Lindelof type of generalization of separability in Banach spaces
aa r X i v : . [ m a t h . F A ] A p r LINDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITYIN BANACH SPACES
JARNO TALPONEN
Abstract.
We will introduce the countable separation property (CSP) of Ba-nach spaces X, which is defined as follows: For each set
F ⊂ X ∗ such thatspan ω ∗ ( F ) = X ∗ there is a countable subset F ⊂ F with span ω ∗ ( F ) = X ∗ .All separable Banach spaces have CSP and plenty of examples of non-separableCSP spaces are provided. Connections of CSP with Markuˇceviˇc-bases, Corsonproperty and related geometric issues are discussed. Introduction
In this article we study a class of relatively small non-separable Banach spaces.If X is a Banach space and X ∗ is its dual, then a subset F ⊂ X ∗ is said to separate X if for all x ∈ X \ { } there is f ∈ F such that f ( x ) = 0. We investigate theBanach spaces X, which have the following countable separation property (CSP):Whenever F ⊂ X ∗ separates X , there exists a countable separating subset F ⊂ F . It turns out that all separable Banach space have CSP but there exist alsonumerous examples of non-separable CSP spaces (see the last section for summary).The definition of the countable separation property resembles that of the Lindel¨ofproperty. In fact it turns out that if the unit sphere S X of a Banach space X isweakly Lindel¨of, then X has CSP. We will point out various connections betweenCSP and the established theory of non-separable spaces. For example, if X is weaklycompactly generated and has CSP, then it follows that X is already separable.This study is also closely related to [4], in which certain generalizations of sepa-rability, the so called Kunen-Shelah properties , are summarized and discussed. Thechain of implicationsseparability = ⇒ KS = ⇒ KS = ⇒ . . . = ⇒ KS was proved there. The only known examples of non-separable spaces having anyof the properties KS − KS are constructed under set theoretic axioms extraneousto ZFC. To mention in this context the most important properties from the list,KS , KS or KS of X means that there is no closed subspace Y ⊂ X admittingan ω -polyhedron, an uncountable biorthogonal system or an uncountable M-basis,respectively, in Y. It turns out that KS = ⇒ CSP = ⇒ KS .In order to motivate the introduction of CSP concept, let us discuss the Kunen-Shelah properties a little further. One could ask whether some of the Kunen-Shelah properties of Banach spaces are in fact equivalent to separability. RecentlyTodorˇcevi´c [19] answered some old questions about the existence of bases in non-separable Banach spaces. In particular, he proved that it is consistent that each Date : November 21, 2018.1991
Mathematics Subject Classification.
Primary 46B26, 46A50; Secondary 46B03. non-separable Banach space admits an uncountable biorthogonal system and henceit is consistent that KS = ⇒ CSP. Actually, by combining the results in [13, p.1086-1099] and [19] one can see that the question about the equivalence of separabilitywith KS is independent of ZFC. This positions the class of CSP spaces interestinglywith respect to the Kunen-Shelah properties. On one hand, CSP is a class of Banachspaces close to KS and still containing absolute non-separable representatives. Onthe other hand, it turns out that many of the known interesting examples of non-separable Banach spaces (both in and outside ZFC) fall into CSP class. General concepts and notations.
Real Banach spaces are typically denoted byX , Y , Z. We denote by B X the closed unit ball of X and by S X the unit sphere ofX. Unless otherwise stated, we will apply cardinal arithmetic notations.See [3], [6] and [13] for the standard notions in set-theory, Banach spaces andtopology, respectively. We refer to Zizler’s survey [21] on the non-separable Banachspaces for most of the definitions and results used here.Recall that F ⊂ X ∗ is a separating subset if and only if for each x ∈ X thereis f ∈ F such that f ( x ) = 0 if and only if span ω ∗ ( F ) = X ∗ (see e.g. [6, p.55]).Let X be a Banach space and let F = { ( x α , x ∗ α ) } α ⊂ X × X ∗ be a biorthogonalsystem, i.e. x ∗ α ( x β ) = δ α,β . If span( x α ) = X and span ω ∗ ( x ∗ α ) = X ∗ , then F is calleda Markuˇceviˇc-basis or M-basis . Recall that each separable Banach space admitsan M-basis, see [6, p.219]. An M-basis { ( x α , f α ) } α ⊂ X × X ∗ is called countably λ -norming if there is λ ≥ x ∈ X there is f ∈ X ∗ satisfying λ || x || ≤ f ( x ) and |{ α : f ( x α ) = 0 }| ≤ ω . If X admits a countably norming M-basis,then X is called Plichko (reformulation according to [20]). It is said that X has the density property (DENS) if ω ∗ − dens(X ∗ ) = dens(X).A compact Hausdorff space K is called a Corson compact if it can be embeddedin a Σ-product of real lines, and a Banach space X is Weakly Lindel¨of Determined(WLD) if ( B X ∗ , ω ∗ ) is a Corson compact. To mention a more general notion, recallthat a compact Hausdorff space K is a Valdivia compact if there is an embedding h : K → R Γ (where R Γ is equipped with the product topology) such that h ( K ) ∩ A is dense in h ( K ), where A ⊂ R Γ is the corresponding Σ-product. We will call atopological space countably perfect if each point is in the countable closure of therest of the space.The following folklore facts will be applied frequently: Suppose that ( T, τ ) is acountably tight topological space, λ is an ordinal with cf( λ ) > ω and { E α } α<λ isa family of closed subsets of T such that E α ⊂ E β for α < β . Then S α<λ E ατ = S α<λ E α . A subset F ⊂ X ∗ separates X if and only if span ω ∗ ( F ) = X ∗ (see e.g.[6, p.55]). Fact 1.1.
