Geometrical representation theorems for cylindric-type algebras
aa r X i v : . [ m a t h . L O ] N ov GEOMETRICAL REPRESENTATION THEOREMS FORCYLINDRIC-TYPE ALGEBRAS
MOHAMED KHALED AND TAREK SAYED AHMED
Abstract.
In this paper, we give new proofs of the celebrated Andr´eka-Resek-Thompson representability results of certain axiomatized cylindric-like algebras. Such representability results provide completeness theo-rems for variants of first order logic, that can also be viewed as multi-modal logics. The proofs herein are combinatorial and we also use sometechniques from game theory. Introduction
Stone’s representation theorem for Boolean algebras can be formulatedin two, essentially equivalent ways. Every Boolean algebra is isomorphic toa field of sets, or the class of Boolean set algebras can be axiomatized bya finite set of equations. As is well known, Boolean algebras constitute thealgebraic counterpart of propositional logic. Stone’s representation theorem,on the other hand, is the algebraic equivalent of the completeness theoremfor propositional logic.Throughout, fix a finite ordinal n ≥
2. Cylindric algebras of dimension n were introduced by A. Tarski [3] as the algebraic counterpart of first orderlogic restricted to n -many variables. Unfortunately, not every abstract cylin-dric algebra is representable as a field of sets, where the extra non-Booleanoperations of cylindrifiers and diagonal elements are faithfully representedby projections and equalities. This is basically a reflection of the essentialincompleteness of the finite variable fragments of first order logics.In the present paper, we consider some variants of these cylindric algebrasthat were shown to have representation theorems. Thus, the correspondinglogics are variants of first order logic that have completeness theorems. Suchlogics can be also viewed as multi dimensional modal logics, c.f. [8] and [9].The following axioms are essentially due to D. Resek and R. Thompson [6]. Definition 1.1.
The class RC n is defined to be the class of all algebrasof the form A = h A, + , · , − , , , c i , d ij i i,j ∈ n that satisfy the axioms (Ax0) Mathematics Subject Classification.
Primary 03G15. Secondary 03G25, 03B45.
Key words and phrases. cylindric algebras, representability, games and networks. through (Ax7) below. Note that for each i, j ∈ n with i = j , we have s ii x def = x and s ij x def = c i ( x · d ij ).(Ax0) h A, + , · , − , , i is a Boolean algebra.(Ax1) c i i ∈ n .(Ax2) x ≤ c i x , for each i ∈ n .(Ax3) c i ( x · c i y ) = c i x · c i y , for each i ∈ n .(Ax4) d ii = 1, for each i ∈ n .(Ax5) d ik · d kj ≤ d ij = d ji = c k d ji , for each i, j, k ∈ n such that k = i, j .(Ax6) c i ( x · d ij ) · d ij ≤ x , for each i, j ∈ n such that i = j .(Ax7) s i m j m c k m · · · s i j c k x · Q { d lτ ( l ) : l ∈ K } ≤ c i x ,for each finite m ≥ i , . . . , i m ∈ n , j , . . . , j m ∈ n , k , . . . , k m ∈ n, i ∈ n such that k t +1 ([ i t /j t ] ◦ · · · ◦ [ i /j ]) ∗ K , t < m , where τ = [ i m /j m ] ◦ · · · ◦ [ i /j ] and K = { i , . . . , i m , k , . . . , k m } \ { i } . Remark 1.2.
Let ψ be any function. We denote its domain and its range by Dom ( ψ ) and Rng ( ψ ), respectively. For any X ⊆ Dom ( ψ ), by ψ ∗ X we meanthe set { ψ ( x ) : x ∈ X } . Let i, j ∈ n . Then, [ i/j ] denotes the transformationon n that fixes everything except that it sends i to j , while [ i, j ] is thetransformation that fixes everything except that [ i, j ]( i ) = j and [ i, j ]( j ) = i .The following axioms are due to R. Thompson and H. Andr´eka [7, 11]. Definition 1.3.
We define DC n = { A ∈ RC n : A | = (Ax8), (Ax9),(Ax10) } ,SC = { A ∈ DC : A | = (Ax11) } and SC n = { A ∈ DC n : A | = (Ax12) } if n ≥
3, where:(Ax8) c i c j x ≥ c j c i x · d jk , for each i, j, k ∈ n and k = i, j .(Ax9) d ij = c k ( d ik · d kj ), for each i, j, k ∈ n such that k = i, j .(Ax10) s ki s ij s jm s mk c k x = s km s mi s ij s jk c k x , for each i, j, k, m ∈ n , k = i, j, m and m = i, j .(Ax11) x · − d ≤ c c ( − d · s c x · s c x ).(Ax12) x ≤ c i c j ( s ij c j x · s ji c i x · Y k ∈ n,k = i,j s ki s ij s jk c k x ), for each i, j ∈ n .In [6], D. Resk and R. Thompson proved a representation theorem for theclass RC n . Despite the novelty of their proof, it was quite long and requiresa deep knowledge of the literature of algebraic logic. A simpler proof wasthen provided by H. Andr´eka (this proof can be found in [10, Theorem 9.4]).Andr´eka’s method could also suggest an elegant proof for the representationof the algebras of DC n [7]. EOMETRICAL REPRESENTATION THEOREMS 3
The representability of the DC n -algebras originally is due to R. Thomp-son, but his original proof was never published. In [11], H. Andr´eka provedthe representability of the algebras in SC n by reducing the problem to thecase of DC n and then applying [7]. Andr´eka’s representing structures werebuild using the step-by-step construction, which consists of treating defectsone by one and then taking a limit where the contradictions disappear.In the present paper, we provide new proofs for the representation the-orems of the classes defined above. We use games (and networks) as intro-duced to algebraic logic by R. Hirsch and I. Hodkinson, c.f. [12] and [13].Our proofs are relatively shorter than all the known proofs. We give directconstructions for all classes, even for SC n , unlike its original proof in [11].The translation from step-by-step techniques to games is not a purelymechanical process. This transfer can well involve some ingenuity, in ob-taining games that are transparent, intuitive and easy to grasp. The realadvantage of the game technique is that games do not only build represen-tations, when we know that such representations exist, but they also tell uswhen such representations exist, if we do not know a priori that they do.Now, we give the formal definition of the representing concrete algebras.We start with the following basic notions. For every i ∈ n and every twosequences f, g of length n , we write g ≡ i f if and only if g = f ( i/u ), forsome u , where, f ( i/u ) is the sequence which is like f except that it’s valueat i equals u . Let V be an arbitrary set of sequences of length n . For each i, j ∈ n and each X ⊆ V , we define C [ V ] i X = { f ∈ V : ( ∃ g ∈ X ) f ≡ i g } .D [ V ] ij = { f ∈ V : f ( i ) = f ( j ) } . When no confusion is likely, we omit the superscript [ V ]. Definition 1.4.
