^ V Infinite ^-decompositions in upper continuous lattices

ROCZNIK.I POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXIX (1990)

A

W

(Warszawa)

Infinite ^-decompositions in upper continuous lattices

Abstract. The present paper is a continuation of the author’s paper [3]. Here we shall study infinite 0-decompositions of the unit element of an upper continuous lattice. In this paper we give a generalization of Theorem 3 from [1].

1. Basic notions. Let L be a complete lattice. Lattice join, meet, inclusion and proper inclusion are denoted respectively by the symbols v , л , < and < . Let 0 be the least element of L, and 1 the greatest element of L. Finally, let K (L) denote the set of all compact elements of L, i.e. c e K (L ) iff, for all T ç L ,

c ^ \ / T imP^es c ^ V ^{T '} for some finite subset T ' of T.

An element u e L is called join-irreducible iff, for all x, y e L, и = x v у implies x = и or у = и. An element и e L is called completely join-irreducible iff, for all T e L , и = \J T implies ueT .

A subset T of the lattice L is said to be redundant iff \J T= \J (T—t)(*) for some te T , otherwise it is irredundant. If a is an element of L and T is a subset of L, we say that a is an irredundant join of T, and we write a = \ / T, if T is irredundant and a = \ f T. If a = \J T and for each t e T , t л \ f (T—t) = 0, then we say that a is a direct join of T, and we write a = \ / T. .

We denote by 0 (L) the set of all congruence relations in the lattice L. Let by 0 e 0 (L ). A subset T of L is said to be О-independent if T is irredundant and for each te T ,

u \ / ( T - t ) = 0 (0).

If a is an element of L and T — {tm: m e M } is a subset of L, then we say that a is a 0-join of T, and we write

a = Y;T or a = £ ( t ra: m eM )

if T is 0-independent and a = \J T. The 0-join of finitely many elements

0 ) If y e X , then X — {y} is also written X —y.

tt , ..., tn is also written t x + . . . + t n. A representation of an element as a 0-join of elements of the lattice L is said to be a в-decomposition of the element.

A nonzero element a of L that cannot be represented as a 0-join of two elements of L, is said to be в-indecomposable. An element b is called a в-summand of a if a — b + x for some element x. We denote by S (L) the set of all 0-summands of the unit element of L.

Let со, i be the congruence relations in L defined by x = у (со) iff x = у,

x = у (i) for all x and y.

Observe that irredundant joins and direct joins are_ special cases of 0-joins.

Indeed, if T ç L — {0}, then a is the i-join of T iff a = \ / T, and a is the w-join of T iff a = \ / T. Furthermore, an element x e L —{0} is j-indecomposable iff it is join-irreducible.

Now, we shall prove the following simple but useful lemma.

L emma 1. Let L be a complete modular lattice. Let 1 = a + b, a = Y j T.

I f the set Tcj{b} is irredundant, then 1 = £ T+b.

P ro o f. Let t be an element of T. Now compute:

t

(b v \ / (T— t)) — t

a

(b v \ / (T— t))

(observe \J (T—t) ^ a and apply modularity)

= t

[(а л b ) v \ f (T—t)]

(observe a л b = 0 ( 0 ))

= t л V ( T - t ) = 0 ( 0 ).

Thus the set T u {b } is 0-independent. Therefore, 1 = ^ T + b . ■ 2. The property (R). Let L be a complete lattice and let 0e<9(L).

D efinition 1. A 0-decomposition a = £ T of an element a e L is said to have the property (R) with respect to a 0-decomposition a — £ T of a iff, for each t e T there is a t' e V such that

a = t'+ Yi ( T - t ) = t+ Y 1( T ' - f ) .

D efinition 2. We say that a 0-decomposition a — ]Г T of a has the property (R) with respect to a 0-decomposition a = £ T if it has property (R) with respect to every 0-decomposition of a of the form a = £ S, where S is a subset of T u T .

Now we shall prove two lemmas.

L emma 2. Let the unit element of the lattice L have two в-decompositions

(1) 1 = £ ( а (: ie /),

(2) 1

I f the 9-decomposition (2) has property (R) with respect to (1), then for every finite subset f = {/j, . . . , j k} £ J there exists a finite subset Г = {il5 ik} ^ I such that

(3) 1 = ain + Z (^ -:

= bjn + bjn- i + ••• + ^ ji + Z ( ai: ie/ —{ il5 in}) for all n = 1, 2 ,..., k.

