and preservation of cardinals we conclude
this section with showing that κ-Sacks forcing preserves GCH
above κ.
Lemma 4.12**.**
V⊨∀λ≥κ(2λ=λ+)⇒V[G]⊨∀λ>κ(2λ=λ+).
Proof.
For the preservation of GCH above κ, we turn to
the notion of a nice name (see ch. VII definition 5.11 in [11]).
A name for a subset of σ∈VP is considered nice
if it is of the form ⋃{{π}×Aπ∣π∈dom(σ)},
where each Aπ is an antichain in P. Every subset
has a nice name.
According to lemma 4.6 there are at most κ+
conditions, and therefore at most κ+ elements in an antichain.
Meaning, there are at most (κ+)κ+=κ++
different possible antichains.
Hence for a given cardinal λ there are at most (κ++)λ
different nice names for subsets of λ˙. Thus for λ≥κ++
we have (κ++)λ=2λ=λ+,
meaning (2λ≤λ+)V[G].
But of course this means V[G]⊩2λ=λ+.
For the case λ=κ+ we derive this instead from λ+=κ++=(κ++)κ+=(κ++)λ.
Thus V[G]⊨∀λ>κ(2λ=λ+),
as required.
∎
We now briefly summarize the attributes we demanded from κ
for the forcing notion and the above theorems to make sense:
κ needed to be regular, for the closure to work (lemma 4.5).
2. 2.
2<κ=κ is necessary to preserve κ+
(lemma 4.9).
3. 3.
2<κ=κ and 2κ=κ+
are necessary for the antichain condition (lemma 4.6).
All of these conditions are necessary to prove the preservation of
cardinals by κ-Sacks forcing. All of them are automatically
true if κ strongly inaccessible. However, for our construction
we don’t want to rely on the existence of large cardinals. Notably,
both conditions 2 and 3 are also implied
by GCH, so in the next section corollary 4.11
and lemma 4.12 will serve us in maintaining enough of
GCH to make the forcing iteration work.
5 κ-Sacks iteration
After defining individual κ-Sacks forcing, it is time to stitch
everything together. We now define the forcing iteration that will
enable us to build a model tower through limit ordinals, and up to
arbitrary height.
For a general introduction to iterated forcing the reader can refer
to Shelah [18].
For the rest of this section, let L be our base model, and let
ζ be the height of the model tower that we wish to build.
Definition 5.1**.**
For α≤ζ, define the forcing iteration
Pα as follows:
Let Q˙α be trivial if α
is a limit ordinal, and the name of ℵα-Sacks forcing
in VPα otherwise.
2. 2.
Pα+1=Pα⋆Q˙α.
3. 3.
At limit stages we use full support, i.e if δ is a limit ordinal
then p∈Pδ⇔∀α<δ(p↾α∈Pα).
Definition 5.2**.**
Denote:
M0=L.
2. 2.
Gα as the generic set in partial order Pα
over M0.
3. 3.
Mα=M0[Gα].
4. 4.
Gα,β as the generic set in partial order \nicefracPβGα
over M0[Gα], so that M0[Gα][G˙α,β]=M0[Gβ].
Lemma 5.3**.**
For every α<β\nicefracPβGα
is ℵα-closed. If α is a limit ordinal, then
\nicefracPβGα is ℵα+1-closed.
Proof.
Every coordinate of the forcing \nicefracPβGα
is either trivial or ℵγ-Sacks forcing for γ≥α.
According to lemma 4.5 each coordinate is therefore
ℵα-closed.
By definition 5.1 we use full support, and therefore
the iteration \nicefracPβGα as a whole
is ℵα-closed.
If α is limit ordinal then Qα is trivial,
and so \nicefracPβGα=\nicefracPβGα+1
which is ℵα+1-closed.
∎
We now show that all cardinals are preserved throughout the entire
forcing iteration.
Definition 5.4**.**
Let OK(α) denote that:
Mα has the same cardinals as M0;
2. 2.
Mα⊨2<ℵα=ℵα;
3. 3.
Mα⊨∀λ≥ℵα(2λ=λ+).
Lemma 5.5**.**
OK(0).
Proof.
M0=L and so satisfies GCH.
∎
Lemma 5.6**.**
If OK(α) then OK(α+1).
Proof.
If Qα is trivial then Mα+1=Mα
and conditions 1 and 3 are trivially
true for α+1. As for condition 2, we know Mα⊨2ℵα=ℵα+1
and so Mα+1⊨2<ℵα+1=2ℵα=ℵα+1
as required.
Otherwise, OK(α) implies Mα has the
same cardinals as M0, Mα⊨2<ℵα=ℵα
and Mα⊨∀λ≥ℵα(2λ=λ+).
According to lemmas 4.5, 4.6
and 4.9Mα+1 has the same cardinals as Mα,
and therefore the same as M0. According to corollary 4.11Mα+1⊨2<ℵα+1=2ℵα=ℵα+1.
And according to lemma 4.12Mα+1⊨∀λ≥ℵα+1(2λ=λ+).
So OK(α+1) is true.
∎
Lemma 5.7**.**
Let δ≤ζ be a limit
ordinal. If for all α<δOK(α), then
Mδ has the same cardinals as M0.
Proof.
For any α<δ we have Mδ=Mα[Gα,δ].
But \nicefracPδGα is ℵα-closed
per lemma 5.3. Thus, all cardinals less than
or equal ℵα in Mα are preserved in Mδ.
But OK(α) implies Mα has the same cardinals
as M0, so all cardinals less than or equal ℵα
in M0 are preserved in Mδ.
As this is true for all cardinals less than ℵδ, it
is true for ℵδ itself.
Also note ∣Pδ∣=∏α<δ2ℵα=∏α<δℵα+1≤ℵδ∣δ∣=2ℵδ=ℵδ+1.
Therefore Pδ is ℵδ+2-c.c, and
Mδ preserves all cardinals greater or equal to ℵδ+2.
It remains to be proven that ℵδ+1 is preserved.
First, assume ℵδ is singular and suppose that the
iteration does collapse it. Denote cf(ℵδ+1M0)Mδ=ℵγ.
The collapse of the cardinal implies ℵγ≤ℵδ,
and due to the latter’s singularity ℵγ<ℵδ,
meaning there is a new set of ordinals A∈Mδ, such that
∣A∣=ℵγ. However according to lemma 5.3\nicefracPδGγ+1 is ℵγ+1-closed,
so no sets of ordinals of cardinality ℵγ are added
when forcing Mδ=Mγ+1[Gγ+1,δ].
Thus A∈Mγ+1, which implies Mγ+1⊨cf(ℵδ+1M0)≤ℵγ<ℵδ+1,
in contradiction to ℵδ+1 being preserved in Mγ+1.
Therefore ℵδ+1 must be preserved as well.
For the case ℵδ is regular we proceed with a variation
of the argument used in lemma 4.9.
Assume ℵδ is regular, meaning ℵδ=δ,
and that F:δ→Ord is a function in Mδ.
We show the existence of a set A∈M0 of cardinality δ,
such that in Mδran(F)⊆Aˇ.
Let p∈Pδ be a condition such that p⊩F˙:δ→Ord.
For any condition q∈Pδ, let q˙α
denote the α coordinate of q, and similarly let q<α
denote the first α coordinates, and q˙>α denote
the name of all higher coordinates. For consistency, if we discuss
a condition q∈\nicefracPδHβ where
Hβ is a generic set in Pβ, then we fix
the first coordinate to be qβ.
Inductively we are going to build an increasing sequence of conditions
⟨p(α)∈Pδ∣α<δ⟩.
Each coordinate is also going to be built inductively.
Start with the first coordinate. Let p0=p, and assume we’ve
defined pn for n<ω. Let Sn be the set of all order
n splitting nodes of pn0. For each s∈Sn let qs⌢0,qs⌢1
and as⌢0,as⌢1 be such that qs⌢i≥⟨pn0↾s⌢i,pn>0⟩
and qs⌢i⊩F˙(n)=as⌢i. Let pn+10=s∈Sn,i=0,1⋃qs⌢i0
be the amalgamation of the qs⌢i0’s, just like we
did in the proof of lemma 4.9. For the rest of the coordinates,
we define pn+1>0 with accordance to the path taken in the
first coordinate. Meaning that qs⌢i0 forces pn+1>0=qs⌢i>0.
From the way we defined the qs⌢i’s it is clear pn+1>0≥pn>0,
and so pn+1≥pn. Also, all splitting nodes of order ≤n
are preserved in the first coordinate, and therefore pn+10≥npn0.
Thus what we have in the first coordinate is a classical fusion sequence,
and in the rest of the coordinates an increasing sequence. So thanks
to lemmas 3.7 and 5.3 we can
conclude pω=n<ω⋂pn∈Pδ.
We define A0={as⌢i∣s∈Sn∧i=0,1}.
Obviously ∣A0∣=ℵ0.
In essence, we used a fusion argument on the first coordinate to create
pω, which is a sort of ’decision tree’ for the first ω
values of F˙. We set p(0)=pω. Next,
we are going to repeat this construction using the higher coordinates.
