Hod up to ADR + Θ is measurable∗†
Rachid Atmai
Department of Mathematics
University of North Texas
General Academics Building 435
1155 Union Circle #311430
Denton, TX 76203-5017
[email protected]
Grigor Sargsyan‡
Department of Mathematics
Rutgers University
110 Frelinghuysen Rd.
Piscataway, NJ 08854
http://math.rutgers.edu/∼gs481
[email protected],
Abstract
Suppose M is a transitive class size model of ADR + “Θ is regular”. We say M is
a minimal model of ADR + “Θ is measurable” if (i) there is µ ∈ M such that M “µ
is a normal R-complete measure on Θ” and (ii) for any transitive class size N ⊆ M ,
N “there is no R-complete measure on Θ”. Continuing Trang’s work in [9], we
compute HOD of a minimal model of ADR + “Θ is measurable”.
The computation of HOD of models of determinacy has been one of the central themes in
descriptive inner model theory. Steel’s seminal [6] jumpstarted the project and a later work
of Steel and Woodin (for instance see [4] or [5]) established connections with the Mouse Set
Conjecture, core model induction and the inner model problem.
The analysis of HOD presented in the above papers, however, only compute VΘHOD .
Woodin computed the full HOD of L(R) under ADL(R) (the proof can be found in [7]).
∗
2000 Mathematics Subject Classifications: 03E15, 03E45, 03E60.
Keywords: Mouse, inner model theory, descriptive set theory, hod mouse.
‡
First author’s work is partially based upon work supported by the National Science Foundation under
Grant No DMS-1352034 and DMS-1201348.
†
1
Trang continued this work in [9]. He presented the exact inner model theoretic structure of
HOD of models of determinacy that are contained inside the minimal model of ADR + “Θ
is regular”. We extend Trang’s work to minimal models of ADR + “Θ is measurable”.
Assume AD+ . We say “Θ is measurable” if there is an R-complete normal measure µ on
Θ. We then say that M is a minimal model of ADR +Θ is measurable if for any N ( M such
that R, Ord ⊆ N , N “Θ is not measurable”. It is then not hard to characterize minimal
models of ADR + “Θ is measurable”.
Lemma 0.1. Suppose that V is a minimal model of ADR + “Θ is measurable”. Let M =
L(P(R)). Let µ be a normal R-complete measure on Θ. Then V = L(P(R))[µ].
Proof. We have that
L(P(R))[µ] ADR + “Θ is measurable”.
Because L(P(R))[µ] ⊆ V , we have that V = L(P(R))[µ].
Acknowledgments. The authors would like to thank Nam Trang for communicating
them the problem considered in this paper.
1. The main theorem
Suppose V is a minimal model of ADR + “Θ is measurable”. Let M = L(P(R)) and working
in M , let
F = {(P, Σ) : (P, Σ) is a hod pair and Σ has branch condensation and is fullness preserving}
Following [4], for (P, Σ), (Q, λ) ∈ F, we let
(P, Σ) (Q, Λ)
if for some α ≤ λQ , (Q(α), ΛQ(α) ) is a tail of (P, Σ), i.e., Q(α) ∈ I(P, Σ) and ΛQ(α) = ΣQ(α) .
It follows from comparison theory of hod pairs (see Chapter 2 of [4]) that is directed. We
Σ
then let M∞ be the direct limit of the system F under the maps πP,Q
: P → Q(α) where
Q
α ≤ λ is such that Q(α) is a Σ-iterate of P.
It is shown in [4] (see Theorem 4.24) that in M , M∞ = VΘHOD . The following is essentially
the generic interpretability result of [4] (see Theorem 3.10 of [4]). Given a hod premouse P
and X generic over P, we let ΣP[X] be the interpretation of ΣP onto P[X] according to the
procedure described in the proof of Theorem 3.10 of [4].
Lemma 1.1. In M (and hence in V ), M∞ is (Θ, Θ)-iterable via a strategy Σ such that
given any T~ according to Σ,
~
Σ(T~ ) = b ⇐⇒ M∞ [T~ ] ΣM∞ [T ] (T~ ) = b.