P ro o f. We prove Lemma 2 by induction on the number of elements in J'.

If |J'j = Ц 1), then this lemma follows from Definition 1. Let us assume this statement for every (k — l)-element subset of J and let J' = { f , ..., j k}. By the induction hypothesis for the subset {jx, . . . , j k- k} °f J' there exists {ix, ..., ik_i} с I such that (3) holds for each n = 1, 2, ..., к — 1. In particular, (4) 1 = bJk_l + ... +bj l +YJ(ai: iE l — {i1, . . . , i k_ 1}).

By Definition 2, the ^-decomposition (2) has property (R) with respect to (4).

Then, for j k there is an index ike l — {i1, ..., ik- x} such that (3) holds for n = k.

Thus, for the subset J' Я J there exists /' = {i1} ..., ik} Ç I such that (3) holds for all n = 1, 2 ,. . . , k, and the proof of Lemma 2 is complete. ■

L emma 3. I f the unit element of the lattice L has two 9-decompositions: (2) and

(5) 1 =£(<!,: ie n + 'Z fr j- .je S )

such that J' is a proper subset of J, at is compact for each i e Г, and (2) has the property (R) with respect to 9-decomposition (5), then there are two countable or finite (with an equal number of elements) subsets I0 = {ilt ..., in, ...} £ Г and

Jo = {/i > • • •, Ли • • •} — J" — J ~ J' sucb that (6) i = « ,„ + ! ( b / . j t j j

= bh +■ ■ ■ + bi, + E (V J e J ">+ 1 (ai: ' e - f a ...'»}) for all n = l , 2 , . . . , and

(7) V (at: i e I 0) < V (bf J e J ' ^ J o)-

P ro o f. Let j x be an arbitrary element of J". By Definition 1, since the 0-decomposition (2) has property (R) with respect to (5), we conclude that for the index j x there is an ixe l' such that

(2) For each set X, |Z | denotes the cardinality of X.

1 = aii+ H (bj : J E j ~ j i)>

(8) 1 = Ьь + ^ ( Ь у . y eJ') + Z ( ai: i e l ' - h ) -

The element ah is compact and hence there is a finite subset

{/2 > • • • > Л } ^ J " ~ У i } ’s u c h t h a t

ah < V(*V 7 ' e i ' u J t), where Л = {/'i Л.

By assumption, the ^-decomposition (2) has property (R) with respect to (5), and, therefore, it has also this property with respect to ( 8 ). Then applying Lemma 2 to the ^-decompositions (2) and ( 8 ), we conclude that for the elements bj2, ..., bjk there exist distinct indices i2, ..., ike f — {it } such that ( 6 ) holds for each n = 2, . . . ,k. In particular,

(9) 1 = bjk + ... + h , J G J ' ) + Z ( ai: ij) .

Again the element ai2v ... v aik is compact, and there exists a finite subset {jk + !>•••', ₇ m} i such that

a i2 v ... v aik ^ V ( bj : 7 ;e J ' v J 2), where J 2 = Л u { ;k + 1, ...,y m}- Now we apply Lemma 2 to the two 0-decompositions (2) and (9), and to the elements bJk+i, ..., bjm. As before, we get the existence of distinct elements ik+1, ime l r —{ i ’ i , ..., ifc} such that ( 6 ) holds for each n = / c + l , ..., m. By continuing this process, we obtain two subsets 7 0 = {il5 ..., in, ...} and J 0 = {Л, ...} such that ( 6 ) and (7) hold. This ends the proof of Lemma 3. ■

3. Upper continuous lattices. A complete lattice L is called upper continuous iff, for every a e L and for every chain C ^ L, а л \J С = \/( a a c : ceC).

We need the following lemma.

L emma 4. Let L be an upper continuous lattice and let T be a subset of K (L).

Let 0 g 0(L), and suppose that 0 has the following property:

(*) for every subset A of L, if a = 0 (0) for each aeA, then \J A = 0 (0).

I f all finite subsets of T are 9-independent, then T is 9-independent.