In each step p(β) will be such a decision tree for
the first ωβ values of F˙.
So assume now that we’ve already defined p(β), and
we shall show how to define p(β+1). We use Qβ+1
to decide the values of F˙ up to ωβ+1. If β
is a limit ordinal we also set Aβ=∅.
Let Hβ+1⊊Pβ+1 be any generic set
such that ⟨p(β)0,p˙(β)1,...,p˙(β)α,...∣α≤β⟩∈Hβ+1.
Now, working in M0[Hβ+1] we repeat the construction.
To start the induction, set pωβ=p(β)>β∈\nicefracPδHβ+1.
Assume pα is defined, we are going to define pα+1.
Remember Qβ+1 is ℵβ+1-Sacks forcing.
So let Sα be the set of all order α splitting nodes
of pαβ+1. For each s∈Sα let qs⌢0,qs⌢1
and as⌢0,as⌢1 be such that qs⌢i≥⟨pαβ+1↾s⌢i,p˙α>β+1⟩
and qs⌢i⊩F˙(α)=as⌢i. Let pα+1β+1=s∈Sα,i=0,1⋃qs⌢iβ+1
be the amalgamation of the qs⌢i’s first coordinate.
As for the rest of the coordinates define p˙α+1>β+1
with accordance to the path taken in the first coordinate. Meaning
that qs⌢iβ+1 forces p˙α+1>β+1=q˙s⌢i>β+1.
From the way we defined the qs⌢i’s it is clear p˙α+1>β+1≥p˙α>β+1,
and so pα+1>β≥pα>β. Also, all
splitting nodes of order ≤α are preserved in pα+1β+1,
and therefore pα+1β+1≥αpαβ+1.
In limit stages τ we simply define pτ=ωβ≤α<τ⋂pα.
In each coordinate of pτ we have ℵβ+1-closure
according to lemma 5.3. So for τ<ωβ+1pτ∈\nicefracPδHβ+1. For the
case τ=ωβ+1 note that we have a fusion sequence
in the first coordinate. So thanks to lemma 4.8
and the ℵβ+2-closure of the higher coordinates we have
pωβ+1∈\nicefracPδHβ+1.
Now, for each as⌢i we used, we pick a Pβ+1-name
a˙s⌢i in M0. Because ∣Pβ+1∣<δ
we can find in M0 a set Aβ+1 of cardinality <δ
such that pωβ⊩Aˇβ+1⊇{a˙s⌢i∣s∈Sα∧i=0,1∧ωβ≤α<ωβ+1}.
We also pick a name for pωβ+1 such that
[TABLE]
We set p(β+1)=⟨p(0)0,p˙(1)1,...,p˙(β)β,p˙ωβ+1β+1,...,p˙ωβ+1η,...∣η<δ⟩∈Pδ.
Obviously p(β+1)≥p(β).
For limit stages τ we define p(τ)=β<τ⋂p(β).
We claim p(τ)∈Pδ. For each coordinate
β<τ note that the condition stabilizes, and so p(τ)β=p(β)β.
For coordinates ≥τ lemma 5.3 provides
at least ℵτ+1-closure, and because we’re using full
support this shows p(τ)∈Pδ.
Therefore by induction we can construct condition p(δ)=⟨p(0)0,p˙(1)1,...,p˙(η)η,...∣η<δ⟩∈Pδ
.
Now define A=α<δ⋃Aα. Because
for all α<δOK(α), we have ∣A∣=δ.
Observe that p(δ)⊩ran(F˙)⊆Aˇ.
Now if ℵδ+1 is not preserved then in Mδℵδ+1M0=δ, and we can take
F to be the bijection between ℵδ+1M0 and
δ. Applying the construction to this F we get p(δ)⊩ℵδ+1M0⊆Aˇ.
But then we get in M0 a surjection from δ onto ℵδ+1,
which is impossible. Therefore p(δ)⊩ℵδ+1M0=ℵδ+1.
We know that p(δ)≥p(0)≥p, and
so by density we can assume without loss of generality that p(δ)∈Gδ.
Therefore ℵδ+1 is indeed preserved.
Thus all cardinals are preserved in all cases, and overall.
∎
Lemma 5.8**.**
Let δ≤ζ be a limit ordinal.
If for all α<δOK(α), then Mδ⊨2<ℵδ=ℵδ.
Proof.
Note that obviously Mδ⊨2<ℵδ≥ℵδ.
Now assume Mδ⊨2<ℵδ>ℵδ.
That means for some α<δMδ⊨2ℵα>ℵδ.
However by lemma 5.3\nicefracPδGα+1
is ℵα+1-closed, and so new subsets of ℵα
are added when forcing Mδ=Mα+1[Gα+1,δ].
Therefore Mα+1⊨2ℵα>ℵδ
and so Mα+1⊨2ℵα+1>ℵα+2,
in contradiction to OK(α+1).
Therefore Mδ⊨2<ℵδ=ℵδ.
∎
Lemma 5.9**.**
Let δ≤ζ be a limit ordinal.
If for all α<δOK(α), then Mα⊨∀λ≥ℵα(2λ=λ+).
Proof.
Recall the proof of lemma 4.12 in the previous section.
As already shown in lemma 5.7∣Pδ∣=ℵδ+1.
Therefore there are at most ℵδ+1ℵδ+1=ℵδ+2
different antichains. With the rest of the proof identical, we get
Mδ⊨∀λ>ℵδ(2λ=λ+).
We now show that Mδ⊨2ℵδ=ℵδ+1.
Let A⊆ℵδ, and assume p∈Pδ
such that p⊨A˙⊆ℵδ∧∀β<δ(A˙∈/Mβ).
We denote individual coordinates like so: p=⟨p0,p˙1,...,p˙α,...∣α<δ⟩.
Let z˙ be the name of the characteristic function of A.
For any condition q∈Pδ stronger than p let
z˙q be the longest initial segment of z˙ that is
decided by q, let γq be the first ordinal for which
z˙ is undecided, let q0 denote the first coordinate
of q, i.e from ℵ0-Sacks forcing over M0, and let
q denote the name of the rest of the coordinates. zq˙
and γq must be well-defined, because if q decides all
of z˙, it decides all of A˙, and then A∈M0,
in contradiction to p. Plainly γq<ωδ.
We now build a fusion sequence. Mark p0=p. For the successor
case, assume that we’ve already chosen pϵ. Note that
pϵ0 is just a perfect tree in Q0, and
pϵ is a name in M0. Let Sϵ denote
the order ϵ splitting nodes of pϵ0. For
every splitting node s∈Sϵ, let’s look at γ⟨pϵ0↾s,pϵ⟩
and conditions ⟨pϵ0↾s⌢i,pϵ⟩.
Suppose that for both i=0,1 we have ⟨pϵ0↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j.
γ⟨pϵ0↾s,pϵ⟩
is undecided, so take q≥⟨pϵ0↾s,pϵ⟩
such that q⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=1−j.
Either q∩⟨pϵ0↾s⌢0,pϵ⟩∈Pδ
or q∩⟨pϵ0↾s⌢1,pϵ⟩∈Pδ.
But q and ⟨pϵ0↾s⌢i,pϵ⟩
are incompatible for i=0,1, therefore our supposition is impossible.
Thus, if for a certain i there is a j such that ⟨pϵ0↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j,
we can define qs⌢i=⟨pϵ0↾s⌢i,pϵ⟩
so that qs⌢i⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j,
and know there will be some qs⌢(1−i)≥⟨pϵ0↾s⌢(1−i),pϵ⟩
such that qs⌢(1−i)⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=1−j.
Alternatively, there is no such i, and we are free to select for
both i=0,1qs⌢i≥⟨pϵ0↾s⌢i,pϵ⟩
such that qs⌢i⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=i.
We now define the first coordinate as an amalgamation of the qs⌢i0’s:
pϵ+10=s∈Sϵ,i=0,1⋃qs⌢i0,
just like we did in the proof of theorem 3.9
of the original Sacks forcing. For the rest of the coordinates, we
define pϵ+1 with accordance to the path taken in
the first coordinate. Meaning that qs⌢i0 forces pϵ+1=qs⌢i.
From the way we defined the qs⌢i’s it is clear pϵ+1≥pϵ,
and so pϵ+1≥pϵ.
In the first coordinate we get a classical fusion sequence ⟨pϵ0∣ϵ<ω⟩.
Therefore we can define pω0=ϵ<ω⋂pϵ0∈Q0.
As for the higher coordinates, we can define pω=ϵ<ω⋂pϵ0
because of the ℵ1-closure, proven in lemma 5.3.
So pω∈Pδ.
So we can now define p(0)=pω. Next we are going
to repeat by induction the construction above, using the higher coordinates.
So for the successor case, assume that we’ve already defined p(β).
For all coordinates 0<α≤β define p˙(β+1)α=p˙(α)α,
and let p(β+1)0=p(0)0.
Let Hβ+1⊆Pβ+1 be any generic set
such that ⟨p(β)0,p˙(β)1,...,p˙(β)α,...∣α≤β⟩∈Hβ+1.
