FORCING AND THE HALPERN-LÄUCHLI THEOREM. 1. Introduction This document is a continuation of [1]. It is intended to be part of a larger paper.

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1 FORCING AND THE HALPERN-LÄUCHLI THEOREM NATASHA DOBRINEN AND DAN HATHAWAY Abstract. We will show the various effects that forcing has on the Halpern-Läuchli Theorem. We will show that the the theorem at an inaccessible κ is preserved by forcings of size < κ and, assuming κ is measurable, by <κ-closed forcings. 1. Introduction This document is a continuation of [1]. It is intended to be part of a larger paper. 2. Basic Definitions For the even more basic definitions, see [1]. We review here the fundamental definitions. Although some of these definitions depend on the cardinal κ, in practice there will be no confusion. Definition 2.1. A tree T <κ κ is regular iff it is a perfect κ-tree in which every maximal branch has length κ. Definition 2.2. Given a set X <κ κ and an ordinal ζ < κ, X(ζ) := {t X : Dom(t) = ζ}. We also write Length(t) for Dom(t). Definition 2.3. Given sets X, Y <κ κ, we say that X dominates Y iff ( y Y )( x X) x y. Definition 2.4. Given t <κ κ, Cone(t) is the set of all t t in <κ κ. Definition 2.5. Given d ω and a sequence X i <κ κ : i < d, define the level product of the X i to be X i := { x i : i < d : ( ζ < κ)( i < d) x i X i (ζ)}. i<d The following is the somewhere dense version of the Halpern-Läuchli Theorem, which we shall denote by SDHL(d, σ, κ). Definition 2.6. For d ω and cardinals 0 < σ < κ with κ infinite, SDHL(d, σ, κ) is the statement that given any sequence T i <κ κ : i < 1

2 2 NATASHA DOBRINEN AND DAN HATHAWAY d of regular trees and any coloring c : i<d T i σ, there exits ζ < ζ < κ, t i T i (ζ) : i < d, and X i T i (ζ ) : i < d such that each X i dominates T i (ζ + 1) Cone(t i ) and c i<d X i = 1. The following is the strong tree version of the Halpern-Läuchli Theorem, which we shall denote by HL(d, σ, κ). By [1], HL(d, σ, κ) is equivlanet to SDHL(d, σ, κ). Definition 2.7. For d ω and cardinals σ < κ with κ infinite, HL(d, σ, κ) is the following statement: given any sequence T i <κ κ : i < d of regular trees and a coloring c : i<d T i σ, there exists a sequence of trees T i : i < d such that (1) each T i is a strong subtree of T i as witnessed by the same set A κ independent of i, and (2) there is some σ < σ such that for each ζ A, c i<d T i (ζ) = {σ }. Finally, the following is the tail cone version of the Halpern-Läuchli Theorem, which we shall denote by HL tc (d, < κ, κ). Definition 2.8. For d ω and κ be an infinite cardinal, HL tc (d, < κ, κ) is the following statement: given a sequence of regular trees T i <κ κ : i < d, a sequence of cardinals σ j < κ : j < κ, and a sequence of colorings c j : i<d T i σ j : j < κ, there exists a sequence of strong subtrees T i : i < d such that (1) each T i is a strong subtree of T i as witnesses by the same set A κ independent of i, where {a j : j < κ} is an increasing enumeration of A, and (2) for each j < κ and ζ j, given any sequence t i T i (a ζ ) : i < d, we have c j ( t i : i < d ) = c j ( t i a j : i < d ). Proposition 2.9. For d ω and κ an inaccessible cardinal, HL tc (d, < κ, κ) is equivalent to the modification where in (2), for each j < κ is repaced by for κ many j < κ. Proof. Assume that the modified version of HL tc (d, < κ, κ) holds. We will show that the unmodified version holds. Let T i <κ κ : i < d be a sequence of regular trees. Let σ j < κ : j < κ be a sequence

