Games and Scott sentences for positive distances between metric structures
aa r X i v : . [ m a t h . L O ] F e b GAMES AND SCOTT SENTENCES FOR LINEARISOMORPHISMS OF BANACH SPACES
ÅSA HIRVONEN AND JONI PULJUJÄRVI
Abstract.
We define Ehrenfeucht–Fraïssé games and an infinitary version ofHeinrich and Henson’s positive bounded logic with approximate semantics, andshow that, using this logic and these games, one can characterise separableBanach spaces up to linear isomorphism.
Introduction
In this paper we define Ehrenfeucht–Fraïssé games for Banach spaces and buildScott sentences characterising separable spaces up to linear isomorphism. We don’tget the sentences in L ω ω but in L ω ω in an infinitary version of the Heinrich–Henson logic presented in [HH86].The main notion of isomorphism used in functional analysis is that of linear iso-morphism, i.e., of being linearly homeomorphic. The notion sits badly in a modeltheoretic context, as it gives rise to a non-transitive notion of type. However, Hein-rich and Henson studied linear isomorphisms in the 1980s [HH86] and developed alogic with approximate semantics that was suitable to handle linear isomorphisms.They used the logic to prove a linear isomorphic version of the Keisler–Shelah iso-morphism theorem, i.e., to characterize the property of having linearly isomorphicultrapowers. The logic was later developed by Henson and Iovino [HI02] and wasone of the seeds for the continuous first-order logic developed by Ben Yaacov andUsvyatsov [BYU10].One of the classical main results of infinitary logic is Scott’s isomorphism the-orem, stating that one can capture countable structures up to isomorphism withsentences in L ω ω . In [BYDNT17] Ben-Yaacov et al. construct Scott sentencesfor separable metric structures in a continuous infinitary logic L ω ω . Inspired byBen Yaacov et al.’s methods for treating isometries of metric spaces via dense se-quences forming gradually improving approximations of an isometry, we set out tosee whether the methods could be used to capture linear isomorphisms of Banachspaces. In addition to being hard to treat model theoretically, linear isomorphisms Date : September 14, 2020.2020
Mathematics Subject Classification.
Key words and phrases.
Banach spaces, Ehrenfeucht–Fraïssé games, infinitary logic, linearisomorphism, metric model theory, Scott sentences.J. Puljujärvi was partially supported by the Academy of Finland, grant 322795. also form a rather complex equivalence relation: Ferenczi et al. [FLR09] haveshown that the linear isomorphism relation of separable Banach spaces is Borelbireducible to the universal relation for analytic equivalence relations, whereas thatof isometry, by [Mel07], is Borel bireducible to the universal relation for relationsinduced by a Borel action of a Polish group.We only treat Banach spaces without additional structure in this paper. Onecould add additional structure to the spaces (to get what Henson and Iovino call
Banach space structures in [HI02]), but we did not have a motivating example forhow one should define linear isomorphisms for them (other than just preservingthe additional structure).Several versions of infinitary continuous logic have been developed. Apart fromthe one used in [BYDNT17], e.g. Eagle [Eag14] has a version that allows formore general structures. However, we return to the Heinrich-Henson logic as ourstarting point for an infinitary logic, as that logic was designed to handle linearisomorphisms.We build our sentences by looking at dynamic Ehrenfeucht–Fraïssé games, fol-lowing the exposition of Väänänen in [Vä11]. However, our game is a peculiarone – it is neither symmetric nor transitive, and it allows the second player toimprove her moves along the game. But we show that it can be used to detectlinear isomorphism by going via an infinite-length game:
Theorem.
For separable Banach spaces A and B , the following are equivalent.(i) II wins the dynamic Ehrenfeucht–Fraïssé game of precision ε and clock α between the spaces, for all α < (density( A ) + density( B )) + .(ii) II wins the Ehrenfeucht–Fraïssé game of length ω and precision ε betweenthe spaces.(iii) There exists an ε -isomorphism A → B .Althought the exposition follows Väänänen, many of our definitions were, in fact,inspired by Ben Yaacov et al.’s methods, e.g., the way they use a weak modulus ofuniform continuity is the basis for our k -good formulas , which are used to capturethe demand on improved moves.By describing a winning condition of the second player, we can then, for each ε ≥ , form a Scott sentence σ A ε , that guarantees a model satisfying it is linearlyisomorphic to A with isomorphism constant e ε : Theorem.
For Banach spaces A and B , II wins the Ehrenfeucht–Fraïssé game oflength ω and precision ε between the spaces if and only if B | = σ A ε .One main question calls for asking: Question.
With our approach we are not able to bound the Scott rank of aseparable Banach space below ω , and as a result our Scott sentences only showup in L ω ω . Can this be improved? AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 3
In the first section, we introduce an infinitary extension of Heinrich and Hen-son’s positive bounded logic and show how it connects to the concept of linearisomorphism. In the second section, we define the Ehrenfeucht–Fraïssé games ofinfinite and dynamic length and prove some properties of them that will be usedlater. In the third section, we define the notion of Scott rank for our games. In thefourth section, we define the Scott sentences of Banach spaces and connect themto our games and thus to linear isomorphism.1.
Language
Here we define an infinitary variant of the positive bounded logic introducedby Heinrich & Henson [HH86] and later developed by Henson & Iovino [HI02].However, the way its semantics is defined causes the logic to be neither positivenor bounded (except on the atomic level), so we get the full expressive power offirst-order logic included in our logic. We will call this logic L κω . Definition 1.
For a regular cardinal κ , we define the syntax of the logic L κω asfollows. We consider the alphabet consisting of a binary function symbol + , unaryfunction symbols f r , r ∈ Q , and two unary predicates P and Q . For t a term, wedenote the term f r ( t ) simply by rt . Then atomic formulas are of the form P ( t ) and Q ( t ) , where t is a term. If ϕ is a formula and v is a variable, then ∃ vϕ and ∀ vϕ are formulas. Whenever ϕ is a set of formulas such that | ϕ | < κ and there isa finite set V of variables such that whenever v is free in some ϕ ∈ ϕ , then v ∈ V ,then V ϕ and W ϕ are formulas. Definition 2.
