DIAGONAL SUPERCOMPACT RADIN FORCING
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
1. Introduction
In this paper we consider two methods for producing models with some global
behavior of the continuum function on singular cardinals and the failure of weak
square. The first method is as an extension of Sinapova’s work [21]. We define a
diagonal supercompact Radin forcing which adds a club subset to a cardinal κ while
forcing the failure of SCH everywhere on the club. The intuition from Sinapova’s
work was that our model should have the failure of weak square at every point in
the club. Unfortunately, the additional closure required to preserve inaccessibility
enforces weak square at some points.
Theorem 1.1. If there are a supercompact cardinal κ and a weakly inaccessible
θ > κ, then there is a forcing extension in which κ is regular and there are a club
C ⊆ κ of singular cardinals where SCH fails, and ∗ν fails for every ν ∈ C \ cof(ω).
In the second method, we use Gitik’s method of iterating Prikry type posets to
iterate Sinapova’s poset and obtain the following theorem.
Theorem 1.2. Let κ be a regular cardinal. If for all regular τ < κ the set {ν < κ | ν
is ν +τ -supercompact } is stationary, then there is a forcing extension in which κ is
inaccessible and there is a club C in κ such that for all κ ∈ C both ∗κ and SCH
fail.
Although the second theorem is formally better than the first, we believe that
the techniques from the proof are unlikely to generalize.
We are motivated by the question of whether in ZFC one can construct a κAronszajn tree for some κ > ω1 . The question is also open if we ask for a special
κ-Aronszajn tree. Forcing provides a possible path to a negative solution by showing
that it is consistent with ZFC that there are no κ-Aronsajn trees on any κ > ω1 .
By a theorem of Jensen [?], ∗µ is equivalent to the existence of a special µ+ Aronszajn tree. So our theorems are partial progress towards a model with no
special Aronszajn trees.
The non-existence of κ-Aronszajn trees (the tree property at κ) and the nonexistence of special κ-Aronszajn trees (failure of ∗ ) are reflection principles which
are closely connected with large cardinals. For example, theorems of Erdös and
Tarski [6], and Monk and Scott [17], show that an inaccessible cardinal is weakly
compact if and only if it has the tree property. Further Mitchell and Silver [16]
showed that the tree property at ℵ2 is consistent with ZFC if and only if the
existence of a weakly compact cardinal is.
Specker [25] showed that if κ<κ = κ, then there is a special κ+ -Aronszajn tree.
This theorem places an important restriction on models where there are no special
Date: February 9, 2016.
1
2
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
Aronszajn trees. From Specker’s theorem, a model with no special κ-Aronszajn
trees for any κ > ℵ1 must be one in which GCH fails everywhere. In particular
GCH must fail at every singular strong limit cardinal, a failure of the Singular
Cardinals Hypothesis (SCH). The consistency of the failure of SCH requires large
cardinals [10] and a model in which GCH fails everywhere was first obtained by
Foreman and Woodin [8].
There are many partial results towards constructing a model in which every
regular cardinal greater than ℵ1 has the tree property. There is a bottom up
approach where one attempts to force longer and longer initial segments of the
regular cardinals to have the tree property, see for example ([1, 3, 19, 26]). We
refer the reader to [?] for some analogous results on successive failures of weak
square. Another aspect of the problem comes from the interaction between cardinal
arithmetic at singular strong limit cardinals µ and the tree property at µ+ . In
the 1980’s Woodin asked whether the failure of SCH at ℵω is consistent with the
tree property at ℵω+1 . More generally one can consider whether this situation is
consistent at some larger singular cardinal. An important result in this direction is
due to Gitik and Sharon [12] who showed that from the existence of a supercompact
cardinal that it is consistent that there is a singular cardinal κ of cofinality ω such
that SCH fails at κ and there are no special κ+ -Aronszajn trees. In fact they show
a stronger assertion (κ+ ∈
/ I[κ+ ]), which we will define later. In the same paper,
they show that it is possible to make κ in to ℵω2 . Cummings and Foreman [4]
showed that there is a PCF theoretic object called a bad scale in the models of
Gitik and Sharon which implies that κ+ ∈
/ I[κ+ ].
The key ingredient Gitik and Sharon’s argument was a new diagonal supercompact Prikry forcing. The basic idea is to start with supercompactness measures Un on Pκ (κ+n ) for n < ω and use it to define a Prikry forcing. This forcing addsSa sequence hxn | n < ωi where each xn is a typical point for Un and
so that n<ω xn = κ+ω . The result is that κ+ω is collapsed to have to have
size κ and κ+ω+1 becomes the new successor of κ. This allows for the property
κ+ω+1 ∈
/ I[κ+ω+1 ] from the ground model to show κ+ ∈
/ I[κ+ ] in the extension.
κ
+ω+2
Moreover if we started with 2 = κ
in the ground model, then we get the
failure of SCH at κ in the extension.
Variations of Gitik and Sharon’s poset have been used construct many related
models. We list a few such results:
(1) (Neeman [18]) From ω supercompact cardinals, there is a forcing extension
in which there is a singular cardinal κ of cofinality ω such that SCH fails
at κ and κ+ has the tree property.
(2) (Sinapova [21]) From a supercompact cardinal κ, for any regular λ < κ,
there is a forcing extension in which κ is a singular cardinal of cofinality
λ, SCH fails at κ and κ carries a bad scale (in particular κ+ ∈
/ I[κ+ ] and
+
there are no special κ -Aronszajn trees).
(3) (Sinapova [22]) From λ supercompact cardinals hκα | α < λi with λ < κ0
regular, there is a forcing extension in which κ0 is a singular cardinal of
cofinality λ, SCH fails at κ0 and κ+
0 has the tree property.
(4) (Sinapova [23]) From ω supercompact cardinals it is consistent that Neeman’s result above holds with κ = ℵω2 .
Woodin’s original question remains open, see [24] for the best known partial
result. A theme in the above results is that questions about the tree property are
DIAGONAL SUPERCOMPACT RADIN FORCING
3
answered by first constructing a model where there are no special κ+ -Aronszajn
trees (or even κ+ ∈
/ I[κ+ ]). To obtain the tree property, one needs to increase the
large cardinal assumption and to give a version of an argument of Magidor and
Shelah [15] who showed that the tree property holds at µ+ when µ is a singular
limit of supercompact cardinals.
In this paper we pursue a global version of the results above. We aim to force
a class club C of singular cardinals such that for all κ ∈ C SCH fails at κ and
there are no special κ+ -Aronszajn trees. From Specker’s theorem, a model with
no Aronszajn trees above ℵ1 (or even no special Aronszajn trees above ℵ1 ) will
have such a club C. Our attempt to construct the natural diagonal supercompact
Radin forcing does not give the desired class C. We succeed in forcing a club C
of singular cardinals where SCH fails, but some successors of elements of C carry
special Aronszajn trees. In the last section of the paper, we sketch the construction
of a model where we do get the class club C that we wanted, but the technique
used looks unlikely to generalize to help answer our motivating question.
