A Non-finitizability Result in Algebraic Logic

We show that if we expand the language of CAω by finitely many substitutions corresponding to bijective maps, then no quantifier free set of formulas containing only finitely many variables axiomatize RCAω. Let U be a set. Then B(U) is the boolean set algebra with unit U . Let τ : ω → ω, i, j < ω and X ⊆ U. Then sτ X = {s ∈ U : s ◦ τ ∈ X}, ci X = {t ∈ U : (∃s ∈ X)[t(j) = s(j) for all j 6= i]}, and dij = {s ∈ U : si = sj}. It is known [1] that the variety RCAω = SP{〈B(U), ci , dij〉i,j<ω : U is a set } of representable cylindric algebras of dimension ω is not axiomatizable by any set Σ of quantifier free formulas containing only finitely many variables. Here and elsewhere SP denotes the operation of forming subdirect products. Consider the following problem: Is there a finite set T of transformations on ω such that K = SP{〈B(U), ci , dij , sτ 〉τ∈T : U is a set } satisfies the following: There is a finite schema Σ of quantifier free formulas axiomatizing K and containing only finitely many variables, such that any 22 Tarek Sayed Ahmed equation valid in RCAω follows from Σ. (This corresponds to a weak form of completeness for the logic corresponding to RCAω.) The Problem an instance of what is referred to in the literature as The Finitization Problem, cf. [6], [7], [4], [9], [5], [8] and [2], admits a (partial) positive solution if we take T = {ssuc, spred} where suc(i) = i+ 1 and pred(0) = 0 and pred(i) = i − 1 for i > 0. Note that succ is not onto while pred is not one to one. We show that this (instance of the Finitization) problem presented herein is not solvable if T is a finite semigroup that contains only bijections. We use the techniques of Andréka in [1]. Theorem. Let T be a finite semigroup of bijective transformations on ω. Let K = SP〈B(U) : ci , dij , sτ 〉τ∈T,i,j<ω : U is a set }. Let Σ be any set of quantifier free formulas valid in K and containing only finitely many variables. Then Σ does not axiomatize the set of equations valid in RCAω, i.e there is an equation valid in RCAω that does not follow from Σ. Proof. Let k < ω we shall construct an algebra Ak with the following properties. (i) Ak / ∈ K. Moreover there is an equation valid in RCAω that does not hold in Ak. (ii) Every k-generated subalgebra of Ak is in K. Here k-generated means generated by k elements. Assume that |T| = n. Let m ≥ 2, m < ω and let 〈Ui : i < ω〉 be a system of disjoint sets each of cardinality m. Let U = ⋃ {Ui : i ∈ ω}, q ∈ ×i<ωUi = {s ∈ U : si ∈ Ui for all i < ω}, and let R = {z ∈ ×i<ωUi : |{i < ω : zi 6= qi}| < ω}. Let A′ be the subalgebra of 〈B(U), ci, dij , sτ 〉τ∈T generated by R. Then R is an atom of A′. Indeed for any two sequences s, z ∈ R there is a permutation σ : U → U of U taking s to z and fixing R, i.e σ ◦ s = z and R = {σ ◦ p : p ∈ R}. σ fixes all the elements generated by R because the operations are permutation invariant. Thus if a ∈ A′ and s ∈ a ∩ R then R ⊆ a showing that R is an atom of A′. We now split R into m atoms. A Non-finitizability Result in Algebraic Logic 23 Consider the group Zm of integers modulo m. (Any finite abelian group with m elements will do.) For each i < ω, let fi : Ui → Zm be a bijection such that fi(qi) = 0. For j < m define Rj = {z ∈ R : ∑ 〈fi(zi) : i < ω〉 = j}. Then {R0, . . . Rm−1} is a partition of R such that ciRj = ciR for all i < ω. Let A” be the subalgebra of 〈B(U), ci, dij , sτ 〉i,j<ω,τ∈T generated by R0, . . . Rm−1. Let R = {sσRj : σ ∈ T, j < m}. Let H = {a+ ∑ X : a ∈ A′, X ⊆ R}. Clearly H ⊆ A” and H is closed under the boolean operations. Also because transformations considered are bijections we have cisσRj = cisσR for all j < m and σ ∈ T. Thus H is closed under ci. Since sσ is additive both A′ and R are closed under sσ, we have that H is closed under sσ. Finally dij ∈ A′ ⊆ H. We have proved that H = A”. This implies that every element of R is an atom of A”. Now we define our final algebra Ak. For any σ ∈ T and j ≤ m, let Xσj be new elements different for different pairs (σ, j). Without loss of generality we assume that Id ∈ T. We write Xj for XIdj . Let H = {Xσj : σ ∈ T, j ≤ m}. Let K = AtA′ \ {sσR : σ ∈ T}. We define Ak = 〈A,+,−, ci, dij , sτ 〉i,j<ω,τ∈T to satisfy • 〈A,+, .〉 is a boolean algebra with atoms K ∪ H. • 〈A′,+,−, ci, dij〉i,j<ω is a subalgebra of the cylindric reduct of Ak (or RdCAAk). Also R = ∑ {Xj : j ≤ m} • All the non boolean operations (with the exception of diagonal elements) are additive. • ck i Xσj = ci sσR, sk τ Xσj = Xτ◦σj and sk σ |\ A′ = sσ |\ A′. 24 Tarek Sayed Ahmed We shall prove RdCAAk is not representable (this will imply that Ak / ∈ K), but its k generated subalgebras are representable as K algebras. For i, j < ω i 6= j, let sjx = ci(dij · x). Let