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Non-extendability of semilattice-valued measures on partially ordered sets
Friedrich Wehrung
To cite this version:
Friedrich Wehrung. Non-extendability of semilattice-valued measures on partially ordered sets. 2006,
pp.191–200. �hal-00012068v2�
ccsd-00012068, version 2 - 28 Feb 2006
ON PARTIALLY ORDERED SETS
FRIEDRICH WEHRUNG
Abstract. For a posetP and a distributiveh∨,0i-semilatticeS, aS-valued poset measureonPis a mapµ:P×P →Ssuch thatµ(x, z)≤µ(x, y)∨µ(y, z), and x ≤ y implies that µ(x, y) = 0, for all x, y, z ∈ P. In relation with congruence lattice representation problems, we consider the problem whether such a measure can be extended to a poset measureµ:P×P→S, for a larger posetP, such that for alla,b∈Sand allx≤yinP,µ(y, x) =a∨bimplies that there are a positive integernand a decompositionx=z0≤z1≤ · · · ≤ zn=yinP such that eitherµ(zi+1, zi)≤aorµ(zi+1, zi)≤b, for alli < n.
In this note we prove that this is not possible as a rule, even in case the posetPwe start with is achainandShas sizeℵ1. The proof uses a “monotone refinement property” that holds inSprovidedSis either a lattice, or count- able, or strongly distributive, but fails for our counterexample. This strongly contrasts with the analogue problem fordistanceson (discrete) sets, which is known to have a positive (and evenfunctorial) solution.
1. Introduction
In the paper [5], the author proved that for any lattice K, any distributive lattice S with zero, and any h∨,0i-homomorphism ϕ from the h∨,0i-semilattice ConcK of all finitely generated congruences of K to S, there are a lattice L, a lattice homomorphismf:K→L, and an isomorphismα: ConcL→S such that ϕ=α◦Concf. In the paper [4], J. T˚uma and the author proved that for ah∨,0i- semilattice S, this statement characterizes S being a lattice. The proof of this negative result strongly uses the lattice structure of the hypothetical lattice L, see the proof of [4, Corollary 1.3].
In the present paper, we show that for a certain semilatticeS of cardinalityℵ1, the poset structure alone is sufficient to get a related counterexample. More pre- cisely, for a h∨,0i-semilattice S, a S-valued poset measure on a poset P is a map µ:P×P →Ssuch thatµ(x, z)≤µ(x, y)∨µ(y, z) (triangular inequality) andx≤y implies thatµ(x, y) = 0, for allx, y, z∈P. We say thatµis aV-measure, if for all x≤yinP and alla,b∈S, ifµ(y, x)≤a∨b, then there are a positive integernand a decomposition x=z0≤z1 ≤ · · · ≤zn =y in P such that eitherµ(zi+1, zi)≤a or µ(zi+1, zi) ≤b for all i < n. In particular, if P is a lattice and S = ConcP, then the map µ defined by µ(x, y) = Θ+(x, y) = Θ(y, x∨y) is a ConcP -valued V-measure onP.
This yields the following poset analogue of the abovementioned lattice-theoretical problem.
Date: February 28, 2006.
2000Mathematics Subject Classification. 06A12, 06A06, 06A05.
Key words and phrases. Semilattice, poset, distributive, isotone, measure, ∆-Lemma, closed unbounded.
1
2 F. WEHRUNG
Problem. LetS be a distributiveh∨,0i-semilattice. Does anyS-valued poset mea- sure on a given poset extend to someS-valued poset V-measure on a larger poset?
A version of this problem for so-calleddistances (instead of measures) on discrete sets (instead of posets) is stated in [3]. The answer to this related question turns out to be positive (and easy). More surprisingly, this positive solution can be made functorial.
