FFoundation ranks and supersimplicity
Santiago Cárdenas-Martín, Rafel FarréVersion 2020-04-11
Abstract
We introduce a new foundation rank based in the relation of divid-ing between partial types. We call DU to this rank. We also introducea new way to define the D rank over formulas as a foundation rank.In this way, SU , DU and D are foundation ranks based in the relationof dividing. We study the properties and the relations between theseranks.Next, we discuss the possible definitions of a supersimple type.This is a concept that it is not clear in the previous literature. Inthis paper we give solid arguments to set up a concrete definition ofthis concept and its properties. We also see that DU characterizessupersimplicity, while D not. We denote by L a language and T a complete theory. We denote by C amonster model of T , that is a κ -saturated and strongly κ -homogeneous modelfor a cardinal κ large enough. Models M, N, . . . are considered elementarysubstructures of C with cardinal less than κ and every set of parameters A, B, . . . is considered as a subset of C with cardinal less than κ .We denote by a, b, . . . tuples of elements of the monster model, possiblyinfinite (of length less than κ ). We often use these tuples as ordinary setsregardless of their order. We often omit union symbols for sets of parameters,for example we write ABc to mean A ∪ B ∪ c . Given a sequence of sets ( A i : i ∈ α ) we use A
Let R be a binary relation defined in a set or class of math-ematical objects. The foundation rank of R is the mapping r assigning toevery element a of the domain of R an ordinal number or ∞ according to thefollowing rules:1. r ( a ) ≥ .2. r ( a ) ≥ α + 1 if and only if there exists b such that aRb and r ( b ) ≥ α .3. r ( a ) ≥ α with α a limit ordinal, if and only if r ( a ) ≥ β for all β < α .One defines r ( a ) as the supremum of all α such that r ( a ) ≥ α . If suchsupremum does not exist we set r ( a ) = ∞ . Now, we define DU and DU f and we will check that really DU does notdepend of the set of parameters. We denote provisionally by DU ( p, A ) the DU rank of the pair ( p, A ) . Definition 2.2. DU and DU f are the foundation ranks of the followingrelations R d and R f : • ( p ( x ) , A ) R d ( q ( x ) , B ) if and only if p ( x ) ⊆ q ( x ) and q divides over A • ( p ( x ) , A ) R f ( q ( x ) , B ) if and only if p ( x ) ⊆ q ( x ) and q forks over A where p is a partial type over A and q is a partial type over B . Remark 2.3.
It is immediate to verify by induction that both ranks areinvariant under conjugation (automorphism).
Lemma 2.4.
Let p ( x ) be a partial type dividing over A . Let B ⊇ A . Then,there exists f ∈ Aut ( C /A ) such that p f divides over B . Proof.
Let p ( x ) = q ( x, a ) for some q ( x, y ) without parameters and a ⊆ A .For λ big enough there exist a set { a i : i ∈ λ } such that a i ≡ A a for any i ∈ λ and (cid:83) i ∈ λ q ( x, a i ) is k -inconsistent. So, we can choose an infinite subsetall having the same type over B , witnessing division over B . Proposition 2.5.
The rank DU does not depend on the set of parameters A .That is, if p ( x ) is a partial type with parameters in A ∩ B then DU ( p, A ) = D ( p, B ) . So, from now on we will use the notation DU ( p ) .Proof. It suffices to prove that given p be a partial type over A and A (cid:48) ⊇ A then DU ( p, A ) = DU ( p, A (cid:48) ) . Obviously DU ( p, A ) ≥ DU ( p, A (cid:48) ) . For theproof of DU ( p, A ) ≤ DU ( p, A (cid:48) ) , we show, by induction on α , DU ( p, A ) ≥ α implies DU ( p, A (cid:48) ) ≥ α .If DU ( p, A ) ≥ α + 1 then there exists q ⊇ p over B such that q dividesover A and DU ( q, B ) ≥ α . By the previous lemma, there exists an A -automorphism f such that q f divides over A (cid:48) . Then, DU ( q f , B f ) ≥ α . Bythe induction hypothesis, DU ( q f , A (cid:48) B f ) ≥ α . As p ⊆ q f and q f divides over A (cid:48) , DU ( p, A (cid:48) ) ≥ α + 1 . DU -rank We begin by setting some basic properties of DU . From the first property,it follows that two equivalent partial types have identical DU -rank. So, the DU -rank of a type-definable set makes sense. Remark 3.1.
