Reasonable non–Radon–Nikodym idealss

arXiv:0806.4760v1 [math.LO] 29 Jun 2008 Reasonable non–Radon–Nikodym idealss Vladimir Kanovei∗ Vassily Lyubetsky† June 9, 2018 Abstract For any abelian Polish σ-compact group H there exist an Fσ ideal Z ⊆ P ( N ) and a Borel Z -approximate homomorphism f : H → H N which is not Z -approximable by a continuous true homomorphism g : H → H N . Introduction Let G, H be abelian Polish groups, and Z be an ideal over a countable set A. We consider H A as a product group. For s, t ∈ H A put ∆s, t = {a ∈ A : s(a) 6= t(a)} . Suppose that Z is an ideal over A. A map f : G → H A is a Z -approximate homomorphism iff ∆f (x)+f (y), f (x+y) ∈ Z for all x, y ∈ G. Thus it is required that the set of all a ∈ A such that fa (x) + fa (y) 6= fa (x + y) belongs to Z . Here fa : G → H is the a-th co-ordinate map of the map f : G → H A . And Z is a Radon–Nikodym ideal (for this pair of groups) iff for any measurable Z -approximate homomorphism f : G → H N there is a continuous exact homomorphism g : G → H N which Z -approximates f in the sense that ∆f (x), g(x) ∈ Z for all x ∈ G. Here the measurability condition can be understood as Baire measurability, or, if G is equipped with a σ-additive Borel measure, as measurability with respect to that measure. The idea of this (somewhat loose) concept is quite clear: the Radon–Nikodym ideals are those which allow us to approximate non-exact homomorphisms by true ones. This type of problems appears in different domains of mathematics. Closer to the context of this note, Velickovic [7] proved that any Bairemeasurable FIN-approximate Boolean-algebra automorphism f of P ( N ) (so that the symmetric differences between f (x) ∪ f (y) and f (x ∪ y) and between f ( N r x) and N r f (x) are finite for all x, y ⊆ N ) is FIN-approximable by ∗ † IITP, Moscow. Partially supported by RFFI grants 06-01-00608 and 07-01-00445. IITP, Moscow. Partially supported by RFFI grant 07-01-00445. 1 a true automorphism g induced by a bijection betveen two cofinite subsets of N . Kanovei and Reeken proved that any Baire measurable Q -approximate homomorphism f : R → R is Q -approximable by a homomorphism of the form f (x) = cx, c being a real constant. See also some results in [1, 4, 5]. The term “Radon–Nikodym ideal” was introduced by Farah [1, 2] in the context of Baire measurable Boolean algebra homomorphisms of P ( N ). Many known Borel ideals were demonstrated to be Radon–Nikodym, see [1, 2, 4, 5]. Suitable counterexamples, again in the context of Boolean algebra homomorphisms, were defined by Farah on the base of so called pathological submeasures. A different and, perhaps, more transparent counterexample, related to homomorphisms T → T N (where T = R / N ), is defined in [5] as a modification of an ideal introduced in [6]. The next theorem generalizes this result. Theorem 1. Suppose that H is an uncountable abelian Polish group. Then there is an analytic ideal Z over N that is not a Radon – Nikodym ideal for maps H → H N in the sense that there is a Borel and Z -approximate homomorphism f : H → H N not Z -approximable by a continuous homomorphism g : H → H N . If moreover H is σ-compact then Z can be chosen to be Fσ . Note that the theorem will not become stronger if we require g to be only Baire-measurable, or just measurable with respect to a certain Borel measure on H — because by the Pettis theorem any such a measurable group homomorphism must be continuous. The remainder of the note contains the proof of Theorem 1. It would be interesting to prove the theorem for non-abelian Polish groups. (The assumption that H is abelian is used in the proof of Lemma 7.) And it will be interesting to find non–Radon–Nikodym ideals for homomorphisms G → H N in the case when the Polish groups G and H are not necessarily equal. 