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}

________________

2000 Mathematical Subject Classiﬁcation. 54B05,54D05.

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

Supported by UGC, New Delhi, INDIA.

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 \cap B = \emptyset

. 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.

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 $α$ -open set [9] if $A \subset τ_{1} -int (τ_{1} -cl (τ_{1} -int (A)))$ , where $τ_{1}$ -int $(A)$ and $τ_{1} - c l (A)$ represent the interior and closure of $A$ with respect to $τ_{1}$ .

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

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 $A \in τ_{1} \cup τ_{2}$ ,

(ii) $τ_{1} τ_{2}$ -closed if $A^{c} \in τ_{1} \cup τ_{2}$ .

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

The set of all $(1, 2) α$ -open sets is denoted by $(1, 2) α O (X)$ . The complement of a $(1, 2) α$ -open set is called a $(1, 2) α$ -closed set. The $(1, 2) α$ -closure 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 $(1, 2) α C L (X)$ .

Deﬁnition 2.6. A function $f : X \to Y$ 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 $(1, 2) α$ g-closed set [10] if $(1, 2) α cl (A) \subseteq U$ , where $A \subseteq U$ and $U \in (1, 2) α O (X)$ .

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

(ii) A bitopological space $X$ is called an ultra- $T_{1}$ [10] space if and only if for each $x, y \in X$ such that $x \neq y$ , 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 $(1, 2) α$ D-set) if there exist two $(1, 2) α$ -open sets $U_{1}$ and $U_{2}$ such that $A = U_{1} ∕ U_{2}$ and $U_{1} \neq X$ .

(iv) A bitopological space $X$ is called an ultra- $D_{1}$ [10] space if for each $x, y \in X$ and $x \neq y$ there exists two $(1, 2) α$ D-sets $G_{1}$ and $G_{2}$ such that $x \in G_{1}$ , $x \notin G_{2}$ and $y \in G_{2}$ , $y \notin G_{1}$ .

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 $x \in X$ . 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 $x \in G \subseteq N_{x}$ .

(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 $(1, 2) α$ -kernel of A is denoted by $(1, 2) α$ -ker(A) and is deﬁned as $(1, 2) α - k e r (A) = \cap {G \in (1, 2) α O (X) ∕ A \subseteq G}$ .

Deﬁnition 3.2. Let $x \in X$ . Then the $(1, 2) α$ -kernel of $x$ is denoted by $(1, 2) α - k e r ({x}) = \cap {G \in (1, 2) α O (X) ∕ x \in G}$ .

Lemma 3.3. Let $X$ be a bitopological space. Then for any non-empty subset $A$ of $X$ , $(1, 2) α - k e r (A) = {x \in X ∕ (1, 2) α - c l ({x}) \cap A \neq \emptyset}$ .

Proof. Let

x \in (1, 2) α - ker (A)

. Suppose

(1, 2) α - c l ({x}) \cap A = \emptyset

. Then

A \subseteq X - (1, 2) α - c l ({x})

and

X - (1, 2) α - c l ({x})

is a

(1, 2) α

-open set containing

A

but not

x

, which is a contradiction.

Conversely, let

x \notin (1, 2) α - k e r ({A})

and

(1, 2) α - c l ({x}) \cap A \neq \emptyset

. Then there exists a

(1, 2) α

-open set D containing

A

but not

x

and a

y \in (1, 2) α - c l ({x}) \cap A

. Hence we get a

(1, 2) α

-nbd of

y

, say,

D

with

x \notin D

, which is a contradiction. Hence

x \in (1, 2) α - k e r (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 $A \subset G$ and $G \cap B = \emptyset$ or $A \cap (1, 2) α - c l (B) = \emptyset$ .

