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COMPLEXITY OF WEAKLY NULL SEQUENCES

arXiv:math/9202204v1 [math.FA] 28 Feb 1992COMPLEXITY OF WEAKLY NULL SEQUENCESDale E. Alspach and Spiros ArgyrosAbstract. We introduce an ordinal index which measures the complexity of aweakly null sequence, and show that a construction due to J. Schreier can be it-erated to produce for each α < ω1, a weakly null sequence (xαn)n in Cωωαwithcomplexity α.

As in the Schreier example each of these is a sequence of indicatorfunctions which is a suppression-1 unconditional basic sequence.These sequencesare used to construct Tsirelson-like spaces of large index. We also show that thisnew ordinal index is related to the Lavrentiev index of a Baire-1 function and usethe index to sharpen some results of Alspach and Odell on averaging weakly nullsequences.0.

IntroductionIn this paper we investigate the oscillatory behavior of pointwise convergingsequences. Our main tool is a new ordinal index which measures the oscillationof such a sequence.

We show that there are weakly null sequences of indicatorfunctions in C(K) with arbitrarily large oscillation index and that the oscillationindex is smaller than other similar ordinal indices.In particular the oscillationindex of a pointwise converging sequence is directly compared to the Lavrentievindex of its limit, the ℓ1-index defined by Bourgain, and the averaging index. Manyof the results are directly related to those in [A-O] where the averaging index isused and we extend some results of that paper.The first example of a weakly null sequence with no subsequence with averagesgoing to zero in norm was constructed by Schreier [Sch].

His example is a sequenceof indicator functions and, as observed by Pelczynski and Szlenk [P-S], can berealized on the space of ordinals less than or equal to ωω in the order topology.Because these are indicator functions the failure of the Banach-Saks property issolely dependent on the intersection properties of the sets. Our construction is alsobased on intersection properties of sets.

Thus one viewpoint on the constructionsin this paper is that there are families of sets with very complicated intersectionproperties. The purpose of the ordinal index is to measure the complexity of theseintersection properties.

The examples of weakly null sequences that we construct,like Schreier’s example, are suppression-1 unconditional basic sequences and can beconsidered as generalizations of Schreier’s construction.Let us note that there is a strong relationship between this work and some unpub-lished results of Rosenthal on the unconditional basic sequence problem. Rosenthal1991 Mathematics Subject Classification.

46B20.Key words and phrases. ordinal, index, complexity, unconditional, L1-predual, C(K)space.Research of the first author was supported in part by NSF grant DMS-8602395.Typeset by AMS-TEX1

2DALE E. ALSPACH AND SPIROS ARGYROSshowed that any weakly null sequence of indicator functions in C(K) has a sub-sequence which is an unconditional basic sequence. In our work we found thatRosenthal’s notion of weakly independent sets fits naturally into our viewpoint andwe have incorporated some of Rosenthal’s work into the exposition.

While we be-lieve that our work explains some difficulties with unconditionality, it is based onintersection properties which cannot provide a complete explanation. For example,in contrast to Rosenthal’s result on sequences of indicator functions, it is knownthat given a weakly null sequence one cannot always find a subsequence which is anunconditional basic sequence [M-R] and [O].

Thus understanding the unconditionalbasic sequence problem is more complicated than just understanding the oscilla-tion properties of weakly null sequences. On the other hand it is clear that theoscillation properties do play a fundamental role in unconditionality.

It would beinteresting if there were some way to incorporate these more subtle properties ofweakly null sequences into an ordinal index.The paper is organized into six sections. In the first we recall the definitions ofsome ordinal indices and trees.

In the second we introduce the oscillation indexand an essentially equivalent index which we call the spreading model index. Inthe third we prove that the oscillation index is essentially smaller than the ℓ1-indexand show that it is related to the Lavrentiev index of a Baire-1 function.

In thefourth section we define for each countable ordinal α a weakly null sequence ofindicator functions and compute the oscillation index of the sequence and the sizeof the smallest C(K) space which can contain the sequence. In the fifth sectionwe use an idea of Odell to show that the construction in Section 4 can be used toconstruct reflexive spaces similar to Tsirelson space with large oscillation index.

Inthe sixth section we show that the averaging index, which has a definition similarto that of the spreading model index, is not smaller than the ℓ1 index. In particulara space is constructed which does not contain ℓ1 but has averaging index ω1.

Wealso extend some results from [A-O] in order to better characterize those sequenceswhich can be averaged a predictable finite number of times in order to get a weaklynull sequence.We will use standard terminology and notation from Banach space theory asmay be found in the books of Lindenstrauss and Tzafriri [L-T,I] and [L-T,II] andDiestel [D1] and [D2]. If α is an ordinal, we will use α rather than α + 1 to denotethe space of ordinals less than or equal to α in the order topology and C(α) todenote the space of continuous functions on α.1.

PreliminariesIn this section we will recall the definitions of the ℓ1-index of Bourgain, theSzlenk index, and the averaging index. In order to define the Bourgain ℓ1-index wefirst need to define trees and some related notions.Definition.

Given a set S a tree T on S is subset of ∪∞n=1Sn∪S∞such that if b ∈Tthen any initial segment of b is also in T , i.e., if b ∈T and b = (s1, s2, . .

. , sn, sn+1)or b = (s1, s2, .

. .

, sn, sn+1, . .

. ), then (s1, s2, .

. .

, sn) ∈T . We will call the elementsof the tree nodes and say that the node b, as above, is a successor of or is below(s1, s2, .

. .

, sn). If b and c are nodes, we will write b > c to indicate that b is belowc.

In particular (s1, s2, . .

. , sn, sn+1) is an immediate successor of (s1, s2, .

. .

, sn).If a node x ∈Sn then n is the length or level of x. A branch of a tree is a maximallinearly ordered subset of the tree under this natural initial segment ordering.

A

COMPLEXITY OF WEAKLY NULL SEQUENCES3subtree of a tree T is a subset of T which is a tree. A tree T is finitely branchingif the number of immediate successor nodes of any node is finite.

It is dyadic ifthis number is at most two for all nodes. Finally a tree is well-founded if all of itsbranches are of finite cardinality.For well-founded trees there is a standard way to define the ordinal index of thetree.Definition.

Suppose that T is a well-founded tree on a set X. Let T0 = T andfor each ordinal α defineT α+1 = ∪∞n=1(x1, x2, .

. .

, xn) ∈T α : there is an x ∈X such that(x1, x2, . .

. , xn, x) ∈T α.For a limit ordinal α let T α = ∩β<αT β.

Let o(T ) be the smallest ordinal γ suchthat T γ = ∅. This is the order of the tree T .Note that if x = (x1, x2, .

. .

, xj) ∈T α\T α+1, the treeTx = ∪∞n=1 {(y1, y2, . .

. , yn) : x + (y1, y2, .

. .

, yn)} .where “+” indicates concatenation, is of order α.Now we are ready to define the ℓ1-index of Bourgain.Definition. Let X be a Banach space and δ > 0.

LetT (X, δ) = ∪∞n=1x1, x2, . .

. , xn∈Xn : ||xj|| ≤1 for all j ≤n and||Xjλjxj|| ≥δX|λj| for all (λj) ∈Rn.The δ ℓ1-index of X is ℓ(X, δ) = o(T (X, δ)).

If T (X, δ)α ̸= φ for all countable αthen ℓ(X, δ) = ω1.It is easy to see that if ℓ1 is isomorphic to a subspace of X then ℓ(X, δ) = ω1 forsome δ > 0 and thus by a result of James [J] for all δ, 0 < δ < 1. The converse isalso true but requires the fact that if T α(X, δ) ̸= ∅for all countable α then the treehas an infinite branch.

(See [Bo].) Bourgain observed that the examples employedby Szlenk [Sz] to show that there are separable reflexive spaces with Szlenk indexgreater than any given countable ordinal also have large ℓ1-index.

Note that thisindicates that the Baire-1 functions in X∗∗provide only a very weak indication asto the ℓ1-index of X.Next we will define Bourgain’s index for Boolean independence.This indexis a technical convenience which we will use in establishing lower bounds on theℓ1-index.Definition. Let K be a set and {(An, Bn)} be a sequence of pairs of subsets ofK.

LetT ({(An, Bn)}) = ∪∞k=1(n1, n2, . .

. , nk) ∈Nk :for all (ǫ1, ǫ2, .

. .

, ǫk) ∈{−1, 1}k, ∩ki=1ǫiAni ̸= ∅,

4DALE E. ALSPACH AND SPIROS ARGYROSwhere (−1)Ai = Bi. Let B({(An, Bn)}) be the order of the tree T ({(An, Bn)}).We will usually take pairs of disjoint sets (An, Bn) when we employ this index.Note that by Rosenthal’s theorem on ℓ1, [R], a bounded sequence (xn) in a Banachspace X has a subsequence equivalent to the unit vector basis of ℓ1 if (and onlyif) it has no weak Cauchy subsequence.

Moreover for any such sequence there is asubsequence (xn)n∈M and two numbers δ > 0 and r such that ifAn = {x∗∈X∗: x∗(xn) ≥r + δ, ||x∗|| ≤1}andBn = {x∗∈X∗: x∗(xn) ≤r, ||x∗|| ≤1}then T ({(An, Bn) : n ∈M}) = ∪∞k=1M k and thus B ({(An, Bn) : n ∈N}) = ω1.Hence if we consider such sets (An, Bn), the index is ω1 for some δ > 0 and r if andonly if (xn) has a subsequence with no weak Cauchy subsequence. In particular if(xn) converges w∗sequentially to some x∗∗∈X∗∗the index is countable.Now we wish to recall some other ordinal indices which are more classical inspirit.

Each of these indices is defined in terms of the oscillation of sequences offunctions on the unit ball of the dual in the w∗topology. First we recall the Szlenkindex.

Let X be a Banach space and for each α < ω1 letPα+1 (ε, BX, BX∗) =x∗: there exists x∗n ∈Pα (ǫ, BX, BX∗) andxn ∈BX such that x∗nw∗−−→x∗, xnw−→0, and lim x∗n(xn) ≥εAt a limit ordinal β we take the intersection, that is,Pβ (ǫ, BX, BX∗) = ∩α<βPα (ǫ, BX, BX∗) .Now fix {ej} a normalized weakly null sequence in X. The next two ordinalindices are defined for each such sequence.

The first occurred in a paper of Zal-cwasser [Z] and in a paper of Gillespie and Hurwitz [G-H] and was used to provethat a pointwise converging sequence of bounded continuous functions on a com-pact metric space has a sequence of convex combinations going to zero in norm.LetZα+1 (ǫ, {ej} , BX∗) =x∗: there exists x∗n ∈Zα (ǫ, {ej} , BX∗) andejn such that x∗nw∗−−→x∗and lim |x∗n(ej2n) −x∗n(ej2n−1)| ≥ǫ.As before Zβ = ∩α<βZα, for a limit ordinal β.In [A-O] the following index was introduced in order to obtain some more preciseinformation about the nature of the convex combinations obtained in these earlypapers in the context of weakly null sequences in a Banach space.Let Aα+1 (ǫ, {ej} , BX∗) =x∗: for every neighborhood N of x∗relative to Aα (ǫ, {ej} , BX∗) there exists an infiniteset L ⊂N such that ℓ1 −SPei|Ni∈L≥ǫ

COMPLEXITY OF WEAKLY NULL SEQUENCES5whereℓ1 −SP ({ei}) =limk limm inf(k−1kXi=1eni : m ≤n1 < n2 < . .

. < nk).As before Aβ = ∩α<βAα, if β is a limit ordinal.

We will refer to this as the averagingindex.In each case the successor set (if non-empty) is a w∗closed nowhere dense subsetof the set. (For the case of the Szlenk sets we need to assume that X∗is separable.

)Thus the Baire Theorem gives in each case a largest ordinal α with the αth set non-empty and the (α + 1)th empty. Denote o(P, ǫ), o(Z, ǫ), and o(A, ǫ) as the indicesfor the Szlenk, Zalcwasser, and average sets, respectively.

Clearly o(P, ǫ) ≥o(Z, 2ǫ)and o(Z, ǫ) ≥o(A, ǫ).2. Weakly Null Sequences and the ℓ1-indexWe wish to introduce an index similar to the Szlenk index which will measureℓ1-ness in a different way than the ℓ1-index of Bourgain.Definition.

