MEASURE-THEORETIC CHARACTERIZATIONS OF HEREDITARILY-NORMAL SPACES

MEASURE-THEORETIC CHARACTERIZATIONS OF HEREDITARILY-NORMAL SPACES

JOSEPH HERTZLINGER

Polytechnic University

(Received March 16, 1991 and in revised form March 30, 1992)

ABSTRACT. Inthispaperwecharacterize hereditarily-normalspacesintermsofthemeasure-theoreticpropertiesof the lattice ofclosedsets.Wethengeneralizefrom thatlatticetoother lattices.We applythe resultstoextremally- disconnectedspaces.

KEYWORDSANDPHRASES. Hereditarily-normal topologicalspaces,two-valued measures, lattice of closed sets, extremally-disconnected topologicalspace.

1991AMS SUBJECTCLASSIFICATIONCODES. 54D15, 54G05.

1. INTRODUCTION

Wecandevisetwomeasure-theoreticcharacterizationsofhereditarily-normalspaces.First, aspaceishereditarily- normal if andonlyifevery pre-measureis a measure.Second,aspaceishereditarily-normalif andonlyif thesetof measuresgreaterthan agivenmeasure form a chain.Wecan thengeneralizethese resultstoother lattices(the^lattices ofopensets, zero sets, cozerosets,

etc.).

Whenweapplytheresultstothe latticeofopensetswecan derive a few conclusions aboutextremally-disconnectedspaces.

We begin by givingafew definitions

(we

^notethat our lattice methods andterminologyare consistentwithstandard usage, e.g.,Camacho[1],Eid[4],Szeto

[8]):

DEFINITION 1.Wecandefine apartial orderonthesetoftwo-valuednontrivial measures on aspaceXas follows:/ -</t2^{if andonly}if/z

(F) _</2(F)

^{forallclosedsetsF. Wedefine_>ina}similar manner.

(We

notethat here we arereferring^tofinitely-additive two-valuedmeasures on thealgebrageneratedbythelattice ofclosed

sets.)

DEFINITION2.A two-valued pre-measureFIon aspaceXis asetfunctionfrom the collection ofclosedsets to {0, suchthatforany pairofclosedsetsF and

F2:

FI(O)

0,

(1.1)

if

F1

^{_CF then}

^FI(FI)

^_<

^{FI(F2), (1.2)}

if

FI(F)

^and

H(F2)

^then

FI(F

^CI

F2)

^1,

(1.3)

ifF UF Xthen either

FI(FI)

^or

H(F2)

^1.

(1.4)

The last condition can becomparedwiththecorrespondingconditionfor measures:

if

H(F

LI

F2)

^theneither

l’l(Fl)

^or

H(F2)

^1.

DEFINITION3.A space

X

has thepre-measure propertyifandonlyifeverypre-measureonXisameasure.

DEFINITION4.A space Xishereditarily normal if andonlyif every subsetYofXisnormal.

Wecanreplacethis withtheequivalent butmoreuseful definition:

DEFINITION4A.A spaceishereditarily normal if and onlyifforevery subsetYofXand every pairofclosed sets

F

^and

^F

the condition

F

^{NF tY O impliesthat thereexistclosedsets}

^F

^3andF4such thatF N

F

^N

^Y

F₄0F NY O^and

Y

C_

F

UF_4.

DEFINITION 5. A regularmeasure/is ameasureonthealgebra generated bythe lattice of closedsetssuch that

/z(M)

^isthesupremumof the measuresofallclosedsetsinside

M,

A regularmeasureisalsoamaximal measure

(i.e.,

thereisno measurestrictly greater thanit). For^atwo-valuedmeasure thesetsofmeasureoneform aclosedultrafiiter.

DEFINITION6.A space

X

has themeasure-treeproperty if andonlyif for all two-valued measures/,/l, and /2 onthealgebra generated bythe latticeofclosedsetstheconditions It -</1

and/

_</2 imply

that/2

-</or/1

-</2" We^notethatthepartial order ^_<is thesame asthatdefinedin Definition 1.

Theterm"measure-tree

property"

requiresexplaining. Inanormalspacethesetofmeasures lessthanagiven measure

(see

Theorem

1)

formapartially orderedsetin whichmeasuresbranch outward from theregularmeasure.

If thespace^doesnothave themeasure-treeproperty those measures also branch inward. If thespacehasthe measure- treeproperty themeasures form atreeinwhich thebranches donotreunite. Thisisequivalenttosayingthat all measuresgreater thanagivenmeasure form a chain.

