# More on vg-separation axioms.

[section]1. IntroductionIn 1967, A. Wilansky has introduced the concept of US spaces. In 1968, C. E. Aull studied some separation axioms between [T.sub.1] and [T.sup.2] spaces, namely, [S.sub.1] and [S.sub.2]. In 1982, S. P. Arya et al have introduced and studied the concept of semi-US spaces and also they made study of s-convergence, sequentially semi-closed sets, sequentially s-compact notions. G. B. Navlagi studied P-Normal Almost-P-Normal, Mildly-P-Normal and Pre-US spaces. Recently S. Balasubramanian and P. Aruna Swathi Vyjayanthi studied v-Normal Almost-v-Normal, Mildlyv-Normal and v-US spaces. vg-open sets and vg-continuous mappings were introduced in 2009 and 2011 by S. Balasubramanian. The purpose of this paper is to examine the normality axioms, vg-US, vg-[S.sub.1] and vg-[S.sub.2] spaces. vg-convergence, sequentially vg-compact, sequentially vg-continuous maps, and sequentially sub vg-continuous maps are also introduced and studied in this paper. All notions and symbols which are not defined in this paper may be found in the appropriate references. Throughout the paper X and Y denote topological spaces on which no separation axioms are assumed explicitly stated.

[section]2. Preliminaries

Definition 2.1. A [subset] X is called

(i) closed [resp: Semi closed; v-closed] if its complement if is open [resp: semi open; v-open];

(ii) r[alpha]-open [v-open] if U [member of] [alpha]O(X)[RO(X)] such that U [subset] A [subset] [alpha][bar.(U)][U [subset] A [subset] [bar.(U)]];

(iii) semi-[theta]-open if it is the union of semi-regular sets and its complement is semi-[theta]-closed;

(iv) r-closed [[alpha]-closed; pre-closed; [beta]-closed] if A = [bar.([A.sup.0])][[([bar.([A.sup.0])]).sup.0] [subset or equal to] A;[bar.([A.sup.0])] [subset or equal to] A;([([bar.(A)]).sup.0])bar. [subset or equal to] A]; (v) g-closed [rg-closed; g"-closed; [??]-closed] if [bar.A] [subset or equal to] U whenever A [subset or equal to] U and U is open [gs-open; semi-open] in X;

(vi) sg-closed [gs-closed] if s[bar.(A)] [subset or equal to] U whenever A [subset or equal to] U and U is semi-open{open} in X;

(vii) pg-closed [gp-closed; gpr-closed] if p[bar.(A)] [subset or equal to] U whenever A [subset or equal to] U and U is pre-open [open; regular-open] in X;

(viii) [alpha]g-closed [g[alpha]-closed; rg[alpha]-closed; rg-closed; [alpha]gs-closed; g[alpha]-closed] if [alpha][bar.(A)] [subset or equal to] U whenever A [subset or equal to] U and U is open [[alpha]-open; r[alpha]-open; r-open; semi-open; gs-open] in X;

(ix) vg-closed if v[bar.(A)] [subset or equal to] U whenever A [subset or equal to] U and U is v-open in X;

(x) The family of all vg-open sets of X containing point x is denoted by vGO(X, x).

Definition 2.2. Let A [subset] X. Then a point x is said to be a

(i) limit point of A if each open set containing x contains some point y of A such that x [not equal to] y;

(ii) [T.sub.0]-limit point [20] of A if each open set containing x contains some point y of A such that cl{x} [not equal to] cl{y}, or equivalently, such that they are topologically distinct;

(iii) v-[T.sub.0]-limit point [11] of A if each open set containing x contains some point y of A such that vcl{x} [not equal to] vcl{y}, or equivalently, such that they are topologically distinct.

Definition 2.3 [10] A function f is said to be

(i) almost-v-irresolute if for each x [member of] X and each v-neighborhood V of f(x), v[bar.([f.sup.-1](V))] is a v-neighborhood of x;

(ii) sequentially v-continuous at x [member of] X if f([x.sub.n])[[right arrow].sup.U]f(x) whenever <[x.sub.n]> [[right arrow].sup.U]X. If f is sequentially v-continuous at all x [member of] X, then f is said to be sequentially v-continuous;

(iii) sequentially nearly v-continuous if for each point x [member of] X and each sequence <[x.sub.n]>[[right arrow].sup.U] x [member of] X, there exists a subsequence <[x.sub.nk]> of <[x.sub.n]> such that <f([x.sub.nk])> [[right arrow].sup.U]f(x).

(iv) sequentially sub-v-continuous if for each x [member of] X and each sequence <[x.sub.n]>[[right arrow].sup.U]x in X, there exists a subsequence <[x.sub.nk]> of <[x.sub.n]> and a point y [member of] Y such that <f([x.sub.nk])>[[right arrow].sup.U]y;

(v) sequentially v-compact preserving if the image f(K) of every sequentially v-compact set K of X is sequentially v-compact in Y.

