The isomorphism class of c 0 is not Borel

Transcription

1 INSTITUTE OF MATHEMATICS THE CZECH ACADEMY OF SCIENCES The isomorphism class of c 0 is not Borel Ondřej Kura Preprint No PRAHA 207

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3 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL ONDŘEJ KURKA ABSTRACT. We show that the class of all Banach spaces which are isomorphic to c 0 is a complete analytic set with respect to the Effros Borel structure of separable Banach spaces. The proof employs a recent Bourgain-Delbaen construction by Argyros, Gasparis and Motais.. INTRODUCTION AND THE MAIN RESULT In a preceding wor [20], we have introduced a new approach to complexity problems in Banach space theory based on the famous Tsirelson space. Using a method from [2], we construct a family of L -spaces that is somehow related to the family of Tsirelson-lie spaces from []. Compared to [20], this provides an analogous but in some sense more powerful method and enables us to determine the complexity of several classes of separable Banach spaces. It can be shown quite easily that the isomorphism class of any separable Banach space is analytic with respect to the Effros Borel structure. B. Bossard ased in [5] whether l 2 is (up to isomorphism) the only infinite-dimensional separable Banach space whose isomorphism class is Borel (see Section 2 for the definitions of the Effros Borel structure and of the related notions used below). Although Bossard s question has been answered negatively (see [4, Theorem 6.2]), it is still not well understood which spaces have a Borel isomorphism class. It is nown that the isomorphism class is not Borel for Pełczyńsi s universal space (see [5, Theorem 2.3]), C(2 N ) (see e.g. [9, (33.26)]) or L p ([0, ]), < p <, p = 2, (see [5, p. 30] or [, Corollary 4.0]), and further examples are provided in [4]. On the other hand, the isomorphism classes of l p, < p <, are Borel (see [5, Theorem 2], see also [7]). It is not nown if the same holds for l. 200 Mathematics Subject Classification. Primary 46B03; Secondary 54H05, 46B25, 46B20. Key words and phrases. Effros Borel structure, complete analytic set, Banach space c 0, Bourgain-Delbaen method, Tsirelson space. The author was supported by grant GAČR S and by RVO:

4 2 ONDŘEJ KURKA G. Godefroy ased in [3] if the isomorphism class of the space c 0 is Borel. In the preceding paper [20], we have found a partial solution, proving that the class of all spaces isomorphic to a subspace of c 0 is not Borel. The main result of the present wor is a full solution of Godefroy s problem. Theorem.. The class of all Banach spaces isomorphic to c 0 is complete analytic. In particular, it is not Borel. A remarable conjecture from [6] states that if a Banach space X has summable Szlen index and its dual X is isomorphic to l, then X is isomorphic to c 0. In [3], G. Godefroy pointed out that validity of the conjecture would imply that the isomorphism class of c 0 is Borel. Therefore, it is not surprising that our proof of Theorem. is based on a counterexample to the conjecture that has been found recently by S. A. Argyros, I. Gasparis and P. Motais [2]. As noticed in [3], we obtain immediately that the class of isomorphic subspaces of c 0 is not Borel (that is the mentioned result from [20]), as the intersection of this class with the Borel class of all infinite-dimensional separable L -spaces is exactly the isomorphism class of c 0 (see [8, Corollary ]). This wors in one direction only. Theorem. is an improvement indeed, since it is basically a statement about the structure of the class of L -spaces. Actually, it follows from our method of proving Theorem. that the isomorphism class of c 0 X is not Borel in the case that X is not big in a sense (in fact, we do not now if there is a space X such that the isomorphism class of c 0 X is Borel). Theorem.2. Let X be a separable Banach space that satisfies any of the following conditions: X does not contain an isomorphic copy of c 0, X does not contain an infinite-dimensional reflexive subspace, X is a subspace of a space with an unconditional basis. Then the class of all Banach spaces isomorphic to c 0 X is not Borel. Furthermore, the class of all Banach spaces isomorphic to a subspace of c 0 X is not Borel as well. Let us remar that the second part of this result has been already established for X without infinite-dimensional reflexive subspaces in [20, Remar 3.9(i)]. The following theorem contains further consequences of Theorem. pointed out by G. Godefroy in [3]. Theorem.3. The following classes of separable Banach spaces are analytic but not Borel:

5 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 3 spaces with an unconditional basis, subspaces of spaces with an unconditional basis, spaces isomorphic to a C(K) space. Let us note that it is not nown whether the class of all spaces with a Schauder basis is Borel. On the other hand, the class of all separable Banach spaces with the bounded approximation property is nown to be Borel (see [2]). Let us mention some achievements of Banach space theory employed in this wor. One of our main tools is the space T introduced by T. Figiel and W. B. Johnson (see [0]), dual to the famous (original) Tsirelson space T (see [22]). The construction of T was later generalized by S. A. Argyros and I. Deliyanni (see []). Their Tsirelson-lie spaces have been used in [20] for proving that the class of isomorphic subspaces of c 0 is not Borel. These spaces play a role also in the present paper, although their application is not so direct this time. The major part of the paper is devoted to the construction of a L - space X M for a compact system M of sets of natural numbers. The gist of this construction is that X M is isomorphic to c 0 if and only if M contains an infinite set. Theorem. follows quite easily, as the set of all compact systems M containing an infinite set is not Borel by a classical result of W. Hurewicz. The space X M is a modification of the space X 0 from [2]. Both spaces are constructed by the Bourgain-Delbaen method introduced in the seminal paper [7]. In fact, the constructions of X M and X 0 are very similar, which enables us to eep most of the used notation. The main difference is that the growth condition n (#Γ ran(ξ) ) 2 from [2, p. 696] is replaced with a condition involving M. 2. PRELIMINARIES Our terminology concerning Banach space theory and descriptive set theory follows [9] and [9]. All Banach spaces in this paper are considered over R. By an isomorphism we always mean a linear isomorphism. Given λ, we say that Banach spaces F and G are λ-isomorphic if there is a surjective linear operator T : F G such that T T λ. A Banach space X is called a L,λ -space if, for any finite-dimensional subspace F of X, there exists a finite-dimensional subspace G of X containing F such that G is λ-isomorphic to l n, where n = dim G. A Banach space X is said to be a L -space if it is a L,λ -space for some λ.

6 4 ONDŘEJ KURKA By c 00 we denote the vector space of all systems x = {x(n)} n= of scalars such that x(n) = 0 for all but finitely many n s. By the canonical basis of c 00 we mean the algebraic basis consisting of vectors e n = {n}, n N. For E N and x c 00, we denote by Ex the element of c 00 given by Ex(n) = x(n) for n E and Ex(n) = 0 for n / E. In the context of Banach spaces, by a basis we mean a Schauder basis. A basis {x i } i= of a Banach space X is said to be unconditional if there is a constant c such that i A a i x i c i B a i x i whenever A B are finite sets of natural numbers and a i R for i B. A basis {x i } i= of a Banach space X is said to be shrining if X = span{x, x 2,... } where x, x 2,... is the dual basic sequence x n : i= a ix i a n. We will need the following standard fact (see e.g. [8, Lemma B.6]), as well as some immediate consequences. Lemma 2.. Let X, X 2 be Banach spaces and let P Xi, i =, 2, denote the projection (x, x 2 ) X X 2 x i. If Y is an infinite-dimensional subspace of X X 2, then there are i {, 2} and an infinite-dimensional subspace Z of Y such that P Xi Z is an isomorphism. Fact 2.2. () If both X, X 2 do not contain an isomorphic copy of c 0, then the same holds for X X 2. (2) If both X, X 2 do not contain an infinite-dimensional reflexive subspace, then the same holds for X X 2. We will need also the following result from [2] uncovering the isomorphic structure of the space c 0. Theorem 2.3 (Rosenthal). Let X be a L -space. If X is isomorphic to a subspace of a space with an unconditional basis, then X is isomorphic to c 0. A Polish space (topology) means a separable completely metrizable space (topology). A set P equipped with a σ-algebra is called a standard Borel space if the σ-algebra is generated by a Polish topology on P. A subset A of a standard Borel space X is called analytic if there are a standard Borel space Z and a Borel mapping g : Z X such that A = g(z). Moreover, a subset A of a standard Borel space X is called a hard analytic set if every analytic subset B of a standard Borel space Y admits a Borel mapping f : Y X such that f (A) = B. A subset of a standard Borel space is called a complete analytic set if it is analytic and hard analytic at the same time.

7 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 5 Let us recall a standard simple argument for a set to be hard analytic. Lemma 2.4. Let A X and C Z be subsets of standard Borel spaces X and Z. Assume that C is hard analytic. If there is a Borel mapping g : Z X such that then A is hard analytic as well. g(z) A z C, By P(N) we denote the set of all subsets of N endowed with the coarsest topology for which {A P(N) : n A} is clopen for every n. Obviously, P(N) is nothing else than a copy of the Cantor space {0, } N. For a topological space X, we denote by F(X) the family of all closed subsets and by K(X) the family of all compact subsets of X. The hyperspace of compact subsets of X is defined as K(X) equipped with the Vietoris topology, i.e., the topology generated by the sets of the form {K K(X) : K U}, {K K(X) : K U = }, where U varies over open subsets of X. If X is Polish, then so is K(X). The set F(X) can be equipped with the Effros Borel structure, defined as the σ-algebra generated by the sets {F F(X) : F U = }, where U varies over open subsets of X. If X is Polish, then, equipped with this σ-algebra, F(X) forms a standard Borel space. It is well-nown that the space C([0, ]) contains an isometric copy of every separable Banach space. By the standard Borel space of separable Banach spaces we mean SE(C([0, ])) = { F F(C([0, ])) : F is linear }, considered as a subspace of F(C([0, ])). By [5, Proposition 2.2], SE(C([0, ])) is a standard Borel space. Whenever we say that a class of separable Banach spaces has a property lie being Borel, analytic, complete analytic, etc., we consider that class as a subset of SE(C([0, ])). The following lemma is needed for proving that the parametrized construction in Section 4 can be realizable as a Borel mapping.