Let p ( x ) , q ( x ) be partial types.1. If p (cid:96) q then DU ( p ) ≤ DU ( q ) .2. DU ( p ∨ q ) = max( DU ( p ) , DU ( q )) .3. DU ( p ) = 0 if and only if p is algebraic.4. Two type-definable sets with a definable bijection between them have thesame DU -rank.Proof. We Assume that p and q are over the same set of parameters A . . We prove DU ( p ) ≥ α implies DU ( q ) ≥ α by induction on α . Assume DU ( p ) ≥ α + 1 . Then, there exists p ⊇ p such that p divides over A and DU ( p ) ≥ α . Now p ∪ q extends q , divides over A , and by the inductivehypothesis, DU ( p ∪ q ) ≥ α . Therefore, DU ( q ) ≥ α + 1 . . By the previous point, DU ( p ∨ q ) ≥ max( DU ( p ) , DU ( q )) . The otherinequality is done by induction on α . Assume DU ( p ∨ q ) ≥ α + 1 . There DU -RANK exists r ( x ) ⊇ p ( x ) ∨ q ( x ) such that r divides over A and DU ( r ) ≥ α . Byinductive hypothesis, as r ≡ ( p ∪ r ) ∨ ( q ∪ r ) , DU ( p ∪ r ) ≥ α or DU ( q ∧ r ) ≥ α .So, DU ( p ) ≥ α + 1 or DU ( q ) ≥ α + 1 . . DU ( p ) ≥ iff p has some extension dividing over A iff p is non-algebraic. . Let p ( x ) , q ( y ) be partial types and let f : p ( C ) → q ( C ) be a definable bijection. We assume p, q are over A and f is defined over A . We prove by induction that DU ( p ( C )) ≥ α implies DU ( q ( C )) ≥ α . If DU ( p ( C )) ≥ α + 1 there is some p (cid:48) ( x ) ⊇ p ( x ) such that p (cid:48) divides over A and DU ( p (cid:48) ( C )) ≥ α . Then f ( p (cid:48) ( C )) is type-definable and, by inductive hypothe-sis, DU ( f ( p (cid:48) ( C ))) ≥ α . It is not difficult to prove that if q (cid:48) ( y ) type-defines f ( p (cid:48) ( C )) then q ( y ) divides over A .We are going to see some equivalences for DU : Proposition 3.2.
Let p ( x ) be a partial type over a set of parameters A and α an ordinal. Denote µ = (cid:0) | T | + | A | (cid:1) + . The following are equivalent:1. DU ( p ) ≥ α + 1 .2. There are ψ ( x, y ) ∈ L and a countable sequence ( a i : i < ω ) such that(a) ( a i : i < ω ) is A -indiscernible.(b) { ψ ( x, a i ) : i < ω } is inconsistent.(c) For every i < ω , we have DU ( p ( x ) ∪ { ψ ( x, a i ) } ) ≥ α .3. There are ψ ( x, y ) ∈ L and a number k ≥ such that for every cardinal λ , there is a sequence ( a i : i < λ ) such that(a) { ψ ( x, a i ) : i < λ } is k -inconsistent.(b) For every i < λ , we have DU ( p ( x ) ∪ { ψ ( x, a i ) } ) ≥ α .4. There are ψ ( x, y ) ∈ L , a number k ≥ and a sequence ( a i : i < µ ) such that(a) { ψ ( x, a i ) : i < µ } is k -inconsistent.(b) For every i < µ , we have DU ( p ( x ) ∪ { ψ ( x, a i ) } ) ≥ α .5. There are a partial type p (cid:48) ( x, y ) over ∅ with | y | ≤ | A | + | T | , a number k ≥ and a sequence ( a i : i < µ ) such that(a) Any set of k types in ( p (cid:48) ( x, a i ) : i ∈ µ ) are inconsistent.(b) p (cid:48) ( x, a i ) (cid:96) p ( x ) for each i < µ . (c) DU ( p (cid:48) ( x, a i )) ≥ α for each i < µ .6. There are a partial type p (cid:48) ( x, y ) over ∅ and a sequence ( a i : i < ω ) suchthat(a) ( a i : i < ω ) is A -indiscernible.(b) (cid:83) i ∈ ω p (cid:48) ( x, a i ) is inconsistent.(c) p (cid:48) ( x, a i ) (cid:96) p ( x ) for each i < ω .(d) DU ( p (cid:48) ( x, a i )) ≥ α for each i < ω .7. There are a partial type p (cid:48) ( x, y ) over the same set of parameters A anda sequence ( a i : i < ω ) such that(a) ( a i : i < ω ) is A -indiscernible.(b) (cid:83) i ∈ ω p (cid:48) ( x, a i ) is inconsistent.(c) p (cid:48) ( x, a i ) (cid:96) p ( x ) for each i < ω .(d) DU ( p (cid:48) ( x, a i )) ≥ α for each i < ω .Proof. ⇒ . If DU ( p ) ≥ α + 1 there exist q ( x ) extending p ( x ) , dividing over A with DU ( q ) ≥ α . Let ψ ( x, a ) ∈ q dividing over A . So, there exist a sequence ( a i : i < ω ) indiscernible over A with a = a such that { ψ ( x, a i ) : i < ω } is inconsistent. By point in Remark 3.1, DU ( p ( x ) ∪ ψ ( x, a )) ≥ α . Byconjugation, conditions ( c ) is satisfied. ⇒ . Extend the indiscernible sequence to an indiscernible sequence of length λ . This sequence satisfies the required conditions. ⇒ . Immediate. ⇒ . Let p ( x ) = p ( x, a ) , where p ( x, y ) is without parameters and a enumer-ates A (we assume the variables y in p ( x, y ) and ψ ( x, y ) are the same). Then p (cid:48) ( x, y ) = p ( x, y ) ∪ { ψ ( x, y ) } and ( b i = aa i : i < µ ) satisfy . ⇒ . Choose an infinite subsequence of ( a i : i ∈ µ ) with all elements havingthe same type over A . Then apply the standard lemma (Lemma 7.1.1 inTent, Ziegler[10]) to obtain a sequence ( a (cid:48) i : i ∈ ω ) indiscernible over A and satisfying the Ehrenfeucht-Mostowski type of the subsequence. Then ( a (cid:48) i : i ∈ ω ) satisfy the conditions of . ⇒ . Immediate. ⇒ . The closure under conjunction of p (cid:48) ( x, a ) divides over A , extends p ( x ) and has DU -rank at least α . Therefore DU ( p ) ≥ α + 1 . DU -RANK Now we want to see that DU may be characterized by the existence ofcertain trees of formula with certain properties. Definition 3.3.