1 Countable subgroup Let us fix a group H as in the theorem, that is, an uncountable abelian Polish group. By 0 we denote the neutral element, by ⊕ the group operation, by d a compatible complete separable distance (and we do not assume it to be invariant). The first step is to choose a certain countable subgroup D ⊆ H of “rational elements”. It is quite clear that there exists a countable dense subgroup D ⊆ H satisfying the following requirement of elementary equivalence type. (∗) Suppose that n ≥ 1, c1 , . . . , cn ∈ D , ε is a positive rational, Ui = {x ∈ H : d(x, ci ) ≤ ε}, and P (x1 , . . . , xn ) is a finite system of linear equations with integer coefficients, unknowns x1 , . . . , xn , and constants in D , of the form: where bi ∈ Z and r ∈ D . b1 x 1 ⊕ . . . ⊕ b n x n = r , 2 Suppose also that this system P has a solution hx1 , . . . , xn i in H such that xi ∈ Ui for all i. Then P has a solution in D as well. (That is, all xi belong to D ∩ Ui .) Let us fix such a subgroup D . 2 The index set Let rational ball mean any subset of H of the form {x ∈ H : d(c, x) < ε}, where c ∈ D (the center), and ε is a positive rational number. Definition 2. Let A, the index set, consist of all objects a of the following kind. Each a ∈ A consists of: − an open non-empty set U a $ H , − a partition U a = U1a ∪ · · · ∪ Una of U a onto a finite number n = na of pairwise disjoint non-empty rational balls Uia ⊆ H , and − a set of points ria ∈ Uia ∩ D such that, for all i, j = 1, 2, . . . , n: (1) either ria ⊕ rja is rka for some k, and (Uia ⊕ Uja ) ∩ U a ⊆ Uka , (2) or (Uia ⊕ Uja ) ∩ U a = ∅. Under the conditions of Definition 2, if 0 ∈ Uia then si = 0 : for take j = i. Lemma 3. A is an infinite (countable) set. Proof. For any ε > 0 there is a ∈ A such that U a a set of diameter ≤ ε: just take na = 1, r1a = 0 , and let U a = U1a be the 2ε -nbhd of 0 in H . The next lemma will be used below. Lemma 4. If y1 , . . . , yn ∈ H are pairwise distinct then there exists a ∈ A such that na = n and yi ∈ Uia for all i = 1, . . . , n. Proof. As the operation is continuous, we can pick pairwise disjoint rational balls B1 , . . . , Bn such that yi ∈ Bi for all i and the following holds: If 1 ≤ i, j ≤ n then either there exists k such that (Bi ⊕ Bj ) ∩ B ⊆ Bk , where B = B1 ∪ · · · ∪ Bn , or just (Bi ⊕ Bj ) ∩ B = ∅. Put Uia = Bi . To obtain a system of points ria required, let P (x1 , . . . , xn ) be the system of all equations of the form xi + xj = xk with unknowns xi , xj , xk , where 1 ≤ i, j, k ≤ n and in reality yi + yj = yk . It follows from the choice of D that this system has a solution hr1 , . . . , rn i such that ri ∈ Uia ∩ D for all i. In other words we have: ri + rj = rk whenever yi + yj = yk . Let ria = ri . This ends S the definition of a ∈ A as required. (An extra care to guarantee that U a = 1≤i≤n Uia is a proper subset of H is left to the reader.) 