[section]3. vg-[T.sub.0] limit point

Definition 3.1. In X, a point x is said to be a vg-[T.sub.0]-limit point of A if each vg-open set containing x contains some point y of A such that vg[bar.{x}] [not equal to] vg[bar.{y}], or equivalently, such that they are topologically distinct with respect to vg-open sets.

Example 3.1. Since regular open [right arrow] v-open [right arrow] vg-open, then r-[T.sub.0]-limit point [right arrow] v-T[0.sup.-] limit point [right arrow] vg-[T.sub.0]-limit point.

Definition 3.2. A set A together with all its vg-[T.sub.0]-limit points is denoted by [T.sub.0]-vg[bar.(A)].

Lemma 3.1. If x is a vg-[T.sub.0]-limit point of a set A then x is vg-limit point of A.

Lemma 3.2 If X is vg-[T.sub.0]-space then every vg-[T.sub.0]-limit point and every vg-limit point are equivalent.

Corollary 3.1.

(i) If X is r-[T.sub.0]-space then every vg-[T.sub.0]-limit point and every vg-limit point are equivalent;

(ii)If X is v-[T.sub.0]-space then every vg-[T.sub.0]-limit point and every vg-limit point are equivalent.

Theorem 3.1. For x [not equal to] y [member of] X,

(i) x is a vg-[T.sub.0]-limit point of {y} iff x[??]vg{y} and y [member of] vg[bar.{x}];

(ii) x is not a vg-[T.sub.0]-limit point of {y} iff either x [member of] vg{y} or vg[bar.{x}] = vg[bar.{y}];

(iii) x is not a vg-[T.sub.0]-limit point of {y} iff either x [member of] vg{y} or y [member of] vg[bar.{x}].

Corollary 3.2.

(i) If x is a vg-[T.sub.0]-limit point of {y}, then y cannot be a vg-limit point of {x};

(ii) If vg[bar.{x}] = vg[bar.{y}], then neither x is a vg-[T.sub.0]-limit point of {y} nor y is a vg-[T.sub.0]-limit point of {x}; (iii) If a singleton set A has no vg-[T.sub.0]-limit point in X, then vg[bar.(A)] = vg{x} for all x [member of] vg[bar.(A)].

Lemma 3.3. In X, if x is a vg-limit point of a set A, then in each of the following cases, X becomes vg-[T.sub.0]-limit point of A({x} [not equal to] A).

(i) vg[bar.{x}] [not equal to] vg[bar.{y}] for y [member of] A, x [not equal to] y;

(ii) vg[bar.{x}] = {x};

(iii) X is a vg-[T.sub.0]-space;

(iv) A-{x} is vg-open.

Corollary 3.3. In X, if x is a limit point [resp: r-limit point; v-limit point] of a set A, then in each of the following cases x becomes vg-[T.sub.0]-limit point of A({x} [not equal to] A).

(i) vg[bar.{x}] [not equal to] vg[bar.{y}] for y [member of] A, x [not equal to] y;

(ii) vg[bar.{x}] = {x};

(iii) X is a vg-[T.sub.0]-space;

(iv) A-{x} is vg-open.

[section]4. vg-[T.sub.0] and vg-[R.sub.i] axioms, i = 0, 1

In view of lemma 3.3(iii), vg-[T.sub.0]-axiom implies the equivalence of the concept of limit point of a set with that of vg-[T.sub.0]-limit point of the set. But for the converse, if x [member of] vg[bar.{y}] then vg[bar.{x}] [not equal to] vg{y} in general, but if x is a vg-[T.sub.0]-limit point of {y}, then vg[bar.{x}] = vg[bar.{y}].

Lemma 4.1. In a space X, a limit point x of {y} is a vg-[T.sub.0]-limit point of {y} iff vg[bar.{x}] [not equal to] vg[bar.{y}].

This lemma leads to characterize the equivalence of vg-[T.sub.0]-limit point and vg-limit point of a set as the vg-[T.sub.0]-axiom.

Theorem 4.1. The following conditions are equivalent:

(i) X is a vg-[T.sub.0] space;

(ii) Every vg-limit point of a set A is a vg-[T.sub.0]-limit point of A;

(iii) Every r-limit point of a singleton set {x} is a vg-[T.sub.0]-limit point of {x};

(iv) For any x, y [member of] X, x [not equal to] y if x [member of] vg[bar.{y}], then x is a vg-[T.sub.0]-limit point of {y}.

Note 4.1. In a vg-[T.sub.0]-space X if every point of X is a r-limit point of X, then every point of X is vg-[T.sub.0]-limit point of X. But a space X in which each point is a vg-[T.sub.0]-limit point of X is not necessarily a vg-[T.sub.0]-space.

Theorem 4.2. The following conditions are equivalent:

(i) X is a vg-[R.sub.0] space;

(ii) For any x, y [member of] X, if x [member of] vg[bar.{y}], then x is not a vg-[T.sub.0]-limit point of {y};

(iii) A point vg-closure set has no vg-[T.sub.0]-limit point in X;

(iv) A singleton set has no vg-[T.sub.0]-limit point in X.

Since every r-[R.sub.0]-space is vg-[R.sub.0]-space, we have the following Corollary.