8 6 ONDŘEJ KURKA Lemma 2.5 ([20, Lemma 2.4]). Let Ξ be a standard Borel space and let {(X ξ, ξ )} ξ Ξ be a system of Banach spaces each member of which contains a sequence x ξ, xξ 2,... whose linear span is dense in X ξ. Assume that the function n ξ Ξ λ x ξ R ξ = is Borel whenever n N and λ,..., λ n R. Then there exists a Borel mapping S : Ξ SE(C([0, ])) such that S(ξ) is isometric to X ξ for every ξ Ξ. 3. TSIRELSON-LIKE SPACES In this section, we recall the definition of Tsirelson-lie spaces introduced by S. A. Argyros and I. Deliyanni []. The structure of these spaces enables us to prove some properties of a compact system of finite sets of natural numbers. For M K(P(N)), a family {E,..., E n } of successive finite subsets of N is said to be M-admissible if an element of M contains numbers m,..., m n such that m E < m 2 E 2 < < m n E n. Definition 3. (Argyros, Deliyanni). For M K(P(N)), the space T[M, 2 ] is defined as the completion of c 00 under the implicitly defined norm { x M, = max x, n 2 2 sup E x M, }, 2 = where the sup is taen over all M-admissible families {E,..., E n }. Lemma 3.2. Let M K(P(N)) consist of finite sets only. Let ω > 0. Then there are N and α,..., α 0 with i= α i = such that A M : i,i A α i < ω. Proof. First of all, since M consists of finite sets, the canonical basis e n = {n} of c 00 is a shrining basis of T[M, 2 ] (to show this, it is possible to adapt the part (a) of the proof of [, Proposition.] if we consider θ = 2 and M = M for every ). In particular, e, e 2,... is not equivalent to the standard basis of l. We can find N and real numbers β,..., β such that M, β i e i < i= 2 2 ω β i. i=

9 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 7 Considering α j = β j / i= β i, we obtain α i 0, i= α i = and i= α i e i M, 2 = i= β i i= β i e i M, 2 = i= β i i= β i e i M, 2 < 2 ω. Given A M, let m < m 2 < < m n be all elements of A with m j and let E j = {m j }. Then the family {E,..., E n } is M-admissible and we can write 2 ω > M, α i e i i= 2 2 n j= E j Therefore, the numbers α,..., α wor. i= α i e i M, 2 = 2 n α mj = 2 j= α i. i,i A Lemma 3.3. Let M K(P(N)) consist of finite sets only. Let ω > 0. Let E, E 2,... be a sequence of intervals of N {0} such that E < E 2 <.... Then there are N and α,..., α 0 with i= α i = such that A M : Proof. Let us consider a mapping α i < ω. i,e i A = A P(N) A = {i : E i A = } P(N). Then M = {A : A M}, being a continuous image of a compact set, is compact itself. It consists of finite sets only, and so Lemma 3.2 can be applied. There are N and α,..., α 0 with i= α i = such that B M : i,i B α i < ω. Clearly, and α,..., α wor. 4. A BOURGAIN-DELBAEN CONSTRUCTION We are going to introduce a construction of a L -space which is the ey ingredient of our proof of Theorem.. This space is based on a recent example from [2], with the novelty that a compact system of sets of natural numbers is involved as a parameter. As well as the authors did in [2, Sect. 5], we fix a natural number N 3 and a constant < θ < N/2. We moreover fix a constant 0 < ω < /θ. Given M K(P(N)), we consider the following objects: 0 = {0},

