Add Journal to My Library
The Quarterly Journal of Mathematics
, Volume 69 (1) – Mar 1, 2018

12 pages

/lp/ou_press/on-coarse-embeddings-into-c0-EPDO2wGwjT

- Publisher
- Oxford University Press
- Copyright
- © 2017. Published by Oxford University Press. All rights reserved. For permissions, please email: journals.permissions@oup.com
- ISSN
- 0033-5606
- eISSN
- 1464-3847
- D.O.I.
- 10.1093/qmath/hax035
- Publisher site
- See Article on Publisher Site

Abstract Let λ be a large enough cardinal number (assuming the Generalized Continuum Hypothesis it suffices to let λ=ℵω). If X is a Banach space with dens(X)≥λ, which admits a coarse (or uniform) embedding into any c0(Γ), then X fails to have non-trivial cotype, i.e. X contains ℓ∞nC-uniformly for every C>1. In the special case when X has a symmetric basis, we may even conclude that it is linearly isomorphic with c0(densX). 1. Introduction The classical result of Aharoni states that every separable metric space (in particular every separable Banach space) can be bi-Lipschitz embedded (the definition is given below) into c0. The natural problem of embeddings of metric spaces into c0(Γ), for an arbitrary set Γ, has been treated by several authors, in particular Pelant and Swift. The characterizations that they obtained, and which play a crucial role in our argument, are described below. Our main interest, motivated by some problems posed in [1], lies in the case of embeddings of Banach spaces into c0(Γ). We now state the main results of this paper. We first define the following cardinal numbers inductively. We put λ0=ω0, and, assuming that n∈N0, λn has been defined, we put λn+1=2λn. Then we let λ=limn→∞λn. (1.1) It is clear that assuming the generalized continuum hypothesis (GCH) λ=ℵω. Theorem 1.1 If X is a Banach space with density dens(X)≥λ, which admits a coarse (or uniform) embedding into any c0(Γ), then X fails to have non-trivial cotype, that is X contains ℓ∞nC-uniformly for some C>1 (equivalently, every C>1). Our method of proof gives a much stronger result for Banach spaces with a symmetric basis. Namely, under the assumptions of Theorem 1.1, such spaces are linearly isomorphic with c0(Γ) (Theorem 4.2). Theorem 1.1 will follow from the following combinatorial result which is of independent interest. For a set Λ and n∈N, we denote by [Λ]n the set of subsets of Λ whose cardinality is n. Theorem 1.2 Assume that Λ is a set whose cardinality is at least λ, n∈N, and σ:[Λ]n→Cis a map into an arbitrary set C. Then (at least) one of the following conditions holds: There is a sequence (Fj)j=1∞of pairwise disjoint elements of [Λ]n, so that σ(Fi)=σ(Fj), for all i,j∈N. There is an F∈[Λ]n−1so that σ({F∪{γ}:γ∈Λ⧹F})is infinite. The above Theorem 1.2 was previously deduced in [4, Lemma 4.3] from a combinatorial result of Baumgartner, provided Λ is a weakly compact cardinal number (whose existence is not provable in ZFC, as it is inaccessible [3, p. 325, 52]). The authors in [4, Question 3] pose a question whether assuming that Γ is uncountable is sufficient in Theorem 1.2. Theorem 1.2 is used in order to obtain a scattered compact set K of height ω0, such that C(K) does not uniformly embed into c0(Γ). It is easy to check that our version of Theorem 1.2 implies a ZFC example of such a C(K) space. It is further shown in [4] that