We define recursively a rooted tree T α,λ for every ordinal α and cardinal λ :1. T ,λ is a tree with a unique node.2. For an ordinal α + 1 , we take λ disjoint copies of T α,λ and add a newnode related with all nodes, that is, a new root.3. For a limit ordinal α , we take a disjoint union of all trees { T β,λ : β ∈ α } and add a new node related with all nodes, that is, a new root. The nodeadded in this step will be called a limit node of the tree. Remark 3.4.
It is immediate that every T α,λ is a tree. That is, the binary re-lation R defined in the tree is a strict partial order (irreflexive and transitive)and for each node t , the set { s : sRt } is well-ordered. We use standard tree terminology: we say that a node s is a child of anode r (or r is the parent of s ) if rRs and there are no nodes t with sRt and tRs . The root of the tree will be the minimum. An end-node is a nodewithout children. We will denote by F α,λ the set of parent nodes in T α,λ which are not limit. P α,λ will denote the set of nodes of T α,λ which are achild of a non-limit.Next lemma characterizes the value of DU using the trees defined above.Compare to the definition of the rank DD in Cárdenas, Farré[2]. Lemma 3.5.
Let p ( x ) be a partial type over A in T , α and ordinal and µ = (cid:0) | T | + | A | (cid:1) + . The following are equivalent:1. DU ( p ) ≥ α .2. There is a sequence of formulas ( ϕ s ( x, y n ) : s ∈ F α,µ ) , a sequence ofnumbers ( k s : s ∈ F α,µ ) and a sequence of parameters ( a s : s ∈ P α,µ ) such that(a) For every s ∈ F α,µ , the set of formulas { ϕ s ( x, a t ) : t is a child of s } is k s -inconsistent.(b) For every end-node s , the set of formulas p ( x ) ∪{ ϕ s ( x, a r ) : tRs, r a child of t } is consistent.Proof. It is easily proved by induction using the equivalence in Proposi-tion 3.2. Proposition 3.6.
Let p ( x ) be a partial type over A . Then, there exists a setof parameters B ⊆ A such that | B | ≤ | T | | DU ( p ) | and DU ( p (cid:22) B ) = DU ( p ) .Proof. We may assume DU ( p ) < ∞ and fix α = DU ( p )+1 . For every partialtype q ( x ) over A consider the type Σ q,ϕ,k in the variables ( y s : s ∈ P α,µ ) expressing the conditions (a) and (b) of Lemma 3.5. Here ϕ = ( ϕ s ( x, y n ) : s ∈ F α,µ ) and k = ( k s : s ∈ F α,µ ) denote sequences of formulas and numbersand µ = (cid:0) | T | + | A | (cid:1) + . That is, DD ( q ) < α if and only if for every ϕ and k , Σ q,ϕ,k is inconsistent.As DD ( p ) < α , for every ϕ and k , by compactness, there is some finite A ϕ,k ⊆ A such that Σ p (cid:22) A ϕ,k ,ϕ,k is inconsistent. Taking B = (cid:83) ϕ,k A ϕ,k weget Σ p (cid:22) B,ϕ,k is inconsistent for every ϕ, k . We are using that p ⊆ q implies Σ p,ϕ,k ⊆ Σ q,ϕ,k . Proposition 3.7.