The class of all relativized cylindric set algebras of di-mension n , denoted by RCs n , is defined to be the class that consists of allsubalgebras of the (full) algebras of the form, P ( V ) def = hP ( V ) , ∪ , ∩ , \ , ∅ , V, C [ V ] i , D [ V ] ij i i,j ∈ n , where V is a non-empty set of sequences of length n and P ( V ) is the familyof all subsets of V . For every A ⊆ P ( V ), the set V is called the unit of A ,while the smallest set U that satisfies V ⊆ n U is called the base of A , where n U is the set of all sequences of length n whose range is subset of U . M. KHALED AND T. SAYED AHMED
Let f, g be two functions such that
Rng ( f ) ⊆ Dom ( g ), then g ◦ f denotesthe function whose domain is Dom ( f ) and g ◦ f ( x ) = g ( f ( x )). Definition 1.5.
The class DCs n of diagonalizable cylindric set algebras, ofdimension n , consists of all algebras A ∈ RCs n whose units V are diagonal-izable sets, i.e. for every f ∈ V and every i, j ∈ n we have f ◦ [ i/j ] ∈ V . Definition 1.6.
The class SCs n of locally squares cylindric set algebras,of dimension n , consists of all algebras A ∈ DCs n whose units V are per-mutable sets, i.e. for every f ∈ V and every i, j ∈ n we have f ◦ [ i, j ] ∈ V .In contrast with the literature, other notations for the classes RCs n ,DCs n and SCs n are Crs n , D n and G n , respectively. Let Λ n def = n n be the setof all transformations on n . Let Ω n def = { τ ∈ Λ n : | Rng ( τ ) | < n } , the setof all transformations on n that are not permutations. Let V be a set ofsequences of length n . The following characterization is well known and canbe verified easily using some simple facts of transformations. • V is diagonalizable if and only if f ◦ τ ∈ V , for every f ∈ V andevery τ ∈ Ω n . • V is diagonalizable and permutable if and only if f ◦ τ ∈ V , for every f ∈ V and every τ ∈ Λ n .For any class K of algebras, I K is the class that consists of all isomorphiccopies of the members of K. As we mentioned before, we aim to reprove thefollowing theorem.
Main Theorem 1.
Let n ≥ be a finite ordinal and let K ∈ { RC , DC , SC } .Then, K n = I Ks n . Preliminary lemmas
Recall the basic concepts of Boolean algebras with operators (BAO) fromthe literature, see e.g. [1]. For any B ∈ RC n , let At ( B ) be the set of allatoms in B .Note that RC n (similarly DC n and SC n ) is defined by positive equations,the negation does not appear in a non-Boolean axiom. Also note that thecylindrifications are defined to be normal and additive operators. Thus, by[1, Theorem 2.18], every algebra in RC n can be embedded into a completeand atomic algebra in RC n . Therefore, it is enough to prove that every com-plete and atomic algebra is representable. We need to prove some auxiliarylemmas that may be interesting in their own. EOMETRICAL REPRESENTATION THEOREMS 5
Lemma 2.1.
Let A ∈ RC n , let i ∈ n and let x, y ∈ At ( A ) . Then, x ≤ c i y ⇐⇒ c i x = c i y. Proof.
We prove the non-trivial direction only. The assumption x ≤ c i y and[3, Theorem 1.2.9] imply that c i x ≤ c i y . Moreover, x ≤ c i y = ⇒ x · c i y = 0= ⇒ c i ( x · c i y ) = 0 by [3, Theorem 1.2.1]= ⇒ c i x · c i y = 0 by axiom (Ax3)= ⇒ c i ( y · c i x ) = 0 by axiom (Ax3)= ⇒ y · c i x = 0 by [3, Theorem 1.2.1]= ⇒ y ≤ c i x by the fact that y ∈ At ( A )= ⇒ c i y ≤ c i x by [3, Theorem 1.2.9] . Note that all the cited theorems uses only the axioms (Ax0) - (Ax3). (cid:3)
To represent an atomic algebra A , we roughly represent each atom a ∈ A by a sequence f . Then, we show that A can be embedded into the full algebrawhose unit consists of all sequences representing atoms. The real challengenow is to arrange that the unit has the desired properties, by adding thesubstitutions f ◦ [ i/j ] or the transpositions f ◦ [ i, j ] whenever it is necessary.Such new sequences need to be associated to some atoms to keep the claimthat each atom is represented by some sequences (maybe more than one),and there are no irrelevant sequences. For example, the following Lemmadefines the substitutions of an atom, if any exists. Notation:
Let i, j ∈ n be such that i = j . Define t ii x def = x and t ij x def = c i x · d ij Lemma 2.2.