Let q : X → X / Y be the quotient map q : x x + Y . Then q ( A ) = { z + Y | z ∈ A + Y } for any subset A ⊂ X .Proof. The condition that x ∈ A + Y is equivalent toinf y ∈ Y , a ∈ A || x − ( a + y ) || = dist X / Y ( q ( x ) , q ( A )) = 0 . (cid:3) INDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITY IN BANACH SPACES 3 General properties
It is easy to see that the following formulations of CSP are equivalent:(2.1) • X has CSP: if
F ⊂ X ∗ separates X, then there is a countable F ⊂ F ,which separates X. • For each ω ∗ -dense linear subspace V ⊂ X ∗ there exists a countable subset F ⊂ V such that span ω ∗ ( F ) = X ∗ . • Each family of closed subspaces with trivial intersection has a countablesubfamily with trivial intersection. • There does not exist an uncountable separating family
F ⊂ X ∗ such thateach separating subfamily F ⊂ F has the same cardinality as F .Note that in particular ω ∗ − dens(X ∗ ) = ω if X has CSP.The spaces c (Γ) , ℓ p (Γ) , ≤ p ≤ ∞ , | Γ | ≥ ω , provide examples of spaceswithout CSP, since { e ∗ γ } γ ∈ Γ is an uncountable minimal separating family.The following fact appeared already in [1] with a different formulation. Proposition 2.1.
Separable Banach spaces have CSP.
This follows also immediately from the following observation.
Proposition 2.2.
Let X be a Banach space such that S X is Lindel¨of in the relativeweak topology. Then X has CSP.Proof. Let { f γ } γ ∈ Γ ⊂ X ∗ be a separating family of functionals. Then S γ ∈ Γ f − γ ( R \{ } ) = X \ { } . In particular, the subsets U γ = f − γ ( R \ { } ) ∩ S X , γ ∈ Γ , define a ω -open cover for S X . According to the Lindel¨of property of ( S X , ω ) thereexists a countable subcover { U γ n } n<ω ⊂ { U γ } γ ∈ Γ . We conclude that { f γ n } n<ω isa separating family of X . (cid:3) We will subsequently give some examples of non-separable CSP spaces. It isstraightforward to verify that CSP is preserved under isomorphisms.
Proposition 2.3.
Let X be a Banach space with CSP and let Y ⊂ X be a closedsubspace. Then Y has CSP.Proof. Assume to the contrary that Y fails CSP. Suppose that
F ⊂ Y ∗ is anuncountable minimal separating set for Y. Let e F ⊂ X ∗ be a set of functionalsobtained from F by Hahn-Banach extension. Then Y ⊥ ∪ e F is an uncountableminimal separating subfamily for X; a contradiction. (cid:3) Example 2.4.
The space ℓ ∞ does not have CSP. The space ℓ ∞ contains an isometric copy of ℓ (2 ω ) (see [6, p.86]). Clearly ℓ (2 ω )does not have CSP. Thus the claim follows by applying Proposition 2.3 that CSPis inherited by the closed subspaces.The preceding example shows that there is a countable F ⊂ ( ℓ ∞ ) ∗ such thatspan( { f | c : f ∈ F } ) is dense in ( ℓ , ω ∗ ) , and span( F ) is not dense in (( ℓ ∞ ) ∗ , ω ∗ ) , even though ℓ is ω ∗ -dense in ( ℓ ∞ ) ∗ by Goldstine’s theorem. JARNO TALPONEN
Example 2.5.
The spaces JL , JL (see [9]) have CSP according to subsequentobservations (see Proposition 3.1), but JL /c = c (2 ω ) and JL /c = l (2 ω ) (seee.g. [21, p. 1757]) clearly do not. We conclude that(i) CSP does not pass to quotients in general.(ii) The dual of a CSP space may contain a ω ∗ -closed subspace, which is not ω ∗ -separable.The spaces JL and JL are not Lindel¨of in their weak topology (see e.g. [21,p.1757,1764]).If X and Y have CSP, does it follow that X ⊕ Y has CSP? If so, is CSP a three-space property, i.e. does X have CSP whenever X/ Y and Y ⊂ X have CSP? Wewill give a partial answer to this problem in Theorem 4.3.Even though CSP is not a sufficient condition for separability, it is still quite astrong condition ’towards separability’.
Proposition 2.6.
Let X be a Banach space such that X ∗ has CSP. Then X isseparable.Proof. Clearly X ⊂ X ∗∗ embedded canonically separates X ∗ . Then accordingto CSP one can find a sequence ( x n ) n ∈ N ⊂ X separating X ∗ . We claim thatspan( x n ) = X. Indeed, if this was not the case, then one could find by the Hahn-Banach theorem a non-zero functional f ∈ X ∗ vanishing on span( x n ). But this isnot possible, since ( x n ) ⊂ X ∗∗ separates X ∗ . Thus X is separable. (cid:3) As a brief remark we would like to mention a mild condition under which sepa-rability and CSP are equivalent.Let us consider the case that X admits a system ( { x α } α , { f β } β ) ∈ P (X) × P (X ∗ )satisfying the following conditions:(2.2) span( { x α } α ) = Xspan ω ∗ ( { f β } β ) = X ∗ For each β we have |{ α | f β ( x α ) = 0 }| ≤ ω ∗ − dens(X ∗ ) . For a Banach space X the DENS property or the existence of an M-basis clearlyprovide a system satisfying (2.2).
Proposition 2.7.
A Banach space X is separable if and only if it has CSP andadmits a system satisfying (2.2) . In particular each CSP space with an M-basis isseparable.Proof. Suppose that X is a CSP space, and the system ( { x α } α , { f β } β ) satisfies(2.2). Then there exists a countable separating subfamily F ⊂ { f β } β , since X hasCSP. Thus |{ α | x α = 0 }| = |{ α | f ( x α ) = 0 for some f ∈ F}| ≤ |F × ( ω ∗ − dens(X ∗ )) | = ω. Hence X is separable. On the other hand, each separable Banach space X has anM-basis (see e.g. [6, p.219]) and hence system (2.2). Recall that, by Proposition2.1, separability of X implies CSP. (cid:3)
Example 2.8.