We define the semantics of L κω in a Banach space the followingway: if A is a Banach space and ¯ a = ( a , . . . , a n − ) ∈ A n , then interpretation ofterms in A with ¯ a is defined the obvious way: + is interpreted as the additionof the space A and each f r as scalar multiplication by r . For an atomic formula ϕ (¯ v ) , ¯ v = ( v , . . . , v n − ) , we define the satisfaction of ϕ in A with ¯ a as follows: if ϕ = P ( t (¯ v )) , then A | = ϕ (¯ a ) if (cid:13)(cid:13) t A (¯ a ) (cid:13)(cid:13) ≤ , and if ϕ = Q ( t (¯ v )) , then A | = ϕ (¯ a ) if (cid:13)(cid:13) t A (¯ a ) (cid:13)(cid:13) ≥ . Satisfaction of more complicated formulas is defined as usual. Remark . We can define negated atomic formulas with the help of infinitaryconnectives already in L ω ω as follows:(i) ¬ P ( t ) := W q ∈ Q ∩ (0 , Q ( qt ) ,(ii) ¬ Q ( t ) := W q ∈ Q ∩ (1 , ∞ ) P ( qt ) .In the syntax of L κω , we defined only unbounded quantifiers but from the usualbounded quantifiers ∀ B and ∃ B of positive bounded logic we could define theunbounded quantifiers as follows:(i) ∀ uϕ (¯ v, u ) := ∀ B y V q ∈ Q ϕ (¯ v, qu ) ,(ii) ∃ uϕ (¯ v, u ) := ∃ B y W q ∈ Q ϕ (¯ v, qu ) .Thus we will not bother working with bounded quantifiers. AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 4
Definition 4.
Let k < ω . A term t is k -good if it is of the form P k − i =0 c i v i , where | c i | ≤ k and each v i is a distinct variable. We say that a formula is k -good if it isan atomic formula containing a k -good term. Remark . We may also think of terms of the form P m − i =0 c i v i , where | c i | ≤ k andeach v i is a distinct variable, as k -good whenever m ≤ k , since if m < k , we mayidentify this term with the term P k − i =0 c i v i , where, for all i ≥ m , c i = 0 and v i is any fresh variable. Note that this implies that k -good formulas are k ′ -good forany k ′ ≥ k . Definition 6.
Let ε ≥ . The ε -approximation ϕ ε of a formula ϕ is defined asfollows: if ϕ = P ( t ) , then ϕ ε = P ( e − ε t ) , and if ϕ = Q ( t ) , then ϕ ε = Q ( e ε t ) . Then ( V ϕ ) ε = V ϕ ∈ ϕ ϕ ε and ( W ϕ ) ε = W ϕ ∈ ϕ ϕ ε and ( ∀ vϕ ) ε = ∀ vϕ ε and ( ∃ vϕ ) ε = ∃ vϕ ε .We say that a space A and a tuple ¯ a approximately satisfy the formula ϕ (¯ v ) , insymbols A | = A ϕ (¯ a ) , if for all ε > , A | = ϕ ε (¯ a ) . Remark . Notice that ( ϕ ε ) ε ′ = ϕ ε + ε ′ and that if ε ≤ ε ′ , then ϕ ε | = ϕ ε ′ . Lemma 8.
Let k < ω and ε ≥ . Let A and B be Banach spaces and ¯ a ∈ A n and ¯ b ∈ B n . Suppose that for all k -good ϕ (¯ v ) , A | = ϕ (¯ a ) = ⇒ B | = ϕ ε (¯ b ) . Then for all k -good terms t (¯ v ) also B | = P ( t (¯ b )) = ⇒ A | = ( P ( t (¯ a ))) ε . Proof.
Let t (¯ v ) be a k -good term. We show the contrapositive of the claim. Sup-pose that A = ( P ( t (¯ a ))) ε , i.e. (cid:13)(cid:13) e − ε t A (¯ a ) (cid:13)(cid:13) > . Then there is ε ′ > such that (cid:13)(cid:13) e − ε t A (¯ a ) (cid:13)(cid:13) ≥ e ε ′ . But then A | = Q ( e − ε − ε ′ t (¯ a )) , and e − ε − ε ′ t is still a k -good term,so by the initial assumption B | = ( Q ( e − ε − ε ′ t (¯ b ))) ε , i.e. (cid:13)(cid:13) t B (¯ b ) (cid:13)(cid:13) ≥ e ε ′ > . Thus B = P ( t (¯ b )) . (cid:3) Lemma 9 (Uniform perturbation distance for k -good formulas) . Let A be a Ba-nach space. Then for any k < ω and ε ≥ there exists δ > such that for any k -good formula ϕ (¯ v ) and for all ¯ a, ¯ b ∈ A k , if d (¯ a, ¯ b ) < δ , then A | = ϕ (¯ a ) = ⇒ A | = ϕ ε (¯ b ) . Such δ will be called a perturbation distance for k and ε in A . Proof.
Let k < ω and ε > be arbitrary and define δ = min { δ P , δ Q } , where δ P = ( e ε − /k and δ Q = (1 − e − ε ) /k . Let ϕ (¯ v ) be k -good and ¯ a, ¯ b ∈ A k .Suppose that d (¯ a, ¯ b ) < δ , i.e. k a i − b i k < δ for i < k . AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 5 (i) First assume ϕ = P ( P k − i =0 c i v i ) and suppose that A | = ϕ (¯ a ) . Now (cid:13)(cid:13)(cid:13)P k − i =0 c i a i (cid:13)(cid:13)(cid:13) ≤ , so (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) e − ε k − X i =0 c i b i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = e − ε (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i ( b i − a i ) + k − X i =0 c i a i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ e − ε (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i ( b i − a i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) + (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i a i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)! ≤ e − ε k − X i =0 | c i | k b i − a i k + 1 ! ≤ e − ε k − X i =0 kδ + 1 ! = e − ε ( k δ + 1) ≤ e − ε ( k δ P + 1)= e − ε (cid:18) k e ε − k + 1 (cid:19) = 1 . Hence A | = ϕ ε (¯ b ) .(ii) Then assume that ϕ = Q ( P k − i =0 c i v i ) . Suppose that B = ϕ ε (¯ b ) , i.e. (cid:13)(cid:13)(cid:13)P k − i =0 c i b i (cid:13)(cid:13)(cid:13) < e − ε . Then (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i a i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i ( a i − b i ) + k − X i =0 c i b i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i ( a i − b i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) + (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i b i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) < k δ + e − ε ≤ k δ Q + e − ε = k − e − ε k + e − ε = 1 , so (cid:13)(cid:13)(cid:13)P k − i =0 c i a i (cid:13)(cid:13)(cid:13) < . Hence A = ϕ (¯ a ) . (cid:3) Definition 10.