To put the theorems of the paper in the context of the results above we need
a brief description of Magidor forcing and Radin forcing. Changing the cofinality of a large cardinal to some prescribed uncountable regular cardinal was first
done by Magidor [14] and the technique has come to be known as Magidor forcing. Sinapova’s results about uncountable cofinalities mentioned above are done by
creating a Magidor forcing version of the Gitik-Sharon poset. Adding a Prikry-like
club in a large cardinal κ while preserving its regularity was first done by Radin
[20]. In this paper we describe a diagonal supercompact Radin forcing which adds
a Prikry-like club of cardinals where SCH fails. Unfortunately, the extra closure
we need to preserve the inaccessibility of our large cardinal κ interfers with the
argument for the non-existence of special Aronszajn trees at (the successors of)
some elements of the club. There are strong reasons to believe that Radin forcing
is the correct technique for solving our motivating question. In particular, Radin
forcing can admit the addition for forcing between the elements of the generic club,
see for example [13, 8, 2, 7].
In the final section of the paper we use Gitik’s idea (See [11] for a reference) of
iterating Prikry type forcing to show that it is consistent to have a class club C
as above. However this technique does not seem to help us with our motivating
question and the ideas are mostly standard, so we only sketch the proof.
The paper is organized as follows. In Section 2 we give some definitions and
background material required for the main result. In section 3 we describe the
main forcing and prove that it gives a model with a class club C of cardinals where
SCH fails. In Section 4 we show that there is a stationary, co-stationary set S ⊆ C
such that for all κ ∈ S κ+ ∈
/ I[κ+ ] and for all κ ∈ C \ S there is a special κ+ Aronszajn tree. In Section 5 we show obtain a class club of cardinals κ where SCH
fails and there are no special κ+ -Aronszajn trees. In Section ?? we make some
concluding remarks and ask some open questions.
2. Background
In this section we will make the notions from the introduction precise and give
some further definitions which are relevant to the rest of the paper.
Definition We define a weak square sequence ∗ν and the approachability ideal
I[ν + ] for a cardinal ν.
4
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
• We say that ν has a weak square sequence (∗ν ) hCγ | γ < ν + i is a partial
square sequence if
(1) for all γ < ν + limit, Cγ ⊂ P(γ) has size |Cγ | ≤ ν such that for every
c ∈ Cγ , c ⊂ γ is club in γ, otp(Cγ ) < ν, and
(2) for all β < γ < ν if β is a limit point of some c ∈ Cγ then c ∩ β ∈ Cβ .
• Let ~z = hzα | α < ν + i be a sequence of bounded subsets of ν + . We say that
a limit ordinal γ is ~z−approachable if there is a closed unbounded set A ⊂ γ
with otp(A) = cf (γ) such that for every β < γ, A∩β ∈ ~z γ = hzα | α < γi.
By arranging that zα+1 is the closure of zα for each α < ν + we may assume
that for every ~z−approachable points γ, there is a witness A ⊂ γ which is
closed unbounded.
The approachability ideal I[ν + ] consists of all subsets S ⊂ ν + for which
there are ~z as above and a club C ⊂ ν + so that every γ ∈ C ∩ S is
~z−approachable.
2.1. Forcing preliminaries. Suppose κ is supercompact and θ > κ. As in [21], we
~ = hUα | α < θi
can arrange that 2κ ≥ θ and that we have a sequence of ultrafilters U
α
and, for all α < θ, a sequence hfη | η < θi such that the following hold.
• For all α < θ, Uα is a normal, fine ultrafilter on Pκ (κ+α ). Let jα : V →
Mα ∼
= Ult(V, Uα ) be the collapsed ultrapower map.
• For all α < β < θ, Uα ∈ Mβ .
• For all α, η < θ, we have fηα : κ → κ and jα (fηα )(κ) = η.
When we write that something happens for most x ∈ Pκ (κ+α ), we mean it
happens for a Uα -measure one set. For α < θ and x ∈ Pκ (κ+α ), let κx denote x ∩ κ.
For most x ∈ Pκ (κ+α ), x ∩ κ is an inaccessible cardinal. We will always assume we
are working with such x and let κx = x ∩ κ. For x, y ∈ Pκ (κ+α ), x ≺ y denotes the
statement that x ⊆ y and otp(x) < κy .
For α < β < θ, let ūβα be a function on Pκ (κ+β ) representing Uα in the ultrapower
+f β (κ )
by Uβ . For most x ∈ Pκ (κ+β ), ūβα (x) is a measure on Pκx (κx α x ). Also, for most
x ∈ Pκ (κ+β ), otp(x∩κ+α ) = fαβ (κx ). For such x, ūβα (x) lifts naturally to a measure
uβα (x) on Pκx (x ∩ κ+α ).
For y ∈ Pκ (κ+β ), let Zyβ = {α < β | κ+α ∈ y}.
Lemma 2.1. For most y ∈ Pκ (κ+β ), the following hold.
(1) Zyβ is < κy -closed.
(2) If cf(β) < κ, then Zyβ is unbounded in β.
(3) otp(Zyβ ) = fββ (κy ) and, if β is a limit ordinal, then so is fββ (κy ). Also, if
cf (β) ≥ κ then cf(fββ (κy )) ≥ κy .
+otp(α∩Zyβ )
(4) For all α ∈ Zyβ , otp(y ∩ κ+α ) = fαβ (κy ) = κy
(5)
(6)
.
+f β (κ )
For all α ∈ Zyβ , ūβα (y) is a measure on Pκy (κy α y ).
1
For all α0 < α1 , both in Zyβ , the function x 7→ ūα
α0 (x) represents
β
the ultrapower by uα1 (y).
ūβα0 (y) in
Proof. Let j = jβ . Note first that, defining g : Pκ (κ+β ) → V by g(y) = Zyβ , we
have j(g)(j“κ+β ) = j“β. Items (1)-(4) then follow easily.
j(β)
To show (5), let j(hūβα | α < βi) = hv̄α | α < j(β)i and j(hfαβ | α < βi) =
j(β)
j(β)
hgα
| α < j(β)i. Let α ∈ j“β, with, say, α = j(ξ). Then v̄α (j“κ+β ) =
DIAGONAL SUPERCOMPACT RADIN FORCING
5
j(β)
j(ūβξ )(j“κ+β ) = Uξ , which is a measure on Pκ (κ+ξ ) = Pκ (κ+gα
(κ)
). Thus, for
β
+fα
(κy )
almost all y ∈ Pκ (κ+β ), ūβα (y) is a measure on Pκy (κy
).
α1
1
|
α0 < α1 ≤ j(β)i and
|
α
<
α
≤
βi)
=
hv̄
We finally show (6). Let j(hūα
0
1
α0
α0
j(β)
β
j(huα | α < βi) = hvα | α < j(β)i. It suffices to show that, in Mβ , for all α0 < α1 ,
j(β)
both in j“β, the function x 7→ v̄αα01 (x) represents v̄α0 (j“κ+β ) in the ultrapower by
j(β)
j(β)
vα1 (j“κ+β ). Note that vα1 (j“κ+β ) is a measure on Pκ (j“κ+ξ1 ) that collapses to
j(β)
Uξ1 . Call this ‘lifted’ measure Ûξ1 . Also note that v̄α0 (j“κ+β ) = Uξ0 . Thus, we
α1
must show that the function x 7→ v̄α0 (x) represents Uξ0 in the ultrapower by Ûξ1 .
Fix α0 < α1 in j“β, with α0 = j(ξ0 ) and α1 = j(ξ1 ). Fix x ∈ Pκ (j“κ+ξ1 ).
There is x̄ ∈ Pκ (κ+β ) such that x = j(x̄). Then v̄αα01 (x) = j(ūξξ10 (x̄)). For most
ξ
+fξ 1 (κx̄)
x̄ ∈ Pκ (κ+ξ1 ), ūξξ10 (x̄) is a measure on Pκx̄ (κx̄
for most x ∈ Pκ (j“κ+ξ1 ), v̄αα01 (x) =
ūξξ10 (x̄).
