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 Authors : Raji George, T. P. Johnson
 Paper ID : IJERTV2IS1489
 Volume & Issue : Volume 02, Issue 01 (January 2013)
 Published (First Online): 30012013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Ultra LTopologies in the Lattice of LTopologies
LTopologies
Raji George
Department of Mathematics, St.Peters College, Kolenchery 682311,Ernakulam Dt ,Kerala State,India.
T. P. Johnson
Applied Sciences and Humanities Division,School of Engineering, Cochin University of Science and Technology, Cochin22,
Kerala State, India.
Abstract
   
   
We study the principal and nonprincipal ultra Lfilters on a nonempty set X,where L is a completely distributive lattice with order reversing involution. Using this notion we study the topological properties of principal and nonprincipal ultra L topologies. If X has n elements and L is a finite pseudo complemented lattice or a Boolean lattice, there are n(n 1)mk principal ultra Ltopoogies, where m is the number of dual atoms and k is the number of atoms. If X is infinite, there are X principal ultra Ltopologies and X nonprincipal ultra
Ltopologies.
Keywords. L – filter, Principal and Nonprincipal Ultra LTopologies, Simple extension.
AMS Subject Classification. 54A40

Introduction
The Purpose of this paper is to identify the Ultra LTopologies in the lattice of LTopologies. For a given topology on X , T. P. Johnson [5] studied the properties of the lattice S,L of LTopologies defined by families of scott
 
 
 
 
continuous functions with reference to on X. In [5] Johnson has proved that S,L is complete, atomic and not complemented. Also he has showed thatS,L is neither modular nor dually atomic in general. In [3] Frolich proved that if X = n there are n(n 1) Principal ultra topologies in the lattice of topologies. In [8] A. K. Steiner studied some topological properties of the ultraspaces. In this paper we showed that if X = n and L is a finite pseudocompleted chain or a Boolean lattice, there are n(n 1)mk Principle Ultra L Topologies, where m and k are the number of dual atoms and atoms
in L respectively. If X is infinite, there are X Principal Ultra L Topologies and X Nonprincipal Ultra Ltopologies. Also we studied some topological properties of the Ultra L Topologies and characterise the T1, T2 Ltopologies.

Preliminaries
/
/
Let X be a non empty ordinary set and L = L( , , ,/ ) be a completely distributive lattice with the smallest element 0 and largest element 1((0 = 1) and with an order reversing involution a a/ . We identify the constant function with value by . The fundamental definition of Lfuzzy set theory and Lfuzzy topoogy are assumed to be familiar to the reader (in the sense of Chang [2] and Goguen [4]). Here we call Lfuzzy subsets as Lsubsets and L fuzzy topology as Ltopology. For a given topology on X, the family S,L of all Ltopologies defined by families of Scott continuous functions from (X, ) to L forms a lattice under the natural order of set inclusion. The least upper bound of a collection of Ltopologies belonging to S,L is the Ltopology generated by their union and the greatest lower bound is their intersection. The smallest and largest elements in S,L are denoted by 0s, and 1s, respectively.
In this paper, Lfilter on X are defined according to the definition given by A. K. Katsaras [6] and P. Srivastava and R. L. Gupta[7], by taking L to be the membership lattice, instead of the closed unit interval[0, 1].
Definition 2.1. A non empty subset U of LX is said to be an Lfilter
if
i.0 / U
ii.f, g U implies f g U and
iii.f U , g LX and g f implies g U .
An Lfilter is said to be an ultra Lfilter if it is not properly contained in any other Lfilter.
Definition 2.2. Let x X, L An Lpoint x is defined by x(y) =
( if y = x
where 0 < 1
0 if y x
Definition 2.3. In a filter U , if there is an L subset with finite support, then U is called a principal Lfilter.
Example 1: {f LX f x, where x is an Lpoint}.
Definition 2.4. In a filter U , if there is no L subset with finite support, then U is called a nonprincipal Lfilter.
{  }
{  }
{  }
{  }
Example 2: f LX f > 0 for all but finite number of points . Let f be a nonzero Lsubset with finite support. Then U (f ) LX defined by U (f ) = g LX g f is an Lfilter on X, called the Principal Lfilter at f .Every Lfilter is contained in an ultra Lfilter. From the definition it follows that on a finite set X, there are only Principal ultra Lfilters.

