Pre-compact families of finite sets of integers and weakly null sequences in Banach spaces

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Pre-compact families of finite sets of integers and weakly

null sequences in Banach spaces

Jordi Lopez Abad, S. Todorcevic

To cite this version:

Jordi Lopez Abad, S. Todorcevic. Pre-compact families of finite sets of integers and weakly null

sequences in Banach spaces. 2007. �hal-00012297v3�

hal-00012297, version 3 - 9 Jun 2007

J.LOPEZ-ABADANDS.TODORCEVIC

1. Introdu tion

The aÆnities b etween the innite-dimen sional Ramsey theory and some problems of the Bana h spa e theory and esp e ially those dealing with S hauder basi sequen es have b een explored forquite some time, starting p erhaps with Farahat's pro of of Rosenthal's `

1

-theorem (see[13℄and[19℄). TheNash-Williams'theorythoughimpli itin allthiswasnotfullyexploited in this ontext. In this pap er we tryto demonstratethe usefulness of this theoryby applying it to the lassi al problem of nding un onditional basi -subsequen e of a given normalized weakly null sequen e in some Bana h spa e E. Re all that Bessaga and Pel zynski [7℄ have shown that every normalized weakly null sequen e in a Bana h spa e ontains a subsequen e forming a S hauder basisfor its losed linear span. However, asdemonstratedby Maureyand Rosenthal [16℄there exist weakly nullsequen es in Bana h spa eswithout un onditional basi subsequen es. So one isleft with atask ofnding additional onditions on a given weaklynull sequen e guaranteeing theexisten e of un onditional subsequen es. Onesu h ondition, given by Rosenthal himself around the time of publi ation of [16℄ (see also [19℄). When put in a prop er ontextRosenthal's ondition reveals the onne tion withthe Nash-Williams theory. It says that if a weakly null sequen e (x

n

) in somespa e of theform ` 1

( ) is su h that ea hx n takesonlythevalues 0or1,then(x

n

) hasanun onditional subsequen e. Toseethe onne tion, onsider thefamily

F =ffn2N:x n

( )=1g: 2 g

andnotethatF isapre- ompa tfamilyofnitesubsetsofN:Asp ointedoutin [19℄,Rosenthal result isequivalent saying thatthere is aninnite subsetM of Nsu h thatthe tra e

F[M℄=ft\M :t2Fg

is hereditary, i.e., it is downwards losed under in lusion. On the other hand, re all that the basi notion of the Nash-Williams' theory is the notion of a barrier, whi h is simply a family F of nite subsets of N no two memb ers of whi h are related under the in lusion whi h has the prop erty that an arbitrary innite subset of N ontains an initial segment in F: Thus, in parti ular, F is a pre- ompa t family of nite subsets of N:Though the tra e of an arbitrary pre- ompa tfamily might b ehardtovisualize, atra eB [M℄ofa barrierB iseasily to ompute asitis simply equal tothe downwards losure ofits restri tion

BM =ft2B:tMg:

Afurtherexamination ofRosenthal'sresultshows thatforeverypre- ompa t familyF ofnite subsetsofNthereisaninnitesetM su hthatthetra eF[M℄isa tuallyequaltothedownwards losureofauniformbarrierBonM,orinotherwordsthatthe-maximalelementsofF[M℄form auniform barrieronM:Asitturns out,this fa tholds onsiderably moreinformationthat the on lusion thatF[M℄is merelyahereditaryfamily whi h isesp e ially noti eableif one need to p erformfurtherrenementsofM whilekeepingtru kontheoriginal familyF:Thisobservation wasthemotivating p oint forourresear h whi h help ed us to realize thatfurther extensions of Rosenthal's result require analysis of not only pre- ompa t families of nite subsets of N but alsomaps frombarriers intopre- ompa t familiesofnite subsetsof N;or,moregenerally, into weakly ompa t subsets of

0

. We have explained this p oint in our previous pap er [14℄, where we have presented various results on partial un onditionality su h as near-un onditionality or onvex-un onditionalityas onsequen es of thestru turetheory ofthis kind ofmappings. This pap erisasa ontinuationofthislineofresear h. InSe tion3weshowhowthe ombinatori son barriers an b eusedtoprovethe

0

-saturationforBana h spa esC(K) whenK isa ountable ompa tum. Re allthatthe

0

-saturationofBana hspa esC(K)over ountable ompa taKis aresult originally duetoPe l zy nski andSemadeni[21℄(see also[5℄and [12℄forre ent a ounts on this result.) More parti ularly, we show that if (x

i

)  C(K) is a normalized weakly-null sequen e, then there is C  1, some innite set M, some uniform barrier B on M of rank at most the Cantor-Bendixson rank of K and some uniform assignment  : B !

+ 00

with the prop erty that supp(s)  s for every s 2 B , and su h that for every blo k sequen e (s

n ) of elements of B ,the orresp onding sequen e (x(s

n ))of linear ombinations, x(s n )= X i2sn ((s n ))(i)x i ;

is anormalized blo ksequen e C-equivalent tothestandardbasisof 0

.

Thelastse tion on erns thefollowing naturalmeasurementof un onditionality presentin a givenweaklynullsequen e(x

n

)in ageneralBana hspa eE:Givenafamily F ofnitesets,we say that(x

n

) is F-un onditional with onstant atmost C 1 i for every sequen e of s alars (a n ), sup s2F k X n2s a n x n kCk X n2N a n x n k:

Thus,ifforsomeinnitesubsetM ofNthetra eF[M℄ ontainsthefamilyofallnitesubsetsof M;the orresp onding subsequen e (x

n )

n2M

isun onditional. Typi ally,one will notb eable to ndsu hatra e,sooneisnaturallyledtostudythisnotion whenthefamilyF ispre- ompa t, or equivalently, when F is a barrier. Sin e for every pair F

0

and F 1

of barriers on N there is an innite set M su h that F

0 [M℄F 1 [M℄or F 1 [M℄F 0

[M℄and sin e the two alternatives dep end on the ranks of F

0

and F 1

; one is also naturally led to the following measurement of un onditionalitythatrefersonlytoa ountableordinal ratherthanaparti ularbarrierofrank :Thus,wesaythatanormalizedbasi sequen e (x

n

)ofaBana hspa e Xis -un onditionally saturated with onstant at most C  1 if there is an -uniform barrier B on N su h that for every innite M Nthereisinnite N M su hthatthe orresp onding subsequen e(x

n )

n2N of (x

n

anybarrier.) Itturnsoutthatonlyinde omp osable ountable ordinals matterforthisnotion. We shall see, extending the well-known example of Maurey-Rosenthal of a normalized weakly-null sequen e without un onditional subsequen es, that every normalized basi sequen e hasa subsequen ewhi his!-un onditionally saturated,andthatthis annotb eextendedfurther. For example,weshowthatforeveryinde omp osable ountable ordinal >! thereisa ompa tum K of Cantor-Bendixsonrank +1and anormalized1-basi weakly-null sequen e(x

n

)C(K) su hthat (x

n

)is-un onditionally saturatedforall < butnot -un onditionallysaturated. More pre isely, the summing basis of

0

is nitely blo k-representable in every subsequen e of (x

n

), andso in parti ular,nosubsequen e of(x n

)is un onditional.

2. Preliminaries

LetNdenotethesetofallnon-negativeintegersandletFINdenotethefamilyofallnitesets ofN. Thetop ologyon FINistheone indu edfromtheCantor ub e

N

2viatheidenti ation of subsets of N with their hara teristi s fun tion. Observe that this top ology oin ides with the oneindu edby

0

,theBana hspa eofsequen es onvergingtozero,withthesameidenti ation of nite setsand orresp onding hara teristi fun tions. Thus, we saythat a family F FIN is ompa t if it is a ompa t spa e under the indu ed top ology. We saythat F  FIN is pre- ompa tifitstop ologi al losureF

top

takenin theCantor ub e N

2 onsistsonlyofnitesubsets of N:Given X ;Y Nwewrite

(1)X <Y imaxX <minY. Wewill usethe onvention;<X andX <; forevery X. (2)X vY iX Y and X <Y nX.