Note that M0[Hβ+1]=the interpretation of M˙β+1 according to H˙β+1.
Working in M0[Hβ+1], we know that ⟨p(β)β+1,p˙(β)β+2,...,p(β)α,...∣α<δ⟩⊩A˙∈/M0[Hβ+1].
Just as before, let z˙ be the name of the characteristic function
of A. For any condition q∈\nicefracPδHβ+1
let z˙q be the longest initial segment of z˙ that
is decided by q, let γq be the first ordinal for which
z˙ is undecided, let qβ+1 denote the first
coordinate of q, i.e from ℵβ+1-Sacks forcing over
M0[Hβ+1], and let q denote the name
of the rest of the coordinates. zq˙ and γq must
be well-defined, because if q decides all of z˙, it decides
all of A˙, and then A∈M0[Hβ+1],
in contradiction to p(β). Plainly γq<ωβ+1.
Just as before, we again build a fusion sequence. This time mark p0=p(β).
For the successor case, assume that we’ve already chosen pϵ.
Note that pϵβ+1 is just a perfect tree in Qβ+1,
and pϵ is a name of a condition. Let Sϵ
denote the order ϵ splitting nodes of pϵβ+1.
For every splitting node s∈Sϵ, let’s look at γ⟨pϵβ+1↾s,pϵ⟩
and conditions ⟨pϵβ+1↾s⌢i,pϵ⟩.
Suppose that for both i=0,1 we have ⟨pϵβ+1↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=j.
γ⟨pϵβ+1↾s,pϵ⟩
is undecided, so take q≥⟨pϵβ+1↾s,pϵ⟩
such that q⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=1−j.
Either q∩⟨pϵβ+1↾s⌢0,pϵ⟩∈\nicefracPδHβ+1
or q∩⟨pϵβ+1↾s⌢1,pϵ⟩∈\nicefracPδHβ+1.
But q and ⟨pϵβ+1↾s⌢i,pϵ⟩
are incompatible for i=0,1, therefore our supposition is impossible.
Thus, if for a certain i there is a j such that ⟨pϵβ+1↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=j,
we can define qs⌢i=⟨pϵβ+1↾s⌢i,pϵ⟩
so that qs⌢i⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=j,
and know there will be some qs⌢(1−i)≥⟨pϵβ+1↾s⌢(1−i),pϵ⟩
such that qs⌢(1−i)⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=1−j.
Alternatively, there is no such i, and we are free to select for
both i=0,1qs⌢i≥⟨pϵβ+1↾s⌢i,pϵ⟩
such that qs⌢i⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=i.
We now define the first coordinate as an amalgamation of the qs⌢iβ+1’s:
pϵ+1β+1=s∈Sϵ,i=0,1⋃qs⌢iβ+1.
For the rest of the coordinates, we define pϵ+1
with accordance to the path taken in the first coordinate. Meaning
that qs⌢iβ+1 forces that pϵ+1=qs⌢i.
From the way we defined the qs⌢i’s it is clear pϵ+1≥pϵ,
and so pϵ+1≥pϵ. In limit stages τ<ωβ+1
we just use the ℵβ+1-closure to define pτ=α<τ⋂pα.
Now, in order to define pωβ+1, note that in the
first coordinate we again get a fusion sequence ⟨pϵβ+1∣ϵ<ω⟩.
Therefore pωβ+1β+1=ϵ<ωβ+1⋂pϵβ+1∈Qβ+1.
As for the rest of the coordinates, we can use the ℵβ+1-closure
proven in lemma 5.3 to define pωβ+1=ϵ<ωβ+1⋂pϵ.
For all α>β we define p˙(β+1)α=pωβ+1α.
We now pick a name ⟨p˙(β+1)β+1,p˙(β+1)β+2,...,p˙(β+1)α,...∣α<δ⟩
for the ⟨p(β+1)β+1,p˙(β+1)β+2,...,p˙(β+1)α,...∣α<δ⟩
that we constructed in \nicefracPδHβ+1,
such that ⟨p(β)0,p˙(β)1,...,p˙(β)β⟩
forces it to be the way it was defined.
Finally, set p(β+1)=⟨p(0)0,p˙(1)1,...,p˙(β+1)β+1,p˙(β+1)β+2,...,p˙(β+1)α∣α<δ⟩.
For limit stages δ we define p(δ)=α<δ⋂p(α).
As each coordinate lesser than δ stabilizes, and by lemma
5.3 each coordinate ≥δ is at least
ℵδ+1-closed, and because we’re using full support,
p(δ)∈Pδ.
As we’ve constructed p(δ) stronger than a general
p∈Pδ, then by density arguments, we may assume
without loss of generality p(δ)∈Gδ.
Just as in corollary 4.11, we can view p(δ)
as a mapping: p(δ) takes as input a sequence of
branches ⟨h˙α∣a<δ⟩,
where each h˙α is a branch of p˙(δ)α,
and interprets A.
For each Gα,α+1, let gα,α+1 be the
generic branch interdefinable with Gα,α+1. Note that
because of the way p(δ) was defined, Mδ⊨p(δ)(⟨gα,α+1∣α<δ⟩)=A.
Therefore, for a given Gδ, there can’t be two different
subsets of ℵδ that produce the same p(δ)∈Pδ
in the construction above.
As ∣Pδ∣=ℵδ+1, there are
at most ℵδ+1 new subsets of ℵδ. Therefore,
(2ℵδ=ℵδ+1)Mδ.
∎
Lemma 5.10**.**
For all β≤ζ, Mβ
has the same cardinals as M0.
Proof.
By induction we prove OK(β). By lemma 5.5OK(0). The successor case is handled in lemma 5.6.
For the limit case, lemmas 5.7, 5.8
and 5.9 show that if OK(α)
is true for all α<δ, then OK(δ).
Therefore by induction OK(β), and so Mβ
has the same cardinals as M0.
∎
Next, we want to verify that during the iteration we don’t create
any inner model of Mζ other than the Mα’s for
α<ζ.
Note that while theorem 4.10 shows that
applying ℵα-Sacks forcing doesn’t add any inner model
between Mα and Mα+1, it says nothing about limit
stages. If Mδ is the limit model, then theoretically there
might be another inner model lurking betweenMδ
and α<δ⋃Mα.
A second type of problem could arise even in the successor stages.
Applying the forcing over Mα with α>0, one might
inadvertantly create some new inner model between M0=L and Mα+1outside the chain. Therefore we need to prove our construction
avoids creating both types of ’accidental’ models.
Lemma 5.11**.**
If N is an inner model of Mζ such
that for all α≤ζN=Mα, and β
is the least ordinal such that N⊉Mβ, then β
is a limit ordinal.
Proof.
It is enough to show that there is no greatest β such that
Mβ⊊N. Working to the contrary, assume β<ζ
is such that Mβ⊊N but Mβ+1⊈N.
Obviously if Qβ is trivial then Mβ+1⊊N
in contradiction to the assumption. Therefore we may assume Qβ
is ℵβ-Sacks forcing.
N is a model of ZFC between Mβ and Mζ.
Therefore according to lemma 15.43 in Jech [9], N=Mβ[A]
for some set of ordinals A∈Mζ.
N=Mβ so obviously A∈N∖Mβ. Let
A˙ be its name in Mβ in the forcing \nicefracPζGβ.
There is an ordinal ν such that 0⊩A˙⊆ν,
and let z˙ be the name of the characteristic function of A,
z:ν→2. A∈/Mβ, so there is a condition
p∈Gβ,ζ that forces A˙∈/Mβ.
For any condition q∈\nicefracPζGβ
stronger than p, let z˙q be the longest initial segment
of z˙ that is decided by q, let γq be the first
ordinal for which z˙ is undecided, let q0 denote the
first coordinate of q, i.e from ℵβ-Sacks forcing
over Mβ, and let q denote the name of the rest
of the coordinates. zq˙ and γq must be well-defined,
because if q decides all of z˙, it decides all of A˙,
and then A∈Mβ, in contradiction to p. Plainly γq<ν.
We’re now going to build a fusion sequence. Mark p0=p. For the
successor case, assume that we’ve already chosen pϵ.
Note that pϵ0 is just a perfect tree in Qβ,
and pϵ is a name in Mβ. Let Sϵ
denote the order ϵ splitting nodes of pϵ0.
For every splitting node s∈Sϵ, let’s look at γ⟨pϵ0↾s,pϵ⟩
and conditions ⟨pϵ0↾s⌢i,pϵ⟩.
Suppose that for both i=0,1 we have ⟨pϵ0↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j.
γ⟨pϵ0↾s,pϵ⟩
is undecided, so take q≥⟨pϵ0↾s,pϵ⟩
such that q⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=1−j.
Either q∩⟨pϵ0↾s⌢0,pϵ⟩∈\nicefracPζGβ
or q∩⟨pϵ0↾s⌢1,pϵ⟩∈\nicefracPζGβ.
But q and ⟨pϵ0↾s⌢i,pϵ⟩
are incompatible for i=0,1, therefore our supposition is impossible.
