ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS

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1 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS JAMES CUMMINGS, SY-DAVID FRIEDMAN, MENACHEM MAGIDOR, ASSAF RINOT, AND DIMA SINAPOVA Abstract. A remarable result by Shelah states that if κ is a singular strong limit cardinal of uncountable cofinality then there is a subset x of κ such that HOD x contains the power set of κ. We develop a version of diagonal extenderbased supercompact Priry forcing, and use it to show that singular cardinals of countable cofinality do not in general have this property, and in fact it is consistent that for some singular strong limit cardinal κ of countable cofinality κ + is supercompact in HOD x for all x κ. 1. Introduction It is a familiar phenomenon in the study of singular cardinal combinatorics that singular cardinals of countable cofinality can behave very differently from those of uncountable cofinality. For example in the area of cardinal arithmetic we may contrast Silver s theorem [11] with the many consistency results producing models where GCH first fails at a singular cardinal of countable cofinality [2]. The results in this paper show that there is a similar sharp dichotomy involving questions about the definability of subsets of a cardinal rather than the size of its powerset. Shelah [12] proved that if κ is a singular strong limit cardinal of uncountable cofinality then there is a subset x κ such that P (κ) HOD x. In this paper we show that the hypothesis is essential, by proving: Theorem. Suppose that κ < λ where cf(κ) = ω, λ is inaccessible and κ is a limit of λ-supercompact cardinals. There is a forcing poset Q such that if G is Q-generic then: The models V and V [G] have the same bounded subsets of κ. Every infinite cardinal μ with μ κ or μ λ is preserved in V [G]. λ = (κ + ) V [G]. For every x κ with x V [G], (κ + ) HODx < λ. From stronger assumptions we can use Q to obtain a model in which κ is a singular strong limit cardinal of cofinality ω, and κ + is supercompact in HOD x for all x κ. Cummings was partially supported by the National Science Foundation, grants DMS and DMS Friedman would lie to than the Austrian Science Fund for its generous support through the research project P Rinot was partially supported by the Israel Science Foundation, grant 1630/14. Sinapova was partially supported by the National Science Foundation, grants DMS and Career

2 2 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA Acnowledgements. The results in this paper were conceived and proved during three visits to the American Institute of Mathematics, as part of the Institute s SQuaRE collaborative research program. The authors are deeply grateful to Brian Conrey, Estelle Basor and all the staff at AIM for providing an ideal woring environment. Notation. Our notation is mostly standard. We write l(s) for the length of a sequence s. When Q is a forcing poset and q Q we write Q q for {r Q : r q}, with the partial ordering inherited from Q. 2. A proof of Shelah s theorem For the reader s benefit we give a short and self-contained proof of the result by Shelah quoted in the Introduction. Let κ be a strong limit singular cardinal of uncountable cofinality μ. Fix a set x κ such that L[x] contains an enumeration (t η ) η<κ of H κ, together with a sequence of ordinals (α i ) i<μ which is increasing and cofinal in κ. To each set X κ we associate a function f X : μ κ, where f X (i) = η for the unique η such that X α i = t η. It is easy to see that if X Y then f X (i) f Y (i) for all large i. We now impose a strict partial ordering on P (κ) by defining X Y if and only if f X (i) < f Y (i) for all large i; since μ = cf(μ) > ω, the ordering is well-founded. Let R α be the set of elements of P (κ) with -ran α, and note that if X and Y are distinct elements of R α then there are unboundedly many i < μ with f X (i) > f Y (i), since otherwise f X (i) < f Y (i) for all large i and it would follow that X Y. We claim that R α 2 μ, so that in particular R α < κ. Supposing for contradiction that R α > 2 μ we fix a sequence (X j ) j<(2μ ) + of distinct elements of R α, and then define a colouring c of pairs from (2 μ ) + by setting c(j, j ) = i for the least i < μ with f Xj (i) > f Xj (i). By the Erdős-Rado theorem c has a homogeneous set of order type μ +, which is impossible because it would yield an infinite decreasing sequence of ordinals. We now define a well-ordering α of R α. To do this we let A α = {range(f X ) : X R α }, note that A α < κ, and define g α to be the order-isomorphism between A α and ot(a α ). If X R α then g α f X H κ, and we define X α Y if and only if the index of g α f X is less than that of g α f Y in the enumeration (t η ) η<κ of H κ. Forming the sum of the orderings α in the natural way, we obtain a wellordering of P (κ) which is clearly definable from x. Now every element of P (κ) is ordinal definable from x as the η th element of P (κ) in the ordering, so that P (κ) HOD x. 3. The main theorem 3.1. Setup. Let κ < λ with λ strongly inaccessible and κ = sup n κ n, where (κ n ) n<ω is a strictly increasing sequence of λ-supercompact cardinals. Fix U n a κ n -complete fine normal ultrafilter on P κn λ, and for κ α < λ let U n,α be the projection of U n to P κn α via the map x x α. We need a form of diagonal intersection lemma. Lemma 1. Let < l < ω, let (α i ) i<l be a -increasing sequence of elements of [κ, λ), and let S i<l P κ i α i be a set of -increasing sequences.