3 FORCING AND THE HALPERN-LÄUCHLI THEOREM 3 of cardinals. Let c j : j < κ be a sequence of colorings such that c j : i<d T i σ j for each j < κ. For each j < κ, let σ j be the product of the cardinals σ l for l < j. Because κ is inaccessible, each σ j is strictly less than κ. For each j < κ, let c j : i<d T i σ j be a coloring which encodes the colorings c l for l < j. That is, given j < κ and t i<d T i, the sequence c l ( t) : l < j can be recovered from knowing c j ( t). Apply the modified version of HL tc (d, < κ, κ) to get a sequence of trees T i : i < d, where each T i is a strong subtree of T i as witnessed by the same set A = {a j : j < κ} independent of i, and let B κ be the size κ set such that for each j B and ζ j, given any sequence t i T i (a ζ ) : i < d, we have c j( t i : i < d ) = c j( t i a j : i < d ). Temporarily fix j B and l < j. Then because c j encodes c l, we have that given any ζ j and any sequence t i T i (a ζ ) : i < d, we have c l ( t i : i < d ) = c l ( t i a j : i < d ). Now shrink each T i to get some T i, where T i is a strong subtree of T i as witnessed by the set {a j : j B} κ. One can check that the trees T i for i < d witness the conclusion of HL tc (d, < κ, κ). 3. Derived Trees Definition 3.1. Let κ be a cardinal, P a forcing, and T a name such that 1 ( T <κ κ is a tree). Then Der( T ), the derived tree of T, is defined as follows. The elements of Der( T ) are equivalence classes of pairs ( τ, α) satisfying 1 ( τ T and Length( τ) = ˇα), where the equivalence relation = is defined by ( τ 1, α 1 ) = ( τ 2, α 2 ) iff 1 ( τ 1 = τ 2 ). The elements of Der( T ) are ordered as follows: [( τ 1, α 1 )] < [( τ 2, α 2 )] iff 1 ( τ 1 τ 2 and τ 1 τ 2 ). We will now show that Der( T ) is a regular tree when 1 ( T is a regular tree), and that all successors of an element named by some ( τ, α) in Der( T ) are named by successors of that ( τ, α). Lemma 3.2. Let κ be strongly inaccessible, P a forcing of size < κ, and T a name for a regular tree. The following are satisfied. 1) Der( T ) is a κ-tree; 2) Der( T ) has no maximal branches of length < κ; 3) Der( T ) is perfect;

4 4 NATASHA DOBRINEN AND DAN HATHAWAY 4) if [( τ, α)] Der( T ) and X is the set of all τ such that [( τ, α + 1)] is a successor of [( τ, α)] in Der( T ), then 1 (every successor of τ in T is named by an element of ˇX). Proof. First note that if [( τ, α)] is in Der( T ) and β < α, then there is a name τ β such that [( τ β, β)] is in Der( T ) and is < [( τ, α)]: simply let τ be a name for the restriction of τ to level β. To verify 1), we must first show that Der( T ) is a tree. Suppose [( τ 1, α)] > [( τ 2, β)], [( τ 3, γ)] are all in Der( T ). Suppose, without loss of generality, that β γ. It is not hard to show that 1 ( τ 2 τ 3 ), because 1 (all initial segments of τ 1 are comparable with respect to ). Thus, [( τ 2, β)] and [( τ 3, γ)] are comparable in Der( T ). This also shows that given [( τ, α)] in Der( T ) and β < α, there is a unique [( τ, β)] on level β of Der( T ) that is < [( τ, α)]. This establishes that Der( T ) is a tree. We now must show that Der( T ) is a κ-tree. That is, we must show that each level of Der( T ) has < κ nodes. Here is where we use the fact that Der( T ) consists of elements [( τ, α)] where 1 (Length( τ) = ˇα). If we drop the α s from the definition of Der( T ), we can verify 2) through 4) but not 1). Since P < κ and 1 ( T is a ˇκ-tree), it is routine to construct a function g : κ κ such that ( α < κ) 1 ( t T ) Length(t) > ˇα implies t(ˇα) < ǧ(ˇα). Now, a pair ( τ, α) such that [( τ, α)] Der( T ) may be construed as a sequence f ξ : ξ < α where each f ξ is a function from some maximal antichain of P to g(ξ). Thus, we can upper bound the size of level α of Der( T ) by the following: g(ξ) P. ξ<α Since κ is strongly inaccessible, this bound is < κ. We have now shown 1). To verify 2), suppose η < κ and S = [( τ α, α)] : α < η is a sequence such that [( τ 0, α 0 )] < [( τ 1, α 1 )] <... Let ṡ be a name such that 1 ṡ : ˇη <κ κ and ( α < η) 1 ṡ(ˇα) = τ α. Note that 1 ( α < β < ˇη) ṡ(α) ṡ(β), and moreover 1 α<ˇη ṡ(α) ˇη κ. Since 1 ( T has no maximal branches of length < κ), if we let τ η be such that 1 τ η = α<ˇη ṡ(α), we have [( τ η, η)] Der( T ), and this node is above each [( τ α, α)] for α < η. To verify 3), consider any [( τ, α)] Der( T ). Let ḃ be such that 1 (ḃ is the leftmost branch of T which extends τ). Let β be the least ordinal such that α β < κ and there is some p P such that

