aa r X i v : . [ m a t h . F A ] A p r α -MINIMAL BANACH SPACES CHRISTIAN ROSENDAL
Abstract.
A Banach space W with a Schauder basis is said to be α -minimal for some α < ω if, for any two block subspaces Z , Y ⊆ W , the Bourgainembeddability index of Z into Y is at least α .We prove a dichotomy that characterises when a Banach space has an α -minimal subspace, which contributes to the ongoing project, initiated by W.T. Gowers, of classifying separable Banach spaces by identifying characteristicsubspaces. Contents
1. Introduction 12. Setup 63. Proof of Theorem 3 74. Proof of Theorem 4 84.1. Generalised α -games 84.2. Ramsey determinacy of adversarial α -games 104.3. A game theoretic dichotomy 124.4. The embeddability index 18References 221. Introduction
Suppose W is a separable, infinite-dimensional Banach space. We say that W is minimal if W isomorphically embeds into any infinite-dimensional subspace Y ⊆W (and write
W ⊑ Y to denote that W embeds into Y ). The class of Banachspaces without minimal subspaces was studied by V. Ferenczi and the author in[2], extending work of W. T. Gowers [3] and A. M. Pelczar [5], in which a dichotomywas proved characterising the presence of minimal subspaces in an arbitrary infinite-dimensional Banach space.The dichotomy hinges on the notion of tightness , which we can define as follows.Assume that W has a Schauder basis ( e n ) and suppose Y ⊆ W is a subspace. Wesay that Y is tight in the basis ( e n ) for W if there are successive finite intervals of Mathematics Subject Classification.
Primary: 46B03, Secondary 03E15.
Key words and phrases.
Ramsey Theory, Infinite games in vector spaces, Isomorphic classifi-cation of Banach spaces.The initial research for this article was done while the author was visiting V. Ferenczi at theUniversity of S˜ao Paulo, Brazil, with the support of FAPESP. The author’s research was likewisesupported by NSF grants DMS 0901405 and DMS 0919700. N , I < I < I < . . . ⊆ N , such that for any isomorphic embedding T : Y → W , if P I m denotes the canonicalprojection of W onto [ e n ] n ∈ I m , thenlim inf m →∞ k P I m T k > . Alternatively, this is equivalent to requiring that whenever A ⊆ N is infinite, thereis no embedding of Y into [ e n (cid:12)(cid:12) n / ∈ S m ∈ A I m ]. Also, the basis ( e n ) is tight if anyinfinite-dimensional subspace Y ⊆ W is tight in ( e n ) and a space is tight in case ithas a tight basis. We note that if W is tight, then so is any shrinking basic sequencein W .Tightness is easily seen to be an obstruction to minimality, in the sense thata tight space cannot contain a minimal subspace. In [2] the following converse isproved: any infinite-dimensional Banach space contains either a minimal or a tightsubspace.J. Bourgain introduced in [1] an ordinal index that gives a quantitative measureof how much one Banach space with a basis embeds into another. Namely, suppose W is a space with a Schauder basis ( e n ) and Y is any Banach space. We let T (( e n ) , Y , K ) be the tree of all finite sequences ( y , y , . . . , y k ) in Y , including theempty sequence ∅ = ( ), such that( y , . . . , y k ) ∼ K ( e , . . . , e k ) . Here, whenever ( x i ) and ( y i ) are sequences of the same (finite or infinite) length inBanach spaces X and Y , we write ( x i ) ∼ K ( y i )if for all a , . . . , a k ∈ R K (cid:13)(cid:13)(cid:13) k X i =0 a i x i (cid:13)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13) k X i =0 a i y i (cid:13)(cid:13)(cid:13) K (cid:13)(cid:13)(cid:13) k X i =0 a i x i (cid:13)(cid:13)(cid:13) . We notice that T (( e n ) , Y , K ) is ill-founded , i.e., admits an infinite branch, if andonly if W = [ e n ] embeds with constant K into Y .The rank function ρ T on a well-founded tree T , i.e., without infinite branches, isdefined by ρ T ( s ) = 0 if s ∈ T is a terminal node and ρ T ( s ) = sup (cid:8) ρ T ( t ) + 1 (cid:12)(cid:12) s ≺ t, t ∈ T (cid:9) otherwise. Then, the rank of T is defined byrank( T ) = sup (cid:8) ρ T ( s ) + 1 (cid:12)(cid:12) s ∈ T (cid:9) , whence rank( T ) = ρ T ( ∅ ) + 1 if T is non-empty. Moreover, if T is ill-founded, welet rank( T ) = ∞ , with the stipulation that α < ∞ for all ordinals α .Then, rank (cid:0) T (( e n ) , Y , K ) (cid:1) measures the extent to which W = [ e n ] K -embedsinto Y and we therefore define the embeddability rank of W = [ e n ] into Y byEmb(( e n ) , Y ) = sup K > rank (cid:0) T (( e n ) , Y , K ) (cid:1) . Since ( e n ) is a basic sequence, there is for any K > δ n ) ofpositive real numbers, such that if y n , z n ∈ Y , k y n − z n k < δ n and ( y , . . . , y k ) ∼ K ( e , . . . , e k ), then also ( z , . . . , z k ) ∼ K +1 ( e , . . . , e k ). Therefore, to calculate the -MINIMAL BANACH SPACES 3 embeddability rank, Emb(( e n ) , Y ), it suffices to consider the trees of all finite se-quences ( y , . . . , y k ) with ( y , . . . , y k ) ∼ K ( e , . . . , e k ), where, moreover, we requirethe y n to belong to some fixed dense subset of Y . We shall use this repeatedly lateron, where we replace Y by a dense subset of itself. This comment also implies thatEmb(( e n ) , Y ) is either ∞ , if W ⊑ Y , or an ordinal < density( Y ) + , if W 6⊑ Y . Inparticular, if Y is separable, then Emb(( e n ) , Y ) is either ∞ or a countable ordinal.Also, note that the embeddability rank depends not only on the space W , but alsoon the basis ( e n ). However, if Y is separable and W 6⊑ Y , then by the BoundednessTheorem for coanalytic ranks (see [4]), the supremum of Emb(( e n ) , Y ) over all bases( e n ) for W is a countable ordinal. In case Emb(( e n ) , Y ) > α , we say that W = [ e n ] α -embeds into Y .Since minimality is explicitly expressed in terms of embeddability, it is naturalto combine it with Bourgain’s embeddability index in the following way. Definition 1.
Let α be a countable ordinal. A Banach space W with a Schauderbasis ( e n ) is α -minimal if any block subspace Z = [ z n ] ⊆ W α -embeds into anyinfinite-dimensional subspace Y ⊆ W . It is easy to check that if W = [ e n ] is a space with a basis and X = [ x n ] and Y = [ y n ] are block subspaces of W such that x n ∈ Y for all but finitely many n , which we denote by X ⊆ ∗ Y , then if Y is α -minimal, so is X . In particular, α -minimality is preserved by passing to block subspaces.Similarly, we can combine tightness with the embeddability index. Definition 2.
