Selective Ultrafilters on FIN

Transcription

1 Selective Ultrafilters on FIN Yuan Yuan Zheng Abstract We consider selective ultrafilters on the collection FIN of all finite nonempty subsets of N. If countablesupport side-by-side Sacks forcing is applied, then every selective ultrafilter in the ground model generates a selective ultrafilter in the extension. We also show that selective ultrafilters localizes the Parametrized Milliken Theorem, and that selective ultrafilters are Ramsey. 0 Introduction In [2] and [11], Baumgartner and Laver showed that selective ultrafilters on N are preserved under both side-by-side and iterated Sacks forcing. That is, if Sacks forcing is applied then every selective ultrafilter on N in the ground model has its upward closure being a selective ultrafilter on N in the extension. Consider the collection FIN of all finite nonempty subsets of N. Milliken [14] introduced the collection FIN [ ] of all infinite block-sequences of members of FIN, a more powerful topological Ramsey space ([20]) than the Ellentuck space N [ ], and strengthened various generalizations of Ramsey s theorem. The original conjecture that many topological Ramsey spaces have an ultrafilter associated to them analogous to the way selective ultrafilters on N are related to the Ellentuck space was made by Todorcevic in the 1990 s when he proved using large cardinals that selective ultrafilters are generic over L(R) for the poset P(N)/FIN ([6]). In search of a notion of selectivity on FIN that captures the preservation under countable-support side-by-side Sacks forcing, we define selective ultrafilters on FIN in such a way that the combinatorial properties of U-trees can be easily applied. The notion of U-trees was introduced by Blass in [3]. Our definition turns out to be equivalent to Blass and Hindman s stable ordered-union ultrafilters in [4], and it is a special case of Mijares general notion of selective ultrafilters for topological Ramsey spaces in [13]. It also coincides with a special case of Di Prisco, Mijares and Nieto s definition in [5], which they used to prove complete combinatorics. It is also worth noting that Fernández-Bretón and Hrušák [9] showed the existence of such selective ultrafilters on FIN in Sacks model assuming a weak diamond principle. To show that selective ultrafilters on FIN are preserved under Sacks forcing, we first prove that a selective ultrafilter localizes the following theorem in [20, Thm. 5.45]: Theorem 0.1 (Parametrized Milliken Theorem [20]). For every finite Souslin-measurable colouring of FIN [ ] (2 ω ) ω there exist B FIN [ ] and a sequence (P i ) i<ω of nonempty perfect subsets of 2 ω such that [B] [ ] i<ω P i is monochromatic. This is done by defining a property of subsets of FIN [ ] (2 ω ) ω, called perfectly U-Ramsey, for selective ultrafilters U, and showing that this property is preserved under the Souslin operation. The result of Mathias [12] and Silver [19], that every analytic set is Ramsey, was strengthened by Ellentuck [8] to all analytic sets with respect to the finer Ellentuck topology. However, Ramsey properties are not often captured by a natural topology. We follow a similar procedure but rely more on the combinatorial properties of U-trees. In Section 1, we recall some definitions related to the Milliken space FIN [ ], perfect subsets of 2 ω and Sacks forcing. In Section 2, we prove the following localized version of the Parametrized Milliken Theorem: Theorem 0.2 (Local Parametrized Milliken Theorem). Let U be a selective ultrafilter on FIN. For every finite Souslin-measurable colouring of FIN [ ] (2 ω ) ω, there exist [X] U and a sequence (P i ) i<ω of nonempty perfect subsets of 2 ω such that [, X] i<ω P i is monochromatic Mathematics Subject Classification. Primary 05D10; Secondary 03E02, 03E40. 1

2 In Section 3, we show that selective ultrafilters are preserved after adding arbitrarily many Sacks reals by countable-support side-by-side Sacks forcing. Theorem 0.3. Let κ be an infinite cardinal, and P κ be countable-support side-by-side Sacks forcing adding κ Sacks reals. Let U be a selective ultrafilter on FIN in the ground model, and V a name for the upward closure {Ẏ FIN : [ ˇX] Ǔ [ ˇX] Ẏ } of U. Then P κ V is a selective ultrafilter on FIN. In Section 4, we prove that every selective ultrafilter on FIN is Ramsey. This is in contrast to the case in the spaces R α, as shown by Trujillo in [21]. The spaces R α, for countable ordinals α, are topological Ramsey spaces constructed by Dobrinen and Todorcevic in [7]. I am grateful to Professor Stevo Todorcevic for his guidance. I would also like to thank Mr. Xiao Qing Zheng for his uncountable support. 