Large cardinals and basic sequences

Annals of Pure and Applied Logic 164 (2013) 1390–1417

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Annals of Pure and Applied Logic www.elsevier.com/locate/apal

Large cardinals and basic sequences Jordi Lopez-Abad Instituto de Ciencias Matematicas (ICMAT), CSIC-UAM-UC3M-UCM, Madrid, Spain

a r t i c l e

i n f o

Article history: Received 11 February 2012 Received in revised form 7 October 2012 Accepted 10 October 2012 Available online 5 July 2013

a b s t r a c t The purpose of this paper is to present several applications of combinatorial principles, well-known in Set Theory, to the geometry of infinite dimensional Banach spaces, particularly to the existence of certain basic sequences. We mention also some open problems where set-theoretical techniques are relevant. © 2013 Elsevier B.V. All rights reserved.

MSC: primary 46B26, 03E55 secondary 03E35, 03E02, 46A35 Keywords: Non-separable Banach spaces Basic sequences Partition calculus

1. Introduction From the beginning of Banach space theory it was well understood how useful was to represent nicely the vectors of a given Banach space as sequences of scalars. This is for example the case when the space has a Schauder basis (xn )n : Every vector is represented as a unique (possibly) infinite linear combination of the vectors (xn )n∈ω . However, not every Banach space, even the separable ones, has a Schauder basis. But fortunately, there are many of them. Indeed every infinite dimensional subspace contains itself a subspace with a Schauder basis, or every non-trivial weakly-null sequence has a Schauder basic subsequence. As it is expected, the more properties the basis has, the better understood the space may be. One of these properties is to be equivalent to the unit basis of the classical sequence spaces p , p  1, or c0 ; another is the unconditionality. It was a central problem in the field to know if every infinite dimensional Banach space has an infinite dimensional subspace isomorphic to one of these classical sequence spaces. This was solved in the 70’s by Tsirelson, who defined a space (interestingly, inspired by the method of forcing) not having subsymmetric sequences, and consequently not having isomorphic copies of the classical sequence spaces. It took a little more time until Gowers and Maurey found in the 90’s a space without subspaces with unconditional bases. In the other direction, Ketonen showed in the 70’s the relationship between the E-mail address: [email protected]. 0168-0072/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.apal.2013.06.016

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existence of unconditional basic sequences and large cardinals by proving that every Banach space whose density is an ω-Erdős cardinal always has a subspace with an unconditional basis. The key combinatorial feature used by Ketonen is the Ramsey property of the ω-Erdős cardinals. In this paper we explain this and some other applications of Ramsey properties of a cardinal κ, like the Polarized property Pl2 (κ), or the Free Set Property Fr(κ, ω), and we see how they force a Banach space of density κ to have a subspace with an unconditional basis. We will also see how Ketonen’s result can be improved to obtain that it is consistent relative to the existence of large cardinals that every Banach space of density at least ωω has a subspace with an unconditional basis. On the other hand, we give details of the constructions, using anti-Ramsey principles, of large sequences without unconditional basic subsequences. At the end of the paper we will also mention problems concerning the existence of certain uncountable sequence, and we will present a general approach to define generic spaces of density ω1 lacking those uncountable sequences. 2. Basic notions and facts Recall that a normed space (X,  · ) is a vector space X (over the real numbers R here) and a norm  ·  : X → R on it, i.e. (N.1) λx = |λ|x for every x ∈ X and λ ∈ R. (N.2) x + y  x + y for every x, y ∈ X. (N.3) x = 0 iff x = 0. The normed space (X,  · ) is a Banach space when the norm  ·  is complete, i.e. Cauchy sequences  1 are convergent. Well-know examples are Rn with the Euclidean norm (ai )i
An isomorphic embedding T : X → Y is a 1–1 operator such that T (X) is a closed subspace of Y and the inverse U : T (X) → X is bounded. The dual X ∗ of a Banach space X is the space of all operators f : X → R. This is a Banach space with the norm    f  := sup f (x): x  1 .

The elements of X ∗ are usually called functionals. The weak topology in X is the topology for which the basic open neighborhoods of a point x ∈ X are {y ∈ X: maxin |fi (x)−fi (y)|  ε} for f0 , . . . , fn ∈ X ∗ and ε > 0. Similarly, the weak∗ topology in X ∗ is the topology with basic open neighborhoods of a functional f , the

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sets {g ∈ X ∗ : maxin |f (xi ) − g(xi )|  ε} for x0 , . . . , xn ∈ X and ε > 0. Notice that these two topologies are in general very much different for dual spaces X ∗ : The Alaoglu Theorem states that the unit ball BX ∗ := {f ∈ X ∗ : f   1} is always compact with respect to the weak∗ topology, but the ball BX ∗ is weakly-compact exactly when X ∗ (or equivalently X) is reflexive. We pass now to introduce well-known types of sequences in a Banach space: (1) A sequence (xγ )γ<κ in a Banach space (X,  · ), indexed in some cardinal number κ, is called a biorthogonal sequence (or system) if for every α < κ there is a functional fα ∈ X ∗ such that fα (xβ ) = δα,β . Notice that in particular biorthogonal sequences are linearly independent sequences. (2) A normalized sequence (xγ )γ<λ in a Banach space (X,  · ), indexed in some ordinal number λ, is called a (Schauder) basic sequence when there is a constant C  1 such that            aγ x γ   C  aγ x γ    γ∈t

γ∈s

for every sequence of scalars (aγ )γ∈s and every initial part t  s ⊆ λ, s finite. The same inequality is true for arbitrary sequences (aγ )γ∈I and J  I ⊆ λ. It easily follows that (xα )α is a biorthogonal sequence. (3) A normalized sequence (xγ )γ<λ in X is a (Schauder) basis of X if it is basic, and the closed linear span of (xα )α is X. Equivalently, every x ∈ X has a unique representation as x=



aγ x γ .

γ<λ

 Recall that y = γ<λ yγ just means that there is a continuous mapping S : λ + 1 → X, where λ + 1 is considered with its usual order topology and X with the norm topology, such that S(1) = y0 , S(λ) = y and S(γ + 1) = S(γ) + yγ for every γ < λ. (4) A normalized sequence (xγ )γ is called an unconditional basic sequence when there is a constant C  1 such that C  1 such that             a x  C a x γ γ γ γ   γ∈t

γ∈s

for every sequence of scalars (aγ )γ∈s and every t ⊆ s ⊆ κ. (5) (xγ )γ<κ is called weakly-null when the set 

   γ < κ: x∗ (xγ )  ε

is finite for every x∗ ∈ X ∗ and every ε > 0. Equivalently, (xγ )γ<κ is weakly-null when for every ε > 0 the set      F (xγ )γ<κ , ε := s ∈ [κ]<ω : there is some f ∈ BX ∗ such that f (xα )  ε for all α ∈ s

is a compact family, i.e. every infinite subset of κ has a finite subset which is not in F((xγ )γ<κ , ε). (6) Two basic sequences (xα )α∈λ and (yα )α∈λ are C-equivalent (C  1) when             1       a x  a y  C a x α α α α α α    C α∈s α∈s α∈s

for every sequence (aα )α∈s indexed in s ⊆ λ finite.

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(7) (xγ )γ is called subsymmetric when there is a constant C  1 such that every two finite subsequences of (xγ )γ of the same length are C-equivalent. (8) (xγ )γ is called symmetric when there is a constant C  1 such that for every finite set s ⊆ γ and every permutation π on s, the finite subsequences (xγ )γ∈s and (xπγ )γ∈s are C-equivalent. (n)

Symmetric bases are always unconditional and subsymmetric. The unit basis (un )n , un = (0, 0, . . . , 0, 1 , 0, 0, . . .), is a Schauder basis of each p or c0 . Moreover, it is a symmetric basis. For p > 1 it is also weakly-null. All these examples are sequence spaces. In fact, every Banach space (X,  · ) with a basis (xγ )γ<λ is  automatically a sequence space: Let Y := {(aγ )γ<λ ∈ Rλ :  γ<λ aγ xγ  < ∞} and for (aγ )γ ∈ Y , let   |||(aγ )γ ||| :=  γ<λ aγ xγ . Then (Y, ||| · |||) is a Banach space and T : Y → X, T ((aγ )γ ) := γ aγ xγ is an isometry. We list now several classical results concerning countable sequences. For a more complete information we recommend the reader the excellent book of Lindenstrauss and Tzafriri [22]. (1) (2) (3) (4)

