ON COMPLEMENTATION PROBLEM IN THE LATTICE OF L-TOPOLOGIES

In this paper, we study the lattice structure of the lattice FX of all L-topologies on a given nonempty set X . It is proved that the lattice FX is complemented and dually atomic when X is any nonempty set and membership lattice L is a complete atomic boolean lattice. Further we introduce the concept of limit point in the membership lattice and prove that if membership lattice L has a limit point, then for any nonempty set X , the lattice FX is not complemented.


Introduction
Lattice theory and topology are two related branches of mathematics, each influencing the other.Many authors have already undertaken the study of the lattice structure of all topologies on a given set.Birkhoff [1] has described comparision of two topologies and noted that the set that the lattice of topologies on an arbitrary set is complemented.An independent proof that the lattice of topologies is complemented is given by Van Rooij [12].Analogously the lattice structure of the set of all L-topologies on a given set came into interest.Johnson [5,6] has investigated lattice structure of the set of all L-topologies on a given set X and proved that this lattice is complete, atomic but not modular, not complemented and not dually atomic in general.
In this paper, we study the lattice structure of the lattice F X of all L-topologies on a given nonempty set X.It has been proved that if membership lattice L is a complete atomic boolean lattice, then for any non-empty set X, the lattice F X is isomorphic to the lattice S X×Y of all (classical) topologies on X × Y , where Y is the set of all atoms in L. Further we introduce the concept of limit point in the membership lattice and prove that if membership lattice L has a limit point, then for any nonempty set X, the lattice F X is not complemented.

Preliminaries
Throughout this paper, X stands for a nonempty set, L for a bounded lattice with the least element 0 and the greatest element 1 and F X stands for the lattice of all L-topologies on X.The constant function in L X , taking value α is denoted by α and x γ where γ( = 0) ∈ L, denotes the Definition 2.4.Let L be a bounded lattice with the least element 0 and the greatest element 1.
An element l( = 1) ∈ L is said to be limit point of L if there exists a subset S ⊂ L such that (i) S = l.
(iii) no element of S can be expressed as arbitrary join or finite meet of members of L \ S.
(iv) if d ∈ S can not be expressed as arbitrary join or finite meet of members of S \ {d}, then Then we say that S is a limit set of l.
Remark 2.5.Clearly, 0 can not be a limit point of L. Then S = {a, b, c} is a limit set of h and h is a limit point of L.
Thus, every element l( = 0, 1) ∈ L is a limit point of L.
Example 2.8.A finite chain has no limit point.
Example 2.9.Let A = {a, b, c} and L = P(A).Then the lattice L, ⊆ has no limit point.

Complementation-I
Theorem 3.1.Let X be a non-empty set and L be a bounded latttice with the least element 0 and the largest element 1.If L has atleast one limit point, then F X is not complemented.
Proof.Let x ∈ X, l ∈ L be a limit point of L and S be a limit set of l.
respect to finite meet and arbitrary join.
We claim that F has no complement.If possible, let F be a complement of F.
Since S is limit set of l and L X = F∨F , therefore for each α ∈ S there exists L-subset g α ∈ F such that g α (x) = α.Then α∈S g α ∈ F and α∈S g α (x) = l.
(iii) no element of S can be expressed as arbitrary join or finite meet of members of L \ S.
(iv) if d ∈ S can not be expressed as arbitrary join or finite meet of members of S \ {d}, then then the lattice F X may or may not be complemented.
Example 3.3.Consider the lattice (L, ≤), where L = [0, 1] and '≤' is the usual relation of 'less than or equal to' on numbers.Then S = {x : .9< x < 1} ⊂ L satisfies all the conditions given in the remark 3.2.The lattice F X is not complemented since every element l( = 0, 1) ∈ L is a limit point of L and then the result follows by theorem 3.1.Let F be an arbitrary L-topology in F X such that F = L X , {0, 1}.
Consider the element x ∈ X and define the sets : There are two cases : In this case, define F x = {g ∈ L X : g(y) = 1 and g(x) = λ where λ ∈ B x }.
In this case, define In the similar way, define F y .
Let A = {0, 1} ∪ F x ∪ F y and F be the L-topology generated by the set A.
Since no element a( = 1) ∈ L can be written as finite meet or arbitrary join of members of L \ {a}, it follows that F ∧ F = {0, 1}.
Clearly, either For any λ ( = 0) ∈ L, either λ ∈ A x or λ ∈ B x and hence the following four cases arise : Case 1 : Then x λ ∈ F.
Remark 3.5.If the membership lattice L has no limit point, then the lattice F X may or may not be complemented.
Example 3.6.Let X be a non-empty set and L = {0, 1}.Then L has no limit point and F X is complemented.
Example 3.7.Let X = {x, y, z} and L be the membership lattice determined by the Hasse diagram given below: Clearly, L has no limit point.
Consider the L-topology If F has a complement say F , then L X = F ∨ F and f (x) = d for any f ( = 0, 1) ∈ F, it follows that there must exist some L-subsets say g 1 , g 2 ∈ F such that g 1 (x) = a and g 2 (x) = b.
Hence F X is not complemented.(iii) no element of S can be expressed as arbitrary join or finite meet of members of L \ S.
(iv) if l ∈ S can not be expressed as arbitrary join or finite meet of members of S \ {l}, then since L \ S is closed with respect to finite meet and arbitrary join.
We claim that F has no complement.If possible, let F be a complement of F.
Since L X = F ∨ F and by the definition of the set S, for each α ∈ L there exists L-subset Hence F X is not complemented.