Suppose that A ∈ RC n is an atomic algebra. Let x ∈ At ( A ) and let i, j ∈ n . Then, t ij x = 0 = ⇒ t ij x ∈ At ( A ) . Moreover, if x ≤ d ij thenwe have t ij x = x .Proof. The statement is obvious if i = j , so we may assume that i = j .Suppose that t ij x = 0. Since A is atomic, one can find an atom y ∈ At ( A )such that y ≤ t ij x = c i x · d ij . Thus, by Lemma 2.1, we have c i x = c i y . Notealso that y ≤ d ij , so axiom (Ax6) implies t ij x = c i x · d ij = c i y · d ij = c i ( y · d ij ) · d ij = y. Therefore, t ij x is an atom in A as desired. The remaining part is obvious. (cid:3) M. KHALED AND T. SAYED AHMED
Definition 2.3.
Let A ∈ DC n and let τ = [ i /j ] ◦ · · · ◦ [ i m /j m ] ∈ Ω n .For each x ∈ A , we define τ A x def = t i m j m · · · t i j x . This is well defined by thefollowing Lemma. Lemma 2.4.
Let A ∈ DC n and let τ, σ ∈ Ω n . Then ( ∀ i ∈ n ) ( τ ( i ) = σ ( i )) = ⇒ ( ∀ x ∈ A ) ( τ A x = σ A x ) . Proof.
See [7] proof of Lemma 1 therein. Axiom (Ax10) is used here. Wenote that And´ekas proof of this lemma is long, but using fairly obviousresults on semigroups a much shorter proof can be provided. (cid:3)
Similarly, we need to define the transpositions of an atom, if A ∈ SC n .This can be done using the axioms (Ax11) and (Ax12). By Lemma 2.6 below,there is at least one choice (maybe many) to define these transpositions Definition 2.5.
Assume that A ∈ SC n . Let i, j ∈ n be such that i = j andlet x ∈ A . We define p ii x def = x and p ij x def = s ij c j x · s ji c i x · Y k ∈ n,k = i,j s ki s ij s jk c k x. If the product is empty (i.e. if n = 2) then it is defined to be 1. Lemma 2.6.
Let A ∈ SC n , let i, j ∈ n and let x ∈ At ( A ) . Then, p ij x = 0 .Proof. If i = j then the statement is trivial. So we assume that i = j . If n ≥ n = 2.If x ≤ − d then the desired follows by axiom (Ax11). If x ≤ d then p x = s c x · s c x = s c ( x · d ) · s c ( x · d )= c ( c ( x · d ) · d ) · c ( c ( x · d ) · d )= c ( x · d ) · c ( x · d ) by axioms (Ax2) and (Ax6)= c x · c x. Hence, x ≤ c x · c x = p x which implies that p x = 0 as desired. (cid:3) We will use the following lemma in the proceeding sections.
Lemma 2.7.
Suppose that A ∈ SC n is complete and atomic algebra. Let x, y, z ∈ At ( A ) be some atoms and let i, j, k ∈ n be such that k = i, j .(1) y ≤ s ki s ij s jk c k x and z = t ik t ji t kj x = ⇒ c k y = c k z .(2) y ≤ s ij c j x and z = t ji x = ⇒ c i z = c i y . EOMETRICAL REPRESENTATION THEOREMS 7
Proof.
Suppose that A ∈ SC n is complete and atomic algebra. Hence, by[3, Theorem 1.2.6 and Theorem 1.5.3], the operations c i ’s and s ij ’s are com-pletely additive.(1) Suppose that y ≤ s ki s ij s jk c k x . Thus, by the assumptions on A , thereare some atoms a , a , a ∈ At ( A ) such that a ≤ c k x, a ≤ c j a , a ≤ c i a , y ≤ c k a ,a ≤ d jk , a ≤ d ij a ≤ d ki . Inductively, Lemma 2.2 implies that a = t kj x, a = t ji a = t ji t kj x, a = t ik a = t ik t ji t kj x. Therefore, z = a and hence y ≤ c k z . The desired follows by Lemma 2.1.(2) Again, by assumptions, there is an atom a such that y ≤ c i a , a ≤ d ij and a ≤ c j x . Hence, by Lemma 2.2, we have a = t ji x = z . Therefore, y ≤ c i a ≤ c i z and we are done by Lemma 2.1. (cid:3) Networks and mosaics
Throughout, let K ∈ { RC , DC , SC } and let A ∈ K n be arbitrary butfixed. Suppose that A is complete and atomic. The networks and mosaicswe define here are approximations of the desired representation of A . Definition 3.1.
A pre-network is a pair N = ( N , N ), where N is a finite(possibly empty) set, and N : n N → At ( A ) is a partial map. We writenodes( N ) for N and edges( N ) for the domain of N . Also, we may write N for any of N , N , N .- We write ∅ for the pre-network ( ∅ , ∅ ).- For the pre-networks N and N ′ , we write N ⊆ N ′ iff nodes( N ) ⊆ nodes( N ′ ), edges( N ) ⊆ edges( N ′ ), and N ′ ( f ) = N ( f ) for all f ∈ edges( N ).- Let α be an ordinal. A sequence of pre-networks h N β : β ∈ α i is said tobe a chain if N γ ⊆ N β whenever γ ∈ β . Supposing that h N β : β ∈ α i is achain of pre-networks, define the pre-network N = S { N β : β ∈ α } withnodes( N ) = [ { nodes( N β ) : β ∈ α } , edges( N ) = [ { edges( N β ) : β ∈ α } and, for each f ∈ edges( N ), we let N ( f ) = N β ( f ), where β ∈ α is anyordinal with f ∈ edges( N β ). Definition 3.2.
Let N be a pre-network and let f, g ∈ edges( N ). A se-quence of edges h , . . . , h m is said to be a zigzag of length m from f to g in N if the following hold: M. KHALED AND T. SAYED AHMED • h = f and h m = g . • For each 0 ≤ t ≤ m , Rng ( f ) ∩ Rng ( g ) ⊆ Rng ( h t ). • For each 0 ≤ t < m , there is i t +1 ∈ n such that h t = h t +1 , h t ≡ i t +1 h t +1 and c i t +1 N ( h t ) = c i t +1 N ( h t +1 ) . Definition 3.3.