Let (K, d) be a Polish space and let (fn) be a pointwise convergentsequence of continuous functions on K. Fix ǫ > 0 and letA+n,m = {k ∈K : fn(k) −fm(k) > ǫ}andA−n,m = {k ∈K : fn(k) −fm(k) < −ǫ} ,For each countable ordinal α we define inductively a subset of K byO0 (ǫ, (fn) , K) = K,Oα+1 (ǫ, (fn) , K) =k ∈Oα (ǫ, (fn) , K) : for every neighborhood N of k there is anN ∈N such that for all n ≥N there exists an M ∈N such that∩m≥M A+n,m ∩Oα (ǫ, (fn) , K) ∩N ̸= ∅or∩m≥M A−n,m ∩Oα (ǫ, (fn) , K) ∩N ̸= ∅}.If β is a limit ordinal, Oβ (ǫ, (fn) , K) =∩α<βOα (ǫ, (fn) , K).Note that if (fn) converges pointwise to 0 and ǫ′ < ǫ, then for large enough M{k : fn(k) ≥ǫ′} ⊃∩m≥MA+n,m ⊃{k : fn(k) ≥ǫ}Thus Oα (ǫ, (fn) , K) is essentially the Szlenk index set, Pα (ǫ, (fn) , K), except thatwe do not allow the use of subsequences. Thus Pα (ǫ′, (fn) , K) ⊃Oα (ǫ, (fn) , K)in this case for all ǫ′ < ǫ.Also it is easy to see that Oα (ǫ, (fn) , K) is alwaysclosed.

If the limit f is not continuous, the relationship with the Szlenk sets isless clear. However in the definition of the Szlenk sets it is possible to use weakCauchy sequences in place of weakly null sequences.

The sets obtained in this waybehave in essentially the same way as the original Szlenk sets provided the dual isseparable. Thus if [fn]∗is separable Oα (ǫ, (fn) , K) will be empty for some α < ω1.Actually the index is always countable but this will be a consequence of our resultrelating this index to the ℓ1-index.This discussion (and some later results) justifies the following.

6DALE E. ALSPACH AND SPIROS ARGYROSDefinition. Suppose (fn) is a pointwise converging sequence of continuous func-tions on a Polish space K. O(ǫ) = O (ǫ, (fn) , K) is the largest ordinal α such thatOα (ǫ, (fn) , K) is non-empty.

We will refer to O(ǫ) as the ǫ oscillation index ofthe sequence (fn). (Some authors, e.g., [K-L], use the term oscillation index for anordinal index defined in terms of the oscillation of a fixed function near a point.Here we only apply the term to sequences so no confusion should result.

)It will be convenient to use a more restrictive index at times.LetO0+ (ǫ, (fn) , K) = K.For each countable ordinal α define inductively a subsetof K byOα+1+(ǫ, (fn) , K) =k ∈Oα+ (ǫ, (fn) , K) : for every neighborhood N of kthere is an N ∈N such that for all n ≥N there existsan M ∈N such that ∩m≥M A+n,m ∩Oα+ (ǫ, (fn) , K) ∩N ̸= ∅.If β is a limit ordinal, Oβ+ (ǫ, (fn) , K) = ∩α<βOα+ (ǫ, (fn) , K). As above the posi-tive ǫ oscillation index O+(ǫ) is the largest ordinal α such that Oα+ (ǫ, (fn) , K) ̸= ∅.In a similar way we define Oα−(ǫ, (fn) , K) and O−(ǫ, (fn) , K).Remark 2.1.

We do not need to have the sequence converging pointwise for thedefinition of the index to make sense. However, if (fn) has no pointwise convergentsubsequence then, by Rosenthal’s ℓ1 theorem [R], there is an infinite set M andreal numbers δ > 0 and r such that ({k : fn(k) ≥r + δ} , {k : fn(k) ≤r})n∈M isBoolean independent.

From this it follows that there is a Cantor set C ⊂K sothat relative to C and taking n and m from M, ∩m>nAn,m is an ǫn-net in C andǫn →0 as n →∞. Hence O1 δ/2, (fn)n∈M , C= C. Thus for this subsequencethe oscillation index is ω1.Definition.

The ǫ-oscillation index of a Banach space is the supremum of the ǫ-oscillation index over all weak Cauchy sequences in the unit ball of the space asfunctions on the dual ball with the w∗topology.The next lemma follows easily from the definitions but we state it for futurereference.Lemma 2.2. If K is a w∗closed subset of the dual ball of X, then Oα (ǫ, (fn) , K) ⊂Oα (ǫ, (fn) , BX∗) for all α and ǫ.

In particular if T is a bounded operator from aBanach space X to a Banach space Y , thenT ∗Oα (ǫ, (T fn) , BY ∗) = Oα (ǫ, (fn) , T ∗BY ∗) ⊂Oα (ǫ/||T ||, (fn) , BX∗) ,and consequentlyO (ǫ, (T fn) , BY ∗) ≤O (ǫ/||T ||, (fn) , BX∗) .On the other hand we do not know if there is an essential equivalence betweensupM Oε, (fn)n∈M , BX∗and supM Oελ, (fn)n∈M , Kif K is a λ norming sub-set of BX∗. Simple examples show that considering subsequences is necessary inorder for the indices to be approximately the same size.Next we will present some examples in which we can compute some bounds onthe oscillation index and compare this index to the Bourgain ℓ1-index.

The firstnontrivial example we wish to present is the Schreier sequence, [Sch].

COMPLEXITY OF WEAKLY NULL SEQUENCES7Example 1. Let F1={F ⊂N : min F ≥cardF} ∪{∅}.We will identifyF1 with {1F : F ∈F1} and note that the later is a countable compact (metric)space in the topology of pointwise convergence on N. Define fn(F) = 1F (n).

Be-cause each F is finite (fn) converges pointwise to 0 on F1. It is easy to see thateach fn is continuous on F1.

A computation shows thatOj(1, (fn), F1) = ∪m>j {F ∈F1 : min F ≥m and card F ≤m −j} ∪{∅}.Thus Oω(1, (fn), K) = {∅} and O(1) = ω. Because these are indicator functionsthe same is true for all ǫ, 1 ≥ǫ > 0.

Note that this is the maximal index because F1is homeomorphic to ωω in the order topology and hence there are only ω non-emptytopological derived sets.It follows from the combinatorial properties of F1 that [fn] contains ℓ1′n s uni-formly and thus ℓ([fn]) ≥ω. (See [Pel-Sz] or Section 4.) Also it is not hard to seethat the ǫ-oscillation index of C(ωω) is essentially the same as the Szlenk index,ω 1ǫ.Example 2.

Tsirelson space T . (See [C-S] or Section 5.) The natural unit vectorbasis of T is a weakly null sequence.

T is reflexive and contains many ℓ1′n s. Thusℓ(T, δ) ≥ω, for δ =14, for example. The computation of Oj (ǫ, (en) , BT ∗) doesnot seem to be easy.

However note that the functionals 1F for F ∈F1 in example1 above act on T by 1fXanen=Xn∈Fan and all ||1F ||T ∗≤2.Moreover if1Fk →1F pointwise on N, then 1Fk(x) →1F (x) for all x ∈T . Therefore themap S : T →C (F1) = C(ωω) defined by Sx(F) = 1F (x) is well defined andbounded by 2 and T en = fn.It follows from Example 1 and Lemma 2.2 thatOω 12, (en) , BX∗̸= ∅.Example 3. ℓ1.

There are no non-trivial weakly null sequences in ℓ1 and thus theoscillation index is 0. On the other hand the ℓ1-index is ω1.The anomalous behavior of the index for ℓ1 can be corrected if we allow generalsequences and modify the definition of Oα+1.

However this seems pointless in viewof Rosenthal’s characterization of ℓ1.Remark 2.3. Suppose that (fn) is a pointwise converging sequence of uniformlybounded continuous functions on a compact metric space K. Then the mappingS from K into c (the space of convergent sequences under the sup norm) S(k) =(fn(k)) is continuous if the range is given the σ(c, ℓ1(N)) topology.

If the functions(fn) separate points on K then S is also one-to-one.From this viewpoint weare investigating σ(c, ℓ1(N)) compact subsets of the ball of c. The oscillation setOα (ǫ, (fn) , K) is mapped by S to the set Oα (ǫ, (en) , S(K)), where en denotesthe functional evaluation at n, and thus the index may be computed in c. If thesequence (fn) converges to 0 then the sets are actually in c0 and the topologyσc0, ℓ1(N)is the weak topology.3. Comparison with the ℓ1-indexThe ℓ1-index defined by Bourgain [Bo] measures the degree to which ℓ1 isomor-phically embeds in the space.

Bourgain showed that this index is related to anordinal index of Baire-1 functions in the second dual. In the next section we will

8DALE E. ALSPACH AND SPIROS ARGYROSshow that the presence of certain weakly null sequences can also raise the ℓ1-index.Thus in contrast to the Baire-1 case the pointwise limit itself provides no infor-mation but rather the sequence carries the information. Our method is to use theoscillation index defined in the previous section.

In this section we will show thatthis oscillation index is essentially bounded above by the ℓ1-index, and prove thatany sequence of continuous functions converging pointwise to a Baire-1 function ofindex α must have a large oscillation index. We then get Bourgain’s result as acorollary.

These ideas are also related to some work of Haydon, Odell, and Rosen-thal [H-O-R] on what they term Baire-1/2 and Baire-1/4 functions in the seconddual.Now we will prove that a large oscillation index implies a large ℓ1-index. For anordinal α = ωγk + β with β < ωγ and k ∈N, we define α/2 = ωγ[(k + 1)/2], where[ ] denotes the greatest integer function.Theorem 3.1.

If (fn) is a pointwise converging sequence on a compact metricspace K and Oα (ǫ, (fn) , K) ̸= ∅then if ǫ′ < ǫ, ℓ([fn] , ǫ′/2) ≥α/2. Moreoverthere is an ℓ1-index tree on (fn) with index α/2.The proof is easier if we assume that Oα+ (ǫ, (fn) , K) ̸= ∅, and in this case we getα rather than α/2 for the ℓ1-index.

The proof naturally divides into two parts: firsta reduction to the case Oα/2+̸= ∅and then the proof of the result in this case. Thereduction is proved as Lemma 3.4.

The main idea in the remainder of the argumentis to construct a tree of Boolean independent pairs of sets with large order wherethe pairs of sets are subsets of the A+n,m′s. Consequently we will actually constructour ℓ1 tree on (fn −fm)n,m∈N with constant ǫ/2.

A final argument is needed toget a ℓ1 tree on (fn).Before we proceed to the proof we need a few lemmas which describe somesufficient conditions for a tree to be a ℓ1-tree. The following is an unpublishedlemma of Rosenthal.

(We actually only use the weaker version in which r does notdepend on n.)Lemma 3.2. Suppose (fn)mn=1 is a finite sequence of norm one functions on K, δ >0, and for each n there is a number rn such that if An = {fn ≥rn + δ} andBn = {fn ≤rn} , {(An, Bn)} is Boolean independent.

Then (fn) is 2/δ equiva-lent to the unit vector basis of ℓ1m.Proof. Suppose that an ∈R for all n. Let F = {n : an ≥0} and G = {n : an < 0}.Let t ∈∩n∈F An ∩∩n∈GBn and t′ ∈∩n∈GAn ∩∩n∈F Bn.

Then|Xanfn(t) −Xanfn(t′)| ≥Xan[fn(t) −fn(t′)] ≥X|an|δ.Indeed, if n ∈F, an ≥0, fn(t) ≥rn + δ and fn(t′) ≤rn and if n ∈G, an < 0,and fn(t) ≤rn and fn(t′) ≥rn + δ. Hence either | P anfn(t)| ≥P |an|δ/2 or| P anfn(t′)| ≥P |an|δ/2.□We would like to construct a Boolean independent tree of pairs of subsets ofK where the pairs are related to the elements of the sequence (fn).

If we could,for example, choose sets of the form An and Bn as in the lemma above, we wouldimmediately get a ℓ1 tree on (fn). Actually somewhat less is sufficient.

COMPLEXITY OF WEAKLY NULL SEQUENCES9Lemma 3.3. Suppose that (xn) is a sequence of functions on a set K with valuesin [-1,1] and T is a tree on (xn) of order α < ω1.

Further assume that there is aδ > 0 such that for each branch B of T there is a mapping ρB : B →2K × 2K suchthata) if ρB(x1, x2, . .

. , xn) = (A, B) then A ⊂{xn ≥r + δ} and B ⊂{xn ≤r} forsome r ∈R.

(r may depend on n and B . )b) ρB(B) is a Boolean independent sequence of sets.Then T is an ℓ1 tree of order α with constant δ/2.Proof.

According to the previous lemma if B = {(x1) , (x1, x2) , . .

. } then (xi) is2/δ equivalent to the unit vector basis of ℓ1.

Thus each branch of T also satisfiesthe requirements for an ℓ1 tree, and thus we have an ℓ1-tree of order α.□Our next lemma allows us to reduce to the case of positive oscillation index.Below p(α) = inf {β + ρ : ρ + β = α}. In particular if α = ωγk + β, where β < ωγ,then p(α) = ωγk.Lemma 3.4.