A sequen e (s i

) of nite setsof integers is alled a blo k sequen e is i

<s j

forevery i<j, anditis alleda
-sequen eithereissomenitesetssu hthatsvs

i

(i2N)and(s i

ns)isa blo ksequen e. Thesetsis alled therootof(s

i

). Notethats i

! i

siforeverysubsequen eof (s

i

)hasa
-subsequen e withro ots. It followsthatthetop ologi al losureF ofapre- ompa t family F ofnite subsets ofN isin luded in its downwards losure

F 

=fst : t2Fg

withresp e t tothein lusion relation and alsoin luded in its downwards losure

F v

=fsvt : t2Fg

withresp e t tothe relation v:We saythat a family F FIN is-hereditary if F =F 

and v-hereditaryifF =F

v

:The-hereditaryfamilieswill simplyb e alled hereditaryfamilies. We shall onsider the following two restri tions of a given family F of subsets of N to a nite or innite subsetX of N

F X =fs2F : sXg;

F[X℄=fs\X : s2Fg:

There are various ways to asso iate an ordinal index to a pre- ompa t family F of nite subsets of N. All these ordinal indi es are based on the fa t that for n 2 N, the index of the family

is smallerorequal from thatof F:Forexample,one may onsider the Cantor-Bendixsonindex r (F),theminimalordinalforwhi htheiteratedCantor-Bendixsonderivative

(F)isequalto ;,then learlyr (F

fng

)r (F)foralln2N:Re allthatF isthesetofallprop era umulation p oints of F and that 

 (F) = T  < ( 

(F)):The rank is well dened sin e F is ountable andthereforeas attered ompa tumsothesequen e



(F)ofiterated derivativesmustvanish. Observe thatif F is anonempty ompa t,thenne essarilyr (F) isa su essorordinal.

We are nowready tointro du e thebasi ombinatorial on epts of this se tion. For this we need thefollowing pie e of notation,whereX andY aresubsets of N



X=XnfminXgand X=Y =fm2X : maxY <mg

The set 

X is alled the shift of X. Given integer n 2 N, we write X=n todenote X=fng = fm2X : m>ng. Thefollowing notionshaveb een intro du ed byNash-Williams.

Denition 2.1. ([15℄)Let F FIN.

(1)F is alled thin if s6vt forevery pair s,t ofdistin t memb ersof F. (2)F is alled Sperner if s*tforevery pair s6=t2F.

(3) F is alled Ramsey if for every nite partition F = F 0

[


[F k

there is an innite set M Nsu h thatatmostone of therestri tions F

i

M isnon-empty.

(4)F is alledafront onM if F P(M),itisthin,and foreveryinniteN M thereissome s2F su h thatsvN.

(5) F is alled abarrier on M if F P(M), itis Sp erner, and forevery innite N M there is somes2F su h thatsvN.

Clearly, every barrier is a front but not vi e-versa. For example, the family N [k ℄

of all k -element subsets of N is a barrier. The basi result of Nash-Williams [15℄ says that every front (andthereforeeverybarrier)isRamsey. Sin easwe willsee so onthere aremanymorebarriers thanthoseoftheformN

[k ℄

this isafarrea hinggeneralizationof the lassi al resultofRamsey. To see a typi al appli ation, let F b ea front on someinnite set M and onsider its partition F =F 0 [F 1 ;whereF 0

isthe family of all-minimal elementsof F. Sin eF isRamseythere is aninnite N M su hthatone oftherestri tions F

i

M isempty. Note thatF 1

N must b e empty. Sin e F

0

N is learly a Sp erner family, it is a barrier on N. Thus we have shown that every front has a restri tion that is a barrier. Sin e barrier are more pleasant to work withone might wonder whyintro du ing thenotion of front atall. Thereasonis thatindu tive onstru tionsleadmorenaturallytofrontsratherthanbarriers. Togetanidea ab outthis,itis instru tive to onsider thefollowing notion intro du ed byPudlak andRodl.

Denition 2.2. ([22℄) Fora given ountable ordinal , a family F ofnite subsets of a given innite set M is alled -uniform on M provided that:

(a)=0implies F =f;g, (b)=+1implies thatF

fng

is-uniform on M=n,

( ) >0 limit implies that there isan in reasing sequen e f n g n2M of ordinals onverging to  su h thatF fng is n

Remark 2.3. (a)If F isafront onM,thenF =F .

(b)If F isuniformon M,thenitis afront (thoughnotne essarily a barrier)on M.

( )IfF is-uniform(front,barrier)on M and :M !N istheunique order-preserving onto mappingb etweenM andN,then "F =f"s : s2Fgis -uniform(front,barrier)on M. (d) If F is -uniform (front, barrier) on M then F  N is -uniform (front, barrier)on N for every N M.

(e) IfF is uniform (front,barrier)on M,thenforevery s2F v

thefamily

F s

=ft : s<t and s[t2Fg

is uniform(front,barrier)on M=s. (e)IfF is-uniformonM,then 

(F)=f;g,hen er (F)=+1. (Hint: usethat  (F fng )= (  (F)) fng

forevery  and every ompa t family F).

(f)It iseasytoprove byindu tion on nthateveryn-uniform family onM is oftheform M [n℄

. Thisis notthe ase in general.

(g)An imp ortant example of a!-uniformbarrier on Nisthefamily S =fs:jsj=min (s)+1g. We all S a S hreier barrier sin e its downwards losure is ommonly alled a S hreier family. Indeed, it an b e proved a B is a !-uniform family on M i there is an unb ounded mapping f :M !! su h thatB=fsM : jsj=f(mins)+1g.

The following result based on Nash-Williams' extension of Ramsey's theorem explains the relationship b etween the on eptsintro du ed ab ove(see [4℄forpro ofsand fuller dis ussion).

Prop osition 2.4. Thefollowing areequivalentfor a familyF of nitesubsetsof N: (a) F isRamsey.

(b) There isan inniteM Nsu h that F M isSperner.

( ) There isan inniteM Nsu h that F M iseither empty or uniformon M. (d) Thereisan inniteM Nsu h that F M iseither empty or a front onM. (e) There isan inniteM Nsu h that F M iseither empty or a barrier onM. (f)Thereis aninniteM N su h that F M is thin.

(g)ThereisaninniteM Nsu h thatfor every inniteN M the restri tion F N annot besplitinto two disjoint families that areuniform onN. 

InthiskindofRamseytheoryonefrequently p erformsdiagonalisation argumentsthat anb e formalized usingthefollowing notion.

Denition 2.5. Aninnitesequen e(M k

) k 2N

ofinnitesubsetsofNis alled afusionsequen e of subsets of M Nif forall k2N:

(a)M k +1 M k M, (b)m k <m k +1 , wherem k =minM k . Theinnite setM

1 =fm

k g

k 2N

is alled thefusion set (or limit)of thesequen e (M k

) k 2N

:

We have also the following simple fa ts onne ting these ombinatorial notions with the top ologi al on epts onsidered attheb eginning of this se tion.

(a) IfF is a barrier on M thenF  =F v =F, and hen eF  isa ompa t family. (b) IfF is a barrier onM then forevery N M, F N

 =F

 N.

( ) Suppose that F is a barrier on M. Then for every N M su h that M nN is innite we have that F[N℄=F N



, and in parti ular F[N℄isdownwards losed. (d) A family F M

[<1℄

is the topologi al losure of a barrier onM i F v max

=F  max

is a barrier on M.