For a Valdivia compact K the space C ( K ) has CSP if and onlyif C ( K ) is separable. If a Banach space X is Plichko and has CSP, then X isseparable. INDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITY IN BANACH SPACES 5
Indeed, if K is Valdivia, then C ( K ) is Plichko (see [7]). Any Plichko spaceadmits an M-basis, so that Proposition 2.7 yields that X is separable. Proposition 2.9.
Let X be a CSP space and C ⊂ X a weakly compact set. Then C is separable.Proof. Observe that Y = span( C ) is a WCG subspace and hence admits an M-basis. According to Proposition 2.3 the space Y has CSP. Thus, by Proposition 2.7Y is separable and so is C . (cid:3) Topological point of view
A topological space T is called dense-separable if each dense subset A ⊂ T isseparable (see [11] for discussion).Recall the following concept due to Corson and Pol. A Banach space X hasproperty ( C ) if, for each family A of closed convex subsets of X having emptyintersection, there exists a countable subfamily A ⊂ A with empty intersection.Pol gave an important characterization of property ( C ) in terms of a dual spaceformulation ( C ′ ), which is a kind of convex version of countable tightness (see [14,Thm.3.4]). This condition appears in the result below: Proposition 3.1.
Let X be a Banach space. Consider the following conditions: (1) (X ∗ , ω ∗ ) is countably tight i.e. for each a ∈ A ω ∗ ⊂ X ∗ there is a subset ( a n ) n<ω ⊂ A such that a ∈ { a n | n < ω } ω ∗ . (2) X ∗ is dense-separable in the ω ∗ -topology. (3) X satisfies property ( C ′ ) : for each A ⊂ X ∗ and f ∈ A ω ∗ there is a countable A ⊂ A such that f ∈ conv ω ∗ ( A ) . (4) For each A ⊂ X ∗ and f ∈ A ω ∗ there is countable A ⊂ A such that f ∈ span ω ∗ ( A ) . (5) X has CSP i.e. for each A ⊂ X ∗ such that span ω ∗ ( A ) = X ∗ there is acountable A ⊂ A such that span ω ∗ ( A ) = X ∗ .If X ∗ is ω ∗ -separable, then ⇒ ⇒ and ⇒ ⇒ ⇒ .Proof. Let us first check that implication (1) = ⇒ (2) holds if X ∗ is ω ∗ -separable.Let X ∗ be ω ∗ -separable space satisfying (1) and let A ⊂ X ∗ be a ω ∗ -dense subset.Fix a ω ∗ -dense subset ( x k ) k<ω ⊂ X ∗ . By using the countable tightness of (X ∗ , ω ∗ ),we may pick { a ( n ) k | n, k < ω } ⊂ A such that x k ∈ { a ( n ) k | n < ω } ω ∗ for each k < ω .Hence { a ( n ) k | n, k < ω } ω ∗ = X ∗ , so that (2) holds, as A was arbitrary.Let us check the implication (4) = ⇒ (5) for ω ∗ -separable X ∗ . First recall (2.1).Let X ∗ be ω ∗ -separable space satisfying (4) and let ( x n ) n<ω ⊂ X ∗ be a ω ∗ -densesubset in X ∗ . Consider a separating family A ⊂ X ∗ . Thus span( A ) is ω ∗ -densein X ∗ . According to condition (4) we can find a countable set C n ⊂ span( A ) foreach n < ω such that x n ∈ span ω ∗ ( C n ). Thus span ω ∗ ( S n C n ) = X ∗ . Note thateach of the sets C n is contained in the linear span of countably many vectors of A .Therefore there is countable A ⊂ A such thatspan ω ∗ ( A ) = span ω ∗ ( [ n C n ) = X ∗ , so that X satisfies (5). Other implications are clear. (cid:3) JARNO TALPONEN
We do not know if (2) = ⇒ (3) above or if some implications can be reversed.For example the following spaces are non-separable and have CSP according toProposition 3.1: JL , JL and C ( K ), where K is the double-arrow space. Indeed,these spaces have property ( C ) and a ω ∗ -separable dual (see e.g. [21, p.1757], [14,p.146]).Assuming CH Kunen constructed an interesting Hausdorff compact K , whichis separable, scattered and non-metrizable (see [13, p.1086-1099] for discussion).Kunen’s C ( K ) space is non-separable and ( C ( K ) ∗ , ω ∗ ) is hereditarily separable(see [5, p.476]), hence separable and countably tight. Example 3.2.
There is a bounded injective linear operator T : JL → c (2 ω ) whoserange is non-separable. In justifying this we will apply the fact that JL contains c and that JL /c = c (2 ω ) (see [21, p. 1757]). Let ( e n ) n<ω ⊂ c be the canonical unit vector basis andlet ( e ∗ n ) n<ω ⊂ ℓ be the corresponding functionals. Let ( f n ) n<ω ⊂ JL ∗ be the Hahn-Banach extension of ( e n ) n<ω . Define S : JL → c by x (( n +1) − f n ( x )) n<ω . Let q : JL → JL /c be the canonical quotient mapping. Then T : x ( S ( x ) , q ( x ))defines the required map JL → c ( ω ) ⊕ c (2 ω ) = c (2 ω ). Problem 3.3.
Suppose that X admits a long unconditional basis and Y ⊂ X is aclosed subspace having CSP. Does it follow that Y is separable? Dual CSP spaces can be characterized as follows:
Theorem 3.4.