If Oα (δ, (fn) , K) ̸= ∅and γ ≤α/2, then for ǫ = + or −,Oγǫδ,fnj, K̸= ∅, for some subsequencefnj.Proof. The proof is by induction on α.We will actually prove that ift ∈Oα (ǫ, (fn) , K) then there is an infinite set L ⊂N and ordinals γ and λ suchthatmin (γ + λ, λ + γ) ≥p(α)andt ∈Oγ+δ, (fn)n∈L , K∩Oλ−δ, (fn)n∈L , K.Suppose that p(α) = ωv · k. Because the result depends only on p(α), we needonly consider ordinals α of the form ωv · k in the induction.If α = 1, let t ∈Oα (δ, (fn) , K) and for ǫ = + or −and i ∈N consider the setN iǫ = {n : there exists an M ∈N such that ∩m≥M Aǫnm ∩Ni ̸= ∅}where Ni is a decreasing sequence of neighborhoods with intersection {t}.

For atleast one choice of ǫ, N iǫ is infinite for all i. For that ǫ let L be an infinite subsetof N such that L\N iǫ is finite for each i.

Clearly t ∈O1ǫδ, (fn)n∈L , K.Now suppose that the lemma is true for all β < α and let t ∈Oα (δ, (fn) , K). Ifα = ωv, let αi ↑α.

The inductive assumption implies that there are sequences γiand λi and Li ⊂N such thatmin (γi + λi, γi + λi) ≥p(αi)andt ∈Oγi+δ, (fn)n∈Li , K∩Oλi−δ, (fn)n∈Li , K.We may assume that Li ⊂Li−1 for all i and that the sequences (γi) and (λi) arenon-decreasing. It now follows that if L is an infinite subset of N such that L\Li isfinite for all i thent ∈Oγ+δ, (fn)n∈L , K∩Oλ−δ, (fn)n∈L , K,where γ = lim γi and λ = lim λi.

We have that min (γ + λ, λ + γ) ≥p(αi) for all iand min (γ + λ, λ + γ) ≥lim p(αi). Thus if p(αi) is increasing to p(α), we are done.

10DALE E. ALSPACH AND SPIROS ARGYROSNow we may assume that p(αi) = ωv · k for all i and p(α) = ωv · (k + 1). (Thisargument will also apply to the successor ordinal case.) We have that for eachs ∈Oωv·k (δ, (fn) , K) there is an infinite set Ls and ordinals γs, λs such thats ∈Oγs+δ, (fn)n∈Ls , K∩Oλs−δ, (fn)n∈Ls , Kandmin (γs + λs, λs + γs) ≥ωv · k.Let S = {s(m) : m ∈N} be a countable dense subset of Oωv·k (δ, (fn) , K).

For eachm ∈N there is an infinite set Lm ⊂Lm−1 ⊂N and ordinals γm and λm such thats(m) ∈Oγm+δ, (fn)n∈Lm , K∩Oλm−δ, (fn)n∈Lm , Kandmin (γm + λm, λm + γm) ≥ωv · k.By a diagonalization argument we may assume that the set L does not depend onm and that γm = ωv · km and λm = ωv · jm. HenceOωv·k (δ, (fn) , K) ⊂∪i+j=kOωv·i+δ, (fn)n∈L , K∩Oωv·j−δ, (fn)n∈L , K.Because each of these sets is closed, we need only consider those sets which containt.

Moreover observe that ift ∈Oωv·i+δ, (fn)n∈L , K∩Oωv·j−δ, (fn)n∈L , Kfor two different pairs (i, j) and (i′, j′) thent ∈Oωv·i′′+δ, (fn)n∈L , K∩Oωv·j′′−δ, (fn)n∈L , Kwhere i′′ = max (i, i′) and j′′ = max (j, j′) and ωv·i′′+ωv·j′′ ≥p(α), as required. Ifthere is only one such pair (i, j), we may assume that a neighborhood of t (relative toOωv·k (δ, (fn) , K)is contained in Oωv·i′′+δ, (fn)n∈L , K∩Oωv·j′′−δ, (fn)n∈L , K.We have thatt ∈Oωv (δ, (fn)) , Oωv·i′′+δ, (fn)n∈L , K∩Oωv·j′′−δ, (fn)n∈L , K and thus the case α = ωv gives thatt ∈Oωvǫδ, (fn)n∈L′ , Oωv·i′′+δ, (fn)n∈L , K∩Oωv·j′′−δ, (fn)n∈L , Kfor some infinite L′ ⊂L and ǫ = + or −.

Hencet ∈Oωv·i+δ, (fn)n∈L′ , K∩Oωv·j−δ, (fn)n∈L′ , K,where i = i′′ + 1 and j = j′′ if ǫ = +, and i = i′′ and j = j′′ +1 if ǫ = −, asrequired.□Now we are ready to begin the proof of Theorem 3.1.

COMPLEXITY OF WEAKLY NULL SEQUENCES11Proof. By Lemma 3.4 we need only prove it in the simpler case indicated above,i.e., suppose that O+(ǫ) ≥α.

Let f(t) = lim fn(t). Fix ρ, ǫ/8 > ρ > 0.

For anyg ∈C(K) let C(g, k0) = {k : |g(k) −f(k0)| < ρ}.The proof is by induction on α. As usual we must actually prove a little strongerstatement to make the induction work.INDUCTIVE HYPOTHESIS: Suppose that k1, k2, .

. .

, kj are a finite number ofpoints in Oα+ (ǫ, (fn) , K) and N1, N2, . .

. , Nj are neighborhoods of k1, k2, .

. .

, kj,respectively, then for every β < α there is a tree T on (fn −fm) of order β suchthat for i = 1, 2, . .

. , j,TNi =nx1|Ni , x2|Ni, .

. .

, xn|Ni: (x1, x2, . .

. , xn) ∈To,is an ℓ1 tree of order β as in Lemma 3.3, i.e., for each i and branch there is amapping into the pairs of subsets of K satisfying a) and b) with δ = ǫ/2 andr = ǫ/4.

Moreover we may assume that for each pair of setsρBfn1 −fm1, . .

. , fnj −fmj= (Aj, Bj) ,Aj ⊂∩j−1i=1 C (fni, k) ∩C (fmi, k) ∩Cfmj, k∩s : |fnj(s) −fnj(k)| < ρ,andBj ⊂∩ji=1C (fni, k′) ∩C (fmi, k′) for some k and k′ in K.Assume the inductive hypothesis holds for all β < α.

First suppose that α is nota limit ordinal and that Oα+ (ǫ, (fn) , K) ̸= ∅. Let (ki) and (Ni) be as above.

Foreach i there is an Ni ∈N such that for each n ≥Ni, there is an M in such that∩m≥MinA+n,m ∩Oα−1+(ǫ, (fn) , K) ∩Ni ̸= ∅.Let n ≥max {Ni} such that |fn (ki) −f (ki) | < ρ, let M = maxM inand choosea point k′i ∈∩m≥MA+n,m ∩Oα−1+(ǫ, (fn) , K) ∩Ni, for each i. We may also assumethat for all p ≥M,|fp (ki) −f (ki) | < ρand|fp (k′i) −f (k′i) | < ρ,for i = 1, 2, .

. .

, j. Fix m ≥M and for each i let N ′i be a neighborhood of k′icontained inNi ∩A+n,m ∩C (fm, k′i) ∩{k : |fn(k) −fn (k′i) | < ǫ/8}and let N ′′i be a neighborhood of ki contained inNi ∩C (fm, ki) ∩C (fn, ki) .Now by the inductive hypothesis for every β < α there is a tree T on (fp −fq)such that TN ′i and TN ′′i are ℓ1 trees of order β for i = 1, 2, .

. .

, j, and satisfy a) andb) of the lemma with δ = ǫ/2 and r = ǫ/4. LetT ′ = {(fn −fm, x1, x2, .

. .

, xn) : (x1, x2, . .

. , xn) ∈T } .

12DALE E. ALSPACH AND SPIROS ARGYROSClearly this is a (β + 1) tree on (fn). We need to check the hypothesis of the lemmafor T ′Ni, i = 1, 2, .

. .

, j. To define ρ′B′ on a branchB′ = {(fn −fm) , (fn −fm, x1) , (fn −fm, x1, x2) .

. . } of T ′Ni,we letρ′B′ (fn −fm, x1, x2, .

. .

, xn) = ρB ((x1, x2, . .

. , xn))andρ′B′ ((fn −fm)) = (N ′i , N ′′i ) ,where ρB denotes the mapping from the branch B = {(x1) , (x1, x2) , .

. . } of TNi.

Ifr = ǫ/4 and δ = ǫ/2, the hypothesis of the lemma is satisfied and thus T ′Ni is anℓ1-tree of order β + 1. Because this is true for all β < α we get an (α + 1) ℓ1-tree.For limit ordinals the conclusion is obvious.

(Note we can actually get an (α+1)ℓ1-tree in this case as well. )The moreover assertion allows us to conclude that the we can construct a ℓ1-treeon (fn) of the same order.

Indeed we claim that the tree obtained by replacingin each coordinate fn −fm by fn is the required tree. First if f is continuouswe can choose a neighborhood N of the point k0 ∈Oα (ǫ, (fn) , K) so that N ⊂{k : |f(k) −c| < ρ}, where c = f (k0), and restrict all of the functions to this setN.

The proof then shows that the sets {k : fni ≥c + ǫ −ρ} and {k : fni ≤c + ρ}are Boolean independent for anyfn1 −fm1, fn2 −fm2, . .

. , fnj −fmjin the treeconstructed.If f is not continuous we use the following lemma.Lemma 3.5.

If (xn) is a uniformly bounded sequence in a Banach space X suchthat || P anxn|| ≥δ P |an| for all (an) ∈RN and y ∈X, ||y|| ≤1, then||Xan (xn + y) || ≥δ′ X|an|,where δ′ = max {δd (y, [xn]) /2, δ −||y||}.We omit the simple proof.Our tree was actually constructed using sequences (fni −fmi)Ni=1 and sets(Ani, Bni) where |fmj(k) −f(k′)| < ρ for all k ∈∩Ni=1Aǫini, for all j, for somek (ǫ1, ǫ2, . .

. , ǫN) ∈∩Ni=1Aǫini.

Thus we can replace each fmi by g whereg(k) = f (k (ǫ1, ǫ2, . .

. , ǫN))if k ∈∩Ni=1Aǫini,0otherwisewith a loss of at most ρ.

Observe that because lim fn = f, limN inf d (f, [fn −f : n ≥N])≥||f||. Thus we get the estimateδ′ ≥maxn ǫ2 −ρ||f||, ǫ −||f|| −ρofrom the lemma.□Let us now examine the relationship between the oscillation index and ℓ1-spreadingmodels.

COMPLEXITY OF WEAKLY NULL SEQUENCES13Proposition 3.6. Let (fn) be a pointwise converging sequence on a Polish space(K, d) with limit 0.

If k ∈Oω (ǫ, (fn) , K) and N is a neighborhood of k, then forevery infinite M ⊂N there is an L ⊂M such thatfn|Nn∈L has ℓ1 spreading modelconstantatleast ǫ.Proof. Fix j ∈N and choose L ⊂M such thatfn|Nn∈L has a spreading model.Because k ∈O2j ((fn) , K, δ), a careful examination of the proof of the previoustheorem shows that there is an ǫ = + or −such that given any N there is anL ⊂{N, N + 1, .

. . } with cardinality j with N ∩∩ℓ∈LAǫℓ,m ̸= ∅, for all large m.Thereforefn|Nn∈L has ℓ1 spreading model constant at least ǫ.□We wish to present two more examples, but before doing so we will introduceanother ordinal index which we call the spreading model index and show that itis closely related to the oscillation index.

This index is defined in terms of thespreading model constant and should be compared with the averaging index.Let S0 (ǫ, (fn) , K) = K and assuming that Sα (ǫ, (fn) , K) has been defined letSα+1 (ε, (fn) , K) =x∗: for every neighborhood N of x∗relative toSα (ε, (fn) , K) and infinite M ⊂N there is an L ⊂Nwith ℓ1 −SPfn|Nn∈L ≥ε.As usual if α is a limit ordinal Sα (ǫ, (fn) , K) = ∩β<αSβ (ǫ, (fn) , K) and theordinal index will be denoted by o(S, ǫ).Next we will show that the oscillation index and the spreading model indexmeasure essentially the same thing.Proposition 3.7. Suppose that (fn) is a weakly null sequence in BC(K) for somecompact metric space K. Theni) if S1 (ǫ, (fn) , K) ̸= ∅, O1 (ǫ′, (fn) , K) ̸= ∅for every ǫ′ < ǫ.ii) if Oω (ǫ, (fn) , K) ̸= ∅, for some ǫ > 0, S1 (ǫ, (fn) , K) ̸= ∅.Proof.