Every infinite dimensional Banach space has a basic sequence (Banach–Mazur). Every normalized weakly-null sequence has a basic subsequence (Mazur). There are separable Banach spaces without bases (Enflo). The structure of subspaces of a Banach space X with a basis (en )n∈ω is determined by block subse quences of the basis (en )n , i.e. sequences (xn )n such that each xn = k∈sn ak ek with sn ⊆ ω finite and such that max sn < min sn+1 for every n. More precisely, for every ε > 0 and every infinite dimensional subspace Y of X there is a normalized block subsequence (xn )n∈ω of (en )n and a normalized sequence  (yn )n in Y such that n xn − yn   ε (Bessaga–Pelczynski). (5) The reflexivity of a space with a basis is determined by the basis. More precisely, a Banach space with a basis (xn )n is reflexive, if and only if (xn )n is shrinking and boundedly complete (James). (6) A space with an unconditional basis is reflexive if and only if it does not contain an isomorphic copy of c0 or 1 (James). (7) c0 or p are finitely block representable in any Banach space, i.e. if (xn )n is a basic sequence in a Banach space, then there is 1  p  ∞ such that for every C > 1 and for every n there is a finite block subsequence (xk )k∈n such that        1 (ak )k
for every sequence of scalars (ak )k∈n (Krivine). (8) There are spaces without isomorphic copies of c0 or p , p  1 (Tsirelson [38]). (n+1)

(9) The summing basis sn = (1, . . . , 1, 0 , 0, . . .) ∈ c0 does not have unconditional subsequences, and it is not weakly-null. (10) Every norm-bounded sequence (xn )n has a subsequence which is either equivalent to the unit basis of 1 or a weakly-Cauchy, i.e., (f (xn ))n is a numerical Cauchy sequence for each f ∈ X ∗ (Rosenthal’s 1 -dichotomy [30]). (11) Suppose that in the dual space X ∗ of X there is an unconditional basic sequence. Then X has a separable quotient with an unconditional basis (Hagler–Johnson). (12) There is a weakly-null basic sequence without unconditional basic subsequences (Maurey–Rosenthal). (13) There is a reflexive Banach space without unconditional basic sequences (Gowers–Maurey). Finally, we introduce the following notation. Given a set I, by c00 (I) we mean the vector space consisting of all f : I → R such that the support supp(f ) := {i ∈ I: f (i) = 0} is finite. Let c00 (I, Q) := c00 (I) ∩ QI .

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Given f ∈ c00 (I) and i ∈ I we write (f )i to denote f (i). The unit basis (ui )i∈I is defined by (ui )j := δi,j for every i, j ∈ I. This is obviously a Hamel basis of c00 (I). Given I ⊆ J, we identify each f ∈ c00 (I) with f  ∈ c00 (J) such that f   I = f and (f  )  (J \ I) = 0. In this way c00 (I) becomes a subspace of c00 (J).  Given θ : I → J, we extend it to θ : c00 (I) → c00 (J) by θ(x) := i∈I (x)i uθ(i) . 3. Existence of certain countable sequences 3.1. Unconditional basic sequences The goal is to understand the following cardinal numbers. Definition 3.1. Given class of Banach spaces C, which is the minimal infinite cardinal number ncC such that every space in C of density at least ncC has an unconditional basic sequence. When C is the class of all Banach spaces, we will write simply nc. The Gowers–Maurey space [15] gives that nc is uncountable, and the Argyros–Tolias space [4] proves that nc > c. Observe that we have not justified yet that ncC is well defined. We pass to do that, assuming the existence of large cardinals. Definition 3.2. We call a basic sequence (xn )n<ω asymptotically unconditional 1 when there is a constant C  1 such that every finite subsequence (xn )n∈s is C-unconditional for every s ⊆ ω with |s|  min s + 1. In this case we say that (xn )n is asymptotically C-unconditional. Remark 3.3. Every subsymmetric and asymptotically unconditional sequence is automatically unconditional. Theorem 3.4 (Schreier unconditionality). (See [29].) For every non-trivial weakly-null sequence (xn )n<ω and every C > 1 there is a subsequence (yn )n<ω of (xn )n<ω which is an asymptotically C-unconditional basic sequence. It readily follows the following. Corollary 3.5. Every non-trivial weakly-null subsymmetric sequence has an unconditional basic subsequence. We need the following standard terminology. Given a set A and a cardinal number λ, let [A]λ := {B ⊆ A: |B| = λ}, [A]λ := {B ⊆ A: |B|  λ}, and [A]<λ := {B ⊆ A: |B| < λ}. Definition 3.6. A cardinal number κ is called ω-Erdős when κ → (ω)<ω 2 , i.e. whenever c : [κ]<ω → 2 there is an infinite c-homogeneous subset A ⊆ κ, i.e. such that c  [A]n is constant for every n ∈ ω. It is not difficult to see that the definition above is equivalent to κ → (ω)<ω c . The following is essentially due to Ketonen in [20], although it is not explicitly stated in there. 1

Often also called Schreier unconditional.

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Proposition 3.7. nc  first ω-Erdős cardinal. Proof. Suppose that κ → (ω)<ω c , and suppose that X is a Banach space with density  κ. Fix a normalized 1-separated sequence (xα )α<κ in X, i.e. xα − xβ   1 for every α < β < κ. It follows then that (xα )α∈s is a basic sequence for every finite subset s ⊆ κ. Fix now a set A of cardinality c such that every finite basic sequence is 1-equivalent to a sequence in A. Now let c : [κ]<ω → A be defined for s ∈ [κ]<ω by choosing c(s) ∈ A which is 1-equivalent to (xα )α∈s . Since κ is ω-Erdős, there is some infinite subset A ⊆ κ such that c  [A]n is constant for every n < ω, or in other words, (xα )α∈A is subsymmetric. Now we use the Rosenthal’s 1 -dichotomy: The first alternative gives that there is a subsequence of (xα )α∈A which is equivalent to the unit basis of 1 , hence unconditional; the second alternative gives a subsequence (xαn )n<ω of (xα )α∈A which is weakly-Cauchy. Since (xα )α<κ is 1-separated, the difference sequence (yn )n<ω , yn = (xα2n+1 − xα2n )/xα2n+1 − xα2n  for n < ω, is a normalized weakly-null sequence, and it is also subsymmetric. By Theorem 3.4, we obtain a subsequence (ynk )k<ω which is an asymptotically unconditional subsymmetric basic sequence, and therefore unconditional. 2 The previous result can be refined to get that in fact such spaces of high density not containing 1 contain a 1-unconditional basic sequence. We see now how a well-known combinatorial property, originally appearing in the problem lists of P. Erdős and A. Hajnal [12], [13, Problem 29] (see also [32]), can be used to improve Ketonen’s result. Definition 3.8. Let κ be a cardinal and d ∈ ω \ 1. By Pld (κ) we shall denote the combinatorial principle asserting that for every coloring c : [[κ]d ]<ω → ω there exists a sequence (Xn )n∈ω of infinite disjoint subsets

m Xn of κ such that for every m ∈ ω the restriction c  n=0 [Xn ]d is constant. Clearly Pld (κ) implies Pld (κ) if d  d  1. From known results one can easily deduce that the principle Pld (expd−1 (ω)+n ) is false for every n ∈ ω and every integer d  1 (see, [14,6,8]). Thus, the minimal cardinal κ for which Pld (κ) could possibly be true is expd−1 (ω)+ω . In fact, Di Prisco and Todorcevic [8] have established the consistency of Pl1 (ωω ) relative the consistency of a single measurable cardinal, an assumption that also happens to be optimal. On the other hand, S. Shelah [32] (see also [26]) proved that GCH and principles Pld (ωω ) (d  1) are jointly consistent, relative to the consistency of GCH and the existence of an infinite sequence of strongly compact cardinals. Theorem 3.9. (See [9].) Let κ be a cardinal and assume that property Pl2 (κ) holds. Then every Banach space E not containing 1 and of density κ contains a 1-unconditional basic sequence. In particular, if E is any Banach space of density κ, then for every ε > 0 the space E contains a (1 + ε)-unconditional basic sequence. We start with the following lemma, which is essentially a multi-dimensional version of Odell’s Schreier unconditionality Theorem 3.4. Lemma 3.10. Let E be a Banach space, m ∈ ω with m  1 and ε > 0. For every i ∈ {0, . . . , m} let (xin ) be a normalized weakly-null sequence in the space E. Then, there exists an infinite subset L of ω such that for every {n0 < · · · < nm } ⊆ L the sequence (xini )m i=0 is (1 + ε)-unconditional. Proof. The first step is the following refinement of Mazur’s Lemma. Claim 3.10.1. For every ε > 0 there exists an infinite subset M of ω such that for every {n0 < · · · < nm } ⊆ M the sequence (xini )m i=0 is a (1 + ε)-Schauder basic sequence.

