Journal of Functional Analysis - Volume 257 - Issues 9, 10, 11, 12

1309 Pages·2009·11.44 MB·English

by  A. Connes

JournalofFunctionalAnalysis257(2009)2723–2741
www.elsevier.com/locate/jfa
On a problem of Rolewicz about Banach spaces that
admit support sets
Piotr Koszmider1
InstytutMatematyki,PolitechnikaŁódzka,ul.Wólczan´ska215,90-924Łódz´,Poland
Received22October2008;accepted25July2009
Availableonline13August2009
CommunicatedbyN.Kalton
Abstract
We construct an example of a nonseparable Banach space which does not admit a support set.2 It is
aconsistent(andnecessarilyindependentfromtheaxiomsofZFC)exampleofaspaceC(K)ofcontinuous
functionsonacompactHausdorffKwiththesupremumnorm.Theconstructiondependsonaconstruction
ofaBooleanalgebrawithsomecombinatorialproperties.ThespaceisalsohereditarilyLindelöfintheweak
topology butitdoesn’thaveanynonseparablesubspacenoranynonseparable quotientwhichisaC(K)
spaceforKdispersed.
©2009ElsevierInc.Allrightsreserved.
Keywords:Banachspace;Biorthogonalsystem;Banachspacesofcontinuousfunctions;Independenceresults
1. Introduction
A nonempty closed convex set C in a Banach space X is called a support set if and only if
foreverypoint z∈C thereisacontinuouslinearfunctionalφ on X suchthat φ(z)=inf φ<
C
E-mailaddress:[email protected].
1 While conducting research leading to the results presented in this paper, the author was partially supported by
a research fellowship Produtividade em Pesquisa from National Research Council of Brazil (Conselho Nacional de
Pesquisa, Processo Número 300369/01-8) and by Polish Ministry of Science and Higher Education research grant
No.N201386234.
2 Similarresults(butusingquiteadifferentcounterexample)wereannouncedroughlyatthesametimebyS.Todorce-
vic.
0022-1236/$–seefrontmatter ©2009ElsevierInc.Allrightsreserved.
doi:10.1016/j.jfa.2009.07.016
2724 P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741
sup φ.In[13],Rolewiczprovedthatassumptionsneededforsolvingacertainnaturalproblem
C
can be weakened in the case of a separable Banach space. Namely that there are no separable
supportsets.Rolewiczalsoprovedin[13]thatinanonseparableHilbertspacethereareconvex
setswhicharesupportsetsandheaskedifeverynonseparableBanachspacehasasupportset.
Thisproblemhasbeenaddressedinseveralpaperswhereextractingcertainuncountablese-
quences in a Banach space is used to construct (necessarily nonseparable) support set. In [8]
a Marcuševicˇ’s basis is used for this purpose, in [9] properties of a compact Hausdorff K like
beingnothereditarilyseparableornothereditarilyLindelöfareemployedinthecorresponding
C(K) and in [2] an equivalent condition in terms of sequences from X×X∗ is found. In [4],
besides more detailed analysis in Banach lattices, it is proved that if K carries a nonseparable
regularmeasure,thenC(K)hasasupportsetandthesameconclusionisobtainedforsomeother
classofBanachspacesin[10].
In this paper we provide an example of a nonseparable Banach space with no support set.
It is a consistent example of a nonseparable C(K) space i.e., a Banach space of continuous
functions on a Hausdorff compact space K with the supremum norm. As S. Todorcevic has
provedin[17]thatitisconsistentthatanynonseparableBanachspacecontainsanuncountable
biorthogonalsystem,itfollowsthatthenonexistenceofourexampleisalsoconsistentwiththe
axioms of ZFC, leaving Rolewicz’s problem undecidable. Todorcevic’s result for C(K) for K
totallydisconnectedfollowsfromBooleanalgebraicresultsof[16].
Letusdescribethecontextofthepaperinmoredetails.