10 8 ONDŘEJ KURKA a sequence, 2,... of pairwise disjoint non-empty finite sets that is defined recursively below, to every γ i= i, we assign age(γ) {,..., N} in a way described below, for every γ i=0 i, we denote by ran(γ) the unique p so that γ p, we put Γ p = p i=0 i for p = 0,,..., a functional c γ (l (Γ p )) is defined below for every p and γ p, an extension operator i p,p : l (Γ p ) l (Γ p ), where p, is defined by i p,p (x)(γ) = { x(γ), γ Γp, c γ(x), γ p, for 0 p < q, we define i p,q = i q,q i p,p+ : l (Γ p ) l (Γ q ) and i p,p as the identity on l (Γ p ), for 0 p q, let r q,p : l (Γ q ) l (Γ p ) be the corresponding restriction operator, for 0 p s q, let us consider the projections on l (Γ q ) given by P q [0,p] = i p,q r q,p and P q (p,s] = Pq [0,s] Pq [0,p]. For q = 0,,..., we define q+ as the set of all tuples of one of the following two forms: (a) (q +, {ε i } i=, {E i} i=, {η i} i=, {h i} i= ), where N, ε i {, } for i =,...,, {E i } i= is a sequence of successive non-empty intervals of {0,..., q}, η i Γ q and h i N {0} for i =,...,, and these numbers satisfy the requirements and or, if q 2, A M : h i q + i= h i < ω (q + ), E i A = (b) (q +, ξ, {ε i } i=, {E i} i=, {η i} i=, {h i} i= ), where N, ξ Γ q \ Γ 0 with age(ξ) < N, ε i {, } for i =,...,, {E i } i= is a sequence of successive non-empty intervals of {ran(ξ) +,..., q}, η i Γ q and h i N {0} for i =,...,, and these numbers satisfy the same two requirements as above.

11 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 9 Moreover, we define age(γ) and c γ (l (Γ q )) for γ q+ as follows (by e η we mean the functional x l (Γ q ) x(η)). For γ q+ of the form (a), we set age(γ) = and define c γ = θ N q + h i ε i e η i P q E i. i= For γ q+ of the form (b), we set age(γ) = age(ξ) + and define c γ = e ξ + θ N q + h i ε i e η i P q E i. i= Finally, let Γ = i=0 i, for p 0, let i p : l (Γ p ) l (Γ) be given by { x(γ), γ Γp, i p (x)(γ) = i p,q (x)(γ), γ q, q > p, (let us point out that i p maps into l (Γ) by Claim 4. below), let d γ = i ran(γ) (e γ ) l (Γ) for γ Γ, where e γ denotes the basic vector {γ} of l (Γ ran(γ) ), we define X M = span {d γ : γ Γ}, for p 0, let r p : X M l (Γ p ) be the corresponding restriction operator, for 0 p s, let us consider the projections on X M given by P [0,p] = i p r p, P (p,s] = P [0,s] P [0,p], P (p, ) = I P [0,p] and P [0, ) = I, we denote by {d γ} γ Γ the system in X M orthogonal to {d γ} γ Γ, that is d γ(x) = P {ran(γ)} (x)(γ) for x X M, for x X M, let range x be the smallest interval E of N {0} such that P E (x) = x. We provide a series of claims which results in a characterization of those M s for which X M is isomorphic to c 0 (Proposition 4.5). The first of them is more or less standard, as it is based on an argument from [7] (see also e.g. [3, Theorem 3.4]). Nevertheless, we include a proof for the convenience of a reader which is not familiar with constructions of this ind. Let us note that it follows from the claim that i p witnesses that l (Γ p ) and span {d γ : γ Γ p } are [N/(N 2θ)]-isomorphic. For this reason, X M is a L -space.

12 0 ONDŘEJ KURKA Claim 4.. We have i p,s N/(N 2θ), and consequently i p N/(N 2θ), P q (p,s] 2N/(N 2θ), P (p,s] 2N/(N 2θ) and d γ 2N/(N 2θ) whenever 0 p s q and γ Γ. Proof. We show by induction on s that i p,s N/(N 2θ) for every p s. Assume that p s and that i t,q N/(N 2θ) whenever t q s. We need to show that i p,s (x)(γ) x N/(N 2θ) for x l (Γ p ) and γ Γ s. If γ Γ p, then i p,s (x)(γ) = x(γ). So, let γ Γ s \ Γ p. There is q with p q s such that γ q+. Note that i p,s (x)(γ) = i p,q+ (x)(γ) = c γ(i p,q (x)). We need to chec the following two simple facts first: (i) P q E (i p,q(x)) x 2N/(N 2θ) for an interval E of {0,,..., q}, (ii) P q E (i p,q(x)) = 0 for an interval E of {p +,..., q}. If p t q, then P q [0,t] (i p,q(x)) = i t,q (r q,t (i p,q (x))) = i t,q (i p,t (x)) = i p,q (x). From this, (ii) follows. Moreover, by the induction hypothesis, P q [0,t] (i p,q(x)) = i p,q (x) i p,q x x N/(N 2θ). To chec (i), it remains to show that P q [0,t] (i p,q(x)) x N/(N 2θ) also for t < p. If 0 t < p, then P q [0,t] (i p,q(x)) = i t,q (r q,t (i p,q (x))) = i t,q (r p,t (x)), and so P q [0,t] (i p,q(x)) i t,q r p,t (x) x N/(N 2θ). So, (i) and (ii) are checed. Now, if γ is of the form (a), then c γ(i p,q (x)) θ N q + h i x i= 2θ 2N N 2θ x N 2θ x N N 2θ. If γ is of the form (b) and ran(ξ) p, then e ξ (i p,q(x)) = x(ξ) and c γ(i p,q (x)) x + θ N q + h i x i= 2θ x + x N 2θ = x If γ is of the form (b) and ran(ξ) > p, then 2N N 2θ N N 2θ. c γ(i p,q (x)) = e ξ (i N p,q(x)) x i p,q x N 2θ. In all three cases, i p,s (x)(γ) = c γ(i p,q (x)) x N/(N 2θ).