Let α be a countable ordinal and W a Banach space with a Schauderbasis ( e n ) . We say that W = [ e n ] is α -tight if for any block basis ( y n ) in W thereis a sequence of intervals of N , I < I < I < . . . ⊆ N such that for any infinite set A ⊆ N , Emb (cid:0) ( y n ) , [ e n (cid:12)(cid:12) n / ∈ [ j ∈ A I j ] (cid:1) α. In other words, if Y = [ y n ] ( α + 1) -embeds into some subspace Z ⊆ W , then lim inf k →∞ k P I k | Z k > . Again, it is easy to see that if W = [ e n ] is α -tight, then so is any block subspaceof W . Also, if W = [ e n ] is α -tight, then no block subspace, Y = [ y n ], is β -minimalfor α < β . And, if Y = [ y n ] is minimal, then Y = [ y n ] is α -minimal for any α < ω .It follows from this that if W = [ e n ] is α -tight, then W = [ e n ] admits no minimalblock subspaces, and thus, as any infinite-dimensional subspace contains a blocksubspace up to a small perturbation, W contains no minimal subspaces either.Our first result says that tightness can be reinforced to α -tightness. Theorem 3.
Let W be a Banach space with a Schauder basis and having no min-imal subspaces. Then there is a block subspace X = [ x n ] that is α -tight for somecountable ordinal α . Our main results, however, provides us with more detailed structural informa-tion.
CHRISTIAN ROSENDAL
Theorem 4.
Let W be Banach space with a Schauder basis and suppose α < ω .Then there is a block subspace X = [ x n ] ⊆ W that is either ωα -tight or ( α + 1) -minimal. Finally, combining Theorems 3 and 4, we have the following refinement of The-orem 3.
Theorem 5.
Let W be a Banach space with a Schauder basis. Then W has aminimal subspace or a block subspace X = [ x n ] ⊆ W that is α -minimal and ωα -tight for some countable ordinal α .Proof. Suppose that W has no minimal subspace and pick by Theorem 3 someblock subspace W ⊆ W that is β -tight for some β < ω . So no block subspace of W is ( β + 1)-minimal. Let now α be the supremum of all ordinals γ such that W is saturated with γ -minimal block subspaces and pick a block subspace W ⊆ W not containing any ( α + 1)-minimal subspace.We claim that W contains a α -minimal block subspace W ∞ . If α is a successorordinal, this is obvious, so suppose instead that α is a limit. Then we can find ordi-nals γ < γ < . . . with supremum α . We then inductively choose block subspaces W ⊇ W ⊇ W ⊇ . . . such that W n is γ n -minimal. Letting W ∞ ⊆ W be a blocksubspace such that W ∞ ⊆ ∗ W n for all n , we see that W ∞ is γ n -minimal for all n , which means that for any block sequence ( z m ) ⊆ W ∞ and infinite-dimensionalsubspace Y ⊆ W ∞ , we have Emb (cid:0) ( z m ) , Y (cid:1) > γ n for all n , whence Emb (cid:0) ( z m ) , Y (cid:1) > sup n γ n = α . So W ∞ is α -minimal and so areits subspaces.Now, W ∞ has no ( α + 1)-minimal subspace, so, by Theorem 4, W ∞ contains an ωα -tight block subspace X , which simultaneously is α -minimal. (cid:3) Since any two Banach spaces of the same finite dimension are isomorphic, oneeasily sees that any space W with a Schauder basis ( e n ) is ω -minimal. On the otherhand, in [2], a space W = [ e n ] is defined to be tight with constants if for any blocksubspace Y = [ y n ] there are intervals I < I < I < . . . such that for any integerconstant K , [ y n ] n ∈ I K K [ e n ] n/ ∈ I K . In this case, it follows that for any infinite set A ⊆ N and any K ∈ A ,rank (cid:0) T (( y n ) , [ e n (cid:12)(cid:12) n / ∈ [ j ∈ A I j ] , K ) (cid:1) max I K , and henceEmb(( y n ) , [ e n (cid:12)(cid:12) n / ∈ [ j ∈ A I j ]) = sup K ∈ A rank (cid:0) T (( y n ) , [ e n (cid:12)(cid:12) n / ∈ [ j ∈ A I j ] , K ) (cid:1) ω. So, if W = [ e n ] is tight with constants, we see that W = [ e n ] is ω -tight and ω -minimal.Following [2], we also define a space W to be locally minimal if there is a constant K > W is K -crudely finitely representable in any infinite-dimensionalsubspace, i.e., if for any finite-dimensional F ⊆ W and infinite-dimensional
Y ⊆ W , F ⊑ K Y . Let us first see local minimality in terms of α -minimality. -MINIMAL BANACH SPACES 5 Proposition 6.
Suppose W is a locally minimal Banach space with a Schauderbasis ( e n ) . Then W = [ e n ] is ω -minimal.Proof. Let K be the constant of local minimality. For any infinite-dimensionalsubspace Y ⊆ W , block sequence ( w i ) ⊆ W and α < ω , we need to show thatEmb(( w i ) , Y ) > α . So choose n such that α < ω · n and find some constant C suchthat if x < . . . < x n and y < . . . < y n are finite block sequences of ( e i ) such that K k x i k k y i k K k x i k , then ( x i ) ∼ C ( y i ). We claim thatrank (cid:0) T (( w i ) , Y , C ) (cid:1) > ω · n. To see this, find some block subspace X such that X ⊑ Y . It suffices to prove thatrank (cid:0) T (( w i ) , X , C ) (cid:1) > ω · n. Let k be given. We shall see that ∅ has rank > ω ( n − k − T (( w i ) , X , C ).So choose by local K -minimality some z , . . . , z k − ∈ X such that( w , . . . , w k − ) ∼ K ( z , . . . , z k − ) . It then suffices to show that ( z , . . . , z k − ) has rank > ω ( n −
1) in T (( w i ) , X , C ),or, equivalently, that for any k , it has rank > ω ( n −
2) + k −
1. So choose z k , . . . , z k + k − in X with support after all of z , . . . , z k − such that( w k , . . . , w k + k − ) ∼ K ( z k , . . . , z k + k − ) . Again, it suffices to show that( z , . . . , z k − , z k , . . . , z k + k − )has rank > ω ( n −
2) in T (( w i ) , X , C ). Et cetera.Eventually, we will have produced z , . . . , z k − < z k , . . . , z k + k − < . . . < z k + ... + k n − , . . . , z k + ... + k n − such that for each l ,( w k + ... + k l − , . . . , w k + ... + k l − ) ∼ K ( z k + ... + k l − , . . . , z k + ... + k l − ) . Since we have chosen the successive sections of ( z i ) successively on the basis, wehave, by the choice of C , that( w , . . . , w k + ... + k n − ) ∼ C ( z , . . . , z k + ... + k n − ) , whereby ( z , . . . , z k + ... + k n − ) ∈ T (( w i ) , X , C ) and hence has rank > ω ( n − n )in T (( w i ) , X , C ). This finishes the proof. (cid:3) In [2], another dichotomy was proved stating that any infinite-dimensional Ba-nach space contains a subspace with a basis that is either tight with constants oris locally minimal. In particular, we have the following dichotomy.