0
), and this is fixed by j. Thus,
Thus, since Ûξ1 is a lifting of Uξ1 ,
x 7→ v̄αα01 (x) represents the same thing in the ultrapower by Ûξ1 as x̄ 7→ ūξξ10 (x)
represents in the ultrapower by Uξ1 , which is Uξ0 . This is true in V and, since Mβ
is sufficiently closed, it is true in Mβ as well.
Lemma 2.2. Suppose β < θ and, for all α ≤ β, Aα ∈ Uα . Let A∗ be the set of all
y ∈ Aβ such that, for all α ∈ Zyβ , {x ∈ Aα | x ≺ y} ∈ uβα (y). Then A∗ ∈ Uβ .
Proof. Let j = jβ . It suffices to show that j“κ+β ∈ j(A∗ ), i.e. for all j(α) ∈ j“β,
{x ∈ j(Aα ) | x ≺ j“κ+β } ∈ Ûα , where Ûα is the lifted version of Uα living on
Pκ (j“κ+α ). Fix such a j(α). Let X = {x ∈ j(Aα ) | x ≺ j“κ+β }, and note that
X = j“A ∈ Ûα .
Lemma 2.3. Suppose γ < θ, z ∈ Pκ (κ+γ ), and z satisfies all of the statements
in Lemma 2.1. Suppose that, for all α ∈ Zzγ , Aα ∈ uγα (z). Fix β ∈ Zzγ , and let A
be the set of y ∈ Aβ such that, for all α ∈ Zyβ , {x ∈ Aα | x ≺ y} ∈ uβα (y). Then
A∗ ∈ uγβ (z).
Proof. For each α ∈ Zzγ , let Āα be the collapsed version of Aα , so Āα ∈ ūγα (z).
γ
Recall that uγβ (z) is a measure on Pκz (z ∩ κ+β ). Let k : V → N ∼
= Ult(V, uβ (z)) be
γ
the ultrapower map. By (6) of Lemma 2.1, for all α ∈ Zz ∩ β, the map y 7→ ūβα (y)
represents ūγα (z) in the ultrapower. Note also that the map y 7→ Zyβ represents
{η < k(β) | k(κ)+η ∈ k“(z ∩ κ+β ) = k“(Zzγ ∩ β). To prove the Lemma, it suffices to
show that k“(z∩κ+β ) ∈ k(A∗ ), i.e. for all α ∈ Zzγ ∩β, {x ∈ k(Aα ) | x ≺ k“(z∩κ+β )}
is in the measure represented by the map y 7→ uβα (y). Call this measure w and note
that it is a lifted version of ūβα (y). Also note that {x ∈ k(Aα ) | x ≺ k“(z ∩ κ+β )} =
k“(Aα ), which collapses to Āα ∈ ūβα (y). Thus, {x ∈ k(Aα ) | x ≺ k“(z ∩ κ+β )} ∈ w,
completing the proof of the lemma.
For β < θ, let Xβ be the set of y ∈ Pκ (κ+β ) satisfying all of the statements in
Lemma 2.1. Fix η < θ.
3. The main forcing
We now define our forcing, PU~ ,η . Conditions of PU~ ,η are pairs (a, A) satisfying
the following requirements.
6
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
(1) a and A are functions, dom(a) is a finite subset of θ \ η, and dom(A) =
θ \ (dom(a) ∪ η).
(2) For all β ∈ dom(a), a(β) ∈ Xβ .
β
.
(3) For all α < β, both in dom(a), a(α) ≺ a(β) and α ∈ Za(β)
(4) For all α ∈ θ \ max(dom(a)) (or, if dom(a) = ∅, for all α ∈ dom(A)),
A(α) ∈ Uα .
(5) For all α ∈ dom(A) ∩ max(dom(a)), if β = min(dom(a) \ α), then A(α) ∈
β
β
.
and A(α) = ∅ if α 6∈ Za(β)
uβα (a(β)) if α ∈ Za(β)
(6) For all β ∈ dom(A) such that A(β) 6= ∅ and dom(a) ∩ β 6= ∅, if α =
max(dom(a) ∩ β), then, for all y ∈ A(β), a(α) ≺ y and α ∈ Zyβ .
If (a, A), (b, B) ∈ PU~ ,η , then (b, B) ≤ (a, A) iff the following requirements hold.
(1) b ⊇ a.
(2) For all α ∈ dom(b) \ dom(a), b(α) ∈ A(α).
(3) For all α ∈ dom(B), B(α) ⊆ A(α).
(b, B) ≤∗ (a, A) if (b, B) ≤ (a, A) and b = a.
Remark In our arguments, for notational simplicity we will typically assume η = 0
and then denote PU~ ,η as PU~ . Everything proved about PU~ can be proved for a
general PU~ ,η in the same way by making the obvious changes. The reason we
introduce the more general forcing is to be able to properly state the factorization
lemma.
In what follows, let P denote PU~ . For any condition p = (a, A) ∈ P, we often
denote (a, A) as (ap , Ap ) and let γ p = max(dom(ap )). We refer to ap as the stem
of p. Note that, if p, q ∈ P and ap = aq , then p and q are compatible. If a is a
non-empty stem, then let γ a denote max(dom(a)), and let a− = a γ a . Suppose a
is a stem, α < θ, and x ∈ Xα . Suppose moreover that either a is empty or γ a < α,
a(γ a ) ≺ x, and γ a ∈ Zxα . Then a_ (α, x) is a stem and (a_ (α, x))− = a. If p ∈ P
and b is a stem, then b is possible for p if there is q ≤ p with aq = b. If p ∈ P and b
is possible for p, then b_ p is the maximal q such that q ≤ p and aq = p. Such a q
always exists.
Lemma 3.1. Suppose (a, A) ∈ P, β ∈ dom(A), and A(β) 6= ∅. Then there is
(b, B) ≤ (a, A) such that β ∈ dom(b).
Proof. Straightforward using Lemmas 2.1, 2.2, and 2.3.
sc
Definition Suppose that G is a P-generic over V . Let Let CG
(sc for super compact) be the set of all points x = a(β) where β ∈ dom(a) for some p = (a, A) in the
sc
} be the generic Radin Club.
generic filter G, and CG = {κx | x ∈ CG
Lemma 3.2. CG is club in κ and the assignment x 7→ κx = x ∩ κ is an increasing
bijection.
Proof. Straightforward by Lemma 3.1 and genericity.
~ y = hūβα (y) | α ∈ Zyβ i. For ξ < f β (y), let
Suppose β < θ and y ∈ Xβ . Let U
β
αξ ∈ Zyβ be such that otp(α ∩ Zyβ ) = ξ. Then ūβα (y) is a measure on Pκy (κ+ξ
y ). Let
β
β
~
Vξ = ūαξ (y). Then Uy = hVξ | ξ < fβ (y)i, and we can define PU~ y as above.
If p ∈ P, then P/p = {q ∈ P | q ≤ p}.
DIAGONAL SUPERCOMPACT RADIN FORCING
7
Lemma 3.3. (Factorization Lemma) Let p = (a, A) ∈ P. Suppose a 6= ∅, β =
max(dom(a)), and y = a(β). Then there is p0 ∈ PU~ y such that P/p ∼
= PU~ y /p0 ×
PU~ ,β+1 /(∅, A (β, θ)).