Barriers des rib e small families ofnite sets,as itisshownin thefollowing.

Theorem 2.7. [14℄ Let F FIN bean arbitrary family. Then thereis an innite setM N su h that either

(a) F[M℄is the losureof a uniformbarrier on M,or (b) M

[1℄ F

 .

Note thatitfollows thatif F is pre- ompa tthen ondition(a) musthold.

We shall follow standard terminology and notation when dealing with sequen es in Bana h spa es(see [13℄). We re allnow fewstandarddenitions we aregoing tousealong this pap er.

Denition 2.8. Let(x i

)b e asequen e in a Bana hspa e E. (a)(x

i

) is alled weakly-null iforevery x 

2E 

,thesequen e of s alars(x  (x i )) i tendsto0. (b)(x i

) is alled aS hauder basis ofE i forevery x2E there is aunique sequen e of s alars (a i ) su h that x = P i a i x i

. This is equivalent to saythat x i

6=0 for every i, the losed linear spanof(x

i

) isX,and thereisa onstant1su hthatforevery sequen eofs alars(a i ),and every intervalI N, k X i2I a i x i k k X i2N a i x i k: (1) ( ) (x i

) is alled a basi sequen e iit is a S hauderbasis of its losed linear span, i.e.,x i

6=0 for every i, and there is   1 su h that for every sequen e of s alars(a

i

), and every interval I N,k P i2I a i x i k k P a i x i

k. Theinmum ofthose onstantsis alled thebasi onstant of (x

i ). (d)(x

i

) is alled  -un onditional (1) iforeverysequen e ofs alars(a i

), andevery subset AN, k X i2A a i x i k k X i2N a i x i k: (2) (x i

) is alled un onditional if it is -un onditional forsome1. Given two basi sequen es (x

i ) i2M and (y i ) i2N

of some Bana h spa es E and F, indexed by the innite sets M;N  N, we say that (x

i ) i2M  E and (y i ) i2N  F are  -equivalent, denoted by (x i ) i2M   (y i ) i2N

, if the order preserving bije tion b etween the two index-sets M and N lifts naturally to an isomorphism b etween the orresp onding losed linear spans of these sequen es sending x

i toy

(i) .

The sequen eof evaluationfun tionals of 0

is thebiorthogonalsequen e (p i ) ofthe natural basis (e i ) of 0 , i.e. if x = P i a i e i 2 0 , then p i (x) = a i

. Note that weakly ompa t subsets K of

0

are hara terized bythe prop ertythat every sequen e in K hasa p ointwise onverging subsequen e to an element of K. It is lear that forevery weakly- ompa t subsetK  the

restri tionsofevaluationmappings(p i

)toKisweakly-nullinC(K). Thesequen eofrestri tions will also b e denoted by (p

i

). Observe that (p i

) asa sequen e in the Bana h spa e C(K) is a monotone basi sequen e i K is losed under restri tiontoinitial intervals.

There are two parti ularly imp ortant examples of weakly- ompa t subsets of 0

naturally asso iated toa normalizedweaklynull sequen e (x

i )

i2M

of aBana h spa e E: (a)the set

R E ((x i ) i2M )=f(x  (x i )) i2M 2 0 : x  2B E  g

is symmetri ,1-b oundedand weakly- ompa tsubsetof 0

. (b)If E=C(K),K ompa tum,then theset

R K ((x i ) i2M )=f(x i ( )) i2M 2 0 : 2Kg

is also1-b ounded and weakly- ompa t. In b oth ases one has that (x

i )

i2M

is 1-equivalent to the evaluation mapping sequen es of C(R E ((x i ) i2M ))and C(R K ((x i ) i2M )). WesaythatasubsetX of

0

isweaklypre- ompa tif its losurerelativetotheweaktop ology of

0

is weakly ompa t. We have thenthefollowing, notdiÆ ult toprove.

Prop osition 2.9. (a)F FINispre- ompa ti thesetf s

: s2FIN g 0

of hara teristi fun tionsof setsin F isweakly-pre- ompa t.

(b) For every weakly-pre- ompa t subset X of 0

and every ">0 one has that

supp "

X=ffn2N : j(n)j"g : 2Xg is pre- ompa t :

Finally, we intro du e few ombinatorial notions on erning mappings from families of nite setsof integersinto

0

. Formoredetails see[14℄.

Denition 2.10. ([14℄) LetF FIN b e anarbitrary family,and let f :F ! 0

. (a)f isinternalif forevery s2F one hasthat suppf(s)s.

(b)f is uniformif for every t2FIN one hasthat

jf'(s)(min(s=t)) : tvs; s2Fgj=1

( ) f isLips hitz if foreveryt2FIN one hasthat

jf'(s)t : tvs;s2Fgj=1

(d)f is alled a U-mappingif F if it isinternal and uniform. (e) f is alled aL-mapping if F if itis internaland Lips hitz.

Remark 2.11. (a) Every uniform mapping is Lips hitz, but the re ipro al is in general false. For example, the mapping f : FIN !

0

dened by f(s)(i) = i if i 2 s and f(s)(i) = 0 is Lips hitz but notuniform.

(b)EveryL-mappingf :F ! 0

anb enaturallyextendedtoa ontinuousmappingf 0 :F v ! 0 bysetting f 0

(t)=f(s)tfor(any)s2F su hthat tvs.

there is N M,a barrier C on N,and an internalmapping g:BN !C su h thatfor every s;t2BN onehas thatf(s)=f(t)ig(s)=g(t).

(d)U-mappingswereusedin[14℄topro du esomeweakly-nullsequen es playingimp ortantrole intheb etterunderstandingofanabstra t on eptofun onditionality(see[14℄formoredetails).

Themain resulton mappingsdened on barriersisthefollowing:

Theorem 2.12. [14℄Supposethat B is a barrier onM,K  0

isweakly- ompa t andsuppose thatf :B!K. Thenforevery ">0thereisN M andthereisaU-mappingg:BN !

00 su h that for every s2BN one has that

kf(s)N g(s)k `

1 ":

Corollary 2.13. Suppose that f : B ! 0

is an internal mapping dened on a barrier B . Suppose that in addition f is bounded, i.e. there is C su h that for every s 2 B one has that kf(s)k

1

C. Then for every ">0 thereexists is a U-mapping g:BN ! 00

su h that for every s2BN one has that

kf(s) g(s)k `

1 ":

Proof. Letusproverstthattheimageoff isweakly-pre- ompa t: Forsupp osethat(f(s n

)) n is an arbitrarysequen e. Let M Nb e su h that (suppf(s

n )) n2M onvergestosome s2B v . This is p ossible b e ause f is internal. Sin e f is b ounded, we an nd N  M su h that (f(s n )) n2N is weak- onvergentin 0 .

Nowthe desiredresult followsfrom 2.12byusing that f isin addition internal. 

3. 0

-saturation of C(K) for a ountable ompa tum K

Re alltheresult of Pel zynski and Semadeni[21℄whi h says that every Bana hspa e of the form C(K) for K a ountable ompa tum is

0

-saturated in the sense that every of its losed innite-dimensi onal subspa es ontains an isomorphi opy of

0

: The purp ose of this se tion is to examine the

0

-saturation using the theory of mappings on barriers develop ed ab ove in Se tion 3. We startwitha onvenient reformulationoftheproblem. Westartwithadenition.

Denition 3.1. Fora given subsetX of 0

,letsuppX=ffi2N : (i)6=0g : 2Xg b e the support set of X. Wesaythat aweakly ompa tsubsetK of

0

issupported by a barrier on M if its supp ortset suppK isthe isthe losure ofa uniform barrieron M.