13 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL Claim 4.2 (cf. [2, Proposition 5.3]). Let q 0 and γ q+. If γ is of the form (a), then e γ = d γ + θ N If γ is of the form (b), then e γ = e ξ + d γ + θ N q + h i ε i e η i P Ei. i= q + h i ε i e η i P Ei. i= (Here, by e η we mean the functional x X M x(η)). Proof. It is sufficient to consider only the form (b) for γ. To distinguish the functionals e η acting on Γ q from those acting on X M, we use the notation e η,γ q. Let us pic x X M. Note first that e γ(p [0,q] (x)) = e γ(i q (r q (x))) = i q (r q (x))(γ) = i q,q+ (r q (x))(γ) = c γ(r q (x)) and e γ(p [0,q+] (x)) = e γ(i q+ (r q+ (x))) = i q+ (r q+ (x))(γ) = x(γ) = e γ(x). We obtain e γ(x) = e γ(p [0,q+] (x)) = e γ(p [0,q] (x))+e γ(p {q+} (x)) = c γ(r q (x))+d γ(x). Note further that, for p q and η Γ q, we have (e η,γ q P q [0,p] )(r q(x)) = i p,q (r q,p (r q (x)))(η) = i p,q (r p (x))(η) It follows that c γ(r q (x)) = e ξ,γ q (r q (x)) + θ N = e ξ (x) + θ N = i p (r p (x))(η) = (e η P [0,p] )(x). q + h i ε i (e η i,γ q P q E i )(r q (x)) i= q + h i ε i (e η i P Ei )(x). i= As this wors for any x X M, the proof is finished. Claim 4.3 (cf. [2, Proposition 5.6]). Let u 0, u,..., u m be elements of X M with u j. If there is A M such that range u 0 < n range u < n 2 < n m range u m for some n,..., n m A, then m u j C := θω j=0 2N(N + θ). N 2θ

14 2 ONDŘEJ KURKA Proof. Let u = m j=0 u j. We prove by induction on ran(γ) that, for every γ Γ and every interval E of N {0}, and even (e γ P E )(u) C (e γ P E )(u) C age(γ) N if ran(γ). This is easy when ran(γ) = 0 (if γ 0, then (e γ P E )(u j ) = 0 for j ). Assume that q N {0} and that the assertion holds for each element of Γ q and each interval of N {0} and fix γ q+ and an interval E of N {0}. If γ is of the form (a), then age(γ) = 0, and if γ is of the form (b), then age(γ) = age(ξ). Using Claim 4.2 and the induction hypothesis, we obtain in both cases that (e γ P E )(u) is less than or equal to C age(γ) N Let us show first that + (d γ P E )(u) + θ N (d γ P E )(u) q + h i (e η i P Ei E)(u). i= 2N N 2θ. If ran(γ) / E, then (d γ P E )(u) = 0. Assuming ran(γ) E, we have (d γ P E )(u) = d γ(u). As ran(γ) belongs to the range of at most one u j, we have d γ(u) = d γ(u j ) for some j. Consequently, (d γ P E )(u) d γ u j [2N/(N 2θ)]. Note that if E i A =, then E i contains no n j, and so it intersects the range of at most one u j. Thus, there is j(i) such that P Ei E(u) = P Ei E(u j(i) ). We obtain h i (e η i P Ei E)(u) i= = h i (e η i P Ei E)(u) + E i A = E i A = h i C + E i A= h i P Ei E u j(i) 2N ω (q + ) C + (q + ) N 2θ. h i (e η i P Ei E)(u j(i) ) E i A=