Suppose that O1 (ǫ′, (fn) , K) = ∅. Then for each k ∈K there is a neigh-borhood Nk of k and a subsequence (fn)n∈Mk such that ||fn|Nk|| ≤ǫ′ for alln ∈Mk.Clearly ||Xn∈Ffn|Nk|| ≤|F|ǫ′ for every finite subset F of Mk.HenceS1 (ǫ, (fn) , K) = ∅for every ǫ > ǫ′.The second assertion follows from Proposition 3.6.□Corollary 3.8.

Suppose that (fn) is a weakly null sequence on a compact metricspace K. Theni) If o(S, ǫ) ≥α, then O(ǫ′) ≥α, for every ǫ′ < ǫ.ii) If O(ǫ) ≥ω1+α, then o(S, ǫ) ≥ωα.Example 4. L1.

Of course ℓL1, 1= ω1. However if (fn) is a weakly conver-gent sequence in L1, then (fn) is uniformly integrable and hence by Dor’s Theorem[Dor] for every K < ∞there is an n such that if F is a set of integers of cardi-nality n, (fj)j∈f is not K equivalent to the unit vector basis of ℓ1n.

Therefore ℓ1−SP (fn)n∈L = 0 for all L and thus Oω (ǫ, (fn) , K) = ∅for every ǫ > 0.

14DALE E. ALSPACH AND SPIROS ARGYROSExample 5. The spaces Szlenk [Sz] used to show that there are reflexive spaceswith arbitrarily large (countable) Szlenk index are defined inductively as X1 = ℓ2,Xα+1 =Xα ⊕ℓ21, and for a limit ordinal α, Xα − Xβ<αXβ2.

The ℓ1-index ofXα = α for ǫ = 1 and increases to αω as ǫ goes to 0.Now let us consider the oscillation index. First suppose that α is a limit ordinal.If (fn) is a weakly null sequence in Xα, then by passing to a subsequence (whichcan only increase the index) we may assume that fn = gn + hn where gn ∈Xβ≤λXβfor all n for some λ < α and hn ∈Xβ∈BnXβ where Bn = {β : βn < β ≤βn+1} forall n and β1 > λ.

Note that [hn] = ℓ2 and thus the ℓ1 spreading model index ofXα is the supremum of the indices of Xβ, β < α. Now suppose that α = β + kfor some integer n. Then Xα =Xβ ⊕ℓ2 ⊕ℓ2 ⊕· · · ⊕ℓ2|{z}k1, and we may writefn = gn + hn where gn ∈Xβ and hn ∈kX1⊕ℓ2 for all n. However we again havethat ℓ1 −SP (fn) = ℓ1 −SP (gn) and hence the spreading model index of (fn) isthe spreading model index of (gn).

Thus the spreading model index of Xα is thesame as the index of ℓ2, for all α, namely 0.This last family of examples illustrates the fact that the oscillation index and thespreading model index are really measuring something stronger than the existenceof many ℓ1′n s in a space.Next we will examine the ordinal index of a Baire-1 function f on a Polish space(K, d). The index is defined by considering two real numbers c and d, c < d, andthe disjoint Gδ sets,C = {k ∈K : f(k) ≤c} and D = {k ∈K : f(k) ≥d} .L(f, c, d) is the smallest ordinal α such that there is a decreasing family of closedsets Fβ, β ≤α, with F0 = K, Fα = ∅, and for all β < α, Fβ\Fβ+1 is disjoint fromC or from D and at a limit ordinal γ, Fγ = ∩β<γFβ.

(See [Bo] where the definitionis given in complementary terms or [K,p452]. )Bourgain shows that if (fn) is a pointwise converging sequence of continuousfunctions with limit f and ǫ < (d −c)/2 then ωℓ([fn],ǫ)+1 is greater than L(f, c, d).

(Actually his result gives a slightly smaller bound.) We wish to show that in factthe oscillation index is also large.

The proof of the following proposition is similarto that of Lemma 5, [H-O-R].Proposition 3.9. Suppose that (fn) is a pointwise converging sequence of contin-uous function on a compact metric space K. Then if d −c > ǫ and if L(f, c, d) =β +m, where β is a limit ordinal and m < ω, then O+ (ǫ, (fn) , K) ≥β +(m−1)/2.Proof.

Consider the following family of closed sets where C = {k : f(k) ≤c} andD = {k : f(k) ≥d}, F0 = K, F1 = D, F2 = F1 ∩C, F3 = F2 ∩D, and in general,Fα+2n+1 = Fα+2n ∩D and Fα+2n+2 = Fα+2n+1 ∩C if α is even and n ∈N. (Limitordinals are even.) If α is a limit ordinal Fα = ∩β<αFβ.

COMPLEXITY OF WEAKLY NULL SEQUENCES15It is easy to see that if α is even (Fα\Fα+1) ∩D = ∅and (Fα+1\Fα+2) ∩C = ∅.Next we will show that Fα+2 ∩D ⊂O1+ (ǫ, (fn) , Fα+1) for all α even.Let d′ ∈Fα+2 ∩D. This implies that d′ ∈Fα+1 ∩C ∩D.

Fα+1 ∩C ∩D containspoints from D which are (non-trivial) limits of points in Fα+1 ∩C and hence thereexists a sequence (ck) in Fα+1∩C with limit d′. Moreover we may assume that no ckis in O1+ (ǫ, (fn) , Fα) (else d′ would be also).

Choose N ∈N such that |fn(d′)−f(d′)|is less than δ/4 for all n ≥N, where 0 < δ < (d −c) −ǫ. For each k ∈N there is anMk such that fm(ck) < δ/4 +f(ck) for all m ≥Mk.

Now if N is a neighborhoodof d′ and n ≥N then for some L ∈N, ck ∈N and fn(ck) > f(d′) −δ/4, for allk ≥L. Becausefn(ck) −fm(ck) > f(d′) −δ/4 −f(ck) −δ/4 ≥d −c −δ/2,for all k ≥L and m ≥Mk, ck ∈N ∩∩m≥MkA+n,m.

Hence d′ ∈O1+ (ǫ, (fn) , Fα+1)and Fα+3 = Fα+2 ∩D ⊂O1+ (ǫ, (fn) , Fα+1), for all α even.Because Oα+1+(ǫ, (fn) , K) = O1+ǫ, (fn) , Oα+(ε, (fn), K. A simple inductionargumentshows that Fα+2m+1 ⊂Om+ (ǫ, (fn) , Fα+1) for any integer m and even ordinal α. Itfollows then that if β is a limit ordinal Fβ ⊂Oβ+ (ǫ, (fn) , K) and for any m ∈N,Fβ+2m+1 ⊂Oβ+m+(ǫ, (fn) , K).□Corollary 3.10. Suppose that f is a Baire-1 function on a compact metric spaceK and that (fn) is a sequence of continuous functions on K with ||fn|| ≥1 whichconverge to f pointwise.

If L(f, c, d) = β + m, where c < d, β is a limit ordinaland m ∈N, then for any ε < d −c there is an ǫ/2 ℓ1-tree of order β + m/2 on(fn −fm), and there is an ǫ/2 ℓ1-tree on (fn) of the same order.Proof. From the above proposition we get that Oβ+(m−1)/2+(ǫ, (fn) , K) ̸= ∅.

Theproof of Theorem 3.1 shows that there is an ℓ1-tree on (fn −fm) of order β + (m −1)/2 + 1 with lower estimate ǫ. The second assertion follows from an examinationof the proofs of Proposition 3.9 and Theorem 3.1.

It is easy to see that in theproof of the theorem we can replace A+n,m by {k : fn(k) −c > ǫ + ρ}, where ρ c + ǫ + ρ} , {k : fns(k) < c + ρ})js=1 are Boolean independent for anynodefn1, fn2, . .

. , fnjof the tree constructed.□4.

Construction of weakly nullsequences with large oscillation indexIn this section we wish to generalize the construction of Schreier of a weaklynull sequence (xn) with no subsequence having the Banach-Saks property.Asobserved by Pelczynski and Szlenk [P-S] the Schreier sequence is a 1-suppressionunconditional basis in C(ωω). Our goal is to proveTheorem 4.1.

For every α < ω1 there is a weakly null sequence (xαn) in CωωαwithOωα 1 −ǫ, (xαn) , ωωα̸= ∅for every ǫ > 0. Moreover for each α, (xαn) is a sequenceof indicator functions and (xαn) is a 1-suppression unconditional basic sequence.To construct these sequences and verify their properties it is useful to have severaldifferent viewpoints.

The first viewpoint is contained in the following result.

16DALE E. ALSPACH AND SPIROS ARGYROSProposition 4.2. Let F be a family of finite subsets of N such thati) if F ∈F and G ⊂F then G ∈F, i.e., the family is adequate.ii) {n} ∈F for all n ∈Niii) if Fj ∈F for j = 1, 2, .

. .

and 1Fj converges pointwise to 1F then F ∈F.Let xn = 1{F ∈F:n∈F } for n = 1, 2, . .

. .Then F is a countable compact metric space under the topology induced by iden-tifying F with {1F : F ∈F} under the topology of pointwise convergence.

(xn)is a weakly null 1-suppression unconditional basic sequence in C(F).Proof. The first assertion is an easy consequence of iii) and we omit the argument.Note that xn(F) ̸= 0 if and only if n ∈F.

Hence||Xanxn|| = sup(|Xn∈Fan| : F ∈F).If G ⊂N, then by i)||Xn∈Ganxn|| = sup(|Xn∈G∩Fan| : F ∈F)= sup(|Xn∈Fan| : F ∈F and F ⊂G)Clearly this is not greater than || P anxn||. Because each F ∈F is finite, xn(F) ̸= 0for only finitely many n. Thus (xn) is weakly null.□This proposition gives us an easy way of defining and verifying the properties ofthe Schreier sequence.

As in the previous section letF1 = {F ⊂N : min F ≥card F}(We consider ∅to be in F1.) If (xn) is defined as in the proposition then it followsthat (xn) is a 1-unconditional basic sequence and is weakly null.

Finally if L ⊂Nis infinite and for each k ∈N, we let Lk be the first k elements of L||Xn∈L2kxn|| ≥kbecause L2k\Lk ∈F1 and||Xn∈L2k+1xn|| ≥kbecause L2k+1\Lk+1 ∈F1. Hence||Xn∈Lkxn||/k ≥(k −1)/(2k) for all k,and therefore (xn)n∈L fails the Banach-Saks property.

(See [D1,p.78]. )The major drawback to this representation of the Schreier sequence is that it isdifficult to understand the topology of F1 and its relation to (xn).

In this sectionwe will prove some results twice. First we will give proofs based on representationslike that above for the Schreier sequence.

The second will be given using trees. We

COMPLEXITY OF WEAKLY NULL SEQUENCES17have found that this second viewpoint is more intuitive (It is easy to draw picturesof the trees.) and we used it to establish these results originally.Our next goal then is to use trees to describe the Schreier sequence and theunderlying topological space and in particular to show that this sequence and itsgeneralizations could be obtained by beginning with the coordinate functions on theCantor set, and then essentially restricting them to suitable subsets of the Cantorset.

More precisely, if we let C = {−1, 1}N and rn ((ǫi)) = ǫn, then xn = (rn + 1) /2is a sequence of indicator functions. If K is a compact subset of C, then the sequence(xn|K) is a sequence of indicator functions in C(K) which will be equivalent to theSchreier sequence if K is properly chosen.Now let us work backwards from a sequence of indicator functions to find anappropriate minimal underlying topological space.

Suppose that (xn) is a sequenceof indicator functions on a set K. Define a tree T on {−1, 1} byT = ∪∞n=1 {(ǫ1, ǫ2, . .

. , ǫn) : ∩ni=1(supp xi)ǫi ̸= ∅}wheresupp xi = (supp xi)1 = {k ∈K : xi(k) = 1}and(supp xi)−1 = K\(supp xi)1.Let T = T ∪{(ǫ1, ǫ2, .

. . ) : ∩ni=1(supp xi)ǫi ̸= ∅for all n}.

T is also a tree and bothare subtrees of the full dyadic treeD = ∪∞n=1{−1, 1}n ∪{−1, 1}N.D has a natural topology given by coordinate-wise convergence. Of course in thistopology {−1, 1}N is just the Cantor set and each finite sequence is an isolated point.Also any infinite sequence is the limit of its restrictions to the first n coordinates,i.e., the nodes above it.

Note also that this tree is closely related to the Booleanindependence tree Tn(supp xn)1 , (supp xn)−1o.Lemma 4.3. T is the closure of T in D.Proof.