the space C[0,ω1] does not uniformly embed into any c0(Γ). Let us point out that a special case of Theorem 1.1 was obtained by Pelant and Rödl [6, Theorem], namely it was shown there that ℓp(λ),1≤p<∞, spaces (which are well known to have non-trivial cotype) do not uniformly embed into any c0(Γ). The paper is organized as follows. In Section 2, we recall Pelant’s [4, 5] and Swift’s [8] conditions for Lipschitz, uniform and coarse embeddability into c0(Γ). In Section 3, we provide a proof for Theorem 1.2. Finally, in Section 4, we provide a proof of Theorem 1.1 as well as the symmetric version of the result. All set theoretic concepts and results used in our note can be found in [3], whereas for facts concerning non-separable Banach spaces, [2] can be consulted. We want to finish this introductory section by thanking the anonymous referee for his or her efforts which improved the paper considerably. 2. Pelant’s and Swift’s criteria for Lipschitz, uniform and coarse embeddability into c0(Γ) In this section, we recall some of the notions and results by Pelant [4, 5] and Swift [8] about embeddings into c0(Γ). For a metric space (M,d), a cover is a set U of subsets of M such that M=⋃U∈UU. A cover U of M is called uniform if there is an r>0 so that for all x∈M, there is a U∈U, so that Br(x)={x′∈M:d(x′,x)<r}⊂U. It is called uniformly bounded if the diameters of the U∈U are uniformly bounded, and it is called point finite if every x∈M lies in only finitely many U∈U. A cover V of M is a refinement of a cover U, if for every V∈V, there is a U∈U, for which V⊂U. Definition 2.1 [4] A metric space (M,d) is said to have the uniform Stone property if every uniform cover U of M has a point finite uniform refinement. Definition 2.2 [8] A metric space (M,d) is said to have the coarse Stone property if every uniformly bounded cover is the refinement of a point finite uniformly bounded cover. Definition 2.3 Let (M1,d1) and (M2,d2) be two metric spaces. For a map f:M1→M2, we define the modulus of uniform continuity wf:[0,∞)→[0,∞], and the modulus of expansion ρ:[0,∞)→[0,∞] as follows: wf(t)=sup{d2(f(x),f(y)):x,y∈M1,d1(x,y)≤t}andρf(t)=inf{d2(f(x),f(y)):x,y∈M1,d1(x,y)≥t}.The map f is called uniformly continuous if limt→0wf(t)=0, and it is called a uniform embedding if, moreover, ρf(t)>0 for every t>0. It is called coarse if wf(t)<∞, for all 0<t<∞ and it is called a coarse embedding, if, furthermore, limt→∞ρf(t)=∞. The map f is called Lischitz continuous if Lip(f)=supx≠yd2(f(x),f(y))d1(x,y)<∞,and a bi-Lipschitz embedding, if, furthermore, f is injective and Lip(f−1) (being defined on the range of f) is also finite. The following result recalls results from [4, Theorem 2.1] (for (i) ⟺(ii) ⟺(v)) and [8, Lemma 2.3, Corollary 3.11] (for (ii) ⟺(iii) ⟺(iv)). Theorem 2.4 For a Banach space X, the following properties are equivalent: X has the uniform Stone property. X is uniformly embeddable into c0(Γ), for some set Γ. X has the coarse Stone property. X is coarsely embeddable into c0(Γ), for some set Γ. X is bi-Lipschitzly embeddable into c0(Γ), for some set Γ. It is easy to see, and was noted in [4, 8], that the uniform Stone property and the coarse Stone property are inherited by subspaces. The equivalence (i) ⟺(ii) was used in [4] to show that C[0,ω1] does not uniformly embed in any c0(Γ). It was