Proof. Let π : y → otp(y) be the unique order-preserving bijection. Define p0 =
(a0 , A0 ) ∈ PU~ y as follows. For ξ < fββ (y), let αξ ∈ Zyβ be such that otp(αξ ∩
Zyβ ) = ξ. Let dom(a0 ) = {ξ < fββ (y) | αξ ∈ dom(a)} and, for ξ ∈ dom(a0 ),
let a0 (ξ) = π“a(αξ ). Then dom(A0 ) = fββ (y) \ dom(a0 ). If ξ ∈ dom(A0 ), let
A0 (ξ) = {π“x | x ∈ A(αξ )}. It is straightforward to verify that p0 thus defined is in
PU~ y and that P/p ∼
= PU~ y /p0 × PU~ ,β+1 /(∅, A (β, θ)).
By repeatedly applying the Factorization Lemma, standard arguments (see, e.g.
[11]) allow us to assume we are working below a condition of the form (∅, A) when
proving the following lemmas about P.
Lemma 3.4. (P, ≤, ≤∗ ) satisfies the Prikry property, i.e. if ϕ is a statement in
the forcing language and p ∈ P, then there is q ≤∗ p such that q k ϕ.
Proof. The proofs of this and the next few lemmas are similar to those for the
classical Radin forcing, which can be found in [11]. Fix ϕ in the forcing language
and p ∈ P. By the Factorization Lemma, it suffices to let p = (∅, A) for some A. Let
a be a stem possible for p, and let α ∈ θ \(γ a +1). Let Ya,α = {x ∈ A(α) | a(γ a ) ≺ x
0
and γ a ∈ Zxα }. Note that Ya,α ∈ Uα . Let Ya,α
= {x ∈ Ya,α | for some B,
_
1
(a (α, x), B) ϕ}, Ya,α = {x ∈ Ya,α | for some B, (a_ (α, x), B) ¬ϕ}, and
i(a,α)
1
0
2
). Fix i(a, α) < 3 such that Ya,α
∪ Ya,α
= Ya,α \ (Ya,α
Ya,α
i(a,α)
∈ Uα , and let
∗
Ya,α
= Ya,α .
For α < θ, let B(α) be the set of x ∈ A(α) such that, for every stem a possible
∗
. We claim that B(α) ∈ Uα .
for p such that a(γ a ) ≺ x and γ a ∈ Zxα , x ∈ Ya,α
+α
∗
Let j = jα . It suffices to show that j“κ
∈ j(B(α)). Let j(hYa,α
| a is a stem
a
∗
possible for p and γ < αi) = hWa,j(α) | a is a stem in j(P) possible for j(p)
and γ a < j(α)i. Suppose that, in j(P), a is a stem possible for j(p) such that
a(γ a ) ≺ j“κ+α and γ a ∈ j“α. Then there is a stem ā possible for p such that
∗
∗
∗
∗
∈ Uα , j“κ+α ∈ Wa,j(α)
and hence
), so, as Yā,α
j(ā) = a. Then Wa,j(α)
= j(Yā,α
j“κ+α ∈ j(B(α)). Thus, (∅, B) ∈ P and (∅, B) ≤∗ p.
Suppose for sake of contradiction that no direct extension of (∅, B) decides ϕ.
Find (a, B ∗ ) ≤ (∅, B) deciding ϕ with |a| minimal. Without loss of generality,
suppose (a, B ∗ ) ϕ. Because of our assumption that no direct extension of (∅, B)
decides ϕ, a is non-empty. Let b = a− and γ = γ a . By our construction of B, we
0
have a(γ) ∈ Yb,γ
, and, for any x ∈ B(γ) such that b_ (γ, x) is a stem, there is B̂x
_
such that (b (γ, x), B̂x ) ϕ. Let p∗ = b_ (∅, B) = (b, B ∗ ). We will find a direct
extension (b, F ) of p∗ forcing ϕ, thus contradicting the minimality of |a|.
We first define F γ b (if b = ∅, then there is nothing to do here). Since
there are fewer than κ possibilities for F γ b and Uγ is κ-complete, we may fix a
function F ∗ on γ b \ dom(b) such that B0 (γ) := {x ∈ B(γ) | b_ (γ, x) is a stem and
B̂x γ b = F ∗ } ∈ Uγ . Then, for all α ∈ γ b \ dom(b), let F (α) = F ∗ (α) ∩ B ∗ (α).
We next define F on the interval (γ b , γ) (or on all of γ, if b = ∅). If α ∈ (γ b , γ),
x ∈ B0 (γ), and α ∈ Zxγ , note that B̂x (α) ∈ uγα (x). Let B̄x (α) be the collapsed
version of B̂x (α). Then B̄x (α) ∈ ūγα (x). Let F ∗ (α) be the set in Uα represented by
8
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
the function x 7→ B̄x (α) in the ultrapower by Uγ , and let F (α) = F ∗ (α) ∩ B ∗ (α).
Let F (γ) be the set of x ∈ B0 (γ) ∩ B ∗ (γ) such that, for all α ∈ Zxγ \ γ b , {y ∈
F (α) | y ≺ x} = B̂x (α). We claim that F (γ) ∈ Uγ . To see this, let j = jγ .
Note that the function x 7→ B̂x (α) represents {j“y | y ∈ F (α)}, which is equal to
{z ∈ j(F (α)) | z ≺ j“κ+γ }. Thus, j“κ+γ ∈ j(F (γ)), so F (γ) ∈ Uγ . We finally
define F on (γ, θ). If α ∈ (γ, θ), let F (α) be the set of y ∈ B ∗ (α) such that
γ ∈ Zyα and, for all x ∈ F (γ) such that x ≺ y, y ∈ B̂x . By now-familiar arguments,
F (α) ∈ Uα . Notice that, by our construction, if (c, H) ≤ (b, F ) and γ ∈ dom(c),
then (c, H) ≤ (b_ (γ, c(γ)), B̂c(γ) ).
Now suppose for sake of contradiction that (b, F ) 6 ϕ. Find (c, H) ≤ (b, F ) such
that (c, H) ¬ϕ. If γ ∈ dom(c), then (c, H) ≤ (b_ (γ, c(γ)), B̂c(γ) ) ϕ, which is
a contradiction. Thus, suppose γ 6∈ dom(c). By our choice of F (α) for α ∈ (γ, θ)
(namely, our requirement that γ ∈ Zyα for all y ∈ F (α)), it must be the case that
H(γ) 6= ∅. But then (c, H) can be extended further to a condition (c0 , H 0 ) such
that γ ∈ dom(c0 ), and this again gives a contradiction.
S
Definition A tree T ⊆ [ α<θ ({α} × Pκ (κ+α ))]≤n is fat if the following conditions
hold.
(1) For all h(αi , xi ) | i ≤ ki ∈ T and all i0 < i1 ≤ k, we have αi0 < αi1 and
xi0 ≺ xi1 .
(2) For all t ∈ T with lh(t) < n, there is αt < θ such that:
(a) for all (β, y) such that t_ (β, y) ∈ T , β = αt ;
(b) {x | t_ (αt , x) ∈ T } ∈ Uαt .
If T is as in the previous definition, then n is said to be the height of T .
Definition Suppose T is a fat tree, α < θ, and x ∈ Xα . T is fat above (α, x) if,
for all h(αi , xi ) | i ≤ ki and all i ≤ k, we have α < αi and x ≺ xi .