3 3 The ideal Let Z be the set of all sets X ⊆ A such that there is a finite set u ⊆ H satisfying the following: for any a ∈ X we have u 6⊆ U a . The idea of this ideal goes back to Solecki [6], where a certain ideal over the set Ω of all clopen sets U ⊆ 2 N of measure 12 (also a countable set) is considered. In our case the index set A is somewhat more complicated. Lemma 5. Z is an ideal containing all finite sets X ⊆ A, but A 6∈ Z . Proof. If a ∈ A then the singleton {a} belongs to Z . Indeed by definition U a is a non-empty subset of H . Therefore there is a point x ∈ H r U a . Then u = {x} witnesses A ∈ Z . To see that Z is closed under finite unions, suppose that finite sets u, v ⊆ H witness that resp. X, Y belong to Z . Then w = u ∪ v obviously witnesses that Z = X ∪ Y ∈ Z . Finally by Lemma 4 for any finite u = {x1 , ..., xn } ⊆ H there is an element a ∈ A such that u ⊆ U a . This implies that A itself does not belong to Z . Proposition 6. Z is an analytic ideal. If H is σ-compact then Z is Fσ . Proof. We claim that X ∈ Z iff there are a natural nS and a partition X = S a 1≤k≤n Xk such that for any k the set Xk ⊆ A satisfies a∈Xk U 6= H . Indeed suppose that X ∈ Z and this is witnessed by a finite set S u = {x1 , . . . , xn } ⊆ H , that is, u 6⊆ U a for all a ∈ X . It follows that X = 1≤k≤n Xk , where Xk = S a a {a ∈ X : xkS6∈ U }. Clearly xk S6∈ a∈Xk U . To prove the converse suppose a that X = 1≤k≤n S Xk ⊆ Aa and a∈Xk U 6= H for all k . Let us pick arbitrary points xk ∈ H r a∈Xk U for all k . Then u = {x1 , . . . , xn } witnesses X ∈ Z , as required. It easily follows that Z Sis analytic. Now suppose that H = ℓ∈ N Hℓ , where all sets HℓSare compact. Then the S inequality a∈Xk U a 6= H is equivalentS to ∃ ℓ (Hℓ 6⊆ a∈Xk U a ). And by the compactness, the non-inclusion Hℓ 6⊆ a∈Xk U a is equivalent to the following S statement: Hℓ 6⊆ a∈X ′ U a for every finite X ′ ⊆ Xk . Fix an enumeration A = {an }n∈ N . Put A ↾ m = {aj : j < m}. Using König’s lemma, we conclude that X ∈ Z iff there exist S natural ℓ, n such that for any m there S exists a partition X ∩ (A ↾ m) = k<n Xk , where for every k we have Hℓ 6⊆ a∈Xk U a . And this is a Fσ definition for Z . 4 The main result Here we prove Theorem 1. Define a Borel map f : H → H A as follows. Suppose that x ∈ H and a ∈ A, na = n. If x ∈ Uia , 1 ≤ i ≤ n, then put fa (x) = x ⊖ ria . (⊖ in the sense of the group H .) If x 6∈ U a then put simply fa (x) = 0 . Finally define f (x) = {fa (x)}a∈A . Clearly f is a Borel map. 4 The maps fa do not look like homomorphisms H → H . Nevertheless their combination surprisingly turns out to be an approximate homomorphism! Lemma 7. f : H → H A is a Borel and Z -approximate homomorphism. Proof. Let x, y ∈ H and z = x ⊕ y . Prove that the set Cxy = {a : fa (x) ⊕ fa (y) 6= fa (z)} belongs to Z . We assert that this is witnessed by the set u = {x, y, z}, that is, if a ∈ Cxy then at least one of the points x, y, z is not a point in U a . Or, equivalently, if a ∈ A and x, y, z belong to U a then fa (x) ⊕ fa (y) = fa (z). To prove this fact suppose that a ∈ A and x, y, z ∈ U a . By definition, a U = U1a ∪ · · · ∪ Una , where n = na and Uia are disjoint rational balls in H . We have x ∈ Uia , y ∈ Uja , z ∈ Uka , where 1 ≤ i, j, k ≤ n. Then by definition fa (x) = x ⊖ ria , fa (y) = y ⊖ rja , fa (z) = z ⊖ rka . Therefore fa (x) ⊕ fa (y) = x ⊕ y ⊖ (si ⊕ sj ). (Here we clearly use the assumption that the group is abelian.) We