3 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 3 Let m [l, ω), let α [α l 1, λ), and let A = (A s ) s S be an S-indexed family with A s U m,α for each s S. Then the following set is in U m,α : D = {x P κm α : (x,..., x l 1 ) S [x l 1 x = x A (x,...x l 1 )]}. Proof. Let j : V M be the ultrapower map computed from U m. To show that D U m,α it suffices to show that j α j(d). Let t = (y,... y l 1 ) j(s) with y l 1 j α. Write B = j( A). We need to show that j α B t. For each i with i < l, y i j α i and y i < κ i < κ m = crit(j), so that y i = j x i = j(x i ) for some x i P κi α i. Write s = (x,... x l 1 ). Then t = j(s) and hence s S. In particular A s U m,α, and so j α j(a s ) = B t. In the setting of Lemma 1 we will refer to D as the diagonal intersection of A and we will write s x as shorthand for s = (x,... x l 1 ) with x l 1 x. With a view to the proof of Lemma 12 below we also need a technical lemma about measure one sets: Lemma 2. Let i < ω and let α i β i β i+1 be ordinals Let B U i,βi and let C U i+1,βi+1. Let C be the set of y C such that for every x B with x α i y there is x B P (y) with x α i = x α i. Then C U i+1,βi+1. Proof. Let j : V M be the ultrapower map computed from U i+1. It suffices to prove that j β i+1 j(c ). Since C U i+1,βi+1, we have j β i+1 j(c). Suppose that X j(b) satisfies X j(α i ) j β i+1 ; we shall find X j(b j β i+1 ) such that X j(α i ) = X j(α i ). Since B P κi β i, κ i < κ i+1 = crit(j) and X j(b), it follows that X j(α i ) = j(t) = j t for some t P κi α i. By elementarity there is x B such that x α i = t. Now j(x ) j(b), j(x ) = j x j β i+1, and j(x ) j(α i ) = j(t) = X j(α i ), so X = j(x ) is as required The poset Q. The poset Q is a diagonal extender-based forcing of the general type introduced by Giti and Magidor [4]. For the experts, we note the main innovations: The extender used here on level n is formed from the approximations U n,α to the supercompactness measure U n, rather than from measures approximating some short extender. The supports of the components of Q are quite large. This is necessary because the proofs below require long diagonalisations, which are only possible because the coordinates of the extender on level n are linearly ordered and all the measures U n,α are normal. Although the supports of the components are large (potentially much bigger than the completeness of the relevant measures) we can avoid some of the complexities of the original extender based forcing [5]. In that context the projections between measures do not commute everywhere, and this complicates matters in several respects: in our context when α < α the natural projection from U n,α to U n,α is just restriction x x α, so that the measures U n,α and the natural projections form a linear Rudin-Keisler directed system with everywhere commutative projection maps. The poset Q is designed so that (Lemma 13) every subset x of κ in the generic extension V [G] lies in some intermediate extension by an explicitly described forcing

4 4 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA poset of size less than λ. This guarantees that λ is preserved by Q (Lemma 14), V [G] and maes possible the analysis of HODx (Lemma 15). A note on notation: The definitions of the poset Q and its ordering contain quite a number of clauses. We will write for example b q as shorthand for Clause b in the definition of q Q and 6 q,r as shorthand for Clause 6 in the definition of q r. Conditions in Q are sequences (q ) <ω such that for some m < ω we have q = f q for < m and q = (a q, Aq, f q ) for m, where: a) f q is a function with dom(f q ) [κ, λ), dom(f q ) < λ, and f q (η) P κ η for all < ω and all η dom(f q ). b) For m we have a q [κ, λ), aq < λ, aq and dom(f q ) are disjoint sets and a q has a maximum element αq. c) For m we have A q U,α q. d) (a q ) m <ω is -increasing with. We call the integer m the length of q and write m = l(q). Remar 1. We note that there is very little interaction among the components q of the condition q: it is only d q that connects entries in q on different levels. Remar 2. Note that by b q and d q, if l(q) < l then α q aq l and α q αq l. Given r and q in Q, r q if and only if: (1) l(r) l(q). (2) For all, f r f q (that is, dom(f r ) dom(f r ) and f r dom(f q ) = f q ). (3) For with l(q) < l(r), a q dom(f r ), f r (αq ) Aq, and f r (η) = f r (αq ) η for all η aq. (4) (f r (αq )) l(q) <l(r) is -increasing. (5) For l(r), we have a q ar, and x αq Aq for all x Ar. (6) For l(r), if l(q) < l(r), then f r l(r) 1 (αq l(r) 1 ) x for all x Ar. Remar 3. Note that if r q then α r αq for all l(r). Lemma 3. The ordering on Q is transitive. Proof. Let r q p be conditions in Q. Note that 1 r,p, 2 r,p and 5 r,p are immediate. To verify 3 r,p, let l(p) < l(r) and distinguish two cases: l(p) < l(q): In this case f r(αp ) = f q (αp ) by 2 r,q and f q (αp ) Ap by 3 q,p, so f r(αp ) Ap. Moreover if η ap then f q (η) = f q (αp ) η by 3 q,p for q p. Also f r(η) = f q (η) and f r(αp ) = f q (αp ) by 2 r,q, so f r(η) = f r(αp ) η. l(q) < l(r): We have α p aq by 5 q,p, so f r(αp ) = f r(αq ) αp by 3 r,q. Now f r(αq ) Aq by 3 r,q, and so f r(αq ) αp Ap by 5 q,p, hence f r(αp ) Ap. Since we will use this information again, we summarise it: (*) f r(αp ) = f r(αq ) αp whenever l(q) < l(r). Moreover if η a p then η, αp aq by 5 q,p, f r (η) = f r (αq ) η and f r (αp ) = f r (αq ) αp by 3 r,q, so f r (η) = f r (αp ) η. To verify 4 r,p, first note that (f q (αp )) l(p) <l(q) is -increasing with by 4 q,p, and also that (f r (αq )) l(q) <l(r) is -increasing by 4 r,q. We now distinguish three cases:

5 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 5 For l(p) < l(q), by 2 r,q, we have f q (αp ) = f r(αp ) so that the sequence (f r(αp )) l(p) <l(q) is -increasing. For l(q) 0 < 1 < l(r), we have that f r 0 (α q 0 ) f r 1 (α q 1 ) and also (by Remar 2) that α p 0 α p 1, so using (*) we get f q 0 (α p 0 ) = f r 0 (α q 0 ) α p 0 f r 1 (α q 1 ) α p 1 = f q 1 (α p 1 ). It follows that (f q (αp )) l(q) <l(r) is -increasing. For l(p) < l(q) < l(r), by 3 r,q we have fl(q) r (αq l(q) ) Aq l(q), so by 6 q,p we have f q l(q) 1 (αp l(q) 1 ) f l(q) r (αq l(q) ) αp l(q) 1. By (*) with = l(q) we have fl(q) r (αp l(q) ) = f l(q) r (αq l(q) ) αp l(q), and we also have f q l(q) 1 (αp l(q) 1 ) = fl(q) 1 r (αp l(q) 1 ) by 2 r,q. Since α p l(q) 1 αp l(q) by Remar 2, we see that f r l(q) 1 (αp l(q) 1 ) = f q l(q) 1 (αp l(q) 1 ) f r l(q) (αq l(q) ) αp l(q) 1 f r l(q) (αq l(q) ) αp l(q) = f r l(q) (αp l(q) ). We thus conclude that (f r(αp )) l(p) <l(r) is -increasing. Finally, to verify 6 r,p, suppose that l(p) < l(r) and let x A r. We distinguish two cases: l(p) < l(q) = l(r). In this case, x α q Aq by 5 r,q. By 6 q,p and the observation that α p l(r) 1 αp l(r) αq l(r) αq, x αp l(r) 1 f q l(r) 1 (αp l(r) 1 ). Finally f q l(r) 1 (αp l(r) 1 ) = f l(r) 1 r (αp l(r) 1 ) by 2 r,q, and so fl(r) 1 r (αp l(r) 1 ) x α p l(r) 1. l(p) l(q) < l(r). In this case, by 6 r,q, fl(r) 1 r (αq l(r) 1 ) x αq l(r) 1. The ordinals α p l(r) 1 and αq l(r) 1 lie in aq l(r) 1 by 5 q,p. Now fl(r) 1 r (αp l(r) 1 ) = fl(r) 1 r (αq lh(r) 1 ) αp l(r) 1 by 3 r,q, and so fl(r) 1 r (αp l(r) 1 ) x αq l(r) 1 α p l(r) 1 = x αp r 1, using the fact that αq r 1 αp r 1 by Remar 3. This completes the proof. It is easy to see that for each m < ω and each α with κ α < λ, the set of conditions q with l(q) > m is dense, as is the set of q with α dom(f q m). It follows that we may view Q as adding a matrix of sets (z m,α ) m<ω,κ α<λ such that z m,α P κm α for all m and α. Lemma 4. The forcing poset Q collapses all cardinals μ with κ < μ < λ and preserves all cardinals μ with μ > λ. Proof. It is easy to see that Q = λ, so that Q trivially has λ + -cc and preserves cardinals greater than λ. Given μ with κ < μ < λ, we will do a density argument to show that μ is collapsed. Let p Q arbitrary, and extend p to obtain q so that there is α μ with α l(q) <ω aq. We claim that q μ m z m,α; to see this let β < μ be arbitrary, choose y A q with β y, and extend q to a condition r such that α q l(q) l(r) = l(q) + 1 and fl(q) r (αq l(q) ) = y, so that β f l(q) r (α) = y α. The argument that the cardinal λ and cardinals μ with μ κ are preserved will require a deeper analysis of the forcing poset Q, which will be given in subsection 3.4. The intuition is that the ω-sequences (z m,α ) m<ω for differing values of α are related, so that no unwanted information can be computed by comparing them.

6 6 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA 3.3. Basic properties of Q. Let us consider a secondary ordering on Q. Let r * q (r is a direct extension of q) if and only if r q and l(r) = l(q). Lemma 5. Let q = (q i ) i<μ be a * -decreasing sequence of conditions where μ < κ l(q0). Then the sequence q has a * -lower bound. Proof. We define a lower bound r with l(r) = l(q 0 ) as follows: For all < ω, let f r = qi i<μ f. Choose some ordinal β < λ such that i<μ (aqi qi dom(f )) β for all l(r), and define a r = i<μ aqi {β} for all such, so that αr = β. For l(r), set A r = {x P κ β : i < μ x α qi Aqi }. It is routine to chec that r is a condition with r * q i for all i. Note that Clauses 3, 4 and 6 from the definition of the ordering are irrelevant here. We will also need a stronger version of *. For i < ω, write r * i q if and only if r * q and in addition (a r, Ar ) = (aq, Aq ) whenever l(q) < l(q) + i. Note that * = * 0 and * i is transitive for all i. Lemma 6 (Fusion). Let (q i ) i<ω be a sequence of conditions such that q i+1 * i q i for all i < ω. Then there exists a condition q such that q * i q i for all i. Proof. Define q with l(q ) = l(q 0 ) as follows: f q = i<ω for all < ω; f qi (a q, Aq ) = (aq +1, A q +1 ) for l(q 0 ). It is routine to verify that q is a condition and q * i q i for all i. Remar 4. Of course Lemma 5 already implies that the decreasing sequence from Lemma 6 has a * -lower bound, the point here is that q * i q i rather than just q * q i. Definition 1. For conditions r q, we let stem(r, q) denote the finite sequence (f r i (αq i )) l(q) i<l(r). Note that by 3 r,q and 4 r,q, stem(r, q) l(q) i<l(r) Aq i, and is -increasing. Definition 2. Let q be a condition, and let s l(q) i<l Aq i be a -increasing sequence for some l (l(q), ω). We define q + s as the ω-sequence (r ) <ω such that: For < l(q), r = f q. For l(q) < l, r is the function with domain dom(f q ) aq such that r (η) = f q (η) for η dom(f q ) and r (η) = s η for η a q. For l, r = (f q, aq, B ) where B = {x A q : s l 1 x}. By convention we also define q + = q. We shall soon establish that q + s is a condition, but let us first point out that q + s may be obtained by adding in one entry of s at a time. Lemma 7. Let q and s = (s i ) l(q) i<l be as in Definition 2. Let n = l l(q). Define a decreasing sequence of conditions (r i ) i n as follows: r 0 = q, and then r i+1 = r i + s i for i < n. Then r n = q + s. Proof. The proof is straightforward by induction on the length of s.