We

alsonotethatthispropertycloselyresemblesanaxiomsuggested bythe noted mathematician LewisCarrol

[2]

^(Sylvieand

Bruno,

Chapter18,p.

425):

^Thingsgreater than the^sameare greater thanoneanother.

2. MAIN RESULTS

Thefollowing

(see

Eid

[4]

andFrolik

[5])

^{is a}measure-theoretic characterization of normalspaces:

THEOREM2.1.A space

X

isnormal if andonlyif for every measure#andeverypair ofregularmeasures1^and

/z_2,if/z

>-

t and t2>_t then t2.

PROOF. LetusassumeXisnormal.Letusconsidermeasure/andregularmeasures/z and #2-

Let/1

^/tand

/2^->/t.

Let/

;/2^{and letusderive a}contradiction.

If/1

;e/2 then there is^aset

M

belongingtothealgebra generated bythetopology^such

that/zl(M 1,/z2(M 0,/l(M’)

^0,

and/z(M’)

^1.

Since/ and/z

areregular measuresthereexistclosedsets

FI

^{andF such}

that/z(F) =/z2(F2) 1,/1(F2) =/z2(Fl)

^0,

F

^_CM,

^F

^{2 M’,and}

Fl ^F

2 O. Wetherefore

have/(F)

^_<

#2(F)

⁰and

t(F2) _</(F2)

0.Since

X,

^isnormal there existopensets G andG₂such that

G

^_C

^F,

^G2C

F2,

^{andG fflG} 0.Lettheircomplementsbe

F G’

^andF4

G2’.

^Since

#1(G1) _>/I(F) and/2(G2) _>/z2(F2)

^wehave

#1(F3) #2(F4)

^0.

Therefore/(F) =/(F2)

^0.

^{Let F}

F

^{OF and}

^F

^{6 F2OF4.}

Since/z(Fs) =/z(F

^LI

F3)

^0,

#(Fs’

^1.Similarly/z(F6’ 1.SinceF

g

F and

F

t::F₆wehaveF

5’

^_CF

3’

^andF

6’

^_C

F4’.

^SinceF

3’

^tF

4’

^Owehave

F’

^ClF

^6’

0. This means

that/(F s’

^t3

F6’

=/z(Fs’ + (F6’

^{2, whichis}impossible.Thenecessarycontradiction has been achieved.

Fortheother direction letXbenotnormal and letus construct ameasure_#andregularmeasures_# and2^such

that/zl

^;₂

but/z

^>_/and/z _> #.SinceXisnotnormal thereare twodisjointclosedsets

F

^{andF whichcannot}

beseparatedbyopensets. Thecollectionof sets{Gisopen

[F

^_CGorF₂C__ G} therefore has thefiniteintersection property. Theopen^filterthatitgenerates can^beextendedtoanopenultrafilter which inducesameasure such that

#(G)

for G in the collection.Let F₀beanarbitraryclosedsetwithmeasureone;

#(Fo’

0 which means

F o’

^is

notinthecollection. ThismeansthatF

o’ F

^andF

o’ ^g

^F2or in other words

F

^F0;@andF tF₀^; O.^Thismeansthat the collectionof sets{Fisclosed

[u(F)

^1}O

{F

^hasthe finite intersection property. Thefilter thatitgeneratescanbeextendedtoaclosedultrafilter.Theregularmeasure basedon it willbe

called/t;/

_</t^and

#(F)

^1.Usingsimilarreasoningwe can seethereis aregularmeasure/2 such that# _<#2^and

#2(F2)

^1.Since F F₂

O,/1

/2^whichmeansthe requiredmeasureshave been constructed.

THEOREM2.2.IfaspaceXhas thepre-measureproperty thenithas themeasure-treeproperty.

PROOF. LetXhavethepre-measurepropertyand let_#,#1, and#2be measures onXsuch that:

and_# #2.

(2.1)

Let FI

min(ul, #2)-

^{Itiseasytoseethat}# _< FI.

The functionFIisalso apre-measure. Conditions

(1.1)-(13)

in Definition 2are obviously fulfilled.Toprove condition

(1.4),

welet

F1

^{UF X.Because}

^#(X)

^{wealso have}

^#(F

^t3

^F2)

^1.The function isalsoa measure and therefore

#(F)

^or

#(F2)

^{1. Since}# _<FI,

FI(FI)

^or

FI(F2)

^1.Therefore condition

(1.4)

^isfulfilled.

