Non-trivial automorphisms from variants of small d

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1 Non-trivial automorphisms from variants of small d Fields October 24, 2012

2 Notation If A and B are subsets of N let denote the equivalence relation defined by A B if and only if A B is finite. Let [A] denote the equivalence class of A modulo. Then P(N)/[N] <ℵ 0 is the same as P(N)/. Notation If f is a function defined on the set A and X A then the notation f (X ) will be used to denote {f (x) x X }. An automorphism Φ of the Boolean algebra P(N)/[N] <ℵ 0 is called trivial if there is f : N N such that f (A) Φ([A]) for each A N.

3 Theorem (W. Rudin) If 2 ℵ 0 = ℵ 1 then there is a non-trivial automorphism of the Boolean algebra P(N)/[N] <ℵ 0. First argument: Let U and V be P-points generated by {U ξ } ξ ω1 and {V ξ } ξ ω1. Choose bijections ψ ξ : U ξ V ξ forming a (partial) coherent family. Define { ψ ξ (A) if A U ξ Ψ([A]) = ψ ξ (N \ A) otherwise. and note that Ψ is a well defined automorphism. How can it be made non-trivial? Ask Hausdorff.

4 Now use that there are 2 2ℵ 0 P-points assuming 2 ℵ 0 = ℵ 1. Second argument: Use the countable saturation of P(N)/[N] <ℵ 0 to inductively construct the automorphism. Use 2 ℵ 0 = ℵ 1 to diagonalize against all possible trivial automorphisms. Definition An automorphism of P(N)/[N] <ℵ 0 is called somewhere trivial if there is an infinite Z N and f : Z N such that f (A) Φ([A]) for each A Z. An automorphism that is not somewhere trivial is called nowhere trivial. The automorphism constructed by the second method can be made nowhere trivial.

5 Theorem (Shelah) It is consistent with set theory that all automorphisms of the Boolean algebra P(N)/[N] <ℵ 0 are trivial. Recall that OCA implies that all coherent families are trivial. This play a key role in the following: Theorem (Velickovic) OCA and MA ℵ1 implies that all automorphism of the Boolean algebra P(N)/[N] <ℵ 0 are trivial. On the other hand, the CH P-point argument can be extended to show: Theorem (Velickovic) It is consistent with MA ℵ1 that there is a non-trivial automorphism of P(N)/[N] <ℵ 0.

6 Just as in the P-point argument, the automorphism constructed by Velickovic is somewhere trivial. However: Theorem (Shelah S.) It is consistent with MA and 2 ℵ 0 > ℵ 1 that there is a nowhere trivial automorphism of P(N)/[N] <ℵ 0. Theorem In the model obtained by adding ℵ 2 Cohen reals to a model of CH there is a nowhere trivial automorphism of P(N)/[N] <ℵ 0. Question Are there nowhere trivial automorphisms of P(N)/[N] <ℵ 0 in the model obtained by adding ℵ 3 Cohen reals to a model of CH?

7 The original model of Shelah is obtained by a finite support iteration of oracle-cc partial orders and hence it shares many properties with Cohen real models; for example d = 2 ℵ 0. Velickovic s argument for getting all automorphisms trivial uses OCA and recall that OCA implies that b = ℵ 2. So one might ask if d > ℵ 1 is necessary for all automorphisms of P(N)/[N] <ℵ 0 to be trivial. However: Theorem (Farah Shelah) It is consistent with d = ℵ 1 that all automorphisms of P(N)/[N] <ℵ 0 are trivial.

8 On the other hand, except for the CH arguments, the other methods mentioned for getting nontrivial automorphisms Cohen model and consistency with MA all use methods that yield d > ℵ 1. So, is it consistent with d = ℵ 1 < 2 ℵ 0 that all automorphisms of P(N)/[N] <ℵ 0 are trivial?

9 Lemma There is a nontrivial automorphism of P(N)/[N] <ℵ 0 provided that there is a partition of N into finite sets {I n } n ω such that: 1 For each ξ ω 1 and n ω there is a Boolean subalgebra B ξ,n of P(I n ) and an automorphism Φ ξ,n of B ξ,n. 2 If ξ η then B ξ,n B η,n and Φ ξ,n = Φ η,n B ξ,n for all but finitely many n ω. 3 For any one-to-one F : N N there are ξ ω 1 and infinitely many n ω such that there is an atom a B ξ,n and j a such that F (j) / Φ ξ,n (a). 4 For any A N there is ξ ω 1 such that A I n B ξ,n for all but finitely many n.