Let X be a Banach space. Then the following are equivalent: (1) X ∗ has CSP (2) X is separable and X ∗ has property (C) (3) X is separable and does not contain ℓ isomorphically.Proof. The equivalence of the last two conditions is known (see [21, Thm.4.2]).If (2) holds, then X ∗∗ is ω ∗ -separable by Goldstine’s theorem, so that Proposition3.1 can be applied together with property ( C ) to obtain that X ∗ has CSP.On the other hand, if X ∗ has CSP, then by Proposition 2.6 we known that Xmust be separable. Now assume to the contrary that X contains an isomorphiccopy of ℓ . Then it is known that X ∗ contains a complemented isomorphic copy of ℓ ∞ . Since ℓ ∞ does not have CSP it follows by Proposition 2.3 that X ∗ fails CSP, acontradiction. Thus X does not contain ℓ . (cid:3) For example the James Tree and the James function space (see [12]) are separablespaces not containing ℓ , and whose duals JT ∗ and JF ∗ are non-separable CSPspaces.3.1. C ( K ) spaces.Proposition 3.5. Let L be a locally compact Hausdorff space such that C ( L ) hasCSP. Then L is dense-separable and the interior of the derived set of L is countablyperfect.Proof. Recall that locally compact Hausdorff spaces are completely regular. If theinterior of the derived set of L is non-empty, then let x ∈ int( D ( L )). If such x exists,it is not an isolated point in L , and in any case we may fix a dense subset Γ ⊂ L such that x / ∈ Γ. Consider the point evaluation maps f δ k f ( k ) in C ( L ) ∗ for k ∈ Γ. INDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITY IN BANACH SPACES 7
Clearly these maps separate C ( L ). Since C ( L ) has CSP, there exists a countablesubset Γ ⊂ Γ such that the associated evaluation maps still separate C ( L ). Weclaim that L = Γ . Indeed, if there exists a point y ∈ L \ Γ , then by the completelyregularity of L there exists f ∈ C ( L ) attaining value 1 at y but vanishing on Γ .This contradicts the fact that the point evaluations associated to the points in Γ separate C ( L ). Thus Γ = L and L is dense-separable as Γ was arbitrary. Notethat x ∈ Γ ∩ int( D ( L )), so that int( D ( L )) is countably perfect. (cid:3) We have not been able to construct a (dense-separable) compact K such that C ( K ) has CSP but fails property (C). Observe that the ˇCech-Stone compactification βω is dense-separable as n ∈ ω are isolated in βω . Since C ( βω ) = ℓ ∞ , we concludethat dense-separability of K is not sufficient for C ( K ) to have CSP.The following result produces plenty of examples of non-separable CSP spaces. Theorem 3.6.
Let K be a scattered and countably tight compact. Then C ( K ) hasCSP if and only if K is separable.Proof. The only if part follows from Proposition 3.5. Towards the other implication,recall that a scattered compact K is countably tight if and only if C ( K ) has property(C) (see [14, Cor.4.1]). If K is separable, then a standard argument using thepoint evaluations gives that C ( K ) ∗ is ω ∗ -separable, so that Proposition 3.1 can beapplied. (cid:3) For example the one-point compactified rational sequence topology (see [17,p.87]) is scattered, countably tight, separable and non-metrizable.Analogous to the open question about preservation of CSP in finite sums is theopen question about the preservation of dense-separability in finite products oftopological spaces. The following fact, however, is easy to verify.
Remark . Let A and B be topological spaces. If A is dense-separable and B hasa countable π -base, then A × B is dense-separable.3.2. Auxiliary results: Intersections of distended subspaces.
In additionto property ( C ) we will treat another type of condition concerning intersections ofconvex sets, which seems to be closely related to ( C ). Throughout this subsectionlet X be a Banach space and κ an uncountable regular ordinal. We denote here by { Z σ } σ<κ a family of closed subspaces of X such that Z α ) Z β for α < β < κ and T σ<κ Z σ = { } and we impose the existence of such family for X. This actuallyexcludes X from CSP class, as it will turn out in the next section.Some subsequent results here depend on the following question: Given X and { Z σ } σ<κ as above, is T σ<κ ( B X + Z σ ) bounded? At first glance the answer may appear to be positive for all Banach spaces X. Forexample, it is easy to see that if X is reflexive, then T σ<κ ( B X + Z σ ) = B X . However,next we present an example of a space for which the answer to the above questionis negative. Define a function ||| · ||| : ℓ ∞ ( ω ) → [0 , ∞ ] by ||| ( x α ) α<ω ||| = || ( x α ) α<ω || ℓ ∞ ( ω ) + X n<ω n lim sup i → ω | x ωi + n | (ordinal arithmetic) . Then (X , ||| · ||| ), where X = { x ∈ ℓ ∞ ( ω ) : ||| x ||| < ∞} , is clearly a Banachspace. Let E σ = { ( x α ) α<ω ∈ X : x α = 0 for α < σ } for σ < ω . We denoteby A : [0 , ω ) → { , } the characteristic function of a given subset A ⊂ ω . Note JARNO TALPONEN that [0 ,σ ] ∈ S (X , |||·||| ) for all σ < ω . Hence { ωi + n | i<ω } ∈ T σ<ω ( B (X , |||·||| ) + E σ )for all n < ω . Consequently T σ<ω ( B (X , |||·||| ) + E σ ) is unbounded.For convenience we denote by ( B ) the class of Banach spaces X satisfying that T σ<κ B X + Z σ is bounded for any { Z σ } σ<κ such as above (or trivially if such { Z σ } σ<κ does not exist at all). Proposition 3.8.
For any X we have that T ǫ> T σ<κ ( ǫ B X + Z σ ) = { } .Proof. Observe that T σ<κ ( ǫ B X + Z σ ) is a symmetric convex set, which contains 0.First we wish to check that T ǫ> T σ<κ ( ǫ B X + Z σ ) does not contain any non-trivial linear subspace L . Indeed, if T σ<κ ( B X + Z σ ) contains L = [ x ] for some x ∈ S X , then dist( kx, Z σ ) ≤ k < ω and σ < κ . Observe that(3.1) ǫ B X + Z σ = ǫ B X + ǫZ σ = ǫ ( B X + Z σ )for all ǫ > σ < κ . Thus, by putting ǫ = k we obtain that dist( x, Z σ ) = dist( kx,Z σ ) k < k for all k < ω and σ < κ . Since Z σ are closed subspaces, we get that x ∈ Z σ for all σ < κ . This contradicts the fact that T σ<κ Z σ = { } 6∋ x and henceX does not contain any non-trivial linear subspace L .Now, let L be a 1-dimensional subspace and write l = L ∩ T σ<κ ( B X + Z σ ).Since T σ<κ ( B X + Z σ ) is convex, symmetric and does not contain L , we obtain that l is bounded. Hence L ∩ T ǫ> ǫ T σ<κ ( B X + Z σ ) = { } . Since L was arbitrary, weobtain that T ǫ> T σ<κ ( ǫ B X + Z σ ) = { } . (cid:3) Note that in the above proof we did not need that κ is uncountable or regular.Next we will give results towards applications of subsequent Lemma 3.12, which isour main technical machinery. Proposition 3.9.
