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Article details
Jeu M. de & Walt J.H. van der (2019), On order continuous duals of vector lattices of continuous functions, Journal of Mathematical Analysis and Applications 479(1): 581-607.
Contents lists available atScienceDirect
Journal
of
Mathematical
Analysis
and
Applications
www.elsevier.com/locate/jmaa
On
order
continuous
duals
of
vector
lattices
of
continuous
functions
Marcel de Jeua,b,1, JanHarm van der Waltb,∗,2
a
MathematicalInstitute,LeidenUniversity,P.O.Box9512,2300RALeiden,theNetherlands
b
DepartmentofMathematicsandAppliedMathematics,UniversityofPretoria,CornerofLynnwood RoadandRoperStreet,Hatfield0083,Pretoria,SouthAfrica
a r t i cl e i n f o a b s t r a c t
Articlehistory: Received14March2019 Availableonline14June2019 SubmittedbyM.Mathieu Keywords:
Vectorlatticeofcontinuous functions
Ordercontinuousdual Resolvabletopologicalspace Normalmeasure
AtopologicalspaceX iscalledresolvableif itcontainsadense subsetwithdense complement.Usingonlybasicprinciples,weshowthatwheneverthespaceX has a resolving subsetthat can bewritten as an at mostcountably infiniteunion of subsets, in such a way that a given vector lattice of (not necessarily bounded) continuousfunctionsonX separateseverypointoutsidetheresolvingsubsetfrom eachofitsconstituents,thentheordercontinuousdualofthislatticeistrivial.In ordertoapplythisresultinspecificcases,weshowthatseveralspaceshaveresolving subsetsthatcanbewrittenasatmostcountablyinfiniteunionsofclosednowhere dense subsets.Anappeal tothemainresultthenyields,forexample,that,under appropriateconditions,vectorlatticesofcontinuousfunctionsonseparablespaces, metricspaces, andtopologicalvectorspaceshavetrivialordercontinuousdualsif theyseparatepointsandclosednowheredensesubsets.Ourresultsinthisdirection extendknownresultsintheliterature.Wealso showthat,underreasonably mild separation conditions,vectorlatticesof continuousfunctionsonlocallyconnected T1 Baire spaces without isolated points have trivial order continuous duals. A discussion of therelation betweenour results andthe non-existence of non-zero normalmeasuresisincluded.
©2019ElsevierInc.Allrightsreserved.
1. Introduction andoverview
LetE beavectorlattice.Werecallthatanorder boundedlinearfunctionalϕ onE isordercontinuous
if inf{|ϕ(fi)| : i∈ I } = 0 whenever(fi)i∈I isa net inE suchthat fi ↓ 0 inE. The vector lattice of all
* Correspondingauthor.
E-mailaddresses:mdejeu@math.leidenuniv.nl(M. de Jeu),janharm.vanderwalt@up.ac.za(J.H. van der Walt).
1 The firstauthor thanksthe Department of Mathematics and Applied Mathematics of the University ofPretoria for their
hospitalityduringavisitinMarch2018.
2 Theresultsinthispaperwereobtained,inpart,whilethesecondauthorvisitedLeidenUniversityfromMaytoJune2017.He
thankstheMathematicalInstituteofLeidenUniversityfortheirhospitality.ThisvisitwassupportedfinanciallybytheGottfried WilhelmLeibnizBasicResearchInstitute,Johannesburg, SouthAfrica,andtheNationalResearchFoundationofSouthAfrica, grantnumber81378.
order continuous functionals on E is called the order continuous dual of E. This paper iscentred around the followingresultonthetrivialityof ordercontinuousduals ofvectorlatticesof continuousfunctions.It will beestablishedinSection3.
Theorem.Let X bea non-empty topological space, andlet E be a vector sublattice of the real-valued con-tinuous functionsonX. Supposethat thereexists anatmost countablyinfinite collection{Γn: n∈ N } of non-empty subsets ofX suchthat
(1) n∈NΓn andX\n∈NΓn are both densein X,and
(2) for every x∈ X \n∈NΓn and every k∈ N, there existsan element u of E+ suchthat u(x)= 1 and u(y)= 0 for ally∈ Γk.
Then 0istheonlyorder continuouslinear functionalon E.
If, in addition,X satisfies thecountablechain condition (in particular,if, inaddition, X isseparable), then 0isalso theonlyσ-ordercontinuous linearfunctional onE.
Letusmakeafewcommentsonthistheorem.Firstofall,wementionexplicitlythatE neednotconsist of bounded functions,and thatthere are no topological properties of X supposed, other than whatis in the theorem.Singletonsneed notevenbe closedsubsets. Inseveralof theexistingresultsintheliterature on thetriviality oforder continuous duals, X issupposed tobe alocally compact Hausdorffspaceand E
is supposed to be avector lattice ofbounded continuous functions onX. Moreoften than not,the Riesz representation theoremisthenused toreducethestudy ofordercontinuouslinearfunctionalsonE tothe analysis of thestructure ofthe measures thatrepresent them.Ourapproachis quitedifferent. Usingonly first principles,the argument withnets and sequencesis completely elementary,and furtherpropertiesof
X orof theelements ofE playnorole.
Let us also mention that it is notsupposed that theΓn are pairwise disjoint, or even different. Ifone
wishes,disjointness canneverthelessalwaysbeobtainedbyreplacingeachΓn withΓn\nj=1−1Γj forn≥ 2,
and disregardingallemptysetsthatmightoccur inthis process.
TherearenoexplicittopologicalconditionsontheΓninthetheorem,butthecouplingwiththecontinuous
functionsinE impliesthatsomepropertiesarestillautomatic.Foronething,ΓkisnowheredenseinX for
all k≥ 1. To see this, fixk ≥ 1. Forevery x∈ X \n∈NΓn,assumption (2) of thetheorem impliesthat
there exists anelement ux ofE such thatux(x)= 1 and ux(y)= 0 forally ∈ Γk.Consequently, theopen
neighbourhoodVx:= u−1[(12,2)] of x iscontainedinX\ Γk.SetV :=
x∈X\∈NΓnVx.ThenV isanopen subset ofX,andX\n∈NΓn⊆ V ⊆ X \ Γk.SinceX \
n∈NΓn isdenseinX,we concludethatX\ Γk
contains theopen densesubsetV ofX. IfV were anon-empty opensubset ofX that is containedinΓk,
thenV∩ V wouldbeanon-emptyopensubsetofX thatiscontainedinΓk\ Γk,whichisimpossible.Hence
Γk hasemptyinterior,i.e.Γk isnowheredense.
If X isaBairespace, thenthis impliesthatonemayaswell assumethattheΓn are allclosednowhere
densesubsetsof X,simplybyreplacing Γn withΓn foralln≥ 1.Indeed,n∈NΓn is thencertainlydense
inX,butX\n∈NΓn isalsostill dense,sincethefactthatX isaBairespacenowimpliesthatn∈NΓn
still has emptyinterior. Furthermore, if k ≥ 1 is fixed, and if x∈/ n∈NΓn, then x ∈/ n∈NΓn, so that
there existsanelementu ofE+ suchthatu(x)= 1 andu(y)= 0 forally∈ Γ
k.Bycontinuity,wethenalso
havethatu(y)= 0 for ally ∈ Γk. HencetheΓn alsosatisfytheassumptions inthetheorem.Of course,if
theoriginal Γn arepairwise disjoint,thisproperty maybe lostwhenreplacing themwiththeirclosures.
Intheapplications thatwegiveinSection5, theΓn are,infact,typicallyclosednowheredensesubsets
of X.Theabovediscussionshowsthatthisisnottotallyunexpected.
Definition.AtopologicalspaceX iscalledresolvable ifthereexistsasubsetD ofX suchthatD andX\ D
arebothdenseinX.WeshallthensaythatsuchasubsetD isaresolvingsubsetofX,orthatitresolves X.
Hencepart (1) of thehypotheses impliesthat X shouldbe resolvable, but notjust that:Combination withpart(2)showsthatitmustberesolvableinaspecialmanner.IfonethinksofE asgiven,then,using thattheΓnmayaswellbetakentobedisjoint,theinterpretationoftheparts(1)and(2)takentogetheris
thefollowing:Thereexists aresolvingsubsetofX thatcanbe splitintoatmostcountablyinfinitelymany non-emptyparts,insuchawaythatE issufficientlyrichwithrespectto eachoftheseparts inordertobe abletosupplyallseparatingfunctionsinpart(2).Ifonewantstocoinamoreorlesssuggestiveterminology, then,althoughitdoesnotcaptureeverything,onecouldcall{Γn: n∈ N } anE-separatedresolutionofX.