Ultra Ltopologies
An Ltopology F on X is an ultra Ltopology if the only Ltopology on X
strictly finer than F is the discrete Ltopology.
{  }
{  }
Definition 3.1. [9] Let (X, F ) be an Ltopological space and suppose that g LX and g / F . Then the collection F (g) = g1 (g2 g) g1, g2 F is called the simple extension of F determined by g.
Theorem 3.1. [9] Let (X, F ) be an Ltopological space and suppose that F (g) be the simple extension of F determined by g. Then F (g) is an Ltopology on X.
Theorem 3.2. [9] Let F and G be two Ltopologies on a set X such that G is a cover of F . Then G is a simple extension of F .
Theorem 3.3. [3] The ultraspaces on a set E are exactly the topologies of the form S(x, U ) = P (E {x}) U where x E and U is an ultrafilter on E not containing {x}.
Analogously we can define ultra Ltopologies in the lattice of Ltopologies according to the nature of Lattices. If it contains Principal ultra Lfilter, then it is called Principal ultra Ltopology and if it contains nonprincipal ultrafilter, it is called nonprincipal ultra Ltopology.
Theorem 3.4. [1] A principal Lfilter at x on X is an ultra Lfilter iff
is an atom in L.
Theorem 3.5. Let a be a fixed point in X and U be an ultra Lfilter
not containing a, 0 L. Define Fa = {f LX f (a) = 0}. Then
S = S(a, U ) = Fa U is an Ltopology.
Proof. Can be easily proved.
Theorem 3.6. If X is a finite set having n elements and L is a finite pseudo complemented chain or a Boolean lattice, there are n(n 1)mk Prin cipal ultra Ltopologies, where m and k are the number of dual atoms and atoms in L respectively. If k = m there are n(n 1)m2 ultra Ltopologies.
Illustration :
{ } { }
{ } { }

Let X = a, b, c , L = 0, , , 1 , a pseudo complemented chain. Here
is the atom and is the dual atom.
1
0
Figure 1:
{  }
{  }
Let S = S(a, U (b)) = {f f (a) = 0} {f f b}, S does not contain the Lpoints a, a , a1 Then S(a, U (b), a ) = S(a ) = simple extension of S by a = f (g a ) f, g S, a / S is an ultra Ltopology, sinceS(a1) is the discrete Ltopology. similarly
if S = S(a, U (c)), thenS(a ) is an ultra Ltopology. if S = S(b, U (a)), thenS(b )
if S = S(b, U (c)), thenS(b )
if S = S(c, U (a)), thenS(c )
if S = S(c, U (b)), thenS(c )
Number of ultra Ltopologies= 6 = 3 2 1 1 = n(n 1)m2, where n =
3, k = m = 1
{ }
{ }

Let X = a, b, c , L = [0, 1], then there is no ultra Ltopology, since [0, 1] has no dual atom and atom.

