Lobachevskii Journal of Mathematics

http://ljm.ksu.ru
ISSN 1818-9962

Vol. 25, 2007, 217–229 \begin{matrix}c \\ ◯ \end{matrix} R.R.Rajeswari, M.L.Thivagar, and S.A.Ponmani

R.RajaRajeswari, M.Lellis Thivagar, and S.Athisaya Ponmani

CHARACTERIZATION OF ULTRA SEPARATION AXIOMS VIA (1, 2)α-KERNEL

(submitted by M. A. Malakhaltsev)

Abstract. In this paper, we introduce the concept of weakly-ultra-separation of two sets in a bitopological space using (1, 2)α-open sets. The (1, 2)α-closure and the (1, 2)α-kernel are defined in terms of this weakly-ultra-separation. We also investigate the properties of some weak separation axioms like ultra-T_{0}, ultra-T_{1}, and ultra-R_{0}.

1. Introduction

In 1962, Kelly initiated the study of a triple (X_{1}, τ_{1}, τ_{2}), where X is a non empty set and τ_{1}, τ_{2} are topologies on X. The notion of (1, 2)α-open sets [7] in a bitopological space was introduced in 1991. In this paper, we define that a set A is weakly-ultra-separated from B if there exists a (1, 2)α-open set G containing A such that G∩B = ∅. Using this concept, we define the (1, 2)α-closure and the (1, 2)α-kernel. We also define the (1, 2)α-derived set and the (1, 2)α-shell of a set A of a bitopological space X. We also offer some new characteristics of the low separation axioms defined and developed in [5] and [10] with respect to (1, 2)α-open sets.

2000 Mathematical Subject Classification. 54B05,54D05.

Key words and phrases. (1, 2)α-kernel and weakly-ultra-separated.

Supported by UGC, New Delhi, INDIA.

218 R.R.RAJESWARI, M.L.THIVAGAR, AND S.A.PONMANI

2. Preliminaries

In this section, let us recall some definitions which are useful in the following sequel.

Definition 2.1. A subset A of a topological space (X, τ_{1}) is called an αopen set [9] if A ⊂ τ_{1}-int(τ_{1}-cl(τ_{1}-int(A))), where τ_{1}-int(A) and τ_{1}-cl(A) represent the interior and closure of A with respect to τ_{1}.

Definition 2.2. A topological space is α-symmetric [2] if for x, y ∈ X, x ∈ α(cl{y}) implies y ∈ α(cl{x}).

Hereafter throughout this paper (X, τ_{1}, τ_{2}) and (Y, σ_{1}, σ_{2}) (or simply X and Y) denote bitopological spaces on which no separation axioms are assumed unless explicitly stated otherwise.

Definition 2.3. A subset A of X is called [7] (i) τ_{1}τ_{2}-open if A ∈ τ_{1} ∪ τ_{2}, (ii) τ_{1}τ_{2}-closed if A^{c} ∈ τ_{1} ∪ τ_{2}.

Definition 2.4. Let A be a subset of a space X. Then τ_{1}τ_{2}-closure [7] of A is denoted by τ_{1}τ_{2}-cl(A) and is defined as the intersection of all τ_{1}τ_{2}-closed sets containing A.

Definition 2.5. A subset A of X is called (1, 2)α-open [7] if

A ⊆ τ_{1}-int(τ_{1}τ_{2}-cl(τ_{1}-int(A))).

The set of all (1, 2)α-open sets is denoted by (1, 2)αO(X). The complement of a (1, 2)α-open set is called a (1, 2)α-closed set. The (1, 2)αclosure of a set A of X is denoted by (1, 2)α-cl(A) and is defined as the intersection of all (1, 2)α-closed sets containing A. The family of (1, 2)α-closed sets is denoted by (1, 2)αCL(X).

Definition 2.6. A function f : X → Y is called a (1, 2)α-irresolute map [10] if the inverse image of every (1, 2)α-open set in Y is a (1, 2)α-open set in X.

Definition 2.7. A subset A of X is called a (1, 2)αg-closed set [10] if (1, 2)αcl(A) ⊆ U, where A ⊆ U and U ∈ (1, 2)αO(X).

Definition 2.8. (i) A bitopological space X is called an ultra-T_{0} space [10] if and only if for each x, y ∈ X such that x ≠ y, there exists a (1, 2)α-open set containing x but not y or a (1, 2)α-open set containing y but not x.

ULTRA SEPARATION AXIOMS VIA (1, 2)α-KERNEL 219

(ii) A bitopological space X is called an ultra-T_{1} [10] space if and only if for each x, y ∈ X such that x ≠ y, there exists a (1, 2)α-open set containing x but not y and a (1, 2)α-open set containing y but not x.