We let Σ be the strategy of M∞ described in Lemma 1.1. Next, we define a model
extending M∞ . Given N D M we say N is good if
2
1. N is sound, and
2. whenever
π : N̄ → N
is elementary and N̄ is countable, then N̄ is ω1 -iterable above π −1 (Θ) as a Σπ premouse.
The next two lemmas are basic lemmas about good mice.
Lemma 1.2. Suppose N0 , N1 are good such that for some η ∈ N0 ∩ N1 , η is a cutpoint of
both N0 and N1 , N0 |η = N1 |η and ρω (N0 ), ρω (N1 ) ≤ η. Then either N0 N1 or N1 N0 .
Proof. We start with clause 1. Suppose that neither N0 N1 nor N1 N0 holds. Let then
π : H → Lξ (P(R))[µ] be elementary such that ξ Θ, Ni ∈ rng(π), Σ ∈ rng(π), η ∈ rng(π)
and |H| = ω.
We let N̄i = π −1 (Ni ) for i = 0, 1. Because Σ has hull condensation, it follows that
Σ H = Σπ . It follows from elementarily that for i = 0, 1, ρω (N̄i ) = π −1 (η). But now
because N̄i are sound Σπ -mice, we have that N̄0 E N̄1 or a vice versa.
Lemma 1.3. Suppose N is good such that ρω (N ) ≤ Θ. Then ρω (N ) = Θ.
Proof. Towards a contradiction assume ρ(N ) < Θ. Let π : H → Lξ (P(R))[µ] be such that
H ∩ Θ ∈ Θ, cp(π) > ρ(N ) and N ∈ rng(π). Let P = π −1 (N ). It follows from the proof of
clause 1 that M “P is OD”. It then follows that P E M∞ . This is a contradiction.
We then let M∗ = ∪{N : M E N , N is good and ρω (M) = N }. Notice that we have
that M∗ ⊆ HOD. Let η = o(M∗ ).
Let now µ ∈ V be a normal R-complete measure on Θ. It follows from Lemma 0.1 that
V = L(P(R))[µ]. Working in HODµ , let πµ be the ultrapower embedding via µ ∩ HODµ .
∗
Let λµ = (Θ+ )πµ (M ) . Notice that the ordinal λµ may depend on µ. Let then
M = πµ (M∗ )|λµ
and let Eµ be (Θ, η)-extender derived from πµ M∗1 . More precisely,
(a, A) ∈ Eµ ⇐⇒ a ∈ (o(M∗ ))<ω , A ∈ [Θ]|a| ∩ M∗ , and a ∈ πµ (A)
We now make the following minimality assumption on µ.
Minimality Assumption on Measures: We say µ satisfies MAM if λµ above is the
least possible.
The following is our main theorem.
Theorem 1.4 (Main Theorem). HOD = J [M, Eµ ].
1
Recall that o(X) = X ∩ Ord.
3
We will present the proof as a sequence of lemmas. Before we go into the proof of the
main theorem, we list some of the complications involved with proving it. First we will show
that Eµ is amenable to M (see Lemma 1.6). It follows from its definition that it coheres M.
It then follows that (M∞ , Eµ ) is a hod premouse.
The next challenge is to show that no level of J [M, Eµ ] projects to or below Θ (see
M
J [M,Eµ ]
Lemma 1.8). A consequence of this is that VΘHOD = VΘ
. This then allows us to
show that P(R) can be symmetrically added to J [M, Eµ ] (which is done as part of proving
Lemma 1.8). We then finish by showing that there is only unique normal R-complete measure over Θ (see Lemma 1.12). It follows from here that µ ∩ HOD ∈ HOD implying that
J [M, Eµ ] ⊆ HOD. Combining the aforementioned results it is then not hard to see that V is
a symmetric extension of J [M, Eµ ], which then easily implies that in fact HOD ⊆ J [M, Eµ ].
1.1. Amenability
We start by showing that
Lemma 1.5. M∗ = πµ (M∞ )|η, M∗ = M|(Θ+ )M and whenever M∗ E N E M is such
that ρ(N ) = η, N is good.