P ro o f. First we shall prove that the set T is irredundant. Suppose on the contrary that for some t0e T we have t0 ^ \ J ( T —t0). But t0 is compact and hence t0 < \J (T ' —10), where V is a finite subset of T containing t0. Thus T is redundant, contrary to the 0 -independence of T .

Let t be an arbitrary element of T. Now, we will prove that

(10) г л \ / ( М = 0 (&)•

P. Crawley ([2], 2.4) has shown that if a is an element of an upper continuous lattice L, A is a subset of L and 91 is the set of all finite subsets of A, then

a /\ \J A = \J (a a \J A ’: A ' e 91).

Therefore,

(11) t A \ / ( T - t ) = \ / ( t a \ / X : X e X ),

where X is the set of all finite subsets of T— {t}. Since every finite subset of T is

^-independent, the set X и {г}, where X e X is 0-independent. Hence, for every X e X, t a \ f X = 0 (0). Then, by property (*), we conclude that

V (t Л V * : X e X ) = 0 (0).

From this and (11) we obtain (10). Therefore, T is 0-independent. ■ We are now ready to prove the following theorem.

T heorem 1. Let L be an upper continuous lattice, and let 0 be a congruence relation on L having the property (*). Let the unit element of L have two O-decompositions (1) and (2) into compact elements. I f the 0-decomposition (2) has property (R) with respect to (1), then there is a one-to-one mapping f of I onto J such that, for each i e I,

(12) l = a ,.+ £ (fv

P ro o f. The idea of the proof comes from [1]. Let be the set of all ordered triples <M, ^ M, f M) where M Я I, is a well-ordering of M ,fM is a one-to-one mapping of M to J and the following conditions are satisfied:

For each m eM ,

(13) 1 = am + X (by. j ф f M (m))

= E (^ / m ( 0 : IG(ml) + Z (ai: i e / —(m]), where (m] = {ieM : i < Mm}, (14) \/( а у ieM ) ^ \f( b fMiiy ieM ).

901 is nonempty since it contains the triple consisting of the empty set, the empty relation, and the empty mapping (here, we are considering relations and functions as sets of ordered pairs). Define a partial order in 90Î by

<M, ^ м, / м> <M', if either M = M' or M = {ieM': i < M'm}

for some m e M ’, ^ M> restricted to M coincides with ^ M, and the restriction of f M, to M coincides with f M.

Let <MA, ^ А,/ л> (ЛеЛ) be a chain of elements in 90Î. Set iV = U (M A: ЛеЛ), = ЛеЛ), f N = {J (fy ЛеЛ).

It is obvious that <ЛГ, ^ , ^ } е Ш and that <iV, ^ N, f N> is an upper bound of the chain <MA, ^ А,/ я> (ЛеЛ). Therefore, by Zorn’s lemma, SCR contains a maximal element <P, ^ p, f P).

We consider the set

A = {bfpd): ie P } v { a y i e l - P } .

V {ay i e P) ^ V (bfp(i): ieP).

By (14) we have

Hence,

1 = \ / ( a i: ie P ) v \J (at: i e l - P ) ^ \J(bfp{i): ie P ) v \/(щ: i e l - P ) . Then 1 = \ / A. By (13), all finite subsets of A are 0-independent. From Lemma 4 we conclude that A is 0-independent. Therefore, 1 = A, and hence, if we set Г = 7 — P and J' = f P{P), then we obtain the 0-decomposition (5).

Now we will prove that P = /. Suppose on the contrary that P ф /, that is Г ф 0 . Consequently, J' ф J. By assumption, the 0-decomposition (2) has property (R) with respect to (1), and, therefore, it has also this property with respect to (5). Then applying Lemma 3 to the 0-decompositions (2) and (5) we get two subsets I 0 = {ilt ..., in, ...} ç Г and J 0 = {j1, £ J — J' such that ( 6 ) and (7) hold.

Set Q = P u I 0. Define the well-ordering ^ Q of Q by the following rules: if i, ï e P, then i ^ Q Г iff i ^ p i', and for every i e P

i < Qi1 < Qi2 < Q... < Qin < Q...

Define the mapping f Q by f Q (i) = f P (i) for every i e P, and f Q(Ü = j n for и = 1 , 2 , . . .

