ON NORMAL AND STRONGLY NORMAL LATTICES

In this paper we will investigate the properties of normality and strong normality of lattices and their relationships to zero-one measures. We will eventually establish necessary and sufficient conditions for lattices to be strongly normal. These properties are then investigated in the case of separated lattices.

3) Hausdorff or T: if z, y X; z # y then IA, B such that z A', B" and 4) Disjunctive if for z Xand L " where z L, =IA such that z A and A

5) is normal if for A, B
where A n B I, =, such that A C ", B C " and 6) /-is compact if any covering of X by " sets has a finite subcovering.

7)
is countably compact if any countable covering of X by/." sets has a finite subcovering.8) /-is Lindel6f if any covering of x by/-" sets has a countable subcovering t() the algebra generated by .() the c-algebra generated by .() the lattice of countable intersections of sets of .r() the lattice of arbitrary intersections of sets of .MEASURE TERMINOLOGY Let be a lattice of subsets of X. M() will denote the set of finite valued bounded finitely additive measures on A().Clearly since any measure in M() can be written as a difference of two non-negative measures there is no loss of generality in assuming that the measures are non-negative, and we will assume so throughout this paper.We will say that a measure # of M() is regular if for any A A()tt(A) sup t,(L).MR(C) represents the set of/.-regular measures of M().LCA L. DEFINITION 2.2. 1) A measure/ e M() is said to be a-smooth on/., if for L n e and L,ltl; then #(L,)-.0.
If/-is a lattice of subsets of X, then we wiI1 denote by: M() the set of c-smooth measures on/. of M(/.) Me(/-) the set of -smooth measures on 4(/-) of M() M(/-) the set of/.-regular measures of M(/.) { 1 0 if z A DEFINITION 2.3.If A t(/.) and if z 6 X then ,x(a)= if t A at z. I(/.) is the subset of M(/.) which consist of non-trivial zero-one valued measures.is the measure concentrated In(/. the set of/.-regular measures of I(/.) Is(/. the set of C-smooth measures on/-of I(/.) IS(/.) the set of C-smooth measures on I(/-) the set of/.-regular measures on IS(/.)DEFINITION 2.4.If , 6 M(/-) then we define the support of g to be: DEFINITION 2.5.We say that the lattice/, is: Replete if S(,) # for any , I().
We now list a few well known facts which will enable us to characterize some previously defined properties in a measure theoretic fashion.The lattice is: 1) Disjunctive if and only if ux 6-In(i,), Yz 6-x.
2) T if and only if $(,) or a singleton for any 3) Compact if and only if 8(#) for any p 6-I 4) Countably compact if and only if In(l. 5) LindelSf if and only if $(p) # for any , 6- 6) Normal if and only if for any 6-I(i,) there exists FILTER AND MEASURE RELATIONSHIPS Let be a lattice of subsets of X. DEFINITION 2. 7 We say that q c I, is an I,-filter if: (3) If L1 L and L 6-L 26-/ DEFINITION 2.8. is said to be a prime I,-filter if: (1) is an I,-filter, and (2) If L,L 6-and L 0 L 6-=L 6_ or L 26- DEFINITION 2.9.If /is an I,-filter we say that is an I,-ultrafilter if is a maximal I,-filter.(1) If p 6-I(i,), then 4 o is an I,-prime filter and conversely any I,-prime filter determines an element # 6-l(i,) and the correspondence is a bijection.
(2) If # 6-In(i,), then 5, is an I,-ultrafilter and conversely any I,-ultrafilter determines an element v 6-In(i,) this correspondence is also bijection.,i s an I,-ultrafilter if and only if # 6-In(i,).