5 FORCING AND THE HALPERN-LÄUCHLI THEOREM 5 p (there are at least two successors of ḃ ˇβ in the tree T ). Let τ 1 be such that 1 ( τ 1 = ḃ ˇβ). Let τ 2 be such that 1 ( τ 2 is the first successor of τ 1 in T ). Let τ 3 be such that 1 ( τ 3 is the second successor of τ 1 in T if it exists, and it is the first successor otherwise). One can see that [( τ 2, β + 1)] and [( τ 3, β + 1)] are successors of [( τ 1, β)] in Der( T ). Since there is some p such that p ( τ 2 τ 3 ), it follows that [( τ 2, β + 1)] [( τ 3, β + 1)] Finally, the verification of 4) follows almost immediately from the definition of Der( T ). That is, fix [( τ, α)] Der( T ). Let G be P-generic over V. Let t be an arbitrary successor of τ G in T G. Let τ be such that τ G = t and 1 τ is a successor of τ. Then [( τ, α + 1)] Der( T ). 4. Small Forcings Preserve SDHL, HL, HL tc In this section we will show that if κ is strongly inaccessible and ( σ < κ) SDHL(d, σ, κ) holds, then this still holds after performing any forcing of size < κ. The same is true if we replace SDHL with HL tc. Theorem 4.1. Let κ be strongly inaccessible. Let 1 d < ω and σ < κ. Let P be a forcing of size < κ. Assume that SDHL(d, σ P, κ) holds. Then SDHL(d, σ, κ) holds after forcing with P. Proof. Let T i : i < d be a sequence of names for regular trees in the extension. Let ċ be such that 1 ċ : T i ˇσ. i<ď We must show that 1 there is a somewhere dense level matrix X i T i : i < ď such that ċ i<ď X i = 1. We will do this by showing that for each q P, there is some q q forcing this statement. Fix q P. Consider the trees Der( T i ) for i < d. Let c : i<d Der( T i ) σ P be a coloring such that for any α < κ and t = [( τ i, α)] Der( T i ) : i < d, c ( t) = σ, p where σ and p satisfy p q and p ċ( τ i : i < d ) = ˇσ. Since SDHL(d, σ P ), κ) holds, there is a somewhere dense level matrix Y i Der( T i ) : i < d that is monocromatic with respect to c.

6 6 NATASHA DOBRINEN AND DAN HATHAWAY Let σ, q be the unique color assigned to each element of i<d Y i by c. For each i < d, let Ẋi be such that 1 Ẋi = { τ : ( α) [(ˇ τ, α)] ˇY i }. For each i < d, let ( τ i, β) be such that Y i dominates the successors of [( τ i, α)] in Der( T i ). Since Y i : i < d is somewhere dense, the same β can be chosen for each i < d. By the definition of c, we have that q ċ i<ď Ẋ i = {ˇσ }. By 4) of the lemma above, we have that 1 ( i < ď) Ẋi dominates the successors of τ i in T i. Thus, we have q Ẋi : i < ď is a somewhere dense level matrix that is monocromatic with respect to ċ. This completes the proof. We have a similar theorem for HL instead of SDHL. Instead of using the fact that HL and SDHL are equivalent, we include the following proof to illustrate how to use derived trees. Theorem 4.2. Let d, σ, κ, and P be as in the above theorem. Assume that HL(d, σ P, κ) holds. Then HL(d, σ, κ) holds after forcing with P. Proof. Let T i : i < d be a sequence of names for regular trees in the extension. Let ċ be such that 1 ċ : i<ď T i ˇσ. Fix q P. Let c : i<d Der( T i ) σ P be as in the previous theorem. Applying HL(d, σ P, κ), we get a sequence of strong subtrees S i Der( T i ) : i < d as witnessed by the same set A κ such that for each ζ A, c i<d S i (ζ) = {(σ, p)}, for some σ < σ and q q. For each i < d, let U i be such that 1 U i = { τ : ( α) [(ˇ τ, α)] Ši}.