Theorem 7 (V. Ferenczi and C. Rosendal [2]) . Any infinite-dimensional Banachspace contains an infinite-dimensional subspace with a basis that is either ω -tightor is ω -minimal. One problem that remains open is to exhibit spaces that are α -minimal and ωα -tight for unbounded α < ω . We are not aware of any construction in theliterature that would produce this, but remain firmly convinced that such spacesmust exist, since otherwise there would be a universal β < ω such that any Banachspace would either contain a minimal subspace or a β -tight subspace, which seemsunlikely. CHRISTIAN ROSENDAL
Problem 8.
Show that there are α -minimal, ωα -tight spaces for unboundedly many α < ω . Out main result, Theorem 5, allows us to refine the classification scheme devel-oped in [3] and [2], by further differentiating the class of tight spaces into α -minimal, ωα -tight for α < ω . Currently, the most interesting direction for further resultswould be to try to distinguish between different classes of minimal spaces, knowingthat these pose particular problems for applying Ramsey Theory.Apart from some basic facts about Schauder bases, the main tools of our paperoriginate in descriptive set theory for which our general reference is the book by A.S. Kechris [4]. In particular, we follow his presentation of trees and games, exceptthat we separate a game from its winning condition and thus talk about playershaving a strategy to play in a certain set , rather than having a strategy to win.2. Setup
For the proof of Theorem 4, we will need to replace Banach spaces with themore combinatorial setting of normed vector space over countable fields, which wewill be using throughout the paper (cf. [6]). So suppose W is a Banach spacewith a Schauder basis ( e n ). By a standard Skolem hull construction, we find acountable subfield F ⊆ R such that for any F -linear combination P mn =0 a n e n , thenorm k P mn =0 a n e n k belongs to F . Let also W be the countable-dimensional F -vector space with basis ( e n ). In the following, we shall exclusively consider the F -vector space structure of W , and thus subspaces etc. refer to F -vector subspaces.We equip W with the discrete topology, whereby any subset is open, and equip itscountable power W N with the product topology. Since W is a countable discreteset, W N is a Polish, i.e., separable and completely metrisable, space. Notice that abasis for the topology on W N is given by sets of the form N ( x , . . . , x k ) = { ( y n ) ∈ W N (cid:12)(cid:12) y = x & . . . & y k = x k } , where x , . . . , x k ∈ W . Henceforth, we let x, y, z, v be variables for non-zero el-ements of W . If x = P a n e n ∈ W , we define the support of x to be the finite,non-empty set supp( x ) = { n (cid:12)(cid:12) a n = 0 } and set for x, y ∈ W , x < y ⇔ ∀ n ∈ supp( x ) ∀ m ∈ supp( y ) n < m. Similarly, if k is a natural number, we set k < x ⇔ ∀ n ∈ supp( x ) k < n. Analogous notation is used for finite subsets of N and finite-dimensional subspacesof W . A finite or infinite sequence ( x , x , x , x , . . . ) of vectors is said to be a blocksequence if for all n , x n < x n +1 .Note that, by elementary linear algebra, for all infinite-dimensional subspaces X ⊆ W there is a subspace Y ⊆ X spanned by an infinite block sequence,called a block subspace . Henceforth, we use variables X, Y, Z, V to denote infinite-dimensional block subspaces of W . Also, denote finite sequences of non-zero vectorsby variables ~x, ~y, ~z, ~v . Finally, variables E, F are used to denote finite-dimensionalsubspaces of W . -MINIMAL BANACH SPACES 7 Proof of Theorem 3
We should first recall a natural strengthening of tightness from [2]. Suppose W is a Banach space with a Schauder basis ( e n ) and find F and W as in section 2.Let also bb ( e n ) ⊆ W N be the closed set of all block sequences in W N . Let I be thecountable set of all non-empty finite intervals { n, n + 1 , . . . , m } ⊆ N and give I N theproduct topology, where I is taken discrete. We say that W = [ e n ] is continuouslytight if there is a continuous function f : bb ( e n ) → I N such that for any block sequence ( y n ) ∈ W N , f (cid:0) ( y n ) (cid:1) = ( I n ) ∈ I N is a sequence ofintervals such that I < I < I < . . . and such that whenever A ⊆ N is infinite,[ y n ] [ e n (cid:12)(cid:12) n / ∈ [ k ∈ A I k ] . In other words, f continuously chooses the sequence of intervals witnessing tight-ness.As in the case of Banach spaces, for any K >
1, block subspace Y ⊆ W , and blocksequence ( x n ) of ( e n ), we define T (( x n ) , Y, K ) to be the non-empty tree consistingof all finite sequences ( y , . . . , y k ) in Y such that( y , . . . , y k ) ∼ K ( x , . . . , x k ) . Similarly define the embeddability index of ( x n ) in Y byEmb(( x n ) , Y ) = sup K > rank (cid:0) T (( x n ) , Y, K ) (cid:1) . Then, if Y denotes the closed R -linear subspace of W spanned by Y , we have, aswas observed earlier, thatEmb(( x n ) , Y ) = Emb(( x n ) , Y ) . We recall the statement of Theorem 3.
Theorem 9.
Let W be a Banach space with a Schauder basis ( e n ) and having nominimal subspaces. Then there is a block subspace X = [ x n ] that is α -tight for somecountable ordinal α .Proof. By the results of [2], we have that, as W has no minimal subspaces, there isa block subspace X = [ x n ] of W = [ e n ] that is continuously tight as witnessed bya function f . So it suffices to show that for some α < ω and any block sequence( y n ) of ( x n ), if ( I n ) = f (cid:0) ( y n ) (cid:1) , thenEmb (cid:0) ( y n ) , [ x n (cid:12)(cid:12) n / ∈ [ k ∈ A I k ] (cid:1) α, for any infinite set A ⊆ N .Note that if D is any countable set, we can equip the power set P ( D ) with thecompact metric topology obtained from the natural identification with 2 D . Let [ N ]denote the space of infinite subsets of N equipped with the Polish topology inducedfrom P ( N ). We define a Borel measurable function between Polish spaces T : bb ( x n ) × [ N ] × N → P ( X < N ) , by setting T (( y n ) , A, K ) = T (( y n ) , [ x n (cid:12)(cid:12) n / ∈ [ j ∈ A I j ] , K ) , CHRISTIAN ROSENDAL where ( I n ) = f (cid:0) ( y n ) (cid:1) .By assumption, the image of T is an analytic set of well-founded trees on X . So,by the Boundedness Theorem for analytic sets of well-founded trees, there is some α < ω such that sup (( y n ) ,A,K ) ∈ bb ( x n ) × [ N ] × N rank (cid:0) T (( y n ) , [ x n (cid:12)(cid:12) n / ∈ [ j ∈ A I j ] , K ) (cid:1) α, whereby, for any block sequence ( y n ) of ( x n ) and any infinite subset A ⊆ N ,Emb (cid:0) ( y n ) , [ x n (cid:12)(cid:12) n / ∈ [ k ∈ A I k ] (cid:1) α, showing that X is α -tight. (cid:3) Proof of Theorem 4