Lemma 3.5. In view of the lemma 3.3 and deﬁnition of 3.4, let us have the followings for $x, y \in X$ 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

\begin{matrix} (1, 2) α - d ({x}) = (1, 2) α -cl ({x}) - {x} = \\ {y : y \neq x and y is not weakly-ultra-separated from x}, \end{matrix}

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

\begin{matrix} (1, 2) α - s h l ({x}) = (1, 2) α - ker ({x}) - {x} = \\ {y : y \neq x and x is not weakly-ultra-separated from y} . \end{matrix}

Deﬁnition 3.7. Let $X$ be a bitopological space. Then we deﬁne

(i) $(1, 2) α - N - D = {x : x \in X and (1, 2) α - d ({x}) = \emptyset}$ ,

(ii) $(1, 2) α - N - s h l = {x : x \in X and (1, 2) α - s h l ({x}) = \emptyset}$ ,

(iii) $(1, 2) α$ - $〈 x 〉 = (1, 2) α - c l ({x}) \cap (1, 2) α - k e r ({x})$ .

Theorem 3.8.

Let $x, y \in X$ . Then the following conditions hold good:

(i) $y \in (1, 2) α$ -ker $({x})$ if and only if $x \in (1, 2) α - c l ({y})$ ,

(ii) $y \in (1, 2) α - s h l ({x})$ if and only if $x \in (1, 2) α$ -d $({y})$ ,

(iii) $y \in (1, 2) α - c l ({x})$ implies $(1, 2) α - c l ({y}) \subseteq (1, 2) α - c l ({x})$ and

(iv) $y \in (1, 2) α - k e r ({x})$ implies $(1, 2) α - k e r ({y}) \subseteq (1, 2) α - k e r ({x})$ .

(iii) Let

z \in (1, 2) α - c l ({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 \cap {y} \neq \emptyset

. Hence

y \in G

and by assumption

G \cap {x} \neq \emptyset

. Hence

z

is not weakly-ultra-separated from

x

. So

z \in (1, 2) α - c l ({x})

. Therefore

(1, 2) α - c l ({y}) \subseteq (1, 2) α - c l ({x})

(iv) Let

z \in (1, 2) α - k e r ({y})

. Then

y

is not weakly-ultra-separated from

z

. So

y \in (1, 2) α - c l ({z})

. Hence

(1, 2) α - c l ({y}) \subseteq (1, 2) α - c l ({z})

. By assumption

y \in (1, 2) α - k e r ({x})

and then

x \in (1, 2) α - c l ({y})

. So

(1, 2) α - c l ({x}) \subseteq (1, 2) α - c l ({y})

. Ultimately

Hence

x \in (1, 2) α - c l ({z})

, that is

z \in (1, 2) α - k e r ({x})

. Therefore

(1, 2) α - k e r ({y}) \subseteq (1, 2) α - k e r ({x})

Let us recall that a subset

A

X

is called a degenerate set if

A

is either a null set or a singleton set.

Theorem 3.9.

Let $x, y \in X$ . Then,

(i) for every $x \in X$ , $(1, 2) α - s h l ({x})$ is degenerate if and only if for all $x, y \in X$ , $x \neq y, (1, 2) α$ -d $({x}) \cap (1, 2) α$ -d $({y}) = \emptyset$ ,

(ii) for every $x \in X$ , $(1, 2) α$ -d $({x})$ is degenerate if and only if for every $x, y \in X, x \neq y, (1, 2) α - s h l ({x}) \cap (1, 2) α - s h l ({y}) = \emptyset$ .

Proof. Let

(1, 2) α

-d

({x}) \cap (1, 2) α

-d

({y}) \neq \emptyset

. Then there exists a

z \in X

such that

z \in (1, 2) α

-d

({x})

and

z \in (1, 2) α

d ({y})

. Then

z \neq y \neq x

and

z \in (1, 2) α - c l ({x})

and

z \in (1, 2) α - c l ({y})

, that is

x, y \in (1, 2) α - k e r ({z})

. Hence

(1, 2) α - k e r ({z})

and so

(1, 2) α - s h l ({z})

is not a degenerate set.