Furthermore,letusobserve that,ifthehypothesesofthetheoremaresatisfiedforE,theyareobviously alsosatisfiedforeverysuperlatticeF ofcontinuousfunctionsonX.Therefore,bypassinganysubtletiesthat willarisewhenconsideringextensionsorrestrictionsofordercontinuouslinearfunctionals,thetrivialityof theorder continuousduals of aparticular vector latticeE ofcontinuous functionsis,when obtainedfrom theabovetheorem,inheritedbyallitssuperlatticesF ofcontinuousfunctions.
Weshallnowcombinetheremainderofthediscussionwithanoverviewofthepaper.
InSection2, weintroducebasicnotation andestablishafew auxiliaryresultsontherole ofthe count-able chaincondition and isolated points regardingorder continuous duals of vector lattices of continuous functions.Thecorestatementof themain trivialitytheorem aboveisthattheordercontinuous dual ofE
is trivial,butif X satisfies thecountable chainconditionthen the order continuousdual and theσ-order
continuousdual coincide.Hencetheσ-ordercontinuousdual isthenalsotrivial. Section3containstheproofofthemaintrivialitytheorem above.
It isclearfrom themain triviality theoremthatattentionshouldbe paid toresolvablespaces, and this is the topicof Section 4. It is easy to see that a non-empty resolvable topological space has no isolated points.Indeed,ifX isresolvableandx∈ X issuchthat{x} isanopensubsetofX,then{x}∩ D = ∅ and
{x}∩ (X \ D) = ∅,sothatx∈ D aswellas x∈ X \ D.IfX avoidsthis obstructionbyhaving noisolated points,thenitisknowntoberesolvableinanumberofcases:ifX isametricspace(see [9,Theorem 41] and [3,Theorem 3.7]), ifX isalocally compactHausdorffspace (see [3,Theorem 3.7]), ifX isfirst countable (see [9,Corollary toTheorem 48]),ifX isaregularHausdorffcountablycompacttopologicalspace(see [3, Theorem 6.7]), and, if oneassumes the Axiom of Constructibility, i.e. if oneassumes that ‘V = L’,then thisisalsotrueifX isaBairetopologicalspace(see [3,Theorem 7.1]).Forausefulapplicationofthemain trivialitytheorem,however,itisneededthatX hasaresolvingsubsetthatcanbedecomposedasn∈NΓn,
insuchamanner thattheΓn are‘naturally’ relatedto thetopologyof X and toseparationpropertiesof E.Theproofsinthepapersjustcitedgivenoinformation aboutthis beingtrueornotinthegeneralityof the pertainingresult, butSection4shows thatthis is indeedthecase inanumberof reasonablyfamiliar situations. The Γn in thatsection are typically closed nowhere dense subsets of X,and inthat case the
requirementunder(2)inthetheoremabovespecialisestoa(mild)separationpropertyforpointsandclosed nowheredensesubsets.
InSection5,thematerialfromSection4onresolvablespacesandthepreparatoryresultsfromSection2
are combinedwith the main triviality theorem from Section3. Wethus obtainresults onthe trivialityof ordercontinuousdualsinanumberofnotuncommoncontexts.Itisourhopethattheresultsinthissection areaccessiblewithoutknowledgeoftherestof thepaper, apartfrom anoccasionalglanceat Section2for notationsanddefinitions.
continuous duals canbe interpreted as theabsenceof non-zeronormalmeasures.Wedonotclaim togive acompleteoverviewhere,andanyomissioninthissectionisentirelyunintentional.
2. Preliminaries
Inthissection,wefirstintroducesomenotationsandconventions.Afterthat,wecollectafewfactsabout thecountablechainconditionforatopologicalspaceanditsrelationwithordercontinuousdualsofvector lattices of continuous functions, and about the role of isolated points for order continuous duals of such vectorlattices.InSection5,thesefactswillbecombinedwiththemaintrivialityresultfromSection3and thematerial onresolvablespacesinSection4.
Weset N :={1,2,3,. . .}.
Allvectorspacesare overtherealnumbers,unless otherwisespecified.If E isavectorlattice, thenE∼
denotes itsorder dual.As in[20,p. 123], ϕ∈ E∼ iscalled σ-ordercontinuous ifinf{|ϕ(fn)|: n∈ N }= 0
whenever(fn) isasequenceinE suchthatfn↓ 0 inE,and,asalreadymentionedinSection1,ϕ iscalled order continuous ifinf{|ϕ(fi)|: i∈ I }= 0 whenever(fi)i∈I isanetinE suchthatfi↓ 0 inE.Wedenote
theσ-ordercontinuouslinearfunctionalsonE andtheordercontinuouslinearfunctionalsonE byEc∼ and
E∼n,respectively.Obviously,E∼n ⊆ E∼c .Furthermore,ifϕ∈ E∼,then,by[20,Lemma 84.1],ϕ∈ Ec∼ ifand onlyifϕ+∈ Ec∼ andϕ− ∈ Ec∼,andϕ∈ En∼ ifandonlyifϕ+∈ En∼ andϕ−∈ E∼n.
A topologicalspace willsimplybe called aspace.There arenoimplicitgeneralsuppositionsconcerning our spaces. AT1 space isaspace inwhichsingletons are closed subsets.An isolatedpoint of aspace isa point suchthat thecorresponding singleton is anopen subset. For a space X and apoint x∈ X, we let
Vx denotethe collectionofopen neighbourhoodsofx in X.Theclosure ofasubsetA of X isdenotedby A. IfX is a space, then 0 denotesthefunction onX that is identicallyzero,and 1 denotes thefunction on X that is identically one. If f is a continuous function on a space X, we denote by Z(f ) the zero
set of f ; i.e. Z(f ) := {x ∈ X : f(x) = 0}. When we speak of a vector lattice of continuous functions
on aspace X,we shall always suppose thatthepartial ordering and the latticeoperations are pointwise. We let C(X), Cb(X), C0(X), and Cc(X) denote the vector lattices of the continuous functions on X, the bounded continuous functions on X, the continuous functions on X that vanish at infinity, and the compactlysupportedcontinuousfunctionsonX,respectively.IfX isametricspace,thenLipb(X) denotes theboundedLipschitzfunctionsonX.Thisislikewiseavectorlatticeofcontinuousfunctions;seee.g.[19, Proposition 1.5.5].
We recall that a space is a Baire space if the intersection of an at most countably infinite collection of dense open subsets is dense as well.Equivalently, a space is aBaire space if the union of an at most countably infinite collection of closed nowhere dense subsets has empty interior. Complete metric spaces and locally compactHausdorffspaces are Bairespaces; there exist metrizable Bairespaces whichare not completely metrizable.
We now turn to the countable chain condition and its relation with order continuous duals of vector latticesofcontinuousfunctions.
Definition2.1. AspaceX issaidtosatisfythecountablechaincondition,ortosatisfyCCC,ifeverycollection of non-emptypairwisedisjoint opensubsetsofX isat mostcountably infinite.
Weincludetheshortproofof thefollowing folkloreresultfortheconvenienceofthereader. Lemma 2.2.Everyseparablespacesatisfies CCC.
Proof. Suppose that X is a separable space with a (possibly finite) dense subset D = {dn : n ∈ N }.
thatU∩ D = ∅ foreveryU ∈ C. ForeachU ∈ C,letnU = min{n∈ N : dn ∈ U }. Sincethemembersof C
arepairwise disjoint,it followsthatthemap thatsends C∈ C to nU ∈ N isinjective.HenceC is at most
countablyinfinite. 2
WerecallthatavectorlatticeE isorderseparable ifeverysubsetofE thathasasupremumcontainsan atmostcountablyinfinitesubsetthathasthesamesupremum.
ForArchimedeanvectorlattices,thereareseveralequivalentalternatecharacterisationsoforder separa-bilityavailable;see[13,Theorem29.3].OneoftheseisthepropertythateverysubsetofE+thatisbounded from aboveand thatconsists ofpairwise disjointelements isat mostcountablyinfinite.It isthisproperty thatisusedtoestablishthefollowingresult,whichisaratherobviousgeneralisationof[5,Theorem 10.3 (i)]. Weincludetheshort proofforthesakeofcompleteness.