15 Therefore, THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 3 (e γ P E )(u) C age(γ) + 2N N N 2θ + θ N q + = C age(γ) N = C age(γ) N. This concludes the proof. + ( ω (q + ) C + (q + ) 2(N + θ) N 2θ + θω C N 2N ) N 2θ Claim 4.4. Suppose that M consists of finite sets only. Let ε > 0 and let x, x 2,... be a sequence in X M such that x j and range x < m < range x 2 < m 2 < range x 3 < m 3 <... for some m, m 2,.... Then there is r N such that r j= x j > θ ε. Proof. We find recursively γ,..., γ N Γ and r(),..., r(n) N such that age(γ n ) = n, ran(γ n ) = m r(n) and e γ n ( r(n) j= x j) > θ ε N n. Let 0 n N and let there be some γ n and r(n) in the case that n. We need to find γ n+ and r(n + ). For practical purposes, we set r(0) = 0. For every i N, let us put E i = range x r(n)+i and let us choose η i Γ and ε i {, } such that ε i x r(n)+i (η i ) = x r(n)+i (η i ) > (ε/2θ). By Lemma 3.3, there are N and α,..., α 0 with i= α i = such that A M : α i < ω. i,e i A = Let q N be large enough that q + > 2θ/ε, E i {0,..., q} and η i Γ q for i =,...,, and q + = m r(n+) for some r(n + ) r(n) +. Let h i = (q + )α i, i =,...,, and γ n+ = (q +, {ε i } i=, {E i} i=, {η i} i=, {h i} i= ) if n = 0, γ n+ = (q +, γ n, {ε i } i=, {E i} i=, {η i} i=, {h i} i= ) if n. It is easy to chec that γ n+ q+. We have age(γ n+ ) = n + and ran(γ n+ ) = q + = m r(n+), and it remains to show that e γ n+ ( r(n+) j= x j ) > θ ε N (n + ).

16 4 ONDŘEJ KURKA Let us denote x = r(n+) j= x j. We realize first that if n, then e γ n (x) > θ ε N n. For j r(n) +, the range of x j is disjoint from [0, m r(n) ] = [0, ran(γ n )], and so e γ n (x j ) = 0. Hence, e γ n (x) = e γ n ( r(n) j= x j) > θ ε N n. Using Claim 4.2, we obtain in both cases n = 0 and n that e γ n+ (x) θ ε N = θ ε N > θ ε N n + d γ n+ (x) + θ N n θ N n + θ N q + q + h i i= q + h i ε i (e η i P Ei )(x) i= h i ε i e η i (x r(n)+i ) i= ( ε 2θ As i= h i i= ((q + )α i ) = q + > (q + )( (ε/2θ)), it follows that e γ n+ (x) > θ ε N n + θ ( N q + (q + ) ε ) 2 θ ε > (n + ). 2θ N Therefore, the choice of γ n+ and r(n + ) wors indeed. Finally, we get r(n) j= x j e γ N ( r(n) j= x j) > θ ε N N = θ ε. Proposition 4.5. () If M contains an infinite set, then X M is isomorphic to c 0. (2) If M consists of finite sets only, then X M does not contain an isomorphic copy of c 0 (cf. [2, Proposition 5.7(iii)]). Proof. () Let c = (N 2θ)/2N, C = 2N(N + θ)/(n 2θ)( θω). If {n, n 2,... } M for an infinite increasing sequence n < n 2 <..., then it follows from Claim 4. and Claim 4.3 that c sup P E u u C sup P E u, u X M, N {0} N {0} where E 0 = {0,..., n } and E = {n,..., n + }. Hence the space X M is isomorphic to the c 0 -sum of a sequence of finitedimensional spaces. It follows that X M is isomorphic to a subspace of c 0. By Theorem 2.3, since it is a L -space at the same time, it is isomorphic to c 0. (Let us remar that a simpler argument is provided in the proof of [8, Corollary ]). ).

17 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 5 (2) Let us fix ε > 0 such that θ ε >. By a simple induction argument, we show that Claim 4.4 holds in fact with the conclusion r j= x j > (θ ε) n. Assume that this statement holds for n. If x, x 2,... is such a sequence as in Claim 4.4, then we can apply the induction hypothesis again and again to find numbers 0 = s < s 2 <... such that s + j=s + x j > (θ ε) n for every N. Applying (unmodified) Claim 4.4 on the sequence (θ ε) n s + j=s + x j, we obtain r N such that (θ ε) n r = s + j=s + x j > θ ε. That is, s r+ j= x j > (θ ε) n+, which proves the statement for n +. Suppose that X M contains an isomorphic copy of c 0. Then there is a sequence u, u 2,... in X M such that u i and range u < range u 2 <... that is equivalent to the standard basis of c 0. The sequence x j = u 2j satisfies the assumption of Claim 4.4. Therefore, for every n N, there is r N such that r j= u 2j = r j= x j > (θ ε) n. As (θ ε) n can be arbitrarily large, the sequence u, u 2,... can not be equivalent to the standard basis of c 0. We introduce here one more result that will be useful later for proving Theorem.2. First, we show that X M has a property called the boundedly complete sipped blocing property in [6]. Claim 4.6. Suppose that M consists of finite sets only. Let a, a 2,... be a sequence in X M such that range a < m < range a 2 < m 2 < range a 3 < m 3 <... for some m, m 2,.... If the sequence of partial sums s i= a i, s =, 2,..., is bounded, then the sum i= a i is convergent. Proof. Assume that the sum i= a i is not convergent. Then there is δ > 0 such that, for every s N {0}, there is r N such that r i=s+ a i > δ. This allows us to find numbers 0 = s < s 2 <... such that s + i=s + a i > δ for every N. The sequence x = δ s + a i, =, 2,..., i=s + satisfies the assumption of Claim 4.4. By the argument from the proof of Proposition 4.5(2), for every n N, there is r N such that r = x > (θ ε) n. That is, s r+ i= a i > δ (θ ε) n. For this reason, the sequence of partial sums i= s a i, s =, 2,..., is not bounded.