Proof. We start by proving the first equality. We have that M∗ ⊆ HOD and therefore,
M∗ ∈ HODµ . It is then enough to show that whenever M∞ E N E M∗ is such that
ρ(N ) = Θ, there is f : Θ → M∞ such that f ∈ HODµ and πµ (f )(Θ) = N . Working in
HODµ , construct a club C and a continuous chain (Nκ : κ ∈ C) along with embeddings
πκ : Nκ → N such that cp(πκ ) = θκ = κ. We have that Nκ ∈ M∞ . It then easily follows
that
πµ (Nκ : κ ∈ C)(Θ) = N .
This shows that M∗ ⊆ πµ (M∞ ), and therefore, M∗ = πµ (M∞ )|η.
For the second equality, it is enough to show that whenever M∞ E N E M is such that
ρω (N ) = Θ then N is good. Let then f : Θ → M∞ be such that πµ (f )(Θ) = N . It then
follows that for a µ-measure one set of κ, there is an elementary embedding πκ : f (κ) → N .
It then also follows that whenever π : N̄ → N is an elementary embedding such that N̄ is
countable, there is a µ-measure one set of κ such that there is an embedding σ : N̄ → f (κ)
with the property that π = πκ ◦ σ. It then follows that N̄ is Σπ -good.
The proof of the third equality is very similar.
In what follows, we will sometimes write M− for M∞ . Next, is our amenability result.
Lemma 1.6. µ is amenable to M, that is µ ∩ M ∈ M
Proof. It is enough to show that if N M is such that ρω (N ) = Θ then µ ∩ N ∈ M. Fix
N and let pN be the standard parameter of N . Construct an elementary chain (Nκ : κ ∈ S)
and elementary embeddings (πκ : κ ∈ S) such that S ⊆ Θ is a club and for κ < λ ∈ S
1. πκ : Nκ → N ,
2. cp(πκ ) = κ, κ = θκ , rng(πκ ) ⊆ rng(πλ ) and
4
3. Nκ ∩ Θ = κ.
Let for κ < λ ∈ S,
πκ,λ =def πλ−1 ◦ πκ : Nκ → Nλ .
For κ ∈ S let pκ be the standard parameter of Nκ . Let
µκ = {A ∈ Nκ : κ ∈ πκ (A)}
be the measure derived from πκ . We claim that
Claim. {κ : πκ [µκ ] ⊆ µ} ∈ µ.
Proof. Let µ∗ = µ ∩ HODµ . Notice that M ∈ HOD and µ ∩ M = µ∗ ∩ M. Moreover, we
can choose S in a way that S ∈ HODµ .
Since S is a club, Θ ∈ πµ (S). Let
πµ (Nκ : κ ∈ S) = (Nκ : κ ∈ πµ (S))
πµ (µκ : κ ∈ S) = (νκ : κ ∈ πµ (S))
πµ (πκ : κ ∈ S) = (σκ : κ ∈ πµ (S)).
Then we have that νΘ is just the measure derived from the embedding πµ NΘ . It follows
that µ ∩ NΘ = νΘ . But because πµ [µ∗ ] ⊆ πµ (µ∗ ), we have that Θ ∈ {κ : πµ [νκ ] ⊆ πµ (µ)} as
πµ [νΘ ] ⊆ πµ (µ∗ ). But then {κ : πκ [µκ ] ⊆ µ} ∈ µ.
Next, let A = {κ : πκ [µκ ] ⊆ µ}. Since
(1) πκ [µκ ] ⊆ µλ , πκ = πλ ◦ πκ,λ and πλ [µλ ] ⊆ µ,
we have that
(2) µκ is just the measure derived from the embedding πκ,λ , i.e., µκ = {A ∈ Nκ : κ ∈ πκ,λ (A)}.