7 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 7 Lemma 8. Let q and s be as in Definition 2. Then q + s is a condition extending q. Moreover r * q + stem(r, q) for all r q. Proof. By definition l(q + s) = l(q) + l(s), so 1 q+s,q holds. Clearly f q+s f q for all, so 2 q+s,q holds. By construction f q+s (α q ) = s for l(q) < l(q) + l(s), so f q+s (α q ) Aq by our assumptions on s. Also by definition f q+s (η) = s η = f q+s (α q ) η for all such, so that 3 q+s,q holds. The sequence (f q+s (α q )) l(q) <l(q+s) is equal to s, which is -increasing by our assumptions on s, so 4 q+s,q holds. We have a q+s = a q and Aq+s A q for all l(q + s), so 5 q+s,q holds. Finally we have by definition that A q+s = {x A q : s l(q+s) 1 x} for l(q + s), and f q+s l(q+s) 1 (αq l(q+s) 1 ) = s l(q+s) 1, so 6 q+s,q holds. Now suppose that r q and s = stem(r, q). If l(r) = l(q) then s = and r * q = q+s, so we may as well assume that l(r) > l(q). We have l(r) = l(q+s), so 1 r,q+s holds. By the definition of q+s and 3 r,q, f r q+s f whenever l(q) < l(r). For l(r), f q+s = f q and f r f q by 2 r,q, so 2 r,q+s holds. Since l(r) = l(q + s), 3 r,q+s, 4 r,q+s and 6 r,q+s hold vacuously. For l(q + s) we have a q+s = a q by definition, and also aq ar by 5 r,q. Finally for all l(r) = l(q + s) and all x A r, x αq+s = x α q Aq, and additionally s l(q+s) 1 x α q l(q+s) 1 ; it follows from the definition that x αq l(q+s) Aq+s l(q+s). Lemma 9. Let q * n p and let r q with l(r) = l(q) + n. Then stem(r, q) = stem(r, p), and q + stem(r, q) * p + stem(r, p). Proof. Let s = stem(r, q), that is s = f r(αq ) for l(q) < l(r). Since q * n p we have that α q = αp for such, and so s = f r(αp ) and hence stem(r, p) = s = stem(r, q). If n = 0 then q = q + s * p = p + s, so we may assume that n > 0. Now l(q + s) = l(r) = l(p + s), so Clauses 1, 3, 4 and 6 are easily seen to hold. For < l(p) or l(r), we have f q+s = f q f p p+s = f by the definitions and 2 q,p. For l(p) < l(r), we have that f q+s f p+s because f q f p by 2 q,p, and f q+s (η) = s η = f p+s (η) for every η a q = ap. For l(r) we have a p+s = a p aq = aq+s and x A q = x αp Ap by 5 q,p. If x A q+s then x A q and x s l(r) 1, so x α p Ap and also x α p s l(r) 1 using the fact that α p αp l(r) Main technical lemma. If τ is a name for an object in the ground model and r is a condition, then we say that r decides τ if and only if there is x V such that r τ = ˇx. We write r τ for r decides τ. We now state and prove the ey technical lemma about Q. Lemma 10. Let p Q and let τ be a name for an element of V. Then there is a direct extension q * p such that for every r q with r τ, we have q +stem(r, q) τ. Proof. We will build an ω-sequence of conditions (p n ) n<ω such that: p 0 = p. p n+1 * n p n for all n. For every r p n+1 with l(r) = l(p)+n, if r τ, then p n+1 +stem(r, p n+1 ) τ.

8 8 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA Before giving the construction we verify that building (p n ) n<ω is sufficient. Appealing to Lemma 6, we find p such that p * n p n for all n. Let r p with r τ and l(r) = l(p) + n. Then p * n p n+1, and so p + stem(r, p ) * p n+1 +stem(r, p n+1 ) by Lemma 9. By construction p n+1 +stem(r, p n+1 ) τ, and so p + stem(r, p ) τ. It follows that p will serve as a witness to the conclusion. To begin the construction we set p 0 = p, and then as whether there is a direct extension of p which decides τ, and set p 1 equal to such an extension if one exists, otherwise p 1 = p. If there is r * p 1 with r τ then r * p, and p 1 = p 1 + stem(r, p 1 ) τ. Suppose now that we have constructed p n for some n > 0. Let (s n j ) j<μ be an enumeration of all sequences s such that l(s) = n and p n + s is well defined, that is to say all -increasing sequences in l(p) <l(p)+n Apn. Since λ is inaccessible, μ < λ. Note to the reader. The objects f j, aj, αj, Aj built during the construction also depend on n, but we have suppressed the index n to lighten the notation. We will construct by recursion for each j < μ: Functions f j for < ω. Sets a j and Aj for l(p) + n < ω. We will maintain the following hypotheses: a pn l(p)+n 1 a0 l(p)+n. For all and all j < j, f j f j and aj aj. For all and j, a j [κ, λ) with aj < λ. For all j, the sequence (a j ) l(p)+n is -increasing with. a j has a maximum element αj and Aj U,α j. For all with l(p) < l(p) + n and all j, dom(f j ) is disjoint from apn. For all with l(p) + n and all j, dom(f j ) is disjoint from aj. To begin the construction we set f 0 pn = f for all, and (a 0, A0 ) = (apn, Apn ) for l(p) + n. Suppose that we have constructed f j, aj and Aj. We define various auxiliary conditions in Q: p j is the condition such that p j = f j for < l(p), pj = (apn l(p) < l(p) + n, and p j = (aj, P κ α j, f j ) for l(p) + n., Apn, f j ) for q j = p j + s n j. If there is r * q j such that r τ, then r j is some such condition r, otherwise r j = q j. We now define: f j+1 = f rj f j+1 (a j+1 for < l(p) or l(p) + n. = f rj rj dom(f ) apn for l(p) < l(p) + n., A j+1 ) = (a rj ) for l(p) + n., Arj For j limit we proceed roughly as in the proof of Lemma 5, with the important caveat that we do not aim to mae the sets A j decrease with j: f j = i<j f i for all. For l(p) + n:

9 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 9 We choose a j as in the final stage of the proof of Lemma 5, that is to say we choose some large enough β j and set a j = i<j ai {βj } for all, so that α j = βj. We set A j = P κ β j. After defining f j, aj and Aj for j < μ, we are ready to define p n+1. f pn+1 = i<μ f i (a pn+1, A pn+1 For l(p) + n: for all. ) = (a pn ) for l(p) < l(p) + n., Apn As at the limit stages below μ, we choose some large enough β * and set a pn+1 = i<μ ai {β* }, so that α pn+1 = β * for all. We define A pn+1 by diagonal intersection, as the set of those x P κ β * such that x α pn A i+1. Note that {x P κ β * : x α i+1 this is the projection of U,β * A pn and, for all i < μ if s n i x then x α i+1 A i+1 } U,β *, because A i+1 U,α i+1. So A pn+1 under the map x x α i+1 U,β * by Lemma 1. It is routine to verify that p n+1 is a condition and that p n+1 * n p n. To finish the proof we must verify that for every r p n+1 such that l(r) = l(p) + n and r τ, p n+1 + stem(r, p n+1 ) τ. Let s = stem(r, p n+1 ), and pic some j < μ such that s = s n j. We claim that p n+1 * n p j. This is a routine verification: note that A pj = P κ α j for l(p) + n, so that there is no problem with Clause 5 for p n+1 p j. By Lemma 8 we have r * p n+1 +s, and by Lemma 9 we have that stem(r, p j ) = s and p n+1 + s * p j + s = q j. So r * q j and r τ, hence we chose r j * q j such that r j τ and used r j in the definitions of f j+1, a j+1, and A j+1. We claim that p n+1 + s r j. The verification is fairly straightforward, the main point is to chec the second part of Clause 5. So let l(p) + n, and recall that by definition A pn+1+s is the set of x A pn+1 such that s x. By the construction of A pn+1 as a diagonal intersection, for every x A pn+1 such that s x we have that x α j+1 A j+1. Since A j+1 = A rj, this is exactly what is needed for Clause Priry lemma. Lemma 11 (Priry lemma). For every condition p and every sentence φ in the forcing language there is q * p such that q φ. Proof. Let τ name an ordinal which is 0 if φ is true and 1 if φ is false. Appealing to Lemma 10 we find q 0 * p which is such that for all r q 0, if r decides φ then q 0 + stem(r, q) decides φ. Let S be the set of sequences s such that q 0 + s is well-defined, and for each s S define F (s) as follows: 0, if q 0 + s φ F (s) = 1, if q 0 + s φ 2, otherwise. and

10 10 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA For each t S we may find a measure one set B t A q0 l(q 0)+l(t) and G(t) such that F (t x ) = G(t) for all x B t. Now for each l(q 0 ), let C be the diagonal intersection of B t for all sequences t of length l(q 0 ), that is the set of x A q0 such that x B t for all t x. Let q 1 be obtained from q 0 by replacing A q0 by C for all l(q 0 ), and note that α q1 = αq0 for all l(q 0) = l(q 1 ). Let q 2 q 1 be a condition deciding φ with l(q 2 ) chosen minimal. We claim that l(q 2 ) = l(q 1 ), so that q 2 * p and we are done. Suppose for a contradiction that l(q 2 ) > l(q 1 ), and let s = stem(q 2, q 1 ). By construction q 1 +s * q 0 + s and q 0 + s decides φ, so q 1 + s decides φ. We will assume that it forces φ; the argument for the case when q 1 + s forces φ is identical. Let x = s(l(s) 1) and let t = s l(s) 1, so that s = t x. By construction x B t, and so by definition G(t) = F (s) = 1. By Lemma 8 it is easy to see that every extension of q 1 + t is compatible with q 1 + t y for some y B t, and since G(t) = 1 we have that q 1 + t y φ for all y B t, so that q 1 + t forces φ. This contradicts the minimal choice of l(q 2 ). Corollary 1. The forcing poset Q adds no bounded subsets of κ. Proof. Let p x μ with μ < κ, and find q p such that l(q) = for some large enough that κ > μ. Appealing to Lemmas 11 and 5 we may build a * -decreasing sequence (q i ) i<μ such that q 0 = q and q i+1 decides whether i x for all i. By Lemma 5 again there is a condition r such that r q i for all i, and r x = ˇy for some y V Projected forcing. Let < ω and let α = (α i ) i<ω be a -increasing sequence from [κ, λ). We define a forcing poset Q α. Conditions in Q α are sequences (q i ) i<ω such that some l : q i P κi α i for i < l and q i U i,αi for i l, and (q i ) i<l is -increasing. We say that l is the length of q and write l = l(q). If r, q Q α then r q if and only if: l(r) l(q). r i = q i for i < l(q). r i A q i for l(q) i < l(r). A r i Aq i for l(r) i < ω. We note that Q α has the λ-cc, because there are fewer than λ possible stems (q i ) i<l(q) and any two conditions with the same stem are compatible. It is also not hard to prove that Q α obeys a version of the Priry Lemma, but we will not need this (it actually follows from the next lemma). It is useful to note that the generic object added by Q α is a -increasing sequence (z i ) i<ω such that z i P κi α i. Lemma 12. Let q Q, and let α = (α i ) l(q) i<ω be a -increasing sequence such that α i a q i for all i. Let π : Q q Q α be the map defined by: π(r) i = fi r(α i) for l(q) i < l(r). π(r) i = {x α i : x A r i } for i l(r). Then π is order-preserving and has the following property: for every r q there is r * r such that for all q 0 π(r * ) there is r r with π(r ) q 0. Proof. It is routine to chec that π is order-preserving. To verify the rest of the conclusion let r q. We obtain a condition r * * r by shrining measure one sets in r as follows:

11 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 11 l(r * ) = l(r). fi r* = fi r for all i. a r* i = a r i for all i l(r). We define A r* for l(r) by recursion on, arranging that Ar* Ar and A r* U,α r. We start by setting A r* l(r) = Ar l(r). For i > 0 we set Ar* l(r)+i to be the set of y A r l(r)+i such that for all x Ar* l(r)+i 1 with x α i y α i+1 there is x A r* l(r)+i 1 with x α i = x α i and x y. By Lemma 2 we have A r* l(r)+i U l(r)+i,β i+1. Let q 0 π(r * ), and choose y B r* such that (q0 ) = y α for l(r) < l(q 0 ). Using the definition of B r* we will choose y l(q 0) i for 0 < i l(r) by induction on i as follows: y l(q = y 0) 1 l(q 0) 1, and y l(q 0) (i+1) Br* l(q with 0) (i+1) y l(q y 0) (i+1) l(q 0) i and y l(q α 0) (i+1) l(q 0) (i+1) = y l(q0) (i+1) α l(q0) (i+1) = (q 0 ) l(q0) (i+1). Let t = (y ) l(r) <l(q 0), and form r + t. By shrining measure one sets appropriately we obtain a condition r r + t r such that π(r ) * q 0. The conclusion of Lemma 12 is a weaening of the standard property of being a projection between forcing posets, and was first isolated by Foreman and Woodin [1]. The following corollary of Lemma 12 is routine: Corollary 2. With the same hypotheses as in Lemma 12, if G is Q-generic with q G then π[g] generates a Q α -generic filter. Remar 5. It is possible to give an alternative argument for Corollary 2 by first proving a characterisation of Q α -generic sequences in the style of Mathias theorem on Priry forcing [8]. This characterisation, which we will not prove, states that an increasing sequence (z i ) i<ω is generic if and only if for every sequence (A i ) i<ω in V with A i U i,αi for all i, we have z i A i for all large i. It is straightforward to verify that the induced filter G α corresponds to a sequence (z i ) i<ω with this property. Lemma 13. Let G be Q-generic and let x κ with x V [G]. Then there exists α such that x V [G α ]. Proof. Note that by Corollary 1 we have x κ n V for all n < ω. Let p Q and let τ n be a name for x κ n. Appealing to Lemmas 10 and 5 we may find q * p such that for all r q, if r decides τ n then then q + stem(r, q) decides τ n. Now let α = (α q ) l(q) <ω. It is routine to verify that q forces that x can be computed from G α. Explicitly, if q G and (z i ) i<ω is the generic sequence added by G α then β x q + (z i ) i< β x. Lemma 14. Let G be Q-generic, then λ = (κ + ) V [G]. Proof. Suppose for a contradiction that λ is not a cardinal in V [G]. By Lemma 4, cardinals μ with κ < μ < λ are all collapsed in V [G], so that if λ is also collapsed then there is a set x κ in V [G] coding a well-ordering of κ with order type λ. By Lemma 13 x V [G α ] for some α, which gives an immediate contradiction since Q α is λ-cc.

12 12 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA 3.7. Definability. Let G be Q-generic. The last step in the proof of our main result is to analyse the class of sets HOD x where x κ with x V [G]. This will require a rather technical result which appears below as Lemma 16. In order to motivate the statement of Lemma 16, we will state a lemma which contains the necessary information about HOD x and prove it modulo one missing fact, which will then be provided by Lemma 16. Lemma 15. Let x κ with x V [G], then there exists α such that (HOD x ) V [G] V [G α ]. In particular (κ + ) (HODx)V [G] < κ +. Proof of Lemma 15, Part One. By Lemma 13 find q G, α with α a q for all V [G] and x a Q α -name such that i G α ( x) = x. Since HODx is a model of ZFC, to V [G] show that HODx V [G α ] it will be sufficient to show that every set of ordinals V [G] y HODx lies in V [G α ]. Let y be such a set and fix a definition of y from x and some ordinal parameter δ, say Y = {γ : V [G] = ψ(γ, δ, x)}. In order to define y in V [G α ], we define an auxiliary set H Q. Let = l(q) and let (z i ) i<ω be the generic sequence corresponding to the filter G α. We define H to be the set of conditions r Q such that r q, and for every there exists r 1 r such that f r1 i (α i ) = z i for i <. Clearly G H and H V [G α ]. We claim that Y = {γ : r H r ψ(γ, δ, x)}. If Y is the set defined on the right hand side of this equation then clearly Y Y and Y V [G α ], so we need only show that Y Y. Suppose for a contradiction that this is not the case, fix γ Y Y, and then choose conditions r H and s G such that r ψ(γ, δ, x) and s ψ(γ, δ, x). We may extend s to find s * s, q such that s * G and l(s * ) l(r), and then use the fact that r H to find r * r (not necessarily lying in H) such that l(r * ) = l(s * ) and fi r* (α i ) = z i = fi s* (α i ) for all i with l(q) = i < l(r * ) = l(s * ). At this point we will be done if we can show that r * and s * can not force contradictory information about ψ(γ, δ, x), and this is exactly the point of Lemma 16. Accordingly we interrupt the proof of Lemma 15 to state and prove Lemma 16. Lemma 16. Let q Q, and let α = (α i ) l(q) i<ω be a -increasing sequence such that α i a q i for all i. Let r* and s * be conditions such that r *, s * q, l(r * ) = l(s * ), and f r* (αq s* ) = f (αq ) for l(q) < l(r* ). Then there exist conditions r * r * and s * s * and ρ : Q r Q s such that: ρ is an isomorphism. If G 0 is Q-generic with r G 0, and G 1 is the Q-generic filter generated by ρ[g 0 Q r ], then G 0, α = G 1, α. Proof. We choose direct extensions r * r and s * s such that: dom(f r ) = dom(f ) for < l(r). dom(f r ) ar = dom(f ) a, αr = α, Ar = A for l(r). To help readability let β be the common value of α r and α, let B be the common value of A r and A, and let d be the common value of dom(f r ) and dom(f r ) for < l(r) and the common value of dom(f ) ar and dom(f ) a for l(r).