also used to prove that certain other C(K)-spaces do not uniformly embed into c0(Γ): let Λ be any set and denote for n∈N by [Λ]≤n and [Λ]n the subsets of Λ which have cardinality at most n and exactly n, respectively. Endow [Λ]≤n with the restriction of the product topology on {0,1}Λ (by identifying each set with its characteristic function). Then define KΛ to be the one-point Alexandroff compactification of the topological sum of the spaces [Λ]≤n, n∈N. It was shown in [4] that if Λ satisfies Theorem 1.2, then C(KΛ) is not uniformly Stone, and thus does not embed uniformly into any c0(Γ). 3. A combinatorial argument We start by introducing property P(α) for a cardinal α as follows. We say that a cardinal number satisfies P(α) if (P(α)) For every n∈N and any map σ:[α]n→C, C being an arbitrary set, (at least) one of the following two conditions hold: (a) There is a sequence (Fj) of pairwise disjoint elements of [α]n, with σ(Fi)=σ(Fj) for any i,j∈N. (b) There is an F∈[α]n−1, so that σ({F∪{γ}:γ∈α⧹F}) is infinite. As remarked in Section 2, if κ is an uncountable weakly compact cardinal number, then P(κ) holds. But the existence of weakly compact cardinal numbers requires further set theoretic axioms, beyond ZFC [3]. In [4, Question 3], the authors ask if P(ω1) is true. Theorem 3.1 For λ defined by (1.1), P(λ)holds. For our proof of Theorem 3.1, it will be more convenient to reformulate it into a statement about n-tuples, instead of sets of cardinality n. We will first introduce some notation. Let n∈N and Γ1, Γ2,…,Γn be sets of infinite cardinality, and put Γ=∏i=1nΓi. For a∈Γ and 1≤i≤n, we denote the ith coordinate of a by a(i). We say that two points a and b in Γ are diagonal, if a(i)≠b(i), for all i∈{1,2,…,n}. Let a∈Γ and i∈{1,2…,n}. We call the set H(a,i)={(b1,b2,…,bi−1,a(i),bi+1,…,bn):bj∈Γj,forj∈{1,…,n}⧹{i}},the hyperplane through the point a orthogonal to i. We call the set L(a,i)={(a(1),…,a(i−1),bi,a(i+1),…,a(n)):bi∈Γi},the line through the point a in direction of i. For a cardinal number γ, we define recursively the following sequence of cardinal numbers (exp+(γ,n):n∈N0):exp+(γ,0)=γ, and, assuming exp+(γ,n) has been defined for some n∈N0, we put exp+(γ,n+1)=(2exp+(γ,n)+)+. Here γ+ denotes the successor cardinal, for a cardinal γ, that is the smallest cardinal number γ′ with γ′>γ. Note that since exp+(γ,1)≤222γ, it follows for the above-defined cardinal number λ, that λ=limn→∞exp+(ω0,n). Secondly, successor cardinals are regular [3], and thus every infinite set of cardinality γ, with γ being a successor cardinal, can be partitioned for n∈N into n disjoint sets Γ1, Γ2,…,Γn, all of them having also cardinality γ, and the map Γ1×Γ2×⋯×Γn→[⋃i=1nΓi]n, (a1,a2,…,an)↦{a1,a2,…,an},is injective. We, therefore, deduce that the following statement implies Theorem 3.1. Theorem 3.2 Let n∈Nand a assume that the sets Γ1, Γ2,…,Γnhave cardinality at least exp+(ω1,n2). For any function, σ:Γ≔∏i=1nΓi→C,where Cis an arbitrary set, at least one of the following two conditions hold: There is a sequence (a(j))j=1∞, of pairwise diagonal elements in Γ, so that σ(a(i))=σ(a(j)), for any i,j∈N. There is a line L⊂Γ, for which σ(L)is infinite. Before proving Theorem 3.2. We need the following observation. Lemma 3.3 Let n∈Nand Γ1, Γ2,…,Γnbe non-empty sets. Let σ:Γ≔∏i=1nΓi→C,be a function that fails