18 6 ONDŘEJ KURKA Proposition 4.7 (cf. [2, Remar 5.9]). If M consists of finite sets only and the dual space X M is separable, then every infinite-dimensional subspace of X M contains an infinite-dimensional reflexive subspace. Proof. This follows from Claim 4.6 and [6, Corollary 2.6()]. 5. CONCLUSION As well as in the previous section, we fix N 3, < θ < N/2 and 0 < ω < /θ. For every M K(P(N)), we consider the space X M constructed for N, θ, ω and M. What remains to show is that the mapping M X M is measurable. Lemma 5.. There exists a Borel mapping S : K(P(N)) SE(C([0, ])) such that S(M) is isometric to X M for every M K(P(N)). Proof. For M K(P(N)), let us denote the objects from Section 4 by M p, ΓM p, c M γ, etc. For every M K(P(N)), it holds that 0 M = Γ0 M = {0} and i0,0 M and P0,M are the identity on l {0} ({0}). We can easily prove by induction on q N {0} the following observation: If M, M 2 K(P(N)) are two compact systems such that { } { } A {0,..., q} : A M = A {0,..., q} : A M2, then we have M q+ = M 2 q+, ΓM q+ = ΓM 2 q+, agem (γ) = age M 2(γ) and c M γ = c M 2 γ for γ M q+, im p,q+ = im 2 p,q+ for 0 p q + and P q+,m E = P q+,m 2 E for an interval E of {0,,..., q + }. Let γ M = 0 (i.e., Γ0 M = {γ M }) for all M K(P(N)). By a recursive procedure, for every q N {0}, we enumerate sets Γ M q+ in the way that we choose the same enumeration M q+ = { γ M Γ M q +, γm Γ M q +2,..., } γm Γ M q+ for M s which have the same set {A {0,..., q} : A M}. We claim that Lemma 2.5 can be applied on Ξ = K(P(N)), X M = X M and x M = d M γ M, N. Let n N and λ,..., λ n R be given. Clearly, γ M Γ M n for n, as Γ M n has at least n elements. We have n ( λ x M = P[0,n] M n ) ( λ x M = (in M r M n ) n ) λ x M, = = =

19 and so n = THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 7 λ x M ( (i M = lim q n,q+ rn M n ) λ x M = It remains to realize that M (in,q+ M rm n )( n = λ x M ) is a Borel function for every q n. This function is continuous in fact. Indeed, due to the above observation and the method of our enumeration, the norm (in,q+ M rm n )( n = λ x M ) depends only on {A {0,..., q} : A M}. For this reason, K(P(N)) can be decomposed into finitely many clopen sets on which this norm is constant. We are going to prove the complexity results proclaimed in the introduction. The following classical result (see e.g. [9, (27.4)]) is behind all of them. ). Theorem 5.2 (Hurewicz). The set { } H = M K(P(N)) : M contains an infinite set is a complete analytic subset of K(P(N)). In particular, it is not Borel. Proof of Theorem.. It is easy to show that the class of all Banach spaces X isomorphic to c 0 (shortly X c 0 ) is analytic (see [5, Theorem 2.3]). Let us show that it is hard analytic. Let S be a mapping provided by Lemma 5.. By Proposition 4.5, we have S(M) c 0 M contains an infinite set. Now, it is sufficient to use Theorem 5.2 and Lemma 2.4. Proof of Theorem.2. We chec first that X M X c 0 X M contains an infinite set, provided that X M is separable (we need to apply Proposition 4.7 in the case that X satisfies the second condition). The implication follows simply from Proposition 4.5(). Concerning the implication, we need to consider the conditions for X separately. Let us assume that M consists of finite sets only, and so that X M does not contain a subspace isomorphic to c 0 by Proposition 4.5(2). If X does not contain a subspace isomorphic to c 0, then X M X has the same property by Fact 2.2(), and thus X M X is not isomorphic to c 0 X. If X does not contain an infinite-dimensional reflexive subspace, then c 0 X has the same property by Fact 2.2(2), and thus c 0 X is not isomorphic to X M X, as X M has an infinitedimensional subspace by Proposition 4.7. If X is a subspace of a