It follows from condensation that Nκ , Nλ M∞ . It then follows that πκ,λ ∈ HODM . This is
simply because πκ,λ is generated by the embedding that sends pκ to pλ and is the identity on
ρω (Nκ ) = κ where pκ and pλ are the standard parameters of Nκ and Nλ respectively. More
precisely,
(3) πκ,λ (x) = τ Nλ (s, pλ ) where s ∈ κ<ω and τ is such that x = τ Nκ (s, pκ ).
It follows from (3) that πκ,λ ∈ HODM . We then have that µκ ∈ HODM .
Let Mκ M be the least such that µκ ∈ Mκ for κ ∈ A. Let
πµ (Mκ : κ < Θ)(Θ) = Q.
Notice that we must have that Q is good and hence, Q M. This follows from the fact
that if τ : R → Q is a countable hull of Q then for some κ there is τ ∗ : R → Mκ such that
∗
Στ = Στ .
We then have that µ ∩ N ∈ Q because
5
N = πµ (Nκ : κ ∈ S)(Θ) and µ ∩ N = πµ (µκ : κ ∈ S)(Θ).
1.2. A strategy for countable submodels of J [M, Eµ ]
Suppose ξ is such that Eµ is a total extender in Jξ [M, Eµ ]. Let π : Q → Jξ [M, Eµ ] be
elementary such that Q is countable. In this section, we show that Q has a π-realizable
iteration strategy. Given any such R, we let E R be the preimage of Eµ . We also let
R− = R|((δ R )+ )R . Recall from [4] that if T~ is a stack on some model M and R is a node in
T~ then T~ ≥R is the portion of T~ after stage R.
Given a stack T~ on Q, it can easily be partitioned into segments by considering when the
image of E Q is used. Thus, we say (M0α , Eα , M1α , T~ α : α < η) are the essential components
of T~ if
1. M00 = Q,
2. E0 , M10 are defined if and only if the first extender used in T~ is E Q in which case
M10 = U lt(M00 , E Q ).
3. T~ 0 is the largest initial segment of T~ that is based on (M11 )− .
0
4. If α + 1 < η then M0α+1 is the last model of T~ α . Then Eα+1 = E Mα+1 and M1α =
U lt(M0α+1 , Eα+1 ). Again, T~ α+1 is the largest portion of T~ ≥M1α+1 that is based on
M1α+1 .
5. If α < η is limit then M0α is the direct limit of (M0β : β < α) under the iteration
embeddings. The rest of the objects are defined as in the successor case.
Suppose now that T~ is a stack on Q with essential components (M0α , Eα , M1α , T~ α : α ≤
η). Suppose that we also have embeddings (πα0 , πα1 : α < η) such that
1. π00 = π.
2. For α < η and i ∈ 2, παi : Miα → Jξ [M, Eµ ].
~
~
T
T
3. For α < β < η and i, j ∈ 2, παi = πβj ◦ πM
and πα1 = πα1 ◦ πM
0 ,M1 .
i ,Mj
α
α
α
β
4. For α < η, T~ α is according to πα1 -pullback of Σ.
We would like to define embeddings (πη0 , πη1 ) such that
1. For i ∈ 2, πηi : M0η → Jξ [M, Eµ ].
~
~
T
T
2. For α < η and i, j ∈ 2, παi = πηj ◦ πM
and πη1 = πη1 ◦ πM
0 ,M1 .
i ,Mj
η
η
α
η
We say T~ is π-realizable if there is ~π =def (πα0 , πα1 : α < η) witnessing the above clauses.
Suppose first η = α + 1. Let (S, Λ) ∈ F be such that
6
1
1
sup(πα+1
[δ Mα+1 ]) = δ M∞ (S,Λ) .
The above equality simply says that the direct limit of all Λ-iterates of S reaches the ordinal
mentioned on the left side. Since M1α+1 is countable, we can also require that
Λ
1
).
[M1α ] ⊆ rng(πS,∞
πα+1
Λ
1
Let then k =def (πS,∞
)−1 ◦πα+1
: M1α → S. We then have that T~ α is according to k-pullback
of Λ. It follows then that letting R be the last model of k T~ α there is an embedding l such that
~
~
(1) l : M0η → R and π kT α ◦ k = l ◦ π T α .