Corollary 4.1. The following conditions are equivalent:

(i) X is a r-[R.sub.0] space;

(ii) For any x, y [member of] X, if x [member of] vg[bar.{y}], then X is not a vg-[T.sub.0]-limit point of {y};

(iii) A point vg-closure set has no vg-[T.sub.0]-limit point in X;

(iv) A singleton set has no vg-[T.sub.0]-limit point in X.

Since every v-[R.sub.0]-space is vg-[R.sub.0]-space, we have the following Corollary.

Corollary 4.2. The following conditions are equivalent:

(i) X is a v-[R.sub.0] space;

(ii) For any x, y [member of] X, if x [member of] vg[bar.{y}], then X is not a vg-[T.sub.0]-limit point of {y};

(iii) A point vg-closure set has no vg-[T.sub.0]-limit point in X;

(iv) A singleton set has no vg-[T.sub.0]-limit point in X.

Theorem 4.3. In a vg-[R.sub.0] space X, a point x is vg-[T.sub.0]-limit point of A iff every vg-open set containing x contains in finitely many points of A with each of which x is topologically distinct.

If vg-[R.sub.0] space is replaced by r[R.sub.0] space in the above theorem, we have the following corollaries.

Corollary 4.3. (a) In an r[R.sub.0]-space X,

(i) If a point x is r[T.sub.0]-limit point of a set then every vg-open set containing X contains infinitely many points of A with each of which X is topologically distinct;

(ii) If a point x is vg-[T.sub.0]-limit point of a set then every vg-open set containing X contains infinitely many points of A with each of which X is topologically distinct.

(b) In an v-[R.sub.0]-space X,

(i) If a point x is r[T.sub.0]-limit point of a set then every vg-open set containing x contains infinitely many points of A with each of which x is topologically distinct;

(ii) If a point x is vg-[T.sub.0]-limit point of a set then every vg-open set containing x contains infinitely many points of A with each of which x is topologically distinct.

Theorem 4.4. X is vg-[R.sub.0] space iff a set A of the form A = [union]vg[bar.{[x.sub.i] [:.sub.i = 1ton]}], a finite union of point closure sets has no vg-[T.sub.0]-limit point.

Corollary 4.4. (a) If X is r[R.sub.0] space and

(i) If A = [union]vg[bar.{[x.sub.i][:.sub.i = 1ton]}], a finite union of point closure sets has no vg-[T.sub.0]-limit point;

(ii) If X = [union]vg[bar.{[x.sub.i][:.sub.i = 1ton]}], then X has no vg-[T.sub.0]-limit point.

(b) If X is v-[R.sub.0] space and

(i) If A = [union]vg[bar.{[x.sub.i][:.sub.i = 1ton]}], a finite union of point closure sets has no vg-[T.sub.0]-limit point;

(ii) If X = [union]vg[bar.{[x.sub.i][:.sub.i = 1ton]}], then X has no vg-[T.sub.0]-limit point.

Theorem 4.5. The following conditions are equivalent:

(i) X is vg-[R.sub.0]-space;

(ii) For any X and a set A in X, x is a vg-[T.sub.0]-limit point of A iff every vg-open set containing x contains infinitely many points of A with each of which X is topologically distinct.

Various characteristic properties of vg-[T.sub.0]-limit points studied so far is enlisted in the following Theorem for a ready reference.

Theorem 4.6. In a vg-[R.sub.0]-space, we have the following:

(i) A singleton set has no vg-[T.sub.0]-limit point in X;

(ii) a finite set has no vg-[T.sub.0]-limit point in X;

(iii) a point vg-closure has no set vg-[T.sub.0]-limit point in X;

(iv) a finite union point vg-closure sets have no set vg-[T.sub.0]-limit point in X;

(v) for x, y [member of] X, x [member of] [T.sub.0]-vg[bar.{y}] iff x = y;

(vi) for any x, y [member of] X, x [not equal to] y iff neither X is vg-[T.sub.0]-limit point of {y} nor y is vg- [T.sub.0]-limit point of {x};

(vii) for any x, y [member of] X, x [not equal to] y iff [T.sub.0]-vg[bar.{x}] [intersection] [T.sub.0]-vg[bar.{y}] = [phi];

(viii) any point x [member of] X is a vg-[T.sub.0]-limit point of a set A in X iff every vg-open set containing

X contains infinitely many points of A with each which X is topologically distinct.

Theorem 4.7. X is vg-[R.sub.1] iff for any vg-open set U in X and points x, y such that x [member of] X-U, y [member of] U, there exists a vg-open set V in X such that y [member of] V [subset] U, x [not member of] V.

Lemma 4.2. In vg-[R.sub.1] space X, if x is a vg-[T.sub.0]-limit point of X, then for any non empty vg-open set U, there exists a non empty vg-open set V such that V [subset] U, x [not member of] vg[bar.(V)].

Lemma 4.3. In a vg-regular space X, if X is a vg-[T.sub.0]-limit point of X, then for any non empty vg-open set U, there exists a non empty vg-open set V such that vg[bar.(V)] [subset] U, x [not member of] vg[bar.(V)].