The functionFIisthusapre-measure.SinceXhas thepre-measureproperty, 1-I^{is a}measure.Thisispossible only^if either

2 </l^or# ^_<#2"

(2.2)

(If

^{thisweren’t}the case then there would beclosedsetsF ^andF such

that/I(FI) #2(F2) and/z(F2) #2(FI)

0. This meansthat

#(F

^U

F2) #2(F

^O

F2)

^{1. SinceH}

min(gp #2),

^wehave

H(F

^O

F2)

^but

H(F) H(F2)

^{0. This is}incompatiblewithH beinga

measure.)

Wehave beenabletoderive

(2.2)

^from

(2.1).

^ThereforeX^{has the}measure-tree

property.,

THEOREM2.3. Ifspace Xishereditarilynormal thenithas thepre-measureproperty.

PROOF. LetHbe apre-measureonX.In^orderto show that itis ameasure we let

F

^andF2beclosedsetssuch that

H(F

U

F2)

^{1.LetY X}

(F!

^fl

F2).

^{Thismeansthat}

F!

^CI

^F

^{CIY 9.SinceXis}hereditarilynormalthis impliesthat there exist closedsetsF andF₄such that

F

CI

F

f’l

Y

F₄f’l

F

₂f’lY gDandYC

F 30 ^F

^{4.Let F and}

F₆beclosedsetssuchthatF F

30 ^(F

^CI

^F2)

^{andF F}

40 ^(F

^N

F2).

^{F UF}6

X

and sinceFIis apre-measure

FI(Fs)

^or

FI(F6)

^1.Without loss ofgeneralitylet

I’I(Fs)

^{1.Let F F N}

(F 10 F2).

Letus nowprove

F

₂ F_7.Wecandothispoint by point.

First, let pointx//F

2.Eitherx F or x F

1’

^{Ifx F thenx F fF whichmeans x/F andthusx}

F7.If,ontheother hand,x F

1’

^{thenx F F}1.SinceF isdisjointfromF F1,x F

4’.

^{Sincex YandF}

30

F₄ Y,x F3.Thereforex F and thusx F_7.

Second, letxF

7.F

⁷

^{(F fF2)}

^LI

(F F1).EitherxF

^CIF

2orxF

^CIF

1.1fxF

^F

2thenx/

F_2.Onthe other hand ifx F f’lF thenx

F

_5.If x F then eitherx

(F

f

F2)

^{orx F NF or x F}3.If x F F thenx F_2.Nowexamine x F_3.Ifx F then, sinceF isdisjointfromF Y,x

(F Y)’.

Since F ^C_

y,

x Y.Thereforex F

1’.

^{Onthe other}hand, sincex

//FT,

^{x/F U}

^F

2.This meansthatx

F

_2.

Wehave thusshownthatF F and thereforeF F CF UF_2.Since

FI(Fs)

^and

FI(F

U

F2)

therefore

H(F2)

^1.

Similarlyif

FI(F6)

^then

rI(Ft)

^1.Inotherwords, either

FI(Ft)

^or

FI(F2)

^1.

Wehave shown that if

1-I(F

U

F2)

^theneither

rI(Fl)

^or

FI(F2)

^{1.Therefore His a}measure. This appliestoallpre-measures whichmeansthatXhasthepre-measure

property.I

LEMMA

2.1.If/is a measure onspace X,

Y

is asubspaceofX,andZis asetinthe algebragenerated bythe topologysuch thatZ

_c

y and

#(Z)

then thesetfunction/ydefined

as/y(F Y)

g(F)for all closedsetsFis well-defined

and/y

^{is ameasure on}X.

PROOF. Let FbeaclosedsetinY. Let

#y(F)

be evaluated intwodifferent ways:

#y(F) #(El)

^where

Fl

^{isclosedandF Y F,}

(2.3)

lq/(F) #(F2)

^whereF isclosed and

F

^{C’lY F.}

^(2.4)

SinceZ Y^andF ^NY F F f’lYweobtainF tZ F ^CIZ.Since

u(Z)

^wealso obtain

u(F1

^N

Z) u(FI)

^and

u(F

₂N

Z) u(F2).

^{This meansthat}

u(F1) u(F2)

^sothat Uyisthesame nomatterhowit isevaluated.

Mostofthepropertieswhichdetermine if a setfunctionis ameasure areinherited

byuy.