ΛR
We then set πη0 = πR,∞
◦ l. Notice now that
~
ΛR
Λ
(2) πS,∞
= πR,∞
◦ π kT α .
(1), (2) and our choice of (S, Λ) imply that
~
T
(3) πα1 = πη0 ◦ πM
1 ,M0 .
α
η
(3) then implies that πη0 is as desired.
To define πη1 we use countable completeness of µ. First let ζ = o((M0η )− ). Thus, ζ is the
0
0
successor of δ Mη in M0η . For each a ∈ ζ <ω , let Aa = ∩{πη0 (A) : (a, A) ∈ E Mη }. Fix now a
fiber f for the set {(πη0 (a), Aa : a ∈ ζ <ω }. We can now define
πη1 ([a, g]E M0η ) = πη0 (g)(f (a)).
It is a standard argument to show that πη1 is as desired. It is now easy to show that
Lemma 1.7. There is an (ω1 , ω1 ) iteration strategy Λ for Q such that whenever T~ is a stack
according to Λ then T~ is π-realizable.
1.3. J [M, Eµ ] is a hod premouse
In this section we show that no level of J [M, Eµ ] projects across or to Θ.
Lemma 1.8. N = J (M, Eµ ) is a hod premouse such that for every α, ρω (N |α) > Θ.
Proof. We start with the following claim. Later we will show that in fact for every α,
ρω (Jα (M, Eµ )) 6= Θ.
Claim 1. For every α, ρω (Jα (M, Eµ )) ≥ Θ
Proof. Suppose not and fix the least α such that ρω (Jα (M, E)) < Θ. Now fix a β such that
β < λM and θβ ≤ ρω (Jα (M, E)) < θβ+1 . Fix a hod pair (Q∗ , Λ∗ ) such that M∞ (Q∗ , Λ∗ ) =
M(β + 2) and an elementary hull σ : H → Lξ (P(R))[µ] such that:
1. ξ Θ,
7
2. |H| = ℵ0
3. (Q∗ , Λ∗ ) ∈ rng(σ),
4. Jα (M, E) ∈ rng(σ).
Now let Q̄ = σ −1 (Jα (M, E))) and let γ = σ −1 (β). Then by elementarity we have that
Q̄
. Let
ρω (Q̄) < θγ+1
Q = CωQ̄ (p, Q̄(γ + 1))
Q
-sound. Let l : Q → Q̄ be the
where p is the standard parameter of Q̄. Notice that Q is δγ+1
core embedding. Let Λ be a σ ◦ l-realizable strategy of Q (see Lemma 1.7).
It follows from the branch condensation of Λ∗ that
(1) ΛQ(γ+1) = Λ∗Q(γ+1) .
Consider the pointclass generated by (Q, Λ):
~ R) ∈ B(Q, Λ)(A <W Code(Λ ~ ))}.
Γ(Q, Λ) = {A : ∃(U,
R,U
Q
Since Q is δγ+1
-sound
(2) Q is ODΛQ(γ+1) .
This is because Q is the unique ΛQ(γ+1) -hod mouse generating the pointclass Γ(Q, Λ). But
Q̄
since Q̄(γ + 2) is ΛQ(γ+1) -full, we must have Q ∈ Q̄(γ + 2). But ρ(Q) < δγ+1
and this is a
contradiction.
The following claim is easier and finishes the proof of the lemma.
Claim 2. For every α, ρ(Jα (M, E))) > Θ
Proof. Suppose not. Let α be the least such that ρ(Jα (M, E))) = Θ. Then let A ⊆ Θ be
definable over Jα (M, E) such that A ∈
/ Jα (M, E). Notice that for every κ < Θ, A ∩ κ ∈ M.
Let S be the set of κ < Θ such that θκ = κ. Given κ ∈ S we let Mκ M be the least
such that A ∩ θκ ∈ Mκ .
Let j : M → U lt(M, µ) be the ultrapower embedding. Let j((Mκ : κ ∈ S)) = (Nκ : κ ∈
j(S)). But then A = j(A) ∩ Θ ∈ NΘ . Notice that NΘ implying that NΘ M. It follows
that A ∈ M, contradiction!