Lemma2.3. LetX beanon-emptyspacethatsatisfiesCCC.Theneveryvectorlatticeofcontinuousfunctions on X isorderseparable.
Proof. It is clearly sufficient to prove that C(X) is order separable. Suppose that G ⊆ C(X)+\ {0} is
bounded from above and thatit consists of pairwise disjoint elements. It is sufficientto provethatevery suchG isatmostcountablyinfinite.Foreachv∈ G,thesetUv= X\ Z(v) isopenandnon-empty.SinceG
consistsofpairwise disjointelements,Uv1∩ Uv2 =∅ ifv1,v2∈ G aredifferent.ThisimpliesthatUv1 = Uv2
if v1,v2 ∈ G are different. Consequently, the map thatsends v ∈ G to Uv is a bijection between G and UG = {Uv : v ∈ G}. Since UG consists of non-empty pairwise disjoint open subsets of X, and since X
satisfiesCCC,UG isat mostcountablyinfinite.HencethesameholdsforG. 2
IfE isanorderseparablevectorlattice,thenEn∼= Ec∼;see[20,Theorem 84.4 (i)].Combiningthiswith Lemmas 2.2and2.3wehavethefollowing.
Proposition2.4. LetX beanon-emptyspace,andletE beavectorlatticeof continuousfunctionsonX.If X satisfiesCCC (in particular,if X isseparable), thenEn∼= Ec∼.
Weconcludewitharesultpointingoutthespecialroleofisolatedpointsforordercontinuousdualsof vec-torlatticesofcontinuousfunctions,whichthereadermayreadilyverifyuponnotingthatthecharacteristic functionofanisolatedpointiscontinuous.
Lemma 2.5.Let X be a space that has an isolated point x0, and suppose that E is a vector lattice of continuous functions on X that contains the characteristic function χ{x0} of {x0}. Define the evaluation
mapδx0 : E → R bysetting δx0(f ):= f (x0). Then{0}= {δx0}⊆ (En∼)
+⊆ (E∼ c )
+ .
3. Maintriviality theorem
Thefollowingresultisthetechnicalheartofthecurrentpaper.Theproofisultimatelyinspiredbyideas in[21,Example 21.6(ii)],where itis establishedthatC([0, 1])∼c ={0}.
Theorem 3.1. Let X be a non-empty space, and let E be a vector lattice of continuous functions on X. Supposethat there existsan atmostcountably infinite collection {Γn: n∈ N } of non-emptysubsets of X suchthat
(1) n∈NΓn resolvesX, and
Then En∼={0}.
If, inaddition,X satisfies CCC (inparticular, if,inaddition,X isseparable), then alsoEc∼ ={0}. Proof. Asapreparatoryreduction,wenotethatwemay(andshall)assumethatalsoΓ1⊆ Γ2⊆ Γ3⊆ · · · . Indeed, for n∈ N,set Γn :=
n
j=1Γj. Then
n∈NΓn =
n∈NΓn resolves X. Ifthe point x is suchthat x∈ X \n∈NΓn = X\n∈NΓn,then, for eachj ∈ N, there existsujx ∈ E+ suchthatujx(x)= 1 and ujx(y)= 0 forally∈ Γj.Fork∈ N,setukx:=
k
j=1ujx.Thenukx ∈ E+,ukx(x)= 1,and ukx(y)= 0 for
ally∈ Γk.HencewemayreplacetheΓn withtheΓn, andthelatterform anon-decreasingchain.
After thisreduction,westartwiththetrivialityofEn∼.
Arguingbycontradiction,supposethatEn∼= {0}.Thenthereexist ϕ∈ En∼ande∈ E suchthatϕ≥ 0, e≥ 0,andϕ(e)= 1.Choose andfixsuchϕ ande fortheremainderoftheproof.
The first step of the proof consists of introducing anauxiliary set function onthe power set of X, as follows.
Foreveryn∈ N and everysubsetA of X,set
ρn(A) := 1 if A∩ Γn= ∅; 0 if A∩ Γn=∅, and set ρ(A) = ∞ n=1 1 2nρn(A).
Sincen∈NΓnisdenseinX,itfollowsthatρ(U )> 0 foreverynon-emptyopensubsetU ofX.Furthermore, ρ: 2X → [0,1] ismonotone.Boththesepropertiesareessentialinthesequeloftheproof;weshallencounter
asimilarauxiliaryfunctiononthenon-emptyopen subsetsof aspaceinProposition4.14.
ThesecondstepoftheproofconsistsofconstructingelementsofE thatareindexedbym∈ N andthat aresuitablyrelatedtotheΓn,toρ,andtoϕ.Weshallusetheordercontinuityofϕ here.Theconstruction
of theseelements,whichwill bedenotedbyvmFm below,isasfollows. Fixm∈ N.Wechooseand fixNm∈ N such that
∞ n=Nm+1 1 2n ≤ 1 m, and set Am= X\ ΓNm. (3.1)
Due tothenon-decreasingnatureofthechain{Γn : n∈ N }, wehave
ρ(Am)≤ ∞ n=Nm+1 1 2n ≤ 1 m. (3.2)
Letx beanarbitraryelementofthenon-emptysetX\n∈NΓn.Asaconsequenceofthesecondpartof
thehypothesesandthefactthate(x)≥ 0,thereexistsumx ∈ E+suchthatumx(x)= e(x) andumx(y)= 0
0≤ vmx≤ e, vmx(x) = e(x),
vmx(y) = 0 for all y∈ ΓNm.
(3.3)
Denote byF thecollectionof non-emptyfinite subsets ofX\n∈NΓn. Then F isnon-empty,sinceit
containsthesingletons{x} forallx∈ X \n∈NΓn.Weintroduceapartialordering onF by inclusion;it
isthen directed.ForF ∈ F, setvmF :=
x∈Fvmx ∈ E.Withm stillfixed, thesubset {vmF : F ∈ F } of E isobviouslyupward directed.Furthermore,(3.3) impliesthat,forallF ∈ F,
0≤ vmF ≤ e,
vmF(x) = e(x) for all x∈ F,
vmF(y) = 0 for all y∈ ΓNm.
(3.4)
We see from (3.4) that e isan upperbound of{vmF : F ∈ F } in E. Weclaim thatactually vmF ↑ e
in E. To see this, let f ∈ E be an upper bound of {vmF : F ∈ F } in E. For every x ∈ X \n∈NΓn,
taking F ={x}∈ F in(3.4) showsthat f (x)≥ vm{x}(x)= e(x). Since X\n∈NΓn is dense inX,and
sincetheelementsofE areactuallycontinuous(whichhasnotbeenusedsofar),itfollowsthatf ≥ e.This establishesourclaimthatvmF ↑ e inE.
Bytheorder continuityof ϕ,it followsthatϕ(e− vmF)↓ 0. Therefore, wecanchooseand fix Fm∈ F
suchthat
ϕ(e− vmFm) < 1
2m+1. (3.5)
Inthethirdstepoftheproof, wecombinethevmFm as theyhavebeenfound forallm∈ N,as follows. Foreachk∈ N, set
wk := k
m=1
vmFm. (3.6)
Then(wk) isdecreasingandboundedbelowby0.Weclaimthatactuallywk↓ 0 in E.Toseethis,suppose
thatw∈ E issuch thatw≤ wk for allk∈ N.Then w+ ≤ wk forallk∈ N. Fixk∈ N. Ifx∈ X is such
thatw+(x)> 0,then(3.6) showsthatcertainlyv
kFk(x)> 0.Inviewof(3.4),wemustthen havex∈ Γ/ Nk, or,equivalently (see (3.1)),we must have x∈ Ak. We conclude that{x∈ X : w+(x) > 0}⊆ Ak. Since ρ is monotone, we haveρ({x∈ X : w+(x)> 0})≤ ρ(A
k). Anappeal to(3.2) allows us to concludethat ρ({x∈ X : w+(x)> 0})≤ 1/k. Sincethis holdsfor allk∈ N,we seethatρ({x∈ X : w+(x)> 0})= 0. Asρ isstrictlypositiveonnon-emptyopensubsetsofX,itfollowsthat{x∈ X : w+(x)> 0}=∅;herewe usethecontinuityofelements ofE again.Hencew+= 0,andthenw≤ 0.This establishesourclaimthat
wk↓ 0 inE.