Let p ( x ) be a partial type over A such that DU ( p ) = ∞ .Then there exists a partial type q ( x ) such that p ⊆ q , q divides over A and DU ( q ) = ∞ .Proof. For each α , there is a p α such that p α (cid:96) ϕ α with ϕ α dividing over A , p ⊆ p α and DU ( p α ) ≥ α . We may assume all formulas ϕ α are conjugateover A . This is true because there are only boundedly many formulas andboundedly many types over A .By conjugation over A we may assume all p α contain a formula thatdivides over A . So, q = (cid:84) p α is a partial type dividing over A . Then, q ( x ) isa dividing extension of p ( x ) with DU ( q ) = ∞ . DU and other ranks The SU -rank has traditionally been defined as the foundation rank of theforking relation. In the same way, we can define the rank SU d using dividinginstead of forking. Namely, SU d will be the foundation rank of the relationof dividing extension between complete types. To avoid confusion we willwrite SU f to refer to the ordinary rank SU for forking. Obviously SU f ( p ) ≥ SU d ( p ) .Now, we are going to see that we can define the known D -rank (for theirdefinitions and properties, see for example, Casanovas[4]) for formulas, as afoundation rank. More precisely, as the foundation rank of the relation ofdividing between pairs ( ϕ, A ) of formulas and set of parameters satisfying dom ( ϕ ) ⊆ A . Using Lemma 2.4 one can easily show that D does not dependon the set of parameters. We can define similarly D f using the forking DU AND OTHER RANKS relation instead of dividing. Later, we will check (Proposition 4.5) that bothranks are the same and therefore D f does no depend on the set of parameters. Definition 4.1. D , D f , SU d and SU f are the foundation ranks of the fol-lowing relations R dd , R df , R sd and R sf : • ( ϕ ( x ) , A ) R dd ( ψ ( x ) , B ) if and only if | = ψ → ϕ and ψ divides over A • ( ϕ ( x ) , A ) R df ( ψ ( x ) , B ) if and only if | = ψ → ϕ and ψ forks over A • p ( x ) R sd q ( x ) if and only if q is a dividing extension of p • p ( x ) R sf q ( x ) if and only if q is a forking extension of p where ϕ is a formula over A , ψ is a formula over B and p and q are completetypes. It is not difficult to verify that this definition of D for formulas coincideswith the traditional definition. For indeed, we can proceed as in Proposi-tion 3.2.Next two remarks state well known properties of SU f (and therefore, of SU in the context of simple theories where SU d and SU f coincide). We cancheck that SU d satisfy them in any theory. The proofs are similar to theproofs for DU in Remark 3.1 and Proposition 3.7. Remark 4.2.
Let p ( x ) ∈ S ( A ) and q ( x ) ∈ S ( B ) . The rank SU d satisfies:1. If q ⊆ p then SU d ( p ) ≤ SU d ( q ) .2. For every r completation of p ∨ q , SU d ( r ) ≤ max( SU d ( p ) , SU d ( q )) .3. SU d ( p ) = 0 if and only if p is algebraic. Remark 4.3.
Let p ( x ) ∈ S ( A ) be such that SU d ( p ) = ∞ . Then for some B ⊇ A , p ( x ) has a dividing extension q ( x ) ∈ S ( B ) such that SU d ( q ) = ∞ . It is easy to verify that D and DU coincide for formulas: Lemma 4.4.
For every formula ϕ ( x ) , we have D ( ϕ ) = DU ( ϕ ) .Proof. We only need to prove DU ( ϕ ) ≤ D ( ϕ ) . A proof by induction reducesthe problem to show DU ( ϕ ) ≥ α + 1 implies D ( ϕ ) ≥ α + 1 . Assume ϕ is over A and DU ( ϕ ) ≥ α + 1 . Then, there exists a partial type q such that ϕ ∈ q , q divides over A and DU ( q ) ≥ α . Assuming q closed under conjunction,there exists a formula ψ ∈ q such that ψ divides over A . Obviously ϕ ∧ ψ also divides over A and DU ( ϕ ∧ ψ ) ≥ α . By the induction hypothesis, D ( ϕ ∧ ψ ) ≥ α . So, D ( ϕ ) ≥ α + 1 .A variation of the proof above also shows D f = DU f for formulas. Now,we are going to prove that D f and DU f are the same as D and DU respec-tively (and therefore do not depend on the set of parameters). So, from nowon, we will use only D and DU . Proposition 4.5.
Let p a partial type and ϕ a formula both over A . Then,1. DU ( p ) = DU f ( p, A ) .2. D ( ϕ ) = D f ( ϕ, A ) .Proof. To prove it suffices to show that DU ( p ) ≥ DU f ( p ) . A proof byinduction reduces to prove the following: DU f ( p ) ≥ α + 1 implies DU ( p ) ≥ α + 1 , assuming it is true for α . If DU f ( p ) ≥ α + 1 , there exists q ⊇ p such that q forks over A and DU f ( q ) ≥ α and by the induction hypothesis, DU ( q ) ≥ α . Then, there exists { q i : i ∈ n } such that q ≡ (cid:87) i q i with each q i extending q and dividing over A . Then, DU ( q ) = max { DU ( q i ) : i ∈ n } . So,for some q i , DU ( q i ) ≥ α and therefore, DU ( p ) ≥ α + 1 . follows from , since D f = DU f for formulas. D is extended in a standard way to partial types p as follows: D ( p ) = min { D ( ϕ ) : ϕ is a finite conjunction of formulas in p } As D = DU for formulas, it is obvious that DU ( p ) ≤ D ( p ) for a partial type p , but in some cases they are not equal. In the next example we even seehow D can be ∞ while DU not. Example 4.6.