Nevertheless, we prove in the present paper that the problem above has anega- tive solution. Unlike what is done in [4], we do not reach here a characterization of all lattices among distributiveh∨,0i-semilattices. Our counterexample, denoted by F(ω1) (see Corollary 4.9) is obtained as an application of a certain “free construc- tion” used by M. Ploˇsˇcica and J. T˚uma in [2]. The semilatticeD of [4, Section 2], which is the simplest example of ah∨,0i-semilattice which is not a lattice, does not satisfy the negative property used here. This is because D is countable, while we prove in Proposition 4.10 that no countable distributiveh∨,0i-semilattice can have the required negative property. On the other hand, in relation to [4, Problem 4], the proof of our counterexample uses very little of the Axiom of Choice (namely, only the Axiom of countable choices), while the proof of the negative property of the abovementioned semilatticeDestablished in [4, Corollary 2.4] uses the existence of an embedding fromω1 into the reals.
2. Basic concepts
For posets (i.e., partially ordered sets) P and Q, a map f:P →Q is isotone, ifx≤y implies that f(x)≤f(y), for all x, y ∈ P. In addition, we say that f is join-preserving, if for any subset X ofP, whenever the joinWX ofX exists inP, Wf[X] exists inQ, andW
f[X] =f W X
. For a subsetX of a posetP, we shall put ↓X ={p∈P | ∃x∈ X such thatp≤x}, and then ↓a=↓{a}, for alla∈P. We say thatX is alower subset of P, ifX =↓X.
A h∨,0i-semilattice S is distributive, if c ≤ a∨b in S implies that there are x ≤ a and y ≤ b in S such that c =x∨y. A distributiveh∨,0i-semilattice S is strongly distributive, if every element ofS is the join of a finite set of join-irre- ducible elements ofS; equivalently,S is isomorphic to the semilattice of all finitely generated lower subsets of some poset.
We shall denote by otpP the order-type of a well-ordered set P. Hence otpP is an ordinal. We shall also use standard set-theoretical notation and terminology, referring the reader to [1] for further information. In particular, we shall denote byω1 the first uncountable ordinal. A subsetC ofω1 is closed unbounded, if C is unbounded inω1 and the join of any nonempty bounded subset ofCbelongs toC.
It is well-known that the closed unbounded subsets form a countably complete filterbasis onω1, see [1, Lemma 7.4]. Hence containing a closed unbounded set is a notion of “largeness” for subsets ofω1.
3. Free distributive extension of a h∨,0i-semilattice
There are several non-equivalent definitions of what should be the “free distribu- tive extension” of a givenh∨,0i-semilattice. The one that we shall use is introduced in [2, Section 2]. Let us first recall the construction.
For ah∨,0i-semilatticeS, we shall putC(S) ={hu,v,wi ∈S3|w ≤u∨v}. A finite subsetxofC(S) isreduced, if it satisfies the following conditions:
(1) xcontains exactly one diagonal triple, that is, a triple of the formhu,u,ui;
we putu=π(x).
(2) hu,v,wi ∈xandhv,u,wi ∈ximplies thatu=v=w, for allu,v,w∈S.
(3) hu,v,wi ∈ x\ {hπ(x), π(x), π(x)i} implies that u,v,w π(x), for all u,v,w∈S.
We denote by R(S) the set of all reduced subsets of C(S), endowed with the partial ordering≤defined by
x≤y ⇐⇒ ∀ hu,v,wi ∈x\y, eitheru≤π(y) orw≤π(y). (3.1) Furthermore, we shall identifyxwith the element{hx,x,xi}ofR(S), for allx∈S.
For set-theoretical purists, this can for example be done by replacingR(S) by the disjoint union of S with the set of non-singletons in R(S). The disjointness can easily be achieved by a suitable modification of the standard definition of a triple.
We shall use the symbol⊲⊳to denote the canonical generators ofR(S), so that
⊲⊳(u,v,w) =
w, if eitheru=v orv = 0 orw= 0,
0, ifu= 0,
{h0,0,0i,hu,v,wi}, otherwise,
Observe that the canonical mapπ: R(S) ։S is isotone and that the restriction of π to S is the identity. Furthermore, π(⊲⊳(u,v,w)) = 0, for any non-diagonal hu,v,wi ∈C(S). The following is an easy consequence of (3.1).
x≤y ⇐⇒ x≤π(y), for all hx,yi ∈S×R(S). (3.2) We recall the standard facts established in [2] about this construction.