(iii) A subset A of X is called a (1, 2)α-difference set [10] (briefly (1, 2)αD-set) if there exist two (1, 2)α-open sets U_{1} and U_{2} such that A = U_{1}/U_{2} and U_{1} ≠ X.

(iv) A bitopological space X is called an ultra-D_{1} [10] space if for each x, y ∈ X and x ≠ y there exists two (1, 2)αD-sets G_{1} and G_{2} such that x ∈ G_{1}, x ∉ G_{2} and y ∈ G_{2}, y ∉ G_{1}.

Definition 2.9. A bitopological space X is said to be an ultra-R_{0} space [5] if every (1, 2)α-open set G contains the (1, 2)α-closure of each of its singleton.

Definition 2.10. Let X be bitopological space and x ∈ X. Then a subset N_{x} of X is called a (1, 2)α-nbd ((1, 2)α-neighborhood) of X [10] if there exists a (1, 2)α-open set G such that x ∈ G ⊆ N_{x}.

3. (1, 2)α-Kernel and (1, 2)α-Closure

In this section, we define the operator (1, 2)α-kernel and find its properties.

Definition 3.1. Let A be a non empty subset of a space X. Then (1, 2)αkernel of A is denoted by (1, 2)α-ker(A) and is defined as (1, 2)α-ker(A) = ∩{G ∈ (1, 2)αO(X)/A ⊆ G}.

Definition 3.2. Let x ∈ X. Then the (1, 2)α-kernel of x is denoted by (1, 2)α-ker({x}) = ∩{G ∈ (1, 2)αO(X)/x ∈ G}.

Lemma 3.3. Let X be a bitopological space. Then for any non-empty subset A of X, (1, 2)α-ker(A) = {x ∈ X/(1, 2)α-cl({x}) ∩ A ≠ ∅}.

Proof. Let x ∈ (1, 2)α- ker(A). Suppose (1, 2)α-cl({x}) ∩ A = ∅. Then A ⊆ X − (1, 2)α-cl({x}) and X − (1, 2)α-cl({x}) is a (1, 2)α-open set containing A but not x, which is a contradiction.

Conversely, let x ∉ (1, 2)α − ker({A}) and (1, 2)α-cl({x}) ∩ A ≠ ∅. Then there exists a (1, 2)α-open set D containing A but not x and a y ∈ (1, 2)α-cl({x}) ∩ A. Hence we get a (1, 2)α-nbd of y, say, D with x ∉ D, which is a contradiction. Hence x ∈ (1, 2)α-ker(A).

Definition 3.4. In a space X, a set A is said to be weakly-ultra-separated from a set B if there exists a (1, 2)α-open set G such that A ⊂ G and G ∩ B = ∅ or A ∩ (1, 2)α-cl(B) = ∅.

Lemma 3.5. In view of the lemma 3.3 and definition of 3.4, let us have the followings for x, y ∈ X of a bitopological space,

(i) (1, 2)α-cl({x}) = {y : y is not weakly-ultra-separated from x} and (ii) (1, 2)α- ker({x}) = {y is not weakly-ultra-separated from y}.

Definition 3.6. For any point x of a space X,

(i) the derived set of x is denoted by (1, 2)α-d({x}) and is defined to be the set

(1, 2)α-d({x}) = (1, 2)α-cl({x}) − {x} = {y : y ≠ x and y is not weakly-ultra-separated from x},

(ii) the shell of a singleton set {x} is denoted by (1, 2)α-shl({x}) and is defined to be the set

(1, 2)α-shl({x}) = (1, 2)α- ker({x}) − {x} = {y : y ≠ x and x is not weakly-ultra-separated from y}.

Definition 3.7. Let X be a bitopological space. Then we define (i) (1, 2)α − N − D = {x : x ∈ X and (1, 2)α − d({x}) = ∅}, (ii) (1, 2)α − N − shl = {x : x ∈ X and (1, 2)α − shl({x}) = ∅}, (iii) (1, 2)α-⟨x⟩ = (1, 2)α-cl({x}) ∩ (1, 2)α-ker({x}).

Theorem 3.8. Let x, y ∈ X. Then the following conditions hold good: (i) y ∈ (1, 2)α-ker ({x}) if and only if x ∈ (1, 2)α-cl({y}), (ii) y ∈ (1, 2)α-shl({x}) if and only if x ∈ (1, 2)α-d({y}), (iii) y ∈ (1, 2)α-cl({x}) implies (1, 2)α-cl({y}) ⊆ (1, 2)α-cl({x}) and (iv) y ∈ (1, 2)α-ker({x}) implies (1, 2)α-ker({y}) ⊆ (1, 2)α-ker({x}).

Proof. The proof of (i) and (ii) are obvious from Remark 3.5.