In the fourth and final step of the proof, we use the wk and the order continuity of ϕ to reach a
contradiction,asfollows.
Sincewk ↓ 0 inE,wehaveϕ(wk)↓ 0.Ontheotherhand,using thate− vmFm ∈ E
+ forallm∈ N,we notethat,forallk∈ N,
≤ k
m=1
(e− vmFm).
Since ϕ≥ 0,(3.5) thereforeyieldsthat,forallk∈ N,
ϕ(e− wk)≤ k m=1 ϕ(e− vmFm) < k m=1 1 2m+1 < 1 2. Since ϕ(e)= 1, itfollowsthat
ϕ(wk) >
1 2
forallk∈ N.Thiscontradicts thefactthatϕ(wk)↓ 0, andweconcludethatwemusthaveEn∼={0}.
Thesecond partofthestatementfollowsfrom thefirstpartandProposition2.4. 2 4. Resolvablespaces
In view of Theorem 3.1, it is clearly desirable to seek resolving subsets of resolvable spaces that can be decomposed into at most countably infinitely many components that relate naturally to continuous functions.Thepresentsection isdevotedto suchresults.
It was already observedinSection1 thatanon-empty resolvablespace has noisolated points, and we collectasecondelementaryresultsforfuturereference.
Lemma 4.1.If X isaresolvable non-emptyT1 space, thenevery resolvingsubset of X isinfinite.
Proof. Suppose thatX is a non-emptyT1 space and thatD ⊆ X resolvesX. If D isfinite, then D isa closed subsetofX,sothatX = D = D.ButthenX = X\ D = ∅, whichisnotthecase. 2
Theremainderofthis sectionisdividedintofour (non-disjoint)parts,coveringseparablespaces, metric spaces, topologicalvectorspaces,andlocally connectedBairespaces,respectively.
4.1. Separablespaces
Thesimplestexamplesofresolvablespacesthathavearesolvingsubsetwitha‘good’decompositionare obviouslythespacesthathaveanatmostcountablyinfiniteresolvingsubset.Inthatcase,thecomponents oftheresolvingsubsetasinpart (1)ofTheorem3.1cansimplybetakentobesingletons.Clearly,thespace is thenseparable,andinanumberofcasesthisisalsosufficient.
Proposition 4.2. Let X be a separable space. If X has the property that every non-empty open subset is uncountable, thenevery atmostcountablyinfinite dense subsetof X resolvesX.
Proposition4.2appliestoanumberofspacesofpracticalinterest.Forexample,non-emptydifferentiable manifolds (which are second countable by definition) and non-empty open subsets of separable real or complextopologicalvectorspaces haveanat mostcountablyinfiniteresolvingsubset.Weshallhavemore tosayaboutnotnecessarilyseparablerealandcomplextopologicalvectorspacesinSection4.3.
Wecontinuewithanothercategoryofspaceswithacountableresolvingsubset.Itcoverse.g.non-empty separablecompletemetricspacesandnon-emptyseparablelocallycompactHausdorffspaces,inbothcases withoutisolatedpoints.
Proposition 4.3.Let X be anon-empty separableBaire T1 spacethat has no isolated points.Choose an at mostcountably infinite densesubsetD of X. ThenD isactually countablyinfinite, andD resolvesX.
Proof. Since X isaT1space thathasnoisolatedpoints,thesets X\ {x} areopenand denseinX for all x∈ X.Since X isaBaire space,x∈DX\ {x}= X\ D isdense inX.HenceD resolves X.Lemma4.1
showsthatD isinfinite. 2
IntheproofofTheorem5.1weshallneedthefinalstatementofthefollowingresult,whichis[10,Lemma 3.12].Theproofof[10,Lemma3.12],however,actuallyshowsthatthefirststatementistrue,anditseems worthwhiletorecordthis.
Lemma 4.4. Let X be a resolvable space, and suppose that the subset D of X is dense in X. Then there existsasubset of D thatresolvesX. Inparticular, ifX is resolvableand separable,then thereexistsan at mostcountably infinite subsetof X thatresolves X.
4.2. Metric spaces
Alittlebitmorecomplicatedthanthecasewherethecomponentsofaresolvingsubsetaresingletonsas inSection4.2,is the casewhere these components areclosed nowhere densesubsets. The presentsection containssuchresultsinthecontextofmetricspaces;Section4.3coversrealandcomplextopologicalvector spaces,andSection4.4dealswithlocallyconnectedBairespaces.
Thefollowingresultappliestonon-emptycompletemetricspacesthathavenoisolatedpoints.Werecall, however,thatthere aremetrizableBairespaceswhicharenotcompletelymetrizable.
Proposition 4.5.LetX be anon-empty metricBaire spacethat has no isolatedpoints. Thenthereexist an atmostcountablyinfinite collection{Γn: n∈ N } ofclosednowheredensesubsetsofX suchthat n∈NΓn resolvesX.
Proof. Theproofof[7,Lemmaon p. 249] showsthatthereexistclosednowheredensesubsetsΓ1,Γ2,Γ3,. . . ofX suchthatn∈NΓnisdenseinX.SinceX isaBairespace,n∈NΓnhasemptyinterior,i.e.X\n∈NΓn
isdenseinX.Hencen∈NΓn resolvesX. 2
Euclideann-spacefalls withinthescopeofProposition4.2,so thatithasaverysimplesuitably decom-posableresolvingsubset.A morecomplicated resolvingsubsetthatisstillsuitably decomposablecouldbe obtainedbytakingtheunionofallmetricspheres
Sr(0) ={ x ∈ Rn : x = r }
Proposition 4.6. Let X be a locallyconnected metric space with metric d(·,·).Suppose that X contains at least twopoints. For x0 ∈ X,set Br(x0):={x∈ X : d(x0,x) < r} andSr(x0)={x∈ X : d(x0,x)= r} forr > 0, andletDx0={d(x0,x): x∈ X }.
Suppose thatthere existsapointx0∈ X suchthat (1) Br(x0) ={x∈ X : d(x0,x)≤ r } forevery r > 0,and (2) Dx0 isconnectedin R.
If D ⊆ Dx0 \ {0} is non-empty and dense in Dx0 (such D exist), then
r∈DSr(x0) is dense in X, and if
D ⊆ Dx0\ {0} iscountably infinite anddensein Dx0 (suchD exist), then
r∈ DSr(x0) resolves X. Forallr > 0, thesubsetSr(x0) of X isaclosed nowhere densesubsetof X.
Proof. Since X containsatleast twopoints, itfollowsfrom thesecondassumption thatDx0 is aninterval
of theform [0,M ] or[0,M ) forsomeM > 0,or equalto [0,∞).Since Dx0 doesnotreduceto{0},sets D
and D as inthestatementquiteobviouslyexist.
Clearly, Sr(x0) is closed for all r > 0. Furthermore, the first part of the hypotheses implies that, for r > 0,Sr(x0)= Br(x0)\ Br(x0).HenceSr(x0) hasemptyinteriorforallr > 0,sothatitisindeedaclosed nowheredensesubsetofX.
SupposethatD ⊆ Dx0\ {0} isdenseinDx0.Weshallshowthat
r∈DSr(x0) isdenseinX.
Supposethatthiswerenotthecase.Thenthere existsanon-emptyconnectedopensubsetU of X such
thatU∩ Sr(x0)=∅ foreveryr∈ D.Considerany r∈ D.Since{d(x0,x): x∈ U } isconnectedinR,but doesnotcontainr,wehaveeitherd(x0,x)< r forallx∈ U,ord(x0,x)> r forallx∈ U.
Choose and fix a point x1 ∈ U. From whatwe havejust seen,if x∈ U is arbitrary, then d(x0,x) < r wheneverr∈ D issuchthatd(x0,x1)< r,andd(x0,x)> r wheneverr∈ D issuchthatd(x0,x1)> r.
Ifd(x0,x1) isaninteriorpointofDx0\{0},then,bythedensityofD inDx0\{0},thereexistsasequence
(rn) in D such thatrn ↓ d(x0,x1). From whatwe havejustobserved, itfollows thatd(x0,x) ≤ rn for all x∈ U and alln∈ N.Henced(x0,x)≤ d(x0,x1) for allx∈ U. Similarly,usingasequence (rn) inD such
thatrn↑ d(x0,x1),wehaved(x0,x)≥ d(x0,x1) forallx∈ U.WeconcludethatU ⊆ Sd(x0,x1)(x0).Thisset,
however,hasemptyinterior.Thiscontradiction showsthatd(x1,x0) isnotaninteriorpointofDx0\ {0}.