Corollary 4.5. In a regular space X,

(i) If X is a vg-[T.sub.0]-limit point of X, then for any non empty vg-open set U, there exists a non empty vg-open set V such that vg[bar.(V)] [subset] U, x [not member of] vg[bar.(V)];

(ii) If x is a [T.sub.0]-limit point of X, then for any non empty vg-open set U, there exists a non empty vg-open set V such that vg[bar.(V)] [subset] U, x [not member of] vg[bar.(V)].

Theorem 4.8. If X is a vg-compact vg-[R.sub.1]-space, then X is a Baire space.

Proof. Let {[A.sub.n]} be a countable collection of vg-closed sets of X, each [A.sub.n] having empty interior in X. Take [A.sub.1], since [A.sub.1] has empty interior, [A.sub.1] does not contain any vg-open set say [U.sub.0]. Therefore we can choose a point y [member of] [U.sub.0] such that y [not member of] [A.sub.1]. For X is vg-regular, and y [member of] (X - [A.sub.1]) [intersection] [U.sub.0], a vg-open set, we can find a vg-open set [U.sub.1] in X such that y [member of] [U.sub.1], vg[bar.([U.sub.1])] [subset] (X - [A.sub.1]) [intersection] [U.sub.0]. Hence [U.sub.1] is a non empty vg- open set in X such that vg[bar.([U.sub.1])] [subset] [U.sub.0] and vg[bar.(U1)] [intersection] [A.sub.1] = [phi]. Continuing this process, in general, for given non empty vg-open set [U.sub.n-1], we can choose a point of [U.sub.n-1] which is not in the vg-closed set [A.sub.n] and a vg-open set [U.sub.n] containing this point such that vg[bar.(Un)] [subset] [U.sub.n-1] and vg[bar.(Un)] [intersection] [A.sub.n] = [phi]. Thus we get a sequence of nested non empty vg-closed sets which satisfies the finite intersection property. Therefore [intersection] vg[bar.([U.sub.n])] [not equal to] [phi]. Then some x [member of] [intersection] vg[bar.([U.sub.n])] which in turn implies that x [member of] [U.sub.n-1] as vg[bar.([U.sub.n])] [subset] [U.sub.n - 1] and x [not member of] [A.sub.n] for each n.

Corollary 4.6. If X is a compact vg-[R.sub.1]-space, then X is a Baire space.

Corollary 4.7. Let X be a vg-compact vg-[R.sub.1]-space. If {[A.sub.n]} is a countable collection of vg-closed sets in X, each [A.sub.n] having non-empty vg-interior in X, then there is a point of X which is not in any of the [A.sub.n].

Corollary 4.8. Let X be a vg-compact [R.sub.1]-space. If {[A.sub.n]} is a countable collection of vg-closed sets in X, each [A.sub.n] having non-empty vg-interior in X, then there is a point of X which is not in any of the [A.sub.n].

Theorem 4.9. Let X be a non empty compact vg-[R.sub.1]-space. If every point of X is a vg-[T.sub.0]-limit point of X then X is uncountable.

Proof. Since X is non empty and every point is a vg-[T.sub.0]-limit point of X, X must be infinite. If X is countable, we construct a sequence of vg-open sets {[V.sub.n]} in X as follows:

Let X = [V.sub.1], then for set [x.sub.1] is a vg-[T.sub.0]-limit point of X, we can choose a non empty vg-open [V.sub.2] in X such that [V.sub.2] [subset] [V.sub.1] and [x.sub.1] [not member of] vg[bar.[V.sub.2]]. Next for [x.sub.2] and non empty vg-open set [V.sub.2], we can choose a non empty vg-open set [V.sub.3] in X such that [V.sub.3] [subset] [V.sub.2] and [x.sub.2] [not member of] vg[bar.[V.sub.3]]. Continuing this process for each [x.sub.n] and a non empty vg-open set [V.sub.n], we can choose a non empty vg-open set [V.sub.n + 1] in X such that [V.sub.n + 1] [subset] [V.sub.n] and [x.sub.n] [not member of] vg[bar.[V.sub.n + 1]].

Now consider the nested sequence of vg-closed sets vg[bar.[V.sub.1]] [contains] vg[bar.[V.sub.2]] [contains] vg[bar.[V.sub.3]] [contains]... [contains] vg[bar.[V.sub.n]] [contains] ..., since X is vg-compact and {vg[bar.[V.sub.n]]} the sequence of vg-closed sets satisfy finite intersection property. By Cantors intersection theorem, there exists an x in X such that x [member of] vg[bar.[V.sub.n]]. Further x [member of] X and x [member of] [V.sub.1], which is not equal to any of the points of x. Hence X is uncountable.

Corollary 4.9. Let X be a non empty vg-compact vg-[R.sub.1]-space. If every point of X is a vg-[T.sub.0]-limit point of x, then X is uncountable.

[section]5. vg-[T.sub.0]-identification spaces and vg-separation axioms

Definition 5.1. Let X be a topological space and let R be the equivalence relation on X defined by x R y iff vg[bar.{x}] = vg[bar.{y}].