Conversely, let

x, y \in (1, 2) α - s h l ({z})

. Then we get

x \neq z, x \in (1, 2) α - k e r ({z})

and

y \neq z

and

y \in (1, 2) α - k e r ({z})

and hence

z

is an element of both

(1, 2) α - c l ({x})

and

(1, 2) α - c l ({y})

, which is a contradiction.

Theorem 3.10.

If $y \in (1, 2) α$ - $〈 x 〉$ , then $(1, 2) α$ - $〈 x 〉 = (1, 2) α$ - $〈 y 〉$ .

Proof. If

y \in (1, 2) α

〈 x 〉

, then

y \in (1, 2) α - c l ({x}) \cap (1, 2) α - k e r ({x})

. Hence

y \in (1, 2) α - c l ({x})

and

y \in (1, 2) α

-ker

({x})

and so we have

(1, 2) α - c l ({y}) \subseteq (1, 2) α - c l ({x})

and

(1, 2) α - k e r ({y}) \subseteq (1, 2) α - k e r ({x})

. Then

Hence

(1, 2) α

〈 y 〉 \subseteq (1, 2) α

〈 x 〉

. The fact that

y \in (1, 2) α

c l ({x})

implies

x \in (1, 2) α - k e r ({y})

and

y \in (1, 2) α - k e r ({x})

implies

x \in (1, 2) α - c l ({y})

. Then we have that

(1, 2) α

〈 x 〉 \subseteq (1, 2) α

〈 y 〉

. So

(1, 2) α

〈 x 〉 = (1, 2) α

〈 y 〉

Theorem 3.11.

For all $x, y \in X$ , either $(1, 2) α - 〈 x 〉 \cap (1, 2) α - 〈 y 〉 = \emptyset$ or $(1, 2) α - 〈 x 〉 = (1, 2) α - 〈 y 〉$ .

Proof. If

(1, 2) α

〈 x 〉 \cap (1, 2) α

〈 y 〉 \neq \emptyset

, then there exists

z \in X

such that

z \in (1, 2) α

〈 x 〉

and

z \in (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 $x, y \in X$ , the following statements are equivalent.

(i) $(1, 2) α - k e r ({x}) \neq (1, 2) α - k e r ({y})$ , and

(ii) $(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})$ .

Proof. (i)

\Rightarrow

(ii) Let us assume

(1, 2) α - k e r ({x}) \neq (1, 2) α - k e r ({y})

. Then there exists a

z \in (1, 2) α - k e r ({x})

but

z \notin (1, 2) α - k e r ({y})

. As

z \in (1, 2) α - k e r ({x}), x \in (1, 2) α - c l ({z})

and

(1, 2) α - c l ({x}) \subset (1, 2) α - c l ({z})

. Also we have taken

z \notin (1, 2) α - k e r ({y})

, by lemma 3.3,

(1, 2) α - c l ({z}) \cap {y} = \emptyset

, so

(1, 2) α - c l ({x}) \cap {y} = \emptyset

and so

y

is weakly-ultra-separated from

x

and hence we get that

y \notin (1, 2) α - c l ({x})

. Hence

(1, 2) α - c l ({y}) \neq (1, 2) α - c l ({x})

(ii)

\Rightarrow

(i) Suppose

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

. Then there exists a point

z \in (1, 2) α - c l ({x})

but

z \notin (1, 2) α - c l ({y})

. So, we get a

(1, 2) α

-open set containing

z

and

x

but not

y

. That is

y \notin (1, 2) α - k e r ({x})

. Hence

(1, 2) α - k e r ({y}) \neq (1, 2) α - k e r ({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 $x, y \in X, x \neq y$ , either $x$ is weakly-ultra-separated from $y$ or $y$ is weakly-ultra-separated from x.

