A product of γ-sets which is not Menger.

Transcription

1 A product of γ-sets which is not Menger. A. Miller Dec 2009 Theorem. Assume CH. Then there exists γ-sets A 0, A 1 2 ω such that A 0 A 1 is not Menger. We use perfect sets determined by Silver forcing (see Grigorieff [3]) and construct an Aronszajn-tree of such perfect sets in the style of Todorcevic (see Galvin-Miller [2]). Define p P iff p : D 2 where D ω is co-infinite, i.e., D = ω \ D is infinite. For p P define [p] = {x 2 ω : p x}. Define p q iff 1 p q or equivalently [p] [q]. For n < ω define p n q iff p q and the first n elements of D p are the same as the first n elements of D q. The fusion property is important: and Fusion Lemma. Suppose (p n P : n < ω) has the property that p n+1 n p n for every n < ω. Then the fusion q = n<ω p n is in P and q n p n for every n. Define Q 0 (p) = {x [p] : n D p x(n) = 0} Q 1 (p) = {x [p] : n D p x(n) = 1}. Each of these are countable dense subsets of [p]. Define q 0 n p iff q n p and q restricted to D q \ D p is identically zero. Lemma 1 Given U an ω-cover of Q 0 (p) and n < ω there exist U U and q 0 n p such that [q] U. 1 We follow the convention that p q means that p is stronger than q or equivalently that p extends q. 1

2 To see why this is true let F be the first n elements of D p. For each s 2 F let x s Q 0 (p) such that x s F = s and x s (n) = 0 for every n D p \ F. Take U U with {x s : s 2 F } U. Since U is open there exists N < ω with [x s N] U for all s 2 F. Define q 0 n p by q = p { k, 0 : k < N and k (D p \ F )} Lemma 2 Given p n P and k n < ω for n < N and an ω-cover U of n<n Q0 (p n ) there exists U U and (q n 0 k n p n : n < N) such that [q n ] U. n<n Let F n be the first k n elements of D pn. For s 2 Fn define x n s Q 0 (p n ) as in the proof of Lemma 1. Let H n<n Q0 (p n ) be a finite set containing all such x n s. Choose U U with H U and determine the q n for n < N as in Lemma 1. Remark. Note that if q 0 k p then Q0 (q) Q 0 (p) and hence any ω-cover of Q 0 (p) is still an ω-cover of Q 0 (q). In these two lemmas, the q we obtain are also equal mod finite to the p, which also implies this. Lemma 3 Given (p n, k n : n ω) elements of P ω and (U n : n < ω) which are ω-covers of Q = n<ω Q0 (p n ) there exists (U m U m : m < ω) and (q n kn p n : n ω) such that n < ω m n [q n ] U m. Construct (qn m : n, m < ω) and (U n U n : n < ω) inductively. Given (qn m : n < ω) and (U n : n < m), we construct qn m+1 and U m U m so that 1. q m+1 n = p n for n m + 1, 2

3 2. q m+1 n 0 k n+m qm n for n m, and 3. [q m+1 n ] U m for n m. Then we let q n = m>n qm n be the fusion. We have that q n kn q n n = p n and [q n ] U m whenever m n. Remark. Obviously, the analogue of this Lemma for Q 1 is also true. Lemma 4 Suppose (p n, k n : n ω) are elements of P ω. Then there exists (q n kn p n : n ω) such that for n m, q n and q m are strongly disjoint, i.e., there are infinitely many k (D qn D qm ) with q n (k) q m (k). Given p 1, p 2 and n it is easy to find q 1 n p 1 and q 2 n p 2 which are strongly disjoint. A fusion argument produces a sequence (q n : n < ω) where all pairs have been considered and made strongly disjoint. Now we construct the Aronszajn tree of Silver conditions. Let U α for α < ω 1 list all ω sequences of countable families of open subsets of 2 ω. Make sure that each such sequence occurs as a U α for both α even and α-odd. We can construct a tree T ω <ω 1 and (p s P : s T ) which has the following properties. 1. T ω <ω 1 is a subtree, i.e., s t T implies s T. 2. T α = T ω α is countable for each α < ω s t T implies p t p s. 4. If s, t T are incomparable, then p s and p t are strongly disjoint (as in Lemma 4). 5. For any α < β < ω 1 and any s T α and n < ω there exists t T β with p t n p s. 6. Define Q 0 α = {Q 0 (p t ) : t T α } and Q 1 α = {Q 1 (p t ) : t T α }. 3