13 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 13 We will now define maps ρ : Q r Q s and σ : Q s Q r with ρ(r ) = s, and argue that they are order-preserving and mutually inverse. The reader may find the following slogan useful: ρ alters r on d to mae it loo lie the -level of an extension s of s with stem(s, s ) = stem(r, r ). Let r r. We define ρ(r ) to be the condition s such that: For < l(r ), dom(f dom(f ) dom(f ) = f ) = dom(f ), f dom(f For l(r ) < l(r ), dom(f f (η) = f (β ) η for η a For l(r ) : dom(f ) = dom(f f a A (dom(f = a (a = A., f dom(f ). ) dom(f ) = dom(f ), f dom(f ) (dom(f ) d ) = f (dom(f d ). ) d ), f ) d ). ) d = f ) = f, f dom(f dom(f ) = f dom(f ) = f, ) d. σ(s ) for s s is defined in an exactly similar way with the roles of r, r and s, s reversed. Note in particular that: For l(r ) < l(r ), β = α r = α ar a, and f α( ) (β ) = f (β ). The definitions of ρ and σ are level by level in the sense that ρ(r ) (resp σ(s ) ) depends only on r (resp ). On each level of r (resp s ) ρ (resp σ) only alters a and f ) inside the set d. and and f (resp a It is very easy to verify that ρ(r ) = s. We claim that ρ and σ are orderpreserving, by symmetry we only need verify this for ρ. From Lemmas 8 and 7 we see that it is enough to verify that ρ is order-preserving for extensions of the form r + x r and r * r. Claim Let r r and let x be such that r + x is defined. then ρ(r )+ x is well-defined and ρ(r + x ) = ρ(r ) + x. Proof. We note that A ρ( ) = A for all l(r ), so that ρ(r ) + x is welldefined. Now we do a case analysis: < l(r ): f ρ( ) is obtained by altering the values of f = f ρ( ) agree with f, and by definition f ρ( )+ x and then f ρ( + x ) with f, so easily f ρ( + x ) l(r ) < l(r ): Note that d on dom(f ) to = f ) to agree. Also f + x is obtained by altering its values on dom(f = f ρ( )+ x. obtained by altering the values of f on dom(f f ρ( )+ x = f ρ( ) ) and to agree with η f dom(f ). By definition, f ρ( ) on d, so as to agree with f (β ) η on a By definition. is. Also f + x = f, and f ρ( + x ) is obtained from it by the scheme of alteration described above, so f ρ( + x ) = f ρ( )+ x. = l(r ): This is the most complicated case. Start by recalling that d = dom(f r ) ar = dom(f ) a dom(f ) a = dom(f + x ), and also that r r.

14 14 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA By definition f + x (η) = f (η) for η dom(f ) and f + x (η) = x η for η dom(a ). When we alter values to obtain f ρ( + x ), we obtain: f ρ( + x ) (η) = f (η) for η dom(f ) d. f ρ( + x ) (η) = x η for η a d. f ρ( + x ) (η) = f (η) for η dom(f ). f ρ( + x ) (η) = f + x (β ) η for η a. In the last of these cases we note that r r and l(r ) =, so that β = α r a and so f + x (β ) = x β. It follows that f ρ( + x ) (η) = x η for η a, so we conclude that: f ρ( + x ) (η) = f (η) for η dom(f ) d. f ρ( + x ) (η) = x η for η a (a d ). f ρ( + x ) (η) = f (η) for η dom(f ). By definition again we have a ρ( ) = a (a d ), dom(f ρ( ) ) = dom(f ) (dom(f ) d ) and: f ρ( ) f ρ( ) (η) = f (η) = f Since f ρ( )+ x f ρ( ) that f ρ( + x ) (η) for η dom(f ) d. (η) for η dom(f ) d. and f ρ( )+ x (η) = x η for η a ρ( ), we see = f ρ( )+ x. > l(r ): By definition a + x = a and f + x definition of the operation ρ we have a ρ( + x ) = a ρ( ) f ρ( ). Also a ρ( )+ x = a ρ( ) and f ρ( )+ x = f ρ( ) follows that a ρ( )+ x = a ρ( + x ) and f ρ( )+ x have A ρ( + x ) = A + x = f, so that by the and f ρ( + x ) = f ρ( + x ) =, from which it. We also = {y A : x y} = {y A ρ( ) : x y} = A ρ( )+ x. Claim Let r * r r, then ρ(r ) * ρ(r ). Proof. For < l(r ) we have d dom(f ) f, while for lh( ) we have d dom(f ) a dom(f r ) a. The definition of ρ(r ) involves altering the values of f (for < l(r )) or f and a (for l( )) in a uniform manner inside d, using only the values of f r and (for lh(r ) < l(r )) the values of f (αr ), and similarly for ρ( ). Since f (αr ) = f (αr ), it is now routine to verify that ρ(r ) ρ(r ). Claim ρ and σ are mutually inverse. Proof. Let ρ(r ) = s. To verify that σ(s ) = r, the ey points are that r r and that for lh(r ) < l(r ) we have f (β ) = f (β ). It follows that for such values of and for all η dom(f r ) a we have f σ( ) (η) = f (β ) η = f (β ) η = f (η), which is the only difficult step in the proof that σ( ) = r. Claim If G 0 is Q-generic with r G 0, and G 1 is the Q-generic filter generated by ρ[g 0 Q r ], then G 0, α = G 1, α.