both conditions (a) and (b) in the definition (P(α)). Then there is a set C˜and a function σ˜:Γ≔∏i=1nΓi→C˜,that fails both, (a) and (b), and, moreover, has (c) for all c∈C˜, there is a hyperplane Hc⊂Γso that {b∈Γ:σ˜(b)=c}⊂Hc. Proof We may assume without loss of generality that σ is surjective. Since (a) is not satisfied, for each c∈C, there exists an m(c)∈N and a (finite) sequence (a(c,j))j=1m(c)⊂σ−1({c}), which is pairwise diagonal, and maximal. Hence σ−1({c})⊂⋃j=1m(c)⋃i=1nH(a(c,j),i). Indeed, from the maximality of (a(c,j))j=1m(c)⊂σ−1({c}), it follows that each b∈σ−1({c}) must have at least one coordinate in common with at least one element of (a(c,j))j=1m(c). We define C˜=⋃c∈C({1,2,…,m(c)}×{1,2…n}×{c}),and σ˜:Γ→C˜,b↦(j,i,c), where c=σ(b),j=min{j′:b∈⋃i′=1nH(a(c,j′),i′)},andi=min{i′:b∈H(a(c,j),i′)}. It is clear that σ˜ satisfies (c). Since σ˜(j,i,b)=σ˜(j′,i′,d) implies σ(b)=σ(d) for (j,i,b),(j′,i′,d)∈Γ˜, (a) fails for σ˜. In order to verify that (b) is not satisfied, assume L⊂Γ is a line, and let {c1,c2,…,cp} be the image of L under σ. By construction, σ˜(L)⊂{(j,i,ck),k≤p,j≤m(ck),i≤n},which is also finite.□ Proof of Theorem 3.2 We assume that σ:Γ=Γ1×Γ2×⋯×Γn→C is a map which fails both (a) and (b). By Lemma 3.3, we may also assume that σ satisfies (c). For each a∈Γ, we fix a number h(a)∈{1,2,…,n} so that σ−1({σ(a)})⊂H(a,h(a)). Thus, h(a) is the direction, for which all b∈Γ, with σ(b)=σ(a), lie in the hyperplane through a orthogonal to h(a). It is important to note that since (b) is not satisfied, it follows that each line L, whose direction is some j∈{1,2,…,n}, can only have finitely many elements b for which h(b)=j. Indeed, if h(b)=j, then b is uniquely determined by the value σ(b). To continue with the proof, the following Reduction Lemma will be essential.□ Lemma 3.4 Let β be an uncountable regular cardinal. Assume that Γ˜1⊂Γ1, Γ˜2⊂Γ2,…,Γ˜n⊂Γnare such that ∣Γ˜i∣≥exp+(β,n), for all i∈{1,2,…,n}. Then, for any i∈{1,2…,n}, there are a number Ki∈N, and subsets Γ1′⊂Γ˜1,Γ2′⊂Γ˜2,…,Γn′⊂Γ˜n, with ∣Γi′∣≥β, so that ∀(a1,a2,…,ai−1,ai+1,…,an)∈∏j=1,j≠inΓi′∣{α∈Γi′:h(a1,a2,…,ai−1,α,ai+1,…,an)=i}∣≤Ki. (3.1) Proof We assume without loss of generality that i=n. Abbreviate βj=exp+(β,j), for j=1,2,…,n. We first choose (arbitrary) subsets Γ˜j(0)⊂Γ˜j, for which ∣Γ˜j(0)∣=βn+1−j. Since the βj’s are regular, it follows for each j=1,2…,n−2 that ∣Γ˜j(0)∣=βn+1−j>2βn−j=2∣Γ˜j+1(0)×Γ˜j+2(0)×⋯×Γ˜n−1(0)∣=∣{f:Γ˜j+1(0)×Γ˜j+2(0)×⋯×Γ˜n−1(0)→N}∣. Here we are using for the second equality the assumption that ∣Γ˜n(0)∣<∣Γ˜n−1(0)∣<⋯<∣Γ˜j+1(0)∣=βn−j, and the assumption that βn−j is regular. For the third equality, we are using the fact that ∣{f:Λ→N}∣=∣{f:Λ→{0,1}}∣=2∣Λ∣ for infinite sets Λ. Abbreviate for j=1,…,n−1: Fj={f:Γ˜j(0)×Γ˜j+1(0)×⋯×Γ˜n−1(0)→N}and note that by above estimates ∣Fj+1∣<∣Γ˜j(0)∣, for j=1,2,…,n−1. We consider the function ϕ1:∏j=1n−1Γ˜j(0)→N,(a1,a2,…,an−1)↦∣{α∈Γ˜n(0):h(a1,a2,…,an−1,α)=n}∣.For fixed a1∈Γ˜1(0), ϕ1(a1,·)∈F2, and the cardinality of F2 is by the above estimates smaller than βn, the cardinality of Γ˜1(0), which is regular. Therefore, we can find a function ϕ2∈F2 and a subset Γ1′⊂Γ˜1(0) of cardinality βn so that ϕ1(a1,·)=ϕ2 for all a1∈Γ1′. Now we can apply the same argument to the function ϕ2:∏j=2n−1Γ˜j(0)→N and obtain a function ϕ3∈F3 and Γ2′⊂Γ˜2(0) of cardinality βn−1 so that