Thus, if for a certain i there is a j such that ⟨pϵ0↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j,
we can define qs⌢i=⟨pϵ0↾s⌢i,pϵ⟩
so that qs⌢i⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j,
and know there will be some qs⌢(1−i)≥⟨pϵ0↾s⌢(1−i),pϵ⟩
such that qs⌢(1−i)⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=1−j.
Alternatively, there is no such i, and we are free to select for
both i=0,1qs⌢i≥⟨pϵ0↾s⌢i,pϵ⟩
such that qs⌢i⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=i.
We now define the first coordinate as an amalgamation of the qs⌢i0’s:
pϵ+10=s∈Sϵ,i=0,1⋃qs⌢i0,
just like we did in the proof of theorem 3.9
of the original Sacks forcing. For the rest of the coordinates, we
define pϵ+1 with accordance to the path taken in
the first coordinate. Meaning that if qs⌢i0∈Gβ,β+1
then pϵ+1=qs⌢i. From the way we
defined the qs⌢i’s it is clear pϵ+1≥pϵ,
and so pϵ+1≥pϵ.
In limit stages δ≤ωβ we take pδ=ϵ<δ⋂pϵ,
so in the first coordinate we get a fusion sequence ⟨pϵ0∣ϵ<δ⟩
just like in the proof of theorem 4.10.
Therefore pδ0∈Qβ. As for the rest
of the coordinates, pδ∈\nicefracPζGβ+1˙
because of the ℵβ+1-closure, as shown in lemma 5.3.
So we can now define q=pωβ∈\nicefracPζGβ.
Note that we constructed such a q≥p over anyp∈Gβ,ζ,
so due to density we may assume without loss of generality that q∈Gβ,ζ.
Now let f={s∈q0∣z˙⟨q0↾s,q⟩⊆z˙Gβ,ζ}.
We claim f is a branch of q0.
From density we know that for every α<ωβ there
is an r∈Gβ,ζ such that r0 has a stem with length
at least α. Because Gβ,ζ is generic, there is
some condition t≥q∩r in Gβ,ζ. This t0
has a stem with length at least α, and so there is some node
s in level α of q0 such that t0≥q0↾s.
Obviously z˙⟨q0↾s,q⟩⊆z˙⟨t0,q⟩⊆z˙Gβ,ζ,
and so for every α<ωβ there is some s in that
level of q0 such that s∈f. Also, if s∈f, then it’s
trivial that for all αs↾α∈f.
Next, we show that f has no splitting nodes. Suppose s is a
splitting node of q, then for i=0,1z˙⟨q0↾s⌢0,q⟩(γ⟨q0↾s,q⟩)=z˙⟨q0↾s⌢1,q⟩(γ⟨q0↾s,q⟩)
and therefore either z˙Gβ,ζ=z˙⟨q0↾s⌢0,q⟩(γ⟨q0↾s,q⟩)
or z˙Gβ,ζ=z˙⟨q0↾s⌢1,q⟩(γ⟨q0↾s,q⟩),
so either s⌢0 or s⌢1 is in f, but not both.
Therefore s is not a splitting node in f, and so there are no
splitting nodes in f. We conclude that f is indeed a branch
in q0.
In fact, we claim that f is equal to the generic branch gβ,β+1
derived from the generic set Gβ,ζ. Let s∈gβ,β+1.
Then due to density there is an r∈Gβ,ζ such that
s is part of the stem of r0, and some condition t≥q∩r
in Gβ,ζ. As above, t0≥q0↾s,
and so z˙⟨q0↾s,q⟩⊆z˙⟨t0,q⟩⊆z˙Gβ,ζ.
Therefore s∈f.
Hence f=gβ,β+1 is a branch of q0 that is definable
in Mβ[z˙Gβ,ζ]=Mβ[A˙Gβ,ζ],
so gβ,β+1∈Mβ[A˙Gβ,ζ].
Meaning, using A˙Gβ,ζ=A, we managed to recover
the generic branch gβ,β+1. But remember, the generic
branch gβ,β+1 is in fact interdefinable with the generic
set Gβ,β+1, and so Gβ,β+1∈Mβ[A].
Therefore Mβ+1=Mβ[Gβ,β+1]=Mβ[gβ,β+1]⊆Mβ[A]⊆N,
in contradiction to our assumption that Mβ+1⊈N.
We conclude that if N violates the theorem, there is no greatest
β such that Mβ⊊N. Thus, the least inner
model of the tower that isn’t included in N must be Mδ
for some limit ordinal δ.
∎
Lemma 5.12**.**
If N is an inner model of Mζ, and
δ is a limit ordinal such that for all β<δN⊇Mβ,
then N⊇Mδ.
Proof.
We show this inductively. So let δ be a limit ordinal, and
assume the lemma is true for every limit ordinal ϵ<δ.
Let N be an inner model of Mζ such that for all β<δMβ⊊N. We aim to show that Gδ∈N by
showing that ⟨Gβ,β+1∣β<δ⟩∈N.
To start things off we first want to define a sequence g=⟨gβ∣β<δ⟩∈N
such that for each β<δ if β is not a limit ordinal
then gβ⊊ℵβ and gβ∈/Mβ.
Note that while each Mβ is by itself definable in N using
set parameters, the sequence might not be, so we can’t simply define
Aβ={a⊊ℵβ∣a∈N∖Mβ∧sup(a)=ℵβ}
and then choose some gβ∈Aβ whenever β is
not a limit.
Instead, we build this sequence inductively, working in N. Let
N0=L. Next, for all β<δ, assuming Nβ is
defined, let Aβ={a⊊ℵβ∣a∈N∖Nβ∧sup(a)=ℵβ},
and if Aβ=∅ choose some gβ∈Aβ,
otherwise set gβ=∅. For the successor step, assuming
Nβ is defined, we define Nβ+1=Nβ[gβ].
In the limit step, assuming Nβ is defined for all β<ϵ<δ,
we define Nϵ as the least inner model that includes every
Nβ.
We claim that for all β<δNβ is definable and
equal to Mβ, and that if β is not a limit ordinal
then gβ⊊ℵβ and gβ∈/Mβ.
For the base case, note that N0=L=M0, which is of course
definable in N. Next, assuming Nβ=Mβ, then Aβ={a⊊ℵβ∣a∈N∖Mβ∧sup(a)=ℵβ}.
If β is a limit, then Qβ is trivial, and
so Mβ=Mβ+1. On the other hand, \nicefracPζGβ˙
is ℵβ+1-closed, as shown by lemma 5.3.
Therefore Mβ and Mζ have the same subsets of ℵβ.
Hence Aβ=∅, and therefore gβ=∅.
We get Nβ+1=Nβ[∅]=Nβ=Mβ=Mβ+1
as required.
If β is not a limit, then Qβ is ℵβ-Sacks
forcing, and therefore there is a new subset of ℵβ
in Mβ+1∖Mβ. Hence Aβ=∅.
On the other hand \nicefracPζGβ+1˙
is ℵβ+1-closed, so Mβ+1 has the same subsets
of ℵβ as Mζ. Therefore Mβ⊊Mβ[gβ]⊆Mβ+1.
But Mβ+1 is generated from Mβ using ℵβ-Sacks
forcing, and so according to theorem 4.10
there is no intermediate model. Hence Nβ+1=Nβ[gβ]=Mβ[gβ]=Mβ+1.
In the limit step, assume that for ϵ a limit ordinal we’ve
already shown that Mβ=Nβ for all β<ϵ.
Gϵ∈N, and therefore Mϵ is definable with
set parameters in N. Hence N recognizes that Mϵ
is its inner model. Working towards a contradiction, assume K is
an inner model of N such that for all β<ϵMβ⊊K
but Mϵ⊈K. K is definable with set parameters
in N, which is definable with set parameters in Mζ. Therefore
K is definable with set parameters in Mζ, and therefore
K is an inner model of Mζ with said properties. But by
the induction hypothesis the lemma is true for every ϵ<δ,
so K⊇Mϵ in contradiction to our assumption.
Therefore there is no such inner model K. So every inner model
of N that includes all the Mβ’s for β<ϵ
must necessarily include Mϵ. Therefore Mϵ
is the least inner model that includes every Mβ. But this
exactly coincides with our definition of Nϵ, and so Nϵ=Mϵ.
Thus the induction is now complete and we’ve managed to define Nβ
and show that it is in fact equal to Mβ for all β<δ.
We’ve also shown that if β is not a limit then Aβ=∅
and so gβ⊊ℵβ and gβ∈/Mβ,
as required. Therefore the set ⟨gβ∣β<δ⟩∈N
is exactly the set which we set out to define.
The sequence ⟨gβ∣β<δ⟩∈Mζ.
However, by lemma 5.3\nicefracPζGδ
is ℵδ+1-closed, and for all β<δgβ∈Mδ.
Therefore ⟨gβ∣β<δ⟩∈Mδ.
Hence there exists a condition u∈Gδ that forces
[TABLE]
By the definition of the forcing, we may assume that for each β<δg˙β is a Pβ⋆Qβ-name
of gβ.