Generalised α -games. Suppose X ⊆ W and α is a countable ordinal num-ber. We define the generalised Gowers α -game below X , denoted G αX , between twoplayers I and II as follows: I Y Y Y k ξ < α ξ < ξ ξ k < ξ k − . . . II F ⊆ Y F ⊆ Y F k ⊆ Y k x ∈ F x ∈ F + F x k ∈ F + . . . + F k Here α > ξ > ξ > . . . > ξ k = 0 is a strictly decreasing sequence of ordinals, Y l ⊆ X are block subspaces, the F l ⊆ Y l are finite-dimensional subspaces, and x l ∈ F + F + . . . + F l non-zero vectors. Since I plays a strictly decreasing sequenceof ordinals, the game will end once ξ k = 0 has been chosen and II has respondedwith some x k . We then say that the sequence ( x , . . . , x k ) of non-zero vectors isthe outcome of the game.Similarly, we can define the asymptotic α -game below X , F αX , as follows I n n n k ξ < α ξ < ξ ξ k < ξ k − . . . II n < F n < F n k < F k x ∈ F x ∈ F + F x k ∈ F + . . . + F k Here again, α > ξ > ξ > . . . > ξ k = 0 is a strictly decreasing sequence ofordinals, n l natural numbers, the F l are finite-dimensional subspaces of [ e i ] ∞ i = n l +1 ,and x l ∈ F + F + . . . + F l non-zero vectors. The game ends once I has played ξ k = 0 and II has responded with some x k . The outcome is the sequence of non-zerovectors ( x , . . . , x k ).If ~x is a finite sequence of non-zero vectors, we define the games G αX ( ~x ), F αX ( ~x )as above, except that the outcome is now ~x ˆ( z , . . . , z k ).We also define adversarial α -games by mixing the games above. For this, suppose E, F are finite-dimensional subspaces of W and ~z is an even-length sequence of non-zero vectors. -MINIMAL BANACH SPACES 9 We define A αX ( ~z, E, F ) by n < E n < E n k < E k x x x k I Y Y Y k ξ ξ ξ k . . .n n n II F ⊆ Y F ⊆ Y F k ⊆ Y k y y y k and B αX ( ~z, E, F ) by: E ⊆ Y E ⊆ Y E k ⊆ Y k x x x k I n n n k ξ ξ ξ k . . .Y Y Y II n < F n < F n k < F k y y y k where α > ξ > ξ > . . . > ξ k = 0is a decreasing sequence of ordinals, Y l ⊆ X are block subspaces, and n l naturalnumbers. Moreover, in A αX ( ~z, E, F ), E l ⊆ X ∩ [ e i ] ∞ i = n l +1 and F l ⊆ Y l are finite-dimensional subspaces, while in B αX ( ~z, E, F ), F l ⊆ X ∩ [ e i ] ∞ i = n l +1 and E l ⊆ Y l are finite-dimensional subspaces. Finally, the non-zero vectors x l and y l are chosensuch that x l ∈ E + E + . . . + E l , while y l ∈ F + F + . . . + F l . Both games terminate once I has played ξ k = 0 and II has responded with some y k . The outcome is then the finite sequence of non-zero vectors ~z ˆ( x , y , x , y , . . . , x k , y k ) . Now suppose instead that ~z is an odd-length sequence of non-zero vectors. Wethen define A αX ( ~z, E, F ) by n < E n < E n k < E k x x x k I Y Y Y Y k ξ ξ ξ k . . .n n II F ⊆ Y F ⊆ Y F k ⊆ Y k y y y k and B αX ( ~z, E, F ) by: E ⊆ Y E ⊆ Y E k ⊆ Y k x x x k I n n n n k ξ ξ ξ k . . .Y Y II n < F n < F n k < F k y y y k where α > ξ > . . . > ξ k = 0is a decreasing sequence of ordinals, x l ∈ E + E + . . . + E l ,y l ∈ F + F + . . . + F l , and otherwise the games are identical to those above. The outcome is now the finitesequence ~z ˆ( y , x , y , . . . , x k , y k ).If ~z = ∅ and E = F = { } , we shall write A αX and B αX instead of A αX ( ~z, E, F ),respectively B αX ( ~z, E, F ). Thus, in both games A αX and B αX , one should rememberthat I is the first to play a vector. And in A αX , I plays block subspaces and IIplays integers, while in B αX , II takes the role of playing block subspaces and I playsintegers.We should also mention the degenerate case when α = 0. The games G αX ( ~z ) and F αX ( ~z ) then terminate immediately with outcome ~z and, if ~z is of even length, thesame holds for the games A αX ( ~z, E, F ) and B αX ( ~z, E, F ). On the other hand, if ~z isof odd length, in A αX ( ~z, E, F ) and B αX ( ~z, E, F ), I will play respectively Y and n and II respond with a single y according to the rules, whereby the outcome is now ~z ˆ y .If X and Y are subspaces, where Y is spanned by an infinite block sequence( y , y , y , . . . ), we write Y ⊆ ∗ X if there is n such that y m ∈ X for all m > n . Asimple diagonalisation argument shows that if X ⊇ X ⊇ X ⊇ . . . is a decreasingsequence of block subspaces, then there is some Y ⊆ X such that Y ⊆ ∗ X n for all n . The aim of the games above is for each of the players to ensure that the outcomelies in some predetermined set depending on the player. By the asymptotic natureof the game, it is easily seen that if T ⊆ W < N and Y ⊆ ∗ X , then if II has a strategyin G αX or A αX ( ~z, E, F ) to play in T , i.e., to ensure that the outcome is in T , then IIwill have a strategy in G αY , respectively A αY ( ~z, E, F ), to play in T too. Similarly,if I has a strategy in F αX or B αX ( ~z, E, F ) to play in T , then I also has a strategy in F αY , respectively in B αX ( ~z, E, F ), to play in T .4.2. Ramsey determinacy of adversarial α -games. We are now ready to provethe basic determinacy theorem for adversarial α -games, which can be seen as arefinement of the determinacy theorem for open adversarial games (see Theorem12 in [6]). Theorem 10.
Suppose α < ω and T ⊆ W < N . Then for any X ⊆ W there is Y ⊆ X such that either (1) II has a strategy in A αY to play in T , or (2) I has a strategy in B αY to play in ∼ T . -MINIMAL BANACH SPACES 11 Proof.