(ii) $y \in (1, 2) α - c l ({x})$ implies $x \notin (1, 2) α - c l ({y})$ .

(iii) For all $x, y \in X$ if $x \neq y$ , then $(1, 2) α$ - $c l ({x}) \neq (1, 2) α - c l ({y})$ .

Proof. (i) Obvious from the deﬁnitions of ultra-

T_{0}

and weakly-ultra-separation.

(ii) By assumption,

y \in (1, 2) α - c l ({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 \notin (1, 2) α - c l ({y})

(iii) If

X

is ultra-

T_{0}

, then for all

x, y \in X

and

x \neq y

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

as evidenced by (ii). Now let us prove the converse. Let

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

. Then there exists a

z \in X

, such that

z \in (1, 2) α - c l ({x})

and

z \notin (1, 2) α - c l ({y})

. If

x

is not ultra-weakly separated from

y

, then

x \in (1, 2) α - c l ({y})

. So

(1, 2) α - c l ({x}) \subseteq (1, 2) α - c l ({y})

. Then

z \in (1, 2) α - c l ({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 $x, y \in X, y \in (1, 2) α - k e r ({x})$ implies $x \notin (1, 2) α - k e r ({y})$ .

(ii) For all $x, y \in X$ , if $x \neq y, (1, 2) α - k e r ({x}) \neq (1, 2) α - k e r ({y})$ .

Theorem 4.3.

A space $X$ is ultra- $T_{0}$ if and only if $[(1, 2) α - c l ({x}) \cap {y}] \cap [(1, 2) α - c l ({y}) \cap {x}]$ 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} \cap (1, 2) α - c l ({y}) = \emptyset

and

{y} \cap (1, 2) α - c l ({x})

is a degenerate set.

Case (ii) :- If

y

is weakly-ultra-separated from

x

, then we have

{y} \cap (1, 2) α - c l ({x}) = \emptyset

and

{x} \cap (1, 2) α - c l ({y})

is a degenerate set. Hence

[{x} \cap (1, 2) α - c l ({y})] \cap [{y} \cap (1, 2) α - c l ({x})]

is a degenerate set.

Conversely, suppose

[{x} \cap (1, 2) α - c l ({y})] \cap [{y} \cap (1, 2) α - c l ({x})]

is a degenerate set. Then it is either

\emptyset

or a singleton set. If it is

\emptyset

, then there is nothing to prove. If it is a singleton set, its value is either

{x}

{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}) \cap (1, 2) α - s h l ({x}) = \emptyset .

Then let

z \in (1, 2) α

-d

({x})

and

z \in (1, 2) α - s h l ({x})

. Then

z \neq x

and

z \in (1, 2) α - c l ({x})

and

z \in (1, 2) α - k e r ({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}) \cap (1, 2) α - s h l ({x}) = \emptyset

. Then there exists a

z \neq x, z \in (1, 2) α - c l ({x})

and

z \notin (1, 2) α - k e r ({x})

. Hence if we have a

z

, which is not weakly-ultra-separate from

x

, then

x

is weakly-ultra-separated from

z

Corollary 4.5.

If $X$ is ultra- $T_{0}$ , then for any $x \in X, (1, 2) α$ - $〈 x 〉 = {x}$ .

Theorem 4.6.

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

(i) For arbitrary $x, y \in X, x \neq y, x$ is weakly-ultra-separated form $y$ .

(ii) For every $x \in X, (1, 2) α$ - $c l ({x}) = {x}$ .

(iii) For every $x \in X, (1, 2) α$ -d $({x}) = \emptyset$ or $(1, 2) α$ -N-D $= X$ .

(iv) For every $x \in X (1, 2) α - k e r ({x}) = {x}$ .

(v) For every $x \in X, (1, 2) α - s h l ({x}) = \emptyset$ or $(1, 2) α$ -N-shl $= X$ .

(vi) For every $x, y \in X, x \neq y (1, 2) α - c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset$ .