Ifd(x0,x1)= 0,thenconsiderationofasequenceinD thatdecreasesto0showsthatd(x0,x)= 0 forall x∈ U.Thatis, U ={x0}.This,however,implies that{x0} isisolated inDx0, whichisnotthecase. This
contradiction showsthatd(x0,x1)= 0.
IfDx0isanintervaloftheform[0,M ) forsomeM > 0,orequalto[0,∞),thenallpossibilitiesford(x0,x1)
havenowbeenexhausted,andthisfinalcontradictionconcludestheproofofthedensityofr∈DSr(x0) in X in thesetwocases.
If Dx0 = [0,M ] for someM > 0, we stillhave to consider the possibility thatd(x0,x1)= M .Using a
sequenceinD thatincreasestoM thenshowsthatd(x0,x)≥ M forallx∈ U.Sincethereverseinequality isobviouslyalsosatisfied,wefindU ⊆ SM(x0),whichisagainimpossible.Thisexhaustsallpossibilitiesfor d(x0,x1) ifDx0 = [0,M ] forsomeM > 0,and thisfinalcontradictioncompletestheproofofthedensityof
r∈DSr(x0) inX alsointhecasewhereDx0= [0,M ] forsomeM > 0.
Weturn to thestatementconcerningD. SupposethatD ⊆ D x0 \ {0} iscountablyinfiniteanddense in
Dx0. Then
r∈ DSr(x0) isdense inX by whatwe have justshown. Onthe other hand,since non-empty opensubsetsofDx0areuncountable,(Dx0\ {0})\ D isalsodenseinDx0.Againbywhatwehavejustshown,
r∈ Dx0\{0} \ DSr(x0) isdense inX.SinceevidentlyX\
r∈ DSr(x0)={x0}∪
r∈ Dx0\{0} \ DSr(x0), we see thatX\r∈ DSr(x0) isdenseinX.
4.3. Topologicalvector spaces
The ideaof working with metric spheresin Proposition4.6 canbe adaptedto the context of real and complextopologicalvectorspaces,where ametricspherewithradiusr > 0 is replacedwithr(U\ U) fora suitable open neighbourhoodU of0. Ourfinal result inthis veinis Theorem 4.10,and we start with the necessary preparations.
Lemma4.7. LetX bea (notnecessarilyHausdorff )real orcomplex topologicalvectorspace, andletU bea balancedopenneighbourhoodof0 inX.SetW :=r>0rU .ThenW isinvariantunderscalarmultiplication.
Proof. Considerx∈ W andascalarα.Fixr > 0.Sincex∈ W ,itfollows that
x∈ 1+r|α|U.
Then
αx∈ 1+αr|α|U = r 1+α|α|U.
Since U is balanced, we have 1+α|α|U ⊆ U. Hence αx ∈ rU. Since r > 0 is arbitrary, it follows that
αx∈ W . 2
Lemma4.8. LetX bea (notnecessarilyHausdorff )realorcomplextopologicalvectorspace,andsupposethat U is abalancedconvexopen neighbourhoodof 0. Ifr1,r2∈ (0,∞) aresuch that r1= r2,thenr1
U\ U ∩ r2
U\ U =∅.
Proof. ConsidertheMinkowskifunctional associatedwith U ,givenby
μU(x) = inf{ t > 0 : x ∈ tU }.
SinceU isbalanced,convex, andabsorbing,[15,Theorem 1.35(c)] showsthatμU isaseminorm.
Further-more, sinceU is an open subset of X, theproof of [15, Theorem 1.36] shows thatμU is continuous, and
that
U ={ x ∈ X : μU(x) < 1}. (4.1)
Suppose now that r1,r2 ∈ (0,∞) are such that r1 < r2, and suppose that x ∈ r1
U \ U ∩ r2
U \ U . Then there exist x1,x2 ∈ U \ U suchthat r1x1 = x = r2x2. Bythe continuity of μU and (4.1), we have μU(x1)≤ 1.Since x2 = rr12x1, wesee thatμU(x2)= rr12μU(x1)< 1. Then(4.1) implies thatx2∈ U,which isacontradiction. 2
Proposition4.9. LetX bea (notnecessarilyHausdorff )real orcomplextopologicalvectorspacesuchthat0
has anopenneighbourhoodthat isnotthewholespace. SupposethatU isabalancedopenneighbourhoodof X suchthat U− U = X (such U exist),and supposethat S⊆ (0,∞) isdense in(0,∞).
Thenr∈Sr U\ U isdenseinX,and,forallr > 0,thesetr U\ U isaclosednowheredensesubset of X.
Proof. ThereexistsanopenneighbourhoodW of0 suchthatW = X,andthenthecontinuityofthevector space operationsimplies thatthere exists anopen neighbourhoodU of 0 such thatU − U ⊆ W .Since U
balanced.ThisestablishestheexistenceofbalancedopenneighbourhoodsU of0 suchthatU− U = X.We chooseandfix onesuchU .
Foreachr > 0,let
Γr= rU\ rU = r(U \ U).
Each Γr isaclosed nowheredensesubsetofX.Weneedtoshow that
r∈S
Γr is dense in X. (4.2)
Supposethat(4.2) werefalse.NotethatX,beingatopologicalvectorspace,islocallyconnected.Therefore, inthatcase,there existsanon-emptyconnectedopensubsetV ofX such that
V ∩ Γr=∅
forallr∈ S.Thatis,
V ⊆ (rU) ∪ (X \ rU)
forallr∈ S.SinceV isconnected, itfollowsthat,forallr∈ S,
either V ⊆ rU or V ⊆ X \ rU. (4.3)
As U is an open neighbourhoodof 0 and S is dense in(0,∞), there exists r∈ S such that V ∩ rU = ∅.
Hence(4.3) showsthat
{ r ∈ S : V ⊆ rU } = ∅, (4.4)
whichenablesusto set
R = inf{ r ∈ S : V ⊆ rU }.
Considerr1,r2 ∈ (0,∞) such thatr1 < r2.Since U isbalanced,we haver1U ⊆ r2U . Combiningthis with the density of S in (0,∞), we see that V ⊆ rU for all r > R. If R > 0 (as we shall demonstrate in a moment), then, using (4.3) and the density of S∩ (0,R) in (0,R), we also see that V ⊆ X \ rU for all
r∈ (0,R).
Weshow thatR= 0.Arguingbycontradiction,supposethatR = 0.From whatwe havejustobserved, itthen followsthatV ⊆r>0rU , sothat
V − V ⊆ r>0
rU− r>0
rU.
Since V − V isaneighbourhoodof0,weconcludethatr>0rU−r>0rU isabsorbing.ThenLemma4.7
impliesthatr>0rU−r>0rU = X,sothatcertainlyU− U = X.Thiscontradictionwithourchoicefor
U showsthatR > 0.
Weclaimthat
To see this, consider x∈ V . We know thatx/r∈ U foreveryr ∈ (R,∞). Sincex/r converges to x/R as r↓ R, weseethatx/R ∈ U.Hencex∈ RU.
Weclaimthat
V ⊆ X \ RU. (4.6)
To see this, consider x∈ RU, so that x= RxR for some xR ∈ U. Weknow that V ⊆ X \ rU ⊆ X \ rU
for allr∈ (0,R).Hence rU ⊆ X \ V forallr ∈ (0,R),so thatrxR ∈ X \ V forall r∈ (0,R).Since rxR
convergesto RxR= x asr↑ R,weseethatx∈ X \ V .Thisestablishes (4.6).
Itfollowsfrom(4.5) and(4.6) thatV ⊆ RU \RU,contradictingthefactthatRU\RU hasemptyinterior. Hence(4.2) musthold. 2
Theorem4.10.LetX bea (notnecessarilyHausdorff )realorcomplex topologicalvectorspace,andletS be acountablyinfinite dense subsetof (0,∞).
If X is aBaire space and0 has an openneighbourhood that is notthe wholespace, letU bea balanced openneighbourhood of0 suchthat U− U = X (such U exist).
If 0 has a convex open neighbourhood that is not the whole space, let U be a balanced convex open neighbourhoodof X suchthat U− U = X (such U exist).
Then, inboth cases,r∈Sr U\ U resolves X and, forallr > 0, theset r U\ U is aclosednowhere densesubset of X.