1 Preliminaries A (non-principal) ultrafilter on some set S is a collection U of subsets of S with the following properties. For subsets M, N, A, B of S, (a) S U but {x} / U for all x S, (b) M N and M U implies that N U, (c) M = A B and M U implies that A U or B U, (d) M U and N U implies that M N U. 1.1 The Milliken space FIN [ ] Let FIN be the collection of all finite nonempty subsets of N. We will concentrate on the case S = FIN. For elements x, y in FIN, by x < y we mean that max(x) < min(y). Listed below are some definitions and notation for us to refer back to. Definition 1.1. A block-sequence is a sequence of elements in FIN of the form X = (x n ) = {x 1, x 2,... } FIN, where each x n is called a block, and x n < x n+1 for all n. Let FIN [ ] denote the collection of all infinite block-sequences, FIN [< ] the collection of all finite block-sequences. For a FIN [< ], let the length a of a be the number of blocks in a. Let FIN [n] the collection of block-sequences of length n. Notation. For X = (x n ) FIN [ ], k < ω and a FIN [< ], let X/a = {x X : a < x}. X = (x n ) FIN [ ], let [X] denote the sublattice of FIN generated by X, that is, [X] = {x n0 x nk : k < ω, n 0 < < n k < ω}. When we say [X] is an element of an ultrafilter, we tacitly understand that X FIN [ ]. Definition 1.2. For two (finite or infinite) block-sequences X = (x n ) and Y, we say X is a block-subsequence of Y, and write X Y, if x n [Y ] for all n. Let [, Y ] denote the collection of all infinite block-subsequences of Y. For an infinite block-sequence X and a finite block-subsequence a X, let [a, X] = {Y FIN [ ] : a Y X}. Unless stated otherwise, we always assume the metric topology for FIN [ ] with the first-difference metric ρ as follows. For X = (x n ) n=1 and Y = (y n ) n=1, For ρ(x, Y ) = 1 k where k = min{n : x n y n }. This induces a topology on FIN [ ] with open balls of the following form. B(A, 1 k ) = {X FIN[ ] : ρ(x, A) < 1 k } = {X = (x n ) FIN [ ] : x n = a n for all n k} 2

3 for A = (a n ) n=1 FIN [ ]. We may also write the open balls as [a] := {X = (x n ) FIN [ ] : x n = a n for all n k} for a = (a n ) k n=1 FIN [< ]. 1.2 The spaces 2 ω and (2 ω ) ω, and Sacks forcing Let 2 ω have the product topology. We identify (2 ω ) ω and 2 ω ω, each equipped with the product topology. Let Sq be the set of all finite 01-sequences, that is, Sq = { n 2 : n ω}. For s Sq, we denote the length of s by s. For n < ω and f 2 ω, by f n we mean the sequence obtained by restricting f to its first n elements. If n > f, let f n = f. For s Sq and f 2 ω, s f means that s is an initial segment of f. For s, t Sq, we say s, t Sq are comparable if s t or t s; otherwise we say s and t are incomparable. Definition 1.3. [1]. A nonempty set p Sq is a tree if for every s p and n ω, s n p. Moreover, a tree p is perfect if for every s p there exist incomparable t, u p such that s t and s u. In particular, every perfect tree is infinite. Notation. For a tree p Sq, let [p] denote the set of all infinite branches of p: [p] = {f 2 ω : f n p for all n ω}. For a closed set F 2 ω, let T F be the tree whose infinite branches are the elements of F : T F = {s Sq : s f for some f F }. Lemma 1.4. [1]. A closed set F 2 ω is perfect if and only if the tree T F is a perfect tree. Definition 1.5. [18]. Sacks forcing P is the set of all perfect trees, with the ordering p q if p q. In particular, p q if and only if [p] [q]. Definition 1.6. Let p P and s p. The branching level of s in p is the number of branchings below s in the tree p, namely {i < s : t p (( t > i) (t i = s i) (t (i + 1) s (i + 1)))}. The n th branching level l(n, p) of p is the set of all s p which have branching level n and are minimal with this property. Definition 1.7. [1]. For p, q P, we write p n q if p q and l(n, p) = l(n, q). Definition 1.8. [1]. For p P and s p, let p s = {t p : t s or s t}. Notice that p s P. Sacks forcing P adds one Sacks real ([18]). In this paper, we consider countable-support side-by-side Sacks forcings P κ which adds κ Sacks reals, where κ is an arbitrary infinite cardinal. Definition 1.9. [1]. Let κ be an infinite cardinal. Let P κ be the set of all functions p : κ P such that for all but countably many α < κ, p(α) = 2 <ω. The ordering on P κ is given by p q if p(α) q(α) for every α < κ. The support of p P κ is supp(p) = {α < κ : p(α) 2 <ω }. The following definitions lead us to the Fusion Lemma The method of fusion is used throughout this paper. Definition [1]. For p P κ, n ω and a finite set F κ, let l(f, n, p) = {σ : (dom(σ) = F ) (( α F )(σ(α) l(n, p(α)))}. We write p F,n q if p q and l(f, n, p) = l(f, n, q). Definition [1]. Let p P κ, and F be a finite subset of κ. Let σ be a function with dom(σ) = F and σ(α) p(α) for every α F. We define p σ P κ as follows. For α < κ, { p(α) for α / F ; p σ(α) = p(α) σ(α) for α F. 3

4 Lemma 1.12 (Fusion Lemma). [1, Lem. 1.6] Suppose p n : n ω P κ and (F n, m n ) : n ω are sequences such that F n F n+1 κ are finite, m n m n+1 < ω, lim m n = and p n+1 Fn,mn p n for each n. Suppose also that {F n : n ω} {supp(p n ) : n ω}. Define q : κ P as follows: q(α) = p n (α) for α < κ. Then q P κ and q Fn,mn p n for each n. n<ω If p n : n ω P κ, (F n, m n ) : n ω and q are as above, then we say that p n : n ω is a fusion sequence and q is the fusion of the sequence. Definition For p P ω, we define the set [p] (2 ω ) ω to be [p] = i<ω [p(i)]. Recall that 2 ω and (2 ω ) ω are assumed to have the product topology. The space 2 ω has basic open sets of the form [s] = {f 2 ω : s f}. Thus, for p P and f [p], every initial segment s f is associated with an open set [s] in 2 ω. We develop a similar notion for p P ω and (2 ω ) ω. Let p P ω. An element ɛ [p] is a function ɛ : ω ω ω with ɛ(j, i) [p(i)] for every i, j < ω. For a fixed