THE CONSTRUCTION OF SOME FREE m-LATTICES ON POSETS

George Gratzer and David Kelly

Department of MathematicsUniversity of Manitoba

Winnipeg, ManitobaCanada

Let m be an infinite regular cardinal. We call a posetan m-complete lattice (or m-lattice) if every nonempty

subset with fewer than m elements has both a join and ameet. This concept was introduced by P. CraWley and R.A.Dean. For m-lattices, enough of the algebraic flavor ofthe usual (finitary) lattice case has been retained so thatthere are free m-Iattices generated by any poset. Weshall describe the free m-lattice;. on a particular 6-ele­ment poset H, and indicate why the description is correct.The poset H is a natural example because its free m-lat­tice can be used to describe the free m-lattice on manyother pesets.

INTRODUCTION

It is well known that lattices can be considered either as ordered sets oras universal algebras. The algebraic approach guarantees the existence of thefree lattice on any poset; that is, the lattice that is generated "as freely aspossible" starting from the poset. (If no ordering is specified on the gener­ating poset, an antichain is understood.)

Many of the ordered sets that naturally occur in various fields of mathema­tics have a lattice structure.. In fact, they are usually complete lattices ..However, some algebraic techniques no longer apply directly to complete lattices.In particular, there is no free complete lattice on three elements • If such acomplete lattice did exist, then every complete lattice L with three generatorswould be a complete homomorphic image of it; in particular, there would be anupper bound for the cardinality of any such L. However, Figure 1 shows a3-generated complete lattice of any given cardinality.. (In this figure,the three complete generators are denoted by squares.)

Of course , any diagram of an infinite paset requires interpretation. Anextreme position is that any such diagram is useless. We are not of thisopinion, but we do advise extreme care, especially in a proof, when readinginformation from a diagram of an infinite poset. For example, it is not obviouswhen an element is join irreducible. (For a finite lattice, join irreducibleelements are those with a unique lower cover.)

The left-hand chain C in Figure 1 can be any well-ordered chain that has alargest element, denoted by t. The right-hand chain is isomorphic to C, andthe remainder of the poset is a linear sum of finite lattices over C. For eachi .,: t in C, the paset of Figure 1 has an i-th level consisting of 6 elements.Observe that an element on any level i < tcan only be generated after anelement on every lower level has been generated ..

The concept of completeness belongs to order theorY,while free lattices onpesets come from the algeb:aic approach. The above example shows that one must

FIGURE 1. An arbitrarily large complete lattice

be careful when attempting to combine these two concepts. We shall resolve thisdifficulty by weakening the concept of completeness.

We shall always use the symbol m to denote an infinite regular cardinal.An m-comp+ete lattice, orm-lattice, is a poset in which every nonempty subsetwith fewer than m elements has both a join and a meet. We shall prefix eachlattice concept with m to denote the corresponding concept for m-lattices.Some examples are m-join, n.-homomorphism and m -sublattice. We emit mwhen m = ~0 because this is the usual lattice case.

Obviously, any complete lattice is an m-lattice for any m. Moreover,if m exceeds the cardinality of a poset P, then any join that exists in P isa join of a nonempty subset of P whose cardinality is less than mj in short,any join is an m- join. However, there are m-Iattices that are complete inwhich not every join is an m-join. If the left-hand chain C in Figure 1 isthe (ordinal) successor of the initial ordinal corresponding to m, then thisfigure provides such an example L. In this case, note that the top element tof C cannot be written as VS for a nonempty subset .S of L with tiSand lSI < m. In. other words, t is an m-join irreducible element of L.Consequently, the three squares in Figure 1 do not m-generate L.

We emphasize that we only take joins and meets of nonempty sets. (In acomplete lattice L, th~ join of the empty set equals the meet of L.) Althoughm-lattices have infinitary· operations, there is a bound on the number of suchoperations so that the concepts and methods of universal algebra apply. Theimportant point here is that there is a set (not a class) that can simultaneouslyindex all the join and meet operations-of every m-lattice. A particularconsequence· is the existence, for any poset P, of Fm(P), the free m-lattice

on P. F (P) is the m-lattice that contains P as a subposet for whichmm-joins and m-meets are formed "as freely as possible".· (In particular,. Fm(P) is m-generated by P.) More precisely, F nfP) is the unique m-lattice

for which every isotone map from P to any m-lattice L has a unique exten­sion to an m-homomorphism from FmCP) to L. We shall use !!, a boldfacenumber, to denote an n-element chain. A lightface number indicates an anti­chain. As usual, P + Q denotes the disjoint union of the posets P and Q.