Proof. Westart withthecasewhere X isaBairespaceand0 hasanopen neighbourhoodthatis notthe wholespace.Proposition4.9thenshowsthatthereexists abalancedopenneighbourhoodU of0 suchthat
U− U = X andthat,for anysuchU , r∈Sr U\ U isdenseinX.Italso assertsthat,forallr > 0,the set r U\ U is aclosed nowhere dense subset of X. Since X is a Bairespace and theset r U\ U is a closed nowheredensesubsetof X foreachr > 0,itis immediate thatX \r∈Sr U\ U isalso densein
X.Hencer∈Sr U\ U resolves X.
We turn to the case where 0 has a convex open neighbourhood W that is not the whole space. An appeal to [15, Theorem 1.14 (b)] shows that there exists abalanced convex open neighbourhood U such
thatU ⊆ 12W ;althoughtheresultreferredto isstatedinacontext whereX isHausdorff,theproofmakes no useof this fact.Then U− U = U + U sinceU is balanced,and U + U ⊆ 1
2W + 1
2W ⊆ W since W is convex.WehavethusestablishedtheexistenceofabalancedconvexopenneighbourhoodU of 0 suchthat
U− U = X.
IfU isanysuchopenneighbourhoodof0,thenProposition4.9showsagainthatr∈Sr U\ U isdense inX andthat,forallr > 0,thesetr U\ U isaclosednowheredensesubsetofX.Inaddition,Lemma4.8
impliesthat X\ r∈S r U\ U ⊇ r∈(0,∞)\S r U\ U .
Since S is countably infinite, (0,∞)\ S is also dense in (0,∞). Proposition 4.9 therefore also shows that r∈(0,∞)\Sr U\ U is dense in X, and then this is certainly true for X \r∈Sr U\ U . Hence
r∈Sr
U\ U resolvesX. 2 4.4. LocallyconnectedBaire spaces
ThefirstmainresultofthissectionisProposition4.14,whichcanbeviewedasdefininganddescribingthe supportofanordercontinuouslinearfunctionalonasufficientlyrichvectorlatticeofcontinuousfunctions onaT1spacewithoutisolatedpoints. Herethespacedoesnotyetneed tobeaBairespace.
ThesecondmainresultofthissectionisProposition4.15,statingthatanon-emptylocallyconnectedT1 Baire space admits aresolving set of theform n∈NΓn, where each Γn is aclosed nowheredense subset
of X,providedthatEn∼ containsastrictly positiveordercontinuouslinearfunctionalforasufficientlyrich vector latticeE ofcontinuous functionsonX.
Westartwithsomepreparatoryresults.
Lemma 4.11.LetX be aspace, and letx0 bea pointinX that is notisolated. LetE beavector lattice of continuous functions on X, andsuppose that S⊆ E+ has theproperty that, forevery x= x
0 in X, there exists fx∈ S suchthat fx(x)= 0. Theninf S = 0 inE.
Proof. Suppose thatg ∈ E is such that g ≤ f for all f ∈ S. Then g ≤ fx for all x= x0 in X, so that g(x)≤ 0 forallx= x0 inX.Ifg(x0)> 0,thenthereisanopenneighbourhoodofx0 onwhichg isstrictly positive.Since x0 is notisolated, thisneighbourhoodcontainsapointx1 thatis differentfrom x0.Inthat case, however,we alreadyknowthatg(x1)≤ 0,contradicting thatg(x1)> 0.Henceg(x0)≤ 0,andwe see thatg≤ 0.Since 0 isalowerboundforS, weconcludethatinf S = 0 inE. 2
Proposition 4.12.LetX beaspace, andletx0 beapointin X thatisnot isolated.LetE beavectorlattice of continuousfunctionsonX with theproperty that,foreveryx= x0 inX,there existsfx∈ E+ suchthat fx(x)= 0 and fx(x0)= 1.
Supposethat (fi)i∈I ⊆ E+ isan orderboundednet inE withtheproperty that,foreveryU ∈ Vx0,there
exists iU ∈ I suchthat X\ Z(fi)⊆ U for alli≥ iU.
Then there existsan orderbounded net (gj)j∈J ⊆ E+ such that gj ↓ 0 in E and with theproperty that, forallj0∈ J,there existsi0∈ I suchthat 0≤ fi≤ gj0 foralli≥ i0.
Consequently, ifϕ∈ (En∼)+,then inf{ϕ(fi): i∈ I }= 0.
Proof. Chooseandfixanupped boundu for(fi)i∈I inE+.Set
J ={ j ∈ E+: j≤ u and there exists i0∈ I such that j ≥ fi for all i≥ i0}.
Then u∈ J,so thatJ = ∅. Weintroduce apartialorder onJ by sayingthat,for j1,j2 ∈ J,j1 j2 inJ if j1≤ j2 inE. WeclaimthatJ is directed.Indeed,suppose thatj1∈ J andi1 ∈ I aresuchthatj1≥ fi
for alli≥ i1,andthatj2∈ J andi2∈ I aresuchthatj2≥ fi foralli≥ i2.There existsi3∈ I suchthat i3≥ i1 andi3≥ i2,and thenj1∧ j2≥ fi for alli≥ i3.Since also0≤ j1∧ j2 ≤ u,we seethatj1∧ j2∈ J. Theinequalitiesj1∧ j2 j1 andj1∧ j2 j2 inJ thenshowthatJ isdirected.
Consider the net (gj)j∈J in E+ thatis definedby gj := j for j ∈ J. It is clear thatthis net is order
bounded and decreasing. We claim that actually gj ↓ 0 in E. To see this, we shall employLemma 4.11.
Fix x1 = x0. Then there exists fx1 ∈ E
+ such that f
x1(x1) = 0 and fx1(x0) = 1. The subset U =
{x∈ X : fx1(x) >
1
2 and u(x) < u(x0)+ 1} of X is an open neighbourhoodof x0, so, by assumption, there exists iU ∈ I such that {x ∈ X : fi(x) = 0} ⊆ U for all i ≥ iU. If i ≥ iU and x ∈ U, then
2(u(x0)+ 1)fx1(x)> u(x0)+ 1> u(x)≥ fi(x), and trivially2(u(x0)+ 1)fx1(x)≥ 0= fi(x) foralli≥ iU
and x∈ U./ Hence2(u(x0)+ 1)fx1 ≥ fi foralli ≥ iU. Then also[2(u(x0)+ 1)fx1]∧ u≥ fi forall i≥ iU,
and weconcludethat[2(u(x0)+ 1)fx1]∧ u∈ J.
Sinceg[2(u(x0)+1)fx1]∧u(x1)= ([2(u(x0)+ 1)fx1]∧ u)(x1)= [2(u(x0)+ 1)fx1(x1)]∧ u(x1)= 0∧ u(x1)= 0,
Ifj0∈ J is given, thenbytheverydefinitionofJ thereexists i0∈ I such that0≤ fi≤ j0 = gj0 for all
i≥ i0.
Wehavethusestablishedtheexistenceofanet(gj)j∈J with therequiredproperties.
Thefinalstatementofthepropositionisclearsinceϕ(gj)↓ 0 for ϕ∈ (E∼n) +
. 2
Remark4.13.The net (fi)i∈I inProposition 4.12converges inorder to 0 inE in thesense of [1,
Defini-tion 1.18]. If we knew E to be Dedekind complete, then [1, Lemma 1.19 (b)] would imply thatit is also convergent in order to 0 in E in the (non-equivalent) sense of [2, p. 33]. Since we do not know E to be Dedekindcomplete,wehaverefrainedfrom includingany(ambiguous) statementaboutorderconvergence of(fi)i∈I inProposition4.12.
Inthefollowingproposition,weintroduceafunctionρ onthecollectionofnon-emptysubsetsofX that
bearssomeresemblanceto thefunctionρ on2X intheproofofTheorem 3.1.