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We use this claim for ε = 1, to get an infinite subset M of ω as described above. Observe that for every {n0 < · · · < nm } ∈ [M ]m+1 and every choice (ai )m i=0 of scalars we have   m   1  i  ai xni   max |ai |.    4 im i=0

(1)

By a simple use of the Ramsey Theorem, we may assume that for every {n0 < · · · < nm }, {k0 < · · · < i m km } ∈ [M ]m+1 the sequences (xini )m i=0 and (xki )i=0 are (1 + δ)-equivalent, where δ > 0 is such that −1 (1 + δ) · (1 − δ)  (1 + ε). Let now c : [M ]m+1 → 2 be defined as follows. We color s = {n0 < · · · < nm } ∈ [M ]m+1 by c(s) = 0 when the sequence (xini )m i=0 is (1 + ε)-unconditional, and c(s) = 1 otherwise. Let L be an infinite subset of M such that c is constant on [L]m+1 . It is enough to find some s ∈ [L]m+1 such that c(s) = 0. Now since each sequence (xin )n is weakly-null, we can find a sequence s0 < · · · < sm of finite subsets of L / F((xin )n∈L , δ/(8(m + 1))) for each i  m, and where for s < t we mean that max s < min t. such that si ∈ Let ni = min(si ) for all i ∈ {0, . . . , m}. Observe that n0 < · · · < nm . We claim that the sequence (xini )m + ε)-unconditional. Let s ⊆ m + 1 and let (ai )im be a sequence of scalars. We have to i=0 is (1    prove that  i∈s ai xini   (1 + ε) im ai xini . It is clear that we may assume that  i∈s ai xini  = 1.   i Now if  i∈s 2, then the triangle inequality readily gives that  im ai xini   1, and so we are / ai x n i    i done. Suppose that  i∈s / s}  8. Let x∗ ∈ BX ∗ be / ai xni   2. Then by (1), it follows that max{|ai |: i ∈  such that x∗ ( i∈s ai xini ) = 1. Now let (ki )im be defined as ki = ni for i ∈ s and ki ∈ si be such that |x∗ (xini )|  8/(δ(m + 1)). Then         m m            ∗    ∗  i  i i ai xki   x ai xki   x ai xki  − ai x∗ xiki   1 − δ.      i=0

i=0

i∈F

i∈F /

Using that (xini )im and (xiki )im are (1 + δ)-equivalent, we conclude that  m   m       1−δ  1  1    i  i  i    ai x n i   ai xki   a x   i n i .    1 + δ  1+δ 1+ε i=0 i=0

2

i∈F

We are ready to proceed to the proof of Theorem 3.9. Proof of Theorem 3.9. Let κ be a cardinal such that Pl2 (κ) holds. Suppose first that X has an isomorphic copy of 1 . By a classical result of R.C. James (see [22, Proposition 2.e.3]), we obtain that in fact X has a (1 + ε)-isomorphic copy of 1 , hence the copy of the unit basis of 1 is a (1 + ε)-unconditional basic sequence. Now suppose that 1 does not embed into X. Let (xα )α<κ be a 1-separated normalized sequence. Let c : [[κ]2 ]<ω → ω be the coloring defined as follows. Given s = ({α0 < β0 }, . . . , {αm < βm }) ∈ [[κ]2 ]<ω we color s by c(s) = ls if ls is the minimal integer l such that the sequence (xβi − xαi )m i=0 is m not (1 + 1/l)-unconditional, and c(s) = 0 when (xβi − xαi )i=0 is 1-unconditional. By Pl2 (κ), there exist a sequence (Xn )n∈ω of infinite subsets of κ and a sequence (lm )m of integers such that for every m ∈ ω

m the restriction c  i=0 [Xi ]2 is constant with value lm . It follows now from Lemma 3.10 that lm = 0 for

every m ∈ ω, so for every infinite sequence of pairs ({αi < βi }) ∈ i∈ω [Xi ]2 the corresponding sequence ((xβi − xαi )/xβi − xαi )i∈ω is a normalized 1-unconditional basic sequence. 2 Theorem 3.11. (See [32].) Suppose that in a set-theoretic universe V there exists a strictly increasing sequence (κn ) of strongly compact cardinals with κ0 = ω. Then, there is a forcing extension of V in which the principle Pl2 (expω (ω)) holds. Moreover, if GCH holds in V , then GCH also holds in the extension.

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Consequently, combining this Theorem 3.11 with Theorem 3.9, we get the following corollaries. Corollary 3.12. It is consistent relative the existence of an infinite sequence of strongly compact cardinals that for every ε > 0 and every Banach space E of density at least expω (ω), the space E contains a (1 + ε)-unconditional basic sequence. Moreover, this statement is consistent with GCH. Corollary 3.13. It is consistent relative to the existence of an infinite sequence of strongly compact cardinals that every Banach space of density at least expω (ω) has a separable quotient with an unconditional basis. Moreover, this statement is consistent with GCH. Proof. A well-known consequence of a result due to J.N. Hagler and W.B. Johnson [17] asserts that if E is a Banach space such that E ∗ has an unconditional basic sequence, then E has a separable quotient with an unconditional basis (see also [2, Proposition 16]). Noticing that the density of the dual E ∗ of a Banach space E is at least as big as the density of E, the result follows by Corollary 3.12. 2 Problem 1. Is some of the finite exponents expn (ω) an upper bound of nc? To finish this part we give some hints for a proof of Shelah’s Theorem 3.11. In fact in [10] it is proved the existence of certain ideals that readily gives that Pl2 (ωω ) holds. Lemma 3.14. Suppose that κ is a strongly compact cardinal and that λ < κ is an infinite regular cardinal. Let G be a Col(λ, < κ)-generic filter over V . Then, in V [G], for every integer d  1 there exists an ideal Id on [(expd (κ))+ ]ω and a subset Dd of Id+ such that the following are satisfied. (1) Id is κ-complete. (2) Dd is dense in Id+ and is λ-closed in Id+ . (3) For every μ < κ, every coloring c : [(expd (κ))+ ]d+1 → μ and every set A ∈ Id+ there exist a color ξ < μ and an element D ∈ Dd with D ⊆ A and such that for every X ∈ D the restriction c  [X]d+1 is constantly equal to ξ. Let Θ0 = ω

 Θn + ++ and Θn+1 = 2(2 ) .

(2)

Notice that the sequence (Θn ) is strictly increasing and that expn (ω) < Θn  exp5n (ω) for every n ∈ ω with n  1. Hence, sup{Θn : n ∈ ω} = expω (ω). In particular, if GCH holds, then Θn = ω5n for every n ∈ ω. Now iterating the previous construction one can get the following. Theorem 3.15. Suppose that (κn ) is a strictly increasing sequence of strongly compact cardinals with κ0 = ω. κn + For every n ∈ ω set λn = (2(2 ) )+ . Let P=



Col(λn , < κn+1 )

n∈ω

be the iteration of the sequence of Lévy collapses. Let G be a P-generic filter over V . Then, in V [G], for every n ∈ ω we have κn = Θn and there exist an ideal In on [(2Θn+1 )+ ]ω and a subset Dn of In+ such that the following are satisfied.

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(P1) In is Θn+1 -complete. (P2) Dn is (< Θn+1 )-closed in In+ ; that is, Dn is μ-closed in In+ for every μ < Θn+1 . (P3) For every μ < Θn+1 , every coloring c : [(2Θn+1 )+ ]2 → μ and every A ∈ In+ there exist a color ξ < μ and an element D ∈ Dn with D ⊆ A and such that for every X ∈ D the restriction c  [X]2 is constantly equal to ξ. Moreover, if GCH holds in V , then GCH also holds in V [G]. Finally, Pl2 (ωω ) holds in the previous iteration of Levy collapses above, starting with a model where GCH holds, by the following result. Proposition 3.16. Let (Θn ) be the sequence of cardinals defined in (2) above. Suppose that for every n ∈ ω there exist an ideal In on [(2Θn+1 )+ ]ω and a subset Dn of In+ which satisfy properties (P1), (P2) and (P3) described in Theorem 3.15. Then, the principle Pl2 (expω (ω)) holds. 3.2. Reflexive spaces and unconditional subsequences of weakly-null sequences We analyze now the cardinal number ncRefl for the class of Refl of reflexive spaces, and the first thing we do is to relate this cardinal with a “sequential property”. Definition 3.17. Let nc0 be the minimal cardinal number κ such that every normalized weakly-null sequence (xα )α<κ has an unconditional basic subsequence. The relationship between ncRefl and nc0 is given by the following classical result. Theorem 3.18. (See Amir and Lindenstrauss [1].) Every reflexive space of infinite density κ has a normalized weakly-null sequence of length κ. Consequently, ncRefl  nc0 . Now there is the following sequential analogue of Theorem 3.9. Theorem 3.19. (See [9].) Let κ be a cardinal and assume that property Pl1 (κ) holds. Then nc0  κ. In fact, every normalized weakly-null sequence (xα )α<κ has a 1-unconditional subsequence. Proof. The proof is very similar to the one of Theorem 3.9. Consider the coloring c : [κ]<ω → ω defined for s = {α0 < · · · < αm } ∈ [κ]<ω as c(s) = ls if ls > 0 is the minimal l ∈ ω such that the sequence (xαi )m i=0 is not (1 + 1/l)-unconditional, and c(s) = 0 if that sequence is 1-unconditional. Using Pl1 (κ) and Theorem 3.4, the result follows. 2 Clearly, Pl1 (expω (ω)) holds in the generic model exposed in Theorem 3.15. However, as we have already indicated, one can obtain the consistency of Pl1 (expω (ω)) using a considerably weaker (and, in fact, optimal) large-cardinal assumption than the one used for Pl2 (expω (ω)). Theorem 3.20. (See [8].) Assume the existence of a measurable cardinal. Then, there is a forcing extension in which GCH and Pl1 (ωω ) hold. Corollary 3.21. It is consistent relative to the existence of a single measurable cardinal that nc0 = ωω . Proof. The first inequality nc0  ωω follows from Theorem 3.19 and Theorem 3.20. The other inequality nc0  ωω will be treated in Section 3.3. 2

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There is another well-known combinatorial property of a cardinal κ which is implied by Pl1 (κ) and which is in turn sufficient for the estimate ncseq  κ. This property is in the literature called the free set property of κ (see [31,21,8] and the references therein). Definition 3.22. By a structure on κ we mean a first order structure M = (κ, (fi )i∈ω ), where ni ∈ ω and fi : κni → κ for all i ∈ ω. The free set property of κ, denoted by Frω (κ, ω), is the assertion that every structure M = (κ, (fi )i∈ω ) has a free infinite set. That is, there exists an infinite subset L of κ such that every element x of L does not belong to the substructure of M generated by L \ {x}. The following is easy to prove. Proposition 3.23. Let κ be a cardinal. Then the following are equivalent. (a) Frω (κ, ω) holds. (b) For every structure M = (κ, (fi )i∈ω ) there exists an infinite subset L of κ such that for every x ∈ L we have   ni  x∈ / fi (s): s ∈ L \ {x} and i ∈ ω .