Obvious.□Throughout the remainder of this section we will assume that the sequence (xn)is pointwise convergent to 0 on K, K is a compact metric space and each xn iscontinuous.Given such a sequence (xn) the tree T ((xn)) defined above will be called the treeassociated to (xn). Because K is compact and (xn) is weakly null, T contains noelements with infinitely many coordinates equal to 1 and every node is on an infinitebranch.Hence T is countable and therefore homeomorphic to some countableordinal in the order topology.

Unfortunately because the tree is not well founded,one cannot use the order of the tree T to determine the topological type. To getaround this problem we must study the relationship between the topology of T andthe structure of the tree.Before we embark on the study of the topology let us consider a property oftrees is analogous to property i) of Proposition 4.2.

In an unpublished paper H.P.Rosenthal introduced the following notation.

18DALE E. ALSPACH AND SPIROS ARGYROSDefinition. Let us say that a tree Ton {−1, 1} is weakly independent if(ǫ1, ǫ2, ǫ3, .

. .

, ǫj) ∈T implies that for allǫ′1, ǫ′2, ǫ′3, . .

. , ǫ′jsuch that ǫ′i = −1if ǫi = −1 and ǫ′i = 1 or -1 if ǫi = 1,ǫ′1, ǫ′2, ǫ′3, .

. .

, ǫ′j∈T . Say that a sequence ofindicator functions is weakly independent if the associated tree is.Rosenthal also proved the following result in a slightly different form.Proposition 4.4.

A weakly null sequence of (non-zero) indicator functions on acompact metric space K determines a weakly independent tree if and only if it is a1-suppression unconditional basic sequence.Proof. First suppose that (xn) is a sequence of indicator functions which is a 1-suppression unconditional basic sequence and T is the associated tree.

Then forany sequence of real numbers (an) we have||Xanxn|| = supn|Xanxn(k)| : k ∈Ko= supXn:k∈supp xnan: k ∈K.In particular suppose that F ⊂{1, 2, . .

. , j} with\n∈F(supp xn)−1 ∩\n/∈F,n≤jsupp xn ̸= ∅,i.e., the node (ǫ1, ǫ2, .

. .

, ǫj) ∈T , where ǫi = 1 if n /∈F and n ≤j, and ǫi = −1 ifn ∈F. In order to check weak independence it is sufficient to check the conditionon a lower node.

Hence we may assume that if {1, 2, . .

.j} ⊃G ⊃F, then G ̸={1, 2, . .

.j}, and we must show that\n/∈Gn≤jsuppxn ∩\n∈G(suppxn)−1 ̸= ∅.Because (xn) is 1-suppression unconditional(j −card G)(1 + card G) = ||(1 + card G)Xn/∈Gn≤jxn +Xn/∈G(−1)xn||and thus there existsk ∈\n/∈Gn≤j(supp xn) ∩\n∈G(supp xn)−1.This implies that the node (ǫ1, ǫ2, . .

. , ǫj) ∈T where ǫi = 1 if i /∈G and i ≤j andǫi = −1 if i ∈G.

Thus T is weakly independent.Conversely suppose that (xn) is a weakly null sequence of indicator functionsand that the associated tree T is weakly independent. Then for any sequence ofreal numbers (an) and finite subset F of N, we claim that||Xanxn|| = supXn:k∈supp xnan: k ∈K

COMPLEXITY OF WEAKLY NULL SEQUENCES19≥supXn∈F :k∈supp xnan: k ∈K= ||Xn∈Fanxn||.To see the inequality suppose that k is any point in K and H = {n : xn(k) = 1}. Be-cause T is weakly independent there is a point k′ in K such that {n : xn(k′) = 1} =F ∩H.

Hence each sum on the righthand side of the inequality also occurs on theleft.□Corollary 4.5. Suppose that (xn) is a weakly null sequence of non-zero indica-tor functions in C(K) for some compact metric space K. Then the following areequivalent.i) The tree associated to (xn) is weakly independent.ii) F = {F ⊂N : F = {n : xn(k) = 1} for some k ∈K} is adequate.iii) (xn) is a 1-suppression unconditional basic sequence.Proof.

We have already shown that ii)=⇒iii) and i) ⇔iii). i)=⇒ii) isimmediate from the definition of weakly independent.□Remark 4.6.

Rosenthal showed (unpublished) that any weakly null sequence ofindicator functions in a C(K) space has a subsequence which is an unconditionalbasic sequence by showing that there is a weakly independent subsequence.Remark 4.7. Note that if (xn) is weakly independent and T is the associated treethen∪k {(n1, n2, .

. .

nk) : (ǫ1, ǫ2, . .

. , ǫm) ∈T where ǫni = 1 for i = 1, 2, .

. .

, k}= Tsupp xn, (supp xn)−1,the Boolean independence tree.In order to write tree elements more efficiently let us introduce the followingnotational conventions: If x = (ǫ1, ǫ2, . .

. , ǫk) and y =ǫ′1, ǫ′2, .

. .

ǫ′jare two ele-ments in ∪∞n=1Sn, for some set S, then x + y =ǫ1, ǫ2, . .

. , ǫk, ǫ′1, ǫ′2, .

. .

, ǫ′j, theconcatenation of x and y. Let e0 be the empty tuple, e1 = (−1), and inductivelydefine en+1 = en + e1, n = 1, 2, .

. .

. Also let eω = (−1, −1, .

. .

).We will next introduce a derivation on a tree T ⊂D. Letδ1(T ) =t ∈T : there are infinitely many s ∈T \T with t < sand if δα(T ) has been defined let δα+1(T ) = δ1(δα(T )).

If β is a limit ordinal letδβ(T ) = ∩α<βδα(T ). Finally define δ(T ) = inf {α : δα(T ) = ∅}.Proposition 4.8.

Suppose that T is a weakly independent subtree of the dyadictree with no infinite nodes, no infinite nodes in T with infinitely many coordinatesequal to 1, and every node of T is on some infinite branch. Then δ(T ) determinesthe homeomorphic type of T up to the number of points in the last derived set, i.e.,δα(T ) = ∅if and only if T(1+α) = ∅.Proof.

First note that each element of T \T is in the first derived set of T , and,in fact, T \T is the first derived set. Now let us set up a correspondence between

20DALE E. ALSPACH AND SPIROS ARGYROSthe derived sets of T and the subtrees of T . If C is any closed subset of T \T thenthere is a tree T (C) with T (C)\T (C) = C. Indeed let T (C) be the set of all nodesx of T for which there is some element c of C below x.As above let C be a closed subset of T \T .

We claim that T (C(1)) = δ1(T (C)).Suppose that x ∈T (C(1)). Then x is above some element c of C(1).

Say c =x+y+eω where y is possibly the empty tuple. Hence there are distinct points ck inC which converge to c and therefore for large k, ck = y+zk+eω, where the zk ′s aredistinct.

This implies that x ∈δ1(T (C)). Conversely suppose that x ∈δ1(T (C)).Then there is a sequence of distinct points (ci) in T (C)\T (C) = C such that ci > xfor all i.

By passing to a subsequence we may assume that ci −→c ∈C. Clearlyc ∈C(1) and c > x, therefore x ∈T (C(1)).Because T \T = T(1) and T (T \T ) = T , we have thatδ1(T ) = T ((T \T )(1)) = T (hT(1)i(1)) = T (T(2)).We claim that for every α < ω1,δα(T ) = T (hT(1)i(α)) = T (T(1+α)).Indeed, if this is true for α we have by the previous claim thatδα+1(T ) = δ1T (hT(1)i(α))= T (hT(1)i(α)(1)) = T (hT(1)i(α+1)).Also observe that if (Ci) is a decreasing family of closed subsets of T \T thenT (∩∞i=1Ci) = ∩∞i=1T (Ci).

Therefore if αi ↑α and the claim is true for each αi thenδα(T ) = ∩δαi(T ) = ∩T (hT(1)i(αi)) = T (∩hT(1)i(αi)) = T (hT(1)i(α)),establishing the claim. Hence δα(T ) = ∅if and only if T(1+α) = ∅.□Let us now return to the Schreier sequence and examine the associated tree S1.LetS0 = ∪∞n=0(ǫ1, ǫ2, .

. .

, ǫn) :ǫi = 1 for at most one i,ǫi = −1 otherwise, 1 ≤i ≤n.To define S1 we need introduce the extension of one tree by another tree. If T and Sare trees on the same set let T ⊞S denote {x + y : x ∈T and en + y ∈S where n is the lengthof x} ∪T .

If (Ti) is a sequence of trees, we inductively define ⊞nTi =⊞n−1Ti⊞Tn.If n = 0, let ⊞nTi = {ej : j = 1, 2, . .

. }.

Also we will use the notation L(T , n) for thesubtree which is the union of {ej : j = 0, 1, 2, . .

. , n −1} and the set of nodes of Tequal to or below en−1 + (1).

In particular L(T , n) ⊂{ej : j = 1, 2, . .

. , n −1}⊞T .We need a way of forming a new tree T out of an infinite sequence of trees Ti,i = 1, 2, .

. .

, on {−1, 1}.Define P∞i=1 Ti = ∪∞i=1L(Ti, i) Clearly the resulting tree will depend on the orderof the Ti ′s.

COMPLEXITY OF WEAKLY NULL SEQUENCES21We claim that the tree associated to the Schreier sequence isS1 =∞Xi=1⊞iS0.Indeed, the Schreier sequence (xn) is defined by the property that if k, N ∈N andk ≤n1 < n2 < . .

. < nk ≤N, thenk\i=1supp xni ∩\n̸=nin

Note that this is equivalent to say-ing that if m ≤k and k = n1 < n2 < . .

. < nm, ∩mi=1supp xni ∩∩n̸=nin

Thus S1 is the required tree.To see what the topological type of S1 is we need only compute the order ofδ [⊞nS0]. Clearly δ(S0) = 2.

A straight-forward induction argument shows thatδn([⊞nS0]) =δn(L(⊞nS0, n) = {eω}. It then follows easily that δ(S1) = ω + 1.

Therefore S1has exactly ω non-empty derived sets.Actually it is not hard to see that it ishomeomorphic to ωω.Now we are ready to generalize the Schreier example. Our goal is to build foreach α < ω1 a weakly null sequence of indicator functions on a compact metricspace (homeomorphic to ωωα) with oscillation index at least ωα.

The sequence willalso be a 1-unconditional basic sequence.We will begin by defining the sequences in terms of subsets of N as we did withthe Schreier sequence. F1 has been defined.

Suppose that Fβ has been defined forall β < α. Let αi = α −1 if α is not a limit ordinal and αi ↑α if α is a limitordinal.

DefineFα = ∪∞n=1 {∪ni=1Fi : Fi ∈Fαi, for i ≤n, k ≤F1 < . .

. < Fi < Fi+1 < .

. .

Fn}where the notation k ≤F1 < . .

. Fi < Fi+1 < .

. .

Fn means that if Fi and Fj arenonempty and i < j, then k ≤min Fi and max Fi < min Fj.There is some ambiguity here but it will not be of any significance as long asthe choice of the sequence (αi) is fixed for each limit ordinal. For each α < ω1 let(xαn) denote the standard sequence of indicator functions on Fα, i.e., xαn(F) = 1 ifn ∈F and 0 otherwise.Proposition 4.9.

For each α < ω1, Fα is a countable compact metric space underthe topology induced by identifying Fα with {1F : F ∈Fα} under the topology ofpointwise convergence and (xαn) is a weakly null 1-suppression unconditional basicsequence in C(Fα).Proof. It is sufficient to verify that Fα satisfies the hypothesis of Proposition 4.2.Property ii) is obviously inherited by each Fα from F1.

For i) use induction and

22DALE E. ALSPACH AND SPIROS ARGYROSnote that if G ⊂F1 ∪F2 ∪. .

.∪Fn ∈Fα, as in the definition, then G∩Fi ∈Fαi, foreach i. Hence G = (G∩F1)∪(G∩F2)∪.

. .∪(G∩Fn) ∈Fα.

Finally for iii) supposethat iii) holds for all β < α, and that for each k, Fk = ∪nkj=1Fkj is an element of Fα,as in the definition. If Fk converges to a non-empty set F, let n = min F. Thenn ∈Fk for all large k and thus nk ≤n.

We may assume by passing to subsequencesthat for each j ≤n, Fkj converges to some Fj. Hence, by induction, because wehave that Fkj ∈Fαj, for all k, Fj ∈Fαj.

the other properties are obvious and thusF = ∪Fj ∈Fα, as claimed.□Now we want to consider the size of the underlying topological space. First wewill compute the size directly using the families Fα.

For each α < ω1 and k, n ∈Nwith n ≤k letFα,n,k = {∪ni=1Fi : Fi ∈Fαi, k ≤F1 < . .