15 ORDINAL DEFINABLE SUBSETS OF SINGULAR CARDINALS 15 Proof. Let r r and recall that r, s q. we claim that f ρ( ) (α ) = f (α ) for all with l(q) < l(r ). If < l(r ) then f ρ( ) (α ) = f ) = f r ) = f ). If l(r ) < l(r ) then since α a q a and r, we have (α ) = f ) α = f r ). f ρ( ) This concludes the proof of Lemma 16. We are now ready to complete the proof of Lemma 15. Proof of Lemma 15, Part Two. Recall that we need to derive a contradiction from the assumption that there are r * and s * satisfying the hypotheses of Lemma 16 such that r * ψ(γ, δ, x) and s * ψ(γ, δ, x), where (crucially) x is a Q α -name. Let us force with Q below r * to obtain a generic object G 0 containing r *, and then use ρ to obtain a generic object G 1 such that G 1 contains s *. Since ρ V, V [G 0 ] = V [G 1 ] and since G 0, α = G 1, α, i G0 ( x) = i G1 ( x) = y say. This is an immediate contradiction. Combining the results we have obtained, we have proved the Main Theorem. Theorem. Suppose that κ < λ where cf(κ) = ω, λ is inaccessible and κ is a limit of λ-supercompact cardinals. There is a forcing poset Q such that if G is Q-generic then: The models V and V [G] have the same bounded subsets of κ. Every infinite cardinal μ with μ κ or μ λ is preserved in V [G]. λ = (κ + ) V [G]. For every x κ with x V [G], (κ + ) HODx < λ. As we mentioned in the Introduction, we can strengthen the conclusion of the Main Theorem. We start with (κ n ) n<ω and λ as before, but we now assume that the cardinals κ n and λ are supercompact. By doing a suitable preparatory class forcing we may assume in addition that for every set-generic extension W of V, V HOD W. The idea of the preparation, which is essentially due to McAloon [9], is that we will mae the κ n and λ Laver-indestructible, and then do λ-directed closed forcing above λ to arrange that every set of ordinals is coded unboundedly often into the values of the continuum function. We claim that if G is Q-generic then in V [G] we have that λ is supercompact in V [G] HOD x for all x κ. To see this find α as before such that HODx V [G α ], so that we have a chain of inclusions V HOD V [G] V [G] HODx V [G α ]. V [G] By the Intermediate Models Theorem the model HODx is a set-generic extension of V via some complete Boolean subalgebra B of r. o.(q α ). Since B < λ, it follows V [G] by the Levy-Solovay theorem that λ is supercompact in HOD x. Remar 6. Giti [3] informed us that the same effect should be possible using Merimovich s supercompact extender-based forcing [10]. This weaens the hypotheses of the Main Theorem to one supercompact cardinal with an inaccessible cardinal above it. The details then appeared in [6].

16 16 J. CUMMINGS, S.-D. FRIEDMAN, M. MAGIDOR, A. RINOT, AND D. SINAPOVA Remar 7. Woodin [13] pointed out that it should also be possible to prove the Main Theorem using methods of Kafoulis [7]. The idea is to start with a supercompact cardinal with an inaccessible cardinal above, construct a choiceless Solovay-lie model V * generated by subsets of κ arising in initial segments of a supercompact Priry extension, and then force over V * to restore choice. References [1] M. Foreman and W. H. Woodin, The Generalized Continuum Hypothesis can fail everywhere, Annals of Mathematics 133 (1991), [2] M. Giti, Priry-type forcings, in Handboo of set theory (ed. M. Foreman and A. Kanamori), , Springer, Dordrecht, [3] M. Giti, private communication, February [4] M. Giti and M. Magidor, Extender based forcings, Journal of Symbolic Logic 59 (1994), [5] M. Giti and M. Magidor, The singular cardinal hypothesis revisited, in Set theory of the continuum, , Springer, New Yor, [6] M. Giti and C. Merimovich, Some applications of supercompact extender based forcings to HOD, arxiv: , August [7] G. Kafoulis, The consistency strength of an infinitary Ramsey property, Journal of Symbolic Logic 59 (1994), [8] A. Mathias, On sequences generic in the sense of Priry, Journal of the Australian Mathematical Society 15 (1973), [9] K. McAloon, Consistency results about ordinal definability, Annals of Mathematical Logic 2 (1971), [10] C. Merimovich, Supercompact extender based Priry forcing, Archive for Mathematical Logic 50 (2011), [11] J. Silver, On the singular cardinals problem, Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 1 (1975), [12] S. Shelah, Set Theory without choice: not everything on cofinality is possible, Archive for Mathematical Logic 36 (1997), [13] W. H. Woodin, private communication, February Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh PA , USA address: jcumming@andrew.cmu.edu Kurt Gödel Research Center for Mathematical Logic, University of Vienna, A-1090 Vienna, Austria address: sdf@logic.univie.ac.at Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Jerusalem, 91904, Israel address: mensara@savion.huji.ac.il Department of Mathematics, Bar-Ilan University, Ramat-Gan, , Israel address: rinotas@math.biu.ac.il Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago, Chicago, IL , USA address: sinapova@math.uic.edu