i ω, we may think of (ɛ(j, i)) j ω as a function in j, and hence an infinite branch in the i th perfect tree p(i) of p. We define a finite partial function σ : ω 2 <ω to be a pre-initial segment of ɛ if for every i dom(σ), σ(i) is an initial segment of (ɛ(j, i)) j<ω : Definition Let ɛ (2 ω ) ω. A partial function σ : ω 2 <ω with finite domain is a pre-initial segment of ɛ if σ(i) (ɛ(j, i)) j<ω for every i dom(σ). Then (2 ω ) ω has basic open subsets of the form [σ] = {ɛ (2 ω ) ω : σ is a pre-initial segment of ɛ} for finite partial functions σ : ω 2 <ω, or equivalently, for σ 2 (<ω) (<ω). Also throughout this paper, FIN [ ] 2 ω and FIN [ ] (2 ω ) ω have the product topology, where FIN [ ] has the first-difference metric topology unless otherwise stated. Therefore basic open sets in FIN [ ] (2 ω ) ω are of the form [a] [σ] where a FIN [< ] and σ 2 (<ω) (<ω). 2 Selective ultrafilters are localizing Definition 2.1. [4, 13]. An ultrafilter U on FIN is selective if (1) U is generated by elements of the form [X] where X FIN [ ] ; (2) for every set {[X a ] : a FIN [< ] } of elements in U, there exists [X] U such that for all a X, X/a [X a ]. Note that by property (1) of selectivity, if [X] U, then for every a FIN [< ], [X/a] U. Moreover, a selective ultrafilter U is idempotent with operation. Lemma 2.2. Given a selective ultrafilter U on FIN, let U U = {A FIN : {x FIN : {y FIN : x y A} U} U}. Then U = U U. Proof. We prove that U U U. Equality follows since U U is also an ultrafilter. Consider [B] U, and let B = (b n ) n<ω. We show that [B] U U. Let X = {x : {y : x y [B]} U}. It is sufficient to prove that X U. We show this by proving that [B] X and using the property that U is closed under superset. For an arbitrary x [B], let x = b n0 b nl with n 0 < < n l. Then [B/{x}] {y : x y [B]}. [B] U, so [B/{x}] U, and hence {y : x y [B]} U, that is x X. As x [B] was chosen arbitrary, we have [B] X as required. Definition 2.3. An ultrafilter U on FIN is localizing if for every Souslin-measurable set B of FIN [ ] (2 ω ) ω there exist [B] U and p P ω such that [, B] [p] B or [, B] [p] B =. A localizing ultrafilter localizes the Parametrized Milliken Theorem 0.1. In this section we aim to prove that every selective ultrafilter is localizing. We first consider open subsets of FIN [ ] (2 ω ) ω in subsection 2.1, then generalize the result to Souslin-measurable subsets in subsection 2.2. From now on in this section, we fix an arbitrary selective ultrafilter U on FIN. 4

5 2.1 Open subsets of FIN [ ] (2 ω ) ω Ultra-Ramsey theory plays a role in the proof. So we adapt definitions in [20, Section 7.2] for the space FIN [ ]. Consider FIN [< ] as a tree ordered by end-extension with root. From now on in this section, by a tree we mean a subtree of FIN [< ] that is downward closed, namely, if the subtree contains an element a, then it contains every initial segment b a. Notation. For a tree T, we denote the collection of all infinite branches of T by [T ], that is, { } [T ] = A = (a i ) i=1 FIN [ ] : {a 1,..., a n } T for all n ω. Let the maximal node of T that is comparable with every other node of T be denoted by st(t ), called the stem of T. Definition 2.4. [20, Def. 7.28]. A U-tree T is a tree such that, for all t T with st(t ) t, {x FIN : t t {x} T } U. For two U-trees T and T, we say T is a pure refinement of T, and write T 0 T, if T T and st(t ) = st(t ). Definition 2.5. [20, Def. 7.35]. A subset G of FIN [ ] is U-open if for every A G there is a U-tree T such that A [T ] and [T ] G. The collection of all U-open subsets of FIN [ ] is a topology on FIN [ ] with basis [T ] for U-trees T. This U-topology includes the metric topology on FIN [ ]. Definition 2.6. [20, Def. 7.37]. A subset X of FIN [ ] is U-Ramsey if for every U-tree T there is a pure refinement T 0 T such that [T ] X or [T ] X =. The proof of Theorem 7.42 in [20] can be readily adapted to give us the following. Lemma 2.7. [20, Thm. 7.42]. If X FIN [ ] is (metrically) Souslin-measurable, then X is U-Ramsey. Lemma 2.8. For a U-tree T with stem a, there exists [X] U with a X such that [a, X] [T ]. Conversely, for a X with [X] U, there exists a U-tree T with stem a such that [T ] [a, X]. Proof. First we assume T is a U-tree with stem a. Since T is a U-tree, for each b > a or b = with a b T the set A b = {x FIN : a b a b {x} T } is in U. So by property (1) of selectivity, there exists [X b ] U with [X b ] A b. On the other hand, for each b FIN [< ] \{b : a a b T }, let [X b ] = FIN [< ]. Now we have {[X b ] : b FIN [< ] } U. By property (2) of selectivity, there exists an [X ] U such that X /b X b for every b X. Let X = a (X /a). Then [X] U. We check that [a, X] [T ]. Let Y = a (y n ) n<ω [a, X]. We show a {y 1,..., y n } T for every n ω by induction. For n = 0, it is a, the stem of T. Suppose n = k + 1. By the induction hypothesis, a {y 1,..., y k } T. Then y k+1 [X/(a {y 1,..., y k )}] = [X /{y 1,..., y k }] [X {y1,...,y k }] A {y1,...,y k }. So by the definition of A {y1,...,y k }, a {y 1,..., y k, y k+1 } T. Thus, if Y [a, X] then Y [T ]. Hence [a, X] [T ] as required. Conversely, let [X] U and a X. The set T := {b FIN [< ] : b a} {b FIN [< ] : a b X} is a U-tree with stem a. This is because, for every b T with a b, {x FIN : b b {x} T } = [X/b] U. Definition 2.9. For a U-tree T and n ω, let l n (T ) be the n th level of the tree T above its stem: l n (T ) = {b T : b = st(t ) + n}. For [X] U and a X, let l n [a, X] = {b FIN [ a +n] : a b X}. Theorem For every Souslin-measurable subset X of FIN [ ], there exists [X] U such that [, X] X or [, X] X =. Proof. By Lemma 2.7, X is U-Ramsey. So for the U-tree FIN [< ], there exists a pure refinement T 0 FIN [< ] such that [T ] X or [T ] X =. By Lemma 2.8, there exists [X] U with [, X] [T ]. 5