20 8 ONDŘEJ KURKA space with an unconditional basis, then c 0 X has the same property, and thus c 0 X is not isomorphic to X M X, as X M is not isomorphic to a subspace of a space with an unconditional basis (otherwise it would be isomorphic to c 0 by Theorem 2.3). We realize that there is a Borel subset B of K(P(N)) containing the analytic non-borel set H from Theorem 5.2 such that X M is separable for every M B. It is sufficient to apply the Lusin separation theorem (see e.g. [9, (4.7)]), as H is disjoint from {M K(P(N)) : X M is not separable} that is analytic by Lemma 5. and [5, Corollary 3.3]. (In fact, concrete Borel subsets are available, for instance B = {M K(P(N)) : Sz(X M ) ω 0 }). It is easy to find a Borel mapping T : K(P(N)) SE(C([0, ])) such that T(M) is isomorphic to X M X for every M K(P(N)) (if I : C([0, ]) X C([0, ]) is an isomorphism, we can put T(M) = I(S(M) X), where S is a mapping given by Lemma 5.). We have T(M) c 0 X M contains an infinite set for every M B. Hence, it follows finally from Theorem 5.2 that the isomorphism class of c 0 X is not Borel. If X does not contain an infinite-dimensional reflexive subspace or X is a subspace of a space with an unconditional basis, then the same proof wors for the class of all Banach spaces Y that are isomorphic to a subspace of c 0 X (shortly Y c 0 X). Indeed, it follows easily from the above arguments that X M X c 0 X M contains an infinite set, provided that X M is separable. Thus, there is the only remaining case that X does not contain an isomorphic copy of c 0. Let us assume that the class of all Banach spaces isomorphic to a subspace of c 0 X is Borel. Then, by Lemma 5., { } C = M K(P(N)) : X M c 0 X is a Borel subset of K(P(N)). We have H C by Proposition 4.5(). Let us further consider { } D = M C : Z, dim Z =, Z X M, Z X. Using Lemma 5. and [5, Theorem 2.3(ii)], it is straightforward to chec that D is analytic. Let us show that C \ H D.

21 THE ISOMORPHISM CLASS OF c 0 IS NOT BOREL 9 Let M C \ H. By Lemma 2., X M has an infinite-dimensional subspace Z such that Z c 0 or Z X. Since every infinitedimensional subspace of c 0 contains an isomorphic copy of c 0, the possibility Z c 0 (which would imply c 0 Z X M ) is excluded by Proposition 4.5(2). Thus, Z witnesses that M D. By Theorem 5.2, the set C \ H is not analytic (by the Lusin separation theorem, H would be Borel in the opposite case). For this reason, the inclusion C \ H D is proper. Pic some M H D. There is an infinite-dimensional Banach space Z such that Z X M and Z X. Since X M c 0 and every infinite-dimensional subspace of c 0 contains an isomorphic copy of c 0, we have c 0 Z X. This contradicts the assumption on X. (Let us remar that the last part of the proof does not require the nowledge of the spaces X M, as much simpler spaces T [M, 2 ] studied e.g. in [20, Section 3] can be used in the definition of C as well). Proof of Theorem.3. It was shown in [4, 5.7], [3, p. 367] and [3, p. 368] that the classes are analytic. Let us show that they are not Borel. We have already seen in the proof of Theorem. that it is sufficient to chec that M contains an infinite set X M has an unconditional basis X M Z, where Z has an unc. basis X M is isomorphic to a C(K) space. As c 0 satisfies all three properties, implications follow simply from Proposition 4.5(). Considering Proposition 4.5(2), to prove implications, we need to verify that any of those three conditions implies that X M contains an isomorphic copy of c 0. It is sufficient to use Theorem 2.3 and the fact that every infinite-dimensional separable C(K) space contains an isomorphic copy of c 0. REFERENCES [] S. A. Argyros and I. Deliyanni, Examples of asymptotic l Banach spaces, Trans. Amer. Math. Soc. 349, no. 3 (997), [2] S. A. Argyros, I. Gasparis and P. Motais, On the structure of separable L - spaces, Mathematia 62, no. 3 (206), [3] S. A. Argyros and R. G. Haydon, A hereditarily indecomposable L -space that solves the scalar-plus-compact problem, Acta Math. 206, no. (20), 54. [4] B. M. Braga, On the complexity of some inevitable classes of separable Banach spaces, J. Math. Anal. Appl. 43, no. (205), [5] B. Bossard, A coding of separable Banach spaces. Analytic and coanalytic families of Banach spaces, Fund. Math. 72, no. 2 (2002), 7 52.

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