A pre-network N is said to be a network if it satisfies thefollowing conditions for each f, g ∈ edges( N ) and each i, j ∈ n :(a) (i) K = DC = ⇒ edges( N ) is diagonalizable.(ii) K = SC = ⇒ edges( N ) is diagonalizable and permutable.(b) N ( f ) ≤ d ij ⇐⇒ f ( i ) = f ( j ).(c) If 0 < | Rng ( f ) ∩ Rng ( g ) | < n then there is a zigzag from f to g in N . Lemma 3.4.
Let N be a network. Let f, g ∈ edges( N ) , i, j ∈ n and τ ∈ Ω n .The following are true:(1) If f ≡ i g then c i N ( f ) = c i N ( g ) . (2) If f ◦ [ i/j ] ∈ edges( N ) then N ( f ◦ [ i/j ]) = t ij N ( f ) .(3) Suppose that K ∈ { DC , SC } . Then, N ( f ◦ τ ) = τ A N ( f ) .Proof. Let
N, f, g, i, j, τ be as required.(1) Suppose that f ≡ i g . If f = g then we are done. Assume that f = g .So, it follows that 0 < | Rng ( f ) ∩ Rng ( g ) | < n . Thus, we can assumethat h , . . . , h m ∈ edges( N ) is a zigzag from f to g .We will define i , . . . , i m , j , . . . , j m , k , . . . , k m such that N ( g ) ≤ s i m j m c k m · · · s i j c k N ( f ) · Y l ∈ J d lτ ( l ) , where i , . . . , i, τ, J satisfy the conditions of axiom (Ax7). Hence, byaxiom (Ax7) again, N ( g ) ≤ c i N ( f ) and we will be done. We definethe i ’s as follows. For each 1 ≤ t ≤ m , let i t ∈ n be such that h t − ≡ i t h t . Let J = { i , . . . , i m } \ { i } and J + = J ∪ { i } . Note that | J + | > | J | . We will define j t and k t for 1 ≤ t ≤ m by induction on t such that by letting τ t = [ i t /j t ] ◦ · · · ◦ [ i /j ] The proof of this item is distilled from Andr´eka’s proof in [10, Theorem 9.4].
EOMETRICAL REPRESENTATION THEOREMS 9 we will have for all t < m that h ( l ) = h t +1 ( τ t +1 ( l )) for all l ∈ J,N ( h t +1 ) ≤ s i t +1 j t +1 c k t +1 N ( h t ) and k t +1 ∈ J + \ τ ∗ t J. Let t < m and assume that j t ′ , k t ′ have been define for all 1 ≤ t ′ ≤ t with the above properties. Now, we have two cases. Case 1:
Suppose that h t ( i t +1 ) ∈ Rng ( h t +1 ), say h t ( i t +1 ) = h t +1 ( j ),for some j ∈ n . Note that h t +1 = h t and h t ( j ) = h t +1 ( j ) = h t ( i t +1 ),then j = i t +1 . Hence, N ( h t +1 ) ≤ s i t +1 j N ( h t ). We let j t +1 def = j and k t +1 ∈ ( J + \ τ ∗ t J ) be arbitrary, note that ( J + \ τ ∗ t J ) = ∅ . Case 2:
Suppose that h t ( i t +1 ) Rng ( h t +1 ). Let j t +1 def = k t +1 def = i t +1 .Let l ∈ J be arbitrary. Then, by l = i , we have h t ( τ t ( l )) = h ( l ) ∈ Rng ( f ) ∩ Rng ( g ) ⊆ Rng ( h t +1 ) . Hence, i t +1 = τ t ( l ) as desired.It is not difficult to check that the above choices satisfy our re-quirements. Then N ( g ) ≤ s i m j m c k m · · · s i j c k N ( f ). Also, for any l ∈ J , g ( l ) = f ( l ) = h ( l ) = h m ( τ m ( l )) = g ( τ ( l )) . Hence N ( g ) ≤ d lτ ( l ) for all l ∈ J by condition (b) of networks.Therefore, N ( g ) ≤ s i m j m c k m · · · s i j c k N ( f ) · Q l ∈ J d lτ ( l ) ≤ c i N ( f ).(2) If i = j then we are done. So, we may suppose that i = j andwe also assume that f ◦ [ i/j ] ∈ edges( N ). Thus, by the above item, c i N ( f ) = c i N ( f ◦ [ i/j ]). Also N is a network, then N ( f ◦ [ i/j ]) ≤ d ij .Therefore, N ( f ◦ [ i/j ]) ≤ c i N ( f ) · d ij = t ij N ( f ).(3) Let τ ∈ Ω n and assume that τ = [ i /j ] ◦ · · ·◦ [ i m /j m ]. The statementfollows by an induction argument that uses item (2). (cid:3) An arbitrary sequence f (of length n ) is said to be a repetition free sequenceif and only if | Rng ( f ) | = n . Definition 3.5.