S
Definition Suppose γ < θ and z ∈ Xγ . A tree T ⊆ [ α<θ ({α} × Pκ (κ+α ))]≤n is
fat below (γ, z) if the following conditions hold.
(1) For all h(αi , xi ) | i ≤ ki ∈ T and all i ≤ k, we have αi ∈ Zzγ and xi ≺ z.
(2) For all h(αi , xi ) | i ≤ ki ∈ T and all i0 < i1 ≤ k, we have αi0 < αi1 and
xi0 ≺ xi1 .
(3) For all t ∈ T with lh(t) < n, there is αt ∈ Zzγ such that:
(a) for all (β, y) such that t_ (β, y) ∈ T , β = αt ;
(b) {x | t_ (αt , x) ∈ T } ∈ uγαt (z).
Suppose T is a fat tree, γ < θ, and z ∈ Pκ (κ+γ ). T (γ, z) is the subtree of T
consisting of all h(αi , xi ) | i ≤ ki ∈ T such that, for all i ≤ k, αi ∈ Zzγ and xi ≺ z.
If T is a fat tree, let γT = sup({α | for some h(αi , xi ) | i ≤ ki ∈ T and i ≤ k,
α = αi }). Note that, if θ is weakly inaccessible and |Pκ (κ+α )| < θ for all α < θ, we
have γT < θ.
Suppose that S is a set of stems and, for all a ∈ S, Ta is a fat tree above
(γ a , a(γ a )). For γ < θ, let S<γ = {a ∈ S | γ a < γ}. Let C = {γ < θ | for all
a ∈ S<γ , γTa < γ}. Note that C is club in θ. For all γ ∈ C, let Yγ be the set of
z ∈ Xγ such that, for all a ∈ S<γ such that a(γ a ) ≺ z and γ a ∈ Zzγ , Ta (γ, z)
is fat below (γ, z). We claim that Yγ ∈ Uγ . To see this, let j = jγ , and note first
that {a ∈ j(S<γ ) | a ≺ j“κ+γ and γ a ∈ j“γ} = j“S<γ and second that, for all
ā ∈ S<γ , j(Tā ) (j(γ), j“κ+γ ) = j“Tā , which is fat below (j(γ), j“κ+γ ). Thus,
j“κ+γ ∈ j(Yγ ), so Yγ ∈ Uγ .
DIAGONAL SUPERCOMPACT RADIN FORCING
9
Lemma 3.5. Suppose p = (a, A) ∈ P and D ⊆ P is a dense open set. Suppose
a = {(αi , xi ) | i < k} is such that, for all i0 < i1 < k, αi0 < αi1 . There are trees
hTi | i ≤ ki and natural numbers hni | i ≤ ki such that the following hold.
(1) For all i ≤ k, Ti is a tree of height ni .
(2) If k > 0, then T0 is fat below (α0 , x0 ) and, for all 0 < i < k, Ti is fat below
(αi , xi ) and above (αi−1 , xi−1 ).
(3) Tk is fat, and, if k > 0, it is above (αk−1 , xk−1 ).
(4) Suppose that, for all i ≤ k, h(β`i , y`i ) | ` < ni i is a maximal element of Ti .
Then, if b = a∪{(β`i , y`i ) | i ≤ k, ` ≤ ni }, there is B such that (b, B) ≤ (a, A)
and (b, B) ∈ D.
Proof. By the Factorization Lemma, it again suffices to consider p of the form (∅, A).
We thus need to find a single fat tree T . We inductively construct a decreasing
sequence of conditions h(∅, An ) | n < ωi. Intuitively, An will take care of extensions
(b, B) ≤ (∅, A) such that |b| = n. We explicitly go through the first few steps of the
construction.
Let A0 = A. If there is a direct extension of (∅, A) in D, then we are done by
setting T = {∅}. Thus, suppose there is no such direct extension. For every stem
a possible for (∅, A0 ) and every α ∈ (γ a , θ), let Y0,a,α = {x ∈ A0 (α) | a(γ a ) ≺ x
0
= {x ∈ Y0,a,α | for some B, (a_ (α, x), B) ∈ D}, and
and γ a ∈ Zxα }. Let Y0,a,α
i(0,a,α)
1
0
let Y0,a,α
= Y0,a,α \ Y0,a,α
. Find i(0, a, α) < 2 such that Y0,a,α
∗
Y0,a,α
∈ Uα , and let
i(0,a,α)
Y0,a,α .
=
For α < θ, let A1 (α) be the set of x ∈ A0 (α) such that, for all
∗
. As in
stems a possible for (∅, A0 ) such that a(γ a ) ≺ x and γ a ∈ Zxα , x ∈ Y0,a,α
∗
the proof of Lemma 3.4, A1 (α) ∈ Uα for all α < θ, so (∅, A1 ) ≤ (∅, A0 ). Note that
(∅, A1 ) satisfies the following property, which we denote (∗)1 :
Suppose q = (a_ (α, x), B) ≤ (∅, A1 ) and q ∈ D. Then, for every
y ∈ A1 (α) such that a(γ a ) ≺ y and γ a ∈ Zyα , there is By such that
(a_ (α, y), By ) ∈ D.
Now suppose there is a stem a = {(α, x)} possible for (∅, A1 ) and a B such that
(a, B) ∈ D. We can then define a fat tree T of height 1 whose maximal elements
are all h(α, x)i such that x ∈ A1 (α). We are then done, as T easily satisfies the
requirements of the lemma. Thus, suppose there is no such a and proceed to define
(∅, A2 ) as follows.
For every stem a possible for (∅, A1 ) and every α ∈ (γ a , θ), let Y1,a,α = {x ∈
0
A1 (α) | a(γ a ) ≺ x and γ a ∈ Zxα }. Let Y1,a,α
be the set of all x ∈ Y1,a,α such that
α
α
there are βx ∈ (α, θ) and Wx ∈ Uβxα such that, for all y ∈ Wxα :
βα
• x ≺ y and α ∈ Zy x ;
• there is B such that (a_ (α, x)_ (βxα , y), B) ∈ D.
i(1,a,α)
1
0
Let Y1,a,α
= Y1,a,α \ Y1,a,α
. Find i(1, a, α) < 2 such that Y1,a,α
∗
Y1,a,α
∈ Uα , and let
i(1,a,α)
Y1,a,α .
=
For α < θ, let A2 (α) be the set of x ∈ A1 (α) such that, for all
∗
stems a possible for (∅, A1 ) such that a(γ a ) ≺ x and γ a ∈ Zxα , x ∈ Y1,a,α
. Then
∗
(∅, A2 ) ≤ (∅, A1 ), and (∅, A2 ) satisfies the property (∗)2 :
Suppose q = (a_ (α, x)_ (β, y), B) ≤ (∅, A2 ) and q ∈ D. Then,
for every x0 ∈ A2 (α) such that a(γ a ) ≺ x0 and γ a ∈ Zxα0 , there is
βxα0 ∈ (α, θ) and Wxα0 ∈ Uβxα0 such that, for all y 0 ∈ Wxα0 , there is B 0
such that (a_ (α, x0 )_ (βxα0 , y 0 ), B 0 ) ∈ D.