Lemma 3.2. Suppose that K is a ountable ompa tum. Suppose that (x i

) C(K) is a nor-malized weakly null sequen e. Then for every ">0 there is subsequen e(x

i ) i2M and a weakly- ompa tsubsetL 0

supported byabarrieronNofranknotbiggerthantheCantor-Bendixson rank of K su h that (x

i )

i2M

and the evaluationmapping (p i

) i2N

of C(L)are(1+")-equivalent.

Proof. Fix">0. Findrst an stri tly de reasing sequen e (" i ) su h that P i " i "and su h that f" i : i2Ng\fjx i ( )j : 2Kg=;: (3)

This is p ossible b e ause K is ountable. Now dene ' : K ! P(N) by '( ) = fi2N : jx( )j"g. Note that (3) implies that ' is a ontinuous fun tion. Enumerate K =f g .

Sin e (x i

) is weakly-null we an nd a fusion sequen e (M k

) su h that for every k and every i2M

k

one hasthat jx i

( k

)j<" k

. Nowif we setN tob ethe orresp onding fusion setthen for every k one has that fi2M : jx

i ( k )j" i g fn 0 ;:::;n k 1

g. This meansthat themapping =

M

is ontinuous withimagein luded in FIN . Set N = 

M and denotethe immediate prede essorofi2N inM byi . Sin eK isa zero-dimensional ompa tum,we annd lop en setsC i K (i2N) su h that Knx 1 i (( " i ;" i ))C i Knx 1 i ([ " i ;" i ℄)foreveryi2N. Sety i = C i x i

forea h i2N. So one has (i) kx i y i k K <" i ,so(x i ) i2N and (y i ) i2N

are 1+"-equivalent, and (ii)forevery 2K and every i2N,if jy

i ( )j" i ,then y i ( )=0. Sin e forevery 2K,by(ii)ab ove,one hasthat

fi2N : y i ( )6=0g=fi2N : 2C i and jx i ( )j" i g= ( );

itfollowsthatthesupp ortsetF ofR K

((y i

) i2N

) oin idewiththeimageof ,soitisa ompa t family ofN. WeusenowTheorem2.7tondP N su hthatF[P℄isthe losureofauniform barrieronP. ThisimpliesthatR

K ((y

i )

i2P

)issupp ortedbyabarrierBonP. Letb etheunique orderpreserving mappingfromNontoP,and let:

0 P =f 2 0 : suppPg 0 ! 0 b edened by ()(n)=( (n)). Thisisan homeomorphism b etween

0

P and 0

,b othwith the weak top ology,so L ="R

K ((y i ) i2P ) is a weakly- ompa t subsetof 0

and supp orted by the barrier 

1

B = f 1

s : s2B g on N. Now it is easy to see that the evaluation mapping (p

i )

i2N

ofC(L)is a normalizedweakly-null sequen e1+"-equivalent to(x i

) i2P

. 

Theorem 3.3. Suppose that (x i

)C(K) isa normalized weakly-nullsequen efor a ountable ompa tum K. Then there is a onstant C  1, an innite set M, a uniform barrier B on M whose rank is at most the Cantor-Bendixson rank of K, and some U-mapping :B !

+ 00 su h that for every blo k sequen e (s

n

) B the orresponding sequen eof linear ombinations ( P i2s n ((s n ))(i)x i ) n

isa normalizedblo k sequen eC-equivalent tothe unit ve tor basisof 0

.

Proof. Thepro of isbyindu tionon theCantor-Bendixsonrank ofK. Firstofall, byLemma 3.2we mayassumethatK isa weakly- ompa tsubsetof

0

supp orted bya barrierB onNand thatthenormalizedweaklynullsequen e(x

i

)isthe orresp ondingevaluationmappingsequen e (p i ) i2N . If =1,then B=N [1℄ and learly (p i

) is equivalent totheunit ve torbasisof 0

. So assumethat>1. Bygoing toasubsequen eof(p

i

)ifneeded, wemayalsoassumeinthis ase that jsj2 forevery s2B . Forea h integer n set F

n = S mn B fmg . Sin e B is a -uniform family, we have that for every n, 

 F

n

= ;, so its Cantor-Bendixson rank is stri tly smaller than+1. Forea h n2N,let

K n

=ff s : s2F n

g:

Thisisa ompa tumwhosesupp ortisF n

andwhoserankisstri tlysmallerthan+1. So,the evaluationmappingsequen e (p

i

)isa weakly-null sequen eofC(K n

) foreveryn. Observethat foreverysequen e of s alars(a

i ) we havethat k X a i p i k n =k X a i p i k K n =sup fk X a i p i k K : s2F n g: (4)

Using thefa tthatthefamilyF n

ishereditary, weobtain that(p i

) is1-un onditional. Sin e we assumethatallthesingletonsfigb elongtoF

n ,itfollowsthat(p i ) i1 isindeeda1-un onditional normalized weakly nullsequen e in C(K

n ). Fix">0,andlet("

n )

n

b easummablesequen ewith P

n "

n

<"=2. BytheRamseyprop erty of the uniformbarrier B ,we an nd a fusion sequen e (M

k )

k

su h that, setting n k

=minM k forea hk2N,we have thatforeveryk thefollowing di hotomyholds:

(I) Either for every s 2 B  M k there is some  k (s) 2 00 with supp k (s)  s, 0   k

(s)(i) 1 for every i2 s, and su h that for every su h that k P i2s  k (s)(i)p i k K = 1 while k P i2s  k (s)(i)p i k n k <" k ,or else (I I)k P i2s a i p i k K 2" 1 k k P i2s a i p i k n k foreverys2BM k and every(a i ) i2s . Supp ose rst that (I) holds for every k . Let M

1 = fn

k

g b e the orresp onding fusion set. Then let C =B M 1 . For s2C,dene (s)=  k (s), where n k

=mins. This is well dened sin e s2BM

k

. Fora given s2C,let

x(s)= X i2s (s)(i)p i :

Our intention is toshowthat forevery blo ksequen e (s i

) i

in C one hasthat (x(s i )) i is 2+ "-equivalent tothe 0

-basis. So xsu hsequen e(s i

)and let(b i

) i2N

b easequen e ofs alarswith jb

i

j1forevery integeri. Sin eea hx(s i

) isnormalizedand sin e(p i ) ismonotone,weobtain that k X i b i x(s i )k K (1=2)k X i b i e i k 1 :

Supp ose that  2K, and let i 0 =min fi : s i \supp6=;g. Fixi>i 0 ,and let k i b e su h that n ki =mins i . Sin e supp\s i 2F maxsi 0 wehave that jx(s i )()jk X j2s i \suppf a (k i ) i p i k maxs i 0 <" k i : (5) It followsthat j X i b i x(s i )()jjb i 0 j+ X i>i 0 jb i jjx(s i )()jjb i 0 j+ " 2 : (6) So,k P i b i x(s i )k K (1+"=2)k P b i e i k 1

. Finally useCorollary2.13top erturb  andmakeit U-mapping.

Supp ose now that k 0

is the rst k su h that (I I) holds for k . Set M = M k

. It readily followsthatforeveryxin the losed linear spanof(p

i )

i2M

onehasthatkxk K " 1 k0 kxk n k 0 . By indu tive hyp othesis applied to (p

i

) C(K n

k 0

), there is some C  1, some uniform barrier C on someN M of rank notbiggerthan theone ofK

n k

0

and some fullling the on lusions of the Lemma. Fix s 2 C. Then k(s)k

n k 0 = 1, so we an nd some t s  s su h that 1 = k(s)k n k 0 = k(s)  t s k K

. Observe that, by 1-un onditionality of k
k n k 0 , k(s)  tk n k 0 = 1. Dene  : C ! 00 by (s) = (s)  t s

. Finally, let us he k that (x(s i ))  C(K) is C" 1 n k -equivalent tothe 0

-basis forevery blo ksequen e (s i ) i in C. Fix s alars(a i ), ja i j1 (i2N). We obtain the inequality k

P a i (s i )k K  (1=2)k P a i e i k 1

sequen e (p i ). Now, k X i a i (s i )k K  1 " n k 0 k X i a i (s i )k n k 0  1 " n k 0 k X i a i (s i )k n k 0  C " n k 0 k X i a i e i k 1 : (7)  4. Conditionality

We start with the following natural slightly variation on the notion of S 

-un onditionality from [3℄,and whi h isa generalizationof un onditionality (see Denition 2.8(d)).