Let the language contain an infinite set of disjoint unarypredicates { Q i : i ∈ ω } and binary relations {≤ i : i ∈ ω } . Each ≤ i being adense linear order without endpoints defined in Q i . Let p denote {¬ Q i ( x ) : i ∈ ω } . Then DU ( p ) = 1 while D ( p ) = ∞ .Proof. As p is not algebraic, DU ( p ) ≥ . Suppose DU ( p ) ≥ . Then, by theequivalence in Proposition 3.2, there exist ϕ ( x, y ) and ( a i : i ∈ µ ) such thatfor each i ∈ µ , DU ( p ∪ { ϕ ( x, a i ) } ) ≥ and { ϕ ( x, a i ) : i ∈ µ } is k -inconsistentfor some k . Here µ = (cid:0) | T | + | A | (cid:1) + . Any two realizations of p different from a i have the same type over a i , so any realization of p except maybe a i satisfy ϕ ( x, a i ) . This shows that { ϕ ( x, a i ) : i ∈ µ } is realized by every realizationof p , except maybe { a i : i ∈ µ } and therefore { ϕ ( x, a i ) : i ∈ ω } is not k -inconsistent. This shows DU ( p ) = 1 .For each fine subset S ⊆ I , we will check that D ( (cid:86) i ∈ S ¬ Q i ) = ∞ , so D ( p ) = ∞ . Fix j ∈ ω − S and choose { a i , b i : i ∈ ω } in Q j such that a < a < . . . < a n < . . . < b n < . . . < b < b DU AND OTHER RANKS
Then, the formula a n < x < b n divides over { a b , . . . a n − b n − } , so there is aninfinite dividing sequence of formulas and therefore (see 14.3.3 Casanovas[4]) D ( (cid:86) i ∈ S ¬ Q i ) = ∞ .The inequality SU f ( p ) ≤ D ( p ) is well known (see Kim[8]) but a standardproof needs simplicity of the theory and does not work in full generality.Actually, it is true in general: Remark 4.7.
Let p be a complete type. It is immediate by Proposition 4.5that SU d ( p ) ≤ SU f ( p ) ≤ DU ( p ) ≤ D ( p ) . In fact, DU , SU d y SU f are equal when they are finite: Proposition 4.8.
Let p a complete type. If SU d ( p ) is finite then SU d ( p ) = SU f ( p ) = DU ( p ) .Proof. We only need to prove that if DU ( p ) ≥ n then SU d ( p ) ≥ n . If DU ( p ) ≥ n we can build a chain of partial types of length n , ( p i : i ≤ n ) ,each p i a partial type over a set A i . Let a | = (cid:83) i Let p ∈ S ( A ) be a complete type. SU d ( p ) = ∞ , SU f ( p ) = ∞ and DU ( p ) = ∞ are equivalent.Proof. Assume DU ( p ) = ∞ . By Lemma 3.7, we can build a dividing chainof partial types ( p i : i ∈ ω ) and sets of parameters ( A i : i ∈ ω ) such that p = p , A = A and for every i ∈ ω , p i ⊆ p i +1 , A i ⊆ A i +1 , p i +1 dividesover A i . Let a | = (cid:83) i ∈ α p i . Then ( tp ( a/A i ) : i ∈ ω ) is a dividing chain ofcomplete types. It is easy to check by induction over α that for every i ∈ ω , SU d (( tp ( a/A i )) ≥ α .In some cases DU and SU d coincide for complete types: Remark 4.10. Assume DU has extension, i.e. for every partial type p ( x ) over A , there exists q ( x ) ∈ S ( A ) such that p ⊆ q and DU ( p ) = DU ( q ) . Thenfor every complete type p DU ( p ) = SU d ( p ) .Proof. We prove that SU d ( p ) ≥ DU ( p ) by induction on α . Let p ∈ S ( A ) such that DU ( p ) ≥ α + 1 . Then, there exists q over B such that p ⊆ q , q divides over A and DU ( q ) ≥ α . By the extension property, there exists q (cid:48) ∈ S ( B ) such that q ⊆ q (cid:48) and DU ( q (cid:48) ) ≥ α . By the induction hypothesis, SU d ( q (cid:48) ) ≥ α and therefore SU d ( p ) ≥ α + 1 .1Now, we are going to explore the relations between the DU -rank andthe DD -rank defined in Cárdenas, Farré[2]. In that paper we can found adefinition of DD from Shelah trees and several equivalences. Here, we define DD by dividing chains of complete types, which is the equivalence that weare going to use. Definition 4.11. Let p be a partial type over A . A dividing chain ofcomplete types of depth α in p is a sequence of complete types ( p i ( x ) : i ∈ α ) such that p ⊆ p , A ⊆ dom ( p ) , p divides over A and for every < i < α , p i is a dividing extension of p