We say that(a) ( ~x, E, F, β, X ) is good if II has a strategy in A βX ( ~x, E, F ) to play in T .(b) ( ~x, E, F, β, X ) is bad if ∀ Y ⊆ X , ( ~x, E, F, β, Y ) is not good.(c) ( ~x, E, F, β, X ) is worse if it is bad and either(1) | ~x | is even and β = 0, or(2) | ~x | is even, β >
0, and ∀ Y ⊆ X ∃ E ⊆ Y ∃ x ∈ E + E ∃ γ < β ( ~x ˆ x , E + E , F, γ, X ) is bad , or(3) | ~x | is odd and ∃ n ∀ n < F ⊆ X ∀ y ∈ F + F ( ~x ˆ y , E, F + F , β, X ) is bad , (d) ( ~x, E, F, β, X ) is wicked if ∀ y ∈ F ( ~x ˆ y , E, F, β, X ) is bad.One checks that good, bad and wicked are all ⊆ ∗ -hereditary in the last coordinate,that is, if ( ~x, E, F, β, X ) is good and Y ⊆ ∗ X , then also ( ~x, E, F, β, Y ) is good, etc.So, by diagonalising over the countably many tuples of ~x , E , F , and β α , we canfind some Y ⊆ X such that for all ~x , E , F , and β α ,(i) ( ~x, E, F, β, Y ) is either good or bad, and(ii) if there is some Y ⊆ Y such that for all F ⊆ Y , ( ~x, E, F + F , β, Y ) iswicked, then there is some n such that for all n < F ⊆ Y , ( ~x, E, F + F , β, Y ) is wicked. Lemma 11. If ( ~x, E, F, β, Y ) is bad, then it is worse.Proof. Assume first that | ~x | is even. The case when β = 0 is trivial, so assume also β >
0. Since ( ~x, E, F, β, Y ) is bad, we have ∀ V ⊆ Y II has no strategy in A βV ( ~x, E, F ) to play in T. Referring to the definition of the game A βV ( ~x, E, F ), this implies that ∀ V ⊆ Y ∃ E ⊆ V ∃ x ∈ E + E ∃ γ < β II has no strategy in A γV ( ~x ˆ x , E + E , F ) to play in T, (note that the subspace Y ⊆ V also played by I becomes the first play of I in thegame A γV ( ~x ˆ x , E + E , F )). But if V ⊆ Y and II has no strategy in A γV ( ~x ˆ x , E + E , F ) to play in T , then ( ~x ˆ x , E + E , F, γ, V ) is not good and hence must bebad. Thus, ∀ V ⊆ Y ∃ E ⊆ V ∃ x ∈ E + E ∃ γ < β ( ~x ˆ x , E + E , F, γ, V ) is bad,which is just to say that ( ~x, E, F, β, Y ) is worse.Now suppose instead that | ~x | is odd. As ( ~x, E, F, β, Y ) is bad, it is not good andso II has no strategy in A βY ( ~x, E, F ) to play in T . Therefore, for some Y ⊆ Y , wehave ∀ F ⊆ Y ∀ y ∈ F + F II has no strategy in A βY ( ~x ˆ y , E, F + F ) to play in T. i.e., ∀ F ⊆ Y ∀ y ∈ F + F ( ~x ˆ y , E, F + F , β, Y ) is not good and hence is bad.In other words, ∀ F ⊆ Y ( ~x, E, F + F , β, Y ) is wicked. So by (ii) we have ∃ n ∀ n < F ⊆ Y ( ~x, E, F + F , β, Y ) is wicked , that is ∃ n ∀ n < F ⊆ Y ∀ y ∈ F + F ( ~x ˆ y , E, F + F , β, Y ) is bad,showing that ( ~x, E, F, β, Y ) is worse. (cid:3) If ( ∅ , { } , { } , α, Y ) is good, the first possibility of the statement of the theoremholds. So suppose instead ( ∅ , { } , { } , α, Y ) is bad and hence worse. Then, usingthe lemma and unraveling the definition of worse, we see that I has a strategy toplay the game B αY such that at any point in the game, if ~x = ( x , y , x , y , . . . , x l , y l ) E , F , E , F , . . . , E l , F l α > ξ > ξ > . . . > ξ l , respectively, ~y = ( x , y , x , y , . . . , y l − , x l ) E , F , E , F , . . . , F l − , E l α > ξ > ξ > . . . > ξ l , have been played, then( ~x, E + . . . + E l , F + . . . + F l , ξ l , Y ) , respectively ( ~y, E + . . . + E l , F + . . . + F l − , ξ l , Y ) , is worse. Since α > ξ > ξ . . . , we eventually have ξ k = 0, that is, the gameterminates with some worse( ~z, E + . . . + E k , F + . . . + F k , , Y ) , whereby the outcome ~z lies in ∼ T . (cid:3) A game theoretic dichotomy.
We first need a lemma ensuring us a certainuniformity.
Lemma 12.
Let β < ω and suppose that for every X ⊆ W there are K > anda block sequence ( y n ) ⊆ X such that II has a strategy in F βX to play ( x , x , . . . , x k ) satisfying ( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . Then there are K > and Y ⊆ W such that for all X ⊆ Y there is a block sequence ( y n ) ⊆ X such that II has a strategy in F βX to play ( x , x , . . . , x k ) satisfying ( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . In other words, K > can be chosen uniformly for all X ⊆ Y .Proof. Assume toward a contradiction that the conclusion fails. Then, as the games F βX to play in any set T ⊆ W < N are determined, i.e., either I or II has a winningstrategy, we can inductively define W ⊇ Y ⊇ Y ⊇ . . . such that for any blocksequence ( y n ) in Y K , I has a strategy in F βY K to play ( x , x , . . . , x k ) satisfying( x , x , . . . , x k ) K ( y , y , . . . , y k ) . -MINIMAL BANACH SPACES 13 For each N ∈ N , let c ( N ) be a constant such that if ( v , v , . . . , v N − , v N , v N +1 , . . . )and ( u , u , . . . , u N − , v N , v N +1 , . . . ) are two normalised block sequences of ( e n ),then ( v , v , . . . , v N − , v N , v N +1 , . . . ) ∼ c ( N ) ( u , u , . . . , u N − , v N , v N +1 , . . . ) . Now choose a block sequence ( x , x , x , . . . ) such that for every N there are nor-malised v , v , . . . , v N − ∈ Y N · c ( N ) with v < v < . . . < v N − < x N < x N +1 < . . . and, moreover, such that x N , x N +1 , . . . ∈ Y N · c ( N ) . Set also X = [ x n ].By the assumptions of the lemma, we can find some constant N ∈ N and anormalised block sequence ( y , y , . . . ) in X such that II has a strategy in F βX toplay ( w , w , . . . , w k ) with( w , w , . . . , w k ) ∼ N ( y , y , . . . , y k ) . Since min supp( x N ) min supp( y N ), it follows by the choice of ( x n ) that there arenormalised v , v , . . . , v N − ∈ Y N · c ( N ) such that v < v < . . . < v N − < y N < y N +1 < . . . . Moreover, by the definition of c ( N ), we have( v , v , . . . , v N − , y N , y N +1 , . . . ) ∼ c ( N ) ( y , y , . . . , y N − , y N , y N +1 , . . . ) . Thus, if we let v n = y n for all n > N , we see that II has a strategy in F βX to play( w , w , . . . , w k ) with( w , w , . . . , w k ) ∼ N ( y , y , . . . , y k ) ∼ c ( N ) ( v , v , . . . , v k ) . But X ⊆ ∗ Y N · c ( N ) , so II has a strategy in F βY N · c ( N ) to play ( w , w , . . . , w k ) with( w , w , . . . , w k ) ∼ N · c ( N ) ( v , v , . . . , v k ) . On the other hand, ( v n ) ⊆ Y N · c ( N ) and so I has a strategy in F βY N · c ( N ) to play( w , w , . . . , w k ) such that( w , w , . . . , w k ) N · c ( N ) ( v , v , . . . , v k ) , which is absurd. This contradiction proves the lemma. (cid:3) Lemma 13.
Suppose X ⊆ W , ( y , y , y , . . . ) is a sequence of vectors in W , α < ω and K > . Assume that II has a strategy in F ω · αX to play ( x , x , . . . , x k ) such that ( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . Then II has a strategy in B αX to play ( u , v , u , v , . . . , u k , v k ) such that ( u , u , . . . , u k ) ∼ K ( v , v , . . . , v k ) . Proof.