10
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
Suppose there is a stem a = {(α, x)_ (β, y)} possible for (∅, A2 ) and a B such
that (a, B) ∈ D. Using (∗)2 , we can define a tree T of height 2 whose maximal
elements are all h(α, x0 )_ (βxα0 , y 0 )i such that x0 ∈ A2 (α) and y 0 ∈ Wxα0 . We are then
done, as T satisfies the requirements of the lemma. If there is no such stem a, then
continue in the same manner.
In this way, we can construct An such that, if there is a stem a possible for (∅, An )
with |a| = n and a B such that (a,
TB) ∈ D, then there is a fat tree of height n as
desired. For α < θ, let A∞ (α) = n<ω An (α). For all n < ω, (∅, A∞ ) ≤∗ (∅, An ).
Find (a, B) ≤ (∅, A∞ ) such that (a, B) ∈ D. Let n∗ = |a|. Then a is possible for
(∅, An∗ ), so there is a fat tree of height n∗ as required by the lemma.
Theorem 3.6. If θ is weakly inaccessible and |Pκ (κ+α )| < θ for all α < θ, then κ
remains inaccessible in V P .
Proof. It suffices to prove that κ remains regular. Let p = (a, A) ∈ P, let δ < κ,
and suppose f˙ is a P-name forced by p to be a function from δ to κ. We will find
q ≤ p forcing the range of f˙ to be bounded below κ.
For all ξ < δ, let Dξ be the set of (b, B) ∈ P such that (b, B) “f˙(ξ) < κb(γ b ) ”.
Each Dξ is a dense, open subset of P. For ξ < δ, let Sξ be the set of stems b such
that, for some B, (b, B) ≤ p and (b, B) ∈ Dξ . For all b ∈ Sξ , fix a Bbξ witnessing
this. For each β ∈ (γ a , θ), let A∗ (β) be the set of y ∈ A(β) such that, for all ξ < δ
and all b ∈ Sξ such that b(γ b ) ≺ y and γ b ∈ Zyβ , y ∈ Bbξ (β). For β ∈ dom(A) ∩ γ a ,
let A∗ (β) = A(β). Then (a, A∗ ) ≤ (a, A). Let T be the set of stems possible for
(a, A∗ ). For γ < θ, let T<γ = {c ∈ T | γ c < γ}. For all c ∈ T , let pc = c_ (a, A∗ ).
For all c ∈ T and ξ < δ, apply Lemma 3.5 to pc and Dξ to obtain a sequence of
trees hTc,ξ,i | i ≤ |c|i. Let C be the set of γ < θ such that, for all c ∈ T<γ and
all ξ < δ, γTc,ξ,|c| < γ. C is club in θ. Fix γ ∈ C \ (γ a + 1). By the discussion
preceding Lemma 3.5, choose z ∈ A∗ (γ) such that, for all c ∈ T<γ such that
c(γ c ) ≺ z and γ c ∈ Zzγ and for all ξ < δ, Tc,ξ,|c| (γ, z) is fat below (γ, z). Then
q = (a_ (γ, z), A∗∗ ) ≤ (a, A∗ ), where A∗∗ (α) = A∗ (α) for all α ∈ dom(A∗ ) ∩ γ a and
all α ∈ (γ, θ), and A∗∗ (α) = {x ∈ A∗ (α) | x ≺ z} for all α ∈ (γ a , γ).
We claim that q forces the range of f˙ to be bounded below κz . Suppose for sake
of contradiction that there is ξ < δ and r ≤ q such that r “f˙(ξ) ≥ κz .” Let
r = (d, F ), and let c = {(α, x) | α < γ}. Then c ∈ T<γ , c(γ c ) ≺ z, and γ c ∈ Zzγ ,
so Tc,ξ,|c| (γ, z) is fat below (γ, z). Suppose that, for all i ≤ |c|, ni is the height
of Tc,ξ,i . Then, for all i ≤ |c|, we can find maximal elements h(β`i , y`i ) | ` < ni i of
Tc,ξ,i such that, letting c0 = c ∪ {(β`i , y`i ) | i ≤ |c|, ` < ni }, c0 ∪ d is possible for
r. In particular, for all (α, x) ∈ c0 , x ≺ z and α ∈ Zzγ , and so κx < κz . Also,
there is B 0 such that (c0 , B 0 ) ∈ Dξ and, by our construction of A∗ , we may assume
0
that, for α ∈ (γ c , θ), B 0 (α) = A∗ (α). All of this together means that (c0 , B 0 ) and
r are compatible. However, as (c0 , B 0 ) ∈ Dξ , (c0 , B 0 ) “f˙(ξ) < κc0 (γ c0 ) < κz , ”
contradicting the assumption that r “f˙(ξ) ≥ κz .”
4. Approachability
In this section we characterize precisely which successors ν + for ν ∈ C have
reflection properties.
DIAGONAL SUPERCOMPACT RADIN FORCING
11
Notation Suppose that β < θ and that y ∈ Xβ and p is a condition with ap (β) = y.
For notational simplicity we write o(y) for fββ (y) = otp(Zyβ ). Note that o(y) <
ν +θ(κy ) where θ(κy ) is the next inaccessible cardinal above κy .
+o(y)+1
Lemma 4.1. PU~ y as defined above Lemma 3.3 has the κy
-Knaster property.
+o(y)
It is not hard to see that there are just κy
many possible a-parts of conditions
in this poset and that conditions with the same a-part are compatible.
Lemma 4.2. Suppose that we have β < θ, y ∈ Xβ , y 0 ∈ Xβ+1 and p is a condition
such that ap (β) = y and ap (β + 1) = y 0 . If µ is an ordinal of cofinality δ with
o(y)+1
κy
≤ δ < κy0 and Ċ is a P-name for a club subset of µ, then there are p0 ≤∗ p
and D ⊆ µ club such that p0 forces D ⊆ Ċ.
Proof. First we show that it is enough to consider δ = µ. Assume for the moment
that δ < µ. Let π : δ → µ be an increasing continuous and cofinal function. Now
by passing to a name for a subset of Ċ we can assume that it is forced that Ċ is a
subset of the range of π. Now a condition will force that there is a ground model
contained in Ċ if and only if there is a ground model club contained in π −1 Ċ.
So we may assume that δ = µ. By Lemma 3.3 and Lemma 3.4, there is a direct
extension p0 of p which forces Ċ to be in the extension by PU~ y . Now by a standard
+o(y)+1
argument using the κy
-cc of PU~ y , there is a club D ⊆ µ in V such that p0
forces D ⊆ Ċ.
We use the above lemma to show that the approachability property fails at
certain points along our Radin club.
Lemma 4.3. Suppose that β is a limit ordinal with cf V (β) < κ and p is a condition
such that ap (β) = y, then p forces that κ+
/ I[κ+
y ∈
y ].
+
Proof. Assume for a contradiction that (some extension of) p forces κ+
y ∈ I[κy ].
+o(y)+1
i be a name for the approachability witness. We can assume
Let hżγ | γ < κy
that the order type of each żγ is forced to be less than κy .
+o(y)+1
-supercompact using an ultrapower
Let j : V → M witness that κy is κy
which projects to ūβα (y) for some α ∈ Zyβ . In particular if we consider j(PU~ y ), then
+f β (κ )
ŷ = j“κy α y is a possible value for the j(a)(j(fαβ )(κy )).