(iii) Let z ∈ (1, 2)α-cl({y}). Then z is not weakly-ultra-separated from y. So there exists a (1, 2)α-open set G containing z such that G ∩ {y} ≠ ∅. Hence y ∈ G and by assumption G ∩ {x} ≠ ∅. Hence z is not weakly-ultra-separated from x. So z ∈ (1, 2)α-cl({x}). Therefore (1, 2)α-cl({y}) ⊆ (1, 2)α-cl({x}).

(iv) Let z ∈ (1, 2)α-ker({y}). Then y is not weakly-ultra-separated from z. So y ∈ (1, 2)α-cl({z}). Hence (1, 2)α-cl({y}) ⊆ (1, 2)α-cl({z}). By assumption y ∈ (1, 2)α-ker({x}) and then x ∈ (1, 2)α-cl({y}). So (1, 2)α-cl({x}) ⊆ (1, 2)α-cl({y}). Ultimately

(1, 2)α-cl({x}) ⊆ (1, 2)α-cl({z}).

Hence x ∈ (1, 2)α-cl({z}), that is z ∈ (1, 2)α-ker({x}). Therefore (1, 2)α-ker({y}) ⊆ (1, 2)α-ker({x}).

Let us recall that a subset A of X is called a degenerate set if A is either a null set or a singleton set.

Theorem 3.9. Let x, y ∈ X. Then,

(i) for every x ∈ X, (1, 2)α-shl({x}) is degenerate if and only if for all x, y ∈ X, x ≠ y, (1, 2)α-d({x}) ∩ (1, 2)α-d({y}) = ∅,

(ii) for every x ∈ X, (1, 2)α-d({x}) is degenerate if and only if for every x, y ∈ X, x ≠ y, (1, 2)α-shl({x}) ∩ (1, 2)α-shl({y}) = ∅.

Proof. Let (1, 2)α-d({x}) ∩ (1, 2)α-d({y}) ≠ ∅. Then there exists a z ∈ X such that z ∈ (1, 2)α-d({x}) and z ∈ (1, 2)α-d({y}). Then z ≠ y ≠ x and z ∈ (1, 2)α-cl({x}) and z ∈ (1, 2)α-cl({y}), that is x, y ∈ (1, 2)α-ker({z}). Hence (1, 2)α-ker({z}) and so (1, 2)α-shl({z}) is not a degenerate set.

Conversely, let x, y ∈ (1, 2)α-shl({z}). Then we get x ≠ z, x ∈ (1, 2)α-ker({z}) and y ≠ z and y ∈ (1, 2)α-ker({z}) and hence z is an element of both (1, 2)α-cl({x}) and (1, 2)α-cl({y}), which is a contradiction.

The proof of (ii) is simple and hence omitted.

Theorem 3.10. If y ∈ (1, 2)α-⟨x⟩, then (1, 2)α-⟨x⟩ = (1, 2)α-⟨y⟩.

Proof. If y ∈ (1, 2)α-⟨x⟩, then y ∈ (1, 2)α-cl({x}) ∩ (1, 2)α-ker({x}). Hence y ∈ (1, 2)α-cl({x}) and y ∈ (1, 2)α-ker ({x}) and so we have (1, 2)α-cl({y}) ⊆ (1, 2)α-cl({x}) and (1, 2)α-ker({y}) ⊆ (1, 2)α-ker({x}). Then

(1, 2)α-cl({y}) ∩ (1, 2)α-ker({y}) ⊆ (1, 2)α-cl({x}) ∩ (1, 2)α-ker({x}).

Hence (1, 2)α-⟨y⟩ ⊆ (1, 2)α- ⟨x⟩. The fact that y ∈ (1, 2)α-cl({x}) implies x ∈ (1, 2)α-ker({y}) and y ∈ (1, 2)α-ker({x}) implies x ∈ (1, 2)α-cl({y}). Then we have that (1, 2)α-⟨x⟩ ⊆ (1, 2)α-⟨y⟩. So (1, 2)α-⟨x⟩ = (1, 2)α-⟨y⟩.

Theorem 3.11. For all x, y ∈ X, either (1, 2)α-⟨x⟩ ∩ (1, 2)α-⟨y⟩ = ∅ or (1, 2)α-⟨ x ⟩ = (1, 2)α-⟨ y ⟩.

Proof. If (1, 2)α-⟨x⟩ ∩ (1, 2)α-⟨y⟩ ≠ ∅, then there exists z ∈ X such that z ∈ (1, 2)α-⟨x⟩ and z ∈ (1, 2)α-⟨y⟩. So by Theorem 3.10, (1, 2)α- ⟨z⟩ = (1, 2)α-⟨x⟩ = (1, 2)α-⟨y⟩. Hence the result.