Suppose that the dual of X satisfies the condition (4) of Propo-sition 3.1. Then X is a member of ( B ).Proof. Let { Z σ } σ<κ be a strictly nested sequence of closed subspaces of X such that T σ<κ Z σ = { } . Observe that span ω ∗ ( S σ<κ Z ⊥ σ ) = S σ<κ Z ⊥ σ ω ∗ = X ∗ . Fix f ∈ X ∗ .Since κ is regular, condition (4) of Proposition 3.1 yields that there is σ < κ suchthat f ∈ Z ⊥ σ whenever σ ≤ σ < κ . Thus f ( T σ<κ B X + Z σ ) ⊂ [ −|| f || , || f || ]. Notethat T σ<κ B X + Z σ is weakly bounded as f was arbitrary. Hence T σ<κ B X + Z σ is bounded according to the uniform boundedness principle. (cid:3) Proposition 3.10.
Suppose that the dual of X satisfies the condition (4) of Propo-sition 3.1, that { Y σ } σ<κ is a nested family of affine subspaces, where κ is an un-countable regular cardinal, and that T σ<κ Y σ = { y } . If y σ ∈ Y σ for σ < κ , then y σ → y weakly as σ → κ .Proof. Suppose that u + Z = Y ⊃ Y = u + v + Z are closed affine subspaces,where u, v ∈ X and Z , Z ⊂ X are closed subspaces. Then v ∈ Z σ and Z σ ⊂ Z σ .Indeed, if Z Z , then there is no u + v ∈ X such that u + Z ⊃ u + v + Z . Onthe other hand, if v / ∈ Z , then v + Z Z , so that u + v + Z u + Z .Let { Z σ } σ<κ be the nested family of closed subspaces of X corresponding to { Y σ } σ<κ . It follows that y σ ∈ y + Z σ for σ < κ and T σ<κ Z σ = { } .Fix f ∈ X ∗ . Observe that span ω ∗ ( S σ<κ Z ⊥ σ ) = X ∗ and Z ⊥ σ is ω ∗ -closed for σ < κ . By using condition (4) of Proposition 3.1 and the fact that cf( κ ) > ω , weobtain that there is σ < κ such that f ∈ Z ⊥ σ . Then f ( y σ − y ) = 0 for σ ≥ σ , sothat we have the claim as f was arbitrary. (cid:3) INDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITY IN BANACH SPACES 9
Theorem 3.11.
Suppose that X satisfies the following condition: For each nestedfamily of closed affine subspaces { Y σ } σ<κ it holds that T σ<κ Y σ = ∅ . Then X is amember of ( B ). Before giving the proof, we note that sup σ<κ dist(0 , Y σ ) < ∞ , since κ is uncount-able and regular. Proof.
Let κ and { Z σ } σ<κ be as in the beginning of this subsection. Since T σ<κ Z σ = { } , we may define a norm ||| · ||| : X → R by ||| x ||| = sup σ<κ dist( x, Z σ ). Observethat ||| x ||| ≤ || x || for x ∈ X.We aim to show that the norms ||·|| and |||·||| are equivalent. After this has beenestablished, it follows that there is
C > || x || ≤ C whenever dist( x, Z σ ) ≤ σ < κ . Actually, it suffices to check that ||| · ||| is complete. Indeed, in suchcase the Banach open mapping principle yields that I : (X , || · || ) → (X , ||| · ||| ) is anisomorphism.Let ( x n ) n<ω ⊂ X be a ||| · ||| -Cauchy sequence, where x = 0. Denote b x σn = x n + Z σ ∈ X /Z σ for n < ω, σ < κ . Observe that ||| x i − x j ||| = sup σ<κ || b x σi − b x σj || X /Z σ for i, j ∈ ω. Since cf( κ ) > ω , there is for ( i, j ) ∈ ω × ω an ordinal α i,j < κ such that ||| x i − x j ||| = || b x α i,j i − b x α i,j j || X /Z αi,j . Observe that α · = sup i,j ∈ ω α i,j < κ . Next we regard σ ∈ [ α, κ ).Thus the sequence ( b x σn ) n<ω ⊂ X /Z σ is Cauchy for all σ ∈ [ α, κ ). Since X /Z σ isa Banach space, we obtain that there is y σ ∈ X /Z σ such that b x σn → y σ as n → ∞ in X /Z σ . Moreover, by the selection of α we get lim n →∞ ||| x n ||| = lim n →∞ || b σ − b x σn || X /Z σ , since x = 0.We may regard y σ · = Y σ · = v σ + Z σ ⊂ X as affine subspaces, where v σ ∈ X for σ ∈ [ α, κ ). Similar interpretation for b x σn ∈ X yields the following: Since b x σ n = b x σ n /Z σ for n < ω, α ≤ σ ≤ σ < κ, we obtain that y σ = y σ /Z σ by thecontinuity of the quotient map X /Z σ → X /Z σ . Thus Y σ = Y σ + Z σ , where α ≤ σ ≤ σ < κ . Note that dist(0 , Y σ ) = lim n →∞ ||| x n ||| for σ ∈ [ α, κ ) bythe selection of α . According to the assumptions T σ<κ Y σ = ∅ and let us choose x ∈ T σ<κ Y σ (even though it turns out promptly that this set is a singleton).Observe that y σ = b x σ · = x + Z σ for σ ∈ [ α, κ ). Thus sup σ ∈ [ α,κ ) || b x σn − b x σ || X /Z σ → n → ∞ . Hence x n |||·||| −→ x as n → ∞ , which completes the proof. (cid:3) Lemma 3.12.