8
1.4. P(R) is symmetrically generic over J [M, Eµ ]
We start by recalling Vopenka algebra. We work in M = L(P(R)). Let ϕ be a formula and
s ∈ Θ<ω . Given t ∈ Θ<ω , define
Asφ,t = {a : dom(~a) = dom(s), a(i) ⊆ s(i)ω , L(P(R)) φ[a, t]}.
We write
(φ, t) ≡s (ψ, v) if and only if Asφ,t = Asψ,v .
We then let [φ, t]s be the ≡s -equivalence class of (φ, t).
Next define
Q = {(s, [ϕ, t]) : s ∈ Θ<ω , ϕ is a formula , t ∈ Θ<ω }
The ordering on Q is defined as follows:
(v, [φ, s]) ≤ (u, [ψ, t]) if and only if u C v, Auψ,t ⊆ Avϕ,s and Avϕ,s dom(u) ⊆ Auψ,t
where Avϕ,s dom(u) = {a dom(u) : a ∈ Avϕ,s }.
Let
P = {(s, a) : s ∈ Θ<ω , dom(a) = dom(s), ∧∀i ≤ n, a(i) ⊆ s(i)ω }.
We set (s, a) ≤P (t, b) if and only if t E s and b E a. Notice that if (f, h) is P-generic then
f : ω → Θ and h : ω → (Θω )V are surjections such that for each i ∈ ω, f (i) ∈ (h(i))ω .
The following lemma is standard and is really due to Vopenka.
Lemma 1.9 (Vopenka’s lemma). Suppose (f, h) is P-generic. Let
n
G = {(f n, [ϕ, v]) : n < ω, (f n, [ϕ, v]) ∈ Q ∧ h n ∈ Afφ,v
}.
Then G is Q/M-generic. Therefore, it is J [M, Eµ ]-generic over Q.
Proof. We first show that for ξ < Θ, we can bound the ordinal parameters used to define
ODM subsets of P(ξ).
Claim 1. In M , there is a function F : Θ<ω → Θ such that whenever t ∈ Θ<ω and a
are such that dom(a) = dom(t), a(i) ⊆ t(i) and a is ODM then a is OD in L(ΓF (t) ) where
Γβ = {A ⊆ R : w(A) < β}.
Proof. For A ⊆ R2 , we say (R, A) codes an OD structure if (R, A) is a well-founded, extensional model of some fragment of ZF C and its transitive collapse is OD. Notice that by
a standard Skolem hull argument, in M , if A ⊆ R2 and (R, A) codes an ordinal definable
structure then for some β < Θ, (R, A) codes an ordinal definable structure in L(Γβ ). Fix
now α < Θ. Then we must have that
sup{β : ∃A ⊆ R2 (L(Γβ ) “(R, A) codes an OD structure”)} < Θ.
This is because otherwise we will have a function G : Γβ → Θ which is unbounded. As Θ is
regular, this is impossible. Let then
9
F (t) = sup{β + ω : ∃A ⊆ R2 (A ∈ Γsup(rng(t))+ω ∧ L(Γβ ) “(R, A) codes an OD structure”)}.
Let then D ∈ M be a dense subset of Q. We want to see that G ∩ D 6= ∅. Let E be the
set of (s, a) such that dom(s) = dom(a), s ∈ Θ<ω , for every i < dom(s), a(i) ∈ s(i)ω and for
some [φ, v] such that (s, [φ, v]) ∈ D, a ∈ Asφ,v .
We claim that E is dense in P. Fix then (s, a) ∈ P.
Claim 2.
∪{Atφ,v dom(s) : s E t, (t, [φ, v]) ∈ D} = {b : dom(b) = dom(s) and for all i < dom(s),
b(i) ∈ s(i)ω }.
Proof. Notice that D 6∈ M , which makes the claim non-trivial. Consider the set A = {Atφ,v dom(s) : (t, [φ, v]) ∈ D}. Each member of A is ODM . We want to see that A itself is OD in
M.