Let X = {a, b, c}, L =Diamond lattice {0, , , 1}
1
0
Figure2:
Here , are the atoms as well as the dual atoms. Let S = S(a, U (b)) =
{f f (a) = 0} {f f b} , does not contain the Lpoints a, a , a1. Then the simple extension S(a) contains the fuzzy point a also. Let S1 = S(a). Then the simple extension S1(a ) contains all Lpoints and hence it is discrete. So S(a) = S(a, U (b), a) is an ultra Ltoology. Sim ilarly the simple extension S(a ) = S(a, U (b), a ) is an ultra Ltopology. If S = S(a, U (b ) = {f f (a) = 0} {f f b } , Then the simple exten sions S(a) and S(a ) are ultra Ltopologies.That is corresponding to the
elements a and b there are 4 ultra Ltopologies. Similarly corresponding to the elements a and c , there are 4 ultra Ltopologies. So there are 8 ultra Ltopologies corresponding to a. Similaly there are 8 ultra Ltopologies cor responding to b and 8 ultra L topologies corresponding to c . Hence total ultra Ltopologies = 8 + 8 + 8 = 24 = 3 2 2 2 = n(n 1)m2, where n = 3, k = m = 2
4.Let X = {a, b, c}, L = P (X) = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}.1 =
{a}, 2 = {b}, 3 = {c}, 1 = {a, b}, 2 = {a, c}, 3 = {b, c}. Atoms are
1, 2, 3and dual atoms are 1, 2, 3
Let S = S(a, U (b1)) = {f f (a) = 0} {f f b1}, does not con tain the Lpoints a1, a2, a3, a1, a2, a3, a1. Let S1 = Simple extension of S by a1 denoted by S(a1).Then S1 contains more Lsubsets than S,but
1
1 2 3
1 2 3
0
Figure 3:
not discrete Ltopology. Let S2 = S1(a2) , Simple extension of S1 by a2. Then S2 contain more L subsets than S1 but not discrete Ltopology. Let S3 = S2(a3) , Simple extension of S2 by a3, which is a discrete Ltopology. Hence S2 = S1(a2) is an ultra Ltopology, which is the L topology generated by S(a1) and S(a2). Also Ltopology generated by S(a1) and S(a3) and Ltopology generated by S(a2) and S(a3) are ultra Ltopologies. That is if S = S(a, U (b1)), there are 3 ultra Ltopologies. Similarly if S = S(a, U (b2)), there are 3 ultra Ltopologies and if S = S(a, U (b3)), there are 3 ultra Ltopologies. So corresponding to the el ements a, b there are 9 ultra Ltopologies. Similarly corresponding to the elements a, c there are 9 ultra Ltopologies. Hence there are 18 ultraL topologies corresponding to the element a. Similarly corresponding to each element b and c there are 18 ultra Ltopologies. So total number of ultra Ltopologies = 54 = 3 2 3 3 = n(n 1)m2, n = 3, k = m = 3.
5.Let X = {a, b, c, d}, L = P (X) = {, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d},
{b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {b, c, d}, {c, d, a}, X} . Let {a} = 1, {b} =
2, {c} = 3, {d} = 4, {a, b} = 1, {a, c} = 2, {a, d} = 3, {b, c} = 4, {b, d} =
5, {c, d} = 6, {a, b, c} = 1, {a, b, d} = 2, {b, c, d} = 3, {c, d, a} = 4 If
S = S(a, U (b1)), there are 4 ultra L topologies. If S = S(a, U (b2))
If S = S(a, U (b3))
Figure 4:
If S = S(a, U (b4))
 
 
So corresponding to the elements a, b,there are 16 ultra Ltopologies. Simi larly corresponding to the elements a, c,there are 16 ultra Ltopologies and corresponding to the elements a, d, there are 16 ultraLtopologies. Hence there are 48 ultra Ltopologies corresponding to the element a. Similarly corresponding to each elements b, c and d, there are 48 ultra Ltopologies. So total number of ultra Ltopologies = 48 4 = 192 = 4 3 4 4 = n(n 1)m2, n = 4, k = m = 4. In general if X = n and L is a finite pseudo complemented chain or a Boolean lattice, there are n(n 1)mk ultra
Ltopologies where m and k are the number of dual atoms and number of atoms respectively. If k = m, it is equal to n(n 1)m2.
Remark 3.1. If L is neither a finite pseudo complemented Lattice nor a Boolean Lattice, we cannot identify the ultra Ltopologies in this way.
Example:
Let X = {a, b, c}, L = D12 = {1, 2, 3, 4, 6, 12}
1
2
1
1 2
0
Figure 5:
Here the atoms are 1 = 2, 2 = 3 and dual atoms are 1 = 4, 2 = 6 . If S = S(a, U (b1)) = {f f (a) = 0} {f f b1} ,Ltopology generated by S(a1) and S(a2) does not contain the Lpointa2. It is not a discrete Ltopology.So we cannot say that S(a1) is an ultra Ltopology.
   