By ( 6 ) and (7) we obtain that the triple <Q, ^ Q, / Q) belongs to 9JÎ. It is obvious that <Q, < g, / q) is greater than <P, ^ P, / P). This contradiction forces the equality P = I. Then Г = 0 and from (5) we have J' = J. Therefore, f = f P is one-to-one mapping of I onto J such that, for each ie l, 0-decomposition (12) holds, and the proof of Theorem 1 is complete. ■

4. Modular lattices. Preliminary lemmas. Throughout this section L will denote a complete modular lattice. Let 0 be a congruence relation on L. For an element a e L we denote by F (a) the set of all functions cp of L such that a(p = a and from x ^ a, xq> = 0(0) follows x = 0(0). For two elements x, y e L we define y/x = {ueL: x < и ^ y}.

Let a be a 0-summand of the element ceL . An element b is called a в-complement of a in the sublattice c /0 iff

(15) c — a + b.

If an element a e L has a 0-decomposition

(16) a = £ ( a m: m eM )

we define

aij... „ = V (am- m e M - { i , j , ..., n})

for each subset {i, j, ..., n} of M. Denote by am the function of L defined by the

formulaxam = am л (x v am). The maps am, me M are called the в-decomposition

functions related to 0 -decomposition (16); any am is called the в-decomposition

function with respect to the 0 -summand am of 0 -decomposition (16).

Let c be an element of L, and let a be a 0-summand of c. Define the set F (c, a) of maps of L by the rule: <xeF(c, a) iff there exists a 0-complement b of a in c/0 such that ха = а л (x v b) for every x e L (i.e., a is the 0-decomposition function with respect to 0-summand a of 0-decomposition (15)).

D

3. Let a be a 0-summand of c eL . We say that a satisfies the В-condition in c/0 if for every cteF(c, a) and for every 0-decomposition of c with two summands

(17) c = c ,+ c 2,

ау1а е / г(а) or ay2aeF (a), where yx, y 2 are the 0-decomposition functions related to 0-decomposition (17).

D

4. If a belongs to the set S (L), we shall say that a satisfies the В-condition, if for every c e L such that a is a 0-summand of c, a satisfies the Б-condition in c/0.

Now, we shall..prove the following lemma.

L

5. Let the unit element of the lattice L have two O-decompositions:

(18) 1 = a + b,

(19) l = d + e.

I f the elements b and e are comparable, d is O-indecomposable, and a satisfies the В-condition in L, then

1 = d + b — a + e.

P ro o f. Let а, /? and ô, e be the pairs of 0-decomposition functions related to 0-decompositions (18) and (19), respectively. Suppose b ^ e. If e ^ b, then the proof of this lemma is similar. Observe that аса ф F (a).

Indeed, suppose on the contrary that a = aam. Then a = а л (aotB v b) ^ а л e (since b ^ e).

Hence a ^ e, a contradiction. Since a satisfies the В-condition and аеа ф F (а), so а0<xeF(a). Therefore, by Lemma 2 of [3] we obtain l — d + b = a + e. m

We are now going to prove the fundamental lemma.

L

6. Let the unit element of the lattice L have a 0-decomposition (19) such that the element d is O-indecomposable and compact, and let 0-decomposition (1) of 1 be given. Let a{, i e l , and ô, e be the 0-decomposition functions related to (1) and (19), respectively. I f d satisfies the В-condition, then there exists an i0e l such that ôoiiQS e F (d).

P ro o f. We shall consider two cases.

Case 1. Let d ^ aio for some i0 e I. Then

dôacioô = doLioô = [aio л (d v aio)] S (by modularity and d ^ aio)

= [d v (aio л ai0)] Ô = d.

Suppose now that x ^ d and xôaio 3 = 0 (0). By Property II from [3], we have x3aio S = x (0). Consequently, x = 0 (0). Therefore, <5aio 3 e F (d).

Case 2. Let d ^ a{ for each i e /. Since d is compact, there is a finite subset I x ^ I such that d < \J{a{: i e l x). Compute:

У {doit. i e l x) = V ( ai л (d v df): i e l x)

(observe for each i ф Ï and apply modularity)

= \ / ( а*: i e I i) л A ( d v ài: IG /i) ^ d-

If we set c = \ J (daf: i e l j), then we have d < c. Without any loss of generality we can assume that the set {daf: i e / J is irredundant. Thus

( 20 ) c = Je/i).