Problem 5.1. Show that xRy iff vg[bar.{x}] = vg{y} is an equivalence relation. Definition 5.1. The space ([X.sub.0], Q([X.sub.0])) is called the vg-[T.sub.0]-identification space of (X, [tau]), where [X.sub.0] is the set of equivalence classes of [tau] and Q([X.sub.0]) is the decomposition topology on [X.sub.0].

Let [P.sub.X] : (X, [tau]) [right arrow] ([X.sub.0], Q([X.sub.0])) denote the natural map.

Lemma 5.1. If x [member of] X and A [subset] X, then x [member of] vg[bar.A] iff every vg-open set containing X intersects A.

Theorem 5.1. The natural map [P.sub.X] : (X, [tau]) [right arrow] ([X.sub.0], Q([X.sub.0])) is closed, open and [P.sub.X]-1([P.sub.X(O)]) = O for all O [member of] PO(X, [tau]) and ([X.sub.0], Q([X.sub.0])) is vg-[T.sub.0].

Proof. Let O [member of] PO(X, [tau]) and let C [member of] [P.sub.X](O). Then there exists x [member of] O such that [P.sub.X](x) = C. If y [member of] C, then vg[bar.{y}] = vg[bar.{x}], which, implies y [member of] O. Since [tau] [subset] PO(X, [tau]), then [P.sup.-1.sub.X]([P.sub.X](U)) = U for all U [member of] [tau], which implies [P.sub.X] is closed and open.

Let G, H [member of] [X.sub.0] such that G [not equal to] H and let x [member of] G and y [member of] H. Then vg[bar.{x}] [not equal to] vg[bar.{y}], which implies that x [not member of] vg[bar.{y}] or y[??]vg[bar.{x}], say x[??]vg[bar.{y}]. Since [P.sub.X] is continuous and open, then G [member of] A = [P.sub.X]{X-vg[bar.{y}]} [not member of] PO([X.sub.0], Q([X.sub.0])) and H [not member of] A.

Theorem 5.2. The followings are equivalent:

(i) X is vg[R.sub.0];

(ii) [X.sub.0] = {vg[bar.{x}]:x [member of] X};

(iii) ([X.sub.0], Q([X.sub.0])) is vg[T.sub.1].

Proof. (i) [right arrow] (ii). Let C [member of] [X.sub.0], and let x [member of] C. If y [member of] C, then y [member of] vg[bar.{y}] = vg[bar.{x}], which implies C [member of] vg[bar.{x}]. If y [member of] vg[bar.{x}], then x [member of] vg[bar.{y}], since, otherwise, x [member of] X-vg[bar.{y}] [member of] PO(X, [tau]) which implies vg[bar.{x}] [subset] X-vg[bar.{y}], which is a contradiction. Thus, if y [member of] vg[bar.{x}], then x [member of] vg[bar.{y}], which implies vg[bar.{y}] = vg[bar.{x}] and y [member of] C. Hence [X.sub.0] = {vg[bar.{x}]:x [member of] X}.

(ii) [right arrow] (iii). Let A [not equal to] B [member of] [X.sub.0]. Then there exists x, y [member of] X such that A = vg[bar.{x}];B = vg[bar.{y}], and vg[bar.{x}] [intersection] vg[bar.{y}] = [phi]. Then [Amember of]C = [P.sub.X](X-vg[bar.{y}]) [member of] PO([X.sub.0], Q([X.sub.0])) and B [not member of] C. Thus ([X.sub.0], Q([X.sub.0])) is vg-[T.sub.1].

(iii) [right arrow] (i). Let x [member of] U [member of] vGO(X). Let y [not member of] U and [C.sub.x], [C.sub.y] [member of] [X.sub.0] containing x and y respectively. Then x [not member of] vg[bar.{y}], which implies [C.sub.x] [not equal to] [C.sub.y] and there exists vg- open set A such that [C.sub.x] [member of] A and [C.sub.y] [not member of] A. Since [P.sub.X] is continuous and open, then y [member of] B = [P.sub.X]-1(A) [member of] x [member of] vGO(X) and x [not member of] B, which implies y [not member of] vg[bar.{x}]. Thus vg[bar.{x}] [subset] U. This is true for all vg{x} implies [intersection] vg[bar.{x}] [subset] U. Hence X is vg-[R.sub.0]

Theorem 5.3. X is vg-[R.sub.1] iff ([X.sub.0], Q([X.sub.0])) is vg-[T.sup.2].

The proof is straight forward from using theorems 5.1 and 5.2 and is omitted.

[section]6. vg-open functions and vg-[T.sub.i] spaces, i = 0,1,2

Theorem 6.1. X is vg-[T.sub.i], i = 0,1,2 iff there exists a vg-continuous, almost-open, 1-1 function f : X into a vg-[T.sub.i] space, i = 0,1,2, respectively.

Theorem 6.2. If f : X [right arrow] Y is vg-continuous, vg-open, and x, y [member of] X such that vg[bar.{x}] = vg[bar.{y}], then vg[bar.{f(x)]} = vg[bar.{f(y)]}.