Complementation-II
Throughout this section, L stands for a complete atomic boolean lattice and P(A) for power set of the set A. Since every complete atomic boolean lattice is isomorphic to some power set of some set, namely the set of all of its atoms.Assume that L is isomorphic to the power set algebra (P(Y ), ⊆), where Y is the set of all atoms in L.
Φ is one-one : Let f , g ∈ L X such that f = g ⇒ there exists some x ∈ X such that f (x) = g(x).
Sine L is atomic, there exists an atom say y ∈ Y such that f (x) ≥ y but g(x) y ⇒ (x, y) Φ is onto : Let A be any subset of X × Y .Consider the L-subset f ∈ L X defined by f (x) = y∈Y y such that (x, y) ∈ A. Then Φ( f ) = A.
Φ preserves arbitrary join : Let (x, y) Remark 4.3.Let F be an L-topology on X.
Similarly, it can be shown that if F is a (classical) topology on X ×Y , then F is an L-topology on X.
Thus F is an L-topology on X ⇐⇒ F is a (classical) topology on X ×Y .
Theorem 4.4.The map Ψ : F X → S X×Y , defined by Ψ is onto : Since the map Φ (defined in theorem 4.1) is onto, for any topology S ∈ S X×Y , consider S ♠ = { f ∈ L X : Φ( f ) ∈ S}.Then S ♠ is an L-topology and Ψ (S ♠ ) = S.
Remark 4.5.It has been proved that the lattice of all (classical) topologies is complete, atomic, dually atomic, complemented but neither modular nor distributive in general.Therefore, it follows that if membership lattice L is a complete atomic boolean lattice, then for any non-empty set X, the lattice of all L-topologies F X is complete, atomic, dually atomic, complemented but neither modular nor distributive in general.

Conclusion
In this paper, we have determined the lattice structure of the lattice F X on a non-empty set X when membership lattice L is a complete atomic boolean lattice.Complementation problem and lattice structure of the lattice F X on a non-empty set X when membership lattice L is other than a complete atomic boolean lattice, will be discussed in future papers.

Example 2 . 6 .
Consider the lattice determined by the Hasse diagram given below :

Example 3 . 4 .
Let X = {x, y} and L = {1 − 1/n : n ∈ N} ∪ {1} be the lattice determined by the Hasse diagram given below: Then S = {1 − 1/n : n ≥ 10000} satisfies all the conditions given in the remark 3.2.But the lattice F x is complemented as shown below :

Theorem 3 . 8 .
Let X be a non-empty set and L be a bounded lattice with the least element 0 and the largest element 1.If there exists some element d( = 0, 1) ∈ L and a finite subset S ⊂ L such that (i) S = d.(ii) d / ∈ S.