We shall call a paset slender if it does not contain any of the followingthree posets as a subposet:

(i) 3 =1 + 1 +. 1fII/IIt# "",..,. #'W

(ii) g + ~

(iii) ! + .2 ·

I. Rival andR. Wille [16] described the free lattice on any finit·e slenderpaset. In a sequence of papers ([9], [10l, [11 J, [12] ), we have generalized thisresult by describing F (P) for any slender paset. Our techniques also work

mwhen m is countable, and so provide another proof of the Rival-Wille result.However, since the m = :{ 0 case differs significantly from all others, we

henceforth asswne that m is uncountable.

Similarly as in [16], the key ingredient in describing the free m-latticeon any slender poset is a description of the free m-lattice on the paset H ofFigure 2. The present paper is devoted to describing F':ll(H) . We shall see that

Fm(H) is a complete lattice (which would not be true if we allowed m = ~o).

In general, we shall only indicate proofs. We not only describe Fm(H) as a

paset, but we give the formulas for (arbitrary nonempty) joins and meets.

FIGURE 2. The POset H

In order to describe Flit HI), we must first introduce some other lit -lat

tices. In Section 1, we describe C( m), the free m-lattice on ~ + ~, witha new 0 and 1. We also require the m-lattices A and B described inSection 2. In Section 3, we describe D(m ), which is Fm(H) with a new 0

and 1. Our paper [9] contains the proofs of all statements made in the firstthree sections. In Section 4, we discuss the proofs that the m-latticesC( m) and D( m), as described, are what we claimed. One approach will be toapply the Structure Theorem for free· m-lattic·es on posets given by P. Crawleyand R.A. Dean [2]. Complete details for Section 4 can be found in [9] and [10].

Let P be a slender poset. If P is also linearly indecomposable, then weshow in [ 11] that P is countable, and that Fm(P) can be embedded, as anm-sublattice, into Fm(H). Consequently, a POset P is slender iff Fm(P}

does not contain Fm(3) as an m-sublattice.

We conclude this introduction by briefly mentioning some further topicsabout· m-lattices. Free m-lattices, defined as free 'in-lattices overantichains, can also be considered as free m-products of one-element m-lat­tices. For free products, we refer to Gratzer, Lakser and Platt [14] andGratzer and Lakser [13]. In our paper [7], we generalized some of the results ofthese two papers, and also studied what happens when m is varied. Forexample, we proved the Stucture Theorem for free m-products, which is similarto the Structure Theorem for free m-lattices of Crawley and Dean [2] •. Anotherresult of [7] is: If a covers b in some free m-product, then a continuesto cover b when m is increased. (Covers are of interest because A. Day [3]has shown that every proper closed interval of a finitely generated free latticecontains a covering pair. ) We also determined which free m-products arecomplete in [7]. In particular, the free m~lattice on three generators is notcomplete (see Whitman [18] for the m =~0 case) •

When representing an element of an ~lattice in terms of an m -generatingset (using m-joins and m~meets), there is no general way to specify a uniquerepresentation, even in a free m-lattice. For example, there are many examplesin Fm(3) of an m-join reducible element that cannot be expressed as an

irredundant m-join. In [7], we generalized the Normal Form Theorem that B.J6nsson [15] proved for the free m-lattice on a poset. We used the the normalrepresentation of [7] to generalize a result about free products in [13].

A natural question about an ordered set is: What is the largest chain thatit contains? We say that a poset P satisfies the R;..chain condition if itcontains no chain of cardinality n • HencefOrth·;-Tet n be uncountable andregular. M.E. Adams and D. Kelly [1] showed that, if the poset P satisfies

the n-chain condition, then so does the free lattice on P. However, foruncountable m, we contructed , in our paper [6] with A. Hajnal (for anyn > m), a JX)set P with no uncountable chains such that the free m-latticeon P contains a chain of cardinality n. In Adams and Kelly [1], it was alsoshown that the free product construction preserves the n -chain condition. Inour paper [6] with A. Hajnal, we also showed that a free .u -lattice cannotcontain a chain with more than m elements. (In the previous sentence" weassumed the Generalized Continuum Hypothesis to simplify the formulation.) Ifm :: ~1 and n :is the successor cardinal to the continuum, then we do not

know whether the free tn-product construction .preserves the n-chain condition.