(c) Every extended structure N = (κ, (gi )i∈ω ), where gi : κ<ω → [κ]ω for all i ∈ ω, has an infinite free subset. That is, there exists an infinite subset L of κ such that for every x ∈ L we have x∈ /





gi (s).

i∈ω s∈(L\{x})<ω

Theorem 3.24. Let κ be a cardinal and assume that Frω (κ, ω) holds. Then every normalized weakly-null sequence (xα )α<κ has a 1-unconditional subsequence Proof. Let (xα )α<κ be a normalized weakly-null sequence in a Banach space X. For every s ∈ [κ]<ω we select a subset Fs of BX ∗ which is countable and 1-norming for the finite dimensional subspace Xs of X spanned by {xα }α∈s , i.e. for every x ∈ Xs we have that x = sup{x∗ (x): x ∈ Fs }. Define g : [κ]<ω → [κ]ω for s ∈ [κ]<ω as g(s) = {α < κ: there is some x∗ ∈ Fs such that x∗ (xα ) = 0}. g is well defined because (xα )α<κ is weakly-null and Fs is countable. Consider the structure N = (κ, g). Since Frω (κ, ω) holds, there exists an infinite free subset L of κ. Then (xα )α∈L is 1-unconditional: Let s ⊆ t be two finite subsets of L, let (aα )α∈t be a sequence of scalars, and let ε > 0 arbitrary. We fix x∗ ∈ Fs such that  

    ∗   a x  (1 + ε) · x a x α α α α .  α∈s

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α∈s

Since the set L is g-free, one has that x∗ (xα ) = 0 for every α ∈ L \ s. Hence  

 

    ∗ ∗   aα xα   (1 + ε) · y aα xα = (1 + ε) · y aα x α  α∈s

α∈s

      (1 + ε) ·  a x α α .  α∈t

Since ε > 0 was arbitrary, the desired result follows. 2

α∈t

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At this present, there exist few examples of non-separable spaces with no infinite unconditional basic sequences. Two of them are the example of Argyros and Tolias in [4] that has density 2ℵ0 and the example in [3] that has density ω1 . There is a crucial difference between these two spaces. The space of [3] is reflexive while the space of [4] is far from this, as it is a dual of a separable hereditarily indecomposable space. In this section we use some known combinatorial properties of cardinals to show that this difference is in fact essential. More precisely, we show that there are no reflexive spaces of density the continuum without unconditional basic sequences unless additional set-theoretic axioms are assumed such as, for example, the Continuum Hypothesis. Theorem 3.25. It is consistent relative to the consistency of the existence of an ω-Erdős cardinal that every weakly-null sequence of length continuum contains an infinite unconditional basic sequence. The proof also reveals the following interesting connection with the classical problem of Banach about extending the Lebesgue measure to all sets of reals. Theorem 3.26. Suppose that the Lebesgue measure extends to a total countably additive measure on R. Then every weakly-null sequence of length continuum contains an infinite unconditional basic sequence. Unlike the first polarized cardinal, the first cardinal satisfying Fr(κ, ω) does not need to be larger than the continuum. Lemma 3.27. The property Fr(κ, ω) is preserved under forcing by posets satisfying the countable chain condition. Proof. Let P be a forcing notion satisfying the countable chain condition, and let f˙ be a P-name for a function f : κ<ω → κ. Since P satisfies the countable chain condition, for every name τ for an ordinal in κ there is a countable set C(τ ) ⊆ κ such that every p ∈ P forces that τ is an element of C(τ ). So we can find a sequence gn : κ<ω → κ, n < ω, with the property that for every s ∈ κ<ω    C f˙(s) = gn (s): n < ω .

Applying our assumption Fr(κ, ω) to the algebra (κ, gn )n<ω we get an infinite subset X of κ that is free in this algebra. It follows that every p ∈ P forces that X is also f -free in the forcing extension of P. 2 Proof of Theorem 3.25. Every ω-Erdős cardinal κ has the property Fr(κ, ω) (see, for example, [7]). Now the conclusion follows from Theorem 3.24 and Lemma 3.27, by going to the forcing extension were at least κ many real numbers are added by a poset satisfying the countable chain condition. 2 Proof of Theorem 3.26. By a result of [34], the assumption implies the existence of a regular cardinal κ  2ℵ0 with the Jonsson property, i.e., with the property that every algebra with domain κ contains a proper subalgebra of the same cardinality. On the other hand, it is proved in [11] (see also [7]) that every Jonsson cardinal κ has the free-set property Fr(κ, ω). Consequently, Fr(2ℵ0 , ω) holds and the desired result follows from Theorem 3.24. 2 3.3. Examples Our intention now is to present constructions of large weakly-null sequences without unconditional subsequences. In particular we will see that nc0  ωω . The constructions rely on the existence of certain countably chromatic graphs, and they are based on the ideas of Maurey and Rosenthal to build a countable weakly-null sequence without unconditional subsequences.

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Definition 3.28. Let G = (V, E) be a graph. The chromatic number χ(G) is the minimal cardinal number κ such that there is a coloring c : V → κ, called a good coloring of G, of the set of vertexes V into κ-many colors such that no two vertexes v0 = v1 in an edge in E have the same color c. Let Gcard (κ) := (V, Ecard ) be the graph with vertexes, V the set of all finite block sequences (si )i
Then Gcard (κ) has countable chromatic number: Let d : ω <ω → ω be any 1–1 function, and for a finite block sequence (si )i
Then, χ(Gn (κ)) > λ for all n < ω.

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This gives the following relation between the polarized property Pl1 (κ) and the chromatic number of Gn (κ): Theorem 3.31. If κ is the minimal cardinal such that Pl1 (κ), then χ(Gn (κ)) = κ for all n < ω. Proof. Let κ be such cardinal. Then ⎛ ⎞<ω 2 ⎝ κ → 2⎠ .. . λ

for every λ < κ (see [8] for more details). Now the result follows from Proposition 3.30. 2 Consequently: Corollary 3.32. It is consistent relative to the existence of a measurable cardinal that for every n < ω one has that χ(Gn (ωω )) = ωω . In the other direction, there are relatively large families on ωn for which the chromatic numbers of their positional graphs are countable. In order to be more precise, we need the following notion of largeness. Definition 3.33. We call a family B of finite subsets of κ very-large when every infinite subset A ⊆ κ has an infinite subset B ⊆ A such that [B]<ω ⊆ B. Theorem 3.34. (See [24].) For every n < ω there is a very-large family Bn on ωn such that χ(G2n−2 (Bn )) = ω. Note that for n  2 the previous result is somehow sharp, as χ(Gm (ωn )) > ω for every m. The importance of these families is explained by the following. Theorem 3.35. (See [24].) Suppose that for some cardinal κ and some n < ω there is a very-large family B ⊆ [κ]<ω such that χ(Gn (B)) = ω. Then there is a norm  ·  on c00 (κ) such that the sequence (uγ )γ<κ of unit vectors of c00 (κ) is a weakly-null Schauder basis of the completion of (c00 (κ),  · ) with no infinite unconditional basic subsequences. Proof. The example is the natural generalization of the classical construction by Maurey and Rosenthal in [25] of a weakly-null ω-sequence without infinite unconditional basic subsequences. The space is the completion of (c00 (κ),  · K ) for a certain norm  · K on c00 (κ) that we pass to define. Let B be a very-large family on κ such that Gn (B) = ω for some fixed n < ω and let c be a good coloring of the graph (B, Ecard ∪ Epos ). In addition we assume that c takes values on a set M ⊆ ω \ n with the lacunary condition 



m∈M l∈M \{m}



min

l , m



m l



 1,

and such that c(∅)  n. We say that a finite block sequence2 (si )i
I.e. max si < min sj for i < j < d.

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 (a) j
Since the family B is very-large, it follows that for every infinite set A of κ and every d there is a B-special sequence of length d consisting on subsets of A. Now let

K :=

 i
 1  1si : (si )i
On c00 (κ) define the norm  · K for x ∈ c00 (κ) by    xK := max x∞ , sup x, f : f ∈ K .

