Tkadlec, Josef

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Institute of Mathematics AS CR, v. v. i.

Mathematica SlovacaJosef TkadlecA note on a function representation of orthomodular posetsMathematica Slovaca, Vol. 39 (1989), No. 1, 27--29Persistent URL: http://dml.cz/dmlcz/128653Terms of use:© Mathematical Institute of the Slovak Academy of Sciences, 1989Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document must containthese Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech Digital MathematicsLibrary http://project.dml.czMath. Slovaca 39, 1989, No. 1 , 27—29 A NOTE ON A FUNCTION REPRESENTATION OF ORTHOMODULAR POSETS JOSEF TKADLEC In the papers [1], [2] the authors give axioms for a set of functions to characterize an orthomodular poset with "enough" states. In the attempt to improve the characterization, D. S t ro jewski [3] offers (seemingly) more lucid conditions and derives several consequences. However, his crucial auxili-ary result does not seem to be correct. In this note we construct the appropriate counterexample and give the correct version of the representation theorem. Let us first review the basic notions. By an orthomodular poset we mean a triple (L, ^ , ') such that (a) (L, ^ ) is a partially ordered set with a greatest element 1, (b) the operation': L -> L is an orthocomplementation, for every a, b e L we have a" = a and a ^ b implies b' ^ a', (c) the least upper bound exists for every pair of orthogonal elements in L, (d) b — a v (b A a') for every a, beL with a ^ b. By a state we mean a function s: L -• [0, 1] such that s(l) = 1 and s(a v b) = = s(a) + s(b) for each pair of the orthogonal elements a, beL. Recall finally that a subset of states is called full if for every a, b e L, a ^ b the subset contains such a state s that s(a) ̂  s(b). Theorem. Let P be a nonvoid set and let P cz [0, \]s for a set S. Let further P satisfy (1) (V feP , f*0) (3 seS) f ( s )>0 .5 , (2) (V fGP) l - fGP , (3) (VfgeP,f+g^l)f+geP, (4) (Vf geP,f+ g ^ \)(3heP, h >f, g)(VkeP, k >f g)k > h. Then (P, ^ , ' ) with the pointwise ordering ^ and the orthocomplementation given byf = 1 — fis an orthomodular poset with a full set of states S = {s: P -> [0, 1]; (VfeP)s(f) =f(s)}. Moreover, f^g—f+g for the orthogonal elements fgzp-Conversely, each orthomodular poset L with a full set S of states is orthoisomor-phic to some subset P a [0, l]s satisfying axioms (1)—(4) with the ordering and the orthocomplementation given as above. 27 Proof. One can easily verify that (F, ^ , ') is a partially ordered set with orthocomplementation. Since the set P is nonempty, there is a n / e P and we have 1 -feP(thecondition (2)) and 1 = / + (1 -f)eP(thecondition (3)). Let / geP be orthogonal. T h u s / + g ^ 1 and according to the condition (4) there exists/v geP. By the condition (3) we h a v e / + geP. Since/+ g ^ f v g, the conditions (2), (3) give (f+g)- (fvg) = 1 - ( ( / v g) + ( / + g)')eP. As fur-ther ( / + g) - ( / v g) ^ min ( / g) ^ min ( / 1 - / ) ^ 0.5, we obtain ( / + g) -— ( / v S) = 0 (lhe condition (1)). Hence /v g = / + g. L e t / geP and f^g. T h e n / g ' e P are orthogonal and a l s o / ( / v g')'eP are orthogonal. Hence we infer that / v (g A / ' ) = / v (g' v / ) ' = / + (1 — -(0-*)+/)) = *. Let conversely L be an orthomodular poset with a full set S of states. Put P = {/*e[0, l ] 5 ; (VsES)fa(s) = s(a\ aeL). Then av-*fa is obviously an orthoisorriorphism between L and P with respect to the respective orderings and orthocomplementations. Hence the axioms (2), (4) hold. L e t / e F , / f l ^ 0.5. Then 0.5 ^f'a and therefore/ </ f l \ Thus a^a \x\ view of the orthoisomorphism. But it means that a = 0, which gives/ = 0. This establishes the condition (1). L e t / , / G F w i t h / + / ^ 1. T h e n / i s orthogonal t o / and making use of the orthoisomorphism again we see that a is orthogonal tob . But it means that (VsE-S)s(a v b) = s(a) + s(b). H e n c e / + / =-favheP. This completes the proof. In the paper [3] the author states the same representation theorem without the condition (4). It turns out, however, that such a theorem is no longer valid as the following example shows (It should be noted that this example disproves also other results in [3] — Theorems 1, 2, etc.). Example . Put P = l(a0, au a2, a3, a4)e<0, - , 1> x {0, l}4; 4 U Y a{ is odd if and only if a0 = -\ • i = i 2) Then P satisfies the axioms (1)—(3) and it is not an orthomodular poset. Proof . The axioms (1)—(3) verify easily. P is not an orthomodular poset not because the orthogonal elements a = ( - , 1, 0, 0, 0 J, b = ( - , 0, 1, 0, 0 J do have the least upper bound in P (we have three incomparable elements in P greater than a, b). REFERENCES [1] MACZYŃSKI, M. J.: On some numerical characterization of Boolean algebras, Colloquium Mathematicum 27, 1973, 207—210. [2] MACZYŃSKI, M. J. TRACZYK, T.: A characterization of orthomodular partially ordered sets admitting a full set of states, Bull. Acad. Polon. Sci., Ser. Sci. Math., Astronom., Phys. 21, 1973, 3—8. [3] STROJEWSKI, D.: Numerical representations of orthomodular lattices and Boolean algebras with infinite operations, Bull. Acad. Polon. Sci., Ser. Sci. Math., Astronom., Phys. 33, 1985, 341 348. Received November 11, 1986 Katedra matematiky Elektrotechnické fakulty ČVUT Suchbátarova 2 166 27 Praha 6 О ФУНКЦИОНАЛЬНОМ ПРЕДСТАВЛЕНИИ ОРТОМОДУЛЯРНЫХ ЧАСТИЧНО УПОРЯДОЧЕННЫХ МНОЖЕСТВ 1о$еГТкас11ес Р е з ю м е М. И. Мочиньски и Т. Трачик установили условия для ортомодулярного частично упоря­доченного множества чтобы оно имело «достаточное» количество состояний. Д. Строевски попытался улучшить эти условия, но это ему не совсем удалось. В этой статье находится контрпример и исправление теоремы Д. Строевского. 29 

A note on a function representation of orthomodular posets

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A note on a function representation of orthomodular posets

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