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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
A topological space X is called resolvable if it contains a dense subset with dense
complement. Using only basic principles, we show that whenever the space X has a resolving subset that can be written as an at most countably infinite union of subsets, in such a way that a given vector lattice of (not necessarily bounded) continuous functions on X separates every point outside the resolving subset from each of its constituents, then the order continuous dual of this lattice is trivial. In order to apply this result in specific cases, we show that several spaces have resolving subsets that can be written as at most countably infinite unions of closed nowhere dense subsets. An appeal to the main result then yields, for example, that, under appropriate conditions, vector lattices of continuous functions on separable spaces, metric spaces, and topological vector spaces have trivial order continuous duals if they separate points and closed nowhere dense subsets. Our results in this direction extend known results in the literature. We also show that, under reasonably mild separation conditions, vector lattices of continuous functions on locally connected
T1 Baire spaces without isolated points have trivial order continuous duals. A
discussion of the relation between our results and the non-existence of non-zero normal measures is included.
© 2019 Elsevier Inc. All rights reserved.
1. Introduction andoverview
LetE beavectorlattice.Werecallthatanorder boundedlinearfunctionalϕonE 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:[email protected](M. de Jeu), [email protected](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.
https://doi.org/10.1016/j.jmaa.2019.06.039
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 be a 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 uof E+ suchthat u(x)= 1and
u(y)= 0 forally∈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,wementionexplicitlythatEneednotconsist 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
functionsinEimpliesthatsomepropertiesarestillautomatic.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 xiscontainedinX\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 wereanon-empty opensubset ofX thatis containedinΓk,
thenV∩V wouldbeanon-emptyopensubsetofX thatiscontainedinΓk\Γk,whichisimpossible.Hence
Γk hasemptyinterior,i.e.Γk isnowheredense.
If X is aBairespace, 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 existsanelementuofE+ 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 suchthatDandX\D arebothdenseinX.WeshallthensaythatsuchasubsetDisaresolvingsubsetofX,orthatitresolvesX.
Hencepart (1) of thehypotheses impliesthat X shouldbe resolvable, but notjust that:Combination withpart(2)showsthatitmustberesolvableinaspecialmanner.IfonethinksofEas given,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 alsosatisfiedforeverysuperlatticeFofcontinuousfunctionsonX.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 as wellas x∈X\D.IfX avoidsthis obstructionbyhaving noisolated points,thenitisknowntoberesolvableinanumberofcases:ifXisametricspace(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 Eis avectorlattice, 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,ϕiscalled order continuousifinf{|ϕ(fi)|:i∈I}= 0 whenever(fi)i∈I isanetinE suchthatfi↓0 inE.Wedenote
theσ-ordercontinuouslinearfunctionalsonEandtheordercontinuouslinearfunctionalsonE byEc∼ and E∼n,respectively.Obviously,En∼ ⊆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 spaceis aspace 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 neighbourhoodsofxinX. Theclosure ofasubsetA of X isdenotedby
A. IfX isa space, then 0denotes thefunction onX thatis 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. AspaceXissaidtosatisfythecountablechaincondition,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 isorderseparableifeverysubsetofEthathasasupremumcontainsan 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 is orderseparable.
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 suchGisatmostcountablyinfinite.Foreachv∈G,thesetUv=X\Z(v) isopenandnon-empty.SinceG
consistsofpairwise disjointelements,Uv1∩Uv2 =∅ifv1,v2∈Garedifferent.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 (inparticular,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
(2) for every x∈X\n∈NΓn andevery k∈N,thereexists anelement uof E+ suchthat u(x)= 1 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 xis 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∈Esuchthatϕ≥0, e≥0,andϕ(e)= 1.Choose andfixsuchϕandefortheremainderoftheproof.
The first step of the proof consists of introducing anauxiliary set function onthe power set of X, as follows.
Foreveryn∈N and everysubsetAof X,set
ρn(A) :=
1 ifA∩Γn=∅;
0 ifA∩Γn=∅,
and set
ρ(A) = ∞
n=1
1 2nρn(A).
Sincen∈NΓnisdenseinX,itfollowsthatρ(U)>0 foreverynon-emptyopensubsetUofX.Furthermore,
ρ: 2X →[0,1] ismonotone.Boththesepropertiesareessentialinthesequeloftheproof;weshallencounter
asimilarauxiliaryfunctiononthenon-emptyopen subsetsof aspaceinProposition4.14.
ThesecondstepoftheproofconsistsofconstructingelementsofE thatareindexedbym∈N andthat aresuitablyrelatedtotheΓn,toρ,andtoϕ.Weshallusetheordercontinuityofϕhere.Theconstruction
of theseelements,whichwill bedenotedbyvmFm below,isasfollows.
Fixm∈N.Wechooseand fixNm∈N suchthat
∞
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)
Letxbeanarbitraryelementofthenon-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 ally∈ΓNm.