Theorem 3.12. For any two points x, y ∈ X, the following statements are equivalent.

(i) (1, 2)α-ker({x}) ≠ (1, 2)α-ker({y}), and

(ii) (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}).

Proof. (i) ⇒ (ii) Let us assume (1, 2)α-ker({x}) ≠ (1, 2)α-ker({y}). Then there exists a z ∈ (1, 2)α-ker({x}) but z ∉ (1, 2)α-ker({y}). As z ∈ (1, 2)α-ker({x}), x ∈ (1, 2)α-cl({z}) and (1, 2)α-cl({x}) ⊂ (1, 2)α-cl({z}). Also we have taken z ∉ (1, 2)α-ker({y}), by lemma 3.3, (1, 2)α-cl({z}) ∩ {y} = ∅, so (1, 2)α-cl({x}) ∩ {y} = ∅ and so y is weakly-ultra-separated from x and hence we get that y ∉ (1, 2)α-cl({x}). Hence (1, 2)α-cl({y}) ≠ (1, 2)α-cl({x}).

(ii) ⇒ (i) Suppose (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}). Then there exists a point z ∈ (1, 2)α-cl({x}) but z ∉ (1, 2)α-cl({y}). So, we get a (1, 2)αopen set containing z and x but not y. That is y ∉ (1, 2)α-ker({x}). Hence (1, 2)α-ker({y}) ≠ (1, 2)α-ker({x}).

4. Ultra-T_{i} (i = 0,1) and Ultra-D_{1} spaces

In this section, some of the properties of ultra-T_{i} (i = 0,1) and ultra-D_{1} spaces are derived by means of weakly-ultra-separation.

Theorem 4.1. A space X is ultra-T_{0} if and only if any of the following conditions holds good:

(i) For arbitrary x, y ∈ X, x ≠ y, either x is weakly-ultra-separated from y or y is weakly-ultra-separated from x.

(ii) y ∈ (1, 2)α-cl({x}) implies x ∉ (1, 2)α-cl({y}).

(iii) For all x, y ∈ X if x ≠ y, then (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}).

Proof. (i) Obvious from the definitions of ultra-T_{0} and weakly-ultra-separation.

(ii) By assumption, y ∈ (1, 2)α-cl({x}) and so y is not weakly-ultra-separated from x. As X is ultra-T_{0}, x should be weakly-ultra-separated from y, that is x ∉ (1, 2)α-cl({y}).

(iii) If X is ultra-T_{0}, then for all x, y ∈ X and x ≠ y, (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}) as evidenced by (ii). Now let us prove the converse. Let (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}). Then there exists a z ∈ X, such that z ∈ (1, 2)α-cl({x}) and z ∉ (1, 2)α-cl({y}). If x is not ultraweakly separated from y, then x ∈ (1, 2)α-cl({y}). So (1, 2)α-cl({x}) ⊆ (1, 2)α-cl({y}). Then z ∈ (1, 2)α-cl({y}), which is a contradiction.

Corollary 4.2. A space X is ultra-T_{0} if and only if one of the following conditions hold good:

(i) For x, y ∈ X, y ∈ (1, 2)α-ker({x}) implies x ∉ (1, 2)α-ker({y}). (ii) For all x, y ∈ X, if x ≠ y, (1, 2)α-ker({x}) ≠ (1, 2)α-ker({y}).

Theorem 4.3. A space X is ultra-T_{0} if and only if [ (1, 2)α-cl({x}) ∩ {y}] ∩ [(1, 2)α-cl({y}) ∩ {x}] is degenerate.

Proof. Let X is ultra-T_{0}. Then we have any one of the two cases viz, x is weakly-ultra-separated from y and y is weakly-ultra-separated from x.

Case (i) :- If x is weakly-ultra-separated from y, then we have {x} ∩ (1, 2)α-cl({y}) = ∅ and {y} ∩ (1, 2)α-cl({x}) is a degenerate set.

Case (ii) :- If y is weakly-ultra-separated from x, then we have {y} ∩ (1, 2)α-cl({x}) = ∅ and {x} ∩ (1, 2)α-cl({y}) is a degenerate set. Hence [{x} ∩ (1, 2)α-cl({y})] ∩ [{y} ∩ (1, 2)α-cl({x})] is a degenerate set.

Conversely, suppose [{x}∩(1, 2)α-cl({y})]∩[{y}∩(1, 2)α-cl({x})] is a degenerate set. Then it is either ∅ or a singleton set. If it is ∅, then there is nothing to prove. If it is a singleton set, its value is either {x} or {y}. If it is {x}, then y is weakly-ultra-separated from x. If it is {y}, then x is weakly-ultra-separated from y. Hence X is ultra-T_{0}.