Let X be the completion of (c00 (κ),  · K ). Then the sequence (uγ )γ<κ is a Schauder basis of X. Moreover (xγ )γ<κ is weakly-null: This readily follows from the fact that if s ⊆ κ is a finite set such that |s| ∈ M , then    1   1 1s   2.  |s| 2 

(5)

We see now that (uγ )γ<κ does not have infinite unconditional basic subsequences. Fix a subset A ⊆ κ of order type ω, and fix L  1. We see that (uγ )γ∈A is not L-unconditional. Let k < ω be such that k > 8L. Let x :=



(−1)i

i
y :=

 i
where (si )i
1 1s , |si |1/2 i

1 1s , |s2i |1/2 2i  i
1si /|si |1/2 ∈ K, it follows that

yK  f, y  k/2. On the other hand, we are going to see that xK  4. To this end, fix g =

 i
1ti /|ti |1/2 ∈ K. Let

  m0 := max i < min{d, k}: |si | = |ti | .

It follows then that c(s0 , . . . , sm0 −1 ) = c(t0 , . . . , tm0 −1 ), and hence |si | = |ti | for all i < m0 . Then, c(s) = c(t),   where s = i
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Since |si | = c(s0 , . . . , si−1 ), i < k and |tj | = c(t0 , . . . , tj−1 ), j < d, it follows that |si | = |tj | for i = j. Hence,      g, x    

    j     1 (−1)j   1 1ti , (−1)  1ti ,   1sj  + 1 sj   |sj | |sj | |ti | |ti | i


j

1+

|(

m0 −1

∩( |si0 |

i=i0 +1 si )

i

m0 −1

i=i0 +1 ti )|

+ 2  4.

2

Corollary 3.36. nc0  ωω . Remark 3.37. With considerably more effort it is possible to use the fact that χ(G0 (ω1 )) = ω to build a Banach space of density ω1 without unconditional basic sequences (this is essentially the construction in [3]). Unfortunately, as we have already seen, the same ideas cannot be used for ωn with n  2. This leads to the following question. Problem 2. Is there a similar combinatorial condition on uncountable κ that ensures the existence of a reflexive Banach space of density κ with no infinite unconditional basic sequences? We pass now to give a proof of Theorem 3.34. Definition 3.38. Let A, X be two totally ordered sets. We say that f : [A]n → X is min-dependant when f (s) = f (t) implies that

min s = min t

for every s, t ∈ [A]n .

(6)

Proposition 3.39. Suppose that B is a hereditary family on κ such that there is a mapping f : [κ]n → ω with the following two properties: (a) For every s ∈ B the restriction f  [s]n is min-dependant. α} ∪ t ∈ B ∩ [κ]n , then f ({α} ∪ t) = f ({¯ α} ∪ t). (b) If α < α ¯ < t are such that {α} ∪ t, {¯ Then G2n−2 (B) has countable chromatic number. Proof. Given two finite sets s, t ⊆ κ of the same cardinality, let ϑs,t denote the unique order-preserving  n+1 mapping between s and t. Now define c : B → k<ω k ω be the coloring c(s) := f ◦ (ϑ|s|,s , . . . , ϑ|s|,s ). We claim that c is a good coloring for G2n−2 (B). Suppose that s, t ∈ B are such that c(s) = c(t). Without loss of generality we assume that |s ∩ t| > 2(n − 1). Let γ0 < · · · < γ2n−3 be the last 2(n − 1) elements of s ∩ t. We prove first that ϑs,t (γi ) = γi for every i < n − 1. Fix such i < n − 1. Then  f (γi , γn−1 , . . . , γ2n−3 ) = f ϑs,t (γi ), ϑs,t (γn−1 ), . . . , ϑs,t (γ2n−3 ) ,

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and since all the ordinal involved above are in t, it follows that γi = ϑs,t (γi ). Now we see that s ∩ γ0 = t ∩ γ0 = s ∩ t ∩ γ0 , which implies that s and t are in 2(n − 1) − Δ-position. By symmetry it suffices to see that s ∩ γ0 ⊆ t ∩ γ0 . So, let γ ∈ s ∩ γ0 . Then   f (γ, γ0 , . . . , γn−2 ) = f ϑs,t (γ), ϑs,t (γ0 ), . . . , ϑs,t (γn−2 ) = f ϑs,t (γ), γ0 , . . . , γn−2 .

Since {γ, γ0 , . . . , γn−2 } ⊆ s, {ϑs,t (γ), γ0 , . . . , γn−2 } ⊆ t and B is hereditary, it follows from the property (b) of B that γ = ϑs,t (γ) ∈ t. 2 Now the following readily gives that nc0  ωω . Proposition 3.40. (See [24].) For every n < ω there is a very-large family B on ωn and f : [ωn ]n+1 → ω such that (a) and (b) in the hypothesis in Proposition 3.39 hold for f . The construction of the function f depends heavily on the -functions invented by Todorcevic (see [37] for full details). We recall the concept of -function. Definition 3.41. A function  : [κ+ ]2 → κ is called an (injective version of) -function if (a)  is subadditive, i.e. for every α < β < γ < κ+ (a.1) (α, β)  max{(α, γ), (β, γ)}, (a.2) (α, γ)  max{(α, β), (α, γ)}. α, β) for every α = α ¯ < β. (b) (α, β) = (¯ (c) (α, β) = (β, γ) for every α < β < γ. It is proved in [37] (see Definition 3.2.1, Lemma 3.2.2 dealing with the case κ = ω and Chapter 9 for the general version) that such a function  : [κ+ ]2 → κ exists for every regular cardinal κ. We pass now to define the function f : [ωn ]n+1 → ω. This is done by defining first auxiliary functions. For each integer n we fix an injective  function (n) on ωn . Let n ∈ ω. For each i  n we define recursively (n) fi : [ωn ]i+1 → ωn−i as follows: (n)

(1) f0 := Idωn ; (2) fi (α0 , α1 , . . . , αi ) := (n−(i−1)) (fi−1 (α0 , . . . , αi−1 ), fi−1 (α1 , . . . , αi )) for each α0 < · · · < αi in ωn and each 0 < i  n. (n)

Let fn := fn

: [ωn ]n+1 → ω.

Theorem 3.42. There is a very-large family B on ωn such that (a) and (b) in the hypothesis in Proposition 3.39 hold for fn . The proof of this result uses the Classical Ramsey Theorem and the Erdős–Rado Canonization Theorem (see [24] for full details). 3.4. Subsymmetric sequences Recall that a basic sequence is called subsymmetric when there is a constant C  1 such that every two subsequences of it of the same length are C-equivalent. Question 1. What is the minimal cardinal κ, denoted by ns, such that every space of density κ has a subsymmetric basic sequence?

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We know that 2ℵ0 < ns  first ω-Erdős cardinal, by Odell [28] and Ketonen [20] results, respectively. There is also the following relationship between nc and ns. Proposition 3.43. nc  ns. Proof. Now suppose that X has density ns. Then X has a subsymmetric basic sequence. Therefore it has a subsequence equivalent to the unit basis of 1 , hence unconditional, or else a subsequence (xn )n which is weakly-Cauchy. Now the normalized difference sequence (yn )n , yn := (x2n+1 − y2n )/y2n+1 − y2n  for each n < ω, is a normalized weakly-null subsymmetric basic sequence, which is in addition subsymmetric. Now it readily follows from Schreier unconditionality Theorem 3.4 that (yn )n is unconditional. 2 Very little is known on how distinguish these cardinal numbers are. In fact in our understanding it is not known if there is a non-separable space without subsymmetric sequences but which is saturated by unconditional basic sequences (i.e. such that every subspace has an unconditional basic sequence). The first known separable example of this kind is the Tsirelson space T . Its most common presentation is done by using the Schreier family of finite sets S := {s ⊆ ω: |s|  max s + 1}. Given x ∈ c00 (N), one defines 



1 xT := max x∞ , sup Ei xT : E1 < · · · < En and {min Ei }ni=1 ∈ S 2 i=1 n



.

(7)

It can be easily seen, for example by induction on the cardinality of the support supp x := {n ∈ N: (x)n = 0} of x, that the previous quantity is well defined. In fact, it defines a norm on c00 (N). The Tsirelson space T is the completion of (c00 (N),  · T ). Since the unit basis is a 1-unconditional Schauder basis of T and since T does not contain isomorphic copies of c0 or 1 (because their corresponding unit bases are symmetric) it follows that T is reflexive. Trying to extend that construction to the non-separable context, it is natural to extract the key properties of S and define the following notions. Definition 3.44. Let κ be a cardinal number. A family B of finite subsets of κ is called (a) hereditary if s ⊆ t ∈ B implies that s ∈ B; (b) compact if it is compact when B is identified with the subset of 2κ consisting on characteristic functions 1s of sets s ∈ B; (c) large if B ∩ [A]n = ∅ for every infinite subset A of κ and every integer n. So compact and large families can be seen as families having arbitrary large finite sets of every infinite set yet no infinite set is in their closure. Keeping this simple idea in mind, one might imagine that such families can be used to define for example spaces not having copies of 1 , yet 1 will appear asymptotically. This fact is the key to build a space not having subsymmetric basic sequences. Although it is a priori quite difficult to define explicitly uncountable compact hereditary and large families, the following proves that in fact there are many of them. Proposition 3.45. (See [24].) Let κ be an infinite cardinal. The following are equivalent: (1) κ is ω-Erdős. (2) ns0  κ, i.e., every non-trivial weakly-null sequence (xα )α<κ has a subsymmetric basic subsequence. (3) There are no large compact and hereditary families on κ.