Let a ∈ At ( A ) and let f be any sequence of length n suchthat ( ∀ i, j ∈ n ) ( f ( i ) = f ( j ) ⇐⇒ a ≤ d ij ). The mosaic generated by f and a , in symbols M ( f, a ), is defined to be the pre-network N , wherenodes( N ) = Rng ( f ) and:(a) If K = RC then edges( N ) = { f } and N ( f ) = a .(b) If K = DC then edges( N ) = { f ◦ τ : τ ∈ Ω n } , N ( f ) = a and, for each τ ∈ Ω n , N ( f ◦ τ ) = τ A a . (c) If K = SC and | Rng ( f ) | < n then the mosaic is defined in the same wayas in item (b) above.(d) Suppose that K = SC and assume that f is a repetition free sequence.Then edges( N ) = { f ◦ τ : τ ∈ ∆ n } , and the labeling is defined asfollows. First, N ( f ) = a and for each τ ∈ Ω n , N ( f ◦ τ ) = τ A a . So,each non-repetition free sequence is labeled. It remains to label therepetition free sequences. Let g , g , . . . , g N ( N ≥ n ≥
2) besome enumeration of the repetition free sequences such that g = fg i = g j ◦ [ k, l ] for some j < i and k = lg i = g j if i = j Such enumeration is possible. Assume that N ( g j ) is defined for all j < i .Let j < i and k = l be such that g i = g j ◦ [ k, l ] (If there are severalsuch j, k, l then we just select one such triple). Now we choose anyatom b ≤ p kl N ( g j ) (such an atom exist by Lemma 2.6) and we define N ( g i ) = b .We check that the mosaic M ( f, a ) is well defined. The essential part followsfrom Lemma 2.4. Suppose that τ, σ ∈ Ω n are such that f ◦ τ = f ◦ σ . Fixan enumeration τ , . . . , τ m of the set { [ i/j ] : i, j ∈ n and f ( i ) = f ( j ) } . Now,let γ = τ ◦ · · · ◦ τ m . Hence, γ ◦ τ = γ ◦ σ and by Lemma 2.2 we musthave τ A ( γ A a ) = σ A ( γ A a ). But by the condition on the sequence f and byLemma 2.2, γ A a = a . Therefore, τ A a = σ A a . Lemma 3.6.
Let a ∈ At ( A ) and let f be any sequence of length n such that ( ∀ i, j ∈ n ) ( f ( i ) = f ( j ) ⇐⇒ a ≤ d ij ) . Then the mosaic M def = M ( f, a ) is actually a network.Proof. Suppose that K = G n and suppose that f is a repetition free se-quence. We prove the above statement for this case only, the other cases aresimilar so we omit their details. The construction of M guarantees that itsatisfies conditions (a) and (b) of networks. It remains to show that M alsosatisfies condition (c). First, we prove the following.(3.1) ( ∀ g ∈ edges( M )) ( ∀ i, j ∈ n ) M ( g ( i/g ( j ))) = t ij M ( g ) . Let i, j ∈ n and let g ∈ edges( M ) be such that i = j . If f = g then by thedefinition of M we have M ( g ( i/g ( j ))) = M ( f ( i/f ( j ))) = M ( f ◦ [ i/j ]) = t ij M ( f ) = t ij M ( g ) . EOMETRICAL REPRESENTATION THEOREMS 11
Suppose that there is a transformation τ ∈ Ω n such that g = f ◦ τ . Then, g ( i/g ( j )) = g ◦ [ i/j ] = f ◦ τ ◦ [ i/j ]. So, by the construction of M , we have M ( g ( i/g ( j ))) = ( τ ◦ [ i/j ]) A a = t ij τ A a = t ij M ( g ). Recall the enumeration g , g , . . . , g N of the repetition free sequences given in the definition of M .To finish proving (3.1), it remains to show that(3.2) ( ∀ m ∈ N + 1) M ( g m ( i/g m ( j ))) = t ij M ( g m ) . We do this by induction on m . If m = 0 then we are done as g = f . Let1 ≤ p ≤ N . Assume that (3.2) is true for all m < p . We will show that (3.2)is true for p , too. Let g p = g m ◦ [ k, l ] be such that m < p and suppose thatwe choose M ( g p ) ≤ p kl M ( g m ). Case 1: i = k, l and j = k or j = l .Assume first that j = l . Consider the edges: h = g m ◦ [ i/k ] , h = g m ◦ [ i/k ] ◦ [ k/l ] and h = g m ◦ [ i/k ] ◦ [ k/l ] ◦ [ l/i ] . Thus, h ≡ i g p because g p = g m ◦ [ k, l ]. Hence, g p ( l ) = h ( l ) = h ( i ), whichmeans that h = g p ( i/g p ( j )). By induction hypothesis, M ( h ) = t ik M ( g m ).None of the sequences h , h and h is repetition free, so we already showedthat M ( g p ( i/g p ( j ))) = M ( h ) = t li M ( h ) = t li t kl M ( h ) = t li t kl t ik M ( g m ) . By M ( g p ) ≤ p kl M ( g m ) we have M ( g p ) ≤ s ik s kl s li c i M ( g m ). Since in K n the so-called the Merry Go Round equation s ik s kl s li c i x = s il s lk s ki c i x is true,then we also have that M ( g p ) ≤ s il s lk s ki c i M ( g m ). Now, Lemma 2.7 impliesthat M ( g p ( i/g p ( j ))) ≤ c i M ( g p ). By condition (a) of the networks, we have M ( g p ( i/g p ( j ))) ≤ c i M ( g p ) · d ij = t ij M ( g p ). The case j = k is completelysimilar, except that we do not have to use the Merry Go Round equation. Case 2: i = k, l and j = k, l .Then [ i/l ] ◦ [ i/j ] = [ i/j ]. Hence, g p ( i/g p ( j )) = g p ◦ [ i/j ] = g p ◦ [ i/l ] ◦ [ i/j ].We can use the previous case to conclude that M ( g p ◦ [ i/l ]) = t il M ( g p ). Notethat g p ◦ [ i/l ] is not repetition free, so we know that M ( g p ( i/g p ( j ))) = M ( g p ◦ [ i/l ] ◦ [ i/j ]) = t ij t il M ( g p ) = t ij M ( g p ) , the last equality follows from Lemma 2.4. Case 3: i = k