   
Theorem 3.7. If X is infinite and L is a finite pseudo complemented chain or a Boolean lattice, there are X principal ultra Ltopologies and X non principal ultra Ltopologies.
Illustration:
If X is countably infinite, cardinality of X, X = 0. If X is uncountable,
 
 
X > 0
case 1: X is infinite and L is finite
Let X = {a, b, …..}, L = {0, , , 1} a pseudo complemented chain.LetS = S(a, U (b)) = {f f (a) = 0} {f f b}. S does not contain the fuzzy points a, a , a1. Here S(a ) = S(a, U (b), a ) is a principal ultra L topology since S(a1) is the discrete Ltopology, where S(a ) is the simple
 
 
   
   
extension of S by a . Similarly we can identify other Ltopologies. Hence corresponding to the element a, there are X 1 = X principal ultra L topologies. Similarly corresponding to each element b, c, d, …… there are X principal ultra Ltopologies So total number of principal ultra Ltopologies
 
 
= XX = X. If S = S(a, U ) = {f f (a) = 0} U , where U is a non principal ultra filter not containing a, 0 /= L. Then the simple extension of S by a = S(a ) = S(a, U , a ) is a nonprincipal ultra Ltopology since S(a1) is discrete Ltopology. So there are X non principal ultra L topolo
gies.
case 2: X and L are infinite
LetX = {a, b, c, ….}, L = P (X). There are X atoms and X dual atoms.
Number of Principal ultra Ltopologies corresponding to one element= XX(X
1) = X Hence total number of principal ultra Ltopologies= XX = X. Let S = S(a, U ) = {f f (a) = 0} U , where U is a nonprincipal ultra filter not containing a, 0 /= L . There are X nonprincipal ultra Lfilters . Corresponding to a there are XX = X nonprincipal ultra Ltopologies. So total number of nonprincipal ultra Ltopologies = XX = X.