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Proof. (1) implies (2) was proved first by Ketonen: Let C be a set of finite basic sequences in some Banach space (for example C[0, 1]) such that every finite linearly independent sequence in a Banach space is 1-equivalent to some sequence in C, and of cardinality |C| = c. Let c : [κ]<ω → C be defined for s ∈ [κ]<ω by c(s) ∈ C such that (xα )α∈s is 1-equivalent to c(s). So if A ⊆ κ is c-homogeneous, then (xα )α∈A is subsymmetric. Now, by the Mazur’s Lemma, there is some B ⊆ A such that (xα )α∈B is in addition basic. (2) implies (3): Let B be a large compact and hereditary family on κ. Define on c00 (κ) the following norm: xB := sup x, 1s s∈B

where ·,· denotes the scalar product in c00 (κ). Then the unit Hamel basis (uγ )γ<κ is a 1-unconditional Schauder basis of the completion of (c00 (κ), ·B ). Recall that the Ptak’s Lemma states that if F is a compact family on ω then for every ε > 0 there is a finite sequence (an )n∈s of positive scalars summing 1 such that  n∈t an  ε for every t ∈ F. Now, using Ptak’s Lemma it follows easily that (uγ )γ<κ is weakly-null; similarly and also using that B is large it follows that (uγ )γ<κ does not have subsymmetric basic sequences. Fix a strictly increasing sequence (αn )n<ω , C  1, in κ. Then F := {s ⊆ ω: {αn }n∈s ∈ B} is a compact, hereditary and large family on ω. Fixing now C  1, we try to see that (uαn )n is not C-subsymmetric. To this end, we apply Ptak’s Lemma to F and ε = (2C)−1 to find the corresponding convex combination (an )n∈u . Let s := {αn }n∈u , and let t ∈ B, t ⊆ {αn }n∈ω be such that |t| = |s|. Then (uα )α∈s and (uα )α∈t  are subsequences of (uαn )n∈ω which are not C-equivalent: In one hand,  n∈u an uαn B  1/(2C), while    on the other  α∈t aϑt,u (α) uα   1t , α∈t aϑt,u (α) xα = n∈u an = 1. (3) implies (1): Suppose that κ is not ω-Erdős, and let c : [κ]<ω → 2 witnessing that. Let B be the family of c-homogeneous subsets of κ. Then this is a compact and hereditary family, since c does not have infinite homogeneous sets. In addition, it follows from the finite Ramsey Theorem that B is large. 2 Consequently we have the following. Corollary 3.46. nc0 := first ω-Erdős cardinal. Problem 3. Does there exist a reflexive non-separable Banach space which is unconditionally saturated and without subsymmetric basic sequences? We explicitly ask for a space saturated by subspaces with unconditional bases because the non-separable example in [3] is a reflexive space without unconditional basic sequences, and consequently, by Corollary 3.5, without subsymmetric sequences because, since every normalized sequence has a weakly-Cauchy subsequence by the Rosenthal’s 1 -dichotomy. A natural candidate could be a Tsirelson-like space defined on c00 (κ) from a large, compact and hereditary family B in κ as we did in (7) changing S by B. However it is proved in [24] that such spaces contain 1 , and therefore they are not reflexive and they contain subsymmetric sequences. 4. Uncountable sequences Recall the following classical problem. Problem 4. Let X be a Banach space. Does there exist an infinite dimensional subspace Y of X such that the quotient space X/Y is separable infinite dimensional? For separable spaces the answer is affirmative by a result of Johnson and Rosenthal [19]. In fact, the authors prove that there is a quotient with a basis. This result was later extended by Todorcevic as follows.

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Theorem 4.1. (See Todorcevic [36].) It follows from Martin’s Maximum Axiom (MM ) that every Banach space of density ω1 has a quotient with a Schauder basis of length ω1 , and therefore a quotient with a Schauder basis. Its proof is a very nice use of the forcing axioms: Given a space X of density ω1 , one finds a sequence (fγ )α<ω1 in X ∗ such that (fγ (x))γ ∈ c0 (ω1 ) for every x ∈ X. This is a consequence of Martin’s Axiom (MA) when the dual ball BX ∗ with its weak∗ topology is countably determined, and of Martin’s Maximum when it is not. Then using (MA) one extracts an uncountable 1-basic subsequence (gα )α<ω1 from (fα )α<ω1 in a  way that the natural bounded operator π : X → gα∗ α<ω1 , π(x) := α<ω1 gα (x)gα∗ , is a quotient map, i.e. onto. Notice that in particular for each α < ω1 there is xα ∈ X such that gα∗ = δxα , i.e., gα (f ) = f (xα ) for every f ∈ fβ β<ω1 . It readily follows that (xα + Ker π)α<ω1 is a 1-Schauder basis of X/ Ker π. So, if X has density ω1 there is a quotient X/Y with an uncountable biorthogonal sequence, and so it does X. On the other hand several non-separable Banach spaces without uncountable biorthogonal systems are known, of course assuming additional set-theoretical axioms: There is the Todorcevic non-metrizable compacta (see [35]) assuming b = ω1 with the property that its corresponding space of continuous functions does not have uncountable biorthogonal systems, or with similar properties the Kunen compacta (see [27]). Both are scattered compacta, so their spaces of continuous functions on them are c0 saturated. A space richer in subspaces is the Shelah space [33], which is a Gurarij space of density ω1 , and therefore universal for separable Banach spaces. This space is constructed by using the combinatorial principle ♦. Finally, Brech and Koszmider [5] provide a space of density ω2 without uncountable biorthogonal systems. Still quite little is known concerning these sorts of spaces. Problem 5. Does there exist a non-separable Banach space without uncountable biorthogonal sequences not having isomorphic copies of c0 ? Per se, this example will not solve a major open problem, but it will perhaps lead to other constructions solving main problems, as for example the existence of an Asplund space without smooth bump functions. We pass now to explain a general methodology to build examples of spaces of density ω1 by the method of forcing in a relatively simple way. These spaces are direct limits of polyhedral finite dimensional spaces. Recall that a Banach space X is called polyhedral when the unit balls BF of finite dimensional subspaces F of X only have finitely many extremal points, i.e. points in BF which are not middle points of segments in BF . Notice that polyhedral spaces are c0 -saturated. Notice also that direct limits of polyhedral spaces may not be polyhedral. Coming back to the forcing construction, the conditions will be of the form p = (c00 (Dp ),  · p ), where Dp ⊆ ω1 is finite. The extension p  q will simply ensure that Dq ⊆ Dp and that xp = xq for every x ∈ c00 (Dq ). In order to preserve cardinalities between the ground and the generic universes it seems reasonable to demand that these forcing notions are c.c.c. This gives a constraint on the norms we allow, because, for example, (R2 ,  · p ), 1  p  ∞, are continuum many incompatible conditions. A way to overpass this is to work only with finite dimensional spaces (c00 (s),  · ) where the norm  ·  is defined by finitely many vectors {hi }i
There are several interesting families of conditions p providing interesting generic spaces (see [23]). We present only two illustrative examples and some properties and their proofs of the corresponding generic spaces. Example 1. Let P0 be the poset consisting of pairs p = (Dp , Hp ) where (1) Dp ⊆ ω1 is finite.

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(p)

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(p)

(2) Hp = {hγ }γ∈Dp is such that hγ ∈ Dp 2 

h(p) γ

γ

= 1 and



h(p)  γ = 0. γ

(8)

p  q when Dq ⊆ Dp and when (q) h(p) γ  Dq = hγ

h(p) γ

for every γ ∈ Dq and  (q)   Dq ∈ convQ ±hη η∈Dq for every γ ∈ Dp \ Dq .

(9) (10)

It is easy to see that  · p defined on Xp := c00 (Dp ) for x ∈ c00 (Dp ) by    xp := max  h(p) γ ,x γ∈Dp

(p)

is a norm on Xp . A useful fact is that the extremal points of BXp∗ are {±hγ }γ∈Dp , and that, consequently, every f ∈ BXp∗ has a unique representation f=



aγ h(p) γ

γ∈Dp

 with γ∈Dp |aγ |  1. It is also easy to see that xp = xq for every p  q and x ∈ Xq . In fact, this is equivalent to conditions (9) and (10) above.