Let X be a Banach space, Y ⊂ X a closed subspace and κ anuncountable regular cardinal. Let Z σ ⊂ X be closed subspaces, for σ < κ , whichsatisfy Z α ( Z β for β < α < κ and T σ<κ Z σ = { } . Then the following facts hold: (i) If dens(Y) < κ and X ∈ ( B ) , then Y = T σ<κ Y + Z σ . (ii) If dens(Y) < κ , then there exists θ < κ such that Z θ ∩ Y = { } .Proof. Let us treat the claim ( i ). Let x ∈ T σ<κ span(Y ∪ Z σ ) and ǫ >
0. Thus thereis a cofinal sequence { σ α } α<κ ⊂ κ and families { y α } α<κ ⊂ Y and { z α } α<κ ⊂ Xsuch that z α ∈ Z σ α and || x − ( y α + z α ) || < ǫ for each α < κ .Note that we have T β<κ { y α | β < α < κ } 6 = ∅ , because { y α | β < α < κ } β<κ is adecreasing sequence of closed sets in Y, where κ is regular, and the Lindel¨of numberof Y is less than κ . Let y ( ǫ ) ∈ T β<κ { y α | β < α < κ } depending on the choice of ǫ . Hence we maypick a cofinal sequence { α δ } δ<κ ⊂ κ such that || y ( ǫ ) − y α δ || < ǫ for each δ < κ .This means that || x − ( y ( ǫ ) + z α δ ) || ≤ || y ( ǫ ) − y α δ || + || x − ( y α δ + z α δ ) || < ǫ for δ < κ . We get that x − y ( ǫ ) ∈ z α δ + ǫ B X and in particular(3.2) x − y ( ǫ ) ∈ Z σ αδ + ǫ B X for δ < κ. Since { Z σ } σ<κ decreases to { } and sup δ<κ α δ = κ , we obtain that T δ<κ Z σ αδ = { } .Since X belongs to ( B ), it follows by (3.1) thatlim ǫ → + diam( \ δ<κ ( ǫ B X + Z σ αδ )) = 0 . Since ǫ > x, Y) = 0, that is x ∈ Yas Y is closed. This completes the proof of claim ( i ).Let us check claim ( ii ). Since dens(Y) < κ , the Lindel¨of number of Y \ { } isless than κ . It follows by the regularity of κ , that there cannot exist a decreasingsequence { ( Z α ∩ Y) \ { }} α<κ of non-empty closed sets in Y \ { } . (cid:3) We note that in the above lemma the additional assumptions in (i) cannot beremoved. Indeed, consider closed subspaces E α = { ( x i ) ∈ ℓ ∞ ( ω ) | x i = 0 , for i ≤ α } , α < ω of ℓ ∞ ( ω ). Then T α<ω ℓ ∞ c ( ω ) + E α = ℓ ∞ ( ω ). Note that dens( ℓ ∞ c ( ω )) = 2 ω anddens( ℓ ∞ ( ω )) = 2 ω .4. Combinatorial approach to CSP spaces
Recall that if X has CSP, then ω ∗ − dens(X ∗ ) = ω . This in turn implies dens(X) ≤| X | ≤ ω . Recall that 2 ω < ℵ ω is consistent with ZFC. For this reason some of theresults here, which involve assumption about the density of the space, can actuallybe thought of as consistency results.For a given Banach space X we define the cofinality range of X, cr(X) for short,as the set of all infinite regular cardinals κ satisfying that there exists a family { E σ } σ<κ of closed subspaces of X such that T σ<κ E σ = { } and E α ( E β whenever α < β < κ . Observe that each κ ∈ cr(X) satisfies κ ≤ dens(X). Theorem 4.1.
Let X be a Banach space such that dens(X) < ℵ ω . Then X hasCSP if and only if cr(X) = { ω } .Proof. Let us first check the easier ’only if’ part. Suppose that for { E σ } σ<κ , asabove, there does not exist ( σ n ) n<ω such that T n<ω E σ n = { } , or equivalentlysup n<ω σ n = κ . Note that S σ<κ E ⊥ σ ⊂ X ∗ separates X by the Hahn-Banach theo-rem. Clearly this set has no countable separating subset, so that X fails CSP.To check the ’if’ part assume that X satisfies dens(X) < ℵ ω and X fails CSP.We aim to show that in such case cr(X) = { ω } . According to (2.1) there is anuncountable separating family F ⊂ X ∗ , whose all separating subfamilies have thesame cardinality, say κ ≥ ω . Write F = { f α } α<κ . Put F σ = T α<σ Ker f α forall σ < κ . Clearly this gives a (not necessarily strictly) nested family of closedsubspaces. Note that T σ<κ F σ = { } . INDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITY IN BANACH SPACES 11
Let φ (0) = 0 and we define recursively φ ( α ) = min { β < κ : \ γ<α Ker( f φ ( γ ) ) Ker( f β ) } . Thus, putting E = X, E σ = T α<σ Ker( f φ ( α ) ) for 0 < σ < κ gives a strictly nestedfamily and T σ<λ E σ = { } for some ordinal λ ≤ κ . Note that by the constructionof φ it holds that { f φ ( α ) } α<λ is a separating family.Since { f α } α<κ does not have a separating subfamily of cardinality less than κ ,we conclude that λ = κ above. As dens(X) < ℵ ω we obtain that λ = κ < ℵ ω is aregular cardinal and { E α } α<λ is the required family witnessing that ω ≤ λ = κ ∈ cr(X). (cid:3) The above proof yields immediately that CSP = ⇒ ( B ). Note that there areCSP spaces with an uncountable biorthogonal system. For example, it suffices toconsider a non-separable dual space with CSP, such as JT ∗ , since it is known thatnon-separable dual spaces have uncountable biorthogonal systems (see [18, Cor.4]).One could ask if the previous result remains valid if one replaces ’cofinality’ by’cardinality’ in the definition of cr(X). This is not the case as the following exampleshows; the assumption about the regularity of κ in the definition of cr(X) is indeedessential: Example 4.2.