SEPARATION OF LATTICES
We are going to state a few known facts about the separation of lattices.We will use these results later on in the paper.DEFINITION 2.11.Let I, and/-2 be two lattices of subsets of x.We say that I, separates/-2 if A2,B 6-1, and A2B = then there exists A,B 6-I, such that A C A,B C B and A fB =.
PROPOSITION 2.12.Let I, be a lattice of subset of x.I, is compact if and only if ri, is compact, in which case I, separates PROPOSITION 2.13.I, Linde15f if and only if ri, is LindelSf and in this case if I, is also then I, separates r.
The proofs for these propositions are esy and will be omitted.
THEOREM 2.14.Suppose I, c 2 and 1,1 separates 2 then I, is normal if and only if 2 is norn'al. PROOF.
(1.) Suppose that I, is normal and let A2, B 6-1,2; A2fB2 }" Since I, separates 1,2 then there exist A, B 6-L such that A C A, B2 C A and A O B =}. Now since L] is normal there exist A, B 6-1,1 C 1-such that A C A', B C B" and A'n B" and A' B'= i.e., L is normal.
(2.) Suppose that 2 is normal.Let /1 E 1(1) and assume that there exist two measures vl,r .la(1) and v, r on L. Let , v d r the respective extensions of the previous meures.Note that later two extensions e unique d belong to l(L2).Furthermore it c seen since L sepates L that S v2 d S r on t.However, since L2 is normM then va ra therefore v r d thence L is normM.