1• THE COMPLETE LATTICE C( til)

Recall that 11\ always denotes an uncountable regular cardinal. Thecomplete lattice C( m) of Figure 3 is the free m-lattice on 2 + 2 with a new,-,,1' __

o and 1. The diagram is also valid for F(g +~) (see Rolf [9] or Chapter VIof Gratzer [5]). We have denoted the 16-element lattice in the middle of Figure3 by C{ ~O). (The way this finite lattice was used by I. Rival and R. Wille

[16] to construct F{H) is similar to the way we shall use C( m) to constructF m(H) • ) In this section , we only discuss the complete lattice structure ofC( m). We leave until Section lithe question whether C( 11\) is what we' haveclaimed.. Observe that Bm,a~, b Ul, and b:n generate C( ~ 0). We also usethe notationf. :: a A band f t = a t v b'.m m m m m m

For every successor ordinal j < m, there is a lower j-th level of 6elements Lj = raj' bj , c j ' dj , e j , f j }, and for every limit ordinal i < m

(inclUding i =0), there is a lower i-th level of 7 elements Li = {ai' bi , ci '

di , e i , f i , gil. These elements are ordered as shown in Figure 3. There is also

an upper i-th level Ui for each i < m, defined dually and denoted by the same

letters with primes. For convenience, we also label 6 elements of C( m) withGreek letters: a = aO' a' =a(J' ~ =bO' ~' = bO' y =go' y' = gO·

In a more formal approach, one defines C( m) to be

C( ~O) u U (L j I j < m) uU (U j I j < m),

where each finite POset that we have listed is assumed to be a subposet ofC( m). Let i and j be ordinals less than m.. The additional comparabilityu < v holds in C( m ) exactly when one of the following conditions 1s satis­fied:

u =a., v =a., and i < j ;1 J

u = bi I v = bj , and i < j

u ~ L i - {ai' bi }, v ~ Lj , and i < j

u = ai and am S v in C( ~ 0) ;

u = bi and bm s. v in C( ~0) ;

u ~ Li - {ai' bi } and v E: C( ~O) ora previous condition holds with u

interchanged, primed elements,C( ~O);

v e Uj ;

and v interchanged, Land Uand reversal of the inequality in

U'i

a'=~

oi

°i+1

OJ

a=oo

b~=I3'

lin

limit

FIGURE 3. The complete lattice C( m)

U :: ai and v E: Uj , but v;te bj ;u:: bi and v E: Uj , but v:¢

Clear1y, y :: go is the element of C( nr}, and y':: SO is thegreatest. In fact, C( m) is a complete lattice. To sketch the proof of thisstatement, let us explain how to calculate (nonempty) joins. From Figure 3, onesees a natural way to break C( m) up into 1 parts: C( ~O) J the bottom leftchain {ai' i:;; m}, the bottom middle part from go to fm , the bottom

rightchain {bi Ii:;; 1ft}, and the corresponding three top parts. C( NO>

has exactly one point common with each of the other six parts. Observe thateach part a complete lattice, in fact, a complete sublatticeof C( m).Since each infinite part is a linear sum over a complete chain, it should beclear how to calculate the join in each part. It only remains to determine thejoin of pairs of incomparable elements from different parts. We leave thederivation of these formulas to the reader.

Let us indicate why C( m) - {y J y'}f.P. We first generate elements of C(~)

m-generated by a, a'; ,:J, andthe follOWing order:

ao v bO' aO 1\ bO' fm, f~, am (= a V faa ), bm, aait, b~.

We can now generate C( NO) we have its four generators, • We gave the

idea for the rest of the argument in the introduction when we discussed thecomplete lattice of Figure 1. If Li has already been m-generated for somei < m, then it is easy to generate Li+1 using finite joins and meets. If, for

a lim.it ordinal j < m,already been =-generated for each i < j,

thenaj and bj can then be m-generated aj :: V( ai I i < j).