Let A and B be Banach spaces and f : A → B a function. If ε ≥ , we say that f is an ε -isomorphism if f is linear and e ε -bi-Lipschitz. We saythat f is a linear isomorphism if it is an ε -isomorphism for some ε ≥ .Note that a -isomorphism is an isometry. AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 6
Lemma 11. If f : A → B is an ε -isomorphism, then for all ϕ (¯ v ) ∈ L κω and ¯ a ∈ A m , A | = ϕ (¯ a ) = ⇒ B | = ϕ ε ( f (¯ a )) and B | = ϕ ( f (¯ a )) = ⇒ A | = ϕ ε (¯ a ) . Proof.
Let ¯ a ∈ A k . We prove by induction on ϕ that if A | = ϕ (¯ a ) , then B | = ϕ ε ( f (¯ a )) . First suppose that ϕ = P ( P k − i =0 c i v i ) . Now if A | = ϕ (¯ a ) , (cid:13)(cid:13)(cid:13)P k − i =0 c i a i (cid:13)(cid:13)(cid:13) ≤ , so (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i f ( a i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) f ( k − X i =0 c i a i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ e ε (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i a i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ e ε . Hence B | = P ( e − ε P k − i =0 c i f ( a i )) , so B | = ϕ ε ( f (¯ a )) . If ϕ = Q ( P ki =0 c i v i ) and A | = ϕ (¯ a ) , then (cid:13)(cid:13)(cid:13)P k − i =0 c i a i (cid:13)(cid:13)(cid:13) ≥ , so (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i f ( a i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) f ( k − X i =0 c i a i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≥ e − ε (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) k − X i =0 c i a i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≥ e − ε , so B | = Q ( e ε P k − i =0 c i f ( a i )) , i.e. Q ε ( P k − i =0 c i f ( a i )) .The other direction for P and Q is similar, just replace f by f − in the argument.The rest of the proof follows immediately from the induction hypothesis. (cid:3) Lemma 12.
A function f : A → B is an ε -isomorphism if for all k < ω , k -good ϕ (¯ v ) and ¯ a ∈ A k , A | = ϕ (¯ a ) = ⇒ B | = ϕ ε ( f (¯ a )) . Proof.
By Lemma 8, we have for all k < ω and k -good terms t (¯ v ) A | = P ( t (¯ a )) = ⇒ B | = ( P ( t ( f (¯ a )))) ε . and B | = P ( t ( f (¯ a ))) = ⇒ A | = ( P ( t (¯ a ))) ε . Let a, a ′ ∈ A . Whenever r > is rational and k ≥ max { /r, } , the term v /r − v /r is k -good, so we have for all r > k a − a ′ k < r = ⇒ A | = P ( a/r − a ′ /r )= ⇒ B | = P ( e − ε ( f ( a ) /r − f ( a ′ ) /r ))= ⇒ e − ε k f ( a ) − f ( a ′ ) k ≤ r, which shows that e − ε k f ( a ) − f ( a ′ ) k ≤ k a − a ′ k , proving that f is e ε -Lipschitz.On the other hand, k f ( a ) − f ( a ′ ) k < r = ⇒ B | = P ( f ( a ) /r − f ( a ′ ) /r )= ⇒ A | = P ( e − ε ( a/r − a ′ /r ))= ⇒ e − ε k a − a ′ k ≤ r, AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 7 whence e − ε k a − a ′ k ≤ k f ( a ) − f ( a ′ ) k . Hence f is e ε -bi-Lipschitz.Left is to show linearity. Let a , a ∈ A and q , q ∈ Q . Denote by a theelement q a + q a . Let b i = f ( a i ) for i = 0 , , . Now for any n < ω let k n ≥ max { n | q | , n | q | , n, } . Then the term nq v + nq v − nv is k n -good. Also,as a = q a + q a , we have k q a + q a − a k ≤ /n for all n < ω . Thus A | = P ( nq a + nq a − na ) for all n and therefore also B | = P ( e − ε ( nq b + nq b − nb )) ,i.e. k q b + q b − b k ≤ e ε /n for all n . Thus b = q b + q b . As f is continuousand Q -linear, it is also R -linear. (cid:3) Games
In this section we define an Ehrenfeucht–Fraïssé game of infinite length and showthat it can be used to show that two separable Banach spaces are ε -isomorphic.We also define a dynamic variant of the game (which is finite) and connect thegames to the language defined in the previous section. Definition 13.
Let A and B be Banach spaces and D A and D B be dense subsetsof the respective spaces. For ε ≥ , ¯ ε = ( ε , . . . , ε n − ) ∈ ( ε, ∞ ) n and tuples ( a , . . . , a n − ) ∈ A n and ( b , . . . , b n − ) ∈ B n , we denote by EF A , B ω,ε, ¯ ε (( D A , ( a , . . . , a n − )) , ( D B , ( b , . . . , b n − ))) the two-player game of length ω , defined as follows. Denote the players by I and II . At each round k < ω , first I picks an element x k + n ∈ D A ∪ D B and in additiona positive real number ε k + n > ε , and then II responds with some y k + n ∈ D A ∪ D B so that if x k + n was in D A , then y k + n is in D B , and vice versa. We call this gamethe Ehrenfeucht–Fraïssé game of length ω and precision ε , between the sets D A and D B , with starting position (( a , . . . , a n − ) , ( b , . . . , b n − ) , ¯ ε ) .For k ≥ n , we denote by a k the element of the set { x k , y k } that belongs to D A and by b k the one that belongs to D B , hence expanding the original tuples ( a , . . . , a n − ) and ( b , . . . , b n − ) into infinite sequences ( a i ) i<ω and ( b i ) i<ω .Player II wins a play of the game if for all k < ω and all k -good ϕ ( v , . . . , v k − ) A | = ϕ ( a i , . . . , a i k − ) = ⇒ B | = ϕ ε k ( b i , . . . , b i k − ) for all i , . . . , i k − ≥ k .If n = 0 , then we denote the game by EF A , B ω,ε ( D A , D B ) . Remark . The game EF A , B ω,ε, ¯ ε (( D A , ( a , . . . , a n − )) , ( D B , ( b , . . . , b n − ))) corre-sponds to a position in the game EF A , B ω,ε ( D A , D B ) on round n , where a i , b i and ε i , i < n , have been played by the players. Definition 15.