10 Define Φ([A]) = lim ξ ω 1 [ ] Φ ξ,n (A I n ) n ω Why is this is well defined? If A B is finite there is α ω 1 such that for all ξ > α and for all but finitely many n the equation Φ ξ,n (A I n ) = Φ ξ,n (B I n ) holds. From Hypothesis 2 it then follows that if ξ and η are greater than α then Φ η,n (B I n ) n ω Φ ξ,n (A I n ) n ω and, hence, Φ([A]) is well defined. Since each Φ ξ,n is an automorphism it follows that Φ is an automorphism of P(N)/[N] <ℵ 0.

11 Why is Φ is nontrivial? Suppose that there is a one-to-one function F : N N such that F (A) Φ([A]) for all A N. Choose ξ ω 1 and an infinite Z N and atoms a n B ξ,n and j n a n such that F (j n ) / Φ ξ,n (a n ) for each n Z. Let W Z be an infinite subset such that for each n W, if F (j n ) I k and k n then k / W. Let A = n W a n. For any η ξ {F (j n ) j W } Φ η,n (a n ) {F (j n ) j W } Φ ξ,n (a n ) n W and, hence, F (A) / Φ([A]). n W

12 When are the hypotheses of Lemma 1 satisfied? Definition Given functions f and g from ω to ω let d f,g be the least cardinal of a family D n ω [f (n)]g(n) such that for every F n ω f (n) there is G D such that F (n) G(n) for all n. Given a filter F on ω define d f,g (F) to be the least cardinal of a family D n ω [f (n)]g(n) such that for every F n ω f (n) there is G D and X F such that F (n) G(n) for all n X. (So d f,g = d f,g (F) where F is the co-finite filter.) Note that d f,g > d and d f,g < d are both possible. Random and Laver reals provide the relevant models.

13 Lemma If there are functions f : N N and g : N N such that for all k N f (n) lim n g(n)k g(n) = and if d f!,g (F) = ℵ 1 for some filter generated by a -descending tower of length ω 1 then the hypotheses of Lemma 1 hold.

14 First step Given the hypothesis, it may be assumed that there are -descending sets {X ξ } ξ ω1 F and functions {G ξ } ξ ω1 n ω [f (n)!]g(n) such that for every F n ω f (n)! there is ξ ω 1 such that F (n) G ξ (n) for all but finitely many n X ξ. Why? Reindex so that for all ξ ω 1 there are cofinally many η ω 1 such that G ξ = G η.

15 Second step There are functions h ξ : N N and H ξ : N [f (n)!] h ξ(n) for ξ ω 1 such that 1 if ξ η ω 1 then 4h ξ h η g 2 if ξ η ω 1 then H ξ (n) H η (n) for all but finitely many n 3 if F n N f (n)! and F (n) G ξ(n) for all but finitely many n X ξ then also F (n) H ξ (n) for all but finitely many n X ξ. Why? The hypothesis that lim n f (n)/g(n)k g(n) = for all k makes it possible to choose h : N N such that lim n and lim n h(n) =. f (n) g(n)h(n)2 g(n)h(n) =

16 Third step Let H 0 (n) = G 0 (n). Given H ξ satisfying Conditions 2 and 3, define H ξ+1 (n) = H ξ (n) G ξ+1 (n) and note that H ξ+1 (n) H ξ (n) + G ξ+1 (n) h ξ (n) + g(n)/h(n) h ξ+1 (n). On the other hand, if η is a limit ordinal and H ξ satisfying the desired requirements have been chosen for ξ η, then a diagonalization argument yields H η such that H η (n) = h η (n) and H ξ (n) H η (n) for all but finitely many n for each ξ η.

17 Fourth step: The first sequence of finite algebras Now let {I n } n ω partition N such that I n = f (n) and let {θ j,n } j f (n)! enumerate all permutations of I n. Without loss of generality, f (n) is even for each n. Let A 0,n and A 1,n partition I n into two equal sized sets and let ϕ 0,n be an involution of I n interchanging A 0,n and A 1,n. For n X 0 let B 0,n = {, I n, A 0,n, A 1,n } and let Φ 0,n be the automorphism of B 0,n induced by ϕ 0,n. For n ω \ X 0 let B 0,n = P(I n ) and let Φ 0,n be the identity.

18 Fifth step: The ξ th sequence of finite algebras As the induction hypothesis assume Condition 2 of the first lemma holds and that, in addition, A ξ,n are the atoms of B ξ,n and that A ξ,n 2 4h ξ(n) provided that n X ξ for n X ξ there are involutions ϕ ξ,n of I n that induce Φ ξ,n. If B ξ,n, A ξ,n, ϕ ξ,n and Φ ξ,n have been defined for all ξ less than the limit ordinal η then a standard diagonalization yields B η,n, A η,n, ϕ ξ,n and Φ η,n.