ϕ2(a2,·)=ϕ3 for all a2∈Γ2′. We continue the process and find Γj′⊂Γ˜j(0), for j=1,2,…,n−2 of cardinality βn+1−j and functions ϕj∈Fj, for j=1,2,…,n−1, so that for all (a1,a2,…,an−2)∈∏j=1n−2Γj′ and an−1∈Γ˜n−1(0) we have ϕ1(a1,a2,…,an−1)=ϕ2(a2,…,an−1)=…=ϕn−1(an−1). (3.2) Then, since ϕn−1 is N valued, we can finally choose Kn∈N and a subset Γn−1′⊂Γ˜n−1(0), of cardinality at least β, so that ϕn−1(an−1)≤Kn, for all an−1∈Γn−1′, which finishes our argument. Continuation of the proof of Theorem3.2. We apply Lemma 3.4 successively to all i∈{1,2,…,n}, and the cardinals β(i)=exp+(ω1,n(n−i)). We obtain numbers K1,K2,…,Kn in N and infinite sets Λj⊂Γj, for j=1,2,…,n, so that for all i∈{1,2,…,n} and all a=(aj:j∈{1,2,…,n}\{i})∈∏j=1,j≠inΛj ∣{α∈Λi:h(a1,a2,…,ai−1,α,ai+1,…,an)=i}∣≤Ki. In order to deduce a contradiction, choose for each j=1,…,n a subset Aj of Λj of cardinality lj=(n+1)Kj. Then it follows that ∏j=1nlj=∣∏j=1nAj∣=∑i=1n∑a∈∏j=1,j≠inAj∣{α∈Ai:h(a1,a2,…,ai−1,α,ai+1,…,an)=i}∣≤∑i=1nKi∏j=1,j≠inlj≤nn+1∏j=1nljwhich is a contradiction and finishes the proof of the Theorem.□ We can now state the ZFC version of [4, Theorem 4.1], in which it was shown that for weakly compact cardinalities κ0 the space C(Kκ0), where Kκ0 was defined at the end of Section 2, cannot be uniformly (or coarsely) embedded into any c0(Γ), where Γ has any cardinality. Since the only property of κ0, which is needed in [4], is the fact that P(κ0) holds, we deduce Corollary 3.5 C(Kλ)does not coarsely (or uniformly) embed into c0(Γ), for any cardinality Γ. 4. Proof of Theorem 1.1 In this section, we use our combinatorial Theorem 1.2 from Section 3 to show Theorem 1.1. Recall that a long Schauder basis of a Banach space X is a transfinite sequence {eγ}γ=0Γ such that for every x∈X, there exists a unique transfinite sequence of scalars {aγ}γ=0Γ such that x=∑γ=0Γaγeγ. Similarly, a long Schauder basic sequence in a Banach space X is a transfinite sequence {eγ}γ=0Γ which is a long Schauder basis of its closed linear span. Recall that the w*−dens(X*) is the smallest cardinal such that there exists a w*-dense subset of X*. Analogously to the classical Mazur construction of a Schauder basic sequence in a separable Banach space, we have the following result, proved for example in [2, p. 135] (the fact that the basis is normalized, that is ∥eγ∥=1, is not a part of the statement in [2], but it is easy to get it by normalizing the existing basis). Theorem 4.1 Let X be a Banach space with Γ=w*−dens(X*)>ω0. Then X contains a monotone normalized long Schauder basic sequence of length Γ. Proof of Theorem 1.1 Using the Hahn–Banach theorem, it is easy to see that w*−dens(X*)≤dens(X). On the other hand, since every x∈X is uniquely determined by its values on a w*-dense subset of X*, it is clear that λ≤dens(X)≤card(X)≤2w*−dens(X*). (4.1) But this implies that w*−dens(X*)≥λ. Indeed, otherwise we had w*−dens(X*)<λ, and thus w*−dens(X*)≤λn, for some n∈N ( λn was defined before the statement of Theorem 1.1), and thus 2w*−dens(X*)≤2λn=λn+1<λ, which contradicts (4.1). In order to prove Theorem 1.1, we may assume without loss of generality that X has a long normalized and monotone Schauder basis (eμ)μ<λ, of length λ. Suppose that F={γ1,…,γn}∈[λ]n where γ1<⋯<γn is arranged in an increasing order. Consider the corresponding finite set