Proof. Suppose vg[bar.{f(x)}] [not equal to] vg[bar.{f(y)}]. Then f(x)[??]vg{f(y)} or f(y)[??]vg[bar.{f(x)}], say f(x) [not member of] vg[bar.{f(y)}]. Then f(x) [member of] A = Y - vg[bar.{f(y)}] [member of] vGO(Y). If B = Y - [bar.[A.sup.0]], then f(x) [not member of] B, and B [intersection] vg[bar.{f(y)}] [not equal to] [phi], which implies f(y) [member of] B, y [member of] [f.sup.-1](B) [member of] vGO(X), and x[??][f.sup.-1](B) which is a contradiction. Thus [bar.[A.sup.0]] = Y. Since f(y) [not member of] A, then y [not member of] [([f.sup.- 1](A)).sup.0]. If x [member of] [bar.[([f.sup.-1](A)).sup.0]], then {x}[union]([f.sup.-1](A))0 is vg-open containing x and not y, which is a contradiction. Hence x [member of] U = X - [bar.[([f.sup.-1](A)).sup.0]] and [phi] [not equal to] f(U) [member of] vGO(Y). Then C = [(f(U)).sup.0] [intersection] A0 = [phi], for suppose not. Then [f.sup.-1](C) [member of] vGO(X), which implies [f.sup.-1](C) [subset] ([f.sup.-1](C))0 [subset] [([f.sup.-1](A)).sup.0], which is a contradiction. Hence C = [phi], which contradicts [bar.[A.sup.0]] = Y.

Theorem 6.3. The followings are equivalent:

(i) X is vg-[T.sub.0];

(ii) Elements of [X.sub.0] are singleton sets;

(iii) There exists a vg-continuous, vg-open, 1-1 function f:X [right arrow] Y, where Y is vg-[T.sub.0].

Proof.(i) is equivalent to (ii) and (i) [right arrow] (iii) are straight forward and is omitted.

(iii) [right arrow] (i). Let x, y [member of] X such that f(x) [not equal to] f(y), which implies vg[bar.{f(x)]} [not equal to] vg[bar.{f(y)]}. Then by theorem 6.2, vg[bar.{x}] [not equal to] vg[bar.{y}]. Hence X is vg-[T.sub.0].

Corollary 6.1. A space X is vg-[T.sub.i], i = 1,2 iff X is vg-[T.sub.i-1], i = 1,2, respectively, and there exists a vg-continuous, vg-open, 1-1 function f:X into a vg-[T.sub.0] space.

Definition 6.1. f : X [right arrow] Y is point-vg-closure 1-1 iff for x, y [member of] X such that vg[bar.{x}] [not equal to] vg[bar.{y}], vg[bar.{f(x)]} [not equal to] vg[bar.{f(y)]}.

Theorem 6.4. (i) If f:X [right arrow] Y is point vg-closure 1-1 and X is vg-[T.sub.0], then f is 1-1;

(ii) If f:X [right arrow] Y, where X and Y are vg-[T.sub.0], then f is point vg-closure 1-1 iff f is 1-1.

Proof. omitted.

The following result can be obtained by combining results for vg-[T.sub.0]-identification spaces, vg-induced functions and vg-[T.sub.i] spaces, i = 1,2.

Theorem 6.5. X is vg-[R.sub.i], i = 0,1 iff there exists a vg-continuous, almost-open point vg-closure 1-1 function f : X into a vg-[R.sub.i] space, i = 0,1, respectively.

[section]7. vg-normal, almost vg-normal and mildly vg-normal spaces

Definition 7.1. A space X is said to be vg-normal if for any pair of disjoint closed sets [F.sub.1] and [F.sub.2], there exist disjoint vg-open sets U and V such that [F.sub.1] [subset] U and [F.sub.2] [subset] V

Note 7.1. From the above Definition we have the following implication diagram.

[ILLUSTRATION OMITTED]

Example 7.1 Let X = {a, b, c} and [tau] = {[phi], {a}, {6, c}, X}. Then X is vg-normal.

Example 7.2 Let X = {a, b, c, d} and [tau] = {[phi], {b, d}, {a, b, d}, {b, c, d}, X}. Then X is vg-normal and is not normal.

We have the following characterization of vg-normality.

Theorem 7.1. For a space X the followings are equivalent:

(a) X is vg-normal;

(b) For every pair of open sets U and V whose union is X, there exist vg-closed sets A and B such that A [subset] U, B [subset] V and A [union] B = X;

(c) For every closed set F and every open set G containing F, there exists a vg-open set U such that F [subset] U [subset] vg[bar.(U)] [subset] G.

Proof. (a) [right arrow] (b). Let U and V be a pair of open sets in a vg-normal space X such that X = U[union]V. Then X-U, X-V are disjoint closed sets. Since X is vg-normal there exist disjoint vg-open sets [U.sub.1] and [V.sub.1] such that X - U [subset] [U.sub.1] and X - V [subset] [V.sub.1]. Let A = X - [U.sub.1], B = X-[V.sub.1]. Then A and B are vg-closed sets such that A [subset] U, B [subset] V and A [union] B = X.