Theorem 4.4. A space is ultra-T_{0} if and only if

(1, 2)α-d({x}) ∩ (1, 2)α-shl({x}) = ∅.

Proof.Let X be ultra-T_{0}. Suppose we have

(1, 2)α-d({x}) ∩ (1, 2)α-shl({x}) ≠ ∅.

Then let z ∈ (1, 2)α-d({x}) and z ∈ (1, 2)α-shl({x}). Then z ≠ x and z ∈ (1, 2)α-cl({x}) and z ∈ (1, 2)α-ker({x}). Then z is not weakly-ultra-separated from x and also x is not weakly-ultra-separated from z, which is a contradiction.

Conversely, let (1, 2)α-d({x})∩(1, 2)α-shl({x}) = ∅. Then there exists a z ≠ x, z ∈ (1, 2)α-cl({x}) and z ∉ (1, 2)α-ker({x}). Hence if we have a z, which is not weakly-ultra-separate from x, then x is weaklyultra-separated from z.

Corollary 4.5. If X is ultra-T_{0}, then for any x ∈ X, (1, 2)α-⟨x⟩ = {x}.

Theorem 4.6. A space X is ultra-T_{1} if and only if one of the following conditions hold good:

(i) For arbitrary x, y ∈ X, x ≠ y, x is weakly-ultra-separated form y. (ii) For every x ∈ X, (1, 2)α-cl({x}) = {x}. (iii) For every x ∈ X, (1, 2)α-d({x}) = ∅ or (1, 2)α-N-D = X. (iv) For every x ∈ X(1, 2)α-ker({x}) = {x}.

(v) For every x ∈ X, (1, 2)α-shl({x}) = ∅ or (1, 2)α-N-shl = X.

(vi) For every x, y ∈ X, x ≠ y(1, 2)α-cl({x}) ∩ (1, 2)α-cl({y}) = ∅. (vii) For every arbitrary x, y ∈ X, x ≠ y, we have (1, 2)α-ker({x}) ∩

(1, 2)α-ker({y}) = ∅.

Proof.

(i) This is just a reformulation of the definition of ultra-T_{1}.

(ii) If x is weakly-ultra-separated from y, then for y ≠ x, we have y ∉ (1, 2)α-cl({x}), and hence x ∉ (1, 2)α-ker({y}). Therefore we get that (1, 2)α-ker({y}) = {y}. Its converse is just a reformulation of the above proof.

(iii) (iv) and (v)are obvious.

(vi) As X is ultra-T_{1}, (1, 2)α-cl({x}) = {x} and (1, 2)α-cl({y}) = {y} so, when x ≠ y, (1, 2)α-cl({x}) ∩ (1, 2)α-cl({y}) = ∅.

(vii) Obvious from (vi).

Definition 4.7. For a bitopological space (X,τ_{1},τ_{2}),

(i) a point x ∈ X is called a (1, 2)α-neat point if it has X as its only (1, 2)α-nbd, and

(ii) a bitopological space X is (1, 2)α-symmetric if {x} ∈ (1, 2)α-cl({y}) implies y ∈ (1, 2)αcl-({x}).

Theorem 4.8. For a ultra-T_{0} space, the following are equivalent. (i) X is ultra-D_{1}. (ii) X has no (1, 2)α-neat point.

Proof. (i) ⇒ (ii) Assume X is ultra-D_{1}. Then [ by 2.8 (iii) and (iv) ] each point x of X is contained in a (1, 2)αD-set S, where S = U/V and so x ∈ U with U ≠ X. This implies that x is not a (1, 2)α-neat point

(ii) ⇒ (i) If X is ultra-T_{0}, then for any two distinct points x, y of X, there exists a (1, 2)α-open set U ≠ X containing the point x but not y. Thus we have set a U ≠ X as a (1, 2)α-D-set. Given that X has no (1, 2)α-neat point. Then any y ∈ X is not a (1, 2)α-neat point and so there exists a (1, 2)α-nbd V of y such that V ≠ X. Thus y ∈ V/U but x ∉ V/U and V/U is a (1, 2)αD-set. Hence X is an ultra-D_{1} space.

Theorem 4.9. A bitopological space X is (1, 2)α-symmetric if and only if {x} is (1, 2)αg-closed for each x ∈ X.