Assume now that we have a condition p∈Pδ stronger
than u. We denote individual coordinates like so: p=⟨p0,p˙1,...,p˙α,...∣α<δ⟩.
We know that p⊩g˙0⊊ℵ0∧g˙0∈/M0.
Let z˙ be the name of the characteristic function of g˙0.
For any condition q∈Pδ let z˙q be
the longest initial segment of z˙ that is decided by q,
let γq be the first ordinal for which z˙ is undecided,
let q0 denote the first coordinate of q, i.e from ℵ0-Sacks
forcing over M0, and let q denote the name of the rest
of the coordinates. zq˙ and γq must be well-defined,
because if q decides all of z˙, it decides all of g˙0,
and then g0∈M0, in contradiction to p. Plainly γq<ω.
We now build a fusion sequence. Mark p0=p. For the successor
case, assume that we’ve already chosen pϵ. Note that
pϵ0 is just a perfect tree in Q0, and
pϵ is a name in M0. Let Sϵ denote
the order ϵ splitting nodes of pϵ0. For
every splitting node s∈Sϵ, let’s look at γ⟨pϵ0↾s,pϵ⟩
and conditions ⟨pϵ0↾s⌢i,pϵ⟩.
Suppose that for both i=0,1 we have ⟨pϵ0↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j.
γ⟨pϵ0↾s,pϵ⟩
is undecided, so take q≥⟨pϵ0↾s,pϵ⟩
such that q⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=1−j.
Either q∩⟨pϵ0↾s⌢0,pϵ⟩∈Pδ
or q∩⟨pϵ0↾s⌢1,pϵ⟩∈Pδ.
But q and ⟨pϵ0↾s⌢i,pϵ⟩
are incompatible for i=0,1, therefore our supposition is impossible.
Thus, if for a certain i there is a j such that ⟨pϵ0↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j,
we can define qs⌢i=⟨pϵ0↾s⌢i,pϵ⟩
so that qs⌢i⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=j,
and know there will be some qs⌢(1−i)≥⟨pϵ0↾s⌢(1−i),pϵ⟩
such that qs⌢(1−i)⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=1−j.
Alternatively, there is no such i, and we are free to select for
both i=0,1qs⌢i≥⟨pϵ0↾s⌢i,pϵ⟩
such that qs⌢i⊩z˙(γ⟨pϵ0↾s,pϵ⟩)=i.
We now define the first coordinate as an amalgamation of the qs⌢i0’s:
pϵ+10=s∈Sϵ,i=0,1⋃qs⌢i0,
just like we did in the proof of theorem 3.9
of the original Sacks forcing. For the rest of the coordinates, we
define pϵ+1 with accordance to the path taken in
the first coordinate. Meaning that qs⌢i0 forces pϵ+1=qs⌢i.
From the way we defined the qs⌢i’s it is clear pϵ+1≥pϵ,
and so pϵ+1≥pϵ.
In the first coordinate we get a classical fusion sequence ⟨pϵ0∣ϵ<ω⟩.
Therefore we can define pω0=ϵ<ω⋂pϵ0∈Q0.
As for the higher coordinates, we can define pω=ϵ<ω⋂pϵ0
because of the ℵ1-closure, proven in lemma 5.3.
So pω∈Pδ.
So we can now define p(0)=pω. Next we are going
to repeat by induction the construction above, using the higher coordinates.
So for the successor case, assume that we’ve already defined p(β).
For all coordinates 0<α≤β define p˙(β+1)α=p˙(α)α,
and let p(β+1)0=p(0)0. We are now
going to deal with g˙β+1. So let Hβ+1⊆Pβ+1
be any generic set such that ⟨p(β)0,p˙(β)1,...,p˙(β)α,...∣α≤β⟩∈Hβ+1.
Note that M0[Hβ+1]=the interpretation of M˙β+1 according to H˙β+1.
So working in M0[Hβ+1], we know that
[TABLE]
Just as before, let z˙ be the name of the characteristic function
of g˙β+1. For any condition q∈\nicefracPδHβ+1
let z˙q be the longest initial segment of z˙ that
is decided by q, let γq be the first ordinal for which
z˙ is undecided, let qβ+1 denote the first
coordinate of q, i.e from ℵβ+1-Sacks forcing over
M0[Hβ+1], and let q denote the name
of the rest of the coordinates. zq˙ and γq must
be well-defined, because if q decides all of z˙, it decides
all of g˙β+1, and then gβ+1∈M0[Hβ+1],
in contradiction to p(β). Plainly γq<ωβ+1.
Just as before, we again build a fusion sequence. This time mark p0=p(β).
For the successor case, assume that we’ve already chosen pϵ.
Note that pϵβ+1 is just a perfect tree in Qβ+1,
and pϵ is a name of a condition. Let Sϵ
denote the order ϵ splitting nodes of pϵβ+1.
For every splitting node s∈Sϵ, let’s look at γ⟨pϵβ+1↾s,pϵ⟩
and conditions ⟨pϵβ+1↾s⌢i,pϵ⟩.
Suppose that for both i=0,1 we have ⟨pϵβ+1↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=j.
γ⟨pϵβ+1↾s,pϵ⟩
is undecided, so take q≥⟨pϵβ+1↾s,pϵ⟩
such that q⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=1−j.
Either q∩⟨pϵβ+1↾s⌢0,pϵ⟩∈\nicefracPδHβ+1
or q∩⟨pϵβ+1↾s⌢1,pϵ⟩∈\nicefracPδHβ+1.
But q and ⟨pϵβ+1↾s⌢i,pϵ⟩
are incompatible for i=0,1, therefore our supposition is impossible.
Thus, if for a certain i there is a j such that ⟨pϵβ+1↾s⌢i,pϵ⟩⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=j,
we can define qs⌢i=⟨pϵβ+1↾s⌢i,pϵ⟩
so that qs⌢i⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=j,
and know there will be some qs⌢(1−i)≥⟨pϵβ+1↾s⌢(1−i),pϵ⟩
such that qs⌢(1−i)⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=1−j.
Alternatively, there is no such i, and we are free to select for
both i=0,1qs⌢i≥⟨pϵβ+1↾s⌢i,pϵ⟩
such that qs⌢i⊩z˙(γ⟨pϵβ+1↾s,pϵ⟩)=i.
We now define the first coordinate as an amalgamation of the qs⌢iβ+1’s:
pϵ+1β+1=s∈Sϵ,i=0,1⋃qs⌢iβ+1.
For the rest of the coordinates, we define pϵ+1
with accordance to the path taken in the first coordinate. Meaning
that qs⌢iβ+1 forces that pϵ+1=qs⌢i.
From the way we defined the qs⌢i’s it is clear pϵ+1≥pϵ,
and so pϵ+1≥pϵ. In limit stages τ<ωβ+1
we just use the ℵβ+1-closure to define pτ=α<τ⋂pα.
Now, in order to define pωβ+1, note that in the
first coordinate we again get a fusion sequence ⟨pϵβ+1∣ϵ<ω⟩.
Therefore pωβ+1β+1=ϵ<ωβ+1⋂pϵβ+1∈Qβ+1.
As for the rest of the coordinates, we can use the ℵβ+1-closure
proven in lemma 5.3 to define pωβ+1=ϵ<ωβ+1⋂pϵ.
For all α>β we define p˙(β+1)α=pωβ+1α.
We now pick a name ⟨p˙(β+1)β+1,p˙(β+1)β+2,...,p˙(β+1)α,...∣α<δ⟩
for the ⟨p(β+1)β+1,p˙(β+1)β+2,...,p˙(β+1)α,...∣α<δ⟩
that we constructed in \nicefracPδHβ+1,
such that ⟨p(β)0,p˙(β)1,...,p˙(β)β⟩
forces it to be the way it was defined.
Finally, set p(β+1)=⟨p(0)0,p˙(1)1,...,p˙(β+1)β+1,p˙(β+1)β+2,...,p˙(β+1)α∣α<δ⟩.
For limit stages δ we define p(δ)=α<δ⋂p(α).
As each coordinate lesser than δ stabilizes, and by lemma
5.3 each coordinate ≥δ is at least
ℵδ+1-closed, and because we’re using full support,
p(δ)∈Pδ.
As we’ve constructed p(δ) stronger than a general
p∈Pδ, then by density arguments, we may assume
without loss of generality p(δ)∈Gδ.
Now we’re going to use p(δ) and the sequence of
⟨gβ∣β<δ⟩ to recover
⟨Gβ,β+1∣β<δ⟩.
By induction, assume that for some β<δ we already recovered
⟨Gα,α+1∣α<β⟩,
and thus Mβ. Therefore, working in Mβ, let z˙
be as before the name of the characteristic function of g˙β
and for each r∈\nicefracPδGβ let
z˙r be the longest initial segment of z˙ that is
decided by r.
Let q=p(δ), and define f={s∈qβ∣z˙⟨qβ↾s,q⟩⊆z˙Gβ,δ}.
We claim that f is a branch of qβ.