Let A and B be Banach spaces and D A and D B be dense sub-sets of the respective spaces. For an ordinal α and a number ε ≥ , and ¯ ε =( ε , . . . , ε n − ) ∈ ( ε, ∞ ) n and ¯ k = ( k , . . . , k n − ) ∈ ω n , and tuples ( a , . . . , a n − ) ∈ A n and ( b , . . . , b n − ) ∈ B n , we define the game EFD A , B α,ε, ¯ ε, ¯ k (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 8 exactly as EF A , B ω,ε, ¯ ε (( D A , ( a , . . . , a n − )) , ( D B , ( b , . . . , b n − ))) , but on each round i , I also chooses some ε i + n > ε , some k i + n < ω and some ordinal α i < α sothat for all i < j , α j < α i . A play ends after the round i when I choosesthe ordinal α i = 0 . We call this game the dynamic Ehrenfeucht–Fraïssé gameof precision ε and clock α , between the sets D A and D B , with starting position (( a , . . . , a n − ) , ( b , . . . , b n − ) , ¯ ε, ¯ k ) .Player II wins a play if for each i , for all k i -good ϕ ( v , . . . , v k i − ) we have A | = ϕ ( a j , . . . , a j ki − ) = ⇒ B | = ϕ ε i ( b j , . . . , b j ki − ) for all j , . . . , j k i − ≥ i .If n = 0 , we denote the game by EFD A , B α,ε ( D A , D B ) . Remark . The game
EFD A , B α,ε, ¯ ε, ¯ k (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) corre-sponds to a position in the game EFD A , B α + n,ε ( D A , D B ) on round n , where a i , b i , k i and ε i , i < n , have been played by the players, and on round ≤ i < n I hasplayed α i = α + ( n − − i ) .Winning strategies for each player are defined as usual. If I has a winningstrategy in a game G , we write I ↑ G , and if II has a winning strategy, we write II ↑ G . We immediately make the following observations. Lemma 17. (i) Both EF A , B ω,ε, ¯ ε (( D A , ( a , . . . , a n − )) , ( D B , ( b , . . . , b n − ))) and EFD A , B α,ε, ¯ ε, ¯ k (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) are determined.(ii) If II ↑ EFD A , B α,ε, ¯ ε (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) and β ≤ α , then also II ↑ EFD A , B β,ε, ¯ ε (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) .(iii) If α is a limit and II ↑ EFD A , B β,ε, ¯ ε (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) forall β < α , then II ↑ EFD A , B α,ε, ¯ ε (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) . Proof. (i) Clear.(ii) All moves of I in the α -game are also valid moves in the β -game, so byfollowing her strategy for the α -game, II wins the β -game.(iii) In the α -game I plays some α < α , so by following her strategy for thegame of clock α + 1 , II wins the α -game. (cid:3) Lemma 18.
Suppose II ↑ EFD A , B α,ε, ¯ ε, ¯ k (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) and D A ⊆ A and D B ⊆ B are dense. Then there is a winning strategy for II in the game EFD A , B α,ε, ¯ ε, ¯ k (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) whose range is contained in D A ∪D B . Proof.
Let τ be a winning strategy for II and fix for any ε ′ > and k < ω a number δ ( ε ′ , k ) that is a perturbation distance for k and ε ′ in both spaces and in additionincreasing with respect to ε ′ and decreasing with respect to k . We define a new win-ning strategy τ ′ whose range is contained in the union of the dense sets. If x i , ε i , α i AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES 9 and k i are valid moves of I for i ≤ m , we let τ ′ (( x , ε , k , α ) , . . . , ( x m , ε m , k m , α m )) be any such y m that(i) if x m ∈ A , then y m ∈ D B , and if x m ∈ B , then y m ∈ D A ,(ii) k y m − z k < δ (( ε i − ε ) / , ˜ k i ) for all i ≤ m , where z = τ (( x , ε + ( ε − ε ) / , ˜ k , α ) , . . . , ( x m , ε + ( ε m − ε ) / , ˜ k m , α m )) and ˜ k i = k i ⌈ e ε i ⌉ .We now prove that this works. Let (( x i , ε i , k i , α i ) , y i ) i Suppose I ↑ EFD A , B α,ε, ¯ ε, ¯ k (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) and D A ⊆ A and D B ⊆ B are dense. Then there is a winning strategy for I in the game EFD A , B α,ε, ¯ ε, ¯ k (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) whose range is contained in D A ∪D B . AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES10 Proof. Let τ be a winning strategy for I and fix for any ε ′ > and k < ω a number δ ( ε ′ , k ) that is a perturbation distance for k and ε ′ in both spaces and in additionincreasing with respect to ε ′ and decreasing with respect to k . We define a newwinning strategy τ ′ whose range is contained in the union of the dense sets. If y i , i < m , are valid moves of II , we let τ ′ ( y , . . . , y m − ) = ( x m , ε m , k m , α m ) , where(i) τ ( y , . . . , y m − ) = ( z, ˜ ε m , ˜ k m , α m ) ,(ii) ε m = ε + (˜ ε m − ε ) / ,(iii) k m = ˜ k m (cid:6) e ˜ ε m (cid:7) ,(iv) if z ∈ A , then x m ∈ D A , and if z ∈ B , then x m ∈ D B ,(v) k x m − z k < δ ((˜ ε i − ε ) / , k i ) for all i ≤ m ,Now we show that this works. Let (( x i , ε i , k i , α i ) , y i ) i Corollary 22. II ↑ EFD A , B α,ε, ¯ ε, ¯ k (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) if and only if II ↑ EFD A , B α,ε, ¯ ε, ¯ k (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) , whenever D A ⊆ A and D B ⊆ B are dense. Proof. Putting together Corollaries 19 and 21 yields this result. (cid:3) Theorem 23. For separable