^Theremainingproperty iswhether

UY)

^{1. Since}Y X Y,

uy(Y) u(X)

whichimpliesthat Uy is a

measure..

THEOREM2.4.Ifaspace Xhas themeasure-treeproperty thenit ishereditarilynormal.

PROOF. Itis easiesttoshowthisusing the contrapositive.

LetX^benothereditarilynormal. Theremustexist asubset YofXsuch thatYisnotnormal.We musthavea measureU^onY^{and a}pair^ofregularmeasuresUl andU2^onYsuch that U

-<

_Uland U

-<

‘//2butUl U2. We^must thereforehavesetsM ^andM inthealgebragenerated bytheinheritedtopologyon

Y

such that:

U(MI) U2(M2)

^and

U2(M1) uI(M2)

^0.

(2.5)

Sincetheseareregularmeasure wemusthaveclosedsets

F

and

F

in

Y

such that:

uI(F1) u2(F2)

^and

u2(F1) Ul(F2)

^O.

(2.6)

Let_{U3, U4,}andU5^betwo-valued functions ofsetsin thealgebra generated bythetopologyon

X

such that for all Minthealgebra:

u3(M) u(M n _r), u4(M) u(M ⁿ Y),

^and

us(M) u6(M n Y). (2.7)

Theseareobviouslymeasures.Theyalso satisfy the

conditionsu3 ^-<

‘//4andU3

-<

_‘//5-IfX had themeasure-treeproperty thenthey would satisfyeither ‘//4

-<

‘//5 or ‘//5

-<

‘//4- SinceF_1,F ^C_Yitfollows that

F

^{F YandF F flY}

whereF andF₄areclosed.Wetherefore have thefollowing

u4(F3) Ul(F3 Y) Ul(F1)

1,

(2.s)

u5(F4) u2(F4

^f’l

Y) u2(F2)

^1,

(2.9)

u.(F.) ,(F. r) ,(F)

0,

(2.10)

u5(F3) u2(F3

^N

Y) u2(F1)

^0.

(2.11)

The conclusion

u4(F3)

^_<

u5(F3)

^{isnot truefor}F and

us(F4)

^_<

u4(F4)

^isnottrueforF_4.The spaceXtherefore does nothave themeasure-treeproperty.

Wehave thusprovedthecontrapositiveand thus theoriginaltheorem that ifX has themeasure-treeproperty then it ishereditarilynormal.1

THEOREM2.5.(Summingup.)Thefollowing conceptsareequivalent:

a)

hereditary normality;b)the measure- tree

proper,

and

c)

^thepre-measureproperty.

The abovereasoningcan also beappliedtoother lattices thantopologies. For example,lattices ofzero sets are hereditarilynormal.

(This

iseasytosee.Let

Z[

^and

Zg

^{bezero sets}of functions

f

^{and g,}respectively.The functions remain continuouson the subspace. The function

(If[ Igl)/(Ifl

^/

Igl)

is also continuous because

(If[ + [gD

is nonzeroif

z/.

^and

Zg

^aredisjoint in thesubspace.The cozerosetsof the function’spositiveand negative portions producethenecessarycozero

sets.)

The theorems inthispapercanbe usedtoprovethe well-known fact thatpre- measures onsuch latticesare measures.Similarly,Booleanalgebras^areobviously hereditarily normal;theyalso have

the

pre-measure

property.

Itiseasy^toseethatthe lattice ofopensetsisanormallattice if andonlyif thespaceisextremallydisconnected (i.e.,^{in an}extremally-disconnected space anytwodisjointopensetscan beseparated byclosed

sets).

^Itisalso easyto see that

if/1

and/z ^aremeasures on aspace

then/Zl(G </2(G)

^{for allopensetsGifandonly}

if/Zl(F >/2(F)

^for

allclosedsetsF.Ifweput these factstogetherwecaneasilyderivethefollowingtheorems:

THEOREM2.6.Aspace Xisextremallydisconnectedifandonlyif foreverymeasure/zandevery pairof minimal measures

(a

^measureis a minimalmeasure if andonlyifthereis no nontrivial measurestrictlyless than it forevery closedset)/z and #2,

if/zl

^</ and/z </

then/

=/z 2.

THEOREM2.7.A spaceXishereditarilyextremally disconnectedif andonlyif foralltwo-valuedmeasures/z,/z1,

and/z onXtheconditions/t >/t and/t ^>2^{imply thateither}lt2>/lor/ >/2"

These theorems canbeusedtoprove:

THEOREM2.8.Anormalextremally-disconnected space ishereditarily normal if andonly if it ishereditarily extremallydisconnected.