Proposition 3.1.
(1) For anyh∨,0i-semilatticeS,R(S)is ah∨,0i-semilattice, and the inclusion map fromS intoR(S)is ah∨,0i-embedding.
(2) For h∨,0i-semilattices S and T, every h∨,0i-homomorphism f: S → T extends to a unique h∨,0i-homomorphism R(f) :R(S) → R(T) such that R(f)(⊲⊳(u,v,w)) =⊲⊳(f(u), f(v), f(w)), for all hu,v,wi ∈C(S).
(3) The assignmentS7→R(S),f 7→R(f) is a functor.
The extension R(S) is defined in such a way that ⊲⊳(u,v,w) ≤ u and w =
⊲⊳(u,v,w)∨⊲⊳(v,u,w), for all hu,v,wi ∈C(S). Hence, putting R0(S) =S and Rn+1(S) = R(Rn(S)) for each n, we obtain that the increasing union D(S) = S(Rn(S)|n < ω) is a distributive h∨,0i-semilattice, extending S. Furthermore, putting D(f) = S(Dn(f)|n < ω) for each h∨,0i-homomorphism f, we obtain thatD is a functor. The proof of the following lemma is straightforward.
Lemma 3.2. Let S be ah∨,0i-semilattice and let hSi|i∈Iibe a family ofh∨,0i- subsemilattices of S. The following statements hold:
(1) R T
i∈ISi
=T
i∈IR(Si)andD T
i∈ISi
=T
i∈ID(Si).
(2) If I is a nonempty upward directed poset and hSi|i∈Ii is isotone, then R S
i∈ISi
=S
i∈IR(Si)andD S
i∈ISi
=S
i∈ID(Si).
Definition 3.3. For ah∨,0i-semilatticeSand an elementx∈D(S), we define the rank of x, denoted by rkx, as the least natural number nsuch that x ∈Rn(S), and thecomplexity ofx, denoted bykxk, bykxk= 0 ifx∈S, and
kxk=X
(kuk+kvk+kwk+ 1| hu,v,wi ∈x), for allx∈D(S)\S.
4 F. WEHRUNG
4. The semilattices S(Λ)and F(Λ)
For anychain Λ, we shall denote byS(Λ) theh∨,0i-semilattice defined by gen- eratorsa, b, and ci, fori ∈Λ, and relationsci ≤a∨band ci≤cj, for alli≤j in Λ. Hence the elements ofS(Λ) either belong to S(∅) = {0,a,b,a∨b} or have the form ci, a∨ci, or b∨ci, for some i ∈ Λ. We shall identify S(X) with the h∨,0i-subsemilattice ofS(Λ) generated byS(∅)∪ {ci|i∈X}, for anyX ⊆Λ.
For chainsX andY, any isotone mapf:X →Y gives raise to a uniqueh∨,0i- homomorphismS(f) :S(X)→S(Y) fixing a andb and sendingci tocf(i), for all i∈X. Of course, the assignment Λ7→S(Λ),f 7→S(f) is a functor.
We denote byF=D◦Sthe composition of the two functorsD andS. The proof of the following lemma is straightforward.
Lemma 4.1. LetΛ be a chain and lethXi|i∈Iibe a family of subsets ofΛ. The following statements hold:
(1) S T
i∈IXi
=T
i∈IS(Xi).
(2) If I is a nonempty upward directed poset and hXi|i∈Iiis isotone, then S S
i∈IXi
=S
i∈IS(Xi).
As an easy consequence of Lemmas 3.2 and 4.1, we get the following.
Lemma 4.2. Let Λ be a chain. Then for anyx∈F(Λ), there exists a least (with respect to the inclusion)subset X of Λ such thatx∈F(X); this subset is finite.
We denote by supp(x) the subset given by Lemma 4.2, and we call it thesupport ofx.
Notation 4.3. For a chain Λ, well-ordered subsetsX andY of Λ such that otpX≤ otpY, andx∈F(X), we setx[Y /X] =F(eX,Y)(x), whereeX,Y denotes the unique embedding fromX intoY whose range is a lower subset ofY.