2. nm LATTICE A

In this section, we define the second block of DC m), the ...":.lli..·"'.L...... 'IW<

A of Figure 4. This was discovered by I. Rival R.Wille [16]. LetJ be the set of dyadic rationals r that satisfy 0 s r s 1. Every nonzero rE: J

has a unique representation, the normal form, r :: a • 2-n, where a is an odd

integer; n the of r; notation,. n :: ord(r). By convention, ord(O)

:: O. Observe that ord(r):: ord(1 - r). We also define ord(r, s> to be themaximum of order) and ord(s).

We define A as asubposetof J2 with the product order:

A = l<r, s> I r < s and s _ r :: 2-ord<r, s> } •

Note that if <r, s> e: A, thenord(r) I. ord(s) with the exception of <0, 1>.Although A is a lattice, it is not a sublattice of J2. For example,

(0, 1> v (1/4, 1/2> =<1/2, 1>

in Figure 4. A can be in two ways:

-n ( )}A= l<r, s> r < sand s - r :: 2, n t:= ord r .

-n ( )}A :: {<r, s> r < sand s - r :: 2 ,n ~ ord s... •

<0,1 >

FIGURE 4. The lattice A

Moreover, if a E: A ­has a left lower cover,

We shall use 1t1 and 1t2 to denote the first and second projections of J2

onto J. For t e J, let us call the set of a E: A that satisfy 1t1(a) :: t

the x:: t line in A, and define the y:: t line similarly. The alternate

definitions of A yield expressions for the x:: rand y:: S lines in A. Inparticular, <r, r + 2-ord (r) > is the largest element on the x:: r line, and

<s - 2-ord (s), s> is the smallest element on the y:: S line.

Each a e A has a right lower cover a* and a right upper cover a*. Theformulas are:

<r, s>. :: «r + s)/2 , s><r, s>. :: <r, (r + s)/2 >.

For example, <1/4, 1/2>* = <3/8, 1/2> in Figure 4.{<O, 1>} satisfies ordeal = ord(-Jt1(a», then a

denoted by .a, and defined by

-order).<r, s> :: <r - 2 , s>.

Similarly, the left upper cover *<r, s> exists and equals <r, s + 2-ord (s»when order) < ord(s). Observe that <0, 1> has no left (lower or upper) cover,but that every other element of A has exactly one left cover. Moreover, eachcover of an element a E: A lies on the x-line through a or the y-linethrough a.

In Figure 4, A seems to be a very sparse subset of J2. For each a::<r, s> e A, let us call the closed interval [r, s] (in either the reals ordyadics) the shadow of a.. We say that two closed intervals overlap if they ha.ve

an interval of nonzero length common, but neither interval contains the other.The "relative sparseness" of A is expressed by the following shadow principle:

The shadows of two elements of A cannot overlap. Let us indicate an easy wayto visualize shadows. We add a vertical line (the shadow line) that passes

through the intersection of the x:: 1/2 and y = 1/2 lines on the right sideof A in Figure 4; index this line in the obvious way with [0, 1J. If oneplaces an opaque right angle at a e A (forming the x-line and the y-linethrough a), then the intersection of the right angle with the shadow line givesthe shadow of a.

We found the shadow principle to be a very natural tool in our proofs. Letus apply it to "explain" the emptiness of the largest open square in Figure 4:if there was an element a in this square, then the shadows of a and <0, 1/2>would overlap.

In Figure 4, the distance of each element <r, s> € A from the shadow lineis proportional to s - r t the length of its shadow. Consequently, for twoincomparable elements a, b E: A, we can define a to be to the left of b iff

a has a longer shadow than b. Observe that a is to the left of b iff1t1(a) < 1t1(b) iff '11:2(a) > 1t2 (b). With this concept, we can visualize the

finitary join in A. Let a and b be incomparable elements of A,with a tothe left of b. The join of a and b is the least element on the y-linethrough a that is greater than b. Consequently, a Vb:: an for the least n

such that an > b, where an is inductively defined by: aO =a, a i+1 :: (ai >* .This shows that A is a lattice.