MF={∑i=1nεieγi:εi∈{−1,1}},containing 2n distinct vectors of X, and put a linear order ≺ on this set according to the arrangement of the signs εi, setting ∑i=1nεieγi≺∑i=1nε˜ieγiif and only if for the minimal i, such that εi≠ε˜i, it holds εi<ε˜i. In order to prove Theorem 1.1, it suffices to show that if M=∪F∈[λ]n,n∈NMF⊂X has the coarse Stone property, then X fails to have non-trivial cotype. To this end, starting with U={B2(x):x∈M}, we find a uniformly bounded cover V, which is point finite and so that U refines V, and we fix for all x∈M a Vx∈V with B2(x)⊂Vx. Let r>0 be such that each V∈V is a subset of a ball of radius r. Let C be the set consisting of all finite tuples of V. We now define the function σ:[λ]n→C as follows. If F∈[λ]n,F={γ1,…,γn} where γ1<⋯<γn, we let σ(F)=(Vy1,…,Vy2n), (4.2)where y1≺⋯≺y2n are the elements of MF arranged in the increasing order defined above. Applying Theorem 1.2 to the function σ, for a fixed n∈N, yields only two possibilities: Case 1: The set σ({F∪{τ}:τ∈λ⧹F}) is infinite for some F={γ1,…,γn−1}, where γ1<⋯<γn−1. In this case, pick an infinite sequence of distinct {τj}j=1∞ witnessing the desired property. By passing to a subsequence, we may assume without loss of generality that either there exists k,1≤k≤n−1, so that for all j∈N, γk<τj<γk+1, or τj<γ1 for all j∈N, or γn−1<τj for all j∈N. For simplicity of notation, assume the last case, that is γ1<⋯<γn−1<τj holds for all j∈N. Denoting Fj={γ1,…,γn−1,τj}, we conclude that there exists a fixed selection of signs ε1,…,εn such that the set B={Vy:y=∑i=1n−1εieγi+εneτj,j∈N}is infinite. Indeed, otherwise the set of values {σ({γ1,…,γn−1,τj}),j∈N}, which are determined by the definition (4.2), would have only a finite set of options for each coordinate, and would, therefore, have to be finite. This is a contradiction with the point finiteness of the system V, because ∑i=1n−1εieγi∈Vy,forallVy∈B. Case 2: there is a sequence (Fj) of pairwise disjoint elements of [λ]n, with σ(Fi)=σ(Fj), for any i,j∈N. In fact, it suffices to choose just a pair of such disjoint elements (written in an increasing order of ordinals) F={γ1,…,γn}, G={β1,…,βn}, such that σ(F)=σ(G). This means, in particular, that for every fixed selection of signs ε1,…,εn, V∑i=1nεieγi=V∑i=1nεieβi.By our assumption, the elements of V are contained in a ball of radius r, hence ∥∑i=1nεieγi−∑i=1nεieβi∥≤2r (4.3)holds for any selection of signs ε1,…,εn. Let uj=eγj−eβj, j∈{1,…,n}. Because {eγ} is a monotone normalized long Schauder basis, we have the trivial estimate 1≤∥uj∥≤2. Equation (4.3) means that 1≤∥∑i=1nεiui∥≤2r (4.4)holds for any selection of signs ε1,…,εn. Since norm functions are convex, this means that for the unit vector ball BE of E=span(ui:i≤n) it follows that {∑j=1najuj:∣aj∣≤12r}⊂BE⊂{∑j=1najuj:∣aj∣≤2},which means that (uj)j=1n is 4r-equivalent to the unit vector basis of ℓ∞n.