Hencex[Y /X] belongs toF(Y), for allx∈F(X).
Lemma 4.4. Let Λbe a chain and let X,Y be well-ordered subsets ofΛsuch that otpX ≤otpY andX∩Y is a lower subset of bothX andY. Thenx[Y /X] =x, for allx∈F(X∩Y).
Proof. As the setZ=X∩Y is a lower subset of bothXandY, the homomorphism F(eZ,X) (resp.,F(eZ,Y)) is the inclusion map from F(Z) into F(X) (resp.,F(Y)).
In particular,x=F(eZ,X)(x) =F(eZ,Y)(x). Therefore,
x[Y /X] =F(eX,Y)(x) =F(eX,Y)◦F(eZ,X)(x) =F(eZ,Y)(x) =x.
We are now reaching a crucial lemma.
Lemma 4.5 (Interpolation Lemma). Let Λ be a chain, let X, Y be finite subsets ofΛ, and lethx,yi ∈F(X)×F(Y). Ifx≤y, then either there existsz∈F(X∩Y) such that x≤z≤y or (Y 6⊆X and cmin(Y\X)≤y).
Proof. We shall denote byπkl the canonical map fromRlS(Λ) ontoRkS(Λ), for all natural numbers k ≤l. Put m = rkxand n= rky. Observe that supp(x) ⊆X and supp(y) ⊆Y. We argue by induction onkxk+kyk. If either supp(x) ⊆Y or supp(y) ⊆X then either z = x or z = y belongs to F(X∩Y) and satisfies the inequalities x≤ z ≤ y, so we are done. So suppose that supp(x) 6⊆ Y and supp(y)6⊆X. In particular, X 6⊆Y and Y 6⊆X, and both supp(x) and supp(y) are nonempty. We putξ= min(Y \X).
Suppose that m = n = 0, that is, x,y ∈ S(Λ). Pick i ∈ supp(x). As ci≤x≤y, we obtain that eithery=a∨b(a contradiction, as then supp(y) =∅) ory∈ {cj,a∨cj,b∨cj} for somej≥i. If i=j, then supp(x) = supp(y) ={i}, a contradiction. Ifi < j, thenξ=j and socξ≤y.
Suppose now that m < n. Then x ≤ y means that x ≤ πmn(y) (use (3.2)).
As πmn(y) has support contained in Y and rank at most m, it follows from the induction hypothesis that either cξ ≤ πmn(y) (thus, a fortiori, cξ ≤ y) or there existsz∈F(X∩Y) such thatx≤z≤πmn(y) (thus,a fortiori,z≤y).
So suppose from now on thatm >0 (i.e.,x∈/S(Λ)) andm≥n. Ifx=W
i<kxi wherek≥2 and eachxi has support contained inX and complexity less thankxk, then we apply the induction hypothesis to each inequality xi ≤ y, for i < k. If cξy, then for alli < k, there existszi∈F(X∩Y) such thatxi≤zi≤y. Hence x≤z≤y, wherez =W
i<kzi belongs toF(X ∩Y). This reduces the problem to the case wherex=⊲⊳(u,v,w), wherehu,v,wiis a non-diagonal triple of elements ofRm−1S(X) of complexity less thankxk.
Ifm > n, then, as⊲⊳(u,v,w) =x≤ywith supp(x)6⊆Y,u,v,w∈Rm−1S(X), and y ∈ Rm−1S(Y), it follows from (3.1) that either u ≤ y or w ≤ y. If, for example,u≤y, then, by the induction hypothesis, eithercξ ≤y(in which case we are done) or there existsz ∈F(X∩Y) such thatu≤z ≤y. In the second case, x≤z≤y. The argument is similar in casew≤y.
The remaining case ism=n >0. As⊲⊳(u,v,w) =x≤y with supp(x)6⊆Y, u,v,w∈Rm−1S(X), andy∈RmS(Y), it follows from (3.1) that eitheru≤πm−1m (y) or w ≤ πmm−1(y). If u ≤ πm−1m (y), then, by the induction hypothesis, either cξ ≤ πm−1m (y) (thus, a fortiori, cξ ≤y) or there exists z ∈ F(X∩Y) such that u ≤z ≤πmm−1(y) (in which case x≤ z ≤ y). The case where w ≤ πm−1m (y) is
similar.