Denition 4.1. LetF b eafamilyofnitesetsofintegers. Anormalizedbasi sequen e(x n

)of a Bana hspa e E is alled F-un onditional with onstantatmostC 1i forevery sequen e of s alars(a n ), sup s2F k X n2s a n x n kCk X n2N a n x n k:

Thisgeneralizesthenotionofun onditionality overedbythe aseofF=FIN . Thequestion is whethereverynormalized weakly-nullsequen e hasaF-un onditional subsequen e. Observe that thesubsequen e (x

n )

n2M

is F-un onditional i it isF[M℄-un onditional, so theexisten e of an F-un onditional subsequen e is losely related to the form of the tra es F[M℄. If we assume that in additionthe family F is hereditary, then, by theTheorem 2.7,two p ossibiliti es an o ur: The rst one is that some tra e of F onsists on all nite subsets of some innite set M. In this ase, for subsequen es of (x

n )

n2M

the F-un onditionality oin ides with the un onditionality. The se ond ase iswhen sometra eof F is the losure of a uniformbarrier. So one is naturally led to examining the standard ompa t families of nite subsets of N. We b egin with the following p ositive result announ ed in [16℄ and rst proved by E. Odell [20℄ on erning theS hreierfamily S =fsN:jsjmin (s)+1g.

Theorem 4.2. Suppose that (x n

) is a normalized weakly-null sequen eof a Bana h spa e E. For every ">0 thereis a S-un onditional subsequen ewith onstant2+". 

Re allthatifF isabarrier onsomesetM thenitstra eF[N℄onany o-innite subsetN of M is hereditary and that forevery pair F

0

and F 1

of barriers on the samedomain M there is an innite set N M su h that F

0 [N℄F 1 [N℄orF 1 [N℄F 0

[N℄. Sin e thetwo alternatives aredep endent on theranksof F

0 and F

1

;one is naturallyled tothefollowing measurement of un onditionality.

Denition 4.3. Supp ose that is a ountable ordinal. A normalized basi sequen e (x n

) of a Bana h spa e E is alled -un onditionally saturated with onstant at most C  1 if for every -uniform barrier Bon N and forevery innite M there is innite N M su h that the orresp onding subsequen e (x

n )

n2N of(x

n

)is B -un onditional with onstant atmostC. We say that (x

n )

n

is -un onditionally saturated if it is -un onditionally saturated with onstant C forsomeC 1.

Remark 4.4. (a)A sequen e (x n

) n

is -un onditionally saturatedi givena -uniformbarrier B everysubsequen e of(x ) hasafurtherB -un onditional subsequen e. Thereasonforthisis

that giventwo -uniform barriersB and C on a set M wehavethat there is N M su h that eitherBN CN BNN

[1℄

orthesymmetri situationholds,whereFG=fs[t : s2G;t2F and s<tg (see [4℄).

(b)ItfollowsfromTheorem4.2thateverynormalizedweaklynullsequen eis!-un onditionally saturated. Sin ethe!-uniformbarriersareoftheformfs2FIN : jsj=f(mins)+1gforsome unb ounded mappingf :M !None an easily mo dify the pro of of Theorem4.2 toprovethat every normalized weakly-null sequen e is !-un onditionally saturated with onstant at most 2+".

( ) If the normalized basi sequen e (x n

) is monotone, then it is B -un onditional i it is B -un onditional forevery uniform barrierB onN.

(d) An analysis of the Maurey-Rosenthal [16℄example of a weakly-null sequen e (x n

) with no un onditional basi subsequen e (see Example 4.5 b elow) reveals an !

2

-uniform barrier B MR su h thatnoinnite subsequen e (x

n )

n2M is B

MR

-un onditional withanynite onstant C. So this is an example of a normalized weakly-null sequen e with no !

2

-un onditionally saturated subsequen e.

(e) Re allthat an ordinal is alled inde omp osable if for every  < ,!  . Equivalently, =!

forsome. Supp ose that is themaximal inde omp osable ordinalsmaller than some xed ordinal . Then a normalized basi sequen e (x

n

) is -un onditionall y saturated if and only itis -un onditionally saturated.

Example 4.5. First of all, for a xed 0 < " < 1 ho ose a fast in reasing sequen e (m i ) su h that 1 X i=0 X j6=i min f( m i m j ) 1=2 ;( m j m i ) 1=2 g " 2 : (8) Let FIN [<1℄

b e the olle tion of all nite blo k sequen es E 0 <E 1 <


<E k of nonempty nite subsetsof N. Now ho osea 1 1fun tion

:FIN [<1℄ !fm i g (9) su h that'((s i ) n i=0 )>s n for all(s i ) 2FIN [<1℄ Nowlet B MR

b e thefamily of unions s 0 [s 1 [


[s n

ofnite setssu hthat (a)(s i ) is blo kand s 0 =fng. (b)js i j=(s 0 ;:::;s i 1 ) (1in). It turns out that B

MR

is a ! 2

-uniform barrier on N (see Prop osition 4.11 b elow), hen e B

MR =B

MR v

isa ompa t family withrank! 2

+1. Observe thatbydenition, everys2B MR hasaunique de omp osition s=fng[s

1

[


[s n

satisfying (a)and(b)ab ove. Nowdene the mapping:B MR ! 00 , (s)=e n + n X i=1 1 js i j 1 2 X k 2si e k : (10)

It follows that  is a U-mapping dened on the barrier B MR

. Now we an dene the Bana h spa e X

MR

asthe ompletion of 00

under thenorm

ThenaturalHamelbasis(e n

)of 00

isnowanormalizedweakly-nullmonotonebasisofX MR

with-out un onditional subsequen es. Indeed, without !

2

-un onditionally saturated subsequen es. Moreoverthis weakly-nullsequen e has theprop ertythat thesumming basis(S

i

) of , the Ba-na h spa e of onvergent sequen es of reals, is nitely-blo k representable in thelinear spanof every subsequen e of (e

i

) (and sothe summing basisof 0

), morepre isely, forevery M, every n2Nandevery ">0 thereis anormalized blo k subsequen e (x

i ) n 1 i=0 of (e i ) i2M

su hthat for every sequen e of s alars(a

i ) n 1 i=0 , maxfj m X i=0 a i j : m<ngk n 1 X i=0 a i x i k C(K) (1+")maxfj m X i=0 a i j : m<ng:

On the other hand, by Prop osition 4.2 the sequen e (p i

) is !-un onditionally saturated with onstant 2.

Another presentation of this spa e is the following: Sin e  is uniform, it is Lips hitz, so there isaunique extension:B

MR !

00

,naturallydened by(s)=(t)t, wheret2B MR is (any) su h that sv t. Now dene K = "B

MR 

00

. This is a weakly- ompa t subset of

00

whose rank the same than B MR

, i.e., ! 2

+1. Then the orresp onding evaluation sequen e (p

i

)C(K) is 1-equivalent tothebasis(e i ) i of X MR .