(vii) For every arbitrary $x, y \in X$ , $x \neq y$ , we have $(1, 2) α - k e r ({x}) \cap (1, 2) α - k e r ({y}) = \emptyset$ .

(ii) If

x

is weakly-ultra-separated from

y

, then for

y \neq x

, we have

y \notin (1, 2) α - c l ({x})

, and hence

x \notin (1, 2) α - k e r ({y})

. Therefore we get that

(1, 2) α - k e r ({y}) = {y}

. Its converse is just a reformulation of the above proof.

(vi) As

X

is ultra-

T_{1}

(1, 2) α - c l ({x}) = {x}

and

(1, 2) α - c l ({y}) = {y}

so, when

x \neq y, (1, 2) α - c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset

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

(i) a point $x \in X$ 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 ${x} \in (1, 2) α - c l ({y})$ implies $y \in (1, 2) α$ cl- $({x})$ .

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)

\Rightarrow

(ii) Assume

X

is ultra-

D_{1}

. Then [ by 2.8 (iii) and (iv) ] each point

x

X

is contained in a

(1, 2) α

D-set

S

, where

S = U ∕ V

and so

x \in U

with

U \neq X

. This implies that

x

is not a

(1, 2) α

-neat point

(ii)

\Rightarrow

(i) If

X

is ultra-

T_{0}

, then for any two distinct points

x, y

X

, there exists a

(1, 2) α

-open set

U \neq X

containing the point

x

but not

y

. Thus we have set a

U \neq X

as a

(1, 2) α

-D-set. Given that

X

has no

(1, 2) α

-neat point. Then any

y \in X

is not a

(1, 2) α

-neat point and so there exists a

(1, 2) α

-nbd

V

y

such that

V \neq X

. Thus

y \in V ∕ U

but

x \notin V ∕ U

and

V ∕ U

is a

(1, 2) α

D-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 $(1, 2) α$ g-closed for each $x \in X$ .

Proof. Let

{y}

(1, 2) α

g-closed. Assume

{x} \in (1, 2) α - c l ({y})

and

{y} \notin (1, 2) α - c l ({x})

. Hence

{y} \subseteq (X - (1, 2) α - c l ({x}))

. As

{y}

(1, 2) α

g-closed,

(1, 2) α - c l ({y}) \subseteq X - (1, 2) α - c l ({x})

, which is a contradiction to the fact that

x \in (1, 2) α

c l ({y})

. Conversely, let

x \in (1, 2) α

c l ({y})

and

y \in (1, 2) α - c l ({x})

. Suppose that

x \in E

E

(1, 2) α

-open and

(1, 2) α - c l ({x})

is not a subset of

E

. Then

(1, 2) α - c l ({x}) \subseteq X - E

and so

(1, 2) α - c l ({x}) \cap (X - E) \neq φ

. So let

y \in (1, 2) α - c l ({x}) \cap (X - E)

but by assumption

{x} \in (1, 2) α - c l ({y}) \subseteq X - E

and hence

x \in 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) α

-closed set is

(1, 2) α

g-closed. Hence

X

(1, 2) α

-symmetric.

Theorem 4.11.

Let $f : X \to Y$ be a $(1, 2) α$ -irresolute, surjective function. If $S$ is a $(1, 2) α$ D-set in $Y$ , then the inverse image of $S$ is a $(1, 2) α$ D-set in $X$ .

Proof. Let

S

be a

(1, 2) α

D-set in

Y

. Then there exist two

(1, 2) α

-open sets

U_{1}

and

U_{2}

such that

S = U_{1} ∕ U_{2}

and

U_{1} \neq Y

. As

f

(1, 2) α

-irresolute,

f^{- 1} (U_{1})

and

f^{- 1} (U_{2})

are

(1, 2) α

-open sets in

X

and

f^{- 1} (U_{1}) \neq X

. Then

f^{- 1} (S) = f^{- 1} (U_{1}) ∕ f^{- 1} (U_{2})

is a

(1, 2) α

D-set in X.