Lemma 4.6. Let Λ be a chain and let X be a nonempty subset of Λ admitting a supremum, say, ξ, inΛ. Then cξ is the supremum of {ci|i∈X} inF(Λ).
Proof. Letx ∈F(Λ) such that ci ≤x for alli ∈X, we prove that cξ ≤ x. Put n= rkx and y =πn0(x). Leti ∈X. From ci ≤xand ci ∈S(Λ) it follows that ci≤y. This holds for alli∈X, hence, ascξis clearly the supremum of{ci|i∈X} inS(Λ), we obtain that cξ ≤y. Therefore,cξ ≤x. Now we can state the main technical result of the paper. It says thathcξ |ξ < ω1i is the least non-eventually constant isotoneω1-sequence inF(ω1) modulo the closed unbounded filter onω1.
Theorem 4.7. Let σ = hxξ|ξ < ω1i be an isotone ω1-sequence of elements of F(ω1). Then either σ is eventually constant or there exists a closed unbounded subsetC of ω1 such that cξ≤xξ for allξ∈C.
Proof. Assume that σ is not eventually constant. We put Xξ = supp(xξ) and nξ =|Xξ|, for allξ < ω1. Soxξ =x′ξ[Xξ/nξ], for somex′ξ∈F(nξ). As all setsXξ
are finite, it follows from the ∆-Lemma (see [1, Lemma 22.6]) that there are an uncountable subset I of ω1 and a finite subset X of ω1 such that Xξ∩Xη =X for all distinct ξ, η ∈ I. We may further assume without loss of generality that there aren < ω and x∈F(n) such thatnξ =n andx′ξ =x, for allξ∈I. Hence xξ =x[Xξ/n], for allξ∈I. Asσis isotone but not eventually constant, it follows
6 F. WEHRUNG
thatX is a proper subset ofXξ, for allξ∈I. PutYξ =Xξ\X. Defineρ(ξ) as the least element ofYξ.
For subsetsU andV of ω1, letU < V hold, if u < v for allhu, vi ∈U×V. By further shrinking I, we might assume that X < Yξ, for all ξ ∈ I. In particular, observe thatX=Xξ∩Xη is a lower subset of bothXξ andXη, for allξ6=η inI.
Letξ < η in I and suppose that there existsz∈F(X) such thatxξ ≤z≤xη. Applying the embedding F(eXξ,Xη) to the inequality x[Xξ/n] ≤ z and using Lemma 4.4, we obtain the inequality x[Xη/n] ≤ z, so xη = x[Xη/n] = z, a contradiction since the left hand side has supportXη while the right hand side has the smaller support X. Therefore, asxξ ≤xη and by Lemma 4.5, we obtain the inequalitycρ(η)≤xη.
Hence, we may assume thatcρ(ξ)≤xξ for allξ∈I. It follows that
cρ(ξ)≤xξ, for allξ < ω1, (4.1) where we put
ρ(ξ) =_
(ρ(η)|η ∈I, η < ξ), for allξ < ω1.
As the range of ρ is unbounded, so is the range of ρ. Hence, asρ is a complete join-homomorphism from ω1 to ω1, the set C = {ξ < ω1 | ρ(ξ) =ξ} is a closed unbounded subset ofω1. It follows from (4.1) that the inequalitycξ ≤xξ holds for
allξ∈C.
The following corollary expresses thatF(ω1) fails a certain “monotone refinement property”.
Corollary 4.8. There are no positive integern and no finite collection of isotone ω1-sequenceshxi,ξ|ξ < ω1iof elements of F(ω1), for 0≤i≤n, such that
(1) x0,ξ= 0 andxn,ξ =cξ, for all large enoughξ < ω1. (2) xi,ξ≤cξ, for all i≤nand all large enoughξ < ω1.