Building on the idea of Example 4.5, we are now going to nd, for every ountable inde- omp osable ordinal ,aU-sequen e with no un onditional subsequen esbut -un onditionally saturatedforevery< . Beforeembarkingintothe onstru tion,weneed tore allalo alized versionofPtak'sLemma. Forthisweneedthefollowingnotation: GivenafamilyF,andn2N, let Fn=fs 0 [


[s n 1 : (s i ) n 1 i=0 F isblo k g:

It an b eshown thatFnisa n-uniform family if F isan -uniform family. Given  2 00 we will write  1=2 to denote ((i) 1=2 ). Given  2 00

and a nite set s, let h;si=h; s i= P i2s (i).

Denition 4.6. A mean isan element 2 + 00

withthe prop ertythat P i2N (i)=1. We say that:B! + 00

isaU-mean-assignment ifis aU-mappingsu h thatforevery s2Bone has that(s) isa mean.

Lemma 4.7. Suppose that B is an -uniform barrier on M,   1. Let = () be the maximal inde omposable ordinal not bigger than ,and let n= n() 2 N, n 1, be su h that n < (n+1). Then for every k2N, k>1, every ">0, and every -uniform barrier C onM with >k there N M and U-mean-assignment :C N !

+ 00 su h that supfh(s) 1 2 ;ti : t2Bg (1+")(n+1) (nk ) 1 2 (11) for every s2CN.

Noti e that if we prove that for every N  M there is one mean  with supp ort in C  N su h that(11)holds,thentheRamseyprop ertyoftheuniform barrierC givestheexisten e for some N  M of a mean-assignment  :C  N !

00

su h that (s) has the prop erty (11)for every s2CN. Then Corollary2.13givesthedesired U-mapping.

Let D b e a -uniform barrier on M (if n =1 we take D = B ), and x N  M. Find rst P N b e su h that (Dnk ) P C aswell as BP D(n+1). Consider(

i ) i2P su h that D fig P is i

-uniform on P=i. Observe thatforeveryi2P we have that i < ,so, sin e is inde omp osable, i !  . Let  0

b e any mean su h that supp 0

2B P. By indu tive hyp othesis applied to appropriate 

i

's, we an nd a blo k sequen e ( j

) nk 1 j=0

of means with supp ortin BP su h thatforevery 1jnk 1,

sup fh 1 2 j

;ti : t2D ; and mintmaxsupp j 1 g< " 2 j+1 : (12) Let  =(1=(nk )) P nk 1 j=0  j

. Observe thatsupp 2(D(nk ))P C. Then, forevery t2B , by(12), h 1 2 ;ti= 1 (nk ) 1 2 k 1 X j=0 X i2t  j (i) 1 2  1+ " 2 (nk ) 1 2 : (13)

Letus p oint outthatsupp is, p ossibly, notasetin C. Howeveritiseasytoslightly p erturb toa newermean with supp ort in C and satisfying (13)for every t2B : Let s2C b esu h that supp vs, and set u=snsupp. LetÆ >0 b esu hthat

(1+ " 2 )(1 Æ) 1=2 +(nk Æjuj) 1=2 1+": (14) Nowset =(1 Æ)+ Æ juj  u : (15)

is amean whosesupp ort iss2C. It an b eshownnowthat forevery t2B , X i2t (i) 1 2  1+" (nk ) 1 2 ; (16)

bythe hoi eof Æ. Finally, lett2Band letus ompute P

i2t ((i))

1=2

: Firstofallwehavethat P i2t ((i)) 1=2 = P i2u ((i)) 1=2

,where u=t\P. Now,sin e u 2BP D(n+1),we an nd t 0 <


<t n in D su h thatuvt 0 [


[t n ,and hen e h 1=2 ;ti= n X j=0 h 1=2 ;t j i (n+1)(1+") (nk ) 1 2 ; (17) aspromised. 

Corollary 4.8. Suppose that B is an -uniform barrier on M,   1. Then for every "> 0 thereissomek=k (;")su hthat forevery -uniform barrier onM with >k thereN M and someU-mean-assignment :CN !

+ 00 su h that, sup fh(s) 1=2 ;ti : t2B g" (18)

Lemma 4.9. Fix an inde omposable ountable  and a sequen e(" n

) of positive reals. Then: (a) there is a olle tion (B

n ) of 

n

-uniform barriers on N=n and a orresponding sequen e of U-mean-assignments n :B n ! + 00

withthe following properties: (a.1)  n >0, sup n  n =,

(a.2) for every m<n and every s2B n supfh n (s) 1 2 ;ti : t2B m g<" n : (19)

(b) Suppose that in addition  = !

with limit. Let  n

"  be any sequen e su h that 

n !

n+1

(n2N). Thenthere isa doublesequen e(B n i

)su h that for every integersnand i (b.1)B n i isan (n) i

-uniform barrier onN=(n+i), with  (n) i >0 and  (n) i " i  n . (b.2) There are U-mean-assignments 

n;i : B n i ! 00

su h that for every s 2 B n i , and every (m;j)< lex (n;i) sup fh n;i (s) 1 2 ;ti : t2B m j g<" n+i ; (20)

wherewe re allthat < lex

denotes the lexi ographi alorder onN 2

dened by (m;i)< lex

(n;j) i m<n, or m=n and i<j.

Proof. (a): Cho ose  n

" n

 su h thatforeveryn2N, n+1 > n k ( n ;" n ),thatis isp ossible sin e  is inde omp osable. LetC

n

b ean  n

-uniform family on N (n2N). By Corollary4.8 we an nd afusion sequen e (M n )su hthat ( ) C m M m C n if mn, and

(d)foreveryn2N there isa U-mean-assignment  n :C n M n ! + 00 su h that supfh n (s) 1 2 ;ti : t2 [ l<n C l g<" n (21) for every s 2 C n  M n . Let M = fm n

g b e the fusion set of (M n

), and  : M ! N b e the orresp ondingorderpreservingontomapping. ItisnotdiÆ ulttoseethatC

n =("B n )(N=n), and  n :C n ! 00

dened naturallyoutof  n

fulls all therequirements. (b): Supp ose that  =!

with limit. Let  n

"  b e anysequen e su h that  n

!  n+1 (n2N).

Claim. Thereisa fusionsequen e(M n ),M n =fm (n) i g,a doublesequen e(B n i ) of (n) i -uniform barriers onM n =m (n) i and U-mean-assignments n;i :B n i ! + 00 su h that (e)  (n) i " i  n (n2N),and (f)for every (m;j)< lex (n;i),every s2B n i and every t2B m j , h( n;i (s)) 1=2 ;ti<" n+i .

Proof of Claim: First,useCorollary4.8appliedto 0

topro du eaninnitesetM 0 =fm (0) i gand asequen e (B 0 i )of  (0) i -uniform barriersonM 0 =fm (0) i g with (0) i " 0 and U-mean-assignments  0;i : B 0 i ! 00

su h that for every i and every s 2 B 0 i , h 0;i (s) 1=2 ;ti  " i for every t 2 B 0 j with j < i. In general, supp ose we have found for every k  n M

k = fm (k ) i g  M k 1 , (B k i )  (k ) i -uniform barriers on M k =m (k ) i and U-mean-assignments  k ;i :B k i ! 00

su hthat for every (k ;j)< lex (m;i) every s2 B m i and every t 2B k j h m;i (s) 1=2 ;ti " m+i

. Forea h kn dene thefollowing families

B k =fsM k :  s2B k g: (22)

Thisis learly an  k -uniform family on M k . Sin e  n !  n+1

,we an use againCorollary 4.8 and nd an innite subset M

n+1 = fm (n+1) i g  M n and a sequen e (B n+1 i ) of  (n+1) i -uniform barriers on M n+1 =m (n+1) i and U-mean-assignments  n+1;i : B n+1 i ! 00

su h that for every s2B n+1 i , supfh( n+1;i (s)) 1 2 ;ti : t2 [ k n B m [ [ j<i B (n+1) j g<" n+i+1 ; (23)

so,in parti ularforevery knandevery t2B k j ,h( n+1;i (s)) 1 2 ;ti<" n+i+1 . 