6 Now we aim to prove that for an open set O of FIN [ ] (2 ω ) ω in the product topology, the set {X FIN [ ] : p P ω ([, X] [p] O) or ([, X] [p] O = )} is Souslin-measurable, after which we could apply Theorem For this, we define uniform families in FIN [ ]. It is analogous to Pudlák and Rödl s definition of uniform family in the Ellentuck space in [17]. Notation. For S FIN [< ] and a FIN [< ], let S [a] = {y : a y S}. Definition [17]. Let γ be a countable ordinal and S FIN [< ]. Let X FIN [ ] and b X. The notion of γ-uniform is defined recursively as follows. The family S is γ-uniform on [b, X] if γ = 0 and S = {b}; γ = β + 1, b / S and for each a l 1 [b, X], S [a] is β-uniform on [a, X]; or γ is a limit ordinal, b / S and there exists a sequence (γ a ) a l1[b,x] of ordinals with a l 1[b,X] γ a = γ such that for each a l 1 [b, X], S [a] is γ a -uniform on [a, X]. We say S is uniform on [b, X] if it is γ-uniform on [b, X] for some γ. For example, for n < ω, the only n-uniform family on [b, X] is l n [b, X]. On the other hand, for each k < ω, the family S = {y = (y n ) FIN [< ] : (b y X) ( y = max(y b +1 ) + k)} is ω-uniform on [b, X], and T = {y = (y n ) FIN [< ] : (b y X) ( y = max(y b +2 ) + k)} is (ω + 1)-uniform on [b, X]. Recall from [20, Def. 1.12] that a family F of finite block-sequences is Nash-Williams if x y for every distinct pair x, y F. The family F is a front on [b, X] if it is Nash-Williams and every element of [b, X] has an initial segment in F. The definition of rank of a barrier in [20, Def. 1.24] can be used for fronts as well. Definition [20, Def. 1.24]. Let F be a front on [b, X]. The set T (F) = {x : y F x y} is the -downward closure of F. Let ρ F (x) = sup{ρ F (s) + 1 : (t s) (s T (F))}, where sup is defined to be 0. Then the rank of F on [b, X] is ρ(f) = ρ F (b). Lemma If S is uniform on [b, X], then it is a front on [b, X]. Proof. Let S be γ-uniform on [b, X]. We prove by induction on γ. If γ = 0, then S = {b} and the conclusion holds. So we assume γ > 0. By the definition of γ-uniform, for each a l 1 [b, X], S [a] is β-uniform for some β < γ. So by the induction hypothesis, S [a] is a front on [a, X]. Then S = a l 1[b,X] S [a] is a front on [b, X]. Lemma If F is a front on [b, X], then there exists a uniform system S on [b, X] such that every s S has an initial segment in F. Proof. We induct on the rank of F. If ρ(f) = 0 then F = {b} itself is a uniform system on [b, X]. So we assume ρ(f) > 0. For each a l 1 [b, X], the family F [a] = {s : a s F} is a front on [a, X] of a smaller rank. By the induction hypothesis, there exists a countable ordinal γ a and a γ a -uniform family S a on [a, X] such that every element in S a has an initial segment in F [a]. Then the family S = a l 1[b,X] S a is ( a l 1[b,X] γ a)-uniform on [b, X] and every element in S has an initial segment in F. Theorem For every open set O FIN [ ] (2 ω ) ω there exist [X] U and p P ω such that [, X] [p] O or ([, X] [p]) O =. Proof. Let X 0 = {X FIN [ ] : ( p P ω ) (([, X] [p]) O = )}, and X 1 = {X FIN [ ] : ( p P ω ) ([, X] [p] O)}. Let X = X 0 X 1. We claim that X 0 and X 1 are Souslin-measurable, hence so is X. Then applying Theorem 2.10 to the set X FIN [ ], we get [X] U such that [, X] X or [, X] X =. If [, X] X then there exists p P ω such that X and p satisfies the theorem. Otherwise, [, X] X =, we would have the 6

7 following contradiction. The set O ([, X] (2 ω ) ω ) is open in [, X] (2 ω ) ω in the subspace topology. By the Parametrized Milliken Theorem 0.1, there exist Y [, X] and q P ω such that [, Y ] [p] O or ([, Y ] [p]) O =. Then Y X, contradicting Y [, X] and [, X] X =. We first prove that X 0 is Souslin-measurable. Since O is open, for each element (X, ɛ) of FIN [ ] (2 ω ) ω, (X, ɛ) O if and only if there exists a X and a pre-initial segment σ of ɛ such that [a] [σ] O. So X 0 can be written as X 0 = {X FIN [ ] : ( p P ω )( a X)( F ω finite)( σ i F p(i))[a] [σ] O}. Therefore, X 0 is analytic, hence Souslin-measurable. Lastly, we show that X 1 is also analytic. Let Y FIN [ ] and p P ω. Assume {Y } [p] O. As above, this implies that for every ɛ [p] there exists a ɛ Y and a pre-initial segment σ ɛ of ɛ such that [a ɛ ] [σ ɛ ] O. Thus {[σ ɛ ] : ɛ [p]} is an open cover of [p]. Since (2 ω ) ω is compact, the closed subset [p] is also compact. So [p] has a finite subcover {[τ 1 ],..., [τ k ]} {[σ ɛ ] : ɛ [p]}. Correspondingly, there are a 1,..., a k Y such that [a i ] [τ i ] O for every i k. We can find a Y such that a i a for every i k. So [a] [p] O. Thus we have proved that if {Y } [p] O then Y has an initial segment a such that [a] [p] O. Note that the converse of this also holds. Therefore, for X FIN [ ] and p P ω, [, X] [p] O if and only if every Y X has an initial segment in the set F = {a X : [a] [p] O}. By shrinking F to a front F on [, X] and using Lemma 2.13 and Lemma 2.14, we have the following. For X FIN [ ], X X 1 if and only if if and only if ( p P)( S FIN [< ] )((S is uniform on [, X]) (( a S)( b F )(b a))), ( p P)( S FIN [< ] )((S is uniform on [, X]) (( a S)( b X)((b a) ([b] [p] O))), Thus X 1 is also analytic. 