(b) [right arrow] (c). Let F be a closed set and G be an open set containing F. Then X-F and G are open sets whose union is X. Then by (b), there exist vg-closed sets [W.sub.1] and [W.sub.2] such that [W.sub.1] [subset] X - F and [W.sub.2] [subset] G and [W.sub.1][union][W.sub.2] = X. Then F [subset] X - [W.sub.1], X - G [subset] X - [W.sub.2] and (X-[W.sub.1]) [intersection] (X-[W.sub.2]) = [phi]. Let U = X-[W.sub.1] and V = X-[W.sub.2]. Then U and V are disjoint vg-open sets such that F [subset] U [subset] X - V [subset] G. As X - V is vg-closed set, we have vg[bar.(U)] [subset] X - V and F [subset] U [subset] vg[bar.(U)] [subset] G.

(c) [right arrow] (a). Let [F.sub.1] and [F.sub.2] be any two disjoint closed sets of X. Put G = X - [F.sub.2], then [F.sub.1] [intersection] G = [phi].[F.sub.1] [subset] G where G is an open set. Then by (c), there exists a vg-open set U of X such that [F.sub.1] [subset] U [subset] vg[bar.(U)] [subset] G. It follows that [F.sub.2] [subset] X-vg[bar.(U)] = V, say, then V is vg-open and U [intersection] V = [phi]. Hence [F.sub.1] and [F.sub.2] are separated by vg-open sets U and V. Therefore X is vg-normal.

Theorem 7.2. A regular open subspace of a vg-normal space is vg-normal.

Proof. Let Y be a regular open subspace of a vg-normal space X. Let A and B be disjoint closed subsets of Y. By vg-normality of X, there exist disjoint vg-open sets U and V in X such that A [subset] U and B [subset] V, U [intersection] Y and V [intersection] Y are vg-open in Y such that A [subset] U [intersection] Y and B [subset] V [intersection] Y. Hence Y is vg-normal.

Example 7.3. Let X = {a, b, c} with [tau] = {[phi], {a}, {b}, {a, b}, X} is vg-normal and vg-regular.

Now, we define the following.

Definition 7.2. A function f : X [right arrow] Y is said to be almost-vg-irresolute if for each x [member of] X and each vg-neighborhood V of f(x), vg[bar.(f-1(V))] is a vg-neighborhood of X.

Clearly every vg-irresolute map is almost vg-irresolute.

The Proof of the following lemma is straightforward and hence omitted. Lemma 7.1. f is almost vg-irresolute iff [f.sup.-1](V) [subset] vg-int(vgcl([f.sup.-1](V)))) for every V [member of] vGO(Y).

Now we prove the following.

Lemma 7.2. f is almost vg-irresolute iff f(vg[bar.(U)]) [subset] vg(f(U)) for every U [member of] vGO(X).

Proof. Let U [member of] vGO(X). Suppose y[??]vg[bar.(f(U))]. Then there exists V [member of] vGO(y) such that V [intersection] f(U) = [phi]. Hence [f.sup.-1](V) [intersection] U = [phi]. Since U [member of] vGO(X), we have vg[(vg[bar.([f.sup.-1](V))]).sup.0] [intersection] vg[bar.(U)] = [phi]. Then by lemma 7.1, [f.sup.-1](V) [intersection] vg[bar.(U)] = [phi] and hence V [intersection] f(vg[bar.(U)]) = [phi]. This implies that y [not member of] f(vg[bar.(U)]).

Conversely, if V [member of] vGO(Y), then W = X - vg[bar.([f.sup.-1](V))] [member of] vGO(X). By hypothesis, f(vg[bar.(W)]) [subset] vg[bar.(f(W)))] and hence X - vg[(vg([f.sup.-1](V))).sup.0] = vg[bar.(W)] [subset] [f.sup.-1](vg(f(W))) [subset] [f.sup.-1](vg[bar.[f(X-[f.sup.-1](V))]]) [subset] [f.sup.-1][vg[bar.(Y - V)]] = [f.sup.-1](Y - V) = X - [f.sup.-1](V). Therefore, [f.sup.-1](V) [subset] vg(vg([f.sup.-1](V)))0 By lemma 7.1, f is almost vg-irresolute.

Now we prove the following result on the invariance of vg-normality.

Theorem 7.3. If f is an M-vg-open continuous almost vg-irresolute function from a vg-normal space X onto a space Y, then Y is vg-normal.

Proof. Let A be a closed subset of Y and B be an open set containing A. Then by continuity of f, [f.sup.-1](A) is closed and [f.sup.-1](B) is an open set of X such that [f.sup.-1](A) [subset] [f.sup.-1](B). As X is vg-normal, there exists a vg-open set U in X such that [f.sup.-1](A) [subset] U [subset] vg[bar.(U)] [subset] [f.sup.-1](B). Then f([f.sup.-1](A)) [subset] f(U) [subset] f(vg[bar.(U)]) [subset] f([f.sup.-1](B)). Since f is M-vg-open almost vg- irresolute surjection, we obtain A [subset] f(U) [subset] vg[bar.(f(U))] [subset] B. Then by theorem 7.1 Y is vg-normal.

