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Non-extendability of semilattice-valued measures on partially ordered sets
For a distributive join-semilattice S with zero, a S-valued poset measure on a poset P is a map m:PxP->S such that m(x,z) <= m(x,y)vm(y,z), and x <= y implies that m(x,y)=0, for all x,y,z in P. In relation with congruence lattice representation problems, we consider the problem whether such...
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Lenguaje: | eng |
Publicado: |
2005
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Materias: | |
Acceso en línea: | http://cds.cern.ch/record/903521 |
Sumario: | For a distributive join-semilattice S with zero, a S-valued poset measure on a poset P is a map m:PxP->S such that m(x,z) <= m(x,y)vm(y,z), and x <= y implies that m(x,y)=0, for all x,y,z in P. In relation with congruence lattice representation problems, we consider the problem whether such a measure can be extended to a poset measure m*:P*xP*->S, for a larger poset P*, such that for all a,b in S and all x <= y in P*, m*(y,x)=avb implies that there are a positive integer n and a decomposition x=z\_0 <= z\_1 <= ... <= z\_n=y in P* such that either m*(z\_{i+1},z\_i) <= a or m*(z\_{i+1},z\_i) <= b, for all i < n. In this note we prove that this is not possible as a rule, even in case the poset P we start with is a chain and S has size $\aleph\_1$. The proof uses a "monotone refinement property" that holds in S provided S is either a lattice, or countable, or strongly distributive, but fails for our counterexample. This strongly contrasts with the analogue problem for distances on (discrete) sets, which is known to have a positive (and even functorial) solution. |
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