2.2 Perfectly U-Ramsey sets Recall that we have fixed an arbitrary selective ultrafilter U on FIN throughout Section 2. Unless otherwise stated, we are working with the metric topology on FIN [ ], and the product topology on (2 ω ) ω and FIN [ ] (2 ω ) ω. In this subsection, we generalizes Theorem 2.15 to all Souslin-measurable subsets of FIN [ ] (2 ω ) ω, thus proving that every selective ultrafilter on FIN is localizing. For a U-tree T and x T, let T x = {y T : y x or x y}. Recall that l n (T ) denotes the n th level of T above its stem. Recall also that by T 0 T we mean T T and st(t ) = st(t ). This notion is extended as follows. Definition [20, Section 7.2]. For two U-trees T and T, and n ω, we write T n T if T T and l n (T ) = l n (T ). Definition [20, Section 7.2]. A sequence (T n ) n<ω of U-trees is a fusion sequence if T n+1 n T n for every n ω. In this case, the set T = n<ω T n is called the fusion of the sequence. Note that T is also a U-tree. Definition A subset B of FIN [ ] (2 ω ) ω is perfectly U-Ramsey if for every U-tree T and every p P ω there exist T 0 T and p p such that [T ] [p ] B or ([T ] [p ]) B =. We say B is perfectly U-Ramsey null if the second alternative always holds. By Lemma 2.8, the following lemma follows from Theorem Lemma Every open subset of FIN [ ] (2 ω ) ω is perfectly U-Ramsey. Now we aim to show that the collection of perfectly U-Ramsey sets is closed under the Souslin operation, through a series of lemmas. The following lemma is immediate from the definition of perfectly U-Ramsey. 7

8 Lemma A subset B of FIN [ ] (2 ω ) ω is perfectly U-Ramsey if and only if for every U-tree T and p P ω and for every finite subset F ω, n < ω, there exist T 0 T and p F,n p such that σ l(f, n, p) [T ] [p σ] B or ([T ] [p σ]) B =. Lemma The set of all perfectly U-Ramsey null subsets of FIN [ ] (2 ω ) ω is a σ-ideal. Proof. Closure under subsets and finite union is clear. We show that it is closed under countable union. By Lemma 2.20 we notice that a set B is perfectly U-Ramsey null if and only if for every n, F, T, p given there exist T 0 T and p F,n p such that ([T ] [p ]) B =. Now suppose Y k is perfectly U-Ramsey null for every k < ω. We prove that so is k<ω Y k. Given T, p we construct fusion sequences (T k ) k<ω and (p k ) k<ω recursively. Let T 0 = T, p 0 = p and F k = {0,..., k} for k ω. At stage k + 1, enumerate l k (T k ) as (x j ) j<ω. Let q 1 = p k. For each j ω, as Y k is perfectly U-Ramsey null, there exist S j 0 T k x j and q j F k+j,k+j q j 1 such that ([S j ] [q j ]) Y k =. Since j<ω F k+j = ω, (q j ) j<ω is a fusion sequence. Then let T k+1 = j ω S j and p k+1 = j<ω q j. By construction, T k+1 is a U-tree, p k+1 P ω, T k+1 k T k, p k+1 F k,k p k and ([T k+1 ] [p k+1 ]) Y k =. As k<ω F k = ω, we can take the fusions T = k<ω T k and p = k<ω p k. Then [T ] [p ] is disjoint from Y k for each k < ω, so it is disjoint from k<ω Y k. So k<ω Y k is perfectly U-Ramsey null. Lemma The set of all perfectly U-Ramsey subsets of FIN [ ] (2 ω ) ω is a σ-field. Proof. Similarly we only show that it is closed under countable union. Suppose Y k is perfectly U-Ramsey for every k < ω. We check that so is Y = k<ω Y k. Given T, p we construct fusion sequences (T k ) k<ω and (p k ) k<ω recursively. Let T 0 = T, p 0 = p and F k = {0,..., k} for k ω. At stage k + 1, let l k (T k ) = (x j ) j<ω. Let q 1 = p k. For each j ω, by Lemma 2.20, there exist S j 0 T k x j and q j F k+j,k+j q j 1 such that for every σ l(f k+j, k + j) we have [S j ] [q j σ] Y k or ([S j ] [q j σ]) Y k =. Let T k+1 = j<ω S j and p k+1 = j<ω q j. Then T k+1 is a U-tree with T k+1 k T k and p k+1 P ω. Let T = k<ω T k and p = k<ω p k. Then T 0 T, p p and for every k < ω and x l k (T ), σ l(f k+1, k + 1, p ) [T x] [p σ] Y k or ([T x] [p σ]) Y k =. Thus Y ([T ] [p ]) is open in [T ] [p ] with the subspace topology. Then by Lemma 2.19, there exist T 0 T and p p such that [T ] [p ] Y or ([T ] [p ]) Y =. In particular, T 0 T and p p as required. Therefore Y = k<ω Y k is perfectly U-Ramsey. We use combinatorial forcing ([16, 10]) to prove that the field of perfectly U-Ramsey subsets of FIN [ ] (2 ω ) ω is closed under the Souslin operation. Recall that by σ 2 (<ω) (<ω) we mean a function σ : F 2 <ω where F ω is finite. Definition Consider an arbitrary U-tree T, p P ω, X FIN [ ] (2 ω ) ω and (a, σ) FIN [< ] 2 (<ω) (<ω). We say (T, p) X -accepts (a, σ) if [T a] [p σ] X ; (T, p) X -rejects (a, σ) if [T a] [p σ] and there does not exist T 0 T a and p p σ such that (T, p ) X -accepts (a, σ); (T, p) X -decides (a, σ) if it either X -accepts or X -rejects (a, σ). The following lemma is immediate from the definition. Lemma Consider an arbitrary U-tree T, p P ω, X FIN [ ] (2 ω ) ω and (a, σ) FIN [< ] 2 (<ω) (<ω). (1) (T, p) X -accepts every (a, σ) such that [T a] [p σ] =. (2) If (T, p) X -decides (a, σ) and [T ] [p ] [T ] [p], then (T, p ) X -decides (a, σ). (3) If a T and σ(n) p(n) for every n dom(σ), then there exist T 0 T a and p p σ such that (T, p ) X -decides (a, σ). 8