Theorem 4.12.

Let $f : X \to$ 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) α

D-sets

G_{x}

and

G_{y}

Y

containing

f (x)

and

f (y)

respectively such that

f (x) \notin G_{y}

and

f (y) \notin G_{x}

. By theorem 4.11,

f^{- 1} (G_{x})

and

f^{- 1} (G_{y})

are

(1, 2) α

D-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 $x \neq y \in X$ , there exists a $(1, 2) α$ -irresolute, surjective function $f : X \to Y$ , 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) α

-irresolute, surjective function and

Y

is ultra-

D_{1}

. Then for any

f (x) \neq f (y)

there exists

(1, 2) α

D-sets

G_{x}

and

G_{y}

such that

f (y) \in G_{y}

and

f (y) \in G_{y}

. By Theorem 4.11,

f^{- 1} (G_{x})

and

f^{- 1} (G_{y})

are two disjoint

(1, 2) α

D-sets containing

x

and

y

respectively. Therefore the space is ultra-

D_{1}

5. More on ultra-

R_{0}

space

Theorem 5.1.

A space $X$ is ultra- $R_{0}$ if and only if for any $x, y \in X$ , $(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})$ implies $(1, 2) α - c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset$ .

Proof. Necessity: Let

X

be an ultra-

R_{0}

space and if

x \neq y

, then

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

. Then there exists a

z \in X

such that

z \in (1, 2) α

c l ({x})

and

z \notin (1, 2) α - c l ({y})

. Hence we get a

(1, 2) α

-open set G containing

z

and

x

but not

y

. Therefore

x \notin (1, 2) α

c l ({y})

. Then

x \in X - (1, 2) α

c l ({y})

, which is a

(1, 2) α

-open set. Since

X

is ultra-

R_{0}

(1, 2) α - c l ({x}) \subseteq X - (1, 2) α - c l ({y})

and so

(1, 2) α

c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset

Sufficiency : Let

V

be a

(1, 2) α

-open set and

x \in V

. Let

y \notin V

. Then

y \in X - V

. Since

x \neq y

and

x \notin (1, 2) α - c l ({y})

, we get

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

. Then by assumption,

(1, 2) α - c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset

. Hence

y \notin (1, 2) α - c l ({x})

and so

(1, 2) α - c l ({x}) \subset 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 $x, y \in X$ , $(1, 2) α - k e r ({x}) \neq (1, 2) α - k e r ({y})$ implies $(1, 2) α - k e r ({x}) \cap (1, 2) α - k e r ({y}) = \emptyset$ .

Proof. If

(1, 2) α - k e r ({x}) \neq (1, 2) α - k e r ({y})

, then by theorem 3.12,

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

. We need to prove that

(1, 2) α - k e r ({x}) \cap (1, 2) α - k e r ({y}) = \emptyset

. If not, let there exist a

z

X

such that

z \in (1, 2) α - k e r ({x})

and

z \in (1, 2) α - k e r ({y})

. Then

x \in (1, 2) α - c l ({z})

and

y \in (1, 2) α - c l ({z})

. Then by theorem 5.1, we have

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

, which is a contradiction. Hence we get that

(1, 2) α - k e r ({x}) \cap (1, 2) α - k e r ({y}) = \emptyset

Conversely, let

X

be a space such that for any two distinct points

x

and

y

X

(1, 2) α - k e r ({x}) \neq (1, 2) α - k e r ({y})

implies that

(1, 2) α - k e r ({x}) \cap (1, 2) α - k e r ({y}) = \emptyset

. By theorem 3.12, we get that

(1, 2) α - c l ({x}) \neq (1, 2) α - c l ({y})