We shall describe the strategy for II in the game B αX , the idea being that,when playing the game B αX , II will keep track of an auxiliary run of F ω · αX , using hisstrategy there to compute his moves in B αX .Now, in B αX , II will play subspaces Y , Y , . . . all equal to Y = [ y n ], whereby thesubspaces Y , Y , . . . and E , E , . . . lose their relevance and we can eliminate them from the game for simplicity of notation. We thus have the following presentationof the game B αX . u ∈ Y u ∈ Y u k ∈ Y I n n n k ξ < α ξ < ξ ξ k < ξ k − . . . II n < F n < F n k < F k v ∈ F v ∈ F + F v k ∈ F + . . . + F k So suppose u , u , . . . is being played by I in B αX . To compute the answer v , v , . . . ,II follows his strategy in F ω · αX to play ( z , z , . . . , z k ) ∼ K ( y , y , . . . , y k ) as follows.First, as u , u , . . . ∈ Y = [ y n ], we can write each u i as u i = m i − X j =0 λ ij y j , where we, by adding dummy variables, can assume that m < m < m < . . . . Soto compute v and F given u , n and ξ , II first runs an initial part of F ω · αX asfollows I n n n ωξ + m − ωξ + m − ωξ . . . II n < F n < F n < F m x ∈ F x ∈ F + F x m − ∈ F + . . . + F m He then plays F = F + . . . + F m and v = m − X j =0 λ j x j ∈ F in B αX .Next, I will play some u , n and ξ , and, to compute v and F , II will continuethe above run of F ω · αX with I n n ωξ + m − ωξ . . . II n < F n < F m x m ∈ F + F x m − ∈ F + F + . . . + F m He then plays F = F + . . . + F m and v = m − X j =0 λ j x j ∈ F + F in B αX .So at each stage, II will continue his run of F ω · αX a bit further until eventually Ihas played some ξ k = 0. Thus, in the game F ω · αX , I will play ordinals α > ωξ + m − > ωξ + m − > . . . > ωξ > ωξ + m − > . . . > ωξ k = 0and integers n > n > . . . > n > n > . . . > n k , while II will use his strategy toplay ( x , x , . . . , x m k − ) such that( x , x , . . . , x m k − ) ∼ K ( y , y , . . . , y m k − ) . -MINIMAL BANACH SPACES 15 Since the v i and u i have the same coefficients over respectively ( x n ) and ( y n ), itfollows that ( u , u , . . . , u k ) ∼ K ( v , v , . . . , v k ) . (cid:3) By a similar argument, we have the following lemma.
Lemma 14.
Suppose X ⊆ W , ( y , y , y , . . . ) is a block sequence in W , α < ω and K > . Assume that II has a strategy in F ω · αX to play ( x , x , . . . , x k ) such that ( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . Then for any block sequence ( z n ) in [ y n ] , II has a strategy in F αX to play ( v , v , . . . , v k ) such that ( v , v , . . . , v k ) ∼ K ( z , z , . . . , z k ) . Proof.
First, as ( z n ) is a block sequence in [ y n ], we can write each z i as z i = m i − X j = m i − λ j y j , where m − = 0 < m < m < m < . . . .As before, when playing F αX , II will keep track of an auxiliary run of F ωαX , usinghis strategy there to compute his moves in F αX . So the game F αX runs as follows: I n n n k ξ ξ ξ k . . . II n < F n < F n k < F k v ∈ F v ∈ F + F v k ∈ F + . . . + F k To compute v , II first runs an initial part of F ωαX as follows I n n n ωξ + m − ωξ + m − ωξ . . . II n < F n < F n < F m x ∈ F x ∈ F + F x m − ∈ F + . . . + F m He then plays F = F + . . . + F m and v = m − X j = m − λ j x j ∈ F in F αX .Next, I will play some ξ and n and to compute v and F , II will continue theabove run of F ωαX with I n n ωξ + m − m − ωξ . . . II n < F n < F m − m x m ∈ F + F x m − ∈ F + F + . . . + F m − m He then plays F = F + . . . + F m − m and v = m − X j = m λ j x j ∈ F + F
16 CHRISTIAN ROSENDAL in F αX .So at each stage, II will continue his run of F ωαX a bit further until eventually Ihas played some ξ k = 0. Thus, in the game F ωαX , I will play ordinals α > ωξ + m − > ωξ + m − > . . . > ωξ > ωξ + m − m − > . . . > ωξ k = 0and integers n > n > . . . > n > n > . . . > n k , while II will use his strategy toplay ( x , x , . . . , x m k − ) such that( x , x , . . . , x m k − ) ∼ K ( y , y , . . . , y m k − ) . Since the v i and z i have the same coefficients over respectively ( x n ) and ( y n ), itfollows that ( v , v , . . . , v k ) ∼ K ( z , z , . . . , z k ) . (cid:3) Lemma 15.
Suppose X ⊆ W , ( y n ) is a block sequence in W , α < ω , and K, C > . Assume that (a) II has a strategy in F αX to play ( x , . . . , x k ) such that ( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) , and (b) II has a strategy in A αX to play ( u , v , . . . , u k , v k ) such that ( u , u , . . . , u k ) ∼ C ( v , v , . . . , v k ) , Then II has a strategy in G αX to play ( v , . . . , v k ) such that ( v , v , . . . , v k ) ∼ KC ( y , y , . . . , y k ) . Proof.
To compute his strategy in G αX , II will play auxiliary runs of the games A αX and F αX in which he is using the strategies described above. Information is thencopied between the games as indicated in the diagrams below.The game G αX : I Y Y Y k ξ ξ ξ k . . . II F ⊆ Y F ⊆ Y F k ⊆ Y k v ∈ F v ∈ F + F v k ∈ F + . . . + F k The game F αX : I n n n k ξ ξ ξ k . . . II n < E n < E n k < E k x ∈ E x ∈ E + E x k ∈ E + . . . + E k The game A αX : n < E n < E n k < E k x ∈ E x ∈ E + E x k ∈ E + . . . + E k I Y Y Y k ξ ξ ξ k . . .n n II F ⊆ Y F k ⊆ Y k v ∈ F v k ∈ F + . . . + F k -MINIMAL BANACH SPACES 17 By chasing the diagrams, one sees that this fully determines how II is to play in G αX . Moreover, since II follows his strategy in F αX , we have( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) , while the strategy in A αX ensures that( x , x , . . . , x k ) ∼ C ( v , v , . . . , v k ) , from which the conclusion follows. (cid:3) Theorem 16.
Suppose α < ω . Then there is X ⊆ W such that one of thefollowing holds (1) For every block sequence ( y n ) in X and K > , I has a strategy in F ωαX toplay ( x , x , . . . , x k ) satisfying ( x , x , . . . , x k ) K ( y , y , . . . , y k ) . (2) For some K > and every block sequence ( z n ) ⊆ X , II has a strategy in G αX to play ( x , x , . . . , x k ) satisfying ( x , x , . . . , x k ) ∼ K ( z , z , . . . , z k ) . Proof.