3. llIECOMPLETE LATTICE D( m)

We build up D( m) from three disjoint sublattices: A, B, and C. Thelattice A was defined in Section 2. We define

B ={<r, s> I <s, r> e Al,

a subposet of J2• Clearly, B is a lattice and its diagram is obtained byreflecting Figure 4 about a vertical line. In other words, mapping <r,s> e Ato <s, r> e B defines an isomorphism of A with B. Consequently, everyelement b of B has left covers .b and *h, and if b;at <1, 0>, then b

has a right lower cover b. or a right upper cover b*.

Finally, let I be the real interval [0, 1], and recall that J denotesthe subset of I consisting of dyadic rationals. For each t e J, we take acopy ct of C( m), with bounds Yt and rt, and generators at, at, ~t'

"i. For each tel which is not a dyadic rational, Ct = {Yt,Yi} is thetwo-element chain with Yt .< Yt • We define C as the linear sum of the Ct 't £ I. Since I is complete and each Ct is complete', C is a completelattice.

To define D( m), we must describe the partial ordering(see Figures 5 and 6):

A u B u C

If <r, s> e-A, <t, u> £ B, vel, and p e Cv'<r, s> < <t, u> iff s < u;<r, s> > <t, u> iff r > t;<r, s> < p iff s < v holds or s = v and<r, s> > p iff r > v holds or r =v and

<t, u> < p iff t < v holds or t =v and<t, u> > p iff U.> v holds or u =v and

then:

Uv :s p hold;a' .~ p hold-,v~v :s p hold;~' .~ p hold.:v

It is easily seen that D( m) is a poset. However, since <0, 11A u B is not a sUbposet of J2. Note that v was assumedrational wherever we used the notation av' a~, Pv' or p~. Thus,dyadic, then in each of the last four cases, only the first clause

j, <1, 1/2>,be a dyadic

v is not.a apply.

It is not difficult to show that D( m) is a lattice, and tl': each of A,B, and C is a sublattice of D(m). For <r, s> E: A, <t, u> eVE: I, andp E: Cv' we give the formulas for joining pairs:

(a) <r, s> v p is(i) CJs vp e C, where the join is formed in C, if s:s

(ii) <r, s>, if r > v or r =v and p:s a; in Cv;(iii) the least <w, s> such that w > v, if r:s v < s and ~ a~ in Cv;(iv) the least <w, s> such that w ~ v, if r:s v < sand p:s ~ in Cv;

(b) <r,(i)

(ii)(iii)(iv)(v)

s> v <t, u> is<t, u>, if s < u;<r, s>, if t < r;the le'ast <w, s> on the y = s line in A such that w > t, if s >t;the least <t, w> on the x = t line in B such that w > s, if s >t;as vl3s ' if s = t, where the join is formed in Cs.

<0,1>

treolnondyodie

<1.0>

<1,1/2>

i dyadic

FIGURE complete .Q.ilVol.""'.... DC m)

6.

A

Cs a~¢··· ~1as . Ps.' "'.

·······! y;'

~Y:•i......0.... 1.... ~.-..., ,',--..-

af 13:a,· I3r

Details

<5,r> B

r<t<5r, s dyadict real nondyodic

DC m)

<s, r>words, •formulas for

vX foralready know thatseoond projections X, and form

tofor B and we

the first andi.n I.

u < v,y = v

A on the

ncr"emlent the ordernonzero i £

U E:

and u i

ao' ~O' <0, 1>, <1, 0>, a1, ~1.

of Co - {YO' yo}A u B of order 1 are generated.

c

At

u = v

u == v= { :: if

is m-generated byfour

u =v, VX

A u B

of elements A u

D( m) - {YO' Y1}

prove one

C, - fY1' Y;}.One then m-aerlerat.E~S

I t ~ J and t < •

In theD( m). Anreduoible)

coverm).

properties of(oompletelya has a \oilu ...'-i\.olr....