From density we know that for every α<ωβ there
is an r∈Gβ,δ such that rβ has a stem with
length at least α. Because Gβ,δ is generic,
there is some condition t≥q∩r in Gβ,δ. This
tβ has a stem with length at least α, and so there
is some node s in level α of qβ such that tβ≥qβ↾s.
Obviously z˙⟨qβ↾s,q⟩⊆z˙⟨tβ,q⟩⊆z˙Gβ,δ,
and so for every α<ωβ there is some s in that
level of qβ such that s∈f. Also, if s∈f, then
it’s trivial that for all αs↾α∈f.
Next, we show that f has no splitting nodes. Suppose that s
is a splitting node of qβ, then for i=0,1z˙⟨qβ↾s⌢0,q⟩(γ⟨qβ↾s,q⟩)=z˙⟨qβ↾s⌢1,q⟩(γ⟨qβ↾s,q⟩)
and therefore either z˙Gβ,δ=z˙⟨qβ↾s⌢0,q⟩(γ⟨qβ↾s,q⟩)
or z˙Gβ,δ=z˙⟨qβ↾s⌢1,q⟩(γ⟨qβ↾s,q⟩),
so either s⌢0 or s⌢1 is in f, but not both.
Therefore s is not a splitting node in f, and so there are no
splitting nodes in f. We conclude that f is in fact a branch
in qβ.
Moreover, we claim that f is equal to the generic branch hβ,β+1
derived from the first coordinate of the generic set Gβ,δ.
Let s∈hβ,β+1. Then due to density there is an r∈Gβ,δ
such that s is part of the stem of rβ, and some condition
t≥q∩r in Gβ,δ. As above, tβ≥qβ↾s,
and so z˙⟨qβ↾s,q⟩⊆z˙⟨tβ,q⟩⊆z˙Gβ,ζ.
Therefore s∈f.
Hence f=hβ,β+1 is a branch of qβ that is definable
in Mβ[z˙Gβ,δ]=Mβ[gβ˙Gβ,δ],
and therefore in Mβ[g˙Gβ,δ].
Meaning we managed to recover the generic branch hβ,β+1.
But remember, the generic branch hβ,β+1 is in fact interdefinable
with the generic set Gβ,β+1, and so Gβ,β+1∈Mβ[g˙Gβ,δ].
But by our inductive assumption ⟨Gα,α+1∣α<β⟩∈M0[g˙Gδ].
Therefore Gβ,β+1∈M0[g˙Gδ].
Completing the induction, Gδ∈M0[g˙Gδ],
and so Mδ⊆M0[g˙Gβ,δ].
But M0[g˙Gδ]⊆N, and so we
conclude Mδ⊆N as required.
∎
Looking at the proof of lemma 5.12, it becomes evident
why we couldn’t have used bounded support even in regular limits:
condition p(δ) that lies at the heart of the proof
satisfies p˙(δ)α=0 whenever Qα
is non-trivial. Moreover, had we used bounded support, we would have
needed to construct a condition p that is at once bounded, and
so has at most than ℵβ<ℵδ splitting nodes,
yet is still somehow able to distinguish the value of the generic
branch in Qβ+1, even though in the standard forcing
that task requires ℵβ+1 splitting nodes.
At last we arrive at the central theorem for the model tower construction:
Theorem 5.13**.**
K* is an inner model of Mζ if
and only if for some α≤ζK=Mα.*
Proof.
Suppose to the contrary that K is an inner model of Mζ
such that K=Mα for all α≤ζ. Then there
is a minimal ordinal β such that K⊉Mβ.
By lemma 5.11β must be a limit ordinal. So K⊋Mα
for all α<β, but K⊉Mβ. However, according
to lemma 5.12 if K⊋Mα for all
α<β then K⊇Mβ. We arrived at a contradiction.
Meaning that there is no such inner model K.
Thus K is an inner model of Mζ if and only if K=Mα
for some α<ζ.
∎
Corollary 5.14**.**
There exists a well-ordered model universe of arbitrary height.
Proof.
Mζ is a well-ordered model universe of height ζ.
∎
6 Class forcing
In the previous section we defined the iterated forcing notion for
sets, and we used it to construct a well-ordered model universe of
arbitrary height. Because that iteration could successfully go through
strongly inaccessible cardinals, we proved that the existence of well-ordered
model universes with ordinal height is in fact consistent with ZFC.
We could simply take V[Gκ]κ, where κ
is a strongly inaccessible cardinal, and Gκ is the generic
set of Pκ as defined in 5.1.
Now however we want to iterate our model tower ’all the way’ by the
use of class forcing. And to make formal use of class forcing, we
return in this section to the axiomatic framework of BGC,
as expounded upon in the introduction. So for the rest of this section
we shall assume to be working within ⟨V,V,∈⟩
a model of BGC, and we shall use forcing to extend a base
model of BGC to another model thereof.
A basic introduction of class forcing the reader may be found in Friedman
[5]. For a more thorough presentation of class forcing
within the context of BGC the reader may refer to Reitz
(appendix A of [15]).
Before going on, it is important to note the main difficulty with
class forcing, which is that unlike set forcing, the generic extension
of class forcing might actually fail to be a model of BGC
(and its sets a model of ZFC). Specifically, the Power
Set Axiom and the Axiom of Replacement might fail (theorem 91 in [15]).
For BGC and ZFC to be satisfied, we will need
to prove that our forcing iteration is progressively closed,
as will be defined later.
We fix the base of our forcing iteration to be ⟨L,L,∈⟩,
where of course L is the constructible universe, and L
is the collection of classes definable therein (remember fact 1.4).
Definition 6.1**.**
Let P∈L be a partially
ordered class defined as follows:
Let Q˙α be trivial if α is a limit
ordinal, and let it be the name of ℵα-Sacks forcing
in VPα otherwise.
2. 2.
Pα+1=Pα⋆Q˙α.
3. 3.
At limit stages we use full support, i.e if δ is a limit ordinal
then p∈Pδ⇔∀α<δ(p↾α∈Pα).
4. 4.
P=α∈Ord⋃Pα.
That is, every condition in P is bounded in its coordinates.
It should be noted that unlike in the ordinal limit stages, where
we use the indirect limit (i.e full support) all the way through,
in the class limit we employ the direct limit instead.
As explained in the previous section, using indirect limits even for
regular cardinals would have spoiled the construction of the condition
used to simultaneously discover all the generic sets - which was necessary
to prove that no inner model ’squeezes in’ between the ascending chain
of models and the limit model. But as will be shown later, unlike
the ordinal limit stages, if we use a direct limit in the class stage
the generic extension is simply the union of the ascending chain,
and therefore automatically minimal over it. So the entire construction
of theorem 5.13 is unnecessary for the class limit
case.
Definition 6.2**.**
Denote:
M0=L.
2. 2.
Gα as the generic set in partial order Pα
over M0.
3. 3.
Mα=M0[Gα].
4. 4.
G as the generic class in partial order P
over M0.
5. 5.
M0[G] the generic extension of ⟨L,L,∈⟩
by G.
6. 6.
M∞ the restriction of M0[G] to
sets.
Note that we have yet to establish that M0[G]
is a model of BGC, or that M∞ is a model of
ZFC.
Lemma 6.3**.**
The forcing \nicefracPGα
is ℵα-closed.
Proof.
Each coordinate is ℵα-closed, and the limit of a set
of bounded conditions in \nicefracPGα is itself
bounded.
∎
Lemma 6.4**.**
M∞* has the same cardinals as
M0.*
Proof.
Let ℵα be a cardinal in M0. According to lemma
5.10M0 has the same cardinals as
Mα+1.
But by lemma 6.3 the forcing \nicefracPGα+1
is ℵα+1-closed, and so adds no new subsets of ℵα.
Therefore ℵαM0=ℵαMα+1=ℵαM∞,
and so all cardinals are preserved.
∎
Lemma 6.5**.**
M∞* satisfies the Power Set Axiom.*
Proof.
It is enough to prove the Power Set Axiom for cardinals. Let ℵα
be a cardinal in M∞. By lemma 6.4
it is also a cardinal in Mα+1.
By lemma 6.3 the forcing \nicefracPGα+1
is ℵα+1-closed, and so adds no new subsets of ℵα.
Therefore P(ℵα)Mα+1=P(ℵα)M∞.
Thus the power set of ℵα is also a set in M∞.
∎
Definition 6.6**.**
A partially order class R is a chain of complete
subposets if R=α∈Ord⋃Rα,
where each Rα is a partially ordered set, such
that if α≤β then Rα is a complete
suborder of Rβ.
Lemma 6.7**.**
P* is a chain of complete
subposets.*
Proof.
By definition 6.1P=α∈Ord⋃Pα.This
is an iterated forcing, and so the identity map iα,β:Pα→Pβ
is a complete embedding (see ch. VIII lemma 5.11 in [11]).
Therefore Pα is a complete suborder of Pβ.
∎
Definition 6.8**.**
P=α∈Ord⋃Pα
is a progressively closed iteration if P is a chain
of complete subposets, and for arbitrarily large regular cardinals
δ there are arbitrarily large α such that there is
a Pα-name P˙[α,∞)={⟨op(βˇ,P˙[α,β)),0⟩∣β>α}
satisfying:
For every β>α the poset Pβ is isomorphic
to the two-stage iteration Pβ≅Pα⋆P˙[α,β);
2. 2.