If X is a CSP space with a biorthogonal system { ( x α , x ∗ α ) } α<ω ,then by using the proof of the previous theorem we see that(1) F · = T β<ω span( { x α | β < α < ω } ) = { } (2) X /F is not a CSP space.(3) There is a strictly nested family { E σ } σ<κ of closed subspaces of X suchthat T σ<κ E σ = { } , where κ > ω .The family of subspaces in the last condition can be obtained as follows: Let { F θ } θ<λ be a strictly nested family of closed subspaces of F , where λ is an ordinaland T θ<λ F θ = { } . It suffices to put E σ = span( { x α | σ ≤ α < ω } ) for σ < ω , E ω = F , and E ω + θ = F θ +1 for θ < λ .We attempt to bring CSP closer to a three space property in the next result. Theorem 4.3.
Let X be a Banach space satisfying ( B ) and dens(X) < ℵ ω . Let Y ⊂ X be a closed subspace. Then cr(X) ⊂ cr(X / Y) ∪ dens(Y) + . In particular,such X has CSP if Y is separable and X / Y has CSP.Proof. Observe that dens(Y) , dens(X / Y) ∈ ℵ ω are regular cardinals. Denote thequotient map q : X → X / Y; q : x x + Y.Consider a regular cardinal κ > dens(Y) such that κ ∈ cr(X). Let { E σ } σ<κ be the corresponding strictly nested family of closed subspaces of X with trivialintersection. By Lemma 3.12 ( i ) we obtain that T σ<κ span( E σ ∪ Y) = Y. Thus T σ<κ (span( E σ ∪ Y) + Y) = Y. Hence Fact 1.1 yields that \ σ<κ q ( E σ ) = \ σ<κ q (span( E σ ∪ Y)) = 0 ∈ X / Y . Note that E σ Y for any σ < κ , since { E σ } σ<κ is strictly nested, κ is regularand dens(Y) < κ . Hence { } ( q ( E σ ) in X / Y for each σ < κ . Thus by passing to acofinal subsequence { σ α } α<κ ⊂ κ such that { q ( E σ α ) } α<κ ⊂ X / Y is strictly nestedwe obtain that κ ∈ cr(X / Y).
For the latter claim recall that according to Theorem 4.1 it holds that cr(X) = { ω } if and only if X has CSP. (cid:3) The Kunen-Shelah properties.
In [4] the Kunen-Shelah propertiesKS = ⇒ KS = ⇒ ... = ⇒ KS are discussed. To mention the most important Kunen-Shelah properties in thiscontext, a space X is said to have KS , KS or KS , if X admits no uncount-able polyhedron, no uncountable biorthogonal system or no uncountable M-basicsequence, respectively. It follows easily from [4, p.114-119] that a Banach spaceX with KS has CSP. For example, Kunen and Shelah have provided samples ofnon-separable KS spaces by assuming CH or ♦ ( ℵ ), see [13, p.1086-1099] and [16].Since CSP passes to subspaces, we obtain that CSP = ⇒ KS by Proposition 2.7.It follows from recent results of Todorˇcevi´c (see [19]) that it is consistent withZFC that the properties KS − KS are in fact equivalent to separability. In par-ticular, KS = ⇒ CSP is consistent.
Theorem 4.4.
Let X be a Banach space such that each quotient X / Y satisfies ( B ) and dens(X) < ℵ ω . If X has KS , then it has CSP. We will first prove the following result, which will be applied.
Proposition 4.5.
Let X satisfy ( B ) and KS . Suppose that { x α } α<ω ⊂ X \ { } and { Γ σ } σ<ω is a family of cofinal subsets of ω such that Γ α ⊃ Γ β for α < β < ω .Then \ σ<ω span( { x α | α ∈ Γ σ } ) = { } . Proof.
Assume to the contrary that above(4.1) \ σ<ω span( { x α | α ∈ Γ σ } ) = { } . Then according to Lemma 3.12 ( ii ) for each β < ω there is σ < ω such that(4.2) span( { x α | α < β } ) ∩ span( { x α | α ∈ Γ σ } ) = { } . Let σ ( β ) be the least ordinal satisfying (4.2) for σ = σ ( β ).Next we will define recursively an uncountable subfamily of { x α } α<ω by usingthe above notations. Let α = 0 and for each θ < ω let α θ = min { γ ∈ Γ σ (sup ǫ<θ α ǫ ) : γ > sup ǫ<θ α ǫ } . Note that { α θ } θ<ω is an increasing sequence by its construction. Observe that thecorresponding family { y θ } θ<ω = { x α θ } θ<ω satisfiesspan( { y θ | θ < γ } ) ∩ span( { y θ | γ ≤ θ } ) = { } for each γ < ω .The assumption (4.1) yields that T γ<ω span( { y θ | γ < θ < ω } ) = { } . Thus, anapplication of Lemma 3.12 ( i ) for Y = span( { y θ : θ < γ } ) and Z γ = span( { y θ | γ <θ < ω } ) for countable γ yields the following fact: For all γ < δ < ω one can findthe least ordinal η ( γ, δ ) ∈ ( δ, ω ) such that y δ / ∈ span( { y θ | θ ∈ [0 , γ ] ∪ [ η ( γ, δ ) , ω ) } ) . INDEL ¨OF TYPE OF GENERALIZATION OF SEPARABILITY IN BANACH SPACES 13
Put ζ = 0, ζ = 1 and for each α ∈ (1 , ω ) we define recursively ζ α = sup β<α η (sup γ<β ζ γ , ζ β ) + 1 . This defines an increasing sequence { ζ α } α<ω ⊂ ω such that y ζ γ / ∈ span( { y ζ θ | γ = θ < ω } ) for γ < ω .In particular { y ζ α } α<ω is a minimal system. One can select by an applicationof the Hahn-Banach theorem suitable functionals g α ∈ X ∗ to obtain a biorthogonalsystem { ( y ζ α , g α ) } α<ω . This contradicts KS , so that we have obtained the claim. (cid:3) Proof of Theorem 4.4.