Let M b e thefusion set of (M n ),i.e. M =fm (n) 0 g. Observe thatm (n+i) 0 m (n) i forevery n and i, soM=m (n) 0 M n =m (n) i . Set C n i =B n i  (M=m (n+i) 0 ). Thisis an  (n) i -uniform barrier on M=m (n+i) 0 . Consider n;i = n;i C n i :C n i ! 00

hastheprop ertythatforevery(m;j)< lex (n;i), everyeverys2C n i and everyt2C m j ,h( n;i (s)) 1=2 ;ti<" n+i . Nowuse:M !N,(m (n) 0 )=0, todene thedesired mean-assignmentsand families. 

Remark 4.10. Observe that if B is -uniform on M with  > 0, then M [1℄

 B . It readily follows that the mean-assignments 

n and 

n;i

obtainedin Lemma 4.9 have the prop ertythat k n (s) 1=2 k 1 " n andk n;i (s) 1=2 k 1 " n+i

forevery sin the orresp onding domains.

Prop osition 4.11. (a) Suppose that C and B i are  and  i -uniform families on M (i 2 N) with  i ",  i ;1. Let :FIN [<1℄

!Nbe1-1. Then for every n2N the family

D=fs 0 [


[s n : (s i ) is blo k, s 0 2C and s i 2B  ((s0;:::;si 1)) for every 1in 1g

is -uniform on M, where =n+ if 1  <! and n>0, and =n+ if  ! or n=0. (b)Supposethat B i is i -uniform onM (i2N)with i ". Let :FIN [<1℄ !Nbe1-1. Then the family C =ffng[s 0 [


[s n 1 :(fng;s 0 ;:::;s n 1 ) is blo k, and s i 2B  ((fng;s0;:::;si 1)) for every 0in 1g is !-uniform on M.

Proof. (a): The pro of is by indu tion on n. If n = 0, the result is lear. So supp ose that n>0. Nowthe pro of is byindu tion on . Supp ose rst that  =1. Then C =M

[1℄ , and so, foreverym2M D fmg =fs 1 [


[s n :(s 1 ;s 2 ;:::;s n ) isblo k, s 1 2B  ((fmg)) and s i 2B  ((fmg;s1;s2;:::;si 1)) forevery2in 1g;

so,byindu tive hyp othesis, D fmg

is(n 1)+ m

-uniform onM=m,dep ending whether  m

is nite orinnite, but in any ase with

m

". Hen e D isn-uniform on M. The general ase for1 <! is shownin the sameway.

Supp ose nowthat !. Then forevery m2M

D fmg =ft[s 1 [


[s n : (t;s 1 ;:::;s n ) isblo k, t2C fmg and s i 2B  ((fmg[t;s ;:::;s )) for every 1in 1g;

Byindu tivehyp othesis,D fmg isn+ m -uniformonM=m,with m ",soDisn+-uniform on M,as desired.

(b)follows easily from (a). 

The following is ageneralization of Maurey-Rosenthalexample forarbitrary ountable inde- omp osable ordinal.

Theorem 4.12. For every ountable inde omposable ordinal  there is a normalized weakly-null sequen e whi h is -un onditionally saturated for every  <  but without un onditional subsequen es.

Proof. Our example is a slightly mo di ation of a U-sequen e intro du ed in [14℄. So, we are going to dene a -uniform barrier B on N, a U-mean-assignment ' : B !

00

and some GFINFIN and thendene thenorm on

00 by

kk=maxfkk 1

;supfjh'(s)t;ij : (s;t)2Ggg (24)

whereGFINFINissu hthatitsrstproje tionisB . Noti ethatsomesortofrestri tions have to b e needed in the formula(24), sin e it is not diÆ ult to see that that for a ompa t and hereditary family F, a normalized weakly-null sequen e (x

i )

i

is F-un onditional i it is equivalent totheevaluation mappingsequen e (p

i ) i ofa weakly- ompa tsubsetK  0 thatis F- losed,i.e. losed underrestri tion on elements ofF.

Fix " > 0, and let " n

= "=2 n+3

. Supp ose that  = !

. There are two ases to onsider. Supp oserstthat =+1. WeapplyLemma4.9(a)totheinde omp osableordinal!

and(" n

) topro du ethe orresp onding sequen es of barriers(C

n ) and U-mean-assignments n :C n ! 00 (n2N)satisfyingthe on lusions(a.1)and(a.2)oftheLemma. If islimit,thenweusethepart (b) ofthat lemma to pro du ea double sequen e (B

n i ) and U-mean-assignments n;i :C n i ! 00 satisfying (b.1)and (b.2). In order tounify the two ases weset forn;i,

B n i = ( C i if issu essor ordinal C n i if islimit ordinal and  n;i = (  i if issu essor ordinal  n;i if islimit ordinal : Let  : FIN [<1℄

! N b e 1-1 mapping su h that ((s 0

;:::;s n

)) > maxs n

for every blo k sequen e (s

0 ;:::;s

n

) of nite sets. For ea hndene

C n =fs 0 [


[s n 1 : (s i ) isblo kand s i 2B n  ((fng;s 0 ;:::;s i 1 )) forevery 0ing;

So, byProp osition 4.11,if  =! +1

,then C n

isa ! 

(n 1)+-uniform family on N,where  is su h that B

n  ((fng))

is -uniform; while if =!

with limit, then itis  n

(n 1)+ where  issu hthat B

n  ((fng))

is -uniform. Now let

C=fs2FIN : 

s2C mins

g: (25)

n = mins and s(i) 2 B  (s[i℄)

, and where s[i℄ = (fng;s 0

;:::;s i 1

) (0 i n 1). For every s2C and every is, set

(s;i)=( mins; (s[i℄) (s(i))) 1=2 : Dene now:C! 00 foreverys2C by (s)=e mins + n 1 X i=0 (s;i); (26)

It isnotdiÆ ult tosee that:C! 00

isa U-mapping. Nowdene on 00

thenorm

kk=sup fjh(s)(snt);ij : s2C;ts(i); forsomei<mins g=

=sup fjh(s)(unt);ij : uvs2C;ts(i); forsomei<minsg; (27)

the last equality b e ause  is Lips hitz and supp orted by a front. Let X the ompletion of

00

under this norm. Then the Hamel basis (e n

) n

of 00

is a normalized basis of X, moreover monotone(sin eisLips hitzwithdomainafront)andweakly-null: Toprovethis,itisenough toseethat theset

L=f(s)(unt) : s2B ; uvs; and ts(i)forsomei<mins g

is weakly- ompa t. So, let ((s n ) (u n nt n )) n

a typi al sequen e in L. Sin e C is a front, we an ndan innite setM and u2FIN su hthat(u

n )

n2M

onvergestou andsu hthat(s n

)is a
-systemwithro otuvr . Sin e is Lips hitz de, we obtain that((s

n )t n ) n2M onverges to(s m

)tfor(any)m2M. Ifu=;,then((s n )(t n [u)) n2M onvergesto 0. Otherwise, let N M and j<minub e su h thatt

n s

n

(j) foreveryn2N. Now(t n ) n2N isa sequen e in the losureof B mins  (s[i℄)

,hen e, we an ndP N su hthat(t n

) n2P

is onvergentwithlimit t. Itfollows that((s n )(u n nt n )) n2P haslimit (s n

)(unt)2L,wherenis(any)integerin P. Thenext isa ru ial omputation.

