Lobachevskii Journal of Mathematics Vol. 13, 2003, 45 – 50

©Niovi Kehayopulu and Michael Tsingelis

ABSTRACT. In this paper we prove the following: If S is an ordered semigroup, then the set P(S) of all subsets of S with the multiplication ” ∘ ” on P(S) defined by ”A ∘ B := (AB] if A,B ∈P(S), A≠∅, B≠∅ and A ∘ B := ∅ if A = ∅ or B = ∅ is an le-semigroup having a zero element and S is embedded in P(S).

If (S,.,≤) is an ordered semigroup, for A ⊆ S, we define (A] := {t ∈ S ∣ t ≤ a for some a ∈ A}. For A = {a}, we write (a] instead of ({a}]. An element 0 of S is called the zero element of S if 0 ≤ x and 0x = x0 = 0 for all x ∈ S [1]. Let (S,.,≤), (T,∘,≼) be ordered semigroups, f : S → T a mapping from S into T.

The mapping f is called isotone if x,y ∈ S, x ≤ y implies f(x) ≤ f(y). f is called reverse isotone if x,y ∈ S, f(x) ≼ f(y) implies x ≤ y. [Each reverse isotone mapping is (1-1): Let x,y ∈ S, f(x) = f(y). Since f(x) ≼ f(y), we have x ≤ y. Since f(y) ≼ f(x), we have y ≤ x.] f is called a homomorphism if it is isotone and satisfies f(xy) = f(x) ∘ f(y) for all x, y ∈ S. f is called an isomorphism if it is onto, homomorphism and reverse isotone. S and T are called isomorphic if there exists an isomorphism between them [3]. S is embedded in T if, by definition, S is isomorphic to a subset of T, i.e., if there exists a mapping f : S → T which is homomorphism and reverse isotone [4]. An l-semigroup (: lattice ordered semigroup) is a semigroup S at the same time a lattice satisfying the conditions a(b ∨ c) = ab ∨ ac and (a ∨ b)c = ac ∨ bc for all a, b,c ∈ S [1]. By an le-semigroup we mean an l-semigroup having a greatest element ”e” (i.e. e ≥ a for all a ∈ S) [2]. We denote by P(S) the set of all subsets of S.

Theorem. Let (S,.,≤) be an ordered semigroup. We define a multiplication ” ∘ ” on P(S) as follows:

where

Then (P(S),∘,⊆) is an le-semigroup having a zero element and (S,.,≤) is embedded in (P(S),∘,⊆).

Proof. First of all, the set P(S) in non-empty. The multiplication ” ∘ ” on P(S) is well defined. Moreover, we have the following:

1) The multiplication ” ∘ ” on P(S) is associative. In fact:

Let A,B,C ∈P(S). If A = ∅ or B = ∅ or C = ∅, then (A ∘ B) ∘ C = ∅ and A ∘ (B ∘ C) = ∅, so (A ∘ B) ∘ C = A ∘ (B ∘ C).

Let A,B,C ∈P(S)∖{∅}. We have A ∘ B,B ∘ C ∈P(S)∖{∅}. Let now x ∈ (A ∘ B) ∘ C := ((A ∘ B)C]. Then x ≤ yc for some y ∈ A ∘ B, c ∈ C. Since y ∈ A ∘ B := (AB], we have y ≤ ab for some a ∈ A, b ∈ B. Then

so x ∈ (A(B ∘ C)] := A ∘ (B ∘ C). Similarly, A ∘ (B ∘ C)] ⊆ (A ∘ B) ∘ C.

2) (P(S),∘,⊆) is an le-semigroup:
Let A,B,C ∈P(S). Then A ∘ (B ∪ C) = (A ∘ B) ∪ (A ∘ C). Indeed:
If A = ∅, then A ∘ (B ∪ C) = ∅, A ∘ B = ∅, A ∘ C = ∅.

If B = ∅, then A ∘ (B ∪ C) = A ∘ C, (A ∘ B) ∪ (A ∘ C) = A ∘ C.

If C = ∅, then A ∘ (B ∪ C) = A ∘ B, (A ∘ B) ∪ (A ∘ C) = A ∘ B.

Let A,B,C ∈P(S)∖{∅}. We have

Since (AB], (AC] ⊆ (A(B ∪ C)], we have (AB] ∪ (AC] ⊆ (A(B ∪ C)].
Let now t ∈ (A(B ∪ C)]. Then t ≤ ax for some a ∈ A, x ∈ B ∪ C. If x ∈ B, then t ∈ (AB] ⊆ (AB] ∪ (AC]. If x ∈ C, then t ∈ (AC] ⊆ (AB] ∪ (AC].
Similarly, for any A,B,C ∈P(S)∖{∅}, we have (A ∪ B) ∘ C = (A ∘ C) ∪ (B ∘ C).
Finally, S is the greatest element and ∅ the zero element of P(S).