(3.3)
Denote byF thecollectionof non-emptyfinite subsets ofX\n∈NΓn. Then F is non-empty,sinceit
containsthesingletons{x}forallx∈X\n∈NΓn.Weintroduceapartialordering onF byinclusion;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 allx∈F,
vmF(y) = 0 for ally∈ΓNm.
(3.4)
We see from (3.4) that eis an upperbound of{vmF :F ∈ F } inE. 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
sincetheelementsofEareactuallycontinuous(whichhasnotbeenusedsofar),itfollowsthatf ≥e.This establishesourclaimthatvmF ↑einE.
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↓0inE.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∈/ΓNk,
or,equivalently (see (3.1)),we must have x∈ Ak. We conclude that{x∈ X : w+(x) >0}⊆ Ak. Since
ρis 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.
Asρisstrictlypositiveonnon-emptyopensubsetsofX,itfollowsthat{x∈X :w+(x)>0}=∅;herewe usethecontinuityofelements ofE again.Hencew+=0, andthenw≤0.This establishesourclaimthat wk↓0inE.
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 ↓0inE,wehaveϕ(wk)↓0.Ontheotherhand,using thate−vmFm ∈E
+ forallm∈N,we
notethat,forallk∈N,
e−wk =e− k
m=1
vmFm
=
k
m=1
≤
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 resolves X. If D is finite, then D is a
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 ofX.ThenD isactually countablyinfinite, andD resolvesX.
Proof. Since X isaT1space thathasnoisolatedpoints,thesets X\ {x}areopenand denseinX forall
x∈X. Since X isaBaire space,x∈DX\ {x}=X\D isdense inX. HenceD resolves X.Lemma4.1
showsthatDis infinite. 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 that resolvesX.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,. . .
ofXsuchthatn∈NΓnisdenseinX.SinceXisaBairespace,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) ofX isaclosed nowhere densesubsetof X.
Proof. SinceX 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 asinthestatementquiteobviouslyexist.
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 ofX such thatU∩Sr(x0)=∅foreveryr∈ D.Considerany r∈ D.Since{d(x0,x):x∈U}is connectedinR,but
doesnotcontainr,wehaveeitherd(x0,x)< rforallx∈U,ord(x0,x)> rforallx∈U.
Choose and fix a point x1 ∈ U. From whatwe havejust seen,if x∈U is arbitrary, then d(x0,x) < r
wheneverr∈ Dissuchthatd(x0,x1)< r,andd(x0,x)> rwheneverr∈ Dissuchthatd(x0,x1)> r.
Ifd(x0,x1) isaninteriorpointofDx0\{0},then,bythedensityofDinDx0\{0},thereexistsasequence (rn) in Dsuch 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) inDsuch
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,thenconsiderationofasequenceinDthatdecreasesto0showsthatd(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 inthesetwocases.
If Dx0 = [0,M] for someM > 0, we stillhave to consider the possibility thatd(x0,x1)=M. Using a sequenceinDthatincreasestoM 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 ⊆ Dx0 \ {0}is countablyinfiniteanddense in Dx0. Then
r∈DSr(x0) isdense inX by whatwe have justshown. Onthe other hand,since non-empty
opensubsetsofDx0areuncountable,(Dx0\ {0})\DisalsodenseinDx0.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 isreplacedwithr(U\U) fora suitable open neighbourhoodU of 0. Ourfinal result inthis veinis Theorem 4.10,and we start with the necessary preparations.
Lemma4.7. LetX be a(notnecessarilyHausdorff)real orcomplex topologicalvectorspace, andletU bea balancedopenneighbourhoodof0inX.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 =r1+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. LetXbea(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 such that 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 (suchU exist),and supposethat S⊆(0,∞)isdense in(0,∞).
Thenr∈SrU\UisdenseinX,and,forallr >0,thesetrU\Uisaclosednowheredensesubset of X.
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 inX. (4.2)
Supposethat(4.2) werefalse.NotethatX,beingatopologicalvectorspace,islocallyconnected.Therefore, inthatcase,there existsanon-emptyconnectedopensubsetV ofX suchthat
V ∩Γr=∅
forallr∈S.Thatis,
V ⊆(rU)∪(X\rU)
forallr∈S.SinceV isconnected, itfollowsthat,forallr∈S,
eitherV ⊆rU orV ⊆X\rU . (4.3)
As U is an open neighbourhoodof 0 and S is dense in(0,∞), there exists r∈ S suchthat 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=xasr↑R,weseethatx∈X\V.Thisestablishes (4.6).
Itfollowsfrom(4.5) and(4.6) thatV ⊆RU\RU,contradictingthefactthatRU\RUhasemptyinterior. Hence(4.2) musthold. 2
Theorem4.10.LetX bea(notnecessarilyHausdorff)realorcomplex topologicalvectorspace,andletSbe acountablyinfinite dense subsetof (0,∞).