Since 1 = d v e, so c = c л (d v e) and by modularity we obtain c = d v (с л e).

Observe that the set {d, с л ej is irredundant. Indeed, с ф с л e and also с ф d, since otherwise d is not 0 -indecomposable, a contradiction.

Moreover, d л (с л e) = 0 (0). Therefore

( 2 1 ) c = d + (c л e).

Let /б / 15 and <5', e' be the ^-decomposition functions related to (20) and (21), respectively. Since d satisfies the В-condition, by Definition 4 d satisfies the В-condition in the lattice c/0. Applying Lemma 4 of paper [3] to the two ^-decompositions ( 20 ) and ( 2 1 ) we conclude that there exists an i0e l x such that 3' yiQ 3 'e F (d). Then d3' yio 3’ = d, hence dyio 3' = d. Therefore, d л [dy,o v (с л e)] = d and we obtain the inequality d ^ dyio v (с л e). From this, since dyio < daio, we have d ^ daio v e. Consequently, 1 = d v e

< daio v e, and hence 1 = daio v e = d3ccio v e. Then

(22) d = d3ctio3.

Suppose now that x ^ d and x3aio 3 = 0 (9). Hence, since x<5' yio 3'

^ x<5afo 3 we obtain that x3' yio 3' = 0 (0). Therefore, since x ^ d and

3' yio 3' e F (d), we get x = 0 (0). From this and (22) we conclude that

3a.io3eF(d). Thus the proof of Lemma 6 is completed. ■

L

7. Let two в-decompositions (1) and (19) of 1 be given. Suppose that each a{ and d is в-indecomposable and that d is compact. I f d satisfies the В-condition, then there exists an i0e l such that

1 = at + e = d + £ ( a f: i Ф i0).

P ro o f. Let af, ie l, and S, e be the ^-decomposition functions related to (1) and (19), respectively. By Lemma 6 we conclude that there exists an i0e l such that ôaiodeF(d). We consider two ^-decompositions

1 = d + e = aio + aio.

From Lemma 2 of [3] we have 1 = aio + e = d + aiQ. ■

Now we prove that the set A = {d} u {ay i Ф i0} is irredundant. Assume on the contrary that there exists an il e l — {i0} such that

(23) \ = d + ^ .

Applying Lemma 5 to the two 0-decompositions (23) and 1 = aio + aio we conclude that

Infinite 9-decompositions in upper continuous lattices 321

This means that the set {ay ie l} is not 0-independent, contrary to our assumptions. Therefore, the set A is irredundant and hence by Lemma 1 we obtain

1 = d + J j {ay i ф i0).

This completes the proof of the lemma. ■ Finally, we shall prove the following lemma.

L

8 . A в-decomposition ( 2 ) of the unit element of the'lattice L into в-indecomposable compact elements satisfying the В-condition has the property (R) with respect to any other в-decomposition (1) of 1 into в-indecomposable summands.

P ro o f. Let j 0 be an arbitrary element of J. We consider two 0-decom- positions of 1:1 = bjo + bjo and (1). By Lemma 7, there exists an i0e l such that (24) 1 = ai0 + bj0 = bj0 + Z («i; * * го)•

Observe that the set A = {aio} и {by. j Ф j 0} is irredundant. Assume on the contrary that 1 = aio v bjojl, for some j l E J —{j0}. Then we have two 0-decompositions of 1:

1 = b h + b io = a h + b i o . h ■

From Lemma 5 we get

This means that the set {by j e J ] is not ^-independent and we obtain a contradiction. Thus, the set A is irredundant. Now, by Lemma 1, from (24) we get

1 = ah + Z ( by: J * J o ) = bjo + Z ( ai: 1 * *<>)•

Therefore, by Definition 1, the ^-decomposition (2) has property (R) with respect to ( 1 ).

5. Upper continuous modular lattices. Main results. The following theorem holds.

T

2. Let L be an upper continuous modular lattice and let 0 be an element of <9(L) having the property (*). I f the unit element of L has a в-decomposition ( 2 ) into в-indecomposable compact elements satisfying the В-condition, then for any other O-decomposition (1) into в-indecomposable compact summands we have that there exists a one-to-one mapping f on I onto J such that, for each ie l, в-decomposition (12) holds.