InSection2wedescribethepropertiesoftheBooleanalgebraofclopensetsofourK(Kwill
betotallydisconnected,andhencetheStonespaceofaBooleanalgebrai.e.,aparticularcaseof
∗
the structure space with Gelfand’s topology for the abelian C -algebra C(K)). K is in a sense
obtainedfromtheCantorsetbysplittingjustonceeachpointofanuncountableset{r : ξ <ω }
ξ 1
ofpointsoftheCantorset.Alreadyaclassicalexampleofthesplitinterval(i.e., [0,1]×{0,1}
withthelexicographicalordertopology)isofthisform,howeverasprovedin[3,ExampleII5.1],
the corresponding C(K) has a support set. There are of course other ways of splitting single
pointsr thaninthesplitinterval,namelyusingtwoopenU ,U ⊆2ωwithU ∪U =2ω−{r }
ξ 1 2 1 2 ξ
and{r }=U ∩U .Thiscanbeappliedtotheresultingspacesandcanbedoneinmanyways,for
ξ 1 2
examplelikein[14]or[15],[1]orin[6]wherefirstcountablecompactspaceswithcontinuous
imageswithnofirstcountablepointsareobtained,beingthebiggestinthiscontext.
Ourpresentconstructionissomewhereinbetweentheconstructionof[14]wherethesplitting
of r ’s is done completely independently of other splittings and n-Rubin algebras constructed
ξ
in[15]andcompactspacesconstructedin[1]wheresomerepeatedsplittingcanhappenunlike
in our case. A rigorous comparison of the structure of such spaces can be done using certain
treesconsideredforexamplein[6].ItcanbeseenthatC(K)obtainedfromtheBooleanalgebras
of[14]haveuncountablebiorthogonalsystemsforthesamereasonasforthesplitinterval.On
theotherhandthecomplexityofthecompactspacesobtainedfromthealgebrasof[15]and[1],
asitseemstotheauthor,makestheanalysisoftheirRadonmeasuresamuchmorechallenging
task. Actually it is the example from [1] that is claimed in Remark 5 of [17] also not to have
asupportset.
Inourcasetherearesimplecombinatorialproperties,statedinDefinition1andTheorem7of
theBooleanalgebraanditsStonespaceneededfortheproofsofallthepropertiesofthecorre-
spondingC(K).Alsointhissectionweobtainarepresentationofsimplefunctions(Lemma4)
on K which is used to prove that our space C(K) is hereditarily Lindelöf in the weak topol-
ogy(Theorem8).TheonlyexampleofsuchanonseparableBanachspaceC(K)whichappears
intheliteratureisKunen’sspace(see[11]),howeveritisanAsplundspace(equivalently K is
P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741 2725
dispersed)andoursC(K)isnot.ThecharacterizationofBanachspaceswhichadmitasupport
setof[2]easilyimpliesthatsubspacesandquotientsofaspacewhichdoesnotadmitasupport
setalsodonotadmitit.Ontheotherhanditfollowsfromresultsof[9]thateverynonseparable
C(K)forK dispersedadmitsasupportset.ItfollowsthatourC(K)isquitefarfromanAsplund
C(K)sandinparticularfromKunen’sexample.
In Section 3 we prove the consistency of the existence of the algebra. It is done using the
methodofforcing(see[7]),i.e.,weconsiderapartialorderoffiniteapproximationstothefinal
algebra.TheapproximationscanbeamalgamatedtoobtaintherelationsasinTheorem7.This
sectioncanbeskippedbyareadernotfamiliarwithforcing.Thefollowingsectionrefersjustto
thestatementsoftheresultsofSection2.
TheforcingmethodguaranteesthatitisconsistentthatthereisadirectlimitAofsuchfinite
approximationssuchthatinanyuncountablefamilyoffinitesubalgebrasof Atwoofthemare
amalgamatedinAinsuchawaythattherelationsofTheorem7aresatisfied.Thisisatechnically
simplewayofshowingtheconsistency,howeveritisheuristicallycomplexduetotheuseofthe
forcingmethod.Anotherwayoftheconstruction,withreversedadvantagesanddisadvantages,
wouldprobablybetousethe♦principleofJensen,forexamplelikein[15].Theproofdoesnot
gothroughassumingjustthecontinuumhypothesis(recallthat♦ isstrongerthanCH,see[7])
forexamplealongthelinesof[14].
InSection4weembarkonanalyzingright-separatedsequencesinthedualballoftheC(K)
∗
with the weak topology, where K is the Stone space of the algebra obtained in the previ-
ous sections. Recall that a sequence (x : α <ω ) of points of a topological space is called
α 1
right-separated (left-separated) if and only if x ∈/ {x : α>β} (x ∈/ {x : α<β}) for every
β α β α
β <ω . It is well known that a regular topological space has an uncountable right-separated
1
(left-separated)sequenceifandonlyifitisnothereditarilyLindelöf(nothereditarilyseparable)
(see e.g. [12,Theorem3.1]) i.e., all subspaces (equivalently all closed subspaces) are Lindelöf
(separable).Notethatfor(x : α<ω )left-separatedmeansthatthereareopen(basicifneeded)
α 1
U ’sforα<ω suchthatx ∈U andx ∈/U forβ<α.Ontheotherhandfor(x : α<ω )to
α 1 α α β α α 1
beright-separatedmeansthatthereareopen(basicifneeded)U ’sforα<ω suchthatx ∈U
α 1 α α
andx ∈/U forβ>α.