Proof. Let {y} is (1, 2)αg-closed. Assume {x} ∈ (1, 2)α-cl({y}) and {y} ∉ (1, 2)α-cl({x}). Hence {y} ⊆ (X − (1, 2)α-cl({x})). As {y} is (1, 2)αg-closed, (1, 2)α-cl({y}) ⊆ X − (1, 2)α-cl({x}), which is a contradiction to the fact that x ∈ (1, 2)α-cl({y}). Conversely, let x ∈ (1, 2)αcl({y}) and y ∈ (1, 2)α-cl({x}). Suppose that x ∈ E, E is (1, 2)α-open and (1, 2)α-cl({x}) is not a subset of E. Then (1, 2)α-cl({x}) ⊆ X − E and so (1, 2)α-cl({x})∩(X −E) ≠ ϕ. So let y ∈ (1, 2)α-cl({x})∩(X −E) but by assumption {x} ∈ (1, 2)α-cl({y}) ⊆ X − E and hence x ∈ X − E, which is a contradiction.

Corollary 4.10. An ultra-T_{1} space is a (1, 2)α-symmetric space.

Proof. If X is an ultra-T_{1} space, then( by theorem 4.9 of [10]) every singleton set is (1, 2)α-closed. Again by remark 3.3 of [10], every (1, 2)αclosed set is (1, 2)αg-closed. Hence X is (1, 2)α-symmetric.

Theorem 4.11. Let f : X → Y be a (1, 2)α-irresolute, surjective function. If S is a (1, 2)αD-set in Y, then the inverse image of S is a (1, 2)αD-set in X.

Proof. Let S be a (1, 2)αD-set in Y. Then there exist two (1, 2)αopen sets U_{1} and U_{2} such that S = U_{1}/U_{2} and U_{1} ≠ Y. As f is (1, 2)αirresolute, f^{−1}(U_{1}) and f^{−1}(U_{2}) are (1, 2)α-open sets in X and f^{−1}(U_{1}) ≠ X. Then f^{−1}(S) = f^{−1}(U_{1})/f^{−1}(U_{2}) is a (1, 2)αD-set in X.

Theorem 4.12. Let f : X → Y be a (1, 2)α-irresolute and bijective function. If Y is an ultra-D_{1} space then X is also an ultra D_{1} space.

Proof. Suppose Y is an ultra-D_{1} space and x and y be any pair of distinct points in X. Since f is injective and Y is ultra-D_{1}, there exist (1, 2)αD-sets G_{x} and G_{y} of Y containing f(x) and f(y) respectively such that f(x) ∉ G_{y} and f(y) ∉ G_{x}. By theorem 4.11, f^{−1}(G_{x}) and f^{−1}(G_{y}) are (1, 2)αD-sets in X containing x and y respectively. Hence X is ultra-D_{1}.

Theorem 4.13. A bitopological space X is ultra-D_{1} if and only if for each pair of distinct points x ≠ y ∈ X, there exists a (1, 2)α-irresolute, surjective function f : X → Y, where Y is ultra-D_{1} such that f(x) and f(y) are distinct.

Proof. Necessity: Defining f as the identity function we can prove this part.

Sufficiency: Let x, y be distinct points of X. Assume f is a (1, 2)αirresolute, surjective function and Y is ultra-D_{1}. Then for any f(x) ≠ f(y) there exists (1, 2)αD-sets G_{x} and G_{y} such that f(y) ∈ G_{y} and f(y) ∈ G_{y}. By Theorem 4.11, f^{−1}(G_{x}) and f^{−1}(G_{y}) are two disjoint (1, 2)αD-sets containing x and y respectively. Therefore the space is ultra-D_{1}.

5. More on ultra-R_{0} space

In this chapter, we derive some properties of ultra-R_{0} spaces.

Theorem 5.1. A space X is ultra-R_{0} if and only if for any x, y ∈ X, (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}) implies (1, 2)α-cl({x})∩(1, 2)α-cl({y}) = ∅.

Proof. Necessity: Let X be an ultra-R_{0} space and if x ≠ y, then (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}). Then there exists a z ∈ X such that z ∈ (1, 2)α-cl({x}) and z ∉ (1, 2)α-cl({y}). Hence we get a (1, 2)αopen set G containing z and x but not y. Therefore x ∉ (1, 2)α-cl({y}). Then x ∈ X − (1, 2)α-cl({y}), which is a (1, 2)α-open set. Since X is ultra-R_{0}, (1, 2)α-cl({x}) ⊆ X − (1, 2)α-cl({y}) and so (1, 2)α-cl({x}) ∩ (1, 2)α-cl({y}) = ∅.

Sufficiency : Let V be a (1, 2)α-open set and x ∈ V. Let y ∉ V. Then y ∈ X −V. Since x ≠ y and x ∉ (1, 2)α-cl({y}), we get (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}). Then by assumption, (1, 2)α-cl({x}) ∩ (1, 2)α-cl({y}) = ∅. Hence y ∉ (1, 2)α-cl({x}) and so (1, 2)α-cl({x}) ⊂ V. Therefore X is an ultra-R_{0} space.