P ro o f. By Lemma 8 we conclude that the 0-decomposition (2) has property (R) with respect to every 0-decomposition of 1 of the form

1 =£(<*,: i e l ’H U b j i j e J ’),

where Г and J' are subsets of / and J, respectively. Then, by Definition 2, (2) has the property (R) with respect to 0-decomposition (1). The statement now follows from Theorem 1. ■

We need the following lemma.

L

9 . Let L be a modular lattice and let ⁰ e© (L). I f a O-indecomposable element aeS(L) is of finite length (5), then a satisfies the B-condition.

P ro o f. Let c be an element of L such that a is a 0-summand of c. The element a is of finite length in L, and, therefore, in c/0. Then, by Lemma 7 of [3], a satisfies the В-condition in the lattice c/0 (in the proof of Lemma 7, [3], it was not used that L is of finite length only that so is a). Hence, by Definition 4 we obtain the statement. ■

The main result of our paper is expressed in the following theorem.

T

3 . Let L be an upper continuous modular lattice and let 0 be a congruence relation on L having the property (*). I f

1 = X K -: ie l) — J e J )

(5) a is of finite length iff the lattice aj0 is of finite length.

are two в-decompositions of the unit element of L such that each at and each bj is of finite length and в-indecomposable, then there exists a one-to-one mapping f of I onto J such that, for each i e I,

1 = аг + Х (Ь;: j # /(0 ).

P ro o f. P. Crawley shows in [1] (Lemma 3) that every element of finite length (in an upper continuous lattice) is compact. Therefore, from Lemma 9 and Theorem 2 our theorem follows. ■

R em ark. The case в = œ yields the theorem of P. Crawley (cf. Theorem 3 in [1]).

It is obvious that every lattice of finite length is upper continuous.

Therefore, from Theorem 3 we get at once

C

(cf. [3], p. 358). Let Lbe a modular lattice of finite length and let 0e0 (L ). I f the unit element o f L has two в-decompositions

' 1 = a 1+ u 2+ ...+ u „ , l = b1 + b2+ . . . + b m

into в-indecomposable elements, then n — m and for every ai there is a bj such that 1 = bj -f-... + bj_ j -f-a{ + bJ +1 + ... -fbn.

Now, we consider the case when в = i. The following lemma holds.

L

10. Every join-irreducible i-summand of the unit element of a modu­

lar lattice L satisfies the B-condition.

P ro o f. Let a be a join-irreducible г-summand of 1 and let c be an arbitrary element of L such that a is a i-summand of c. By the proof of Corollary 2 ([3]), it is obvious that a satisfies the В-condition in the sublattice c/0, and hence, by Definition 4, we conclude that a satisfies the B-condition. ■

From Theorem 2 we obtain now

T

4 . Let L be an upper continuous modular lattice. I f the unit element of L has two irredundant decompositions

(25) 1 = ÿ ( a ,: ie l) = \/( b t: je J )

into completely join-irreducible elements, then there is a one-to-one mapping f of I onto J such that, for each i e /,

1 = at v V (bf. j # /(0 ).

P ro o f. In the proof of Lemma 3 ([1]) it was shown that every completely

join-irreducible element of an upper continuous lattice is compact. Therefore,

all i-summands of i-decompositions (25) are compact. Moreover, since every

completely join-irreducible element of a complete lattice is join-irreducible, by Lemma 10 we infer that each a, and each bj satisfies the В-condition. Now, from Theorem 2 our theorem follows. ■

References

[1] P. C r a w le y , Direct decompositions with finite dimensional factors, Pacific J. Math. 12 (1962), 457-468.

[2] P. C r a w le y , R. P. D ilw o r t h , Algebraic Theory o f Lattices, Prentice-Hall, Englewood Cliffs, N.J., 1973.

[3] A. W a le n d z ia k , Solution o f Gràtzer’s problem, Comm. Math. 26 (1986), 349-359.

(INSTYTUT MATEMATYKI, WY2SZA SZKOLA PEDAGOGICZNA, OPOLE) INSTITUTE O F MATHEMATICS, PEDAGOGICAL COLLEGE,

OPOLE, POLAND