Proposition 4.14. LetX be a non-empty T1 space without isolated points, and let E be a vector lattice of continuous functionson X such that,for every pointx0∈ X and every U ∈ Vx0,there existV ∈ Vx0 and
f ∈ E such thatV ⊆ U,0≤ f ≤ 1, f (x)= 1 for allx∈ V ,andX\ Z(f)⊆ U. Letϕ∈ (En∼)+ bea positiveordercontinuous linearfunctional onE. Foreach non-emptyopensubset U of X,define
ρ(U ) := sup{ ϕ(f) : f ∈ E, 0 ≤ f ≤ 1, and X \ Z(f) ⊆ U }. (4.7)
Then:
(1) 0 ≤ ρ(U) ≤ ∞ for each non-empty open subset U of X, and ρ(U ) ≤ ρ(V ) for all non-empty open subsets U andV of X suchthatU ⊆ V ;
(2) ifU andV arenon-empty opensubsets ofX suchthat ρ(U )= ρ(V )= 0,then alsoρ(U∪ V )= 0;
(3) forevery x0∈ X andε> 0,there existsU ∈ Vx0 suchthat ρ(U )< ε;
(4) the set Sϕ := {x∈ X : ρ(U) > 0for all U ∈ Vx} is a closed subset of X, and if ϕ = 0, then it has non-empty interior,
(5) forf ∈ E+,ϕ(f )= 0 ifand onlyif f (x)= 0 for allx∈ S ϕ; (6) ϕ is strictlypositiveif andonly ifSϕ= X.
Proof. Let U be a non-empty open subset of X. Since the set in the right hand side of (4.7) contains
ϕ(0)= 0,thefirstpartofthefirstconclusionisclear.Itisalsoclearthatρ ismonotone.
Before proceedingwiththe remaining conclusions,we note thefollowing for later use.Consider a non-empty open subset V of X such that ρ(V ) = 0. We claim that then ϕ(f ) = 0 whenever f ∈ E+ is bounded on V and X \ Z(f) ⊆ V . Indeed, since f is bounded on V and zero outside V , there exists
M > 0 such that0≤ f ≤ M1,or 0≤ f/M ≤ 1. Since X \ Z(f/M) = X\ Z(f) ⊆ V , this implies that 0≤ ϕ(f/M)≤ ρ(V )= 0,sothatϕ(f )= 0.
Wenowturntothesecondconclusion.LetU andV benon-emptyopensubsetsofX withtheproperty thatρ(U )= ρ(V )= 0.Then
ϕ(f ) = 0 whenever f ∈ E, 0 ≤ f ≤ 1, and X \ Z(f) ⊆ U (4.8) and
Set U0 := U \ V and V0 := V \ U0. Then U0 and V0 are disjoint open subsets of X. Consequently, U0∩ V0 =∅ and V0∩ U0 =∅.It canhappenthatU0 =∅, butV0 isnever empty.Indeed,if V0 =∅,then V ⊆ U0= U\ V ⊆ X \ V ⊆ X \ V = X \ V ,whichcontradictsthatV isnon-empty.
It is easy to see that U0∪ V0 and X \ (U ∪ V ) are disjoint subsets of X. Furthermore, their union U0∪ V0∪ (X \ (U ∪ V )) isadensesubsetofX.Toseethis,itissufficienttoshowthatU∪ V ⊆ U0∪ V0.For this, fixx∈ U ∪ V ,andconsideranarbitraryW ∈ Vx.WearetoshowthatW∩ (U0∩ V0)= ∅,andforthis wemayevidentlysupposethatW ⊆ U ∪V .IfW∩U0=∅,thenW ⊆ V ,sothatW∩V = ∅.Butinthiscase alsoW∩U0=∅,asX\W isaclosedsubsetofX thatcontainsU0.ItfollowsthatW∩V0= W∩V = ∅.We concludethatx∈ U0∪ V0.Therefore,U∪V ⊆ U0∪ V0,asdesired,sothatthesubsetU0∪V0∪(X \(U ∪V )) of X isindeeddenseinX.
After these preparations, we consider afixed h∈ E such that 0≤ h≤ 1 and X\ Z(h) ⊆ U ∪ V . We shall show that ϕ(h)= 0, as follows. It follows from the first partof the hypotheses and thefact thatE
is a vector lattice that, for each y ∈ U0 (if any), there exists fy ∈ E+ such that fy(y) = h(y), fy ≤ h,
and X\ Z(fy)⊆ U0. Likewise, foreachz ∈ V0,there exists gz ∈ E+ suchthatgz(z)= h(z), gz ≤ h, and X \ Z(gz)⊆ V0.Denote byA andB the collectionsoffinite subsetsofU0 andV0,respectively.For∅∈ A, we define f∅ := 0; for ∅ ∈ B, we define g∅ := 0. For a non-empty A = {y1,. . . ,yn}∈ A (if any) and a
non-empty B ={z1,. . . ,zk}∈ B,wedefine
fA:= fy1∨ · · · ∨ fyn,
gB := gz1∨ · · · ∨ gzk. Then, forallA∈ A andB ∈ B,wehave
fA(y) = h(y) for all y∈ A (if any), 0 ≤ fA≤ h ≤ 1, and X \ Z(fA)⊆ U0 (4.10) and
gB(z) = h(z) for all z∈ B (if any), 0 ≤ gB≤ h ≤ 1, and X \ Z(gB)⊆ V0. (4.11) Since U0 and V0 are disjoint, we also have, for all A ∈ A and B ∈ B, that fA(x)+ gB(x) = h(x) for
all x ∈ A∪ B, 0 ≤ fA+ gB ≤ h, and X \ Z(fA + gB) ⊆ U0∪ V0. Clearly, the sets {fA : A ∈ A}
and {gB : B ∈ B } are both upward directed; therefore so is {fA+ gB : A ∈ A,B ∈ B }. Furthermore, fA+ gB ≤ h forallA∈ A andB∈ B.Weclaimthat,infact,fA+ gB ↑ h inE.Supposethatw∈ E issuch
thatfA+gB≤ w forallA∈ A andB∈ B.Thenw≥ f∅+g∅= 0.Inparticular,forx∈ X \(U ∪V ),wehave w(x)≥ 0= h(x).Ifx∈ U0∪ V0, thenwecanchooseA∈ A and B∈ B suchthatx∈ A∪ B,inwhichcase w(x)≥ fA(x)+ gB(x)= h(x).Wehavenowestablishedthatw(x)≥ h(x) forallx∈ U0∪V0∪(X \(U ∪V )). SinceU0∪ V0∪ (X \ (U ∪ V )) isdenseinX,itfollowsthatw≥ h.WeconcludethatfA+ gB↑ h,asclaimed.
It follows from (4.8),(4.9),(4.10),and (4.11) thatϕ(fA+ gB)= ϕ(fA)+ ϕ(gB)= 0 for allA∈ A and B ∈ B.Since fA+ gB ↑ h,theordercontinuityofϕ thenimpliesthatϕ(h)= 0,as weintendedtoshow.
Since this istruefor allh∈ E suchthat0≤ h≤ 1 and X\ Z(h)⊆ U ∪ V ,we see thatρ(U∪ V )= 0. Wehavethusestablishedthesecondconclusion.
We turnto thethirdconclusion.Fixx0∈ X andε> 0.Weareto showthatthere existsU ∈ Vx0 such
thatρ(U )< ε.Supposethatthiswerenotthecase.Then,foreveryU ∈ Vx0,there existsfU ∈ E suchthat
0≤ fU ≤ 1,X\ Z(fU)⊆ U,and ϕ(fU)≥ ε/2. Weshalluse Proposition4.12toshow thatthis leadstoa
contradiction.
Thefirstpartofthehypothesesalso impliesthatthere existV0∈ Vx0 andf0∈ E suchthat0≤ f0≤ 1
andf0(x)= 1 forallx∈ V0.LetVx0 denotethecollectionofopenneighbourhoodsofx0thatarecontained
inV0,orderedbyreverseinclusion,andconsider thenet(fW)W∈ Vx0 inE+.
Thisnetisboundedfrom aboveinE byf0.
Furthermore,ifU ∈ Vx0 isgiven,thenX\ Z(fW)⊆ U forallW ∈ Vx0 suchthatW ⊆ U ∩ V0∈ Vx0.
Wecannowapply Proposition4.12tothenet(fW)W∈ V
x0 toconcludethatinf{ϕ(fW): W ∈ Vx0}= 0. This, however,contradictsthefactthatϕ(fW)≥ ε/2 forallW ∈ Vx0.We havethus establishedthethird
conclusion.
We now turn to the fourthconclusion. Supposethat x∈ X \ Sϕ. Bythe definitionof Sϕ, there exists V ∈ Vxsuchthatρ(V )= 0.SinceV isanopensubsetofX,wehaveV ∈ Vyforally∈ V .HenceV ⊆ X \Sϕ.