Let p̂ ≤ j(p) be a condition in j(PU~ y ) such that ap̂ (fαβ (κy )) = ŷ and ap̂ (fαβ (κy ) +
+o(y)+1
1) = ŷ 0 for some ŷ 0 . We set µ = sup j“κy
and δ = cf(µ). We have that p̂
o(y)+1
forces that µ is approachable with respect to j(żγ | γ < κy
i). So there is a
name Ċ for a club subset of µ such that for all γ < µ, Ċ ∩ γ is enumerated as j(ż)γ 0
for some γ 0 < µ.
By the previous lemma there are a club D ⊆ µ in M and a direct extension
o(y)+1
0
p̂ of p̂ such that p̂0 forces D ⊆ Ċ. Let E = {γ < κy
| j(γ) ∈ D}. It is
o(y)+1
o(y)
straightforward to see that E is < κy -club in κy
. Let γ ∗ be the κy th element
in some enumeration of E. We can assume that there is an index γ̂ such that p̂0
forces that Ċ ∩ j(γ ∗ ) is enumerated before stage j(γ̂).
Note that o(y) < κy since cf(β) < κy . Now if x ⊆ E ∩ γ ∗ has ordertype at most
o(y), then there is a condition px which forces x ⊆ żγ for some γ < γ̂. To see this
notice that j of this statement is witnessed by p̂0 .
12
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
By the chain condition of PU~ y , we can find a condition which forces that for
+o(y)+1
κy
many x, px is in the generic. This is impossible, since
each żγ is forced to
S
have order type less than κy and hence in the extension | γ<γ̂ P(żγ )| ≤ κy .
Next we show that weak square holds at points taken from Xβ where cf(β) ≥ κ.
To do so we need a claim about the cofinalities of points in the extension and a few
definitions.
sc
Suppose that ν = κ ∩ x ∈ C with x ∈ CG
. Let p = (a, A) ∈ G so that x = a(β)
for some β ∈ dom(a). By the definition of the sets Xβ , β < θ and P, we have that
• β is a limit ordinal if and only of o(x) is,
• If β is limit then (ν + )V [G] = (ν +o(x)+1 )V ,
• cf(β) ≥ κ if and only if o(x) ≥ ν.
Lemma 4.4. If β is limit and cf(β) ≥ κ, then ν and ν +o(x) change their cofinality
to ω in V [G].
Proof. As the forcing P/(β + 1) does not add new subsets to ν θ(ν) it is sufficient to
focus on the forcing PU~ x which adds a Radin club to ν.
~ =
For notational simplicity suppose that ν = κ and then PU~x ∼
= PU~ where U
hUα | α < βi where β < θ and cf(β) = ρ ≥ κ, and let us show that κ and κ+β
change their cofinality to ω.
~ = hβα | α < ρi cofinal in β. For
Choose an increasing continuous sequence β
sc
every y ∈ CG let α(y) < ρ be the minimal α < ρ so that y = a(β 0 ) for some β 0 ≤ βα
and some p = (a, A) ∈ G.
Since θ is the first inaccessible cardinal above κ then κ+ρ > ρ. Let α0 < ρ be
the first ordinal so that ρ < κ+βα0 . Let β0 = βα0
Note that for every β 0 , β0 ≤ β 0 < β Yβ 0 = {x ∈ Xβ 0 | α(x) ∈ x} belongs to
0
Uβ 0 ⊂ P(Pκ (κ+β )).
sc
, if
It follows that in V [G] there is some ν0 ∈ C such that for every x ∈ CG
0
0
x = a(β ) for some β ≥ β0 and κx = κ ∩ x > ν0 , then x ∈ Yβ 0 . Let y0 be
sc
the minimal x ∈ CG
satisfying the above. Starting from y0 , we define a sequence
sc
. For each n < ω, let yn+1 be the minimal x above yn in
~y = hyn | n < ωi ⊂ CG
sc
CG so that α(x) > sup(yn ∩ ρ) < ρ. Let κω = ∪n κyn . We claim that κω = κ.
sc
. Pick p = (a, A) ∈ G and
Suppose otherwise. Then κω = κy for some y ∈ CG
0
0
β ≥ β0 so that y = a(β ). Let α = α(y) < ρ, then α ∈ y < sup y ∩ ρ. Since
hκyn | n < ωi are cofinal in κω then y ∩ ρ = ∪n (yn ∩ ρ), so there is some m < ω such
that sup(ym ∩ ρ) > α. But α ≥ α(ym+1 ) which means that α(ym+1 ) < sup(ym ∩ ρ),
contradicting the definition of the sequence ~y .
sc
⊂ Pκ (κ+β )
It follows that κ = κω so it changes its cofinality to ω. The set CG
+β
is ⊆ −cofinal in Pκ (κ ). Since hκn | n < ωi is cofinal in κ then the sequence
sc
hyn | n < ωi is ⊂ unbounded in CG
, thus κ+β = ∪n yn . It follows that cf(κβ ) = ω
+β
as each yn is bounded in κ .
It follows easily that cardinals between ν and ν +o(ν) also change their cofinality
to ω. To show that weak square holds we need the definition of a partial square
sequence.
Definition Let λ < δ be regular cardinals and S ⊆ δ ∩ cof(λ). We say that S has
a partial square sequence hCγ | γ ∈ Si is a partial square sequence if
(1) for all γ ∈ S, Cγ is club in γ and otp(Cγ ) = λ and
DIAGONAL SUPERCOMPACT RADIN FORCING
13
(2) for all γ < γ ∗ from S if β is a limit point of Cγ and Cγ ∗ , then Cγ ∩ β =
Cγ ∗ ∩ β.
Next we need a theorem of Dzamonja and Shelah [5].
Theorem 4.5. Let λ be a regular cardinal and µ > λ be singular. If cf([µ]≤λ , ⊆) =
µ, then µ+ ∩ cof(≤ λ) is the union of µ many sets each of which carries a partial
square.
Lemma 4.6. Suppose that β < θ with cf(β) ≥ κ and p is a condition where
ap (β) = y for some y. Then p forces ∗κy .
+o(y)+1
+o(y)+1
Proof. For each regular λ < κy , we have that (κy
)≤λ = κy
using the
+o(y)+1 ≤λ
+o(y)+1
supercompactness of κy and hence cf([κy
] , ⊆) = κy
.
+o(y)+1
+o(y)
So we can write κy
∩cof(≤ λ) as the union of κy
sets which have partial
+o(y)
λ,i
~
squares. We call these sequences C for i < κy
.
By Lemma 4.4, in the extension we have that each cardinal in the interval
+o(y)
+o(y)+1
[κy , κy
] changes its cofinality to ω. So in the extension, we can write κy
+o(y)+1
as the disjoint union of κy
∩ cof(< κy ) which we call T and a set S of ordinals
of countable cofinality.
We define a weak square sequence as follows. For γ ∈ T we let Cγ = {Cγλ,i
0 | λ <
+o(y)
and γ is a limit point of Cγλ,i
κy , i < κy
0 }. For γ ∈ S, we let Cγ = {C} where C
is some cofinal ω-sequence in γ.
The coherence is obvious, so we just have to check that each Cγ is not too large.
+o(y)+1
. Then by the Pigeonhole principle
Suppose that there is γ such that |Cγ | ≥ κy
0
we can find two elements C and C on which the indices λ and i are the same, but
then we have that C = C 0 by the coherence of the partial square sequence with
indices λ and i, a contradiction.
5. Iteration from below
In this section we describe how to iterate Sinapova’s poset from [21] to obtain
many singular cardinals ν where SCH fails and there are no special ν + trees.