Pα⊩δˇ is a regular cardinal and P˙[α,β) is <δˇ-closed;
3. 3.
For β′>β>α the isomorphisms at β and β′
yield complete subposets Pα⋆P˙[α,β)⊆cPα⋆P˙[α,β′)
such that the complete embeddings commute with the isomorphisms.
4. 4.
Pα⊩P˙[α,∞) is a chain of complete subposets.
Lemma 6.9**.**
P* is a progressively closed iteration.*
Proof.
By lemma 6.7P is a chain of
complete subposets. Let ℵδ be a successor cardinal,
and α=δ+1.
By lemma 6.4 all the cardinals are preserved,
and by lemma 6.3P˙[α,β)
is ℵδ+1-closed.
3. 3.
Let p∈Pβ. Then p=⟨p<α,p˙α,p˙α+1,...,p˙γ,...∣γ<β⟩.
Which embeds to ⟨p<α,p˙α,p˙α+1,...,p˙γ,...,0,...∣γ<β⟩∈Pα⋆P˙[α,β′).
Similarly p embeds to ⟨p<β,0,...⟩∈Pβ′,
which through the isomorphism is equal to ⟨p<α,p˙α,p˙α+1,...,p˙γ,...,0,...∣γ<β⟩∈Pα⋆P˙[α,β′).
Hence the complete embeddings commute with the isomorphisms.
4. 4.
Lemma 6.7 applies to the tail of the forcing
as well.
∎
Lemma 6.10**.**
M0[G]⊨BGC* and M∞⊨ZFC.*
Proof.
By theorem 98 of [15] a progressively closed iteration generates
a generic extension that satisfies BGC.
According to lemma 6.7P is a
chain of complete subposets. So applying lemma 88 of [15]
to the sets, we get M∞=α∈Ord⋃M0[Gα].
∎
Theorem 6.12**.**
N* is a proper inner model of M∞
if and only if for some α∈OrdN=Mα.*
Proof.
Working to the contrary, assume there exists N an inner model of
M∞ such that N=Mα for all α∈Ord.
Suppose there exists a greatest ordinal β such that Mβ⊆N.
Mβ⊨AC so according to Vopěnka [19]
there exists a set of ordinals A∈N∖Mβ. By lemma
6.11 there exists an ordinal α such that
A∈Mα. But that means Mβ⊊Mβ[A]⊆Mα,
so according to theorem 5.13Mβ[A]=Mγ
for some β<γ≤α.
Therefore Mβ[A]=Mγ⊆N, in contradiction
to β being the greatest ordinal such that Mβ⊆N.
So there is no such greatest β.
Next, suppose that for a limit ordinal δ, Mα⊊N
for all α<δ.
As shown in the proof of lemma 5.12, there is a set
g=⟨gα∣α<δ⟩∈N such
that if α is not a limit ordinal gα∈Mα+1∖Mα
and gα⊆ℵα, and if α is a limit
ordinal then gα=∅. Define gα′={ωα+β∣β∈gα}.
Obviously gα′⊆ωα+1∖ωα.
Define g′=α<δ⋃gα′.
We get g′∈N a set of ordinals. If g′∈Mα for some
α<δ then gα′=(g′∩ωα+1)∖ωα∈Mα
and then gα∈Mα in contradiction to its definition.
Therefore g∈/Mα for all α<δ.
By lemma 6.11 there exists an ordinal β
such that g′∈Mβ, so M0[g′]⊆Mβ.
According to theorem 5.13 this means M0[g′]=Mγ
for some δ≤γ≤β, and so M0[g′]⊇Mδ.
On the other hand, because g′ is a set of ordinals and M0⊨AC,
M0[g′] is the smallest model of ZFC such
that g′∈M0[g′], and so M0[g′]⊆N.
Hence Mδ⊆N.
As a result, by induction for all α∈OrdMα⊆N,
and so M∞=α∈Ord⋃Mα⊆N,
in contradiction of N being a proper inner model of M∞.
We conclude that N is a proper inner model of M∞ if
and only if N=Mα for some α∈Ord.
∎
At last, we arrive at what we set out to prove:
Corollary 6.13**.**
The existence of well-ordered model universes with the height of the
ordinals is consistent with ZFC.
Proof.
Theorem 6.12 shows that M∞ is a
well-ordered model universe with ht(M∞)=∞.
∎
We conclude this section with the observation that in M∞
the class of inner models M(M∞) (as
defined in 1.7) is in fact definable in
M∞: M0=L, for all αMα+1=L(P(ℵα)M∞),
and for all limit δMδ=L(P(ℵδ)M∞).
Thus M∞ in a sense ’knows’ that it is a well-ordered model
universe.
7 Open questions
In the previous section we constructed an example of a nice well-ordered
model universe of height equal to Ord. We did this by
an iteration of progressively increasing κ-Sacks forcing.
In this section we discuss some remaining open questions regarding
well-ordered model universes:
Can we construct a well-ordered model universe that isn’t nice?
2. 2.
What can we say about models when the inner models are just totally-ordered,
not well-ordered by inclusion?
3. 3.
What if we consider all inner models of ZF, not just inner
models of ZFC?
7.1 Non-nice well-ordered model universes
For the first question, recall definition 1.11. A well-ordered
model universe is considered nice if its underlying order is equivalent
to some ordinal or to Ord. This is essentially a limit
on the length of the well-ordering. Any well-ordered set is order-isomorphic
to some ordinal, so if a well-ordered model universe isn’t nice then
the underlying order must be a proper class, but one which is not
order-isomorphic to Ord.
Can we define such a well-ordering? Of course - just take A={α∣α∈Ord∨α={1}},
and extend the natural ordering by defining {1}>α
for all α∈Ord. It is easy to see that this is indeed
a well-ordering: if B⊆A is a non-empty class, then if
it contains any ordinal, then the least ordinal it contains is its
least element according to our extended ordering, and if not then
{1} is the least element. It is also obvious that
our extended ordering is not order-isomorphic to Ord -
our ordering has a greatest element, whereas Ord clearly
does not.
So such a well-ordering is very much definable. Could we extend our
construction further then we did in the previous section?
For the rest of the subsection, let M∞ be as defined in
6.2. In general, there is no obstacle to applying
ℵβ-Sacks forcing to M∞. The normal properties
of Sacks-forcing would still hold, i.e there won’t be any intermediate
model between M∞ and M∞[G]. Moreover,
there would be a chain of inner models of M∞[G]
that would be ’longer’ than Ord. However, regardless of
the ℵβ-Sacks forcing we use, M∞[G]
would invariably contain some new inner model that is not on the chain.
Lemma 7.1**.**
Let ℵβ be a regular cardinal,
let S be the ℵβ-Sacks forcing notion over
M∞, and let H be a generic set in S. Then
M∞[H] is not a well-ordered model universe.
Proof.
By lemma 6.11M∞=α∈Ord⋃.
Therefore P(S)∈Mα for
some α∈Ord. As all the dense sets of S
in M∞ are already in Mα, H is also a generic
set of S∈Mα, and so Mα[H]
is a generic extension generated by ℵβ-Sacks forcing
over Mα.
Obviously Mα[H]⊆M∞[H],
but Mα[H]⊈M∞. So for all γ∈OrdMα[H]=Mγ and Mα[H]=M∞.
By theorem 4.10 there are no intermediate
models between M∞ and M∞[H]. So either
Mα[H]=M∞[H], or Mα[H]
is ’off-chain’.
But Mα[H] is ℵβ-Sacks forcing
over Mα, and so has no intermediate model between Mα
and Mα[H], whereas Mα⊊M∞⊊M∞[H].
Therefore Mα[H]=M∞[H],
so Mα[H] is ’off-chain’, and M∞[H]
is not a well-ordered model universe.
∎
Okay, but lemma 7.1 only shows that we can’t use
κ-Sacks forcing to produce the next step of the construction.
Could some other set forcing notion do the trick for us?
Looking back at theorem 5.13, we proved that we
could take any set created by the iteration and use it to completely
recover all the preceding generic sets. So essentially, each generic
set must code all the preceding generic sets. But because we used
class-many generic sets to construct M∞, we need our new
generic set to encode class-many previous generic sets, which is a
tall order. In fact, it is impossible:
Lemma 7.2**.**
Let S be some minimal set forcing
notion over M∞, and let H be a generic set in S.
Then M∞[H] is not a well-ordered model universe.
Proof.
Let ℵβ=∣S∣. By lemma 6.11
there exists an ordinal α such that P(S)∈Mα.
Take γ=max(α,β+1).
Obviously S has the ℵβ+1-c.c property.
By lemma 6.3\nicefracPGγ
is at least ℵβ+1-closed, and by lemma 6.7\nicefracPGγ is a chain of complete subposets.
Because S∈Mγ, M∞[H] is
actually the result of product forcing, where the first forcing is
the tail of class forcing P and the second set forcing
S, so M∞[H]=Mγ[\nicefracGGγ][H].