. < Fi < Fi+1 < .

. .

< Fn, and Fn ∈Fαn} ,where αi = α −1 if α is not a limit ordinal and αi ↑α if α is a limit ordinal.Proposition 4.10. For each α < ω1, F(ωα)α= {∅}.Proof.

The result will follow fromCLAIM: If ρ ≤ωαn, thenF(ρ)α,n,k =n∪ni=1Fi : Fi ∈Fαi, k ≤F1 < . .

. < Fi < Fi+1 < .

. .

< Fn, and Fn ∈F(ρ)αnoThe proof of the claim is by induction on α, n, and ρ.If α = 0, F0 = {{n} : n ∈N} ∪{∅}. Clearly F(1)0= {∅}.Let α = 1 and k ≥n ≥1.

Then if G ∈F(1)1,n,k, there is a non-trivial sequence(Gm) in F1,n,k which converges to F. Observe that we need only show that cardG < n. However this is obvious because card Gm ≤n for all m and some portionof the Gm ′s must go to ∞.Conversely, if G ∈F1,n,k and card G < n, thenG ∪{m} ∈F1,n for m = k, k + 1, . .

. .

Hence G ∈F(1)1,n,k. Finally note that eachF1,n,k is closed, F(1)1,n,k = F1,n−1,k, and if Gm ∈F1,m,m then Gm −→∅.

ThereforeF(ω)1= {∅}.Now assume the result holds for all γ < α. Fix ρ < ωαn and k ≥n ≥1 andassume that the result has been proved for ρ.

(Note that it always holds for ρ = 0. )Let F = F(ρ)α,n,k.

Suppose that (Gm) is a non-trivial sequence in F which convergesto G. Suppose that Gm = ∪ni=1Gm,i. By passing to a subsequence if necessarywe may assume that for each i, Gm,i −→Gi.

If (Gm,n) is a non-trivial sequence,then Gn ∈F(ρ+1)αnand thus G = ∪ni=1Gi where Gi ∈Fαi for i = 1, 2, . .

. , n andGn ∈F(ρ+1)αn, that is,G ∈n∪ni=1Fi : Fi ∈Fαi, k ≤F1 < .

. .

< Fi < Fi+1 < . .

. < Fn, and Fn ∈F(ρ+1)αno,If (Gm,n) is trivial (eventually constant) then (Gi,n) is eventually constant for alli.

However this contradicts the non-triviality of the sequence (Gm).Conversely ifG ∈n∪ni=1Fi : Fi ∈Fαi, k ≤F1 < . .

. < Fi < Fi+1 < .

. .

< Fn, and Fn ∈F(ρ+1)αno,

COMPLEXITY OF WEAKLY NULL SEQUENCES23Then there is a non-trivial sequence (Gm) in F(ρ)αn which converges to Fn. Clearly wemay assume that min Gm > max ∪n−1i=1 Fi for all m. Then G′m = ∪n−1i=1 Fi ∪Gm ∈F(ρ)α,n,kfor all m and (G′m) converges to G. Therefore G ∈F(ρ+1)α,n,k .Clearly if ρj ↑ρ, F(ρ)α,n,k =∩∞j=1n∪ni=1Fi : Fi ∈Fαi, k ≤F1 < .

. .

< Fi < Fi+1 < . .

. < Fn, and Fn ∈F(ρj)αno=n∪ni=1Fi : Fi ∈Fαi, k ≤F1 < .

. .

< Fi < Fi+1 < . .

. < Fn, and Fn ∈F(ρ)αno.Thus the formula holds for all ordinals ρ ≤ωαn.Finally observe that by induction we have that F(ωαn)α,n,k = Fα,n−1,k and hence thatF(ωαn+ωαn−1+...+ωα1)α,n,n= {∅}.

To see that F(ωα)α= {∅} note that Fα = ∪∞n=1Fα,n,nand that if Gn ∈Fα,n,n for all n then Gn −→∅.□Next we will prove the result again but using trees. First we will translate theconstruction into the tree representation.Observe that there is a simple correspondence between{∪ni=1Fi : Fi ∈Fαi, k ≤F1 < .

. .

< Fi < Fi+1 < . .

. < Fn}and ⊞ni=1Sαi, where Sαi is the tree corresponding to Fαi.

Indeed,\n∈Fisupp xαn ∩\n/∈Fin≤m(supp xαn)−1 ̸= ∅,for all m, if and only if Fi ∈Fα. Thus if yi =ǫmi+1, ǫmi+2, .

. .

, ǫmi+1, wheremi = min Fi−1, and ǫj = 1 if j ∈Fi, ǫj = −1, otherwise, then we have y1+y2+. .

.+yn+em in ⊞nSα, for all m. Conversely if we have y1+y2+. .

.+yn in ⊞nSα, let mi bethe sum of the lengths of the yj′s, j = 1, 2, . .

. , i −1, and Fi = {mi + k : ǫk = 1}for yi =ǫ1, ǫ2, .

. .

, ǫmi+1−mi. Then Fi ∈Fα, and max Fi < min Fi+1, for i =1, 2, .

. .

, n.Next we need to take care of the n ≤min F1 condition. Observe that we coulddefineFα = ∪∞n=1 {∪∞n=1Fi : Fi ∈Fαi, n ≤F1 < .

. .

< Fi < Fi+1 < . .

. < Fn and n ∈F1}and the set would remain the same.

We claim that the tree associated to xα+1nisSα+1 =∞Xi=1⊞iSα.We have already shown above that ⊞iSα corresponds to the union of i ordered setsfrom Fα. Now note that the nodes corresponding to ⊞iSα in the sum are all belowor equal to ei−1 + (1), and thus correspond exactly to the unions of i orders setsfrom Fα with smallest element of the first set equal to i.

Thus Sα+1 is the correcttree. A similar argument shows that for a limit ordinal αSα =∞Xi=1⊞ij=1Sαj

24DALE E. ALSPACH AND SPIROS ARGYROSwhere αi is the defining sequence for Fα.Now that we have the trees Sα, α < ω1, we can determine the underlying topo-logical spaces by using Proposition 4.7.Proof of Proposition 4.10. Inductively assume that δ(L(Sβ, j)) is ωβ, for all β < αand j ∈N.

Let αi = α −1 if α is a successor and αi ↑α otherwise.CLAIM: δ(L(⊞ni=1Sαi, j)) = ωαn + ωαn−1 + . .

. + ωα1 + 1 for all j ∈N.Indeed if x ∈L⊞n−1i=1 Sαi, j, x = y1 + y2 + .

. .+ yn−1 and the length of x is k ≥j,then{ek + y : x + y ∈L (⊞ni=1Sαi, k + 1)} ∪{ei : i = 1, 2, .

. .

, k} ⊃L (Sαn, k + 1)Thusx ∈δωαn (L (⊞nSαi, j)) ,by the inductive hypothesis. Because k is arbitrary the claim follows by inductionon n. (Obviously δ (L (⊞ni=1Sαi, j)) ≤ωαn + ωαn−1 + .

. .

+ ωα1 + 1. )It now is easy to see that δ (L (Sα, j)) = ωα + 1, because the order is largerthan λn = ωαn + ωαn−1 + .

. .

+ ωα1, for all n and the elements ei, i ∈N, are inδλnL (⊞nSαi, j) for all n.□Our next task is to compute the oscillation index of (xαn). As before we willcompute this in two ways first by using the family Fα and then by using trees.Proposition 4.11.

Oωα (ǫ, (xαn) , Fα) ̸= ∅, for all α < ω1, ǫ < 1.Proof. We will show that Oλ (ǫ, (xαn) , Fα) = F(λ)αfor all λ.

In view of Proposition4.10, it is sufficient to show that if F ∈F(λ+1)αthen there is a N ∈N such that forall n ≥N, F ∪{n} ∈F(λ)α . We will use induction on α and λ.If α = 1, then for any λ ∈N the claim is immediate from the definition of F1and Proposition 4.10.Now assume that α > 1 and that the claim is true for all β < α and all λ. IfF ∈F(1)αthen either F = ∅and the claim is obvious or F ∈F(1)α,k,k for some k ∈N.By Proposition 4.10,F(1)α,k,k =n∪ki=1Fi : Fi ∈Fαi, k ≤F1 < .

. .

< Fi < Fi+1 < . .

. < Fk, and Fk ∈F(1)αko.Suppose that F = ∪ki=1Fi as above.

By the inductive hypothesis Fk ∪{n} ∈Fαk forall n ≥N, for some N ∈N, and thus for n > max{N}∪Fk, ∪k−1i=1 Fi ∪Fk ∪{n} ∈F(0)α,k,k.Next assume the claim for all γ ≤λ and let F ∈F(λ+1)α.If F = ∅then{n} ∈F(λ)αfor all sufficiently large n. Otherwise F ∈F(λ+1)α,k,kfor some k. Letλ = ωαk + ωαk−1 + . .

. + ωαj + ρ for some j ≤k + 1 and ρ < ωαj−1.

By the claimin the proof of Proposition 4.10F(λ+1)α,k,k = F(ρ+1)α,j−1,k=n∪j−1i=1 Fi : Fi ∈Fαi, k ≤F1 < . .

. < Fi < Fi+1 < .

. .

< Fj−1, and Fj−1 ∈F(ρ+1)αj−1o.

COMPLEXITY OF WEAKLY NULL SEQUENCES25Suppose that F = ∪j−1i=1 Fi as above. By the inductive hypothesis Fj−1 ∪{n} ∈Fαj−1 for all n ≥N, for some N ∈N, and thus for n > max{N} ∪Fj−1,∪j−2i=1 Fi ∪Fj−1 ∪{n} ∈F(ρ)α,j−1,k.□In the proof above we established that we can always use the special sequence(F ∪{n})n≥N to reach a set F from a smaller derived set.

The next definitiondescribes this same property for the associated tree.Definition. Let T be a tree on {−1, 1} with no nodes in T with infinitely manycoordinates equal to 1.

We will say that T has property FB (fully branching) if x+eω ∈T implies that there is an N ∈N such that eitheri) x + ej + (1) ∈T for all j ≥N orii) x + ej + (1) /∈T for all j ≥N.Lemma 4.12. Suppose that T and U are weakly independent trees with propertyFB and all nodes of T and U are on branches with limit of the form x + eω, thenT ⊞U has property FB, and for all n, L (T , n) has property FB.Proof.

Suppose that x + eω ∈T .Because T ⊞U ⊃T , if i) occurs in T , thesame is true in T ⊞U.If ii) occurs in T but not in T ⊞U, then there is asequence of incomparable nodes of the form en + yj in U where n is greater thanthe length of x and does not depend on j and yj has at least one coordinate equalto one. By passing to a subsequence we may assume that en + yj converges toen + z + eω.

The assumption that the en + yj′s are incomparable guarantees that(en + z) + eω does not satisfy ii). Because U has property FB there is an N suchthat en + z + ej + (1) ∈U for all j > N. Because U is weakly independent thisimplies that ek + ej + (1) ∈U for all j > N, where k equals n plus the length of z.Hence x +em + (1) ∈T ⊞U for all m > N + k, i.e., x + eω satisfies i).If x + y + eω ∈T ⊞U\T , where en + y ∈U, n is the length of x and x ∈T , thenx + y + eω will satisfy i), respectively ii), if en + y + eω satisfies i), respectively ii).The second assertion is obvious.□Proposition 4.13.

Suppose that (xn) is a weakly independent sequence of contin-uous indicator functions on a compact metric space K which converge pointwise to0 and that the associated tree T has property FB. Then for any ǫ < 1, and α < ω1,Oα (ǫ, (xn) , K) ̸= ∅if and only if T(1+α) ̸= ∅.Proof.

For each n and (ǫi) ∈T \T , let ˆxn ((ǫi)) = 1 if ǫn = 1, and 0, otherwise.In this way we have defined a sequence of indicator functions on Q = T \T withspan isometric to the span of (xn). (Actually T \T is homeomorphic to the naturalquotient of K determined by the xn ′s.

)Clearly Oα (ǫ, (xn) , K) ̸= ∅if and only if Oα (ǫ, (ˆxn) , Q) ̸= ∅. Now observethat (ǫi) + ǫω ∈O1 (ǫ, (ˆxn) , Q) if and only if there is an N ∈N such that for allj ≥N, (ǫj) + ej + (1) ∈T .

(Use the weak independence of (ˆxn) .) Clearly thislatter condition implies that (ǫi) + eω ∈T(2) = Q(1).

Conversely if (ǫ1) ∈Q(1),then weak independence and property FB imply that (ǫi) + ej + (1) ∈T , for allj ≥N, for some N ∈N. Finally note that weak independence and property FBare inherited by T (Q(1)) which is the tree associated toˆxn|Q(1).