β α
Notethat,ifK iscompactandnonmetrizable,andhencewithoutacountablebase,applying
theHahn–Banachtheoremonecanchooseasequence(f ,μ ) ⊆C(K)×C(K)∗suchthat
α α α<ω1
μ (f )=1andμ (f )=0forallα<β<ω i.e.,thedualballofanonseparableC(K)cannot
α α β α 1
∗
behereditarilyLindelöfintheweak topology(thisappliestoanynonseparableBanachspace).
Weshowthatinanysuchsequence(inourC(K)),thereareα<β<ω suchthatμ (f )<0.
1 α β
RecalladefinitionofBorweinandVanderwerfffrom[2].LetXbeaBanachspace.Asequence
(x ,φ ) ofelementsofX×X∗ iscalledsemibiorthogonalifandonlyif
α α α<κ
(i) φ (x )=1,
α α
(ii) φ (x )=0ifα<β,
β α
(iii) φ (x )(cid:2)0ifα<β.
α β
Hence, we show that our C(K) has no uncountable semibiorthogonal sequences. As it is
proved in [2] that X has a support set if and only if it has a semibiorthogonal sequence of
lengthω ,weconcludethatC(K)hasnosupportsets,providinganegativeanswertoRolewicz’s
1
question.
Thenotationandterminologyisquitestandard.WeoftenidentifythedualtoaC(K)withthe
spaceM(K)ofRadonmeasuresonK withthevariationnorm.
2726 P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741
2. ABooleanalgebra,itsStonespaceanditscontinuousfunctions
Let us fix the terminology concerning Boolean algebras and the Stone duality. The comple-
mentofa=1a inaBooleanalgebraisdenotedby−a or−1a,thesupremum,theinfimum,the
leastelementandthebiggestelementaredenoted∨,∧,0,1,respectively.Thebasicclopensets
of the Stone space of a Boolean algebra A consisting of all ultrafilters of A containing a are
denotedbya∗.Wehave(a∧b)∗=a∗∩b∗,etc.WewillconsidertheBooleanorderwherea(cid:3)b
if and only if a∧b=a. Recall that any subset of the algebra which has the finite intersection
property(i.e.,everyfinitesubsethasinfimumdifferentthan0)extendstoanultrafilter.
Afamily(U : i∈ω)includedinaBooleanalgebraAiscalledindependentifandonlyif
i
(cid:8) U ∧···∧(cid:8) U (cid:10)=0
0 0 n n
for all (cid:8) ,...,(cid:8) ∈ {−1,1} and every n ∈ ω. By U we denote (cid:8) U ∧ ··· ∧ (cid:8) U where
0 n s 0 0 n n
s∈{−1,1}n+1issuchafinitefunctions:{0,...,n}→{−1,1}thats(i)=(cid:8) foralli(cid:3)n.Since
i
any element of the Boolean algebra generated by an independent family as above is the supre-
mum of a finite family of elements of the form U , the ultrafilters of the algebra are generated
s
bythesets{U : x|n=s}forallx∈{−1,1}ω,andthetopologyoftheStonespaceistheusual
s
topologyoftheCantorsetidentifiedwith{−1,1}ω withtheproducttopology.
ConsideranotherexampleoftheStonespaceofaBooleanalgebraincludedinthepowerset
of{−1,1}ω×{−1,1}generatedbythesetsoftheform
(cid:2) (cid:3) (cid:2) (cid:3)
a = (x,i): x<r, i=−1,1 ∪ (r,−1) ,
r
(cid:2) (cid:3) (cid:2) (cid:3)
−a = (x,i): x>r, i=−1,1 ∪ (r,1) ,
r
(cid:2) (cid:3)
U = (x,i): x|n=s, i=−1,1 ,
s
wherer∈{−1,1}ω,<isthelexicographicalorderands∈{1,−1}n forsomen∈ω.Anyultra-
filter of this Boolean algebra intersected with the algebra generated by the independent family
must be an ultrafilter of the latter i.e., of the form {U : x|n=s} for some x ∈{−1,1}ω. Now
s
noteforanyr (cid:10)=x,thereisannsuchthateitherUr|n⊆ar orUr|n⊆−ar.Sinceultrafiltersare
closed under bigger elements of the algebra, we conclude that all ultrafilters of the algebra are
generatedbythesetsoftheform{U : x|n=s}∪{a }or{U : x|n=s}∪{−a }.Onecancheck
s x s x
thattheStonetopologyisexactlyof“thesplitCantorset”i.e.,thesubspaceofthesplitinterval
correspondingtotheCantorset.