Let p be a partial type. Then,1. DD ( p ) ≤ DU ( p ) .2. If DD ( p ) is finite then DD ( p ) = DU ( p ) .3. DD ( p ) ≥ ω + if and only if DU ( p ) = ∞ .Proof. The proof of is as in 4.8, the proof of is as in 4.9 and followsfrom and .Observe that in the case of a complete type, by Propositions 4.8 and 4.9the results of Proposition 4.12 also hold replacing DU by SU f and SU d .We are going to use a result about DD in Cárdenas, Farré[2] to provethat DU , SU d and SU f have a bounded number of different values. We takenext lemma from Proposition 3.12 in Cárdenas, Farré[2]: Lemma 4.13. Let p ( x ) be a partial type over A . Then, there exists a set ofparameters B ⊆ A such that | B | ≤ | T | | DD ( p ) | and DD ( p (cid:22) B ) = DD ( p ) . Proposition 4.14. There is some ordinal α such that DU ( p ) ≥ α implies DU ( p ) = ∞ . Proof. Observe that, as DU takes the same values over conjugate sets ofparameters and there are boundedly many non-conjugate sets of parametersof size ≤ | T | ℵ , the DU -values on types over a set of parameters of size ≤ | T | ℵ is upper bounded. By Proposition 4.12 and the previous lemma,for any partial type p over A with DU ( p ) < ∞ there is some B ⊆ A with | B | ≤ | T | ℵ and DU ( p ) ≤ DU ( p (cid:22) B ) < ∞ . Therefore the set of non-infinitevalues taken by DU is bounded. Remark 4.15. The same α as in 4.14 satisfies: SU d ( p ) ≥ α implies SU d ( p ) = ∞ and SU f ( p ) ≥ α implies SU f ( p ) = ∞ . We have seen so far that D = D f and DU d = DU f , but we do not knowif, in general, SU d = SU f or SU f = DU . We know that the three ranksare equal for finite values and the value ∞ , but in all intermediate cases,when DD ( p ) = ω − , we do not have the answer. So, we have these two openquestions: Question 4.16. Is there a complete type p such that SU d ( p ) < SU f ( p ) ? Question 4.17. Is there a complete type p such that SU f ( p ) < DU ( p ) ? In Cárdenas, Farré[3] we prove that in an N T P theory, for any stablecomplete type p , SU d ( p ) = SU f ( p ) . We also prove that if SU d has extensionthen SU d = SU f . The following are two equivalent definitions of a simple type (see in Hart,Kim, Pillay[7] and Chernikov[6]). Definition 5.1. Let p ( x ) be a partial type over A . p is simple if and onlyif one of the following two equivalent conditions are satisfied:1. for every B ⊇ A and every realization a of p ( x ) , there is some B ⊆ B with | B | < | T | + such that a | (cid:94) dB B .2. for every B ⊇ A and every realization a of p ( x ) , there is some B ⊆ B with | B | < | T | + such that a | (cid:94) dAB B . From this, one might think in defining a supersimple type in two differentways, replacing in both definitions the bound | T | + by ℵ . In fact, in Hart,Kim, Pillay[7], they suggest to define a supersimple type through the firstalternative, although they do not develop the implications of this possibility.3We will see through the Example 5.9 that these possible definitions arenot equivalent and that the first one depends on the set of parameters whileby Corollary 5.6 the second not. In addition, in this example we show also asuperstable complete type not satisfying the first possible definition of super-simple. All of that indicate us that the correct way of defining a supersimpletype will be the second: Definition 5.2. Let p ( x ) be a partial type over A . p is supersimple if andonly if for every B ⊇ A and every realization a of p ( x ) , there exists a finiteset B ⊆ B with a | (cid:94) dAB B . Remark 5.3. It is obvious that the discarded definition of supersimple typeimplies our definition of supersimple type. The notion of supersimple type satisfies the following expected properties: Remark 5.4. The following are satisfied:1. If p ( x ) is supersimple and p ( x ) ⊆ q ( x ) , then q ( x ) is supersimple.2. If p ( x, y ) is supersimple, then the type ∃ yp ( x, y ) is supersimple.3. tp ( ab/A ) is supersimple if and only if tp ( a/A ) and tp ( b/Aa ) are super-simple. More generally, tp (( a i : i ∈ n ) /A ) is supersimple if and only if tp (( a i : i ∈ n ) /A ) is supersimple.4. Assume that x and y are disjoint. p ( x ) and q ( y ) are supersimple if andonly if p ( x ) ∪ q ( y ) is supersimple.5. Let p ( x ) be over A . Then p is supersimple if and only if every q ( x ) ∈ S ( A ) extending p ( x ) is supersimple.6. p ( x ) , p ( x ) are supersimple if and only if p ∨ p is supersimple.Proof. 