and j = l .Then g p ( i/g p ( j )) = g p ◦ [ i/j ] = g p ◦ [ k/l ] = g m ◦ [ l/k ]. Thus, by theinduction hypothesis, we have M ( g p ( i/g p ( j ))) = t lk M ( g m ). We also have M ( g p ) ≤ s ij c j M ( g m ) because of the fact that M ( g p ) ≤ p kl M ( g m ). Hence, by Lemma 2.7, M ( g p ( i/g p ( j ))) ≤ c i M ( g p ). Again, since M ( g p ( i/g p ( j ))) ≤ d ij , M ( g p ( i/g p ( j ))) = c i M ( g p ) · d ij = t ij M ( g p ). The case i = k and j = l is asabove. The case i = l is completely analogous.Therefore, (3.2) is true and thus we have shown that (3.1) is also true.We need also to show that the following holds:(3.3) ( ∀ τ ∈ Ω n ) there is a zigzag from f ◦ τ to f in M. Let τ ∈ Ω n . So, we can suppose that τ = [ i /j ] ◦ · · · ◦ [ i m /j m ] with thesmallest possible m . Consider the edges: h = f, h = h ◦ [ i /j ] , h = h ◦ [ i /j ] , . . . , h m = h m − ◦ [ i m /j m ] = f ◦ τ. By (3.1), it is easy to see that h , . . . , h m is a zigzag from f ◦ τ to f in M . Hence, (3.3) is proved. Now we are ready to show that the mosaic M satisfies condition (c) in Definition 3.3. Let g, h ∈ edges( M ) be such that0 < | Rng ( g ) ∩ Rng ( h ) | < n .(I) Suppose that none of g and h is repetition free. By (3.3) we have thefollowing:- There is a zigzag h , . . . , h m from g to f in M .- There is a zigzag w , . . . , w d from h to f in M .It is not hard to see that h , . . . , h m = f = w d , . . . , w is a zigzag from g to h in M . This true because of the facts Rng ( g ) ⊆ Rng ( f ) and Rng ( h ) ⊆ Rng ( f ).(II) Suppose that one of g and h is repetition free, say g . Thus, h mustnot be repetition free, by the assumption 0 < | Rng ( g ) ∩ Rng ( h ) | < n .Also, we can find an element u ∈ Rng ( g ) \ Rng ( h ). Let i ∈ n besuch that g ( i ) = u and let j ∈ n \ { i } (this j exists because n ≥ Rng ( g ◦ [ i/j ]) ∩ Rng ( h ) = Rng ( g ) ∩ Rng ( h ). Hence, by (I)above, there is a zigzag h , . . . , h m from g ◦ [ i/j ] to h in M . But (3.1)and Lemma 2.1 imply c i M ( g ) = c i M ( g ◦ [ i/j ]). Thus, g, h , . . . , h m isa zigzag from g to h in M .Therefore, the mosaic M is indeed a network. (cid:3) Games and representability
The games we use here are games of infinite lengths between two players ∀ and ∃ . These games are basically Banach-Mazur games (see [4, Problem43] and [2]) in disguise. EOMETRICAL REPRESENTATION THEOREMS 13
Definition 4.1.
Let α be an ordinal. We define a game, denoted by G α ( A ),with α rounds, in which the players ∀ and ∃ build a chain of pre-networks h N β : β ∈ α i as follows. In round 0, ∃ starts by letting N = ∅ . Suppose thatwe are in round β ∈ α and assume that each N λ , λ ∈ β , is a pre-network.If β is a limit ordinal then ∃ defines N β = S { N λ : λ ∈ β } . If β = γ + 1 is asuccessor ordinal then the players move as follows:(a) ∀ chooses an atom b ∈ At ( A ), ∃ must respond with a pre-network N β ⊇ N γ containing an edge g with N β ( g ) = b .(b) Alternatively, ∀ chooses an edge g ∈ edges( N γ ), an index i ∈ n and anatom b ∈ At ( A ) such that N γ ( g ) ≤ c i b . In this case, ∃ must respondwith a pre-network N β ⊇ N γ such that for some u ∈ nodes( N β ) we have g ( i/u ) ∈ edges( N β ) and N β ( g ( i/u )) = b . ∃ wins if each pre-network N β , β ∈ α , played during the game is actually anetwork. Otherwise, ∀ wins. There are no draws.What we have defined are the rules of the game. There are many differentmatches of the game that satisfy the rules. The idea of the game is that thecurrent network is refined in the current round of the game. ∀ can challenge ∃ to find a suitable refinement in any of the two ways he likes, and she musteither do or lose. Since the game then continues from there, her responsemust itself be refinable in any way, if she is not to lose. Proposition 4.2.
Let α be an ordinal. ∃ has a winning strategy in the game G α ( A ) .Proof. Let α be an ordinal and let β ∈ α . Clearly, ∃ always wins in theround β if β = 0 or β is a limit ordinal. So, we may suppose that β = γ + 1is a successor ordinal. We also may assume inductively that ∃ has managedto guarantee that N γ is a network. We consider the possible moves that ∀ can make.