To see this let N E M be such that ρω (N ) = Θ and D ∈ N . Fix ζ Θ and let
π : H → Lζ (P(R))[µ] be such that
1. N , F, D ∈ rng(π),
2. cp(π) = κ = θκ .
H
Let (N̄ , F̄, D̄) = π −1 (N , F, D). We have that F̄ = F θκ and D̄ = {(t, ([φ, v])L(P(R)) ) :
M
(t, [φ, v]) ∈ D}. Let Ā = π −1 (A). Then N E M∞ = VΘHOD . We then have that D̄ ∈ ODM .
Thus Ā is ODM .
Notice next that Ā = A. Indeed, fix B ∈ A. We have that B is ODM . It follows that B is
ODL(ΓF (s) ) and hence, B is ODL(ΓF (s) ) . Let then (φ, v) be such that v ∈ (ΘL(ΓF (s) ) )<ω and B is
definable over L(ΓF (s) ) via (φ, v). Since D̄ is dense in H, we must have (t, [ψ, u]) ≤Q (s, [φ, v])
such that (t, [ψ, u]) ∈ D̄. But then B = Atψ,u ∈ Ā.
It now follows that A ∈ ODM implying the claim.
It follows from Claim 2 that there is (t, [φ, v]) ∈ D such that a ∈ Atφ,v dom(s). Let then
b ∈ Atφ,v dom(s) be such that b dom(a) = a. Then (t, b) ∈ E and (t, b) ≤P (s, a).
Since we now have that E is dense, we can fix n < ω such that (f n, h n) ∈ E. Let
n
[φ, v] be such that h n ∈ Afφ,v
. We then have that (f n, [φ, v]) ∈ G ∩ D. We leave it to
the reader to verify that G is a filter.
The following is the main lemma of this section.
Lemma 1.10. V is a symmetric extension of J [M, Eµ ]. In fact, V = J [M, Eµ ](Θω ).
Proof. Let (f, h) be P-generic. Let G be as in Lemma 1.9. Then Θω ∈ J [M, Eµ ][G]. It is
a consequence of AD+ that every set of reals A is (ODt )M for some t ∈ Θω . Therefore, we
have that
P(R) ⊆ HODM (Θω ) ⊆ J [M, Eµ ](Θω ).
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The reader can find more on the above equality is proved by consulting Section 2 of [1]. It
then follows that J [M, Eµ ](P(R)) = J [M, Eµ ](Θω ).
Because every set A ∈ P(Θ) ∩ J [M, Eµ ](Θω ) is added to J [M, Eµ ] by a small forcing
(in fact by the Vopenka algebra at some θα < Θ), we have that Eµ has a canonical extension
Eµ+ to J [M, Eµ ](Θω ) (see Theorem 2.4 of [1]). Let ν = {A : (Θ, A) ∈ Eµ+ }. We then have
that
J [M, Eµ ](Θω ) “ν is a normal R-complete measure on Θ”.
It then follows that V = J [M, Eµ ](Θω )2 .
1.5. The uniqueness of µ
In this section we show that there is a unique normal R-complete measure µ. We start with
the following lemma, which is a simple consequence of Lemma 1.10 and the fact that small
forcing doesn’t create new measures.
Lemma 1.11. Suppose ν is a normal R-complete measure on Θ. Then ν ∩ J (M, Eµ ) ∈
J (M, Eµ ).
Lemma 1.12. There is a unique R-complete measure on Θ that satisfies MAM.
Proof. The proof is a combination of the comparison argument and the argument that shows
that there are no bicephali. Suppose that there is a normal R-complete measure ν such that
µ 6= ν and λν = λµ . We then have two normal measures µ0 , ν0 ∈ J [M, Eµ ] such that letting
+
+
+
ω
µ+
0 , ν0 be the extensions of µ0 and ν0 in J [M, Eµ ](Θ ), µ = µ0 and ν = ν0 . It then follows
that µ0 6= ν0 . Let then F = (Θ, (Θ+ )M ))-extender derived from πν0 (in J [M, Eµ ]). Let
E = Eµ . We now have that J [M, E, F ] is a bicephali.