THE WALLMAN SPACE
If L is a disjunctive lattice of subsets of abstract t X then there is a WMIm spe siated with it.We will briefly review the fundentM prorties of this WMlm space.For y A in (L), define W(A) to be W(A)= { I(L): W(L) is a compact lattice, d the topologicM sp&ce I(L) with closed sets W(L) is a compact T spe cMled the WMlm space siated with X d L. Since L is disjunctive, it will be T if d only if L is normM.In addition to eh v q M(L) there corrpondence a unique M(W(L)), where (W(A)) u(A) for A (L) d conversely.Also, v M(L) if d only if q M(W(L)).Since W(L) is compact is rW(L), d W(L) sepates W(L) (s Proposition 2.11).Furthermore M(W(L)) h a uque extension to G M(rW(L)).
Next we consider the space I(L) d its topolo.DEFINITION 2.15.Let L be a disjunctive lattice of subsets of X,L L d A (L).
The following properties hold d e not dict to prove.
PROPOSITION 2.16.Let L be a disjunctive lattice then for A,B () For eh v M(L) there corresnds a unique v' M(W,()), where V'(W(A)) v(A) for A (L) d conversely.
v Ms(L It c be sho that the lattice W,(L) is replete d hat I(L) with W(L) the tolo of closed sets is disjunctive d T. It will be T if we further sume that property (P1) is satisfied; where (P1) is defined follows: (P1)" For each t, I(L) there exists at most one v I() such that/ < v on L. A proof of the last statement can be found in [8].
3. NORMAL AND STRONGLY NORMAL LATTICES PROPOSITION 3.1.L is normal if and only if for each L L where L C Li O L and L1,L L; then there exists AI, A .sucht hat A1 C L and A C L and L A 13 A2. PROOF.
(1.) Assume that is normal and that L CL'OL then L rLlt3L2= or equivalently (Lf3L)3(LnL)=$.Since L is normal there exist ft, 2 such that LnL C [4,LL2C [4' and r t2 .Let A1 L f3 1 and A L f'l 12. Clearly A C L and A C L.  (2.) Let L C L and L,Lx L then X L L and by the condition there exist AI,A L such that A C L, A C L and A 10 A X, clearly L C Ai, L C A and A Cl A t} and thence is normal.
DEFINITION 3.2.Let ,r: L-{0,1}; r will be called a premeasure on L if t(X) 1, is monotonic and multiplicative i.e., r(L3L2)=r(L).r(L2)for LI,L2 L. II(L) denotes all such premeasures defined on .It can be shown that there is a one-to-one correspondence between elements of II(L) and L-filters.DEFINITION 3.3.Let I(L) {t II(L): if L10 L X then =(L) or r(L2) 1} Clearly, lit(L) C I() C (L) C II() Let *J" {L L:LA # for all A L such that t(A) 1,r i(L)} THEOREM 3.4.L is normal if and only if is an -ultrafilter.PROOF.
(1.) Assume that is normal we have to show that: (a We have to show that LL2rA#$ for all AL such that t(A)= 1. Assume otherwise i.e., LnLA= for some AL and =(A)=I where =I(L) then (LICA)N(LA)=$.Since L is normal there exist A,A2 such that LfA C Ai, L2rA A' and AI A=.
Clearly A O A X=vt(A 113 A2) l=>r(A1) or r(A2) 1. Say r(A) 1. then =(A t3 A1) and L A t3 A $ which is a contradiction since L (d) Now assume that c } where 1 is an L-ultrafilter.Assume their 6xists L } but L , hence there exists A L such that =(A)= but At3 L I}. However since '(A)= then L t3 A # $ for all L {J D ty D {A L:r(A) 1} which is a contradiction.Therefore is an L-ultrafilter.
(2.) Now assume that 3" is an L-ultrafilter we have to show that L is normal i.e., if p I() there exists a unique vlit(L) such that tt<v on L. Suppose there exist v,v lit(L) and v<vl<v2 on L. Let $,= {L L:p(L)= 1} and "J'v= {L L:Lf3A #for all A such that t,(A)= 1}.u and are ultrafilters and we have _< v u c "=Yu c "Y=u "Y, for 1,2.Therefore .Furthermore we have that q c i and hence v ql " Finally since all the ultrafilters are equal we get that v v which proves that t is normal.
Let v I(L).Define for any E C X, (E) in] /(L').Then it is easily seen that is a finitely EcL subadditive outer measure.L PROPOSIT!ON 3.5.is normal if and only if ; {L L: (L)= 1} is a prime -fi]ter.
PROOF.Suppose L is normal.If L1,L then (L) (L2) 1.Now if p (L f'l L2) 0 then there exists A t such that L n L 2 C A" and (A') 0. But then L n L2 C A', and since t is normal, by Proposition 3.1, we have A A] U A where A],A ,A] C L, d A C L[. Now p(A)= then p(A) or p(A)= 1. Say p(A), then p(A)= 0 which is a contriction since ] c A d y (L)= 1.
Thus , L implies L L .
The rt of the prf is cle.THEOREM 3.6.t H() then: x i() if d only if there exists I() such that PROOF.Supse =i() d let = {L'':=(L)=O}.d h the finite intersection prorty.The interaction of elements of form :filter be.Now sume that c and is :ultrlter.Then p I(').If =(L)=0 then L' C p(L')= lp(L)= 0 hence p S on d therefore I() such that p S The second pt of the prf is ey d shl omitted.
Let c , X d if L, L then L L .Consider the set of l t-filters that exclude , (i.e., =).We ptiMly order by set inclusion.Since {X) is t-filter that exclude , then there exists at let one .Furthermore, sin {, is a ptiM ordeng, which is inductive ordering then by rn's le there must exist a mimM element.Suppose that F0 such that AFo= then Fo(AB)(FoA)U(FoB) F 0 B B which is a contradiction thus we may now sume that A for 1 F .DEFINITION 3.9.We say that is strongly norm then p S p or a2 S p on t.THEOREM 3.10.t is strongly norm if d only if I(t)= i(t) If 6-I(i,) let , {L 6-I,:#(L)= 1}.PROPOSITION 2.10. Now

t
the filter generated by M1 {A F F }. Since A d A C , similly let the filter generated by MI {BFFO},OC.So there ests H6,H6 such that A F C H for some F d similly there ests H , H such that A F C H for some FO.t FF2=F3 then HIOHD(AF1)O(BF)D(AF3)U(BF3)HIOH2D(AOB) F however since H H2 , it is a contriction.Thus A e or B or equivently is a prime filter; d I(L).COROLLARY 3.8.Let  e H(t)then / PROOF.be e Z-filet epresening c mim -filter ntning d excluding filter d A (, wch is a contradiction; since A longs to Therefore , d hence = H(t).Thus e z(t)