□ In fact, our proof gives a much stronger condition than just failing cotype, because our copies of ℓ∞k are formed by vectors of the type eα−eβ. This fact can be used to obtain much stronger structural results for spaces with special bases. Recall that a long Schauder basis {eγ}γ=1Λ is said to be symmetric if ∥∑i=1naieγi∥=∥∑i=1naieβi∥for any selection of ai∈R, and any pair of sets {γi}i=1n⊂[1,Λ), {βi}i=1n⊂[1,Λ). It is well-known (cf. [7, Proposition II.22.2]) that each symmetric basis is automatically unconditional, that is there exists K>0 such that 1K∥∑i=1n∣ai∣eγi∥≤∥∑i=1naieγi∥≤K∥∑i=1n∣ai∣eγi∥.In particular, 1K∥∑i∈Aaieγi∥≤∥∑i∈Baieγi∥whenever A⊂B. Theorem 4.2 Let X be a Banach space of density Λ≥λ, with a symmetric basis {eγ}γ=1Λ, which coarsely (or uniformly) embeds into some c0(Γ). Then X is linearly isomorphic with c0(Λ). Proof By the proof of the above results, if X embeds into c0(Γ), there exists a C>0, such that for each k∈N, there are some vectors {vi}i=1k of the form vi=eγi−eβi satisfying the conditions 12maxj∣ai∣≤∥∑i=1kaivi∥≤Cmaxj∣aj∣. (4.5) Using the fact that the basis {eγ} is unconditional and symmetric, we obtain that there exist some constants A,B>0 such that A∥∑i=1kaieγi∥≤∥∑i=1kaivi∥≤B∥∑i=1kaieγi∥. (4.6) Combining (4.5) and (4.6), we finally obtain that for some D≥1, and any k∈N, 1Dmaxj∣ai∣≤∥∑i=1kaieβi∥≤Dmaxj∣aj∣ (4.7)for all {β1,…,βk}⊂[1,Λ), which proves our claim.□ 5. Final comments and open problems Let us mention in this final section some problems of interest. First of all, we do not know whether or not Theorem 1.1 is true if we replace λ by smaller cardinal numbers. Problem 5.1 Assume that X is a Banach space with dens(X)≥ω1, and assume that X coarsely embeds into c0(Γ)for some cardinal number Γ. Does X have trivial co-type? If, moreover, X has a symmetric basis, must it be isomorphic to c0(ω1)? Of course, Problem 5.1 would have a positive answer if the following is true. Problem 5.2 Is Theorem1.2true for ω1? Connected to Problems 5.1 and 5.2 is the following. Problem 5.3 Does ℓ∞coarsely embed into c0(κ)for some uncountable cardinal number κ? Another line of interesting problems asks which isomorphic properties do non-separable Banach spaces have which coarsely embed into c0(Γ). Problem 5.4 Does a non-separable Banach space which coarsely embeds into some c0(Γ), Γ being uncountable, contain copies of c0, or even c0(ω1)? Funding The first named author was supported by GAČR 16-07378S and RVO: 67985840. The second named author was supported by the National Science Foundation under the Grant no. DMS-1464713. References 1 G. Godefroy, G. Lancien and V. Zizler, The non-linear geometry of Banach spaces after Nigel Kalton, Rocky Mtn. J. Math. 44 ( 2014), 1529– 1583. Google Scholar CrossRef Search ADS 2 P. Hájek, V. Montesinos, J. Vanderwerff and V. Zizler, Biorthogonal Systems in Banach Spaces, CMS Books in Mathematics , Springer, New York, 2008. 3 Th. Jech, Set Theory, Springer Monographs in Mathematics , 4th edn, Springer Verlag, Berlin, Heidelberg, New York, 2006. 4 J. Pelant, P. Holický and O. Kalenda, C(K) spaces which cannot be uniformly embedded into c0(Λ), Fund. Math. 192 ( 2006), 245– 254. Google Scholar CrossRef Search ADS 5 J. Pelant, Embeddings into c0, Topol. Appl. 57 ( 1994), 259– 269. Google Scholar CrossRef Search ADS 6 J. Pelant and V. Rödl, On coverings of infinite-dimensional metric spaces, Discrete Math. 108 ( 1992), 75– 81. Google Scholar CrossRef Search ADS 7 I. Singer, Bases in Banach spaces I, Grundlehren Math. Wiss. 154 , Springer-Verlag, Berlin, 1970. Google Scholar CrossRef Search ADS 8 A. Swift, On coarse Lipschitz embeddability into c0(κ), preprint, arXive:1611.04623v1. © 2017. Published by Oxford University Press. All rights reserved. For permissions, please email: journals.permissions@oup.com