WeconcludethatSϕ isaclosed subsetofX.
We now show thatSϕ has non-empty interior ifϕ = 0. Suppose, to the contrary, thatthe interior of Sϕ were empty. Since ϕ > 0, there exists u ∈ E+ such thatϕ(u) > 0. Choose and fix such u.Since Sϕ
isaclosed subsetof X withemptyinterior, X\ Sϕ is anon-emptyopen anddense subsetof X.Foreach z∈ X \ Sϕ, there exists V ∈ Vz suchthatρ(V )= 0 andu is boundedon V .Denote byK thenon-empty
set of allpairs(z,V ),where z ∈ X \ Sϕ and V isanopen neighbourhoodof z such thatρ(V )= 0 and u
isbounded onV , andbyL the non-empty collectionofnon-emptyfinite subsetsof K.Itfollows from the firstpartofthe hypothesesand thefactthatE isavectorlatticethat,foreach(z,V )∈ K, thereexists a functionf(z,V )∈ E+ suchthat
f(z,V )≤ u, f(z,V )(z) = u(z), and X\ Z(f(z,V ))⊆ V.
If(z,V )∈ K,thenu isbounded onV .Hencef(z,V )isalsoboundedonV .Sinceρ(V )= 0,thepreliminary remark inthebeginning ofthis proof showsthatϕ(f(z,V ))= 0. For eachT ={(z1,V1),. . . ,(zk,Vk)}∈ L,
set
fT := f(z1,V1)∨ · · · ∨ f(zk,Vk)∈ E
+.
Then
fT ≤ u, fT(zi) = u(zi) for i = 1, . . . , k, and X\ Z(fT)⊆ V1∪ . . . ∪ Vk.
Bythesecondconclusion,ρ(V1∪ . . . ∪ Vk)= 0.Furthermore,u andthereforefT isboundedonV1∪ . . . ∪ Vk.
Itfollows fromthepreliminaryremark inthebeginningoftheproofthatϕ(fT)= 0 forallT ∈ L.
The set {fT : T ∈ L} is upward directed, and fT ≤ u for all T ∈ L. Weclaim that fT ↑ u in E. To
see this,suppose thatv∈ E issuchthatfT ≤ v for allT ∈ L. Ifx∈ X \ Sϕ,then u(x)= fT(x) for some T ∈ L, whichimpliesthatu(x)= fT(x) ≤ v(x).Since X\ Sϕ is denseinX,it followsthatu≤ v.Hence fT ↑ u,as claimed.
Bytheordercontinuityofϕ,wehaveϕ(fT)↑ ϕ(u).Henceϕ(u)= 0,contrarytothefactthatϕ(u)> 0.
Thiscontradiction concludestheproofofthefactthatSϕhasnon-emptyinterior.
Wenowturnto thefifthpartoftheconclusion.
Let f ∈ E+ be suchthatϕ(f ) = 0,and suppose that there exists x
0 ∈ Sϕ such thatf (x0)= 0.Since f ≥ 0,itfollowsthatthereexistarealnumberε0> 0 andV ∈ Vx0 suchthatε0< f (x) forallx∈ V .Since
x0∈ Sϕ,wehaveρ(V )> 0.Hencethere existsu∈ E+ suchthat0≤ u≤ 1,X\ Z(u)⊆ V ,andϕ(u)> 0.
Thenε0u≤ f.Indeed, ifx∈ V ,then ε0u(x)≤ ε0< f (x),and ifx∈ X \ V ,then ε0u(x)= 0≤ f(x).This impliesthat0< ε0ϕ(u)= ϕ(ε0u)≤ ϕ(f),sothatϕ(f )> 0.This isacontradiction,and weconcludethat f (x)= 0 for allx∈ Sϕ.
Conversely,suppose thatf ∈ E+ issuchthatf (x)= 0 for allx∈ S
open neighbourhoodof z such that V ⊆ X \ Sϕ, ρ(V )= 0, and f is bounded on V . Then there exists a
functionf(z,V )∈ E+suchthatf(z,V )≤ f,f(z,V )(z)= f (z),andX\ Z(f(z,V ))⊆ V .Wenotethat,sincef is bounded onV ,soisf(z,V ).Asρ(V )= 0,itfollowsthatϕ(f(z,V ))= 0.Employingamethodasintheproof ofthefourthconclusionabove,wefind,usingthatf (x)= 0 forallx∈ Sϕ,thatthereexistsanetfα↑ f in E suchthatϕ(fα)= 0 for eachα.Onceagain, theordercontinuityofϕ impliesthatϕ(f )= 0.
It remainstoestablishthesixthandfinalconclusion.
Supposethatϕ isstrictlypositive.Chooseandfixx0∈ X,andletV ∈ Vx0 bearbitrary.Bythefirstpart
of thehypotheses, thereexists afunction fV ∈ E such that0≤ fV ≤ 1, f (x0)= 1, and X\ Z(fV)⊆ V .
Bythestrict positivityofϕ and thedefinitionofρ,itfollows thatρ(V )≥ ϕ(fV)> 0. Hencex0∈ Sϕ.We
concludethatSϕ= X.
Conversely,supposethatSϕ= X,andthatu∈ E issuchthatu> 0 andϕ(u)= 0.Chooseandfixapoint x0 inX suchthatu(x0)> 0.Byscalingu,wemaysupposethatu(x0)> 1.LetV0={x∈ X : u(x)> 1}. Then V0 ∈ Vx0. If f ∈ E is such that0≤ f ≤ 1 and X\ Z(f) ⊆ V0, then 0≤ f ≤ u. Since ϕ(u) = 0,
this implies thatϕ(f )= 0. Thisshows thatρ(V0)= 0.Hencex0 ∈ S/ ϕ= X, which isacontradiction.We
concludethatϕ isstrictlypositive. 2
Proposition4.14isusedtoestablishthefollowing result,whichmakescontactwith resolvability. Proposition 4.15.LetX beanon-emptylocallyconnectedT1 Bairespacewithoutisolatedpoints, andletE be avectorlatticeof continuousfunctionson X suchthat,forevery pointx0∈ X andeveryU ∈ Vx0,there
existV ∈ Vx0 andf ∈ E such thatV ⊆ U,0≤ f ≤ 1,f (x)= 1 forallx∈ V ,andX\ Z(f)⊆ U.
Suppose thatthere existsastrictlypositiveorder continuouslinear functionalon E.
Then thereexistsan atmostcountably infinite collection{Γn: n∈ N } ofclosed nowheredense subsets of X suchthat n∈NΓn resolves X.
Proof. Chooseandfixastrictlypositiveordercontinuouslinearfunctionalϕ onE.Forallnon-emptyopen subsetsU of X,letρ(U ) beas definedinProposition4.14. Weobservethat,dueto thestrictpositivityof
ϕ,ρ isstrictly positiveonthenon-emptyopensubsetsofX,bythesixthconclusionofProposition4.14. Fix n∈ N. Bythe thirdconclusionof Proposition4.14, there exists anon-empty open subset U ofX
such that0< ρ(U ) < n1, and Zorn’s Lemmathen yields amaximal (with respect to inclusion) collection
Fn of non-empty mutually disjoint open subsets U of X such that 0 < ρ(U ) < n1 for all U ∈ Fn. Let Gn =U∈FnU .Clearly,Gn isanopen subsetofX.
We claim that Gn is dense in X. To see this, consider a non-empty open subset V of X. If x is a
point in V , then there exists an open neighbourhood W of x such that 0< ρ(W ) < 1n. Since then also 0< ρ(V ∩ W )<n1, themaximalityofFn impliesthat(V ∩ W )∩ Gn= ∅. HenceGn isindeeddenseinX.
We set Γn := X\ Gn. Then Γn is a closed nowhere dense subset of X, and we shall now show that
n∈NΓn resolves X.
Firstly,sinceX isaBairespace,itfollowsthatX\n∈NΓn =
n∈NGn isdenseinX.
Secondly,weclaimthatn∈NΓnisdenseinX.Arguingbycontradiction,supposethatV isanon-empty
open subsetofX suchthat
V ⊆ X \ n∈N Γn= n∈N Gn.
SinceX islocallyconnected,wemaysupposethatV isconnected.Fixn∈ N.WehaveV ⊆ Gn=U∈FnU .
Since V isconnectedandthemembersofFn aremutually disjointopensubsetsofX,thereexists U ∈ Fn