If X is aBaire space and0 has an openneighbourhood that is notthe wholespace, letU bea balanced openneighbourhood of0suchthat U−U =X (suchU 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∈SrU\Uresolves X and, forallr >0,theset rU\Uis 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∈SrU\UisdenseinX.Italso assertsthat,forallr >0,the set rU\U is aclosed nowhere dense subset of X. Since X is a Bairespace and theset rU\U is a closed nowheredensesubsetof X foreachr >0,itis immediate thatX \r∈SrU\Uisalso densein X.Hencer∈SrU\Uresolves 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 sinceW is
convex.WehavethusestablishedtheexistenceofabalancedconvexopenneighbourhoodU of0 suchthat U−U =X.
IfU isanysuchopenneighbourhoodof0,thenProposition4.9showsagainthatr∈SrU\Uisdense inXandthat,forallr >0,thesetrU\UisaclosednowheredensesubsetofX.Inaddition,Lemma4.8
impliesthat
X\
r∈S
rU\U⊇
r∈(0,∞)\S
rU\U.
Since S is countably infinite, (0,∞)\S is also dense in (0,∞). Proposition 4.9 therefore also shows that r∈(0,∞)\SrU\U is dense in X, and then this is certainly true for X \r∈SrU\U. Hence
r∈Sr
U\UresolvesX. 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.TheninfS =0inE.
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 for allx= 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 0isalowerboundforS, weconcludethatinfS=0inE. 2
Proposition 4.12.LetX beaspace, andletx0 beapointin X that isnot isolated.LetE beavectorlattice
of continuousfunctionsonX with theproperty that,foreveryx=x0 inX,there existsfx∈E+ suchthat
fx(x)= 0and fx(x0)= 1.
Supposethat (fi)i∈I ⊆E+ isan orderboundednet inE withtheproperty that,foreveryU ∈ Vx0,there exists iU ∈I suchthat X\Z(fi)⊆U foralli≥iU.
Then there existsan orderbounded net (gj)j∈J ⊆E+ such that gj ↓0in E andwith 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 boundufor(fi)i∈I inE+.Set
J ={j∈E+:j≤uand there existsi0∈Isuch that j≥fi for alli≥i0}.
Then u∈J, so thatJ = ∅. Weintroduce apartialorder onJ bysayingthat,for j1,j2 ∈J,j1j2 inJ
if j1≤j2 inE. WeclaimthatJ is directed.Indeed,suppose thatj1∈J andi1 ∈I are suchthatj1≥fi
for alli≥i1,andthatj2∈J andi2∈I are suchthatj2≥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∧j2j1 andj1∧j2j2 inJ then showthatJ 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 ↓ 0in 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 andu(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, and sincex1 is anarbitrarypoint inX differingfrom x0,Lemma4.11shows thatinf{gj :j ∈J}=0in
Ifj0∈J isgiven, thenbytheverydefinitionofJ thereexists i0∈I suchthat0≤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 0inE 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 0in 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,weintroduceafunctionρonthecollectionofnon-emptysubsetsofX that bearssomeresemblanceto thefunctionρon2X 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 suchthatV ⊆U,0≤f ≤1,f(x)= 1 forallx∈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, andX\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 such thatU ⊆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 allU ∈ 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 forallx∈S
ϕ; (6) ϕ isstrictlypositiveif 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.Itisalsoclearthatρismonotone.
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 be non-emptyopensubsetsofX withtheproperty thatρ(U)=ρ(V)= 0.Then
ϕ(f) = 0 wheneverf ∈E,0≤f ≤1, andX\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 isaclosedsubsetofXthatcontainsU0.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 suchthat 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 byAandB thecollectionsoffinite 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∈ AandB ∈ B,wehave
fA(y) =h(y) for all y∈A(if any), 0≤fA≤h≤1, andX\Z(fA)⊆U0 (4.10)
and
gB(z) =h(z) for all z∈B (if any), 0≤gB≤h≤1, andX\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 ≤hforallA∈ AandB∈ B.Weclaimthat,infact,fA+gB ↑hinE.Supposethatw∈Eissuch
thatfA+gB≤wforallA∈ AandB∈ B.Thenw≥f∅+g∅=0.Inparticular,forx∈X\(U∪V),wehave
w(x)≥0=h(x).Ifx∈U0∪V0, thenwecanchooseA∈ Aand B∈ Bsuchthatx∈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,theordercontinuityofϕthenimpliesthatϕ(h)= 0,as weintendedtoshow.
Since this istruefor allh∈E suchthat0≤h≤1and 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.
Firstof all,ifx∈X issuchthatx=x0,then, sinceX isT1,X\ {x}is anopenneighbourhoodof x0.