assert that ria ⊕ rja = rka — then obviously fa (x) ⊕ fa (y) = fa (z) by the above, and we are done. Note that z = x ⊕ y ∈ U a , hence (Uia ⊕ Uja ) ∩ U a 6= ∅. We conclude that (2) of Definition 2 fails. Therefore (1) holds, ria ⊕ rja = rka′ for some k′ and (Uia ⊕ Uja ) ∩ U a ⊆ Uka′ . But the set (Uia ⊕ Uja ) ∩ U a obviously contains z , and (Lemma ) z ∈ Uka . It follows that k′ = k , rka′ = rka , ria ⊕rja = rka , as required. Lemma 8. The approximate homomorphism f is not Z -approximable by a continuous homomorphism g : H → H A . Proof. Assume towards the contrary that g : H → H A is a continuous homomorphism which Z -approximates f . Thus if x ∈ H then the set ∆x = {a : fa (x) 6= ga (x)} belongs to Z , where, as usual, ga (x) = g(x)(a). Note that all of these projection maps ga : H → H are continuous group homomorphisms since such is g itself. Thus if x ∈ H then ∆x ∈ Z , and hence there is a finite set ux ⊆ D satisfying the following: if a ∈ A and ux ⊆ U a then a 6∈ ∆x , that is, fa (x) = ga (x). Put Xu = {x ∈ H : ∀ a ∈ A (u ⊆ U a =⇒ fa (x) = ga (x))} for every finite u ⊆ D . These sets S are Borel since so are maps f, g (and g even continuous). Moreover H = u⊆D finite Xu since every x ∈ H belongs to Xux . Thus at least one of the sets Xu is not meager, therefore, is comeager on a certain rational ball B ⊆ H . Fix u and B . By definition for comeager-many x ∈ B and all a ∈ A satisfying u ⊆ U a we have fa (x) = ga (x). 5 Arguing as in the proof of Lemma 4, we obtain an element a ∈ A satisfying the following properties: u ⊆ U a , U a ∩ B 6= ∅, but the set B r U a is non-empty and moreover is not dense in B . Fix such a. Thus there exists a non-empty rational ball B ′ ⊆ B that does not intersect U a . By definition fa (x) = 0 for all x ∈ B ′ , and hence ga (x) = 0 for comeager-many x ∈ B ′ by the choice of B . We conclude that ga (x) = 0 for all x ∈ B in general, because g is continuous. Now, let na = n. Then U a = U1a ∪ · · · ∪ Una . Recall that the intersection B ∩ U a of two open sets is non-empty by the choice of a. It follows that there exists an index i, 1 ≤ i ≤ n, and a non-empty rational ball B ′′ ⊆ B ∩ Uia . Then by definition fa (x) = x ⊖ r for all x ∈ B ′′ , where r = ria . Therefore ga (x) = x ⊖ r for comeager-many x ∈ B ′′ , and then ga (x) = x ⊖ r for all x ∈ B ′′ since g is continuous. To conclude, ga , a continuous group homomorphism, is constant 0 on a nonempty open set B ′ , and is bijective on another non-empty open set B ′′ . But (Lemma ) this cannot be the case. Lemmas 7 and 8 complete the proof of Theorem 1. References [1] I. Farah, Approximate homomorphisms. II: Group homomorphisms, Combinatorica, 2000, 20, No.1, pp. 47-60. [2] I. Farah, Analytic quotients. Theory of liftings for quotients over analytic ideals on the integers, Mem. Am. Math. Soc., 2000, 702, 171 p. [3] V. Kanovei, M. Reeken, On Baire measurable homomorphisms of quotients of the additive group of the reals, Math. Log. Q., 2000, 46, No.3, pp. 377–384. [4] V. Kanovei, M. Reeken, On Ulam stability of the real line, Abe, Jair Minoro (ed.) et al., Unsolved problems on mathematics for the 21st century. A tribute to Kiyoshi Iséki’s 80th birthday. Amsterdam: IOS Press, 2001, pp. 169-181. [5] V. Kanovei, M. Reeken, On Ulam’s problem of stability of non-exact homomorphisms, Grigorchuk, R. I. 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