PROOF. LetXbe normaland hereditarilyextremallydisconnected andletusproveitishereditarily normal.Let /Ul,/t2,and/.t ^{bethreemeasures}suchthat/z _</t and/.t ^_<_3"Let/Z ^{be a}regularmeasuregreaterthan/t2.and let lt5^{be a}regularmeasuregreaterthatt3.Both_it and/a ^aregreater

than/zl

and therefore, sincethe spaceisnormal, /4 It5"^{This meansthatit _<}t ^andI3

-<

lt4" The space^ishereditarilyextremallydisconnectedandtherefore12^and

/z arecomparable (i.e.,^eithert ^_<It ^orIt ^_<

/t2).

^Wehave shown thatanytwomeasuresgreaterthan the same measure arecomparable. The spaceisthereforehereditarily^normal.

Wecan use similarreasoningtoshow thatifaspaceisextremally disconnected and hereditarilynormal then it is hereditarilyextremallydisconnected.1

If we examine the lattice of zerosetsinaTychonoff space,wecan seethat since it ishereditarily normal,if the lattice ofcozerosetsis normal(i.e.,^{if the}spaceis an

F-space

seeGillman andJerison

[6],

Chapter14)^thenit is hereditarilynormal.Wecan alsoeasily provethe well-known results thataz-ideal in an

F-space

belongstoachain fromaminimal idealtoamaximal ideal.

The obvious conclusion thepropertyofbeinganF-spaceishereditary isfalse.Dow

[3]

constructedanF-space with anopen subspace whichisnot an

F-space.

This can be reconciled with the fact that the lattice of cozerosetsin an

F-space

ishereditarilynormal. The members of the normaldescendantlattice on thesubspace consistof those cozerosetsinheritedfrom theoriginal space.However,noteverycozerosetinthesubspacehas been inherited from the original space.The cozerosetsofasubspacecaninclude additionalsetsiftherearemorecontinuous functions on thesubspacethan on theoriginal space.

Atthispointwehavetwoquestions:

1)

Are there anyspaceswhicharenormal andextremallydisconnected but which are not hereditarily extremally disconnected?

2)

Are there any spaceswhich are hereditarily normal and extremallydisconnected but whichare notdiscrete?

The answertothe firstquestionisyes.TheStone-(ech compactification^{of the} integers(fiN, Example 111in SteenandSeebach

[7])

^istheStone spaceof acomplete algebra. Itistherefore normal andextremally disconnected.

Thepowersetof theintegersmodulo the ideal of finitesetsisincompletesothereitsStone spaceisnotextremally disconnected eventhoughtit is asubspaceof

fiN.

^{This meansthat}

fin

^isnothereditarilyextremallydisconnected even thoughit isnormal andextremallydisconnected.

The answertothe second questionisalsoyes. Ifweconsider thesingle ultrafilter(Example 114 inSteenand Seebach

[7])

wecan seethatthisspacemustbeextremallydisconnectedbecause if there aretwodsjointopensets, atleastoneofthem doesnotbelongtotheparticularultrafilterand itmusttherefore be closed. Thetwo setscan therefore beseparated byclopensets.Thespacemustalso behereditarilynormal.Itiscountable and thushereditarily Lindel0f.Aregularhereditarily-Lindelofspaceishereditarily normal.

REFERENCES

1.CAMACHOJr., J. Onmaximalmeasureswithrespecttoa lattice,Int. J.Math. Math. Sci. 14

(1991),

93.

2.CARROL, L.TheCompleteLewis Carrol,VintageBooks,NewYork, 1976.

3.DOW,A.CHandopen subspacesof

F-spaces, Proc.

Amer. Math.Soc.,89

(1983),

341.

4.EID G. Onnormal lattices and Wallmanspaces, Int. J.Math.Math.Sci.,13

(1990),

31.

5.FROLIK, Z.Prime filters withtheC.I.P., Comm.Math. Univ. Carolinae, 13

(1972),

553.

6.GILLMAN L.andJERISON M.Rings of Continuous Functions VanNostrand,NewYork, 1960.

7.STEENL.andSEEBACHJr,J.A.CounterexamplesinTopology,Springer-Verlag,New York-Heidelberg-Berlin, 1978.

8.SZETO M. Measure repleteness^andmapping preservations, J.IndianMath.Soc.,43

(1979),

^35.