7 FORCING AND THE HALPERN-LÄUCHLI THEOREM 7 By Lemma 3.2, 1 each U i is a strong subtree of T i as witnessed by the same set Ǎ ˇκ. Furthermore, if ζ A, then q c i<d This is what we wanted to show. U i (ζ) = {σ }. Theorem 4.3. Let d, σ, κ, and P be as in Theorem 4.1. Assume that HL tc (d, < κ, κ) holds. Then HL tc (d, < κ, κ) holds after forcing with P. Proof. The proof is similar to the above theorem. Fix q P. The sequence of colorings ċ j : j < κ induces a sequence of colorings c j : j < κ where for each j < κ, 1 ċ j : i<ď T i ˇσ j and c j : i<d Der( T i ) σ j P. We may get strong subtrees S i Der(T i ) for i < d as witnessed by the same size κ set A κ. We may find a size κ set B κ such that the colorings c j for j B all use the same condition q. This shows that the modified HL tc (d, < κ, κ) holds in the extension. By proposition 2.9, this is equivalent to the full HL tc (d, < κ, κ) holding. Remark 4.4. In [3], Zhang shows that a certain polarized partition relation follows from HL tc (d, < κ, κ). Since there is a model in which HL tc (d, < κ, κ) holds for κ strongly inaccessible, it follows that in that model, the relevant polarized partition relation cannot be forced false by any poset of size < κ. 5. Reflection First, we will show that if SDHL (equivalentely HL) holds on a stationary set below κ, then it holds at κ. We do know that ZFC does not prove the analogous result for HL tc, because otherwise we would have, by the argument in the next section, a model in which HL tc holds at a cardinal that is not weakly compact, which contradicts [3]. Proposition 5.1. Assume that SDHL(d, σ, α) holds for a stationary set S of α < κ. Then SDHL(d, σ, κ) holds. Proof. Let T i <κ κ : i < d be a sequence of regular trees and let c : i<d T i σ be a coloring. If we can find an α < κ such that each T i <α κ is an α-tree and SDHL(d, σ, α) holds, then we will be

8 8 NATASHA DOBRINEN AND DAN HATHAWAY done. An elementary argument shows that for each i < d, there is a club C i κ such that T i <α κ is an α-tree for each α C i. The set i<d C i is a club, so it must intersect S. An α < κ in the intersection is as desired. When it comes to measurable cardinals, we have an even stronger form of reflection using a normal ultrafilter. The same argument works for HL and HL tc as well. Theorem 5.2. Let κ be a measurable cardinal and U be a normal ultrafilter on κ. Then SDHL(d, σ, κ) {α < κ : SDHL(d, σ, α)} U. Proof. Let j : V M be the ultrapower embedding. Because V κ+1 M, SDHL(d, σ, κ) SDHL(d, σ, κ) M. By Los s Theorem, SDHL(d, σ, κ) M {α < κ : SDHL(d, σ, α)} U. 6. SDHL at a Cardinal That is Not Weakly Compact Theorem 6.1. Fix d ω. If GCH holds and κ is (κ + d)-strong, then there is a forcing extension in which ( σ < κ) SDHL(d, σ, κ) holds, but κ is not weakly compact. Proof. First force over V to get a model V [G 1 ] in which SDHL holds at κ, which is also measurable [1]. By a theorem of Hampkins [2], any nontrivial forcing of size < κ followed by a non-trivial <κ-closed forcing will make κ not weakly compact. Perform any non-trivial forcing of size < κ over V [G 1 ] to get V [G 1 ][G 2 ]. This will preserve SDHL at κ by the previous theorem. Since κ is measurable in V [G 1 ][G 2 ], SDHL holds on a stationary (in fact, measure one) subset of κ. Now perform any non-trivial <κ-closed forcing over V [G 1 ][G 2 ] to get V [G 1 ][G 2 ][G 3 ]. Stationary subsets of κ are preserved, so inside V [G 1 ][G 2 ][G 3 ], SDHL holds on a stationary subset of κ. Thus, SDHL holds at κ in this model. References [1] N. Dobrinen and D. Hathaway, The Halpern-Lauchi Theorem at a Measurable Cardinal. To appear in Journal of Symbolic Logic. [2] J. Hamkins Small forcing makes any cardinal superdestructible. The Journal of Symbolic Logic 63: 51-58, 1998.

9 FORCING AND THE HALPERN-LÄUCHLI THEOREM 9 [3] J. Zhang, A Tail Cone version of the Halpern-Luchli theorem at a large cardinal, eprint arxiv: , April address: Daniel.Hathaway@du.edu