9 (4) If (T, p) X -decides (a, σ), then (T, p) X -decides every element of the form (a, σ ) in the same way, where σ (j) p σ(j) for every j dom(σ ). Lemma Given a Souslin scheme X s : s ω <ω of subsets of FIN [ ] (2 ω ) ω, a U-tree T and p P, there exist T 0 T, p p and an enumeration (b n ) n<ω of {b T : st(t ) b} such that for every s ω <ω and every k max(s), if τ 2 (<ω) (<ω) and τ(j) k for every j dom(τ), then (T, p ) X s -decides (b k, τ). Proof. Let {b T : st(t ) b} be enumerated as (b n ) n<ω such that b m b n when n < m. In particular, b 0 = st(t ). We recursively shrink T and p cone by cone, starting from T 1 = T and p 1 = p. At stage k + 1, let F k+1 = {0, 1,..., k + 1}. Let l(f k+1, k + 1, p k ) {0, 1,..., k + 1} <ω be enumerated as (τ l, s l ) l m where m is a finite number. Let q 1 = p k. By Lemma 2.24 (3), for every l m there exists S l 0 T k b k+1 and q l Fk+1,k+1 q l 1 with q l τ l q l 1 τ l such that (S l, q l ) X sl -decides (b k+1, τ l ). Let p k+1 = q m, then p k+1 Fk+1,k+1 p k and p k+1 τ l q l τ l for every l m. Let T k+1 = (T k \ T k b k+1 ) l m S l. Note that T k+1 is a U-tree. Then (T k+1, p k+1 ) X s -decides (b k+1, τ) for every τ l(f k+1, k + 1, p k ) and every s {0,..., k + 1} <ω. Re-enumerate T k+1 \ {b 0,..., b k+1 } as (b k+1+n ) n<ω such that b m b n when n < m. Let T = k<ω T k and p = k<ω p k. Since k<ω F k = ω and p k+1 Fk+1,k+1 p k by construction, (p k ) k<ω is a fusion sequence and p P ω. We check that T is a U-tree. Let b T. There exists n such that b is enumerated as b n+1 since stage n. Then, as T n+1 is a U-tree, {x FIN : b b {x} T } = {x FIN : b b {x} T n+1 } U. It remains to check that T and p satisfy the theorem. Suppose n ω and s {0,..., n} <ω. Let b T and τ 2 (<ω) (<ω) with [T b] [p τ] be given. Assume further that b is enumerated as b k in T for some k n, and τ(j) k for every j dom(τ). Let us check that (T, p ) X s -decides (b k, τ). We can find τ l(f k, k, p k ) = l(f k, k, p ) such that τ(j) p k τ (j) for every j dom(τ). Then [T k ] [p k ] X s -decides (b k, τ ). By Lemma 2.24 (4), [T k ] [p k ] also X s -decides (b k, τ). Hence by Lemma 2.24 (2), [T ] [p ] X s -decides (b k, τ) as well. On the other hand, if [T b] [p τ] =, then by Lemma 2.24 (1), [T ] [p ] X s -accepts (b, τ). Theorem The field of perfectly U-Ramsey subsets of FIN [ ] (2 ω ) ω is closed under the Souslin operation. Proof. Let Y s : s ω <ω be a Souslin scheme of perfectly U-Ramsey sets. Assume without loss of generality that Y t Y s whenever s t. Let Y = f ω ω n<ω Y f n. For a set X FIN [ ] (2 ω ) ω, let X c denote its complement FIN [ ] (2 ω ) ω \ X. Given T, p, we aim to show that there exist T 0 T and p p such that [T ] [p ] Y or Y c. Let Ys = f s n<ω Y f n. Note that Ys Y s and Y = Y. By Lemma 2.25, we can find T 0 T and p p where the elements in T above the stem are enumerated as (b n ) n<ω with b m b n when n < m such that for every s ω <ω and every k max(s), if τ 2 (<ω) (<ω) and τ(j) k for j dom(τ), then [T ] [p ] (Ys ) c -decides (b k, τ). For each s ω <ω, let O s = {(b, τ) : [T b] [p τ] (Ys ) c }. We define an envelope for each Ys by Φ(Ys ) = ([T ] [p ] Y s ) \ [T b] [p τ]. (b,τ) O s Since [T ] [p ], Y s and [T b] [p τ] are perfectly U-Ramsey, Φ(Y s ) is also perfectly U-Ramsey by Lemma Clearly, [T ] [p ] Y s Φ(Y s ) [T ] [p ] Y s. We say a set B is perfectly U-Ramsey null inside [T ] [p ] if for every U-tree S and q P ω with [S] [q] [T ] [p ] there exist S 0 S and q q such that [S ] [q ] B c. Claim For s ω <ω, if B Φ(Ys ) \ Ys is perfectly U-Ramsey, then B is perfectly U-Ramsey null inside [T ] [p ]. Suppose B Φ(Ys ) \ Ys is perfectly U-Ramsey. Given [S] [q] [T ] [p ], we can find S 0 S and q q such that [S ] [q ] B or B c. Assuming [S ] [q ] B, we aim for a contradiction. Since [S ] [q ] [T ] [p ], we can pick k max(s), b k S and τ such that, for each j dom(τ), τ(j) q (j) and τ(j) k. Since we chose T and p to satisfy the conclusion of Lemma 2.25, (T, p ) (Ys ) c - decides (b k, τ). By Lemma 2.24(2), (S, q ) also (Ys ) c -decides (b k, τ). But [S ] [q ] B and B is disjoint 9

10 from Ys, we must have that (S, q ) (Ys ) c -accepts (b k, τ). Thus (T, p ) must also (Ys ) c -accept (b k, τ). Hence (b k, τ) O s and [S b k ] [q τ] (b,τ) O s [T b] [p τ]. This contradicts the assumption that [S ] [q ] B which is disjoint from (b,τ) O s [T b] [p τ]. Therefore [S ] [q ] B c. This finishes the proof of Claim For s ω <ω, let M s = Φ(Ys )\ n<ω Φ(Y s n ). Note that Y s = n<ω Y s n and Φ(Y s n ) [T ] [p ] Y s n. So [T ] [p ] Ys n<ω Φ(Y s n ). Hence M s Φ(Ys ) \ ([T ] [p ] Ys ) = Φ(Ys ) \ Ys since Φ(Ys ) [T ] [p ]. As M s is perfectly U-Ramsey, by Claim M s is perfectly U-Ramsey null inside [T ] [p ]. The set of all perfectly U-Ramsey null sets form a σ-ideal (Lemma 2.21), so similarly s ω M <ω s is also perfectly U-Ramsey null inside [T ] [p ]. Hence there exist T 0 T and p p such that [T ] [p ] s ω M <ω s =. Claim ([T ] [p ]) Φ(Y ) = ([T ] [p ]) Y. The left inclusion is clear since Φ(Y ) Y. To prove the right inclusion, pick