19 Sixth step: The successor step Assume that B ξ,n, A ξ,n, ϕ ξ,n and Φ ξ,n have been defined. Let A ξ+1,n be the atoms generated by A ξ,n and {A n (j), ϕ ξ,n (A n (j))} j Hξ+1 (n). Then A ξ+1,n A ξ,n 4 h ξ+1(n) 2 g(n). Since f (n) > g(n)2 g(n) there must be some a n A ξ+1,n such that a n > g(n) for each n X ξ. Let ϕ : a n ϕ ξ,n (a n ) be any bijection such that for each n X ξ+1 and each j H ξ+1 (n) there is some k j,n a n such that ϕ(k j,n ) θ j,n (k j,n ).

20 Now for n X ξ+1 let A ξ+1,n = A ξ+1,n {{k j,n} j H ξ+1 (n)} and let ϕ ξ+1,n be defined by ϕ ξ,n (z) if z / a n ϕ ξ,n (a n ) ϕ ξ+1,n (z) = ϕ(z) if z a n ϕ 1 (z) if z ϕ ξ,n (a n ) and let Φ ξ+1,n be induced by ϕ ξ+1,n. Let B ξ+1,n be the Boolean algebra whose atoms are A ξ+1,n. On the other hand, for n ω \ X ξ+1 let B ξ+1,n = P(I n ) and let Φ ξ+1,n be induced by ϕ ξ,n Then B ξ,n B ξ+1,n and that Φ ξ+1,n B ξ,n = Φ ξ,n. Moreover, A ξ+1,n A ξ+1,n + h ξ+1(n) 2 3h ξ+1(n) + h ξ+1 (n) 2 4h ξ+1(n) for all but finitely many n X ξ+1 as required.

21 Why is this non-trivial? Let F : N N be one-to-one. If there are infinitely many n such that there is z n I n such that F (z n ) / I n then let ξ = 0 and, without loss of generality, it may be assumed that z n belongs to the atom A 0,n of B 0,n for infinitely many n. Since ϕ 0,n (A 0,n ) = A 1,n I n it is clear that F (z n ) / ϕ 0,n (A 0,n ). If F (I n ) I n for all but finitely many n then F I n = θ J(n),n for some J(n) also or all but finitely many n. There is some ξ ω 1 such that J(n) H ξ (n) for all but finitely many n X ξ. By construction, for all but finitely many n X ξ there is a singleton {k} A ξ,n such that ϕ ξ,n ({k}) = {ϕ ξ,n (k)} and ϕ ξ,n (k) θ J(n),n (k).

22 Why is the automorphism defined on all of P(N)/[N] <ℵ 0? This is the same argument using that 2 n n!. Corollary If d f!,g = ℵ 1 then there is a nontrivial automorphism of P(N)/[N] <ℵ 0. Let F be the co-finite filter. Corollary If there is an ℵ 1 -generated filter F such that d f!,g (F) = ℵ 1 d then there is a nontrivial automorphism of P(N)/[N] <ℵ 0. Let F be generated by {X ξ } ξ ω1. Use Rothberger s argument and ℵ 1 d to construct a -descending sequence {Y ξ } ξ ω1 all of whose terms are F positive and such that Y ξ X ξ. Let F be generated by {Y ξ } ξ ω1 and note that d f!,g (F ) = ℵ 1.

23 Recall that Farah Shelah showed it is consistent that d = ℵ 1 and all automorphisms of P(N)/[N] <ℵ 0 are trivial. Hence the assumption of ℵ 1 d in Corollary 2 is essential. (To be precise, one should check that u = ℵ 1 in their model.) Question Is ℵ 1 d essential in Corollary 2?

24 It is worth observing that the automorphism of Lemma 1 is trivial on some infinite sets indeed, if ξ ω 1 and X N are such that {x} belongs to some B ξ,n for each x X then Φ is trivial on P(X ). However, if T (Φ) is defined to be the ideal {X N Φ P(X ) is trivial } then T (Φ) is a small ideal in the sense that the quotient algebra P(N)/T (Φ) has large antichains, even modulo the ideal of finite sets in the terminology of Farah, the ideal T (Φ) is not ccc by fin. One should not, therefore, expect to get a nowhere trivial automorphism by these methods. Hence the following result:

25 Theorem Let P be the product of κ Sacks partial orders. Assuming 2 ℵ 0 = ℵ 1, there is an automorphism Θ of P(N/[N] <ℵ 0 such that 1 P Θ is a nowhere trivial automorphism of P(N)/[N] <ℵ 0 and note that, in particular, this means Θ has a natural definition in the generic extension by P.

CCC Forcing and Splitting Reals

CCC Forcing and Splitting Reals Boban Velickovic Equipe de Logique, Université de Paris 7 2 Place Jussieu, 75251 Paris, France Abstract Prikry asked if it is relatively consistent with the usual axioms