. If

(1, 2) α - c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset

, then by theorem 5.1,

X

is ultra-

R_{0}

. If not, let there be a

z

X

such that

z \in (1, 2) α - c l ({x})

and

z \in (1, 2) α - c l ({y})

. Then

x \in (1, 2) α - k e r ({z})

and

y \in (1, 2) α - k e r ({z})

. So we obtain that

which is a contradiction. Therefore

(1, 2) α - c l ({x}) \cap (1, 2) α - c l ({y}) = \emptyset

. 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 $A \neq \emptyset$ and $G \in (1, 2) α O (X)$ such that $A \cap G \neq \emptyset$ there exists a $F \in (1, 2) α C L (X)$ such that $F \cap G \neq \emptyset$ and $F \subset G$ .

(iii) For any $G \in (1, 2) α O (X), G = \cup {F \in (1, 2) α C L (X) ∕ F \subset G}$ .

(iv) For any $(1, 2) α$ -closed set $F$ , $F = \cap {G \in (1, 2) α O (X) ∕ F \subset G}$ .

(v) For any $x \in X, (1, 2) α - c l ({x}) \subset (1, 2) α - k e r ({x})$ .

Proof. (i)

\Rightarrow

(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 \cap G \neq \emptyset

. So there exists a

x \in A \cap G

. Then

x \in A

and

x \in G

. As

X

is ultra-

R_{0}

(1, 2) α - c l ({x}) \subset G

. Take

F = (1, 2) α - k e r ({x})

and then

F \subset G

and

A \cap G \neq \emptyset

(ii)

\Rightarrow

(iii) Let

G \in (1, 2) α O (X)

and

F \in (1, 2) α - C L ({X})

such that

F \subset G

. Then

⋃ {F ∕ F \in (1, 2) α - C L ({X}) and F \subset G} \subset G

. To prove the other inclusion, let

x \in G

. Then by assumption, there exists a

(1, 2) α

-closed set

F

such that

x \in F

and

F \subset G

. Hence

(iv)

\Rightarrow

(v) Let

x \in X

and there exists a

y \in X

such that

y \notin (1, 2) α - k e r ({x})

. Then there exists a

(1, 2) α

-open set

V

containing

x

but not

y

. That is

(1, 2) α - c l ({y}) \cap V = \emptyset

, which implies that

(1, 2) α - c l ({y}) \subset X - V

. Again, by our assumption,

\cap {G \in (1, 2) α O (X) ∕ (1, 2) α - c l ({y}) \subset G} \cap V = \emptyset

. Hence there exists a

(1, 2) α

-open set G such that x

\notin

G and

(1, 2) α - c l ({y}) \subset G

. Therefore

(1, 2) α

c l ({x}) \cap G = \emptyset

and

y \notin (1, 2) α - c l ({x})

. Consequently,

(1, 2) α - c l ({x}) \subset (1, 2) α - k e r ({x})

(v)

\Rightarrow

(i) Let

G \in (1, 2) α O (X)

and

x \in G

. Suppose that

y \in (1, 2) α - k e r ({x})

. Then

(1, 2) α - c l ({y}) \cap {x} \neq \emptyset

. So

{x} \in (1, 2) α - c l ({y})

and

y \in G

which implies that

(1, 2) α - c l ({x}) \subset (1, 2) α - k e r ({x}) \subset 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) $(1, 2) α - c l ({x}) = (1, 2) α - k e r ({x})$ for all $x \in X$ .

Proof. Suppose that

X

is an ultra-

R_{0}

space. Then by theorem 5.3,

(1, 2) α - c l ({x}) \subset (1, 2) α - k e r ({x})

for each

x \in X

. Assume that

y \in (1, 2) α - k e r ({x})

. Then

x \in (1, 2) α - c l ({y})

and so

(1, 2) α - c l ({x}) = (1, 2) α - c l ({y})

. Therefore

y \in (1, 2) α - c l ({x})

and so

(1, 2) α

k e r ({x}) \subset (1, 2) α - c l ({x})

. Thus

(1, 2) α - k e r ({x}) = (1, 2) α - c l ({x})

Theorem 5.5.