Suppose that there is no X ⊆ W for which (1) holds. Then, using that thegame F ωαX is determined, for every X ⊆ W there is a block sequence ( y n ) in X andsome K > F ωαX to play ( x , x , . . . , x k ) satisfying( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . So, by Lemma 12, there is some K > Y ⊆ W such that for all X ⊆ Y there is some block sequence ( y n ) in X such that II has a strategy in F ωαX to play( x , x , . . . , x k ) satisfying( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . If thus follows from Lemma 13 that for all X ⊆ Y , II has a strategy in B αX toplay ( u , v , u , v , . . . , u k , v k ) such that( u , u , . . . , u k ) ∼ K ( v , v , . . . , v k ) . Therefore, there is no X ⊆ Y such that I has a strategy in B αX to play a sequence( u , v , u , v , . . . , u k , v k ) satisfying( u , u , . . . , u k ) K ( v , v , . . . , v k ) , and thus, by Theorem 10, we can find some X ⊆ Y such that II has a strategy in A αX to play ( u , v , u , v , . . . , u k , v k ) satisfying( u , u , . . . , u k ) ∼ K ( v , v , . . . , v k ) . Let ( y n ) be the block sequence in X such that II has a strategy in F ωαX to play( x , x , . . . , x k ) satisfying( x , x , . . . , x k ) ∼ K ( y , y , . . . , y k ) . Then, using Lemma 14, we see that for any block sequence ( z n ) ⊆ [ y n ], II has astrategy in F αX to play ( x , x , . . . , x k ) such that( x , x , . . . , x k ) ∼ K ( z , z , . . . , z k ) . In other words, there is some block sequence ( y n ) in X such that for any blocksequence ( z n ) ⊆ [ y n ] (a) II has a strategy in F αX to play ( x , . . . , x k ) satisfying( x , x , . . . , x k ) ∼ K ( z , z , . . . , z k ) , and(b) II has a strategy in A αX to play ( u , v , . . . , u k , v k ) satisfying( u , u , . . . , u k ) ∼ K ( v , v , . . . , v k ) , So finally, by Lemma 15, for any block sequence ( z n ) ⊆ [ y n ], II has a strategy in G αX to play ( v , . . . , v k ) such that( v , v , . . . , v k ) ∼ K ( z , z , . . . , z k ) . Replacing X by the block subspace [ y n ] ⊆ X and K by K , we get (2). (cid:3) The embeddability index.Lemma 17.
Suppose α < ω , K > , X ⊆ W and ( z n ) ⊆ W is a block sequencesuch that II has a strategy in G αX to play ( y , . . . , y k ) satisfying ( y , . . . , y k ) ∼ K ( z , . . . , z k ) . Then for any subspace Y ⊆ X , rank (cid:0) T (( z n ) , Y, K ) (cid:1) > α .Proof. Let Y ⊆ X and suppose toward a contradiction that rank (cid:0) T (( z n ) , Y, K ) (cid:1) = ξ + 1 α , where ξ is the rank of the root ∅ in T (( z n ) , Y, K ). Now, let I play Y, ξ in G αX and let II respond using his strategy I Yξ II E ⊆ Yy ∈ E Then the rank of ( y ) ∈ T (( z n ) , Y, K ) is some ordinal ξ < ξ , so in G αX , I continuesby playing Y, ξ and II responds according to his strategy I Y Yξ ξ II E ⊆ Y E ⊆ Yy ∈ E y ∈ E + E Again, the rank of ( y , y ) ∈ T (( x n ) , Y, K ) is some ordinal ξ < ξ , so in G αX , Icontinues by playing Y, ξ and II responds according to his strategy I Y Y Yξ ξ ξ II E ⊆ Y E ⊆ Y E ⊆ Yy ∈ E y ∈ E + E y ∈ E + E + E Etc.Eventually, we will have constructed some ( y , y , . . . , y k − ) whose T (( z n ) , Y, K )-rank is ξ k = 0, while I Y Yξ ξ k − . . . II E ⊆ Y E k − ⊆ Yy ∈ E y k − ∈ E + . . . + E k − has been played according to the strategy of II. -MINIMAL BANACH SPACES 19 It follows that if I continues the game by playing
Y, ξ k = 0, I Y Y Yξ ξ k − ξ k = 0 . . . II E ⊆ Y E k − ⊆ Yy ∈ E y k − ∈ E + . . . + E k − using his strategy, II must be able to respond with some E k and y k ∈ E + . . . + E k I Y Y Yξ ξ k − ξ k = 0 . . . II E ⊆ Y E k − ⊆ Y E k ⊆ Yy ∈ E y k − ∈ E + . . . + E k − y k ∈ E + . . . + E k Since II played according to his strategy, we have ( y , y , . . . , y k ) ∼ K ( z , z , . . . , z k )and thus ( y , y , . . . , y k ) ∈ T (cid:0) ( z n ) , Y, K (cid:1) , contradicting that ( y , . . . , y k − ) has T (cid:0) ( z n ) , Y, K (cid:1) -rank 0 and hence is a terminal node. (cid:3) Lemma 18.
Suppose ( x n ) ⊆ W is a block sequence, β < ω , and that for everynormalised block sequence ( y n ) in X = [ x n ] and K > , I has a strategy in F βX toplay ( z , z , . . . , z k ) such that ( z , z , . . . , z k ) K ( y , y , . . . , y k ) . Then, for every normalised block sequence ( y n ) in X and K > , there is a sequence ( J m ) of intervals of N with min J m → ∞ , such that if A ⊆ N is infinite, contains and Z = [ x j (cid:12)(cid:12) j / ∈ S m ∈ A J m ] , then rank (cid:0) T (( y n ) , Z, K ) (cid:1) β. Proof.