Theorem 5.2. A space X is ultra-R_{0} if and only if for any two points x, y ∈ X, (1, 2)α-ker({x}) ≠ (1, 2)α-ker({y}) implies (1, 2)α-ker({x}) ∩ (1, 2)α-ker({y}) = ∅.

Proof. If (1, 2)α-ker({x}) ≠ (1, 2)α-ker({y}), then by theorem 3.12, (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}). We need to prove that (1, 2)α-ker({x})∩ (1, 2)α-ker({y}) = ∅. If not, let there exist a z in X such that z ∈ (1, 2)α-ker({x}) and z ∈ (1, 2)α-ker({y}). Then x ∈ (1, 2)α-cl({z}) and y ∈ (1, 2)α-cl({z}). Then by theorem 5.1, we have (1, 2)α-cl({z}) = (1, 2)α-cl({x}) = (1, 2)α-cl({y}), which is a contradiction. Hence we get that (1, 2)α-ker({x}) ∩ (1, 2)α-ker({y}) = ∅.

Conversely, let X be a space such that for any two distinct points x and y in X, (1, 2)α-ker({x}) ≠ (1, 2)α-ker({y}) implies that (1, 2)α-ker({x}) ∩ (1, 2)α-ker({y}) = ∅. By theorem 3.12, we get that (1, 2)α-cl({x}) ≠ (1, 2)α-cl({y}). If (1, 2)α-cl({x}) ∩ (1, 2)α-cl({y}) = ∅, then by theorem 5.1, X is ultra-R_{0}. If not, let there be a z in X such that z ∈ (1, 2)α-cl({x}) and z ∈ (1, 2)α-cl({y}). Then x ∈ (1, 2)α-ker({z}) and y ∈ (1, 2)α-ker({z}). So we obtain that

(1, 2)α-ker({x}) = (1, 2)α-ker({z}) = (1, 2)α-ker({y}),

which is a contradiction. Therefore (1, 2)α-cl({x}) ∩ (1, 2)α-cl({y}) = ∅. Hence X is ultra-R_{0}.

Theorem 5.3. For a space X, the following statements are equivalent: (i) X is ultra-R_{0}. (ii) For any A ≠ ∅ and G ∈ (1, 2)αO(X) such that A ∩ G ≠ ∅ there

exists a F ∈ (1, 2)αCL(X) such that F ∩ G ≠ ∅ and F ⊂ G.

(iii) For any G ∈ (1, 2)αO(X), G = ∪{F ∈ (1, 2)αCL(X)/F ⊂ G}. (iv) For any (1, 2)α-closed set F, F = ∩{G ∈ (1, 2)αO(X)/F ⊂ G}.

(v) For any x ∈ X, (1, 2)α-cl({x}) ⊂ (1, 2)α-ker({x}).

Proof. (i) ⇒ (ii) Assume X is ultra-R_{0}. Let A be a non empty subset of X and G be a (1, 2)α-open set such that A ∩ G ≠ ∅. So there exists a x ∈ A∩G. Then x ∈ A and x ∈ G. As X is ultra-R_{0}, (1, 2)α-cl({x}) ⊂ G. Take F = (1, 2)α-ker({x}) and then F ⊂ G and A ∩ G ≠ ∅.

(ii) ⇒ (iii) Let G ∈ (1, 2)αO(X) and F ∈ (1, 2)α-CL({X}) such that F ⊂ G. Then ⋃{F/F ∈ (1, 2)α-CL({X}) and F ⊂ G} ⊂ G. To prove the other inclusion, let x ∈ G. Then by assumption, there exists a (1, 2)α-closed set F such that x ∈ F and F ⊂ G. Hence

G ⊂ U{F ∈ (1, 2)α-CL({X}) and F ⊂ G}.

Hence we get the result.

(iii) ⇒ (iv) Obvious.

(iv) ⇒ (v) Let x ∈ X and there exists a y ∈ X such that y ∉ (1, 2)α-ker({x}). Then there exists a (1, 2)α-open set V containing x but not y. That is (1, 2)α-cl({y})∩V = ∅, which implies that (1, 2)α-cl({y}) ⊂ X −V. Again, by our assumption, ∩{G ∈ (1, 2)αO(X)/(1, 2)α-cl({y}) ⊂ G} ∩ V = ∅. Hence there exists a (1, 2)α-open set G such that x ∉ G and (1, 2)α-cl({y}) ⊂ G. Therefore (1, 2)α-cl({x}) ∩ G = ∅ and y ∉ (1, 2)α-cl({x}). Consequently, (1, 2)α-cl({x}) ⊂ (1, 2)α-ker({x}).