The Quarterly Journal of Mathematics – Oxford University Press

**Published: ** Mar 1, 2018

Loading...

personal research library

It’s your single place to instantly

**discover** and **read** the research

that matters to you.

Enjoy **affordable access** to

over 12 million articles from more than

**10,000 peer-reviewed journals**.

All for just $49/month

Read as many articles as you need. **Full articles** with original layout, charts and figures. Read **online**, from anywhere.

Keep up with your field with **Personalized Recommendations** and **Follow Journals** to get automatic updates.

It’s easy to organize your research with our built-in **tools**.

Read from thousands of the leading scholarly journals from *SpringerNature*, *Elsevier*, *Wiley-Blackwell*, *Oxford University Press* and more.

All the latest content is available, no embargo periods.

## “Hi guys, I cannot tell you how much I love this resource. Incredible. I really believe you've hit the nail on the head with this site in regards to solving the research-purchase issue.”

Daniel C.

## “Whoa! It’s like Spotify but for academic articles.”

@Phil_Robichaud

## “I must say, @deepdyve is a fabulous solution to the independent researcher's problem of #access to #information.”

@deepthiw

## “My last article couldn't be possible without the platform @deepdyve that makes journal papers cheaper.”

@JoseServera

- Read unlimited articles
- Personalized recommendations
- No expiration
- Print 20 pages per month
- 20% off on PDF purchases
- Organize your research
- Get updates on your journals and topic searches

**$49/month**

Start Free Trial

14-day Free Trial

Best Deal — 39% off
### Annual Plan

- All the features of the Professional Plan, but for
**39% off**! - Billed annually
- No expiration
- For the normal price of 10 articles elsewhere, you get one full year of unlimited access to articles.

~~$588~~

**$360/year**

Start Free Trial

14-day Free Trial

Read and print from thousands of top scholarly journals.

System error. Please try again!

or

By signing up, you agree to DeepDyve’s Terms of Service and Privacy Policy.

Already have an account? Log in

Bookmark this article. You can see your Bookmarks on your DeepDyve Library.

To save an article, **log in** first, or **sign up** for a DeepDyve account if you don’t already have one.