Claim. For every s;t2C and every imins andj mins,we havethat

0h(t;j);(s;i)i ( " maxfmins;mintg ift[j℄6=s[i℄ 1 ift[j℄=s[i℄:

Proof of Claim: Set n = mins, m = mint, and assume that t[j℄ 6= s[i℄. Supp ose rst that  = !

+1

. Then, by denition of the mean assignments, h(t;j);(s;i)i  "

maxf (t[j℄); (s[i℄)g , but (u 0 ;:::;u k )  maxu k

for every blo k sequen e (u i

), whi h derives into the desired in-equality. Assume now that  = !

, limit ordinal. If mins = mint, then h(t;j);(s;i)i "

mins+maxf (t[j℄); (s[i℄)g "

mins

. Whileifmint6=mins,saymint<mins,thenh(t;j);(s;i)i "

mins+ (s[i℄) "

mins .

If(s[i℄)=(t[j℄)=l , thenmins=mint=n, and

h(s;i);(t;j)ik( n;l (s(i))) 1=2 k ` 2 k( n;l (t(j))) 1=2 k ` 2 1; (28)

sin e b oth aremeans. 

Proof of Claim: Fix aninnite set M ofintegers,and l2N. Letv2BM=l , v=fng[v(0)[


[v(n 1)its anoni al de omp osition,and set

x i = X j2v (i) (v;i)(j)e j : (29)

Observe that h(v);x(v;i)i= h(v;i);(v ;i)i= 1, so from the previous laim we obtain that kx

i

k=1. Now onsider s alars(a i ) in 1 with k P in 1 a i S i k 1

=1. Observe thatthis implies thatmax

in 1 ja

i

j2. Weare goingtoshowthat

1k X 0in 1 a i x i k3+": (30)

To get the left hand inequality, supp ose that 1 = k P 0in 1 a i S i k 1 = j P im a i j, where mn 1. Lett=fng[s(0)[


[s(m). By (27)it follows that

k X in 1 a i x i kh(v)t; X in 1 a i x i i=j X im a i j=1: (31)

Next, x s 2 C and t  s(k ) for some k < mins. Supp ose rst that minv = mins. Let i

0

=maxfin 1 : v(i)=s(i)g. If k>i 0

thenbytheprevious laim we obtain

jh(s)(snt); X in 1 a i x i i)j j X ii0 a i j+ X i0<i;jn 1 2jh(s;i);(t;j)ji  k X in 1 a i S i k 1 +2n 2 " n (1+")k X in 1 a i S i k 1 : (32)

Supp ose that ki 0 . Then jh(s)(snt); X in 1 a i x i i)jj X ii 0 ;i6=k a i +a k h(v;k );(v;k )(s(k )nt)ij+ + X i 0 <i;jn 1 2jh(s;i);(t;j)ji 3k X in 1 a i S i k 1 +2n 2 " n (3+")k X in 1 a i S i k 1 : (33)

Supp ose nowthat n= minv 6=mins, say mins<minv. Let i 0

<n, if p ossible, b e su h that mins2v(i 0 ). Then, jh(s)(snt); X in 1 a i x i i)j ja i0 jk(v;i 0 )k 1 +2 X i 0 i<n X 0j<mint h(t;j);(s;i)i  2" n +2n 2 " n ": (34) 

Finally,itreststoshowthatthesequen e(e n

)is-un onditionallysaturatedforevery <. We onsider thetwoobvious ases:

Case 1. =! +1 . Let D=fsN :  s2B 0 mins g: This is an ! 

-uniform family on N sin e ea h family B 0 m

is  m

-uniform and sup m  m = !  . Therefore, thenext laimgives that(e ) is -un onditionally saturatedforevery  <.

Claim. (e n

) n

isD -un onditionalwith onstant at most2+".

Proof of Claim: Fixt2D ,and let(a i

) i2N

b es alarssu hthatk P i2N a i e i k=1. Fixalso s2C. Supp ose rst that mins2t. Then sin e (s[i℄)>minsmintand

 t2B 0 mint weobtain that jh(s); X i2t a i e i ijja mins j+"(1+")k X i a i e i k: (35)

Nowsupp ose thatmins2= t, but s\t6=;(otherwiseh(s); P i2t a i e i i=0). Let i 0

=minfi<mins : s(i)\t6=;g:

Then forevery i 0

<i<minswehave that (s[i℄)>maxs i 0 mint, so j X j2t a j (s;i)(j)j<"  (s[i℄) ; (36) hen e jh(s)u; X i2t a i e i ij j X j2t\s(i 0 ) a j (s;i 0 )(j)j+ X i 0 <i<mins j X j2t a j (s;i)(j)j= = jh(s)(fng[s(0)[


[(s(i 0 )\t)); X imint a i e i ij+ + X i 0 <i<mins j X j2t a j (s;i)(j)jk X imint a i e i k+"k X i2N a i e i k  (2+")k X i a i e i k; (37)

thelastinequality b e ause(e i

)is monotone. 

Case2. =!

, a ountablelimit ordinal. Thedesired resultfollows fromthefollowing fa t.

Claim. For every n2N; the sequen e(e i

) is B n 0

-un onditional with onstantat most 2n+1.

Proof of Claim: Fix n2 Nand t2B n 0 . Let (a i ) i2N

b e s alars su h that k P i2N a i e i k =1. Fix s2 C. Supp ose rst that n mins. Then in a similar manner that in Case 1 one an show that jh(s); X i2t a i e i ijja mins j+"(1+")k X i a i e i k: (38)

Supp ose that m=mins<n, then

jh(s); X i2t a i e i ijja mins j+ m 1 X i=0 j X j2s(i)\t a j (s;i)(j)j= =ja mins j+ m 1 X i=0 jh(s)u i ; X jmin(s(i)\t) a j ij (2m+1)k X i a i e i k: (39) whereu i =s(0)[


[(s(i)\t). 

Corollary 4.13. For every inde omposableordinal  there is a weakly- ompa t K  00 su h that (a) K  B 0

is point-nite (i.e. f(n) : 2Kg is nite for every integer n) supported by a -uniform barrier on N,

(b) the evaluation mapping sequen e(p i

) i

of C(K) is a normalized weakly-null monotone basi sequen e,and

( ) The summing basis of is 4-nitely representable in every subsequen e of (p i ) i ; hen e no subsequen eof (p n

) is un onditional, but (d) (p

i )

i

is -un onditionally saturated for every  <.

Proof. Let C b e the -uniform family on N and let  :C ! 00

b e the U-mapping given in pro ofofTheorem4.12. LetM Nb esu hthatCN isa-uniformbarrieronN. Letb ethe order-preserving mappingfrom M ontoN. Let B= "C =f "(s) : s2Cg and let ':B!

00 b e naturally dened by '(s)=(

1

(s)). B is a uniform barrier on N and 'is a U-mapping. Observe thateverys2B hasa unique de omp osition,given bytheone of

1 s. Let

K =f'(s)(unt) : uvs2B ; ts(i) forsomeig:

Thisisaweakly- ompa tsubsetof 0

,andthe orresp onding evaluationmappingsequen e(p i

) i is 1-equivalent to thesubsequen e (e

n )

n2M

of theweakly-null sequen e (e i

) i

given in thepro of of Theorem4.12. So K fullls alltherequirements. 

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Equipe de Logique Ma th

ematique,

Universit

e Paris 7, 2 pla e Jussieu, 75251 Paris Cedex 05, Fran e.

E-mailaddress:abadlogique.jussieu.fr

C.N.R.S.,U.M.R.7056et

Universit

eParis7,U.F.R.de Math

ematiques,Case7012,2Pla eJussieu, 75251Paris,Cedex5,Fran e.

DepartmentofMathemati s,UniversityofToronto,Toronto,Canada,M5S3G3 E-mailaddress:stevomath.toronto.edu