Also if αk ↑α,∩TQ(αn)= TQ(α)and it is straight-forward to check that weak independenceand property FB are inherited by the intersection. Thus transfinite induction maybe used to complete the proof.□

26DALE E. ALSPACH AND SPIROS ARGYROSProposition 4.11 follows as a corollary of this result, that is, Oωα (ǫ, (xαn) , K) ̸= ∅,for all α < ω1, ǫ < 1.Because the underlying space for (xαn) is ωωα, ωα is the maximal possible oscilla-tion index. Also note that because each (xαn) is an unconditional basic sequence, allof the spaces are isomorphic to complemented subspaces of the Pelczynski universalspace U1, [L-T,I,p92].5.

Reflexive spaces with large oscillation indexIn the previous section we constructed weakly null sequences with oscillationindex ωα. Because these sequences were in C (Fα) it follows that the span of (xαn)contains c0 and thus is not reflexive.

In this section we will explore an idea of E.Odell for constructing Tsirelson-like spaces with large oscillation index.To define these spaces we begin with the space [xαn] in place of c0 in the Tsirelsonconstruction, [C-S,p 14]. Suppose that x = P antαn where (tαn) is the unit vectorbasis of the space of sequences with only finitely many nonzero coordinates.

Let||x||0 = ||Xanxαn||and inductively define||x||m+1 = max(||x||m, 2−1max{pi}∈F1kXi=1||pi+1−1Xn=piantαn||m),where k is the cardinality of {pi}. Let ||x|| = lim ||x||m.

Let Tα be the completionof span {tαn } under || · ||. Observe that Tsirelson space is T0 in this construction,i.e., let F0 = {{n} : n ∈N}, thenx0n=1{{n}}is equivalent to the c0 basis.Also observe that for each α the norm on Tα satisfies(*)||x|| = max(||x||0, 2−1max{pi}∈F1kXi=1||pi+1−1Xn=piantαn||).Proposition 5.1.

For each α < ω1, Tα is a reflexive Banach space with uncon-ditional basis with no subspace isomorphic to ℓp, 1 ≤p < ∞, or c0. MoreoverOωα ǫ, (tαn) , B(Tα)∗̸= ∅for all ǫ < 1.Proof.

Fix α < ω1. Clearly ||x||0 ≥||x||c0.

Therefore ||x|| ≥||x||T for all x ∈Tαwhere || · ||T denotes the norm on Tsirelson space. (This is T0 above.) It followsfrom (∗) that if {ui} is a sequence of k normalized blocks of the basis in Tα withsupport beyond k thatkXi=1|ci| ≥||kXi=1ciui||T ≥2−1kXi=1|ci|,for all choices of scalars (ci).

Therefore Tα does not contain c0 or ℓp, for any p > 1.To see that ℓ1 is not isomorphic to a subspace of Tα we need only examine theproof that T does not contain ℓ1 as given in [C-S,p 17]. The only properties of Tthat are used in the argument are that the norm satisfies equation (∗) and thatthe c0 norm of a long average is small.

The proof then for T will carry over to Tα

COMPLEXITY OF WEAKLY NULL SEQUENCES27provided we use a sequence with small || · ||0. To do this let (ui) be a normalizedblock basis of (tαn) such thatX|ai| ≥||Xaiui|| ≥89X|ai|.Because (xαn) is weakly null in C(Fα), so is (ui) and thus there is a sequence ofdisjoint convex combinations of (ui),yj =Xi∈Ejλiui,where E1 < E2 < .

. .

,Xi∈Ejλi = 1, and λi ≥0 for all i, such that ||yj||0 ≤2−j forj = 0, 1, . .

. .

It follows then thatX|ai| ≥||Xaiyi|| ≥89X|ai|,and||y0 + r−1(y1 + y2 + . .

. + yr)||0 ≤1 + r−1.Using these yi ′s the remainder of the proof carries over without change to Tα.Obviously Tα has an unconditional basis and thus by a result of James [L-T,I,p.97], Tα is reflexive.Finally to see that the oscillation index is large we use the observation that theoperator S from Tα to [xαn] defined by S(P cntn) = P cnxαn is bounded by 1 andthus Oλ ǫ, (tαn) , BT ∗α⊃S∗Oλ (ǫ, (xαn) , Fα) ̸= ∅, for every λ ≤ωα, by Lemma2.2.□While these space Tα share important properties with T let us note that theydo not possess the property that every block basis dominates a subsequence of thebasis.

In particularProposition 5.2. For every α < ω1 there is a block basis (ui) of (tαn) and anincreasing sequence of integers (ki) such that||Xaiti||T ≤||Xaiui|| ≤2(1 + ǫ)||Xaitki||Tfor any sequence of scalars (ai).Proof.

Fix α < ω1. Let (vi) be a normalized block basic sequence of (xαn) whichis (1 + ǫ)-equivalent to the usual unit vector basis of c0 and let ui = vi/||vi||,i = 1, 2, .

. .

. The idea is to show that for any sequence of scalars (ai)(**)||Xaiti||T,m ≤||Xaiui||and(***)||Xaiui||m ≤2(1 + ǫ)||Xait2i||T

28DALE E. ALSPACH AND SPIROS ARGYROSm = 0, 1, . .

. , where||Xaiti||T,m+1 = max(||Xaiti||T,m, 2−1 max{pi}∈F1kXi=1||pi+1Xn=pi+1antn||T,m),and||Xaiti||T,0 = sup |ai|.Once this is accomplished we use the fact that there is a constant K, such that||Xaiti||T ≤||Xait2i||T ≤K||Xaiti||Tand hence [ui] is isomorphic to T .

(See [C-S,p.26]. In fact the argument given hereis derived from the arguments of Casazza, Johnson and Tzafriri, [C-S,p.34-38].

)We will establish (∗∗) by induction on m. For m = 0, (∗∗) is immediate. Nowassume the inequality holds for m and we will prove it for m + 1.Let kn be the first element in the support of un.

Then||Xaiti||T,m+1 = max(||Xaiti||T,m, 2−1max{pi}∈F1kXi=1||pi+1−1Xn=piantn||T,m),≤max(||Xaiui||, 2−1max{pi}∈F1kXi=1||pi+1−1Xn=pianun||),≤max(||Xaiui||, 2−1max{pi}∈F1kXi=1||PiXanun||)= ||Xaiui||where PiP bntαn =qi+1−1Xn=qibntαn. The last inequality above holds because {qi} ={kpi} ∈F1 if {pi} ∈F1.For the inequality (∗∗∗) we need to work a little harder.||Xanun||m+1 = max(||Xaiui||m, 2−1 max{qi}∈F1kXi=1||PiXanun||m)where Pi denotes the basis projection onto [tj : qi ≤j < qi+1].

Fix {qi} ∈F1 andfor each i let Gi = {n : qi ≤kn < kn+1, and n /∈Gi}. For each n letHn = {i : kn ≤qi < kn+1 or kn < qi+1 ≤kn+1, and n /∈Gi} .Consider the sum corresponding to the qi ′s.

COMPLEXITY OF WEAKLY NULL SEQUENCES292−1kXi=1||PiXanun||m ≤2−1kXi=1||Xn∈Gianun||m + 2−1∞Xn=1Xi∈Hn||Pianun||m≤2−1kXi=1||Xn∈Gianun||m + 2−1Xn:Hn̸=∅2||anun||m+1≤2−1kXi=12(1 + ǫ)||Xn∈Giant2kn+1||T +Xn:Hn̸=∅||ant2kn+1||T≤(1 + ǫ)kXi=1||Xn∈Giant2kn+1||T +Xn:Hn̸=∅||ant2kn+1||TObserve that there are at most k integers n such that Hn ̸= ∅and that k ≤q1

. .

, k. Then{2kni+1 : i = 1, 2, . .

. , k} ∪{2kn+1 : Hn ̸= ∅}is a set of at most 2k integers greater than 2k.

Hence this sum in brackets is atmost 2|| P ant2kn+1||T . Thereforemax(||Xaiui||m, 2−1 max{qi}∈F1kXi=1||PiXanun||m)≤max(||Xait2ki+1||T , 2(1 + ǫ) max{qi}∈F1kXi=1||PiXant2kn+1||T)=2(1 + ǫ)||Xait2ki+1||T ,as claimed.□Remark 5.3.

Argyros [A] has modified the construction to obtain spaces Xα, α <ω1, such that all of the subspaces of Xα have index at least α. To accomplish thishe uses the sets Fα in the definition of the norm instead of starting with the space[xαn].6.

Comparison with the averaging indexIn [A-O] the averaging index (See Section 1 for the definition.) was used to geta somewhat more constructive version of Mazur’s Theorem.

In this section we willshow that the averaging index is much larger than the spreading model index byshowing that there exists a Banach space X such that for every α < ω1 there is aweakly null sequence (xn) in X with averaging index at least α, and yet ℓ1 is notisomorphic to a subspace of X. We will also show that the spreading model indexcan be used to strengthen some of the results in [A-O].The construction of the example will be based on the infinite branching Jamestree construction [J] in combination with ωω.

LetT = ∪∞n=1 {(α1, α2, . .

. , αn) : αi < ωω for each i} .

30DALE E. ALSPACH AND SPIROS ARGYROSLet X0 be the linear subspace of the functions from T into C0(ωω) which arenonzero at only finitely many points of T . We will use the notation (ft) whereft ∈C(ωω) for all t in T to denote an element of X0 with the understanding thatthe index t runs over T .

We will also use ft to denote the element of X0 which is0 except at t and ft at t.Next we will introduce some linear functionals on X0. For each s ∈T and i < |s|(the length of s) defineL(s,i)(ft) = 0if s ̸> t or |t| < ift(s(|t| + 1))if s > t and i ≤|t| < |s| ,where s > t denotes that s is below t, and extend linearly to X0.We will refer to a pair (s, i) as a segment and define it to be the set of nodes oflength at least i which are above s. We also want to have a notion of incomparablesegments.

Suppose that j ≤i and that (s, i) and (t, j) are segments then (t, j)and (s, i) are incomparable if they are disjoint and t(m) ̸= s(m) for some m ≤j,i.e., they are on branches that split by level j. Note that if {(sk, ik)} are pairwiseincomparable nodes then L(sk,ik)(ft) ̸= 0 for at most one k for each t ∈T .Now we will introduce a norm on X0.

For F ∈X0 define||F|| = supXjL(sj,ij)(F)212: {(sj, ij)} pairwise incomparableLet X be the completion of X0 under this norm. Clearly C0(ωω) is isometric toXt = [ft : f ∈C0(ωω)] for each fixed t. For each k ∈N define a projection Pk onX byPk((ft))s = fsif |s| ≤k0if |s| > k .It is easy to see that ||Pk|| = 1 for all k.Proposition 6.1. ℓ1 is not isomorphic to a subspace of X.The example is similar to an example of Odell [O] and his arguments can bemodified to prove Propositon 6.1.

However we will give a slightly different proofwhich does not directly use the branch functionals.Proof. Suppose that (yn) is a normalized sequence in X which is K-equivalent tothe usual unit vector basis of ℓ1.

Because ℓ1 is not isomorphic to a subspace ofC0(ωω) and range (Pk −Pk−1) is isometric to [P C0(ωω)]ℓ2, it can be shown byinduction that ℓ1 is not isomorphic to a subspace of range Pk for any k. Thereforeby passing to a subsequence we may assume that (Pk(yn))n is weakly Cauchy foreach k. Because for each k there are convex combinations of (Pk(y2n −y2n−1)) withsmall norm we can find a sequence of disjointly supported (relative to the yn ′s)convex combinations of (y2n −y2n−1), (zj), and an increasing sequence of intergers(kj) such that∞Xj=m||Pkmzj|| < 2−m,

COMPLEXITY OF WEAKLY NULL SEQUENCES31for m = 1, 2, . .

. .

Moreover because X0 is dense in X we may assume (by passingto a subsequence) thatm−1Xj=1||(I −Pkm)zj|| < 2−m,for m = 1, 2, . .

. .

In this way we get a sequence equivalent to the unit vectorbasis of ℓ1 which is essentially supported on disjoint levels of T . By a standardperturbation argument we may assume that (I −Pkm+1)zm = 0 = Pkmzm, and thatzm|t ̸= 0 for only finitely many t for all m.Next note that by a theorem of James [J] we may assume that (zj) is (1 + ǫ)-equivalent to the basis of ℓ1.

For any node s let W(s) =t : t > s, the wedgedetermined by s. For each i there are finitely many nodes s(i, j) of length ki suchthat if t /∈∪jW(s(i, j)) then zi|t = 0. Let N be the number of nodes in the supportof z1.