Topological Properties
PRINCIPAL ULTRA LTOPOLOGIES
Let X be a nonempty set and L is a finite pseudo complemented chain. If S = S(a, U (b) = {f f (a) = 0} {f f y}, then a principal ultra Ltopology = S(a, U (b), a ) = S(a ) , which is the simple extension of S
by a = {f (g a ), f, g, S, a / S}, where a, b X, and are the
atom and dual atom in L respectively.
Let X be a nonempty set and L is a Boolean lattice. If S = S(a, U (b)) =
{f f (a) = 0} {f f b} where a, b X, is an atom, then a Principal
ultra Ltopology = S(a, U (b), aj ) = Ltopology generated by any(m 1),
S(ai) among m, S(ai), i = 1, 2, …, m, j = 1, 2, …, m, i j if there are m
dual atoms 1, 2, …m, where S(ai) = S(a, U (b), ai)
Theorem 4.1. Let X be a nonempty set and L is a finite pseudo complemented chain or a Boolean Lattice. Then every principal ultra L topology is L T0 but not L T1.
Example:
Let X is a non empty set
Let L is a finite pseudo complemented chain and a, b X, , are atom and dual atom in L respectively. Take two distinct L points a1, b. b is an open L subset contain b but not a1. SinceU (b) = {f f b}, any open set contain a1 must contain b. So S(a, U (b), a ) is L T0 but not L T1. Let L is a Boolean lattice and a, b X, is an atom and 1, 2, ….. are dual atoms in L. Take two distinct Lponints a1, b.b is an open L subset contain b but not a1 . Since U (b) = {f f b}, any open set contain a1
must contain b. So S(a, U (b), aj ) is L T0 but not L T1.
p/>
Definition 4.1. An Ltopological space (X, F ), F LX is called door
Lspace if every Lsubset g of X is either Lopen or Lclosed in F .
Example : Let X = {a, b} and L = {o, .5, 1}.Define f1(a) = 0, f1(b) = 0, f2(a) = 0, f2(b) = .5, f3(a) = 0, f3(b) = 1, f4(a) = .5, f4(b) = 0, f5(a) =
.5, f5(b) = .5, f6(a) = .5, f6(b) = 1, f7(a) = 1, f7(b) = 0, f8(a) = 1, f8(b) =
.5, f9(a) = 1, f9(b) = 1 . Let F = {f1, f9, f2, f3, f4, f5, f6}. Thenf7 and f8 are closed Lsubsets. So (X, F ) is a door Lspace. Let X = {a, b, c}, L = [0, 1] and the the Ltopology F = {0, Âµ{a}, Âµ{b,c}, 1}.Then (X, F ) is not a door L space sinceÂµ{b} is neither an L open set nor an Lclosed set. In a principal ultra Ltopology S(a, U (b), aj ) every L subset of X is either open or closed
if L is a finite pseudocomplemented chain or a Boolean lattice . So the principal ultra Ltopological space is a door L space.
Definition 4.2. An Ltopological space (X, F ) is said to be regular at an Lpoint a if for every closed L subset h of X not containing a, there exists disjoint open sets f, g such that a f and h g.(X, F ) is said to be regular Ltopology if it is regular at each of its Lpoints.
Theorem 4.2. Let X be a non empty set and L = P (X). Then the principal ultra Ltopology S(a, U (b), aj ) is not regular if X 3.
Example: let X = {a, b, c}, L = P (X) = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}.1 =
{a}, 2 = {b}, 3 = {c}, 1 = {a, b}, 2 = {a, c}, 3 = {b, c}. Atoms are
1, 2, 3 and dual atoms are 1, 2, 3 . Take = 1 in the principal ultra Ltopology S(a, U (b), a3). Consider a1. a3 is a closed L subset not con taining a1.Consider the open sets f, g such that f (a) = 1, f (b) = 1, f (c) = 1, g(a) = 2, g(b) = 0, g(c) = 0.f is an open set containing a1 andg is an open set containing a3 but f g /= 0. That is f and g are not disjoint.
Definition 4.3. An L topological space (X, F ) is said to be Normal if for every two disjoint closed L subsets h and k , there exists two disjoint open L subsets f, g such that h f and k g.
Theorem 4.3. Let X be a nonempty set and L = P (X). Then the principal ultra Ltopology S(a, U (b), aj ) is not a normal Ltopology if
X 3.
Example: Let X = {a, b, c}, L = P (X) = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, X}.1 =
{a}, 2 = {b}, 3 = {c}, 1 = {a, b}, 2 = {a, c}, 3 = {b, c}. Atoms are
1, 2, 3 and dual atoms are 1, 2, 3.Take = 1 in the principal ultra Ltopology S(a, U (b), a3).a2 and a3 are disjoint closed L subsets. There are no disjoint open L subsets containing a2 and a3.
NON PRINCIPAL ULTRA L TOPOLOGY
Let X be an infinite set and L is a finite pseudocomplemented chain. If
S = S(a, U ) = {f f (a) = 0} U where U is a non principal ultra Lfilter
not containing a, 0 L .Then the non principal ultra Ltopology=
S(a, U , a ) = S(a ),is the simple extension of S by a = {f (g a ), f, g,
S, a / S},where a, b X, is the dual atom in L.
Let X be an infinite set and L is a Boolean lattice. If S = S(a, U ),
Then a non principal ultra Ltopology S(a, U , aj ) = Ltopology generated by any(m 1), S(ai) among m, S(ai), i = 1, 2, …, m, j = 1, 2, …, m, i /= j if there are m dual atoms 1, 2, …m where S(ai) = S(a, U , ai). Here m can be assumed infinite values.
Theorem 4.4. Every non principal ultra Ltopology S(a, U , aj ) is
L T1.
Proof. Let a, b be any two distinct Lpoints. Since U is a nonprincipal ultra Lfilter,there exists L open sets containing each Lpoints but not the other.
Theorem 4.5. Every non principal ultra L topology S(a, U , aj ) is
L T2.
Proof. Let a, b be any two distinct Lpoints. Since U is a nonprincipal ultra Lfilter, we can find disjoint open sets f and g such that a f, b / f and b g, a / g
Theorem 4.6. Every non principal ultra Ltopology is a door Lspace. Proof. In S(a, U , a ) every L subset of X is either L closed or L open,
if L is a finite pseudo complemented chain . If L is a Boolean Lattice,in
S(a, U , aj ) every L subset is either L cosed or L open
Theorem 4.7. If X is an infinite set and L is a finite pseudo comple mented chain or a diamond lattice, then the non principal ultra Ltopology S(a, U , a ) is a regular Ltopology.
Proof. It is trival.
Theorem 4.8. Let X is an infinite set and L = P (X).Then the non principal ultra Ltopology S(a, U , aj ) is not a regular Ltopology.
Proof. Let X = {a, b, c….}, L = P (X). Let 1, 2, …. are atoms and 1, 2, … are dual atoms in L. Consider a1. Then there exists a closed L subset ai for some i not containing a1. But we cannot find disjoint open L subsets f and g such that f contains a1 and g contains ai.
Theorem 4.9. If X is infinite set and L is a finite pseudo complemented chain or a diamond lattice, the non principal ultra L topology S(a, U , a ) is a normal L topology.
Proof. It is trivial
Theorem 4.10. If X is an infinite set and L = P (X), then the non principal ultra L topology S(a, U , aj ) is not a normal Ltopology.
Proof. Let X = {a, b, c, ….}, L = P (X). Let 1, 2, … are atoms and 1, 2, ….. are dual atoms in L. Then there exists two closed L subsets ai and aj for some i and j . But there doesnot exists disjoint open L subsets f and g such that f contains ai and g contains aj .
Theorem 4.11. Let X is an infinite set and L is a finite pseudo com plemented chain or a Boolean lattice. An ultra Ltopology F is a T1 L topology if and only if it is a non principal ultra Ltopology.
Proof. Suppose that the ultra Ltopology F is a T1 L topology. We have to show that F is a non principal ultra Ltopology. F is a principal
ultra Ltopology implies F is not a T1 L topology. So we can say that F
is a T1 L topology implies F is a non principal ultra Ltopology.
Next assume that F is a non principal ultra Ltopology. Then by theorem
4.4 F is a T1 L topology.
Theorem 4.12. An Ltopology F on X is a T1 L topology if and only if it is the infimum of non principal ultra Ltoologies.
Proof. Necessary part
Any Ltopology finer than a T1 L topology must also be a T1 L topology. So a T1 L topology can be the infimum of only non principal ultra L topologies.
Sufficient part
Each non principal ultra Ltopology on X contains non principal ultra L filter. So there exists distinct L points a, b where a, b X; , Land L opensets f, g such that a f, b / f and a /, b g. This is also true in the infimum of any family of non principal ultra Ltopologies since every L points are closed in non principal ultra Lfilters. So infimum of any family of non principal ultra Ltopologies is a T1 L topology.
Theorem 4.13. Let X is an infinite set and L is a finite pseudo com plemented chain or a Boolean lattice . Then an ultra Ltopology is a T2 L topology if and only if it is a non principal ultra Ltopology.
Proof. Suppose that an ultra Ltopology is a T2L topology. This implies that the ultra Ltopology is a T1 L topology. Hence it is a non principal ultra Ltopology.
Conversely suppose that the ultra Ltopology is a non principal ultra L topology. Since a non principal ultra Ltopology contains a non principal ultra Lfilter, for any two distinct L points in the non principal ultra L topology there exists disjoint L open sets contains each L point but not the other. So it is a T2 Ltopology.