Our examplewill be obtainedby splitting,in a more complicatedway,some (arbitrary) un-
countablesequence {r : ξ <ω } ofdistinctelementsoftheCantorset {−1,1}ω.However,the
ξ 1
Stone space of the algebra can be interpreted as a subspace of {−1,1}ω ×{−1,1} with some
compactHausdorfftopology.Thefollowingdefinitionliststhepropertiesofthegeneratorsand
the following lemma concludes the form of the ultrafilters of the algebra i.e., the form of the
neighbourhoodbasesofthepoints.
Definition 1. Suppose that (r : ξ <ω ) is a sequence of distinct elements of {−1,1}ω. Let A
ξ 1
beaBooleanalgebrageneratedbyanindependentfamily(U : i∈N)andadditionalgenerators
i
(a : ξ <ω )whichsatisfy:
ξ 1
P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741 2727
(a) Foreveryξ <ω1and(cid:8)∈{−1,1}thefamily{Urξ|i: i∈N}∪{(cid:8)aξ}hasthefiniteintersection
property.
(b) Ifδ∈{−1,1}andη<ξ <ω ,thenthereisi∈N suchthat
1
Urη|i∧δaη(cid:3)aξ or Urη|i∧δaη(cid:3)−aξ.
(c) Ifr∈{−1,1}ω−{r : η(cid:3)ξ},thenthereisi∈N suchthat
η
Ur|i (cid:3)aξ or Ur|i (cid:3)−aξ.
Then the Stone space of A is called an unordered split Cantor set. For such a space a , r , U
ξ ξ i
willrefertotheobjectsasabove.
Remark.(a)saysthata splitsr foreveryξ <ω ;(b)saysthatifa splitsr forη<ξ,thenit
ξ ξ 1 ξ η
doesit,locally,thesamewayasa ;(c)saysthata doesnotsplitr ’sforη>ξ northepoints
η ξ η
of{−1,1}ω outsideofthesequence{r : η<ω }.
η 1
Lemma 2. Suppose that A is the algebra whose Stone space is an unordered split Cantor set.
Then,allultrafiltersofA,thatisthepointsoftheStonespace,areofthefollowingtwoforms:
(1) generated by (Ux|n: n∈N) if and only if x ∈({−1,1}ω −{rξ: ξ <ω1}), which will be
denotedby[x],
(2) generatedby(Urξ|n: n∈N)∪{(cid:8)aξ}forξ <ω1 andany(cid:8)∈{−1,1},whicharedenotedby
[r ,1],[r ,−1],respectively.
ξ ξ
Theset{[r ,1],[r ,−1]}willbedenotedbyR .
ξ ξ ξ
Proof. Provetheabovepropertiesbyinductionon0(cid:3)α(cid:3)ω foralgebrasA generatedbythe
1 α
sequence(U : i∈N)andgenerators(a : ξ <α)satisfying(a)–(c)ofDefinition1forη,ξ <α
i ξ
insteadofη,ξ <ω .Thelimitstep,inparticularforα=ω istrivialsinceifasetgeneratesan
1 1
ultrafilter in algebras belonging to an increasing chain, it generates an ultrafilter in the algebra
whichistheunionofthechain.
Tomakethesuccessorstep,notethatanultrafilterinabiggeralgebraAβ+1 intersectedwith
the subalgebra A is an ultrafilter in the subalgebra. If this intersection is like in (1) or in (2)
β
for ξ (cid:10)=β, the conditions (b), (c) of Definition 1 imply that the ultrafilter (of Aβ+1) is already
generatedfromA .Otherwise(a)impliesthatbothofthepossibleextensionsareultrafilters. (cid:2)
β
Lemma3.SupposethatAisaclopenalgebraofanunorderedsplitCantorset.LetA denote
α
thesubalgebraofAgeneratedby(U : i∈ω)and{a : ξ <α}forα(cid:3)ω .Foreveryn∈ω we
i ξ 1
have
aα−Urα|n∈Aα.