We claim that for each i there is a set F = F(i) of cardinality at most Nsuch that||zi −zi|∪j∈F W(s(i,j))|| < 4ǫ.Indeed, if not let S = {(s, i)} be a family of incomparable segments which computethe norm of z1 + zi. Let S′ be the set of segments in S which intersect the supportof z1 and S′′ = S\S′.

Clearly S′ contains at most N segments. We have that4(1 + ǫ)−2 ≤||z1 + zi||2=X(s,j)∈S′L(s,j)(z1 + zi)2 +X(s,j)∈S′′L(s,j)(z1 + zi)2≤X(s,j)∈S′L(s,j)(z1)212+X(s,j)∈S′L(s,j)(zi)212 2+X(s,j)∈S′′L(s,j)(z1 + zi)2≤(1 + (1 −4ǫ))2 + (4ǫ)2 = 4(1 −4ǫ + 8ǫ2).Clearly this is impossible for small enough ǫ.It follows that by another perturbation argument that we may assume that eachzi is supported in at most N wedges.

As above let s(i, j), j = 1, 2, . .

. , N be thenodes of length ki so that zi is supported in ∪jW(s(i, j)).We will next refineour sequence (zi) to get a subsequence such that there are branches b1, .

. .

, bk,k ≤N such that if s is any branch then s > s(i, j) for at most N nodes not onsome bm and bj > s(i, j) for j = 1, 2, . .

. , k for all i.

Such a subsequence is easilydetermined by induction on N. Indeed, if N = 1, either there are infinitely manyi and incomparable branches (ti, ki) such that ti > s(i, 1) and ti ̸> s(m, 1) for anym ̸= i, or there is a branch b1 which contains all but finitely many of the nodess(i, 1). Now suppose that b1, .

. .

, bk are branches such that if s is any branch thens > s(i, j) for at most N-1 nodes not on some bm, and bj > s(i, j) for j = 1, 2, . .

. , kfor all i.

As above if there is some branch which contains all but finitely many ofthe nodes s(i, N) we add that branch to our list as bk+1 and pass to a subsequence

32DALE E. ALSPACH AND SPIROS ARGYROS(zi)i∈M such that bk+1 > s(i, N) for all i ∈M. If this is not the case then wecan find a subsequence such that there is at most one of the nodes s(i, N) on anybranch.

Clearly this subsequence has the required properties.To complete the argument we need to make a few observations about the norm onX. First observe that if s is any branch and for each node t on s, ft is a fixed functioninC0(ω ω) then [ft]s>t is isomorphic to c0.

Second if the norm of P aizi is com-puted using only segments which intersect at most N of the supports of the zi ′s,i.e., s(i, j) for at most N j′s, then the norm is at mostP a2i 12 N12 . Finally observethat for the sequence (zi) a segment can intersect more than N of the supports ofthe zi ′s only if it lies along one of the branches bk.

A straightforward computationusing these observations shows that the zi ′s are not equivalent to the unit vectorbasis of ℓ1.□Our next goal is to show that the averaging index of X is uncountable. Thebasic idea is to construct for each α < ω1 a weakly null sequence with averagingindex α by using well-founded subtrees of T of order α.Proposition 6.2.

Let S be a well-founded subtree of T and for each s ∈S letfnsbe a normalized weakly null sequence in X s. Thenfnsn∈N,s∈S (reordered) isa weakly null sequence in X.Proof. X∗is the closed linear span of the functionalshL(t,i)it∈T ,i∈N.

Because Sis well founded, for any t ∈T , s < t for only finitely many s ∈S. Hence for anyǫ > 0, t ∈T , and i ∈N, |L(t,i)fns| ≥ǫ for only finitely many n and s. Thereforefnsn∈N,s∈S is a weakly null sequence in X.□In order to estimate the averaging index we need to have some information aboutthe w∗topology on the functionals L(t,i).Lemma 6.3.

Let t ∈T and let (αi) be a sequence in ωω which converges to α.Theni) If α < ωω, L(t+(αi),j)w∗−→L(t+(α),j)ii) If α = ωω, L(t+(αi),j)w∗−→L(t,j).Proof. We need only consider the values of the functionals at fs for those s ∈Tsuch that s ≤t and |t| −1 ≤|s| ≤|t|.

For all others the values do not depend on(αi). If s = t thenL(t+(αi),j) [fs] = fs(αi) −→fs(α) = L(t+(α),j) [fs] , if αi −→α < ωω,andL(t+(αi),j) [fs] = fs(αi) −→fs(α) = 0 = L(t,j) [fs] , if αi −→ωω.If t = s + (β), thenL(t+(αi),j) [fs] = fs(β) −→fs(β) = L(t+(α),j) [fs] , if αi −→α < ωω,andL(t+(αi),j) [fs] = fs(β) −→fs(β) = L(t,j) [fs] , if αi −→ωω.□We will need to use well-founded subtrees of T of a special type, so as a technicalconvenience we introduce the following.

COMPLEXITY OF WEAKLY NULL SEQUENCES33Definition. A well-founded tree S ⊂T is said to be complete ifi) s + (β) ∈S for some β < ωω and s ∈T then s + (α) ∈S for all α < ωω ands ∈S,ii) (α) ∈S for all α < ωω.Next we will verify that such things exist.Proposition 6.4.

For every α < ω1 there exists a complete well-founded treeS ⊂T of order at least α.Proof. For α = 1 this is obvious.

Suppose for all β < α there is a complete well-founded subtree S of T with order β. If α = β+1, we define U = {(η) + s : s ∈S ∪{()}andη < ωω}.Clearly o(U) = o(S) + 1 and it is obvious that U is complete.If αis a limit ordinal and αi ↑α, for each i let Sαi be a complete well-founded subtreeof T of order αi.

Let U = {(η) + s : s ∈∪iSαi ∪{()} and η < ωω}. Clearly U iscomplete, o(U ) ≥o(Sαi) for all i and thus o(U) ≥α.□We are now ready to show that there are weakly null sequences in X with largeaveraging index.Proposition 6.5.

Suppose that S is a complete well-founded subtree of T and foreach s ∈S,fnsn∈N is the Schreier sequence. Then for every β < o(S) and s ∈S(β),L(s,1)∈Aβh12,fntn∈N,t∈S , BX∗i.Proof.

We proceed by induction on β. It is sufficient to prove the result for β + 1assuming it for β.

If s ∈S(β+1), then because S is complete s + (α) ∈S(β) for allα < ωω. Moreover L(s+(α),1)fns= fns (α).

In Section 4 it was shown that ωω ∈O1 12, (f n) , ωω⊂A1 12,fn, ωωwherefnis the Schreier sequence. Thereforeby Lemma 6.3, L(s,1) ∈A1h12,fnsn∈N , Aβh12,fntn∈N,t∈S , BX∗ii.

Because thedefinition of the averaging index only requires that a subsequence have ℓ1 −SP ≥12,it follows thatL(s,1) ∈A112,hfntin∈N,t∈S , Aβ12,hfntin∈N,t∈S , BX∗= Aβ+112,hfntin∈N,t∈S , BX∗.□Corollary 6.6. For every α < ω1 there is a weakly null sequence in X such thato(A, 12 ) ≤α.Next we want to consider the relationship between the spreading model indexand the constructive version of Mazur’s theorem as presented in [A-O].

One purposeof that paper was to try to use the averaging index to determine if given a Banachspace X there is an integer k such that any weakly null sequence needs to beaveraged at most k times in order to get a norm null sequence. In what follows wewill use the notation of [A-O] and refer the reader there for the relevant definitions.The definition of the spreading model index makes some arguments of the typeused in [A-O] difficult because of its sensitivity to passing to subsequences.

On theother hand we are going to consider all weakly null sequences in the space X so itseems natural to look for sequences which are in some sense extremal.

34DALE E. ALSPACH AND SPIROS ARGYROSProposition 6.7. Suppose X is a subspace of C(K) for some compact metric spaceK and that (xn) is a weakly null sequence in the unit ball of X.

Then for everyǫ > 0 there is a subsequence (xn)n∈M of (xn) such that for any L ⊂M, infinite,and α < ω1,Sα ǫ, (xn)n∈L , K= Sα ǫ, (xn)n∈M , K.Proof. The idea is to use repeatedly the following lemma [A-O].Lemma 6.8.

Let K be a second countable compact Hausdorffspace and let (xn)be a weakly null sequence in C(K). Then there is a subsequence (xn)n∈M such thatfor every t ∈K and every neighborhood N of t there is a neighborhood N ′ of t, N ′⊂N, such that (xn|N ′) has a spreading model.We will prove the proposition by induction on α.

Note that there is an ordinalα0 such that Sα0 (ǫ, (yn) , K) = ∅for every ǫ > 0 and weakly null sequence (yn) inC(K), and thus the induction is over a countable set. Fix ǫ > 0.Induction Hypothesis.

If (xn) is a weakly null sequence in the unit ball of X.Then for every ǫ > 0 there is a subsequence (xn)n ∈M of (xn) such that for anyL ⊂M, infinite, and β ≤α,Sβ ǫ, (xn)n∈L , K= Sβ ǫ, (xn)n∈M , K.Observe that the lemma proves the result for α = 1. Now suppose that the resultis true for all β < α.

If α = β + 1, for some β, then let (xn)n∈L be a subsequencesuch thatSλ ǫ, (xn)n∈J , K= Sλ ǫ, (xn)n∈L , K,for all infinite J ⊂L and λ ≤β. Apply the lemma toxn|Sβ(ǫ,(xn)n∈J ,K)n∈L toget a subsequence (xn)n∈M and note that the case α = 1 applies to show thatSβ+1 ǫ, (xn)n∈J , K= S1ǫ,xn|Sβ(ǫ,(xn)n∈L,K)n∈J , Sβ ǫ, (xn)n∈L , K= S1ǫ,xn|Sβ(ǫ,(xn)n∈L,K)n∈M , Sβ ǫ, (xn)n∈L , K= Sβ+1 ǫ, (xn)n∈M , K.for all infinite J ⊂M.If α is a limit ordinal, let αi ↑α and choose infinite sets M1 ⊂M2 ⊂.

. .

⊂Mi ⊂. .

. , such thatSλ ǫ, (xn)n∈J , K= Sλ ǫ, (xn)n∈Mi , K,for all infinite J ⊂Mi and λ ≤αi.

Then if M is an infinite set such that M\Mi isfinite for all i, it follows thatSλ ǫ, (xn)n∈J , K= Sλ ǫ, (xn)n∈Mi , K,for all infinite J ⊂M and λ < α. However because Sα = ∩λ<αSλ, the equalityholds for λ = α as well.□

COMPLEXITY OF WEAKLY NULL SEQUENCES35Corollary 6.9. Suppose that (xn) is a weakly null sequence in the ball of C(K)for some compact metric space K. Then there is a subsequence (xn)n∈M such thatAλ ǫ, (xn)n∈M , K= Sλ ǫ, (xn)n∈M , K,for all λ ≤α and ǫ > 0.Proof.

The proof above gives us a subsequence so that passing to subsequences hasno effect on ℓ1 −SPxn|Nfor some N ↓{t} for all t, thereforeA1 ǫ, (xn)n∈M , K= S1 ǫ, (xn)n∈M , K.Induction completes the proof.□Corollary 6.10. Let X be a subspace of C(K), K compact metric, such thatsup {o (S, (xn) , ǫ) : ǫ > 0 and (xn) ⊂BX is weakly null}<ωk+1,forsomek ∈{−1, 0, 1, 2, .

. .

}. Then X has property-A(k + 2).Proof.

To establish property A(m) it is sufficient to show that every weakly nullsequence has a subsequence with a norm null convex block subsequence of m-Averages.According to the previous corollary every weakly null sequence hasa subsequence for which the averaging index and spreading model index agree.Therefore by [A-O, Theorem 4.1], X has property-A(k + 2).□Because the spreading model index is in general smaller than the averaging indexthis corollary gives a real strengthening of [A-O, Corollary 4.2]. In particular thespace constructed above has averaging index ω1 but spreading model index at mostω.Acknowledgement.

The authors were visitors at the University of Texas duringthe time that a portion of this work was done and would like to thank the Depart-ment of Mathematics at University of Texas for its hospitality. In particular we aregrateful to H.P.

Rosenthal and E. Odell for many helpful comments.References[A] S. Argyros, Banach spaces of the type of Tsirelson, preprint. [A-O] D. Alspach and E. Odell, Averaging weakly null sequences, Lecture Notes in mathematics Vol.1332, Springer-Verlag, Berlin 1988.

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[Z] Z. Zalcwasser, Sur une propriet´e der champ des fonctions continues, Studia Math. 2 (1930),63-67.Department of Mathematics, Oklahoma State University, Stillwa-ter, OK 74078-0613Department of Mathematics, University of Crete, Heraklion, Crete

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