For a space $X$ , the following properties are equivalent:

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

(ii) $x \in (1, 2) α - c l ({y})$ if and only if $y \in (1, 2) α - c l ({x})$ , for any points $x$ and $y$ in $X$ .

Proof. (i)

\Rightarrow

(ii) Assume

X

is ultra-

R_{0}

. Let

x \in (1, 2) α - c l ({y})

and

D

be any

(1, 2) α

-open set such that

y \in D

. As

X

is ultra-

R_{0}

(1, 2) α - c l ({y}) \subset D

and hence

x \in D

. Therefore every

(1, 2) α

-open set containing

y

contains

x

. Hence

y \in (1, 2) α - c l ({x})

(ii)

\Rightarrow

(i) Let

U

be a

(1, 2) α

-open set and

x \in U

. If

y \notin U

, then

x \notin (1, 2) α - c l ({y})

and hence

y \notin (1, 2) α - c l ({x})

. This implies that

(1, 2) α - c l ({x}) \subset 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 $F = (1, 2) α - k e r ({F})$ .

(iii) If $F$ is $(1, 2) α$ -closed, then $(1, 2) α - k e r ({x}) \subset F$ , where $x \in F$ .

(iv) If $x \in X$ then $(1, 2) α - k e r ({x}) \subset (1, 2) α - c l ({x})$ .

Proof. (i)

\Rightarrow

(ii) Let

F

be a

(1, 2) α

-closed set and

x \notin F

. Then

X - F

is a

(1, 2) α

-open set containing

x

. Since

X

is ultra-

R_{0}

(1, 2) α - c l ({x}) \subset X - F

. Thus

(1, 2) α - c l ({x}) \cap F = \emptyset

. Hence, by lemma 3.3,

x \notin (1, 2) α - k e r ({F})

and so

(1, 2) α - k e r ({F}) \subset F

. Also by the deﬁnition of kernel of a set,

F \subset (1, 2) α - k e r ({F})

. Hence

F = (1, 2) α - k e r ({F})

(ii)

\Rightarrow

(iii) Let

x \in F

. So

(1, 2) α - k e r ({x}) \subset (1, 2) α - k e r ({F})

. As

F

(1, 2) α

-closed,

(1, 2) α - k e r ({F}) = F

(iii)

\Rightarrow

(iv) Since

x \in (1, 2) α - c l ({x})

and

(1, 2) α - c l ({x})

is a

(1, 2) α

-closed set, by (iii)

(1, 2) α - k e r ({x}) \subset (1, 2) α - c l ({x})

(iv)

\Rightarrow

(i) Let

x \in (1, 2) α - c l ({y})

. Then

y \in (1, 2) α - k e r ({x})

and

(1, 2) α

c l ({y})

is a

(1, 2) α

-closed set. By assumption, we have

y \in (1, 2) α - k e r ({x}) \subset (1, 2) α - c l ({x})

, which implies that

y \in (1, 2) α - c l ({x})

. Hence, by Theorem 5.5,

X

is ultra-

R_{0}

Department of Mathematics, Sri Parasakthi College,Courtallam-627802,Tirunelveli Dt., TamilNadu, India

Department of Mathematics, Arul Anndar College, Karumathur-625514, Madurai Dt., TamilNadu, India

Department of Mathematics, Jayaraj Annapackiam College for Women, Periyakulam-625 601,Theni Dt.,TamilNadu, India

1. Introduction

2. Preliminaries

3. (1, 2)α-Kernel and (1, 2)α-Closure

4. Ultra-Ti (i = 0,1) and Ultra-D1 spaces

5. More on ultra-R0 space

3. $(1, 2) α$ -Kernel and $(1, 2) α$ -Closure

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

5. More on ultra- $R_{0}$ space