x [T ] [p ] Φ(Y ). [T ] [p ] M =, so there exists n 0 such that x Φ(Y n ). [T 0 ] [p ] M n0 =, so there exists n 1 such that x Φ(Y n ). By repeating this process we find f = n 0,n 1 k k ω ω such that x Φ(Yf k ) Y f k for all k < ω. So x k<ω Y f k Y = Y. This finishes the proof of Claim Since [T ] [p ] Φ(Y ) is perfectly U-Ramsey, there exist T 0 T and p p such that [T ] [p ] ([T ] [p ]) Φ(Y ) or (([T ] [p ]) Φ(Y ))c. By Claim , ([T ] [p ]) Φ(Y ) = ([T ] [p ]) Y = ([T ] [p ]) Y. So, in particular, we have T 0 T, p p and [T ] [p ] Y or Y c. Therefore Y is perfectly U-Ramsey. Thus we have proved that the field of perfectly U-Ramsey subsets of FIN [ ] (2 ω ) ω is closed under the Souslin operation. Corollary Every Souslin-measurable subset of FIN [ ] (2 ω ) ω is perfectly U-Ramsey. Thus, we have proved that every selective ultrafilter on FIN is localizing. Therefore, as promised in the introduction, we see that a selective ultrafilter localizes the Parametrised Milliken Theorem 0.1 as follows. Theorem 0.2 (Local Parametrized Milliken Theorem). Let U be a selective ultrafilter on FIN. For every finite Souslin-measurable colouring of FIN [ ] (2 ω ) ω, there exist [X] U and a sequence (P i ) i<ω of nonempty perfect subsets of 2 ω such that [, X] i<ω P i is monochromatic. 3 Selectivity preserved under Sacks forcings Let U be an arbitrary selective ultrafilter on FIN. We fix an arbitrary infinite cardinal κ, and prove that U is preserved after adding κ Sacks reals using countable-support side-by-side Sacks forcing P κ. Recall that P κ is the set of all functions p : κ P with countable support, ordered by p q if p(α) q(α) for every α < κ. Forcing with P κ adds a Sacks real for each α κ. From now on in this subsection, is used to denote the forcing relation with respect to P κ. Lemma 3.1. [1, Lem. 1.9]. Suppose p P κ, n ω and that F κ is finite. If q p then there exists σ l(f, n, p) such that q and p σ are compatible. Corollary 3.2. [1, Cor. 1.10]. Suppose p P κ, n ω and that F κ is finite. If p τ V then there exists q F,n p such that for each σ l(f, n, q) there exists a σ V such that q σ τ = a σ. If q and τ are as in this Corollary then we say q determines τ relative to (F, n). We say q determines τ if there exist F, n such that q determines τ relative to (F, n). For q τ : FIN V, we say q determines τ if q determines τ(x) for every x FIN. As FIN is countable, we enumerate it as FIN = {x n : n ω}. Corollary 3.3. If p P κ and p τ : FIN 2 then there exists q p such that q determines τ. Proof. We construct a fusion sequence p n : n < ω recursively, starting from p 0 = p and F 0 =. For n < ω, let F n+1 = F n min(supp(p n ) \ F n ). By Corollary 3.2, there exists p n+1 Fn,n p n such that for each σ l(f n, n, q), p n+1 σ determines τ(x n ). Then the fusion q = n<ω p n satisfies the statement. 10

11 Let V be a name with respect to P κ for the upward closure of U. Our aim is to prove that, forcing with P κ, V is a selective ultrafilter in the extension. It is straightforward to check the properties of V for being a selective ultrafilter except for the following. Theorem 3.4. If p P κ and p τ FIN, then there exist q p and [B] U such that q [B] τ or q [B] τ =. Proof. We abuse notation and use τ to denote the characteristic function of the set, so p τ : FIN 2. We will find q p and [B] U such that q τ is constant on [B]. By the proof of Corollary 3.3 we can construct a fusion sequence p n : n ω with (F n, n) : n ω such that the fusion p of the sequence satisfies the following condition. n < ω σ l(f n, n, p ) i σ 2 such that p σ τ(x n ) = i σ. Note that p P κ, so supp(p ) is countable. We may assume without loss of generality that supp(p ) ω. Then [p ] (2 ω ) ω. We consider the following subset F of FIN [p ]: F = {(x n, ɛ) : ( σ l(f n, n, p ))( i F n )((σ(i) (ɛ(j, i)) j<ω ) (p σ τ(x n ) = 0))}. Let X = {((y n ) n=1, ɛ) FIN [ ] [p ] : (y 1, ɛ) F}. Then X is open in FIN [ ] [p ] where FIN [ ] [p ] FIN [ ] (2 ω ) ω has the subspace topology. Since U is localizing, there exists [B] U and q p such that [, B] [q] X or ([, B] [q]) X =, hence [B] [q] F or ([B] [q]) F =. We check that q and [B] satisfy the theorem. Firstly, suppose [B] [q] F. We prove that if x n [B], then q τ(x n ) = 0. Assuming there exists r q with r τ(x n ) = 1, we aim for a contradiction. By Lemma 3.1, there exists σ l(f n, n, q) such that r is compatible to q σ. Let ɛ [q] be such that σ is a pre-initial segment of ɛ, that is, σ(i) (ɛ(j, i)) j<ω for every i F n. Then, as (x n, ɛ) F and by the definition of F, there exists a pre-initial segment σ l(f n, n, p ) of ɛ such that p σ τ(x n ) = 0. Since q p, we have σ (i) σ(i) for every i F n, so q σ p σ. Therefore q σ τ(x n ) = 0. This contradicts that r τ(x n ) = 1 and r, q σ are compatible. If [B] [q] F =, then we similarly have q τ(x n ) = 1 for every x n [B]. This completes the proof of Theorem 0.3 in the Introduction. Theorem 0.3. Let κ be an infinite cardinal, and P κ be countable-support side-by-side Sacks forcing adding κ Sacks reals. Let U be a selective ultrafilter on FIN in the ground model, and V a name for the upward closure {Ẏ FIN : [ ˇX] Ǔ [ ˇX] Ẏ } of U. Then P κ V is a selective ultrafilter on FIN. 4 Selective ultrafilters are Ramsey In [13], Mijares defined a notion of Ramsey ultrafilters in topological Ramsey spaces. Definition 4.1. [13, Def. 3.2]. An ultrafilter U on FIN is Ramsey if for [X] U, a X and n ω, and for every partition FIN [ a +n] = F 0 F 1, there exists [Y ] U with Y [a, X] such that the set {b FIN [ a +n] : a b Y } is contained in one of F 0, F 1. Here we consider the following version used in [7], but we call it Nash-Williams instead of Ramsey. The change of name is because, although this property of Nash-Williams and Mijares property of Ramsey coincide in many topological Ramsey spaces, we do not know if there exists a space that distinguishes the two. For a family F FIN [< ] and an infinite block-sequence X FIN [ ], let F X = {a F : a X}. Definition 4.2. [7, Def. 5.1]. An ultrafilter U on FIN is Nash-Williams if for every Nash-Williams family F FIN [< ] and every partition F = F 0 F 1 there exists [X] U and i 2 such that F i X =. It is immediate that every Nash-Williams ultrafilter is Ramsey. In this section, we prove that every selective ultrafilter is Nash-Williams. Let U be an ultrafilter on FIN. Note that we do not assume U is selective here. Recall from Definition 2.5, the U-topology on FIN [ ] is generated by sets of the form [T ] where T is a U-tree. It is a topology finer than the metric topology on FIN [ ]. Recall from Definition

12 that a set X FIN [ ] is U-Ramsey if every U-tree T has a pure refinement T such that [T ] is contained in or disjoint from X. First we note that every (metrically) open set in FIN [ ] is U-Ramsey. The proofs of the following lemmas can be readily obtained by modifying the corresponding proofs found in [20] for the Ellentuck space. The proofs do not require U to be selective. Definition 4.3. [20, Def. 7.32]. A set G FIN [< ] is U-open if for every a G there exists a U-tree T with stem a such that {b T : a b} G. Lemma 4.4. [20, Lem. 7.33]. A set G FIN [< ] is U-open if and only if for every s G the set G s := {x FIN : s x G} is in U. Lemma 4.5. [20, Lem. 7.38]. Every U-open set is U-Ramsey. Now we show that every selective ultrafilter is Nash-Williams. Theorem 4.6. If U is a selective ultrafilter on FIN, then U is Nash-Williams. Proof. Let F FIN [< ] be a Nash-Williams family and F = F 0 F 1 be a partition. Then the set X := a F 0 [a] FIN [ ] is metrically open, which is therefore U-Ramsey. So there exists a U-tree T 0 FIN [< ] with stem such that [T ] X or [T ] X =. Since U is selective, by Lemma 2.8, we can find [X] U such that [, X] [T ]. If [, X] X, then every Y X has an initial segment in F 0, and cannot have an initial segment in F 1 since F is Nash-Williams. Therefore F 1 X =. Otherwise, [, X] X =, hence F 0 X =. References [1] J. E. Baumgartner. Sacks forcing and the total failure of Martin s axiom. Topology Appl., 19(3): , [2] J. E. Baumgartner and R. Laver. Iterated perfect-set forcing. Ann. Math. Logic, 17(3): , [3] A. Blass. Selective ultrafilters and homogeneity. Ann. Pure Appl. Logic, 38(3): , [4] A. Blass and N. Hindman. On strongly summable ultrafilters and union ultrafilters. Trans. Amer. Math. Soc., 304(1):83 97, [5] C. A. Di Prisco, J. G. Mijares, and J. E. Nieto. Local ramsey theory. an abstract approach. Preprint. [6] C. A. Di Prisco and S. Todorcevic. Perfect-set properties in L(R)[U]. Adv. Math., 139(2): , [7] N. Dobrinen and S. Todorcevic. A new class of Ramsey-classification theorems and their applications in the Tukey theory of ultrafilters, Part 2. Trans. Amer. Math. Soc., 367(7): , [8] E. Ellentuck. A new proof that analytic sets are Ramsey. J. Symbolic Logic, 39: , [9] D. J. Fernández-Bretón and M. Hrušák. A weak diamond principle and union ultrafilters. Unpublished notes. [10] F. Galvin and K. Prikry. Borel sets and Ramsey s theorem. J. Symbolic Logic, 38: , [11] R. Laver. Products of infinitely many perfect trees. J. London Math. Soc. (2), 29(3): , [12] A. R. D. Mathias. On a generalisation of Ramsey s theorem. Notices of the American Mathematical Society, 15:931, [13] J. Mijares. A notion of selective ultrafilter corresponding to topological Ramsey spaces. MLQ Math. Log. Q., 53(3): , [14] K. R. Milliken. Ramsey s theorem with sums or unions. J. Combinatorial Theory Ser. A, 18: ,

13 [15] Y. N. Moschovakis. Descriptive set theory, volume 155 of Mathematical Surveys and Monographs. American Mathematical Society, Providence, RI, second edition, [16] C. St. J. A. Nash-Williams. On well-quasi-ordering transfinite sequences. Proc. Cambridge Philos. Soc., 61:33 39, [17] Pavel Pudlák and Vojtěch Rödl. Partition theorems for systems of finite subsets of integers. Discrete Math., 39(1):67 73, [18] G. E. Sacks. Forcing with perfect closed sets. In Axiomatic Set Theory (Proc. Sympos. Pure Math., Vol. XIII, Part I, Univ. California, Los Angeles, Calif., 1967), pages Amer. Math. Soc., Providence, R.I., [19] J. Silver. Every analytic set is Ramsey. J. Symbolic Logic, 35:60 64, [20] S. Todorcevic. Introduction to Ramsey Spaces (AM-174). Annals of Mathematics Studies. Princeton University Press, [21] T. Trujillo. Selective but not Ramsey. Topology Appl., 202:61 69, Yuan Yuan Zheng, Department of Mathematics, University of Toronto, Room 6290, 40 St. George Street, Toronto, Ontario, Canada M5S 2E4 address: 13