A and B , II ↑ EF A , B ω,ε ( A , B ) if and only if there existsan ε -isomorphism A → B . Proof. If there exists an ε -isomorphism f , II wins the game by playing moves f ( a n ) when I chooses a n ∈ A and f − ( b n ) when he chooses b n ∈ B .For the converse, let D A ⊆ A and D B ⊆ B be countable and dense. Let ( ε i ) i<ω be a sequence converging to ε and let ( x i ) i<ω enumerate D A ∪ D B withoutrepetition. Let I play ( x i , ε i ) on round i in the game EF A , B ω,ε ( D A , D B ) , and let y i be the choices of II following a winning strategy.We show that whenever ( i n ) n<ω is an increasing sequence of indices, ( a i n ) n<ω isCauchy iff ( b i n ) n<ω is. Suppose ( a i n ) n<ω is Cauchy and let r > be rational. Let k ≥ be large enough so that e ε +1 /r ≤ k , ε k − ε ≤ and for all m and n suchthat i m , i n ≥ k , we have and k a i m − a i n k ≤ re − ( ε +1) . Then the formula P ( e ε +1 r v − e ε +1 r v ) is k -good, so A | = P ( e ε +1 r a i m − e ε +1 r a i n ) implies B | = P ( e − ε k ( e ε +1 r b i m − e ε +1 r b i n )) , whence k b i m − b i n k ≤ e ε k − ε − r ≤ e r = r for large enough m and n ,proving that ( b i n ) n<ω is Cauchy. In the other direction one can use exactly thesame argument after noticing that the winning condition of the game also implies,by Lemma 8, that for all k and k -good term t ( v , . . . , v k − ) and j , . . . , j k − ≥ k ,we have B | = P ( t ( b j , . . . , b j k − )) = ⇒ A | = P ε k ( t ( a j , . . . , a j k − )) . This shows that the function ϑ : A → B defined by ϑ (lim n →∞ a i n ) = lim n →∞ b i n is a well-defined partial injection. In fact it is a total bijection because it mapslimit points of ( a i ) i<ω to limit points of ( b i ) i<ω , and these sequences go throughthe dense sets D A and D B respectively.Next we show that ϑ is an ε -isomorphism. By Lemma 12, it is enough to showthat for all k < ω , for all k -good ϕ (¯ v ) and for all ¯ c ∈ A k , A | = ϕ (¯ c ) = ⇒ B | = ϕ ε ( ϑ (¯ c )) . So let k < ω , ϕ (¯ v ) k -good and ¯ c ∈ A k and assume that A | = ϕ (¯ c ) . Fix some ( i mn ) n<ω , m < k , such that a i mn → c m . AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES12 First suppose that ϕ = P ( P k − j =0 q j v j ) and denote η ( n ) = max m AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES13 Proof. The idea of II in the game EFD A , B α,ε ( A , B ) is to pretend that she is playing EF A , B ω,ε ( A , B ) against a player that plays I ’s moves but in rate slow enough so thatthe round number k catches up with the k i ’s I plays in the α -game. To be moreprecise, let τ be a winning strategy for II in the long game. Define a strategy τ ′ for II in the α -game by setting τ ′ (( x , ε , k , α ) , . . . , ( x n , ε n , k n , α n )) = τ ((˜ x i , ˜ ε i ) i ≤ m ) , where(i) m = k n ,(ii) ˜ x k i = x i and ˜ ε k i = ε i for i ≤ n and(iii) ˜ x j = ˜ x k i and ˜ ε j = ˜ ε k i for k i < j < k i +1 .We now show that II wins any play using this strategy. Suppose for a contra-diction that (( x i , ε i , k i , α i ) , y i ) i Let D A ⊆ A and D B ⊆ B be dense sets of minimal cardinality and let κ = |D A | + |D B | . The strategy for II in EF A , B ω,ε ( A , B ) is to play such moves thatthe following condition holds at every round: ( ∗ ) If the position in EF A , B ω,ε ( A , B ) is (( x , ε ) , y , . . . , ( x n − , ε n − ) , y n − ) , thenfor all α < κ + II ↑ EFD A , B α,ε, ( ε ,...,ε n − ) , (0 ,...,n − (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) . We show that if this kind of a strategy exists, it is winning. We may assumethat I has played a decreasing sequence of ε i ’s. Suppose for a contradiction that II loses a play using this strategy. Then there are k < ω , k -good ϕ ( v , . . . , v k − ) and k ≤ i < · · · < i k − , such that A | = ϕ ( a i , . . . , a i k − ) and B = ϕ ε k ( b i , . . . , b i k − ) .But consider the game EFD A , B ,ε, ( ε ,...,ε ik − ) , (0 ,...,i k − ) (cid:0) ( A , a , . . . , a i k − ) , ( B , b , . . . , b i k − ) (cid:1) . As by ( ∗ ) II wins this game and A | = ϕ ( a i , . . . , a i k − ) , we must also have B | = ϕ ε k ( b i , . . . , b i k − ) , a contradiction.Next we show that the strategy really exists. We show that II can always playsuch moves that ( ∗ ) holds by induction on the round n . If n = 0 , then no moves Here we, for simplicity of notation, assume that k < · · · < k n and ε n ≤ · · · ≤ ε . AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES14 have been played yet and by initial assumption, ( ∗ ) is satisfied. Suppose that ( ∗ ) is true until round n and, on round n + 1 , let I play the element x n and ε n . Now II will have to respond with some y n so that ( ∗ ) will remain true for the nextposition. By Lemma 18, if II has any chance of survival, she can pick her movefrom the dense set D A ∪ D B . Let Y be the set of all possible moves of II in thedense set and assume that no y n ∈ Y will make ( ∗ ) true. This means that forevery y n ∈ Y there is some ordinal α y n < κ + such that II EFD A , B α yn ,ε, ( ε ,...,ε n ) , (0 ,...,n ) (( D A , a , . . . , a n ) , ( D B , b , . . . , b n )) , where a n and b n are obviously these x n and y n . Let α = sup y n ∈ Y ( α y n + 1) . As | Y | ≤ |D A ∪ D B | = κ , we get α < κ + . By the induction hypothesis, II ↑ EFD A , B α +1 ,ε, ( ε ,...,ε n − ) , (0 ,...,n − (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) . Let, in this game, I play x n , ε n , n and α . Then the winning strategy gives us some y n ∈ Y . But now II ↑ EFD A , B α,ε, ( ε ,...,ε n ) , (0 ,...,n ) (( D A , a , . . . , a n ) , ( D B , b , . . . , b n )) , which, as α y n ≤ α , implies II ↑ EFD A , B α yn ,ε, ( ε ,...,ε n ) , (0 ,...,n ) (( D A , a , . . . , a n ) , ( D B , b , . . . , b n )) , a contradiction. (cid:3) Corollary 26. If I ↑ EF A , B ω,ε, ¯ ε (( A , a , . . . , a n − ) , ( B , b , . . . , b n − )) and D A ⊆ A and D B ⊆ B are dense, then I ↑ EF A , B ω,ε, ¯ ε (( D A , a , . . . , a n − ) , ( D B , b , . . . , b n − )) . Proof. If I wins the long game between the models, he wins, by Theorem 25, somedynamic game between the models, so by Corollary 19 he wins the same dynamicgame between the dense sets, so by Theorem 24 he wins the long game betweenthe dense sets. (cid:3) Scott rank In this section, we define the Scott rank with respect to our games and provesome of its properties that will be used to construct the Scott sentences in thenext section. Definition 27. Let ε ≥ , ¯ ε ∈ ( ε, ∞ ) n and ¯ k = ( k , . . . , k n − ) ∈ ω n , with k < · · · < k n − . For Banach spaces A and B and tuples ¯ a ∈ A n and ¯ b ∈ B n , wedefine the ε -Scott watershed SW ¯ ε, ¯ kε (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) of the pair (( A , ¯ a ) , ( B , ¯ b )) withparameters ¯ ε and ¯ k to be the least ordinal α such that I ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) . Remark . Since the dynamic EF-game is determined, the watershed always existsunless II wins the game for all α . Note that it is also a successor ordinal, as II having a winning strategy for every ordinal below some limit implies II having awinning strategy at the limit. AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES15 Definition 29. The ε -Scott rank SR ε ( A ) of a Banach space A is the supremumof all ordinals sup n<ω sup ¯ ε ∈ ( ε, ∞ ) n sup ¯ k ∈ ω n sup ¯ a ∈ A n sup ¯ b ∈ B n SW ¯ ε, ¯ kε (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) over all Banach spaces B such that density( B ) ≤ density( A ) . Lemma 30. Let κ be an infinite cardinal with density( A ) ≤ κ . Then SR ε ( A ) ≤ κ + . Proof. By Theorem 25, if I wins any dynamic game between A and some otherspace B with density( B ) ≤ density( A ) , he will win one with the clocking ordinalbelow κ + . Thus SR ε ( A ) is a supremum of ordinals < κ + and hence SR ε ( A ) ≤ κ + . (cid:3) Question 31. By Lemma 30 the Scott rank (for linear isomorphism) of a separablemodel is at most ω . Can this be improved?For classical (discrete) structures, the Scott rank of a countable model is al-ways countable. This is essentially because the corresponding dynamic game istransitive, which guarantees that it is enough to consider the wathershed of pairs (( A , ¯ a ) , ( A , ¯ b )) whereas here we seem to need to go through all (linearly isomorphic) copies B of A . Lemma 32. For a fixed Banach space A and ε ≥ , let ( ∗ ) denote the followingproperty of ordinals α :for any other Banach space B such that density( B ) ≤ density( A ) and for all n < ω , ¯ a ∈ A n , ¯ b ∈ B n , ¯ ε ∈ ( ε, ∞ ) n and ¯ k ∈ ω n , if II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) , then II ↑ EFD A , B α +1 ,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) . Then the following hold:(i) If α is an ordinal with property ( ∗ ) , then every β ≥ α has ( ∗ ) .(ii) If α has ( ∗ ) , then for all B , n < ω , ¯ a ∈ A n , ¯ b ∈ B n and ¯ ε and ¯ k , if II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) , then II ↑ EFD A , B β,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) for all β ≥ α .(iii) SR ε ( A ) is the least ordinal with property ( ∗ ) . Proof. Suppose α has ( ∗ ) . We prove (i) and (ii) by simultaneous induction on β ≥ α . If β = α , the claim is trivial. Suppose that β = γ + 1 . The inductionhypothesis is that γ satisfies ( ∗ ) and if, given some parameters, II wins the α -gamewith these parameters, she also wins the γ -game with the same parameters. AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES16 (i) Let B , ¯ a ∈ A n , ¯ b ∈ B n and ¯ ε and ¯ k be arbitrary and suppose that τ is a winning strategy for II in the β -game with these parameters. Wedescribe the winning strategy of II in the β + 1 -game. Suppose I plays ( x n , ε n , k n , α n ) . If α n < β , II can win the play following τ , so we mayassume that α n = β . Now we let II respond with y n = τ ( x n , ε n , k n , γ ) .Since τ is a winning strategy, II ↑ EFD A , B γ,ε, ¯ ε ⌢ ε n , ¯ k ⌢ k n (( A , a , . . . , a n ) , ( B , b , . . . , b n )) . Since γ has ( ∗ ) , now II ↑ EFD A , B β,ε, ¯ ε, ¯ k (( A , a , . . . , a n ) , ( B , b , . . . , b n )) . Butthis means that she can survive the rest of the β + 1 -game. Thus β has ( ∗ ) .(ii) Suppose that II wins the α -game with given parameters. Thus, by induc-tion hypothesis, she wins the γ -game. But then again, as γ has ( ∗ ) , shealso wins the β -game.Then suppose β is a limit. The induction hypothesis is that every α ≤ γ < β has ( ∗ ) and if II wins the α -game, she wins the γ -game.