Definition 4.2. We say that an uncountable sequence (pα )α<ω1 is well-placed Δ-system when (DS1) (Dpα )α<ω1 forms a Δ-system with root R such that Dpα \ R < Dpβ \ R for all α < β < ω1 . (DS2) |Dpα | = |Dpβ | for all α, β < ω1 . p (DS3) hϑβα,β (γ) = ϑα,β (hpγα ) for every α, β < ω1 and γ ∈ Dpα , where ϑα,β is the unique order-preserving bijection between Dpα and Dpβ . We define the type of (pα )α as (t) (p0 ) ) for every k ∈ D and where ϑ : D → Dp0 the condition t = (D, H) where D = |D0 |, and hk := ϑ−1 (hϑ(k) is the unique order-preserving bijection. It is clear that every uncountable sequence of conditions (pα )α<ω1 has a further uncountable subsequence which is a well-placed Δ-system. Proposition 4.3. P0 has the countable chain condition. Proof. Suppose that (pα )α<ω1 is an uncountable sequence in P0 . To simplify the notation, let us write pα = (Dα , Hα ). By going to an uncountable subsequence if needed, we may assume that (pα )α<ω1 is a well-placed Δ-system. Now given α0 < · · · < αn we define the basic amalgamation p = (Dp , Hp ) of pα0 , . . . , pαn as follows: (d) Dp =

 in

Dαi . (p)

(p)

(pα )

(e) Let γ ∈ Dp . If γ ∈ Dα0 , then we define hγ piecewise by hγ  Dαi = hϑα i,α (γ) for every i  n. This 0 i is well defined because of property (DS3) above. Suppose now that γ ∈ Dαi \ Dα0 for some 1  i  n. (pα ) (p) Note that then i is unique. Then we define hγ := hγ i . It readily follows that p ∈ P0 and that p extends pα0 , . . . , pαn . 2

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 Let G be a generic filter for P0 . A simple density argument proves that p∈G Dp = ω1 , so now we can define on the linear space c00 (ω1 ) the following norm  · G : Given x ∈ c00 (ω1 ), let p ∈ G be such that x ∈ Xp , and declare xG = xp . This is well defined, independently of the choice of p, since G is a filter. Let now XG be the completion of (c00 (ω1 ),  · G ).

Proposition 4.4. XG has density ω1 . Proof. First, notice that ω1 is the same in the ground model and in the generic extension, by the countable chain condition of P0 . It is clear by construction that the density is at most ω1 . Now given α < β, let us see that uβ − uγ G  1. Let p ∈ G be such that α, β ∈ Dp . Then  (p)  uβ − uα G = uβ − uα p  hβ , uβ − uα = 1.

So, (uα )α<ω1 is a 1-separated sequence in XG , and consequently the density of XG is at least ω1 .

(11) 2

Proposition 4.5. XG does not have uncountable biorthogonal systems. Proof. Suppose otherwise that (xα , fα )α<ω1 is a biorthogonal system. By going to a subsequence and normalizing the xα ’s we may assume that (a) there is some ε > 0 such that   n   1   x αi   ε xα0 −   n i=1

(12)

for every α0 < · · · < αn . (b) xα ∈ c00 (ω1 ) and xα G = 1 for every α < ω1 . Choose for each α < ω1 a condition pα ∈ G such that xα ∈ Xpα for every α < ω1 . Again, by going to an uncountable subsequence if needed, we may assume that (pα )α is a well-placed Δ-system such that in addition the unique order-preserving bijections ϑα,β : Dpα → Dpβ satisfy that ϑα,β (xα ) = xβ for every α < β < ω1 . Fix now α0 < · · · < αn , and let p be the basic amalgamation of pα0 , . . . , pαn . We claim that   n   1 1   x αi    x α0 −   n i=1 n

(13)

p

which, together with a simple density argument, contradicts (12). So, fix γ ∈ Dp . Suppose first that γ ∈ Dα0 . Then 1 − xα n i=1 i n

hpγ , xα0

!

n  (p )  1  (pαi )  = hγ α0 , xα0 − hϑα ,α (γ) , xαi 0 i n i=1 n  (p )  1  (pα0 )  = hγ α0 , xα0 − hγ ◦ ϑαi ,α0 , xαi n i=1 n  (p )  1  (pα0 )  = hγ α0 , xα0 − hγ , ϑαi ,α0 (xαi ) = 0. n i=1

J. Lopez-Abad / Annals of Pure and Applied Logic 164 (2013) 1390–1417

1411

If γ ∈ Dαi \ Dα0 , then  !   n     1 1   pαi 1  p  x . h , x − x = h ,  γ α0 γ αi  αi      n i=1 n n

2

(p)

Given γ < ω1 , let hγ ∈ 2ω1 be defined piecewise by hγ  Dp := hγ for every p ∈ G. Given a limit ordinal (λ) number λ < ω1 , let XG be closed linear span of (uα )α<λ in XG . (λ)

Proposition 4.6. XG is c0 -saturated. In fact, for every limit ordinal α < ω1 the sequence (hα  XG )α<λ (λ) (λ) is a Schauder basis of the dual space (XG )∗ of XG which is 1-equivalent to the unit basis of 1 (λ). Consequently, the dual of XG is isometric to 1 (ω1 ). (λ)

Proof. Given a limit ordinal number λ < ω1 we set Xλ := XG proof has two parts, which we present as two claims.

(λ)

and gα := hα  Xλ for every α < λ. The

(λ)

Claim 4.6.1. (gα )α<λ is 1-equivalent to the unit basis of 1 (λ). That is, for every finite sequence of scalars   (λ) (aγ )γ∈s one has that  α∈s aα gα Xλ∗ = α∈s |aα |. Proof. This is a typical density argument: Given a finite subset s ⊆ ω1 , and given a condition p with s ⊆ Dp (q) there exists q  p such that (hγ  c00 (Dq ∩ λ))γ∈s is 1-equivalent to 1 (s). To see this, let t ⊆ λ be such (q) (q) (p) that Dp < t and with |t| = 2|s| . Define Dq = Dp ∪ t, hγ := uγ for γ ∈ t, hγ := hγ for γ ∈ Dp \ s. Let  : t → {−1, 1}s be a bijection. Then for every γ ∈ s let (p) h(q) γ := hγ +



(η)(γ)uη .

η∈t (q)

It is easy to see that (hγ  c00 (Dq ∩ λ))γ∈s is 1-equivalent to the unit basis of 1 (s). 2 (λ)

In particular, the sequence (gα )α<λ is a 1-unconditional basic sequence. Claim 4.6.2. gα α<λ is norm-dense in Xλ∗ . (λ)

Proof. This is done by induction on λ. The first case to consider is λ = ω. Fix f ∈ BXλ∗ . Choose a sequence (pn )n∈N in G such that pn+1  pn and pn ∩ ω = kn ∈ ω for every n. Then for each n one has the unique representation f  Xpn =



(pn ) a(n) γ hγ

(14)

γ∈Dpn

with

 γ∈Dpn

(n)

(p)

|aγ |  1. Since (hγ )η = 0 for every η < γ, it follows that f  c00 (kn ) =



(pn ) a(n) γ hγ

(15)

γ∈kn (n)

(m)

for every n. Hence, by the uniqueness of the coefficients, it follows that aγ = aγ for every m  n and (n) γ < km . So we can define the infinite sequence (ak )k<ω by declaring ak := ak for n such that k < kn .   Since k<ω |ak |  1, it follows that g := k<ω ak hk ∈ BXG∗ , and clearly f = g  Xω . So, f is in the closed linear span of {hk  Xω }k<ω .

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If λ is limit of limits, then using the inductive hypothesis and the uniqueness of the coefficients we obtain the desired result. Finally suppose that λ = γ + ω with γ limit. Fix f ∈ BXλ∗ . Then, f  Xγ ∈ BXγ∗ , so,   by inductive hypothesis, we get that f  Xγ = η<γ aη hη  Xγ with η<γ |aη | = f  Xγ . Now let  g := f − η<γ aη hη  Xλ , and let h := g/g. A very similar argument that the one used for the case λ = ω gives the desired result, first for h, and consequently for f . 2 Theorem 4.7. Let X ⊆ XG be a subspace. Then every operator T : X → XG is the sum of a multiple of the inclusion and a separable range operator. We need the following lemmas. Lemma 4.8. Let (yα )α<ω1 and (zα )α<ω1 be two sequences in c00 (ω1 , Q) such that (yα )α is  · G -normalized and such that  for every α < β < ω1 one has that d zβ − zα , yβ − yα :=

inf

y∈ yβ −yα

  y − (zβ − zα ) > ε. G

Then for every m ∈ N there are α0 < · · · < α2m−1 such that       (y − y ) α2i+1 α2i   1  i
G

i
G

and

     (zα2i+1 − zα2i )    mε.

Proof. Let (pα )α<ω1 be an uncountable sequence in G such that yα , zα ∈ Xpα for every α < ω1 . Without loss of generality, we assume that (pα )α is a well-placed Δ-system of type t = (D, H) and root R (see Definition 4.2). We write pα = (Dα , Hα ) for every α. Let ϑα,β : Dα → Dβ and ϑα : D → Dα be the  , and z = ϑ−1 corresponding order-preserving bijections, and let y := ϑ−1 α (yα ) = α (zα ) for γ∈Dα (y)γ uϑ−1 α (γ) any α. Claim 4.8.1. dXt (z, {y} ∪ {uk }k∈|R| ) > ε. Proof. Otherwise, we would have that dXG (y1 − y0 , z1 − z0 ) > ε, a contradiction. The idea behind this is the following: Suppose that the statement in the claim is false. Let p be the basic amalgamation of p0 and p1 (see the proof of Proposition 4.3). Then  dXp z1 − z0 , y1 − y0  ε.