(3) Either xi+1,ξ≤a∨xi,ξ orxi+1,ξ≤b∨xi,ξ, for alli < nand all ξ < ω1. Proof. We prove that for all i≤ n, there existsηi < ω1 such that xi,ξ ≤cηi for all ξ < ω1. We argue by induction oni. For i = 0 it holds by assumption, with η0 = 0. Suppose thatxi,ξ ≤cηi, for all ξ < ω1. Let ξ > ηi. Assume, for example, that xi+1,ξ ≤a∨xi,ξ; so xi+1,ξ ≤ a∨cηi. Observing that cξ a∨cηi, we get that cξ xi+1,ξ. As this holds for all ξ > ηi and by Theorem 4.7, we obtain that hxi+1,ξ|ξ < ω1iis eventually constant, and hence, by (2), below some cηi+1, therefore completing the induction step.
In particular, fori=n, we obtain that theω1-sequencehcξ|ξ < ω1iis eventually
dominated by the constantcηn, a contradiction.
Hence we get a negative extension property for posets.
Corollary 4.9. There are a poset measure µ: (ω1+ 1)×(ω1+ 1)→ F(ω1)such that µ(ω1,0) =a∨bbut there are no posetP containingω1+ 1, no poset measure µ:P ×P →F(ω1) extendingµ, no positive integer n, and no decomposition 0 = z0≤z1≤ · · · ≤zn =ω1inP such that eitherµ(zi+1, zi)≤a orµ(zi+1, zi)≤bfor alli < n.
Proof. Defineµ: (ω1+ 1)×(ω1+ 1)→F(ω1) by
µ(ξ, η) =
0, ifξ≤η, cξ, ifη < ξ < ω1, a∨b, ifη < ξ=ω1.
for allξ, η≤ω1.
It is straightforward to verify thatµis a poset measure onω1+ 1. Suppose thatP, µ, n, z0, . . . , zn satisfy the given conditions. We put xi,ξ = µ(ξ, zn−i), for all ξ < ω1. It is not hard, using the triangular inequality, to verify that the elements xi,ξ satisfy the assumptions (1)–(3) of Corollary 4.8, a contradiction.
As the following result shows, more “amenable” semilattices do satisfy a certain
“monotone refinement property”.
Proposition 4.10. LetSbe a distributiveh∨,0i-semilattice. IfSis either a lattice, or strongly distributive, or countable, then for alla,b∈S, every chainΛ, and every isotone Λ-sequencehci|i∈Λiof elements of S such thatci≤a∨bfor alli∈Λ, then there are isotone Λ-sequences hai|i∈Λi and hbi|i∈Λi of elements of S such that ai≤a,bi≤b, andci=ai∨bi, for alli∈Λ.
Proof. IfS is a lattice the conclusion is trivial: putai=a∧ci andbi=b∧ci, for alli∈Λ.
Now assume that S is strongly distributive. Denote by Ci the (finite) set of all maximal join-irreducible elements of S below ci, for all i ∈ Λ. Observe that Ci⊆ ↓Cj, for alli≤j in Λ. For every finite subsetI of Λ, denote byXI the set of all familieshhAi, Bii |i∈Iisuch that
(1) Ai⊆ ↓a,Bi⊆ ↓b, andAi∪Bi=Ci, for alli∈I.
(2) For all i≤j inI,Ai⊆ ↓Aj andBi⊆ ↓Bj.
We claim thatXIis nonempty, for every finite subsetIof Λ. We argue by induction on|I|. The conclusion is obvious forI =∅. ForI ={i}, putAi =↓a∩Ci and Bi = ↓b∩Ci. Now suppose that I = {i} ∪J, where i < j for all j ∈ J and J is nonempty. By induction hypothesis, there exists an element hhAk, Bki |k∈Ji in XJ. Putj = minJ,Ai=↓Aj∩Ci, and Bi=↓Bj∩Ci. It is straightforward to verify thathhAk, Bki |i∈Iibelongs toXI. This completes the induction step.