Suppose that X fails CSP and that
F ⊂ S X ∗ is a separatingset without any countable separating subset. Then by using the proof of Theorem4.1 we obtain the following: There exists a family { E σ } σ<λ of closed subspacessuch that E β ( E α whenever α < β < λ . Here λ is uncountable but unlike inthe proof of Theorem 4.1, here we do not need any control over the intersection T σ E σ or the cofinality of λ . Moreover, similarly as in the proof of Theorem 4.1,we may choose { E σ } σ in the following manner. Namely, that there exists a family { f φ ( γ ) } γ<ω ⊂ F such that E σ = T γ<σ Ker f φ ( γ ) for each σ ≤ ω . We may assignfor each σ < ω such x σ ∈ E σ \ E σ +1 that E σ +1 + [ x σ ] = E σ and f σ ( x σ ) = 1.By the selection of { E σ } σ and canonical identifications we get that E ω = \ γ<ω Ker f φ ( σ ) and f φ ( γ ) ∈ (X /E ω ) ∗ = E ⊥ ω , for γ < ω . Put F = \ β<ω span( { x σ | β < σ < ω } )and observe that F ⊂ E ω , since span( { x σ | β < σ < ω } ) ⊂ T σ<β Ker( f φ ( σ ) ) for β < ω . Write ˆ x σ = x σ + F ∈ X /F for σ < ω . Note that(4.3) \ β<ω span( { ˆ x σ | β < σ < ω } ) = { } ⊂ X /F. Since X /F satisfies ( B ) by the assumptions, we may apply the proof of Propo-sition 4.5 with Γ σ = [ σ, ω ] for σ < ω to obtain that X /F admits a biorthogonalsystem of length ω . This can be lifted to obtain a corresponding biortohogonalsystem in X. (cid:3) Regarding the assumption dens(X) < ℵ ω , note that if dens(X) > ω , then ω ∗ − dens(X ∗ ) > ω and X fails KS (see e.g. [4, p.97-98]).5. Conclusions: examples, remarks and renormings
Let us briefly recall the list of CSP spaces mentioned here: Separable spaces, theJohnson-Lindenstrauss spaces JL and JL , the duals JT ∗ , JF ∗ due to Lindenstrauss-Stegal, Shelah’s space S under ♦ ( ℵ ) and C ( K ) spaces, where K is the double-arrowspace, Kunen’s compact under CH, or any scattered separable countably tight com-pact.Observe that the non-separable spaces above do not admit a system (2.2). Proposition 5.1.
Let X be a Banach space with CSP. Then X admits an equivalentuniformly Gateaux (UG) norm if and only if X is separable. Proof.
Each separable space X admits an equivalent UG norm (see [21, Cor.6.3]).Conversely, each space X with an equivalent UG norm is weakly countably deter-mined and thus WLD (see [21, Thm.6.5,Thm.3.8]). As WLD spaces are Plichko([8, Thm.1]) we obtain by Example 2.8 that X is in fact separable. (cid:3)
A similar result does not hold for LUR renormings even for dual spaces. Thespace JT ∗ is a non-separable CSP space, which admits an equivalent LUR norm bythe three space property of LUR renormings (see [21, p.1758,1785]). On the otherhand, for Kunen’s compact K the space C ( K ) does not admit a Kadets-Klee andin particular not a LUR norm (see [21, p.1794]).Often the dual spaces behave better than their underlying spaces, so it is rea-sonable to restate the following known question. Problem 5.2.
Does any dual space X ∗ with CSP, i.e. a space X ∗ such that X isseparable and does not contain ℓ isomorphically, admit an equivalent (not neces-sarily dual) LUR norm? A positive answer to the previous question would partly generalize the followingresult. If X is an Asplund space then X ∗ has an equivalent (not necessarily dual)LUR norm (see [21, Thm. 7.13]). Note that any non-separable CSP dual space is not LUR by ω ∗ -Kadets-Klee property of X ∗ .If X ∗ has the RNP and CSP, then X is a separable Asplund space by Theorem3.4 and the duality of the Radon-Nikodym and Asplund properties. This meansthat X ∗ must be separable. Problem 5.3.
Are all spaces X with CSP and the RNP in fact separable? We would like to emphasize the significance of the following problems.
Problem 5.4. If X and Y have CSP, does it follow that X ⊕ Y has CSP? If so, isCSP a three-space property, i.e. does X have CSP whenever X/ Y and Y ⊂ X haveCSP? Problem 5.5.
Does there exist a CSP space without property ( C ) ? Proposition 5.6. If X ∗ and Y ∗ are dual spaces with CSP, then X ∗ ⊕ Y ∗ has CSP.Proof. First we note that the property of being a predual of a CSP space is athree-space property. Indeed, the property of simultaneous separability and non-containment of ℓ isomorphically is a three-space property (see [2, 2.4.h,3.2.d]).Thus we may apply Theorem 3.4.Now, let X and Y be some preduals of X ∗ and Y ∗ , respectively. Since being apredual of a CSP is a three space property, we get that X ⊕ Y is a predual of a CSPspace. The dual (X ⊕ Y) ∗ is thus a CSP space, which is isomorphically X ∗ ⊕ Y ∗ . (cid:3) Acknowledgements.
I am grateful to Heikki Junnila for invaluable advice.
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