Let now ξ = (Θ+ω )J [M,E,F ] and let π : Q → Jξ [M, E, F ]. It follows from the construction
of Subsection 1.2 that there is an iteration strategy Λ for Q. The rest of the argument is
as follows. We compare (Q, Λ) with itself. The goal is to iterate Q to some R such that
Λ
Λ
(F ). The complete proof of this comparison process is beyond the scope
(E) = πQ,R
R πQ,R
of this paper. It can be proved using Theorem 1.4 of [3] and Theorem 9.2 of [2].
1.6. Computation of HOD
We first show that
Lemma 1.13. VΘHOD = M∞ . Moreover, if A ⊆ R2 is such that (R, A) codes an OD structure
then in M , (R, A) codes an structure.
Proof. Fix A ⊆ θα < Θ such that A is OD over L(P(R))[µ]. It follows from Lemma 1.10
that J (M, E)[Θω ] “A is OD”. Fix ζ > Θ such that Jζ (M, E)[Θω ] A is OD. Let
M̄ = HullJζ (M,E) (Θ). We then have that
M̄(Θω ) “A is OD”.
2
In fact, ν = µ but we do not need this.
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Let Q E M be such that M̄ ∈ Q. Let π : H → Lζ (P(R))[µ] be such that cp(π) = κ = θκ >
θα , |H| = κ and M̄, Q ∈ rng(π). Let ξ = o(H), N = π −1 (M̄ ), and P = π −1 (Q). Then it
follows from elementarity that
N (κω ) “A is OD” and N ∈ P.
However, P E M∞ as ρω (P) = θκ . It then follows that N is ODM implying that A ∈ ODM .
Hence, A ∈ M∞ .
The proof of the second clause of the lemma is very similar.
To finish we will use the following two lemma due to Trang and Woodin.
Lemma 1.14 (Trang-Woodin, Theorem 1.1 of [9]). Suppose Q is the Vopenka algebra of M .
Then HODM = L[Q] = L[M∞ ].
Finall we show that
Lemma 1.15. HOD = J [M, Eµ ]
Proof. We have that J [M, Eµ ] ⊆ HOD. It is then enough to show that HOD ⊆ J [M, Eµ ].
Because µ is ordinal definable, we have that HOD = L[P ](µ) where P is the Vopenka algebra
(see for instance Theorem 3.1.9 of [8]). However, notice that it follows from Lemma 1.13 and
the proof of Lemma 1.9 that if Q is the Vopenka algebra of M , then in fact L[Q](µ) = HOD.
Since we also have that Q ∈ M it follows that HOD ⊆ J [M, Eµ ].
References
[1] Andres Caicedo, Paul Larson, Grigor Sargsyan, Ralf Schindler, John R. Steel, and Martin
Zeman. Square in Pmax extensions, in preparation.
[2] William J. Mitchell and John R. Steel. Fine structure and iteration trees, volume 3 of
Lecture Notes in Logic. Springer-Verlag, Berlin, 1994.
[3] Grigor Sargsyan. The analysis of HOD below the theory AD+ +“the largest Smuslin
cardinal is a member of the Solovay sequence, available at math.rutgers.edu/∼gs481.
[4] Grigor Sargsyan.
Hod mice and the Mouse Set Conjecture.
http://math.rutgers.edu/∼gs481/.
Available at
[5] Grigor Sargsyan. Descriptive inner model theory. Bull. Symbolic Logic, 19(1):1–55, 2013.
[6] John R. Steel. HODL(R) is a core model below Θ. Bull. Symbolic Logic, 1(1):75–84, 1995.
[7] John R. Steel. An outline of inner model theory. In Handbook of set theory. Vols. 1, 2,
3, pages 1595–1684. Springer, Dordrecht, 2010.
[8] Nam Trang. Generalized Solovay Measures, the HOD Analysis, and the Core Model
Iinduction. PhD Thesis, UC Berkeley, 2013.
[9] Nam Trang. HOD in natural models of AD+ . Ann. Pure Appl. Logic, 165(10):1533–1556,
2014.
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