4 (a) For α an even ordinal, if U α = (U α n : n < ω) is a sequence of ω-covers of Q 0 α then there exists a sequence (U n U α n : n < ω) which is a γ-cover of Q 0 α {[p s ] : s T α+1 }. (b) For α odd, the analogous statement is true but with Q 1 α in place of Q 0 α. 7. Let D be the family of f ω ω such that for some s T the function f : ω D ps is the unique order preserving bijection. Then D is a dominating family, i.e., for all g ω ω there exists f D such that g(n) < f(n) for all n < ω. To construct T λ and p s for s T λ where λ is a countable limit ordinal, proceed as follows. For any s T <λ and N < ω choose a strictly increasing sequence λ n cofinal in λ with s = s 0 T λ0. By inductive hypothesis we can find s n T λn with p sn+1 N+n p sn. Take p t for t = n<ω s n to be the fusion of this sequence. Repeat countably many times, to take care of all s T <λ and N < ω. At successor stages for α even, check to see if U α is an ω-sequence of ω-covers of Q 0 α. If it is not, we need never worry about it since the set we are building will contain Q 0 α. If it is, let {x n : n < ω} = Q 0 α and let U n = {U U α n : {x i : i < n} U}. Let (p n, k n : n < ω) list all elements of {p s : s T α } ω with infinite repetitions and apply Lemma 3 followed by Lemma 4. From the resulting sequence we may find for each s T α and n < ω a distinct condition q n p s which we label p sˆ n. Obtaining the last condition (7), is easy (in fact, it seems hard to avoid), and is left to the reader. The γ sets are the sets: A 0 = s T Q 0 (p s ) and A 1 = s T Q 1 (p s ). 4

5 For x A 0 and y A 1 we note that there are infinitely many n with x(n) y(n). To see this note that if x Q 0 (p s ) and y Q 1 (p t ), and s and t are incomparable, then p s and p t are strongly disjoint. On the other hand, if they are comparable, for example, s t, then since p t p s, we have that D pt D ps. So for all but finitely many n D pt we will have that y(n) = 1 and x(n) = 0. Condition (7) gives us a continuous map from A 0 A 1 onto a dominating family D ω ω. Namely, if x 0 Q 0 (p s ) is identically zero on D ps and x 1 Q 1 (p s ) identically one on D ps, then D ps = {n : x 0 (n) x 1 (n)}. So the continuous map is φ : A 0 A 1 ω ω defined by φ(x, y) = f where f is the unique order preserving map from ω to the infinite set {n : x(n) y(n)}. Remark. Our result also shows that assuming CH there are Borel-cover γ-sets whose product is not Menger. To see this note the following: Lemma 5 Suppose p P, n < ω, and B 2 ω a Borel set. Then there exists q n p such that [q] B is clopen in [q]. Let F be the first n elements of D p and let φ : ω (D p \F ) be a bijection. For X ω let ψ X : (D p \ F ) 2 be the restriction of the characteristic function of φ(x). For each s 2 F define C s = {X [ω] ω : (p s ψ X ) B}. Since these are Borel sets by the Galvin-Prikry Theorem [1] there exists H [ω] ω such that for each s 2 F either [H] ω C s or [H] ω C s =. Let H 1 H be infinite such that H \ H 1 is also infinite. Let q = p (φ(h) {0}) (φ(h 1 ) {1}). Note that D q = F φ(h \ H 1 ). We claim that given any x, y [q] if x F = y F = s, then x B iff y B. Letting H x = φ 1 (x 1 (1)) we see that H 1 H x H and so H x is an infinite subset of H. Similarly for H y. By choice of H we have that H x C s iff H y C s and so the claim follows. Modify our construction in the open cover case as follows. Let B β for β < ω 1 list all Borel sets. At the end of each stage β apply Lemma 5 accross the level to make sure that [p s ] B β is relatively clopen for each s T β+1. 5

6 Let B α = (Bα n : n < ω) for α < ω 1 all ω sequences of countable families of Borel sets. We may assume that each element of B α has already been listed as a B β for some β < α. At successor stages for α even, check to see if B α = (Bn α : n < ω) is an ω-sequence of countable Borel ω-covers of Q 0 α. If it is not, we need never worry about it since the set we are building will contain Q 0 α. If it is, let {x n : n < ω} = Q 0 α and let B n = {U B α n : {x i : i < n} U}. Since the elements of each B n are relatively open in the [p s ] the rest of the argument is the same as in the open case. References [1] Galvin, Fred; Prikry, Karel; Borel sets and Ramsey s theorem. J. Symbolic Logic 38 (1973), [2] Galvin, Fred; Miller, Arnold W.; γ-sets and other singular sets of real numbers. Topology Appl. 17 (1984), no. 2, [3] Grigorieff, Serge; Combinatorics on ideals and forcing. Ann. Math. Logic 3 (1971), no. 4,