(a) Suppose that ∀ chooses an atom b ∈ At ( A ). If there is an edge in N γ with the required conditions, then ∃ lets N β = N γ . Otherwise, she picksbrand new nodes f , . . . , f n − such that( ∀ i, j ∈ n ) ( f i = f j ⇐⇒ b ≤ d ij ) . Let f = ( f , . . . , f n − ). She defines N β = N γ ∪ M ( f, a ), where M ( f, a )is the mosaic generated by f and a . Note that our construction guaran-tees that nodes( N γ ) ∩ nodes( M ( f, a )) = ∅ , so N β is well defined. Now,Lemma 3.6 guarantees that N β is a network. (b) Alternatively, suppose that ∀ chooses an edge f ∈ edges( N γ ), an index i ∈ n and an atom b ∈ At ( A ) such that N γ ( f ) ≤ c i b . Again, if there is anode u ∈ nodes( N γ ) such that f ( i/u ) ∈ edges( N γ ) and N γ ( f ( i/u )) = b ,then ∃ lets N β = N γ . Suppose that such u does not exist. ∃ ’s strategygoes as follows. Let a = N γ ( f ), so we have c i a = c i b by Lemma 2.1.Suppose that there is some j ∈ n \ { i } such that b ≤ d ij . Then b ≤ c i a · d ij and thus b = t ij a . If f ( i/f j ) ∈ edges( N γ ) then we have N γ ( f ( i/f j )) = t ij a = b , which contradicts the assumptions. So, we mayassume that f ( i/f j ) edges( N γ ) (hence, K n = Crs n because N γ is anetwork). ∃ defines N β = N γ ∪ M ( f ( i/f j ) , b ). Thus, N β is indeed welldefined and it is not hard to see that it is a network.Suppose that b d ij for every j ∈ n \ { i } . In this case, ∃ picks brandnew node u and then she lets N β = N γ ∪ M ( f ( i/u ) , b ). We need tocheck that N β is well defined. Let g ∈ edges( N γ ) ∩ edges( M ( f ( i/u ) , b )).Remember that u was a brand new node, so the existence of such g means that K ∈ { D , G } and that there are j ∈ n \ { i } and τ ∈ Ω n suchthat g = f ( i/u ) ◦ [ i/j ] ◦ τ . Hence, g = f ◦ [ i/j ] ◦ τ . By Lemma 3.4, itfollows that N γ ( g ) = τ A t ij a = τ A ( c i a · d ij ). Let M def = M ( f ( i/u ) , b ), by theconstruction of the mosaic M , we also have M ( g ) = τ A t ij b = τ A ( c i b · d ij ).It remains to see that N γ ( g ) = M ( g ), which is true because c i a = c i b .Thus, ∃ could manage to guarantee that N β is well defined. Now, weshow that N β is actually a network.Lemma 3.6 and the induction hypothesis guarantee that N β satisfiesconditions (a) and (b) of networks. For the same reasons, it is apparentthat N β satisfies condition (c) for any two edges that both lie in N γ orin the mosaic M . Let g ∈ edges( N γ ) and let h ∈ edges( M ) \ edges( N γ ).Suppose that 0 < | Rng ( g ) ∩ Rng ( h ) | < n . We need to find a zigzag from g to h in the pre-network N β .Note that u ∈ Rng ( h ). Let j ∈ n be such that h ( j ) = u and choosean index k ∈ n \ { j } . Let h ′ = h ◦ [ j/k ]. Thus, h ′ ∈ edges( N t ) and Rng ( h ) ∩ Rng ( g ) = Rng ( h ′ ) ∩ Rng ( g ). By the induction hypothesis,there is a zigzag w , . . . , w m from g to h ′ in N β . Remember the fact that N β ( h ′ ) = M ( h ′ ) = t jk M ( h ) = t jk N β ( h ). Hence, by Lemma 2.1, we have c j N β ( h ′ ) = c j N β ( h ). Therefore, w , . . . , w m , h is a zigzag from g to h in N β as required.Therefore, if ∃ plays according to the strategy above she can win any playof the game G α ( A ), regardless of what moves ∀ makes. (cid:3) EOMETRICAL REPRESENTATION THEOREMS 15
Roughly, ∃ ’s winning strategy goes as follows. In each step, she adds awhole mosaic, and she uses the axioms to label the elements of the mosaicsby relevant atoms. Now we will use this to show that the algebra A can berepresented into a full algebra whose unit is a union of mosaics. Proposition 4.3. A is representable, i.e. there is B ∈ Ks n such that A ∼ = B .Proof. Let α be an ordinal (large enough) and consider a play h N β : β ∈ α i of G α ( A ) in which ∃ plays as in Proposition 4.2, and ∀ plays every possiblemove at some stage of play. That means,(G 1) each atom a ∈ At ( A ) is played by ∀ in some round, and(G 2) for every b ∈ At ( A ), every β ∈ α , and each f ∈ edges( N β ) and every i ∈ n with N β ( f ) ≤ c i b , ∀ plays b , i , f in some round.Let U = S { nodes( N β ) : β ∈ α } and let V = S { edges( N β ) : β ∈ α } ⊆ n U .By condition (a) of Definition 3.3, we can guarantee that P ( V ) ∈ Ks n . Thus,it remains to show that A is embeddable into P ( V ). For this, we define thefollowing function. For each x ∈ A , letΨ( x ) = { f ∈ V : ∃ β ∈ α ( f ∈ edges( N β ) and N β ( f ) ≤ x ) } . It is not hard to see that Ψ is a Boolean homomorphism. Also, (G 1) aboveimplies that Ψ is one-to-one. We check cylindrifications and diagonals. Let x ∈ A . Then, for each f ∈ V , we have f ∈ Ψ( c i x ) ⇐⇒ ∃ β ∈ α ( f ∈ N β and N β ( f ) ≤ c i x ) ⇐⇒ ∃ g ∈ V ∃ β ∈ α ( f, g ∈ N β , f ≡ i g, N β ( g ) ≤ x ) ⇐⇒ ∃ g ∈ V ( f ≡ i g and g ∈ Ψ( x )) ⇐⇒ f ∈ C [ V ] i Ψ( x ) . The second ⇐⇒ follows by (G 2) and Lemma 3.4 (1). For the diagonals,let i, j ∈ n , then by the second condition of networksΨ( d ij ) = { f ∈ V : ∃ β ∈ α ( f ∈ edges( N β ) and N β ( f ) ≤ d ij ) } = { f ∈ V : f ( i ) = f ( j ) } = D [ V ] ij . Therefore, A is isomorphic to a subalgebra B ⊆ P ( V ) as desired. (cid:3) Proof of Main Theorem 1.
Let K ∈ { RC , DC , SC } . It is easy to see thatevery concrete algebra in Ks n satisfies the axioms defining K n . Conversely,let A ∈ K n , we need to show that A ∈ I Ks n . By [1, Theorem 2.18], A can be embedded into a complete and atomic A + ∈ K n . Hence, Proposition 4.3implies that A + is representable. Therefore, A is representable as well. (cid:3) Main Theorem 2.