X £ t £ x m}

x =Yt t i. 0; ort ~ J and x =eO or x ~ , , } for 0 <: i <: m; orx a joinreduoible .....""...u ...... ,l .... of C(~0); or

x :: or x =bm; orx € " for i <: m.

m)

thex

xd s y

m-joinand x= f to the reader. The formulasm

4. JUSTIFYING THE DESCRIPTION OF D( m)

Let D' denote D( m) - {yo Y1}. We shall comment on our two proofsthat D' is F41 (H) • Certainly, D' contains H as a subposet, and H m -gen

erates D'. Moreover, for any x € H, x f: V ( Y € H I x f: y) in D'. By theStructure Theorem of Crawley and Dean [2] for free m-lattices on posets, itonly remains to show that D' satisfies- the condition (Wm). An m-lattice Lsatisfies (Wm) iff whenever S and Tare nonempty m-subsets of L ( ISI ,ITI < m), then I\S -sV T implies s -s VT for some S € S, or AS -s t forsome t ~ T.

Let us call an interval [a, b] of D' a (W )-failure if there are non­m

empty m-subsets S, T of D' such that a =1\ S, b =VT, but S n [a,b]and T n [a, b] are both empty. One way to show that D' is F. (H) is to

mprove that D' does not contain- any (Wm)-failures. We have done this in [9J.For example, suppose there is a (Wm)-failure [a, b] with both a and b inA. Let x be the greatest element on the x-line through a, and let y bethe least element on the y-line through b. Since some element of S must beon. the X-line through a, x $. b. Consequently, ~(x) > ~(b) = ~(y).

Similarly, 1t1(y) < 1t1(a) = 1t1(x). Since this means that the shadows of x

and y overlap, such a (Wm)-failure is impossible. Of course, C( m) must

also satisfy (Wm>. In fact, our proof that there is no (W~-failure with a

or b in C is essentially an analysis of why C( m) satisfies (lla). Thefinal case, that a € A and b € B , required the most complicated argument.

For our second proof in [10], we assumed the description of F(H) given byI. Rival and R. Wille [16]. We then developed some elementary universalalge­braic leumas to show that D' is F (H). The Structure Theorem was not used butmwe did reqUire the fact that C( m) -{y, y'} is Fm,(2, + 2,). This proof takes

advantage of the fact that the elements of C labeled with a, Ct', p, and (3'freely m-generate C - {yo' Yl}.

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Some free m-lattices on posets 117

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(2] P. Crawley and R.A. Dean, Free lattices with infinite operations, Trans.Amer. Math. Soc. 92 (1959), 35-47.

(3) A. Day, Splitting lattides all lattices, Algebra Universalis 7(977), 163-170.

N. Funayama, Notes on lattice (semi-) lattices,Bull. Yamagata Univ. (Nat. Sci.)

G. Gratzer, General Lattice Pure and Applied Mathematics Series,Academic Press, New York, Y. ; Mathematische Reihe, Band 52,Birkhauser Verlag, Basel; Akademie Verlag, Berlin, 19~8.

[6J G. Gratzer, A. Hajnal and 01 Kelly, Chain conditions in free products oflattices with infinitery operations, Pacific J. Math. 83 (1979),107-1'5.

[7J G. Gratzer and D. Kelly, t:'ree m-products of lattices I and II, Colloq.Math., to appear.

[8] G. Gratzer and D. Kelly, A normal form theoremgenerated by a SUbset, Proc. Amer. Math. Soc.

G. Gratzer and D. Kelly, The free m-lattice on the

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H, to appear.

G. Gratzer and D. Kelly, A tecHnique to generate m-ary free lattices fromfinitary ones, to appear.

(11] G. Gratzer and D. Kelly, An embedding theorem for free ITt-lattices onslender posets, to appear.

[12J G. Gratzer and D. Kelly, A description of free m-lat tices on slenderposets, to appear.

[13] G. Gratzer and H. Lakser, Free-lattice like sublattices of fre·e products oflattices, Proc. Amer. MC:ith. Soc. 44 (1974), 43-.l.l5.

[1L1J G. Grat.zer, H. Lakser and C.R. Platt, Free products of lattices, Fund.Math. 69 (1970 ), 233-240.

(15) B. J6nsson, Arithmetic properties of freely a-generated lattices, Canad.J. Math. 14 (1962), Ll76-!.I81.

(16) I. Rival and R. Wille, Lattices freely generated by partially ordered sets:which can be "drawn"?, J. Reine. Angew. Math. 310 (1979), 56-80.

H.L. Rolf, The free lattice generated by a set of chains, Pacific J. Math.8 (1958), 585-595.

Ph. M. Whitman, Free lattices Ann. of Math. (2) 43 ( , 104-115.