Proof. Let K denote the Stone space of A. Any point of K −R has a neighbourhood a∗
α
included in a∗ or disjoint from a∗ for a∈A by Definition 1 (b) and (c). Since a∗ −U∗ is
acompactsubαspaceofthisset,weαhaveafinitαesubcoverconsistingofsubsetsi.e.,aαα−Urrαα||nn is
thesupremumofafinitefamilyofelementsofA asrequired. (cid:2)
α
2728 P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741
LetA beasinthepreviouslemmaandletK beitsStonespace.Itiswellknownthatthere
α α
is a canonical isometric embedding of C(K ) into C(K ). We will identify the image of this
α ω1
embeddingwithC(K ),inparticularforα=0wehaveacopyofC({−1,1}ω)insideC(K ).
α ω1
Letusseethegeneralformofcontinuousrationalsimplefunctionsonunorderedsplitinterval.
Byarationalsimplefunctionwemeanafunctionassumingonlyfinitelymanyrationalvalues.
Lemma4.SupposethatK isanunorderedsplitCantorset,ε>0,μisa(regular)Radonmea-
sure on K and that f is a continuous rational simple function on K. Then there is a simple
rational function g ∈C({−1,1}ω), distinct ξ ,...,ξ <ω and there are rationals q , non-
1 k 1 i
negativeintegersmi andsi ∈{−1,1}mi suchthatsi =rξi|mi for1(cid:3)i(cid:3)k∈ωsuchthat
(cid:4)
f =g+ qiχa∗∩U∗
ξi si
1(cid:2)i(cid:2)k
andsuchthat
(cid:4) (cid:5) (cid:6)
|q ||μ| U∗ −R (cid:3)ε,
i si ξi
1(cid:2)i(cid:2)k
whereR ={[r ,1],[r ,−1]}.
ξ ξ ξ
Proof. ConsidersubalgebrasA ofAforξ (cid:3)ω generatedbyU ’sanda ’sforη<ξ andtheir
ξ 1 i η
Stone spaces K . By inductionon ξ we prove that anycontinuoussimple rational functionsin
ξ
C(K )canbewrittenintheformasinthelemma.
ξ
It is clear that the limit stage is trivial. So suppose we are done for A and we are given
ξ
acontinuoussimplerationalfunctionf onKξ+1.Notethat
(cid:7)
U∗ =R
rξ|m ξ
m∈N
since, by Lemma 2, the only ultrafilters in Aξ+1 which extend the filter generated by {Urξ|n:
n ∈ N} are {[r ,1],[r ,−1]}. Hence, the regularity of the Radon measures, implies that
ξ ξ
|μ|(U∗ −R )convergeto0.Letn besuchthat
rξ|n ξ 1
(cid:5) (cid:6)
ε
|μ| U∗ −R (cid:3)
rξ|n ξ 4(cid:12)f(cid:12)
forn(cid:2)n .
1
Notealso,thatasimplefunctionisalinearcombinationofcharacteristicfunctionsofclopen
sets,i.e.,setscorrespondingtoelementsof A,hencethereisafiniteset ξ1,...,ξk−1<ξ <ω1
such that preimages under f belong to the subalgebra of Aξ+1 generated by Ui’s for i <n2
andaξ1,...,aξk−1,aξ.Nowletn(cid:2)n1,n2 besuchthatforevery1(cid:3)i<k eitherUrξ|n(cid:3)aξi or
UrξI|tnf(cid:3)oll−owaξsi tfhoart1f(cid:3)iis(cid:3)cokns−ta1ntwohnicah∗c∩anUb∗eobatnadineisdcboynsDtaenfitnoitnioUn1∗(c)−. a∗. Let q , q be the
ξ rξ|n rξ|n ξ 1 2
correspondingvalues,notethat|q −q |(cid:3)2(cid:12)f(cid:12).So
1 2
(cid:8) (cid:5) (cid:6) (cid:9)
f = f| Kξ −Ur∗ξ|n +q2χUr∗ξ|n +(q1−q2)χaξ∗∩Ur∗ξ|n.
P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741 2729
Notethatf|(K −U∗ )belongstoC(K )byLemma3,andso
ξ rξ|n ξ
(cid:5) (cid:6)
ε
f =h+qkχaξ∗∩Ur∗ξ|n, |qk||μ| Ur∗ξ|n−Rξ (cid:3) 2
whereq =q −q andh∈C(K ).Hencetheinductiveassumptionforε/2canbeusedwhich
k 1 2 ξ
completestheproofofthelemma. (cid:2)
Lemma5.Supposen∈N andμl for1(cid:3)l(cid:3)nandα<ω arenonatomicRadonmeasureson
α 1
anunorderedsplitCantorsetK.SupposeU forα<ω areopensetsinM(K)n whereM(K)
α 1
∗
isconsideredwiththeweak topologyandsupposethatforallα<ω wehave
1
(cid:5) (cid:6)
μ1,...,μn ∈U .