(i) We again describe the winning strategy for II in the β + 1 -game with givenparameters in the case that II wins the β -game with the same parameters.Again, let I play ( x n , ε n , k n , β ) . Let τ be a winning strategy of II in the β -game and respond with y n = τ ( x n , ε n , k n , α ) . Then we know that II ↑ EFD A , B α,ε, ¯ ε, ¯ k (( A , a , . . . , a n ) , ( B , b , . . . , b n )) , so by the induction hypothesis II ↑ EFD A , B γ,ε, ¯ ε, ¯ k (( A , a , . . . , a n ) , ( B , b , . . . , b n )) for all α ≤ γ < β . Next round in the β + 1 -game, let I play the sequence ( x n +1 , ε n +1 , k n +1 , α n +1 ) , where α n +1 < β . Now II survives the rest of thegame by following her winning strategy for α n +1 + 1 -game.(ii) Suppose that II wins the α -game with some parameters. Then by inductionhypothesis she wins every γ -game for γ < β , so then se wins also the β -game.This finishes the proof of (i) and (ii).For (iii), let α = SR ε ( A ) , and let B , ¯ a ∈ A n , ¯ b ∈ B n and ¯ ε and ¯ k bearbitrary. Suppose that II ↑ EFD A , A α,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) . If we suppose fora contradiction that II EFD A , B α +1 ,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) , then this means that I ↑ EFD A , B α +1 ,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) , and thus α + 1 = SW ¯ ε, ¯ kε (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) . Butby the definition of Scott rank, now α + 1 ≤ α which is impossible. Thus II ↑ EFD A , B α +1 ,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) , so α has ( ∗ ) .Last we show that α is the least ordinal that has ( ∗ ) . Let β be the leastsuch ordinal and suppose for a contradiction that β < α . Then β + 1 ≤ α .Now by the definition of Scott rank there exist B , ¯ ε , ¯ k , ¯ a and ¯ b such that SW ¯ ε, ¯ kε (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) ≥ β + 1 . Now SW ¯ ε, ¯ kε (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) = γ + 1 for some AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES17 γ ≥ β . As β has ( ∗ ) , by (i) also γ has ( ∗ ) . But by the definition of Scottwatershed, II ↑ EFD A , B γ,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) but II EFD A , B γ +1 ,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) , acontradiction. (cid:3) Scott sentences In this section we define the ε -Scott sentences of Banach spaces and show that,for separable spaces, they characterise the space up to ε -isomorphism. Definition 33. Let A be a Banach space, ε ≥ and D A ⊆ A a dense set ofcardinality < κ . Given n < ω , ¯ ε ∈ ( ε, ∞ ) n ¯ k ∈ ω n and ¯ a ∈ A n , define the formulas ε σ A , ¯ aα, ¯ ε, ¯ k ( v , . . . , v n − ) ∈ L κω , α ∈ On, by induction on α as follows:(i) For α = 0 , ε σ A , ¯ aα, ¯ ε, ¯ k ( v , . . . , v n − ) = ^ { ϕ ε i ( v j , . . . , v j ki − ) | i < n, ϕ k i -good, j , . . . , j k i − ≥ i, A | = ϕ ( a j , . . . , a j ki − ) } , (ii) for α = β + 1 , ε σ A , ¯ aα, ¯ ε, ¯ k ( v , . . . , v n − ) = ^ ε n ∈ ( ε,ε n − ) ∩ Q ^ k n <ω ^ a n ∈D A ∃ v nε σ A , ¯ a ⌢ a n β, ¯ ε ⌢ ε n , ¯ k ⌢ k n ( v , . . . , v n ) ∧ ∀ v n _ a n ∈D A ε σ A , ¯ a ⌢ a n β, ¯ ε ⌢ ε n , ¯ k ⌢ k n ( v , . . . , v n ) ! , and(iii) for α limit, ε σ A , ¯ aα, ¯ ε, ¯ k ( v , . . . , v n − ) = ^ β<α ε σ A , ¯ aβ, ¯ ε, ¯ k ( v , . . . , v n − ) . Whenever n = 0 , we leave out ¯ ε , ¯ k and ¯ a (who all are the empty tuple) from thenotation. Lemma 34. For Banach spaces A and B , and ¯ a ∈ A n and ¯ b ∈ B n , the followingare equivalent:(i) B | = ε σ A , ¯ aα, ¯ ε, ¯ k (¯ b ) ,(ii) II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) . Proof. We prove this by induction on α . For the case α = 0 , first assume that B | = ε σ A , ¯ a , ¯ ε, ¯ k (¯ b ) . Then for all i < n , k i -good ϕ (¯ v ) and j , . . . , j k i − ≥ i such that A | = ϕ ( a j , . . . , a j ki − ) , we have B | = ϕ ε i ( b j , . . . , b j ki − ) . But this exactly meansthat II wins the game EFD A , B ,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) . Conversely, if we assume that II ↑ EFD A , B ,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) , then for any i and k i -good ϕ we have A | = ϕ ( a j , . . . , a j ki − ) = ⇒ B | = ϕ ε i ( b j , . . . , b j ki − ) AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES18 for all j , . . . , j k i − ≥ i , so B | = ε σ A , ¯ a , ¯ ε, ¯ k (¯ b ) .If α is limit, the equivalence of the two claims follows directly from the inductionhypothesis.So suppose that α = β + 1 . Suppose that II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) . Then,given any ε n , k n and a n ∈ D A , II can respond with some b n ∈ B so that II ↑ EFD A , B β,ε, ¯ ε ⌢ ε n , ¯ k ⌢ k n (cid:0) ( D A , ¯ a ⌢ a n ) , ( B , ¯ b ⌢ b n ) (cid:1) , and by induction hypothesis this meansthat for such b n we get B | = ε σ A , ¯ a ⌢ a n β, ¯ ε ⌢ ε n , ¯ k ⌢ k n (¯ b, b n ) . On the other hand given any b n ∈ B , II can play some a n ∈ D A so that B | = ε σ A , ¯ a ⌢ a n β, ¯ ε ⌢ ε n , ¯ k ⌢ k n (¯ b, b n ) . Hence B | = ε σ A , ¯ aα, ¯ ε, ¯ k (¯ b ) .Conversely, suppose that B | = ε σ A , ¯ aα, ¯ ε, ¯ k (¯ b ) . Then for any given ε n and k n ,whenever a n ∈ D A there is b n ∈ B with B | = ε σ A , ¯ a ⌢ a n β, ¯ ε ⌢ ε n , ¯ k ⌢ k n (¯ b, b n ) and viceversa. By the induction hypothesis this means that for any ε n and k n , II ↑ EFD D A , B β,ε, ¯ ε ⌢ ε n , ¯ k ⌢ k n (cid:0) ( A , ¯ a ⌢ a n ) , ( B , ¯ b ⌢ b n ) (cid:1) . Thus II has a winning strategy in EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) . (cid:3) Remark . By Corollary 22, if D A and D ′ A are two different dense sets in A , weget II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) ⇐⇒ II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( A , ¯ a ) , ( B , ¯ b ) (cid:1) ⇐⇒ II ↑ EFD A , B α,ε, ¯ ε, ¯ k (cid:0) ( D ′ A , ¯ a ) , ( B , ¯ b ) (cid:1) , so constructing ε σ A , ¯ aα, ¯ ε, ¯ k (¯ x ) using two different dense sets of cardinality κ gives logi-cally equivalent formulas. Definition 36. The ε -Scott sentence σ A ε of a Banach space A is ε σ A SR ε ( A ) ∧ ^ n<ω ^ ¯ ε ∈ (( ε, ∞ ) ∩ Q ) n ^ ¯ k ∈ ω n ^ ¯ a ∈D n A ∀ v . . . ∀ v n − (cid:16) ε σ A , ¯ a SR ε ( A ) , ¯ ε, ¯ k → ε σ A , ¯ a SR ε ( A )+1 , ¯ ε, ¯ k (cid:17) . Theorem 37. For Banach spaces A and B , B | = σ A ε ⇐⇒ II ↑ EF A , B ω,ε ( A , B ) . Proof. First assume B | = σ A ε . By Corollary 26, it is enough for II to win the game EF A , B ω,ε ( D A , B ) . The strategy of II in this game is to play such moves y i that atall times the following holds: ( ∗ ) If the position in EF A , B ω,ε ( D A , B ) is (( x , ε ) , y , . . . , ( x n − , ε n − ) , y n − ) , then B | = ε σ A , ( a ,...,a n − )SR ε ( A ) , ( ε ,...,ε n − ) , (0 ,...,n − ( b , . . . , b n − ) .This strategy guarantees the win in the long game, as the same argument that weused in the proof of Theorem 25 works also here (after the observations that isLemma 34).So we show that there is such a strategy. In the beginning when no moves havebeen made, we know that B | = ε σ A SR ε ( A ) by the definition of the Scott sentence. So AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES19 suppose we are at round n and the position is (( x , ε ) , y , . . . , ( x n − , ε n − ) , y n − ) .We let I play x n and ε n . Assume that x n ∈ B (the other case is similar). By theinduction hypothesis, B | = ε σ A , ( a ,...,a n − )SR ε ( A ) , ( ε ,...,ε n − ) , (0 ,...,n − ( b , . . . , b n − ) . But then from the Scott sentence it follows that B | = ε σ A , ( a ,...,a n − )SR ε ( A )+1 , ( ε ,...,ε n − ) , (0 ,...,n − ( b , . . . , b n − ) . Thus B | = ^ ε n ^ k n ∀ v n _ a n ε σ A , ( a ,...,a n )SR ε ( A ) , ( ε ,...,ε n ) , (0 ,...,n ) ( b , . . . , b n − , x n ) , so we can find some y n ∈ D A such that B | = ε σ A , ( a ,...,a n )SR ε ( A ) , ( ε ,...,ε n ) , (0 ,...,n ) ( b , . . . , b n ) . This completes this direction of the proof.Then assume that II ↑ EF A , B ω,ε ( A , B ) . By Lemma 24, II ↑ EFD A , B α,ε ( A , B ) for all α ∈ On. In particular, she wins the game EFD A , B SR ε ( A ) ,ε ( A , B ) . By Lemma 34, B | = ε σ A SR ε ( A ) . Left is to show that B | = ^ n<ω ^ ¯ ε ∈ (( ε, ∞ ) ∩ Q ) n ^ ¯ k ∈ ω n ^ ¯ a ∈D n A ∀ v . . . ∀ v n − (cid:16) ε σ A , ¯ a SR ε ( A ) , ¯ ε, ¯ k → ε σ A , ¯ a SR ε ( A )+1 , ¯ ε, ¯ k (cid:17) . So let n , ¯ ε , ¯ k , and ¯ a ∈ A n and ¯ b ∈ B n be arbitrary, and suppose that B | = ε σ A , ¯ a SR ε ( A ) , ¯ ε, ¯ k (¯ b ) . Again by Lemma 34, this means that II ↑ EFD A , B SR ε ( A ) ,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) . By Lemma 32, we get that II ↑ EFD A , B SR ε ( A )+1 ,ε, ¯ ε, ¯ k (cid:0) ( D A , ¯ a ) , ( B , ¯ b ) (cid:1) . Again, by Lemma 34, we get that B | = ε σ A , ¯ a SR ε ( A )+1 , ¯ ε, ¯ k (¯ b ) , showing that B | = ε σ A , ¯ a SR ε ( A ) , ¯ ε, ¯ k (¯ b ) → ε σ A , ¯ a SR ε ( A )+1 , ¯ ε, ¯ k (¯ b ) . As ¯ b was arbitrary, we get that B | = ∀ v . . . ∀ v n − (cid:16) ε σ A , ¯ a SR ε ( A ) , ¯ ε, ¯ k → ε σ A , ¯ a SR ε ( A )+1 , ¯ ε, ¯ k (cid:17) , and since all the other parameters were also arbitrary, we can conclude that B | = ^ n<ω ^ ¯ ε ∈ (( ε, ∞ ) ∩ Q ) n ^ ¯ k ∈ ω n ^ ¯ a ∈D n A ∀ v . . . ∀ v n − (cid:16) ε σ A , ¯ a SR ε ( A ) , ¯ ε, ¯ k → ε σ A , ¯ a SR ε ( A )+1 , ¯ ε, ¯ k (cid:17) , finishing the proof. (cid:3) Notice that while I plays any ε n , the conjunction is taken over rational ε n . This, however,does not matter as one can pick a slightly smaller rational number and choose y n according tothis stricter ε n and this will not be a worse choice for II . AMES AND SCOTT SENTENCES FOR LINEAR ISOMORPHISMS OF BANACH SPACES20 Corollary 38. For separable Banach spaces A and B , B | = σ A ε if and only if thereexists an ε -isomorphism A → B . Proof. Combine Theorems 23 and 37. (cid:3) Remark . (i) If A is separable, then SR ε ( A ) ≤ ω , so σ A ε ∈ L ω ω . Hence we have shownthat there exists a sentence of L ω ω that characterises a separable Banachspace up to ε -isomorphism.(ii) Now the sentence W ε ≥ σ A ε characterises a separable Banach space A up tolinear isomorphism.(iii) The Banach–Mazur distance of two Banach spaces A and B is defined as d ( A , B ) = log(inf {k T k (cid:13)(cid:13) T − (cid:13)(cid:13) : T : A → B is an isomorphism } ) . 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