3) We consider the mapping

The mapping f is well defined. Moreover,

A) The mapping f is a homomorphism. Indeed:
Let a,b ∈ S. We have (a], (b] ∈P(S)∖{∅} (since a ∈ (a],b ∈ (b]).
Thus we have

Let a,b ∈ S, a ≤ b. Then f(a) := (a] ⊆ (b] := f(b).

B) The mapping f is reverse isotone: Let a,b ∈ S, f(a) ⊆ f(b). Then
a ∈ (a] ⊆ (b], and a ≤ b.

Remark. More generally, we have the following: If A, Bi ∈P(S), i ∈ I, then

In fact,

A) If A = ∅, then A ∘ (⋃ i∈I Bi) = ∅, and A ∘ Bi = ∅ for all i ∈ I, so ⋃ i∈I(A ∘ Bi) = ∅. Thus A ∘ (⋃ i∈I Bi) = ⋃ i∈I(A ∘ Bi).

B) If A≠∅, then

I) If ⋃ i∈I Bi = ∅, then A ∘ (⋃ i∈I Bi) = ∅. Since ⋃ i∈I Bi = ∅, we have Bi = ∅ for all i ∈ I, then A ∘ Bi = ∅ for all i ∈ I, and ⋃ i∈I(A ∘ Bi) = ∅. Then
A ∘ (⋃ i∈I Bi) = ⋃ i∈I(A ∘ Bi).

II) Let ⋃ i∈I Bi≠∅. We put J := {i ∈ I ∣ Bi = ∅}, K := {i ∈ I ∣ Bi≠∅}. Clearly I = J ∪ K and J ∩ K = ∅. If K = ∅, then I = J, Bi = ∅ for all i ∈ I, and ⋃ i∈IBi = ∅. Impossible. Thus K≠∅.

   α) Let J = ∅. Then I = K, Bi ≠ ∅ for all i ∈ I. Since A≠∅ and Bi ≠ ∅ for all i ∈ I, we have A ∘ Bi := (ABi] for all i ∈ I. Then ⋃ i∈I(A ∘ Bi) = ⋃ i∈I(ABi].

Besides, ⋃ i∈I(ABi] = (A(⋃ i∈I Bi)]. Thus we have

Since A≠∅ and (⋃ i∈I Bi)≠∅, we have A ∘ (⋃ i∈I Bi) = (A(⋃ i∈I Bi)]. Then, by (*), A ∘ (⋃ i∈I Bi) = ⋃ i∈I(A ∘ Bi).

   β) Let J≠∅. Then Bi = ∅ for all i ∈ I, ⋃ i∈IBi = ∅, and

Since A≠∅ and ⋃ i∈I Bi≠∅, we have

Since A≠∅ and Bi≠∅ for all i ∈ K, we have A ∘ Bi := (ABi] for all i ∈ K, and ⋃ i∈K(A ∘ Bi) = ⋃ i∈K(ABi]. Thus we have

Since Bi = ∅ for all i ∈ J, we have A ∘ Bi = ∅ for all i ∈ J, then ⋃ i∈J(A ∘ Bi) = ∅. Then

By (**) and (***), we have

Example. We consider the ordered semigroup

S = {x,y,z} defined by the multiplication and the figure below:

Applying the Theorem of this note, the ordered semigroup (S,.,≤) is embedded into the le-semigroup L = {a,b,c,d,e,f,g,h}, defined by the multiplication ”.”and the order ” ≤L” below:

≤L :={(a,a), (a,d), (a,e), (a,g), (b,b), (b,d), (b,f), (b,g), (c,d), (c,e), (c,f), (c,g), (d,d), (d,g), (e,e), (e,g), (f,f), (f,g), (g,g), (h,a), (h,b), (h,c), (h,d), (h,e), (h,f), (h,g), (h,h)}.

We give the covering relation ”≺’ and the figure of S.

≺={(a,d), (a,e), (b,d), (b,f), (c,e), (c,f), (d,g), (e,g), (f,g), (h,a), (h,b), (h,c)}.

The embedding is given by the mapping:

This research was supported by the Special Research Account of the University of Athens (Grant No. 5630).

References

UNIV. OF ATHENS, DEPT. OF MATHEMATICS;
HOME ADDRESS: NIOVI KEHAYOPULU, NIKOMIDIAS 18, 161 22 KESARIANI, GREECE

E-mail address: nkehayop@cc.uoa.gr

Received September 30, 2003