1 is obvious assuming p and q are over the same set of parameters. . Let a | = ∃ yp ( x, y ) , then ab | = p ( x, y ) for some b . Let B ⊇ A . As p ( x, y ) is supersimple, there exists a finite B ⊆ B with ab | (cid:94) dAB B . So, a | (cid:94) dAB B . ⇒ ). tp ( a/A ) = ∃ ytp ( ab/A ) and tp ( b/A ) = ∃ xtp ( ab/A ) are supersimpleby and therefore by tp ( b/Aa ) is also supersimple. ⇐ ). Let B ⊇ A and a (cid:48) b (cid:48) ≡ A ab . By the first condition, as a (cid:48) ≡ A a , thereexists B ⊆ B finite such that a (cid:48) | (cid:94) dAB B . As tp ( b (cid:48) /Aa (cid:48) ) is supersimple, thereexists B ⊆ B finite such that b (cid:48) | (cid:94) dAa (cid:48) B Ba (cid:48) . Now, taking B = B B wehave a (cid:48) | (cid:94) AB B and b (cid:48) | (cid:94) Aa (cid:48) B B . By left transitivity we obtain a (cid:48) b (cid:48) | (cid:94) AB B .4 . Assume p ( x ) and q ( y ) are supersimple over A and let B ⊇ A and ab | = p ( x ) ∪ q ( y ) . Then tp ( a/A ) and tp ( b/Aa ) are supersimple by . By , tp ( ab/A ) is supersimple. So, there is some finite B ⊆ B with ab | (cid:94) dAB B . ⇒ ) is trivial by . ⇐ ). Let B ⊇ A and a | = p . As tp ( a/A ) is supersimple, there exists B ⊆ B finite such that a | (cid:94) dAB B . follows from , since any completion of p ∨ q is either a completion of p or a completion of q .We remember the following result for DD from Cárdenas, Farré[2]: Lemma 5.5. Let p be a partial type over a set of parameters A . Let κ beany regular cardinal number. The following are equivalent:1. DD ( p ) < κ + .2. For every B ⊇ A and a | = p , there exists a set B ⊆ B with | B | < κ such that a | (cid:94) dAB B . Corollary 5.6. The definition of supersimple does not depend on the set ofparameters. Moreover, the following are equivalent for a partial type p over A : . p is supersimple, . DD ( p ) < ω + , . DU ( p ) < ∞ , . For everycompletion q ∈ S ( A ) of p , SU d ( q ) < ∞ . . For every completion q ∈ S ( A ) of p , SU f ( q ) < ∞ .Proof. By previous Lemma, the fact that DD ( p ) does not depend on the setof parameters, Remark 4.7 and Proposition 4.12.We remember the definition of the Lascar rank U and the definition of asuperstable type of Poizat[9]: Definition 5.7. The U -rank for a complete type p ( x ) ∈ S ( A ) is defined asfollows:1. U ( p ) ≥ .2. U ( p ) ≥ α + 1 if and only if for each cardinal number λ there is a set B ⊇ A and there are at least λ many types q ( x ) ∈ S ( B ) extending p and such that U ( q ) ≥ α .3. U ( p ) ≥ α with α a limit ordinal, if and only if U ( p ) ≥ β for all β < α . U ( p ) is the supremum of all α such that U ( p ) ≥ α . If such supremum doesnot exist we set U ( p ) = ∞ . Definition 5.8. Let p ∈ S ( A ) . p is superstable if and only if U ( p ) < ∞ . In Cárdenas, Farré[3] we prove that a complete type is stable and super-simple if and only if is superstable. Example 5.9. There is an example of a superstable and supersimple type p ∈ S ( A ) not satisfying the discarded definition of supersimple. However,for some b , p considered over Ab satisfies the alternative definition, so thediscarded definition depend on the set of parameters.Proof. Consider the theory of infinitely many refining equivalence relations,which is a stable non supersimple theory with quantifier elimination. Thelanguage consists in ω equivalence relations { E i : i ∈ ω } , E has infinitemany classes, E i +1 refines E i and each E i -class is partitioned into infinitelymany E i +1 -classes. Given d and C , denote ρ ( d/C ) = ∞ if d ∈ C , ρ ( d/C ) = sup { n : dE n c for some c ∈ C } otherwise. Here we consider ω < ∞ . One canverify that tp ( d/BC ) divides over C if and only if ρ ( d/C ) < ρ ( d/BC ) . So, D | (cid:94) C B ⇔ for every d ∈ D : ρ ( d/C ) = ρ ( d/BC ) Now we choose and fix a and b such that aE i b for every i ∈ ω and take A = { a i : i ∈ ω } such that aE i a i and a (cid:8)(cid:8)(cid:8) E i +1 a i for every i ∈ ω . The exampleis p = tp ( a/A ) . p is supersimple: Given B ⊇ A and a (cid:48) | = p , ρ ( a (cid:48) /A ) = ω and ρ ( a (cid:48) /B ) ≥ ω .So, taking B = { a (cid:48) } ∩ B , we have a (cid:48) | (cid:94) AB B . p is superstable: p is not algebraic, so U ( p ) ≥ . On the other hand, itis immediate to check that for any parameter set, p only has a non algebraicextension, so applying the definition of U , U ( p ) (cid:54)≥ . p does not satisfy the discarded definition of supersimple but p consideredas a partial type over Ab satisfies the discarded definition of supersimple: forevery finite B ⊆ A , we have ρ ( a/B ) < ω , So, a (cid:54) | (cid:94) B A . But given B ⊇ Ab and a (cid:48) | = p , we have ρ ( a (cid:48) /b ) = ω and ρ ( a (cid:48) /B ) ≥ ω . So, taking B = { a (cid:48) b }∩ B ,we have a (cid:48) | (cid:94) B B .Although for a particular type the discarded definition is not equivalentto supersimplicity, for a fixed theory the fact that all types satisfy one of thedefinitions is equivalent to all types satisfy the other. Remark 5.10. The following are equivalent: : T is supersimple. : { x = x } is supersimple. : Every complete type is supersimple. : Every partialtype is supersimple. : For every p ∈ S ( A ) , there exists a finite subset A ⊆ A such that p does not divide over A . : For every p ∈ S ( A ) , every B ⊇ A and every realization a of p , there exists a finite subset B ⊆ B suchthat a | (cid:94) dAB B . Proof. The equivalence between and follows from remark 5.4. The otherare standard, see . , and . in Casanovas[4].Now we improve slightly for SU d and SU f the known fact that in simpletheories SU is preserved by non-forking extensions. We recall that a theory iscalled Extensible if forking has existence, that is every complete type doesnot fork over its parameter set. For instance, simple theories are extensible. Proposition 5.11. In a extensible theory, let p ( x ) ∈ S ( A ) and q ( x ) ∈ S ( B ) be such that p ( x ) ⊆ q ( x ) and tp ( B/A ) is simple. If q does not fork over A then SU d ( q ) = SU d ( p ) and SU f ( q ) = SU f ( p ) .Proof. We will prove that SU d ( p ) ≤ SU d ( q ) . The proof is similar for SU f .If SU d ( q ) = ∞ the result is immediate, so we can assume without loss ofgenerality that q is supersimple. We use induction on α to prove that if SU d ( p ) ≥ α then SU d ( q ) ≥ α . This is clear for α = 0 or limit ordinal.Assume SU d ( p ) ≥ α + 1 . Now, by the definition of SU d , there is a dividingextension p ∈ S ( C ) of p such that SU d ( p ) ≥ α .Let d | = q and d (cid:48) | = p . As d (cid:48) ≡ A d , there exists C (cid:48) such that d (cid:48) C ≡ A dC (cid:48) .Using T extensible we can choose C (cid:48)(cid:48) such that C (cid:48)(cid:48) ≡ Ad C (cid:48) and C (cid:48)(cid:48) | (cid:94) fAd B .As d | (cid:94) fA B and C (cid:48)(cid:48) | (cid:94) fAd B , by left transitivity, C (cid:48)(cid:48) d | (cid:94) fA B . As tp ( B/A ) is simple, using symmetry (Proposition in Casanovas[5]), B | (cid:94) fA C (cid:48)(cid:48) d andtherefore B | (cid:94) fC (cid:48)(cid:48) d . Since tp ( d/B ) and tp ( B/A ) are simple, tp ( dB/A ) issimple and therefore tp ( d/C (cid:48)(cid:48) ) is simple. Using symmetry again d | (cid:94) fC (cid:48)(cid:48) B .Since tp ( B/A ) is simple tp ( B/C (cid:48)(cid:48) ) is simple and therefore tp ( C (cid:48)(cid:48) B/C (cid:48)(cid:48) ) isalso simple. By induction hypothesis, SU d ( d/C (cid:48)(cid:48) B ) ≥ α . By B | (cid:94) fA C (cid:48)(cid:48) d weget B | (cid:94) dA C (cid:48)(cid:48) . With d (cid:54) | (cid:94) dA C (cid:48)(cid:48) , we obtain d (cid:54) | (cid:94) dB C (cid:48)(cid:48) and therefore tp ( d/C (cid:48)(cid:48) B ) divides over B . So, finally SU d ( q ) ≥ α + 1 . Corollary 5.12. In an extensible theory, let p be a complete type over A and q be a partial type over B such that q is a non forking extension of p and tp ( B/A ) is simple. If q is supersimple then p is supersimple. By Proposition in Cárdenas, Farré[2], we have a similar corollaryusing DD with somewhat different hypotheses: Corollary 5.13.