(v) ⇒ (i) Let G ∈ (1, 2)αO(X) and x ∈ G. Suppose that y ∈ (1, 2)α-ker({x}). Then (1, 2)α-cl({y})∩{x} ≠ ∅. So {x} ∈ (1, 2)α-cl({y}) and y ∈ G which implies that (1, 2)α-cl({x}) ⊂ (1, 2)α-ker({x}) ⊂ G. So X is ultra-R_{0}.

Theorem 5.4. For a bitopological space X, the following properties are equivalent:

(i) X is ultra-R_{0}.

(ii) (1, 2)α-cl({x}) = (1, 2)α-ker({x}) for all x ∈ X.

Proof. Suppose that X is an ultra-R_{0} space. Then by theorem 5.3, (1, 2)α-cl({x}) ⊂ (1, 2)α-ker({x}) for each x ∈ X. Assume that y ∈ (1, 2)α-ker({x}). Then x ∈ (1, 2)α-cl({y}) and so (1, 2)α-cl({x}) = (1, 2)α-cl({y}). Therefore y ∈ (1, 2)α-cl({x}) and so (1, 2)α-ker({x}) ⊂ (1, 2)α-cl({x}). Thus (1, 2)α-ker({x}) = (1, 2)α-cl({x}).

(ii) ⇒ (i) is obvious by Theorem 5.2.

Theorem 5.5. For a space X, the following properties are equivalent: (i) X is ultra-R_{0}. (ii) x ∈ (1, 2)α-cl({y}) if and only if y ∈ (1, 2)α-cl({x}), for any points

x and y in X.

Proof. (i) ⇒ (ii) Assume X is ultra-R_{0}. Let x ∈ (1, 2)α-cl({y}) and D be any (1, 2)α-open set such that y ∈ D. As X is ultra-R_{0}, (1, 2)α-cl({y}) ⊂ D and hence x ∈ D. Therefore every (1, 2)α-open set containing y contains x. Hence y ∈ (1, 2)α-cl({x})

(ii) ⇒ (i) Let U be a (1, 2)α-open set and x ∈ U. If y ∉ U, then x ∉ (1, 2)α-cl({y}) and hence y ∉ (1, 2)α-cl({x}). This implies that (1, 2)α-cl({x}) ⊂ U. Hence X is ultra-R_{0}.

Theorem 5.6. For a space X, the following properties are equivalent: (i) X is ultra-R_{0}. (ii) If F is (1, 2)α-closed, then F = (1, 2)α-ker({F }). (iii) If F is (1, 2)α-closed, then (1, 2)α-ker({x}) ⊂ F, where x ∈ F. (iv) If x ∈ X then (1, 2)α-ker({x}) ⊂ (1, 2)α-cl({x}).

Proof. (i) ⇒ (ii) Let F be a (1, 2)α-closed set and x ∉ F. Then X −F is a (1, 2)α-open set containing x. Since X is ultra-R_{0}, (1, 2)α-cl({x}) ⊂ X − F. Thus (1, 2)α-cl({x}) ∩ F = ∅. Hence, by lemma 3.3, x ∉ (1, 2)α-ker({F }) and so (1, 2)α-ker({F }) ⊂ F. Also by the definition of kernel of a set, F ⊂ (1, 2)α-ker({F }). Hence F = (1, 2)α-ker({F }).

(ii) ⇒ (iii) Let x ∈ F. So (1, 2)α-ker({x}) ⊂ (1, 2)α-ker({F }). As F is (1, 2)α-closed, (1, 2)α-ker({F }) = F.

(iii) ⇒ (iv) Since x ∈ (1, 2)α-cl({x}) and (1, 2)α-cl({x}) is a (1, 2)αclosed set, by (iii) (1, 2)α-ker({x}) ⊂ (1, 2)α-cl({x}).

(iv) ⇒ (i) Let x ∈ (1, 2)α-cl({y}). Then y ∈ (1, 2)α-ker({x}) and (1, 2)α-cl({y}) is a (1, 2)α-closed set. By assumption, we have y ∈ (1, 2)α-ker({x}) ⊂ (1, 2)α-cl({x}), which implies that y ∈ (1, 2)α-cl({x}). Hence, by Theorem 5.5, X is ultra-R_{0}.

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Department of Mathematics, Sri Parasakthi College,Courtallam-627802,Tirunelveli Dt., TamilNadu, India E-mail address: raji arul2000@yahoo.co.in

Department of Mathematics, Arul Anndar College, Karumathur-625514, Madurai Dt., TamilNadu, India E-mail address: mlthivagar@yahoo.co.in

Department of Mathematics, Jayaraj Annapackiam College for Women, Periyakulam-625 601,Theni Dt.,TamilNadu, India E-mail address: athisayaponmani@yahoo.co.in

Received October 10, 2006; Revised version: December 26, 2006