α α α
Then there exists an uncountable set A⊆ω such that for every α,β ∈A we have (μ1,...,
1 β
μn)∈U .
β α
Proof. Wemayw.l.o.g.assumethatU =U ×···×U where
α α,1 α,n
(cid:7) (cid:2) (cid:5) (cid:6) (cid:3)
U = μ: μ−μl (h )<ε
α,l α α,l,j
1(cid:2)j(cid:2)pl
where h ∈C(K) for 1(cid:3)l (cid:3)n and 1(cid:3)j (cid:3)p , p ∈N and ε>0. We may also w.l.o.g.
α,l,j l l
assumethatthenormsofallthemeasuresμl areboundedbysomeδ>0.Letf besimple
α α,l,j
rationalfunctionssatisfying
(cid:12)f −h (cid:12)<ε/6δ.
α,l,j α,l,j
By the previous lemma and the fact that there are countably many rational simple functions
in C({−1,1}ω) as well as there are countably many clopen subsets of {−1,1}ω and countably
manyrationals,thereisanuncountablesetA(cid:13)⊆ω andthereexistg ∈C({−1,1}ω),distinct
1 l,j
α,l,j α,l,j
ξ ,...,ξ <ω and there are rationals q , non-negative integers m and k and
1 kl,j 1 l,j,i l,j,i l,j
sequences sl,j,i ∈{−1,1}ml,j,i such that sl,j,i =rξα,l,j|ml,j,i for 1(cid:3)i (cid:3)kl,j ∈ω such that for
everyα∈A(cid:13) wehave i
(cid:4)
fα,l,j =gl,j + ql,j,iχa∗ ∩U∗
1(cid:2)i(cid:2)kl,j ξiα,l,j sl,j,i
andsuchthat
(cid:4) (cid:10) (cid:10)(cid:5) (cid:6)
|q |(cid:10)μl (cid:10) U∗ −R (cid:3)ε/3,
l,j,i α sl,j,i ξα,l,j
1(cid:2)i(cid:2)kl,j i
whereR ={[r ,1],[r ,−1]}.Sincethemeasuresμl arenonatomicandR scon-
ξα,l,j ξα,l,j ξα,l,j α ξα,l,j
i i i i
2730 P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741
sistoftwopointsofK weobtainthat
(cid:4) (cid:10) (cid:10)(cid:5) (cid:6)
|q |(cid:10)μl (cid:10) U∗ (cid:3)ε/3.
l,j,i α sl,j,i
1(cid:2)i(cid:2)kl,j
MoreoverthereexistsanuncountableA⊆A(cid:13) suchthatforeveryα,β∈Aandevery1(cid:3)j (cid:3)p
l
for1(cid:3)l(cid:3)nwehave
(cid:10)(cid:5) (cid:6) (cid:10)
(cid:10) μl −μl (g )(cid:10)<ε/3.
β α l,j
Henceforα,β∈A
(cid:10)(cid:5) (cid:6) (cid:10) (cid:10)(cid:5) (cid:6) (cid:10)
(cid:10) μl −μl (h )(cid:10)(cid:3)(cid:10) μl −μl (f )(cid:10)+2δ.ε/6δ
β α α,l,j β α α,l,j
(cid:10)(cid:5) (cid:6) (cid:10)
(cid:3)(cid:10) μl −μl (g )(cid:10)+ε/3+ε/3<ε
β α l,j
which means that μl ∈ U which means that (μ1,...,μn) ∈ U which completes the
β α,l β β α
proof. (cid:2)
Proposition6.SupposethatK isanunorderedsplitCantorset.ThedualofC(K)(inthenorm)
isisomorphictothedualofC({−1,1}ω).Foreverypositiveintegern∈N then-thpowerMn(K)
ofthedualofC(K)intheweak∗topologyishereditarilyseparableifandonlyifforeveryn∈N
thespaceln(K)ishereditarilyseparableintheweak∗ topology.