By lemma 121 in [15] we have that \nicefracGGγ
is \nicefracPGγ-generic over Mγ[H].
Therefore Mγ[H]=M∞[H].
Also, because H∈Mγ[H] we have Mγ[H]⊈M∞.
And because S is minimal, so there is no intermediate
model between M∞ and M∞[H].
Therefore Mγ[H] is a proper inner model of M∞[H]
that is off the chain, so M∞[H] is not a well-ordered
model universe.
∎
And what about class forcing? Could that be used to somehow lengthen
our well-ordered model universe?
For class forcing of chain of complete subposets, this is again impossible.
Lemma 7.3**.**
Let S be a chain of complete subposets over M∞,
and let H be a generic class in S
such that M∞[H]⊨ZFC.
Then M∞[H] is not a well-ordered model
universe.
Proof.
According to lemma 88 of [15] M∞[H]=α∈Ord⋃M∞[Hα],
so A∈M∞[Hα] for some α∈Ord.
But by lemma 7.2M∞[Hα]
is not a well-ordered model universe, so the inner models of M∞[Hα]
are not well-ordered by inclusion.
As M∞[Hα] is definable with set parameters
in M∞[H], so are the inner models of
M∞[Hα] similarly definable, and so they
are inner models of M∞[H].
Therefore the inner models of M∞[H]
are not well-ordered by inclusion.
∎
What about some more general form of class forcing?
At first thought this might also appear impossible, because even for
class forcing to minimally extend M∞, we would still need
every new set in M∞[H]
to somehow encode the entire class of generic sets! However,
the remarkable Jensen’s Coding Theorem [1] actually uses
class forcing to achieve something similar: the existence of class
forcing notion P such that if G is P-generic
over V then V[G]⊨ZFC+V=L[A]+A⊆ω.
This set A in effect ’codes the universe’. Applying the theorem
to M∞, all we really need is for every new set in M∞[H]
to code this A, which sounds far more reasonable. So we are left
with the following open question:
Problem 7.4**.**
Is the existence of a well-ordered model universe with an underlying
order longer than Ord consistent with ZFC?
7.2 Totally-ordered model universes
So far in this article we focused exclusively on models where all
the inner models are well-ordered by inclusion. However, a natural
weakening of the definition is to demand the ordering to only be total,
i.e for any two inner models of V, either M1⊆M2
or M2⊆M1. We’ll call this a totally-ordered
model universe.
Prima facie, this concept is far weaker than a well-ordered model
universe. For one, our proof that V has no measurable cardinals
in theorem 2.2 immediately fails, because in theory
there could be an infinite descending sequence of inner models. However,
upon closer inspection we find that the proof of theorem 2.3
actually still holds, because it hinges on the fact if both G0
is generic over L[G1] and vice-versa, then the inner
models aren’t totally-ordered. Therefore:
Theorem 7.5**.**
If V is a totally-ordered model universe, 0♯ doesn’t
exist.
If V is a totally-ordered model universe, then V has no measurable
cardinal.
Proof.
By Gaifman [6], the existence of a measurable cardinal
implies the existence of 0♯.
∎
So some of the basic properties of well-ordered model universes extend
to totally-ordered model universes, and the total-ordering property
by itself is sufficient to prove that V is inherently small and
quite ’close’ to L.
Therefore for each result proven about well-ordered model universes,
we should ask ourselves whether it extends to totally-ordered model
universes as well.
7.3 Inner models of ZF
Another natural extension of the definition of a well-ordered model
universe is the consideration of general inner models of ZF,
not just inner models that satisfy Choice.
Returning to the framework we introduced at the beginning, we give
the following expanded definition, which is almost verbatim definition
1.7:
Definition 7.7**.**
Let ⟨V,V,∈⟩
be a model of BG. We call a model N⊆V of ZF
where all its inner ZF models are well-ordered with respect
to inclusion a well-ordered ZF model universe.
Formally, we postulate the existence of a class M in ⟨V,V,∈⟩,
which is the sequence of all proper inner ZF models of
N ordered by inclusion. This means:
M⊆I×N;
2. 2.
M is a proper inner ZF model
of N if and only if there exists a unique a∈I such that M=Ma={x∣(a,x)∈M};
3. 3.
I is a well-ordered class;
4. 4.
If a<Ib then (a,x)∈M→(b,x)∈M.
In summary, applying the convention that lower-case letters indicate
sets and upper-case letters indicates classes, we demand the following
be true:
[TABLE]
We define the height of V in exactly the same way we did for the
original definition, using the order type of I.
Obviously, if N⊨ZFC is a well-ordered ZF
model universe then it is also a well-ordered model universe. Given
the sequence of proper ZF inner models we can directly
define the sequence of proper ZFC inner models as defined
in 1.7. However, by our definition well-ordered
ZF model universes are not required themselves to satisfy
AC, and therefore not every well-ordered ZF
model universe is necessarily a well-ordered model universe.
Next we outline a few of the basic properties of well-ordered ZF
model universes. For the rest of the subsection, assume ⟨V,V,∈⟩⊨BG,
V is a well-ordered ZF model universe, and M⊆I×V
is its sequence of proper inner ZF models ordered by inclusion.
The proof of theorem 2.4 doesn’t work for
well-ordered ZF model universes, as it involves heavy
use of the Axiom of Choice. In general, it is very much possible to
have an infinite chain of inner models that satisfy AC,
but that the least inner model to include them all does not. So even
though we can carry out the successor stage of the proof, we cannot
prove Mω⊨AC, meaning it is very possible
Mω=L[A] for all A∈V.
Note that because we can still carry out the successor stages of the
induction, we have the following corollary:
Corollary 7.10**.**
If V⊨¬AC then ht(V)≥ω.
Proof.
Applying the successor steps in the proof of theorem 2.4,
we get that for all n<ωMn=L[A] for some
A∈V.
∎
This result can actually be strengthened, even without the well-ordering
property:
Theorem 7.11**.**
If V is not of the form L[A]
for some A∈V, then V has an infinite number of inner models.
Proof.
By induction we prove every model with a finite number of inner models
is of the form L[A]. The case for [math] proper inner
models is trivially true because L=L[∅].
Now assume that we’ve proven the induction for models with n proper
inner models. Assume K is a model that has n+1 proper inner
models. Every inner model of an inner model of K is an inner model
of K itself, so all inner models of K have at most n proper
inner models, and so they are all of the form L[Am]
for some m≤n.
Let’s consider two possibilities: either K has a greatest proper
inner model R⊊K, or it doesn’t. If R exists, then
by Vopěnka [19] there is a set of ordinals A∈K∖R,
and so K=L[A] as required.
Otherwise, just as we did in the proof of theorem 2.4,
we can arrange a family of mutually disjoint sets of ordinals Bm
such that for all mL[Am]=L[Bm].
Take L[m≤n⋃Bm]. For all mL[m≤n⋃Bm]⊇L[Bm].
But the only model that includes all proper inner models of K is
K itself. Therefore L[α⋃Bα]=K.
Therefore if V is not a model of the form L[A], V
must have an infinite number of inner models.
∎
Corollary 7.12**.**
If V is a well-ordered ZF
model universe and ht(V)<ω then V is a well-ordered
model universe.
Proof.
According to theorem 7.11V and all of its
inner models must satisfy the Axiom of Choice. Therefore by definition
7.7V is a well-ordered model universe.
∎
Despite the failure of theorem 2.4 for well-ordered
ZF model universes, we have a small consolation prize:
Lemma 7.13**.**
Let α+1∈I denote the successor
of α in the well-ordering of I. Then Mα+1=L(A)
for some A∈V.
Proof.
Take A∈Mα+1∖Mα. L(A)
is the smallest inner ZF model containing A, so obviously
L(A)⊆Mα+1. But because V is a well-ordered
universe Mα⊊L(A), and there are no
intermediate inner models between Mα and Mα+1.
Therefore L(A)=Mα+1.
∎
Corollary 7.14**.**
If V has a greatest proper inner ZF model K, there
exists A∈V such that V=L(A).
Proof.
Take A∈V∖K. We get K⊊L(A)⊆V,
and so for the same reasons as lemma 7.13L(A)=V.
∎
In conclusion, there isn’t much we know about well-ordered ZF
model universes.
As for actually constructing a well-ordered ZF model universe,
corollary 7.12 shows that using the iteration defined
in 5.1 up to finite height, would generate a well-ordered
ZF model universe.
A well-ordered ZF model universe of height ω is
achievable by iterating the forcing up to Mω, and then
taking N=HOD(⟨Gn∣n<ω⟩)⊊Mω.
This N will be the minimal inner model of ZF that includes
Mn for all n<ω.
However, we can’t use the same construction to build well-ordered
ZF model universes of arbitrary height, because we can’t
tell what’s going on between N and Mω. The intermediate
inner models there might not even be totally-ordered.
So we are left with one glaring open question:
Problem 7.15**.**
Is the existence of a well-ordered ZF model universe the
height of the ordinals consistent with ZF?
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