Mixed L topologies
In[8] Steiner studied the mixed topologies. Analogously we ca say that a mixed Ltopology on X is not a T1 L topology and does not have a principal representation. Thus a mixed Ltopology is the intersectionof a T1 L topology and a principal Ltopology.
The representation of a mixed Ltopology as the infimum of a T1 L
topology and a principal L topology need not be unique.
Example:
Let C = {ÂµAX A is finite} 0 , is a T1 L topology and and / be the principal Ltopologies given by = aX{b,c}S(a, U (b), aj ) ,
/ = aX{b}S(a, U (b), aj ) , is an atom and j s are dual atoms in L.
C = {ÂµAb A, X A is finite or f = 0} = C / is a mixed Ltopology.
Here c andc / /. That is the representation of a mixed topology as
the infimum of T1 L topology and principal Ltopology need not be unique.

Conclusion
In this paper we identified the principal and non principal ultra Ltopologies and studied some topological properties. Also we introduced mixed Ltopologies.

Future scope
Study some properties of mixed Ltopologies.

Aknowledgement
We would like to thank Dr. T. Thrivikraman, Former Head, Dept. of Math ematics, Cochin University of Science and Technology, Cochin, Kerala State, India for discssions and suggestions. The first author wishes to thank the University Grant Commission, India for giving financial support.
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(On FIP)
Department of Mathematics,
Cochin University of Science and Technology, Cochin27, India.