1
Proof. ItiswellknownthatforanyKthedualM(K)toC(K)isisomorphictoNA(K)⊕l (|K|)
1
whereNA(K)isthesubspaceofM(K)consistingofnonatomicRadonmeasures.Sincethecar-
dinalitiesofKand{−1,1}ωarethesamewewillobtaintherequirednormisomorphismbetween
the dual spaces of C({−1,1}ω) and C(K) if we can show that NA({−1,1}ω) is isomorphic to
NA(K).Theisomorphismwilljustbetherestriction.
First,wewillshowthatifμisanonatomicRadonmeasureonK whichisnullonA i.e.,the
0
clopensetsgeneratedbytheindependentfamily,thenμ=0.
Indeed,fix μ asaboveandtakeany f ∈C(K) whichisrationalsimplefunctionand ε>0.
ByLemma4wehave
(cid:4)
f =g+ qiχa∗∩U∗
ξi si
1(cid:2)i(cid:2)k
with
(cid:4) (cid:5) (cid:6)
|q ||μ| U∗ −R (cid:3)ε,
i si ξi
1(cid:2)i(cid:2)k
(cid:11)
wherealltheobjectsarelikeinLemma4.AsμisnullonA0,wehave gdμ=0.Th(cid:11)efactthatμ
isnonatomicimpliesthatμ(Rξi)areall(cid:11)zero.So,theaboveinequalityimpliesthat| f dμ|(cid:3)ε.
Sinceε>0wasarbitrarywehavethat f dμ=0,foranysimplerationalfunctionf.Butsuch
functionsarenormdenseinC(K)bytheStone–Weierstrasstheorem,soμ=0asrequired.
P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741 2731
Bytheaboveobservationtoobtainanisomorphismbetweenthespaceofnonatomicmeasures
onC({−1,1}ω)andthespaceofnonatomicmeasuresofC(K),weneedtoshowthattherestric-
tionsofnonatomicmeasuresonC(K)toC({−1,1}ω)(treatedasfunctionals)arenonatomicand
thateverynonatomicmeasureonC({−1,1}ω)hasanonatomicextensiontoC(K).Thesefollow
fromthefactthatthepointsof{−1,1}ω aresplitjustintwopointsofK.
∗
NowletusturntothedualofC(K)consideredwiththeweak topology.Asl (K)withthe
1
∗
weak topologyisitssubspace,toprovethesecondpartofthepropositionitisenoughtoprove
thatMn(K)ishereditarilyseparableforeveryn∈N assumingthatln(K)hasthispropertyfor
1
everyn∈N.
First see that every finite power of NA(K)×l (K) is hereditarily separable where NA(K)
1
∗
is the space of nonatomic measures on K with the weak topology. If not, there would be an
uncountableleft-separatedsequence(fordefinitionandrelevanceseetheintroduction)insome
finitepowerofNA(K)×l (K).ByLemma5,itwouldreducetoanuncountableleft-separated
1
sequenceinafinitepowerofl (K)contradictingthehereditaryseparabilityofthatspace(note
1
thattheproductoftwohereditarilyseparablespacesdoesnothavetobehereditarilyseparable).
ToconcludethateveryfinitepowerofthedualtoC(K)ishereditarilyseparableitisenoughto
notethatM(K)isacontinuousimageofNA(K)×l (K),additionbeingthemappingwhereall
1
spacesareconsideredintheweak∗ topology. (cid:2)
Theorem7.ItisconsistentthatthereisaBooleanalgebraAwhoseStonespaceisanunordered
splitCantorsetsuchthatgivenanycollectionofpairwisedisjointsetsF ={ξ1,...,ξk}⊆ω
α α α 1
forα<ω ,(cid:8) ∈{−1,1},δ ∈{−1,1}forall1(cid:3)i(cid:3)k anda⊆{1,...,k}thereareα<β such
1 i i
that
R ⊆a∗ , ifi∈a,
ξi ξi
β α
R ⊆−a∗ ifi∈/a
ξi ξi
β α
forall1(cid:3)i(cid:3)k,andthereareα<β suchthat
[r ,1]∈(cid:8) a∗ ,
ξαi i ξβi
[r ,−1]∈δ a∗
ξαi i ξβi
for all 1(cid:3)i (cid:3)k. The Stone space of such an algebra will be called a generic unordered split
Cantorset.
Theproofoftheabovetheoremisthesubjectofthefollowingsection.Now,however,wecan
provesomepropertiesofagenericunorderedsplitCantorsets.