We relativise the notions of support of vectors et cetera to the basis ( x n )for X . So the reader can assume that ( x n ) is the original basis ( e n ) and X = W .Assume ( y n ) is a normalised block sequence in X and K >
1. Let also ∆ = ( δ j )be a sequence of positive real numbers such that whenever z j , v j ∈ X , k z j − v j k < δ j ,and ( v , . . . , v k ) ∼ K ( y , . . . , y k ) , then ( z , . . . , z k ) ∼ K ( y , . . . , y k ) . We choose sets D i ⊆ X such that for each finite set d ⊆ N , the number of z ∈ D i such that supp( z ) = d is finite, and for every v ∈ X with k v k K there is some z ∈ D i with supp( z ) = supp( v ) and k z − v k < δ i . This is possible since the K -ballin [ x j ] j ∈ d is totally bounded for all finite d ⊆ N .The strategy for I in F βX in the game for ( y n ) with constant 2 K can be seen asa pair of functions ξ and n that to each legal position ( z , E , . . . , z j , E j ) of II in F βX provide the next play ξ ( z , E , . . . , z j , E j ) ∈ Ord and n ( z , E , . . . , z j , E j ) ∈ N by I.We define a function p : N → N by letting p ( m ) be the maximum of m andmax (cid:0) n ( z , [ x l ] l ∈ d , . . . , z i , [ x l ] l ∈ d i ) (cid:12)(cid:12) d j ⊆ [0 , m − z j ∈ [ x l ] l ∈ d ∪ ... ∪ d j ∩ D j (cid:1) . By assumption on the sets D j , p is well-defined and so we can set J m = [ m, p ( m )] ⊆ N . We claim that if A ⊆ N is an infinite set containing 0 and Z = [ x n | n / ∈ [ m ∈ A J m ] , then rank (cid:0) T (( y n ) , Z, K ) (cid:1) β. To see this, we define a monotone function φ , i.e., ~v ≺ ~w ⇒ φ ( ~v ) ≺ φ ( ~w ), associatingto each ~v = ( v , v , . . . , v i ) ∈ T (( y n ) , Z, K ) some φ ( ~v ) = ( z , z , . . . , z i ) ∈ D × D × . . . × D i such that for all j i , k z j − v j k < δ j and supp( z j ) = supp( v j ), whereby, inparticular, z j ∈ Z . Also set T = φ (cid:2) T (( y n ) , Z, K ) (cid:3) and note that T is a subtree of Z < N with rank( T ) > rank (cid:0) T (( y n ) , Z, K ) (cid:1) . Suppose toward a contradiction that rank( T ) > β , whereby the rank of ∅ in T is > β . We describe how II can play against the strategy for I in F βX to play( z , . . . , z k ) such that ( z , . . . , z k ) ∼ K ( y , . . . , y k ) , which will contradict the assumption on the strategy for I. The case β = 0 is trivial,so we assume that β > ξ ( ∅ ) < β and n ( ∅ ). Since, a = 0 ∈ A , we have n ( ∅ ) p ( a ) =max J a < Z and thus there is some n ( ∅ ) < z ∈ T whose rank in T is > ξ ( ∅ ).Find also a ∈ A such that z < J a and let E = [ x j (cid:12)(cid:12) J a < x j < J a ]. So let IIrespond by I n ( ∅ ) ξ ( ∅ ) II n ( ∅ ) < E z ∈ E Now, by his strategy, I will play some ξ ( z , E ) < ξ ( ∅ ) and n ( z , E ) p ( a ) =max J a . So find some z such that ( z , z ) ∈ T and has rank > ξ ( z , E ) in T .Find also a ∈ A such that z < J a . Then, as a , a ∈ A , if we set E = [ x j (cid:12)(cid:12) J a Suppose ( x n ) ⊆ W is a normalised block sequence, β < ω , and thatfor every normalised block sequence ( y n ) in X = [ x n ] and K > , I has a strategyin F βX to play ( z , z , . . . , z k ) such that ( z , z , . . . , z k ) K ( y , y , . . . , y k ) . Then, for every normalised block sequence ( y n ) in X there is a sequence I < I < I < . . . of intervals of N , such that if A ⊆ N is infinite and Z = [ x j (cid:12)(cid:12) j / ∈ S m ∈ A I m ] , then Emb (cid:0) ( y n ) , Z (cid:1) β. Proof. Fix a normalised block sequence ( y n ) in X and relativise again all notionsof support et cetera to the block basis ( x n ). By Lemma 18, we can for every K finda sequence ( J Kn ) of intervals of N with min J Kn −→ n →∞ ∞ such that for any infinite set A ⊆ N containing 0, we haverank (cid:0) T (( y n ) , [ x j (cid:12)(cid:12) j / ∈ [ n ∈ A J Kn ] , K ) (cid:1) β. Also, for every N , we let c ( N ) ∈ N be a constant such that any two subsequencesof ( x j ) differing in at most N terms are c ( N )-equivalent.We construct intervals I < I < I < . . . such that each I n contains an intervalfrom each of the families ( J i ) , . . . , ( J ni ) and, moreover,min I n < max I n − max J n · c (min I n )0 . We claim that if A ⊆ N is infinite and Z = [ x j (cid:12)(cid:12) j / ∈ S m ∈ A I m ], thenEmb (cid:0) ( y n ) , Z (cid:1) β. Suppose towards a contradiction that this fails for some A and pick some N suchthat rank (cid:0) T (( y n ) , Z, N ) (cid:1) > β . Choose a ∈ A such that a > N and note thatmin I a < max I a − max J a · c (min I a )0 . Thus, by changing only the terms x j for j < min I a of the sequence( x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m ) =( x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m & j < min I a ) ∪ ( x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m & j > max I a ) , we find a subsequence of( x j (cid:12)(cid:12) max J a · c (min I a )0 < j max I a ) ∪ ( x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m & j > max I a ) that is c (min I a )-equivalent with( x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m ) . Since N · c (min I a ) a · c (min I a ), it follows that if Y = [ x j (cid:12)(cid:12) max J a · c (min I a )0 < j max I a ] + [ x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m & j > max I a ] , then Z ⊑ c (min I a ) Y , and so β < rank (cid:0) T (( y n ) , Z, N ) (cid:1) rank (cid:0) T (( y n ) , Y, a · c (min I a )) (cid:1) . But, by the choice of the I n , we see that there is an infinite subset B ⊆ N containing0 such that Y is outright a subspace of [ x j (cid:12)(cid:12) j / ∈ S m ∈ B J a · c (min I a ) m ], whereby, bychoice of the intervals J a · c (min I a ) m , we haverank (cid:0) T (( y n ) , Y, a · c (min I a )) (cid:1) β, which is absurd. This contradiction shows that the intervals I n fulfill the conclusionof the lemma. (cid:3) By combining Theorem 16 and Lemmas 17 and 19, we obtain Theorem 20. Suppose α < ω . Then there is a block subspace X = [ x n ] ⊆ W such that one of the following holds (1) For every normalised block sequence ( y n ) in X there is a sequence I < I < I < . . . of intervals of N , such that if A ⊆ N is infinite, then Emb (cid:0) ( y n ) , [ x j (cid:12)(cid:12) j / ∈ [ m ∈ A I m ] (cid:1) ωα. (2) For any subspace Y ⊆ X and any block sequence ( z n ) ⊆ X , Emb (cid:0) ( z n ) , Y (cid:1) > α. And by replacing the normed F -vector subspaces X and Y in Theorem 20 bytheir closures X and Y in W , we obtain Theorem 4. Theorem 21. Let W be Banach space with a Schauder basis and suppose α < ω .Then there is a block subspace X = [ x n ] ⊆ W that is either ωα -tight or ( α + 1) -minimal. References [1] J. Bourgain, On separable Banach spaces, universal for all separable reflexive spaces , Proc.Amer. Math. Soc. 79 (1980), no. 2, 241–246.[2] V. Ferenczi and C. Rosendal, Banach spaces without minimal subspaces , Journal of FunctionalAnalysis, Volume 257, Issue 1, 1 July 2009, 149–193.[3] W. T. Gowers, An infinite Ramsey theorem and some Banach-space dichotomies , Ann. ofMath. (2) (2002), no. 3, 797–833.[4] A. S. Kechris, Classical descriptive set theory , Springer-Verlag, New York, 1995.[5] A. M. Pelczar, Subsymmetric sequences and minimal spaces , Proc. Amer. Math. Soc. 131(2003) 3, 765–771.[6] C. Rosendal, An exact Ramsey principle for block sequences , Collectanea Mathematica, Vol.61, no. 10 (2010) 25–36. -MINIMAL BANACH SPACES 23 Department of Mathematics, Statistics, and Computer Science (M/C 249), Universityof Illinois at Chicago, 851 S. Morgan St., Chicago, IL 60607-7045, USA E-mail address : [email protected] URL : ˜˜