(16)

Let λ ∈ R and v ∈ uk k<|R| be such that z − λy + vt  ε. Then we claim that z1 − z0 − λ(v1 − v0 )  ε. (p) (p ) (p) (p1 ) Let γ ∈ Dp = D0 ∪ D1 . Suppose first that γ ∈ D0 . Then hγ  D0 = hγ 0 and hγ  D1 = hϑ0,1 γ . Hence hγ , z1 − z0 − λ(v1 − v0 ) = 0. If γ ∈ D1 \ D0 , then hγ := ϑ1 (hk ) for k = ϑ−1 1 (γ) ∈ D \ |R|. Hence, (p)

(p)

(t)

 (p)     (t)  hγ , z1 − z0 − λ(y1 − y0 ) = h(p) γ , z1 − λy1 = hk , z − λy + v  z − λy + vt  ε.

2

So, we may assume that vt = 1, and that     dt,H w, c00 |R| ∪ {v} > ε.

(17)

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1413

Let h ∈ convQ (±H) be such that    h  c00 |R| ∪ {v} = 0 and h(w) = ε.

Note that h exists because of the assumption (17). Without loss of generality, we assume that there is  (p) R < η < D0 \ R. Let p = (Dp , Hp ) be the following condition: Dp = {η} ∪ i<2m Di and hγ is defined as follows: (p)

(p )

(a) hγ  Di = hγ i for every γ ∈ R and i < 2m.  (p) (b) hη := uη + i
i
On the other hand,  

        (w2i+1 − w2i )  ϑ2i+1 (h) w2i+1 − w2i = m · h(w) = mε.  i
p

i
2

i
Lemma 4.9. Let (xα )α<ω1 be a normalized separated sequence in XG , and let X be its closed linear span. Then for every bounded operator T : X → XG there are α ¯ < ω1 and λ ∈ R such that for every α  α ¯ one has that (α) ¯

T (xα ) − λxα ∈ XG .

(18)

Proof. Fix all given data. Claim 4.9.1. There is α0 < ω1 such that for every α  α0 one has that  (α )  T (xα ) ∈ XG 0 ∪ {xα } .

(19)

Proof. We give the proof for the case that xα , T xα ∈ c00 (ω1 , Q) for every α. The general case needs a non-difficult approximation argument. Working towards a contradiction, suppose that such α0 < ω1 does not exist. Using that (xα )α<ω1 is a separated sequence, it is possible to find an uncountable subsequence (yα )α<ω1 of (xα )α and 0 < ε < 1 such that for every α < ω1 one has that    d yα , yβ β<α + T (yβ ) β<α > ε,    d T (yα ), T (yβ ) β<α + yβ βα > ε.

and

We can use now Lemma 4.8 for m such that mε > T , to find {αi }i<2m such that      yα2i+1 − yα2i     1,

(20)

i
     T yα2i+1 − T yα2i     mε. i
(21)

J. Lopez-Abad / Annals of Pure and Applied Logic 164 (2013) 1390–1417

1414

But this is impossible, since            T yα2i+1 − T yα2i   T  yα2i+1 − yα2i     T . i
2

i
We fix some α0 < ω1 and for each α  α0 a scalar λα ∈ R such that T (xα ) − λα xα ∈ Xα0 .

(22)

The proof of the lemma will be finished once we establish the following. Claim 4.9.2. There is some α0  α ¯ < ω1 such that λα = λβ for all α ¯  α, β < ω1 . Proof. Otherwise, we can find βξ < δξ < βξ+1 , ξ < ω1 such that: (a) λβξ = λδξ for every ξ < ω1 . (ξ)

(ξ)

(b) xβξ ∈ / XG and xδξ ∈ / XG ∪ {xβξ } for every ξ < ω1 . Let Y ⊆ X be the closed linear span of {xδξ − xβξ }ξ<ω1 , and T0 = T  Y . By applying Claim 4.9.1 to the sequence ((xδξ − xβξ )/xδξ − xβξ )ξ<ω1 , we obtain that there is some α0  α1 < ω1 , and for each ξ  α1 a scalar ηξ such that (α1 )

T0 (zξ ) − ηξ zξ ∈ XG

.

(23)

In particular, if we take ξ = α1 , it follows from (22) and (23) that (α1 )

ηα1 (xδα1 − xβα1 ) − (λδα1 xδα1 − λβα1 xβα1 ) ∈ XG

.

(24)

Hence  (α )  (ηα1 − λδα1 )xδα1 ∈ XG 1 ∪ {xβα1 } .

(25)

It follows from (b) that ηα1 = λδα1 . Hence, from this and (24), we obtain that (α1 )

(ηα1 − λβα1 )xβα1 ∈ XG

.

(26)

Again using (b), it follows that λβα1 = ηα1 = λδα1 , which is contradictory with (a). 2 Proof of Theorem 4.7. Without loss of generality, we assume that X is non-separable. Let (xα )α<ω1 be a separated normalized sequence of X such that xα α is dense in X. We use Lemma 4.9 to find β < ω1 and

J. Lopez-Abad / Annals of Pure and Applied Logic 164 (2013) 1390–1417

(β)

1415

(γ)

λ ∈ R such that T xα − λxα ∈ XG for every α  β. Let β  γ be such that T ( xα α<β ) ⊆ XG , and set U := T − λi, where i : X → XG is the inclusion map. We claim that (γ)

Im(U ) ⊆ XG . So, fix x ∈ X. Let ε > 0, and let y ∈ xα α<ω1 be such that y − x  ε, and let v ∈ xα α<β and w ∈ xα αβ be such that y = v + w. Hence, (γ)

(β)

(γ)

U (y) = U (v) + U (w) ∈ XG + XG = XG . (γ)

Since U is bounded, and ε > 0 is arbitrary, we get that in fact U (x) ∈ XG .

2

Example 2. Let P1 be the set of triples p = (Dp , Fp , Hp ) with the following properties: (C.1) (Dp , Hp ) ∈ P0 . (C.2) Fp ⊆ c00 (Dp , Q ∩ [−1, 1]) is finite and symmetric. The ordering p b q is defined by: (O.1) (Dp , Hp )  (Dq , Hq ) in P0 . (O.2) Fq ⊆ Fp  Dq = {f  Dq : f ∈ Fp } ⊆ convQ (Fq ). Similarly as for P0 , the poset P1 also has a basic amalgamation for conditions satisfying (DS1), (DS2) and (DS3). This readily gives that P1 has the countable chain condition. Given a generic filter G, one can define the norm  · G on c00 (ω1 ) as the G-limit of the norms  · p , xp := maxf ∈Fp | f, x |, and then XG as the completion of (c00 (ω1 ),  · G ). In a similar way that for the first generic example, one can see this space has density ω1 , it does not have uncountable biorthogonal systems and it has few operators. But differently now, the generic space XG is far from c0 -saturated: It is universal for separable Banach space, indeed it is a Gurarij space; that is, a space such that given two finite dimensional spaces F ⊆ G, an isometry T : F → XG , and ε > 0, there is an extension U : G → XG of T such that (1 − ε)T x  x  (1 + ε)T x for every x ∈ G. The proof of this fact is done by a density argument. It is a remarkable fact that the separable Gurarij space G is unique up to isometry. Of course this is not the case (not even up to isomorphism) for non-separable Gurarij spaces. We give a list of more involved examples, built by the general method. Theorem 4.10. (See [23].) 1. There are X ⊆ Y non-separable L∞,1+ such that: (a) X is Asplund and c0 -saturated, Y is Gurarij and Y /G ≡ X (we say that (X, Y ) is a pair of spaces). (b) Both (X, w)n and (Y, w)n are HL for every integer n. (c) Both X and Y have no supported sets. 2. There are spaces X ⊆ Y satisfying 1.(a) above and such that X has uncountable fundamental ε-biorthogonal sequences for every ε > 0 but it does not have uncountable biorthogonal sequences. 3. There is a non-separable space X that has uncountable (1 + ε)-Schauder basic sequences for every ε > 0 but it does not have uncountable monotone basic sequences. 4. All of the spaces above have few operators. 5. There exist a non-metrizable Poulsen simplex and a non-metrizable Bauer simplex such that the corresponding space of probability measures is hereditarily separable in all finite powers.

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We finish by mentioning two more problems. Recall that a biorthogonal system (xi , fi )i∈I in X is called a Markushevich system when xi i∈I is dense in X, and when (fi )i∈I separates points of X, i.e. if supi∈I |fi (x)| = 0 then x = 0. We refer the reader to [16] for more complete information about these and other similar problems. Problem 6. Find a space with an uncountable Markushevich system but without uncountable basic sequences. Recall that given two isomorphic spaces X and Y , the Banach–Mazur distance dBM (X, Y ) := inf{T  · T −1 : T : X → Y is an isomorphism}. Problem 7. Suppose that X is a Banach space such that sup

dBM (X, Y ) < ∞.

(27)

Y isomorphic to X

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