It follows that the set ΩI of all familieshhAi, Bii |i∈Λiof elements of the Carte- sian product Ω =Q
(P(Ci)×P(Ci)|i∈Λ) (whereP(X) denotes the powerset of a setX) whose restriction toIbelongs toXI is nonempty, for every finite subsetI of Λ. Endow Ω with the product topology of the discrete topologies on all (finite) setsP(Ci)×P(Ci). By Tychonoff’s Theorem, Ω is compact. Hence the intersection of all ΩI, forIa finite subset of Λ, is nonempty. LethhAi, Bii |i∈Λibe an element of that intersection. Then the collection of all elementsai=WAi andbi=WBi, fori∈Λ, satisfies the required conditions.
Assume, finally, that S is countable. Define an equivalence relation≡on Λ by i≡j iffci =cj, for alli, j∈Λ, and denote by [i] the ≡-equivalence class of i, for any i∈Λ. Puttingc[i]=ci makes it possible to replace Λ by Λ/≡. In particular, as S is countable, Λ becomes countable as well. Now write Λ = S
(Λn|n < ω), wherehΛn|n < ωiis an increasing sequence of finite subsets of Λ with|Λn|=n, for alln < ω. Denote by Yn the set of all familieshhal,bli |l∈Λnisuch that ai≤a, bi≤b,ci=ai∨bi, andi≤j implies thatai≤aj andbi≤bj, for alli≤jin Λn. Suppose that we are given an element ofYn as above, and denote byk the unique
8 F. WEHRUNG
element ofYn+1\Yn. Suppose, for example, that min Λn< k <max Λn, and denote byi(resp.,j) the largest (resp., least) element of Λn belowk(resp., abovek). As ck ≤ cj = aj ∨bj, there are a′ ≤ aj and b′ ≤ bj such that ck = a′∨b′. Put ak =ai∨a′ andbk =bi∨b′. Thenhhal,bli |l∈Λn+1iis an element ofYn+1. So every element ofYnextends to an element ofYn+1. The proof is even easier in case eitherk <min Λn ork >max Λn. Hence we have constructed inductively a family hhai,bii |i∈Λi whose restriction to Λn belongs to Yn, for all n < ω. Therefore, the elementsai andbi, fori∈Λ, are as required.
The “monotone refinement property” described above fails in F(ω1). Indeed, consider the isotone ω1-sequencehcξ|ξ < ω1itogether with the inequalities cξ ≤ a∨b, for ξ < ω1. Suppose that haξ |ξ < ω1i and hbξ|ξ < ω1i are isotone ω1- sequences in F(ω1) such that cξ = aξ ∨bξ while aξ ≤ a and bξ ≤ b, for all ξ < ω1. Set x0,ξ = 0, x1,ξ =aξ, andx2,ξ =cξ, for all ξ < ω1. Then the isotone ω1-sequenceshxi,ξ|ξ < ω1i, for i∈ {0,1,2}, satisfy (1)–(3) of Corollary 4.8 (with n= 2), a contradiction.
Nevertheless we do not know whether the monotone refinement property of S either implies or is implied by the statement that every S-valued poset measure extends to a V-measure.
References
[1] T. Jech, “Set Theory”, Academic Press, Harcourt Brace Jovanovich, New York - San Francisco - London, xi + 621 p, 1978.
[2] M. Ploˇsˇcica and J. T˚uma,Uniform refinements in distributive semilattices, Contributions to General Algebra10, Proceedings of the Klagenfurt Conference, May 29 – June 1, 1997. Verlag Johannes Heyn, Klagenfurt 1998.
[3] P. R˚uˇziˇcka, J. T˚uma, and F. Wehrung, Distributive congruence lattices of congruence- permutable algebras, preprint.
[4] J. T˚uma and F. Wehrung,Unsolvable one-dimensional lifting problems for congruence lattices of lattices, Forum Math.14, no. 4 (2002), 483–493.
[5] F. Wehrung,Forcing extensions of partial lattices, J. Algebra262, no. 1 (2003), 127–193.
CNRS, UMR 6139, D´epartement de Math´ematiques, BP 5186, Universit´e de Caen, Campus 2, 14032 Caen cedex, France
E-mail address: wehrung@math.unicaen.fr URL:http://www.math.unicaen.fr/~wehrung
