ON M-OPERATORS OF q-LATTICES

It is well known that every complete lattice can be considered as a complete lattice of closed sets with respect to appropriate closure operator. The theory of q-lattices as a natural generalization of lattices gives rise to a question whether a similar statement is true in the case of q-lattices. In the paper the so-called M -operators are introduced and it is shown that complete q-lattices are q-lattices of closed sets with respect to M -operators.


Introduction
The idea of introducing lattice-like structure on a quasiordered set is due to I. Chajda in [1].
Having a quasiordered set (A; Q) with a quasiorder relation Q (i.e.Q is both reflexive and transitive relation on A), denote by is known to be a partial order relation on A/E Q .To simplify notation we shall write ≤ instead of Q/E Q .

R. Halaš
A mapping χ : A/E Q −→ A with the property χ(B) ∈ B for each B ∈ A/E Q is called a q-function on A.
If for each B, C ∈ A/E Q there exist sup ≤ (B, C) and inf ≤ (B, C), then the triple (A, Q, χ) is called an L-quasiordered set.The equivalence class [a] E Q will be denoted simply by [a].
L-quasiordered sets give rise to lattice-like operations on A in the following manner [1]: Lemma 1.Let (A, Q, χ) be an L-quasiordered set.Let us define for x, y ∈ A the operations Then the algebra (A; ∨, ∧) satisfies the identities An algebra A = (A; ∨, ∧) satisfying the axioms of Lemma 1 is called a q-lattice.
Conversely, having a q-lattice A = (A; ∨, ∧), the relation Q on A defined by (x, y) ∈ Q iff x ∨ y = y ∨ y is a quasiorder relation, the so-called induced quasiorder on A.
Let us note that (x, y) The set SkA = {x ∈ A : x ∨ x = x} of all idempotent elements of A, the so-called skeleton of A, forms a lattice with respect to the induced operations ∨ and ∧; this lattice is called the induced lattice of a q-lattice A.
Hence a q-lattice A = (A; ∨, ∧) is a lattice if and only if A = SkA.
The set C(a) = {x ∈ A; a ∨ a = x ∨ x} for a ∈ A is called the cell of a.It is clear that every q-lattice is a disjoint union of cells and every cell contains exactly one element from the skeleton.
When visualizing a q-lattice A = (A; ∨, ∧), we firstly draw the lattice skeleton SkA and then we add the corresponding cells.For example, the diagram a b c d represents a q-lattice with a skeleton SkA = {a, c} and with two cells

M -operators
Since the join (the meet) of two (not necessarily distinct) elements of a q-lattice A is always a skeletal element, A is complete iff {a; a ∈ X} (or {a; a ∈ X}) exists for an arbitrary subset X of A.
By an operator on A we mean a mapping C : The set of all C-closed sets will be denoted by L(C).
The set P(A) can be quasiordered in a natural way as follows:

R. Halaš
Then ≤ is a quasiorder relation on P(A) and, moreover , P(A) is a q-lattice with respect to the operations with SkP(A) = P(M ).
P roof.Easy.
The q-lattice from Lemma 2 will be called a set-M -q-lattice on A. It is easy to see that set-A-q-lattice on A is just a set-lattice on A. (i.e.lettice of all subsets of A) We are ready to formulate our natural problem: Given a complete q-lattice A, does there exist an operator C on A and M ⊆ A such that the set L(C) of all C-closed sets on A is closed under the operations ∧ and ∨ (as introduced in Lemma 2) and the set-M -q-lattice L(C) is isomorphic to A?
In the following we give a positive answer to the above problem.Remember that an operator C : For a singleton a ∈ A, we shall write C(a) instead of C({a}).
We start from the following definition: otherwise.
On M -operators of q-lattices 123 The C M -closure does not have the properties (C1)-(C3) of a closure operator.Its properties are listed in the following proposition.
Proposition 1. C M -closure operator on A has the following properties for X, Y ⊆ A : (1) (1) is easily seen from the definition of C.
Since C is a closure operator on A, (2) follows from the fact that X ⊆ C(X) for each X ⊆ A.
Further we have this proves the property (3).
Let us verify the property (4).Since X ∩ M ⊆ M , we have by (1) C M (X ∩ M ) ⊆ M , and, moreover by (C3) and (C2) of the operator C.
Proposition 1 leads us to the following definition: for each X, Y ⊆ A, is called an M -operator on A.

R. Halaš
Let us note that for an M -operator C * on A, the set L(C * ) of all C * -closed sets is non-empty.Indeed, by (MC4) we have {C * (X) : X ⊆ M } ⊆ L(C * ), and, by (MC1) and (MC2) M ∈ L(C * ).
Next we will show that L(C * ) can be endowed by a set-M-q-lattice structure: Proposition 2. Let C * be an M-operator on A, let X α ∈ L(C * ), α ∈ Λ.Then L(C * ) is a complete q-lattice w.r.t.operations where ≤ is the quasiorder on A induced by ∧.Moreover , SkL(C * ) = {X ∈ L(C * ) : X ⊆ M }.
P roof.Firstly we have to prove that the operations are well-defined, i.e. that X α ∩ M ∈ L(C * ) whenever X α ∈ L(C * ) for each α ∈ Λ.By (MC2) we have Conversely, X α ∩ M ⊆ X α for each α ∈ Λ, hence using (MC1) and (MC3) one gets for each α ∈ Λ.But this yields also verifying the closedness of the set X α ∩ M.
The operation ∧ on L(C * ) is then well defined and induces a quasiorder relation ≤ on L(C * ) as follows: We show that X α ∩ M is the greatest lower bound of X α ′ s w.r.t.induced quasiorder.Indeed, let X ∈ L(C * ) and suppose that X ≤ X α for each α It is immediately seen that X α is the least upper bound of X α ′ s w.r.t.≤ and, altogether, L(C * ) is a complete q-lattice.

R. Halaš
Let us verify that the mapping φ : L −→ L(C M ) defined by Injectivity of φ is easily seen from its definition, surjectivity then yields from the fact that the elements of L(C M ) are of the form L Sk (x) for x ∈ SkL or L Sk (y ∨ y) ∪ {y} for y ∈ SkL.Now let x, y ∈ L. To verify that φ is a homomorphism, we distinguish three cases: By the definition of join in L(C M ) we have To prove the converse inclusion, we have to show that If Y = L Sk (z) for some z ∈ SkL, we get i.e. x ≤ z, y ≤ z, and since SkL is the lattice, x ∨ y ≤ z.But then In the remaining case, we have Y = L Sk (u ∨ u) ∪ {u} for some u ∈ SkL.This yields x ≤ u ∨ u, y ≤ u ∨ u and hence x ∨ y ≤ u ∨ u.Finally, we get veryfying that, in the Case 2, φ is ∧-preserving.
The join of φ(x) and φ(y) is of the form Similarly as in the Case 1, we have to prove