Let n ≥ be a finite ordinal and let K ∈ { RC , DC , SC } .Then, every A ∈ K n is completely representable into an algebra B ∈ I Ks n .That means, A is embeddable into B via an embedding that preserves infinitesums and infinite products.Proof. See the proofs of Proposition 4.3 and Main Theorem 1. (cid:3) Other networks If n = 2 then one can easily see that (Ax7) follows from (Ax0)-(Ax6).Suppose that n ≥
3. H. Andr´eka and I. N´emeti showed that the class RCs n can not be characterized by finitely many equations, c.f. [5, Theorem 5.5.13]and [10, Theorem 9.3], thus in this case (Ax7) can not be replaced with afinite set of axioms. However, in [7] and [11], axiom (Ax7) was omitted fromthe characterizations of the classes DCs n and SCs n , and hence it was shownthat these classes are finitely axiomatizable.Although, axiom (Ax7) is essential in the proof presented in the previoussections, this axiom can be omitted from the definitions of DC n and SC n .It could be possible to give a syntactical proof for this fact, but we preferto introduced some modifications of the networks that would allow us toproceed without the need of (Ax7). Let DC − n and SC − n be the resultingclasses after deleting axiom (Ax7) from the definitions of DC n and SC n ,respectively. Suppose that K ∈ { DC − , SC − } and let A ∈ K n be completeand atomic. We show that A is representable. Definition 5.1.
A ‘modified’ network N is a pre-network that satisfies thefollowing conditions for each f ∈ edges( N ) and each i, j ∈ n .(a) edges( N ) is diagonalizable, and edges( N ) is permutable iff K = SC − .(b) N ( f ) ≤ d ij ⇐⇒ f ( i ) = f ( j ).(c) N ( f ◦ [ i/j ]) = t ij N ( f ).Let N be a modified network. Let i ∈ n and let f, g ∈ edges( N ) be suchthat f ≡ i g . We prove that c i N ( f ) = c i N ( g ). Let j ∈ n be such that j = i .Note that g ◦ [ i/j ] = f ◦ [ i/j ] ∈ edges( N ). By condition (c) of the abovedefinition and Lemma 2.1, it follows that c i N ( f ) = c i N ( f ◦ [ i/j ]) = c i N ( g ◦ [ i/j ]) = c i N ( g ) . EOMETRICAL REPRESENTATION THEOREMS 17
Hence, we have shown that N satisfies all the items of Lemma 3.4. Note thatwe did not use axiom (Ax7). Let α be an ordinal. We also define the modifiedgame G ′ α ( A ) to be as same as G α ( A ) except that the players now have tobuild modified networks. It is not hard to see that if ∃ plays with the samestrategy given in Proposition 4.2 then she will win any game regardless ofwhat moves ∀ can make. It is quite straightforward to see that the mosaicsin this case are also modified networks (see the proof of (3.1)). Now theproof of Proposition 4.3 works verbatim to show that A is representable.We note that the modified networks and the modified games can beused also to prove the representability of some variants of Pinter’s algebrasand the quasi polyadic algebras. For more details, one can see [14]. Gamesand networks were also used to reprove similar representability results forvariants of relation algebras [13, Theorem 7.5] and [15]. References [1] B. J´onsson and A. Tarski (1951). Boolean algebras with operators I. American Jour-nal of Mathematics 73, pp. 891–939.[2] J. C. Oxtoby (1957). The BanachMazur game and Banach category theorem, Con-tribution to the Theory of Games III. Annals of Mathematical Studies 39, Princeton,pp. 159163[3] L. Henkin, J. D. Monk and A. Tarski (1971). Cylindric Algebras I. Studies in logicand the foundation of mathematics, volume 64. North Holland.[4] R.D. Mauldin (1981). The Scottish Book: Mathematics from the Scottish Caf´e.Birkhauser-Verlag, Boston-Basel-Stuttgart.[5] L. Henkin, J. D. Monk and A. Tarski (1985). Cylindric Algebras II. Studies in logicand the foundation of mathematics, volume 115. North Holland.[6] D. Resek and R. J. Thompson (1988) Characterizing relativized cylindric algebras.In: H. Andr´eka, D. Monk and I. N´emeti, eds, (1988). Algebraic Logic. ColloquiaMathematica Societatis J´anos Bolyai, volume 54, pp. 519-538.[7] H. Andr´eka and R. J. Thompson (1988). A stone type representation theorem foralgebras of relations of higher rank. Transaction of the American MathematicalSociety, 309 (2), pp. 671–682.[8] M. Marx (1995). Algebraic relativization and arrow logic. Ph.D thesis. University ofAmsterdam, ILLC dissertation Series.[9] M. Marx and Y. Venema (1997). Multi-dimensional modal logic. Applied Logic Se-ries, volume 4. Kluwer Academic Publishers.[10] J. D. Monk (2000). An introduction to cylindric set algebras. Logic Journal of theIGPL, 8 (4), pp. 451-492.[11] H. Andr´eka (2001). A finite axiomatization of locally square cylindric-relativized setalgebras. Studia Scientiarum Mathematicarum Hungarica 38, pp. 1-11.[12] R. Hirsch and I. Hodkinson (1997). Step by step-building representations in algebraiclogic. Journal of Symbolic Logic, 62 (1), pp. 225-279.[13] R. Hirsch and I. Hodkinson (2002). Relation algebras by games. Studies in logic andthe foundation of mathematics, volume 147. Elsevier, North Holland.[14] M. Khaled and T. Sayed Ahmed (2013). Building relativized representations usinggames. Preprint, arXiv:1304.1404 [math.LO].[15] T. Aslan and M. Khaled (2018). Stone type representation theorems via games.Colloquium Mathematicum, in press.
Mohamed Khaled, Bahcesehir University, Faculty of Engineering andNatural Sciences, Istanbul, Turkey & Alfr´ed R´enyi Institute of Math-ematics, Hungarian Academy of Sciences, Budapest, Hungary.
E-mail address : [email protected] Tarek Sayed Ahmed, Department of Mathematics, Faculty of Science,Cairo University, Giza, Egypt.
E-mail address ::