(1) Suppose that κ is Mahlo and U = hUα,τ | α < κ, τ < oU (α)i is a coherent
sequence of supercompact measures so that for each α < κ and τ < oU (α),
Uα,τ is a Pα (α+τ ) supercompact measure.
We further assume o(α) < α for each α < κ and that for every τ < κ, the
set ∆τ = {α < κ | oU (α) = τ } is stationary in κ.
(2) We define a Gitik-iteration Pκ = hPα , Qα | α < κi of Sinapova’s poset from
[21].
By induction on α, we define Pα and Qα so that the following holds:
(a) Qα will be nontrivial if and only if o(α) > 0.
(b) (Qα , ≤α , ≤∗α ) is a Prikry type forcing notion and ≤∗α is < α closed.
(c) If o(α) > α then a Qα generic filter assigns a sequence ~xα = hxα
i | i <
o(β)i which generic for the poset in [21] with respect to the measures
hUβ,τ | τ < o(β)i. In particular, if o(α) is a limit ordinal then α+o(α) =
S β
i xi V [Gα ]
(d) We use the Gitik support from [9] to define Pα . Conditions p ∈ Pα are
of the form hpβ | β ∈ gi where g ⊂ α is an Easton support set.
14
OMER BEN-NERIA, CHRIS LAMBIE-HANSON AND SPENCER UNGER
Suppose that α ≤ κ and that Pα has been defined. Let G ⊂ Pα be a
generic filter. Suppose that o(α) = λ. We define posets Qτα for τ ≤ λ by
induction on τ . Qτα will add a generic sequence ~xα,τ = hxα
i | i < τ i which
is Prikry/Magidor generic for the sequence hUα,γ | γ < τ i.
(a) If λ = 0 then Q0α = Qα is trivial.
(b) Suppose 0 < τ ≤ λ and Qα,γ has been defined for every γ < τ .
Suppose that γ < τ and x ∈ Pα (α+γ ). Following Gitik’s notations
from [9], we say that x is γ−coherent if αx = α ∩ x < α and o(αx ) = γ.
We say that x is coherent if it is γ−coherent for some γ. For each such
x, let πx : x → x̄ be the transitive collapse.
Conditions q ∈ Qα,τ will be of the form hx, T i where x is γ−coherent
for some γ < τ and T is a tree whose splitting sets belong to ultrafilters
of the form Uα,γ 0 (y) where γ < γ 0 < τ and y is coherent.
More precisely, T ⊂ [Pα (α+τ )]≤ω satisfies that
• stem(T ) = ∅,T
• SuccT (∅) ∈ {Uα,γ 0 (x) | γ < γ 0 < τ } and x ≺ y for every
y ∈ SuccT (∅),
• For every s = hx0 , . . . , xk i ∈ T , s is a ≺ increasing sequence of
subsets in Pα (α+τ ) and the sequence ho(αx0 ), . . . , o(αxk )i is an
increasing sequence of ordinals below τ .
• If τ = τ 0 + 1 is successor ando(αxk ) = τ 0 then s is a maximal
branch in T .
0
0
• If
T τ is limit or τ = τ + 10 but o(αxk ) < τ , then SuccT (s) ∈
{Uα,γ 0 (xk ) | o(αxk ) < γ < τ } and xk ≺ y for every y ∈
SuccT (s).
It remains to define the ultrafilters Uα,γ 0 (x) for every γ−coherent x
with γ < γ 0 < τ . They are defined by induction on γ 0 . We note
0
that Uα,γ 0 (x) will concentrate on the set of all y ∈ Pα (α+γ ) so that
α
x̄ = x¯γ y .
Let jα,γ 0 : V → Mα,γ 0 ∼
= Ult(V, Uα,γ 0 ) be the ultrapower embedding.
Working in Mα,γ 0 , we use the fact that the iteration satisfies the Prikry
condition (see [9]) and that the end tail of the iteration R = jα,γ 0 (Pα )\
(α + 1) is (2α )+ −closed to define a sequence of master conditions
0
hrη | η < 2α i deciding all the R−statements jα,γ 0 “α+γ ∈ jα,γ 0 (Ẋ)
0
for every Pα name Ẋ for a subset of Pα (α+γ ).
0
Define Uα,γ 0 (x) as follows. Let X ⊂ Pα (α+γ ) and Ẋ, a Pα name for
X. X ∈ Uα,γ 0 (x) if and only if there is some p ∈ G ⊂ Pα , η < 2α and
a tree T so that
0
p_ hx, T i_ rη jα,γ 0 “α+γ ∈ jα,γ 0 (Ẋ)
Extensions and direct extensions are defined as usual. The arguments
0
in [9] show that Uα,γ 0 (x) is a α−complete fine ultrafilter on Pα (α+γ )
and that the poset Qα,τ satisfies the Prikry property.
Let Hα,τ ⊂ Qα,τ be a generic filter. For every γ < τ there exists a
unique γ−coherent x so that hx, T i ∈ Hα,τ for some tree T . Denote x
by xα
xα,τ = hxα
γ . Let ~
γ | γ < τ i be the induced generic sequence. Since
all the measures Uα,γ 0 (x) are fine, a density argument shows that if τ
DIAGONAL SUPERCOMPACT RADIN FORCING
15
S
is limit then α+τ = γ<τ xα
γ.
We finally define Qα = Qλα .
(3) Suppose that α < κ and Gα+1 ⊂ Pα+1 is a generic filter. The arguments
in [21] show that if o(α) is a limit ordinal then ∗α fails in V [Gα+1 ]. Since
the rest of the iteration does not add new subset to α+ , it follows there is
no ∗α sequence in a Pκ generic extension. S
(4) Let G ⊂ Pκ be a generic filter and let ∆ = {∆τ | τ < κ limit }. It follows
that ∆ is a fat stationary set in V [G] and ∗α fails at every α ∈ ∆. Let R be
the forcing which adds a club to ∆ by initial segments. By Abraham-Shelah
[?] R is < κ−distributive and therefore does not collapse cardinals nor does
it add a weak square to a cardinal α ∈ ∆. Furthermore, κ remains Mahlo
V [G∗H]
in a R generic extension V [G][H] of V [G]. Let N = (V [G][H])κ = Vκ
.
Then N is a model of GB (Godel-Bernays) which contains a club C through
its ordinals α which doe not carry ∗α sequence.
6. Conclusion
In a forthcoming paper of the third author [?], a model is constructed where ℵω2
is strong limit and weak square fails for all cardinals in the interval [ℵ1 , ℵω2 +2 ]. In
particular, it is shown that one can put collapses between the Prikry points of the
Gitik-Sharon [12] construction which enforce the failure of weak s quare at cardinals
below κ while making κ into ℵω2 .
It is reasonable to believe that this construction could be combined with the
forcing from the present paper, but we are left with the unsatisfactory result that
weak square will hold at some successors of singulars in the extension. To make
this precise we formulate a question which seems to capture the limit of a naive
combination of the two techniques.
Question 6.1. Suppose that κ is a singular cardinal of cofinality ω such that ∗λ
fails for all λ ∈ [ℵ1 , κ) and |{λ < κ | λ is singular strong limit }| = κ. Is there a
∗κ -sequence?
Question 6.2. Are there versions of Theorems 1.1 and 1.2 where the failure of ∗ν
is replaced with the tree property at ν + ?
Question 6.3. Let CG ⊂ κ be a generic Radin club added by the poset P defined
in Section 3. Does ν,ω fail at every ordinal ν ∈ CG ?
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