Theorem8.SupposethatK isagenericunorderedsplitCantorset.Foreveryn∈N,theprod-
uctKn ishereditarilyseparable.K ishereditarilyLindelöfandK2 isnot.C(K)ishereditarily
∗
LindelöfintheweaktopologyandthedualspaceofC(K)withtheweak topologyishereditarily
separable.
2732 P.Koszmider/JournalofFunctionalAnalysis257(2009)2723–2741
Proof. NotethattoprovethatKnishereditarilyseparableorthatK ishereditarilyLindelöfitis
sufficient(bythinningoutthesequences)toworkwithpointsofK oftheform[r ,(cid:8)]sincethe
ξ
remainingpointsformasubspaceofthemetrizableCantorset.
Soassumethat(xα,...,xα)∈Knforα<ω areofthisform.Makinganinductiveargument
1 n 1
onecanassumewithoutlossofgenerality(againthinningoutthesequence)thatxα(cid:10)=xαifi(cid:10)=j.
i j
Now,considersomeneighbourhoodsofthesepointsinKn,theycouldbebasicneighbourhoods,
ii.me.p,loiefsththeaftotrhmerUesaiαre∩α(cid:8)i<aξβiα fsourchditshtiantcxtαfix∈e(cid:8)dasiα∗’sii.en.,sothmeese2qmu.eBnucettihsensoetcloenftd-speapratroafteTdh.eorem7
i i ξβ
i
ToprovethatKishereditarilyLindelöfusethesameargumentandthefirstpartofTheorem7.
Finally to see that the square K2 is not hereditarily Lindelöf. Consider the sequence
([r ,1],[r ,−1]) withtheneighbourhoods(a∗×(−a∗)) .Asa doesnotsplitthesetR
forξξ >ηξby Defiξn<iωti1on 1(c), we have {[r ,1],[rξ,−1]}⊆ξ aξ∗<ωo1r {[r ,η1],[r ,−1]}∩a∗ =∅ iξf
ξ ξ η ξ ξ η
η<ξ.Thisimpliesthat
(cid:5) (cid:6) (cid:5) (cid:6)
[r ,1],[r ,−1] ∈/a∗× −a∗
ξ ξ η η
foreveryη<ξ i.e.,K2 hasanuncountableright-separatedsequence.
NowweturntothepropertiesoftheBanachspaceC(K)anditsdualM(K).BesidesProposi-
tion6wewillneedtwofolklorefactsthateveryfinitepowerofaBanachspaceXishereditarily
∗
LindelöfintheweaktopologyifandonlyifeveryfinitepowerofX ishereditarilyseparablein
∗
theweak topologyandthateveryfinitepowerof l (K) ishereditarilyseparableifandonlyif
1
everyfinitepowerofK ishereditarilyseparable.Bothofthemwewillimplicitlyreprovedbelow
intheneededdirections.
WewillbeusingagainthefactthataregulartopologicalspaceXisnothereditarilyseparable
ifandonlyifithasanuncountableleft-separatedsequenceanditisnothereditarilyLindelöfif
andonlyifithasanuncountableright-separatedsequence.
∗
Firstwewillprovethatthedualishereditarilyseparableintheweak topology.Thefirststep
is to note that the fact that Kn is hereditarily separable for every n∈N implies that for every
n∈N thespacel (K)n ishereditarilyseparableintheweak∗ topology.
1
Suppose that {μl : α <ω , 1(cid:3)l (cid:3)n} is a collection of measures in l (K) such that for
α 1 1
x∗=(μ1,...,μn)thesequence(x∗) isleft-separatedi.e.,asbeforethereareopensetsU s
suαchthaαtwehavαex∗∈U andx α∈/αU<ωf1orβ<α.Usingthisnotationwemayw.l.o.g.assumαe
α α β α
thatU =U ×···×U where
α α,1 α,n
(cid:7) (cid:2) (cid:10)(cid:5) (cid:6) (cid:10) (cid:3)
U = μ: (cid:10) μ−μl (h )(cid:10)<ε
α,l α α,l,j
1(cid:2)j(cid:2)pl
where h ∈C(K) for 1(cid:3)l (cid:3)n and 1(cid:3)j (cid:3)p , p ∈N and ε>0 such that x∗ ∈U and
α,l,j l l α α
x∗∈/U .Wemayassumewithoutlossofgeneralitythatallh ’sareboundedbysomeM>0.
β α α,l,j
Nowforeachα<ω and1(cid:3)l(cid:3)nfindνl ∈l (K)suchthat
1 α 1
(cid:12) (cid:12)
(cid:12)νl −μl (cid:12)<ε/3M
α α