Lemma 7.3. A mapping f is M-vg-closed if and only if for each subset B in Y and for each vg-open set U in X containing [f.sup.-1](B), there exists a vg-open set V containing B such that [f.sup.-1](V) [subset] U.

Now we prove the following.

Theorem 7.4. If f is an M-vg-closed continuous function from a vg-normal space onto a space Y, then Y is vg-normal.

Proof of the theorem is routine and hence omitted.

Now in view of lemma 2.2 [19] and lemma 7.3, we prove the following result.

Theorem 7.5. If f is an M-vg-closed map from a weakly Hausdorff vg-normal space X onto a space Y such that [f.sup.-1](y) is S-closed relative to X for each y [member of] Y, then Y is vg-[T.sup.2]. Proof. Let [y.sub.1] and [y.sub.2] be any two distinct points of Y. Since X is weakly Hausdorf f, [f.sup.-1]([y.sub.1]) and [f.sup.-1]([y.sub.2]) are disjoint closed subsets of X by lemma 2.2 [19]. As X is vg-normal, there exist disjoint vg-open sets [V.sub.1] and [V.sub.2] such that [f.sup.-1]([y.sub.i]) [subset] [V.sub.i], for i = 1,2. Since f is M-vg-closed, there exist vg-open sets [U.sub.i] containing [y.sub.i] such that [f.sup.-1]([U.sub.i]) [subset] [V.sub.i] for i = 1,2. Then it follows that [U.sub.1] [intersection] [U.sub.2] = [phi]. Hence Y is vg-[T.sup.2].

Theorem 7.6. For a space X we have the following:

(a) If X is normal then for any disjoint closed sets A and B, there exist disjoint vg-open sets U, V such that A [subset] U and B [subset] V;

(b) if X is normal then for any closed set A and any open set V containing A, there exists an vg-open set U of X such that A [subset] U [subset] vg[bar.(U)] [subset] V.

Definition 7.2. X is said to be almost vg-normal if for each closed set A and each regular closed set B such that A [intersection] B = [phi], there exist disjoint vg-open sets U and V such that A [subset] U and B [subset] V.

Note 7.2. From the above definition we have the following implication diagram.

[ILLUSTRATION OMITTED]

Example 7.4. Let X = {a, b, c} and [tau] = {[phi], {a}, {a, b}, {a, c}, X}. Then X is almost vg-normal and vg-normal.

Now, we have characterization of almost vg-normality in the following.

Theorem 7.7. For a space X the following statements are equivalent:

(a) X is almost vg-normal;

(b) for every pair of sets U and V, one of which is open and the other is regular open whose union is X, there exist vg-closed sets G and H such that G [subset] U, H [subset] V and G [union] H = X;

(c) for every closed set A and every regular open set B containing A, there is a vg-open set V such that A [subset] V [subset] vg[bar.(V)] [subset] B.

Proof. (a) [right arrow] (b). Let U be an open set and V be a regular open set in an almost vg-normal space X such that U[union]V = X. Then (X - U) is closed set and (X - V) is regular closed set with (X-U) [intersection] (X - V) = [phi]. By almost vg-normality of X, there exist disjoint vg-open sets [U.sub.1] and [V.sub.1] such that X - U [subset] [U.sub.1] and X - V [subset] [V.sub.1]. Let G = X - [U.sub.1] and H = X - [V.sub.1]. Then G and H are vg-closed sets such that G [subset] U, [Hsubset]V and G[union]H = X.

(b) [right arrow] (c) and (c) [right arrow] (a) are obvious.

One can prove that almost vg-normality is also regular open hereditary. Almost vg-normality does not imply almost vg-regularity in general. However, we observe that every almost vg-normal vg-[R.sub.0] space is almost vg-regular.

Next, we prove the following.

Theorem 7.8. Every almost regular, v-compact space X is almost vg-normal.

Recall that f : X [right arrow] Y is called rc-continuous if inverse image of regular closed set is regular closed.

Now, we state the invariance of almost vg-normality in the following.

Theorem 7.9. If f is continuous M-vg-open rc-continuous and almost vg-irresolute surjection from an almost vg-normal space X onto a space Y, then Y is almost vg-normal.

Definition 7.3. A space X is said to be mildly vg-normal if for every pair of disjoint regular closed sets [F.sub.1] and [F.sub.2] of X, there exist disjoint vg-open sets U and V such that [F.sub.1] [subset] U and [F.sub.2] [subset] V.

Note 7.3. From the above Definition we have the following implication diagram.

[ILLUSTRATION OMITTED]

We have the following characterization of mild vg-normality.

Theorem 7.10. For a space X the following are equivalent.

(a) X is mildly vg-normal;

(b) for every pair of regular open sets U and V whose union is X, there exist vg-closed sets G and H such that G [subset] U, H [subset] V and G [union] H = X;

(c) for any regular closed set A and every regular open set B containing A, there exists a vg-open set U such that A [subset] U [subset] vg[bar.(U)] [subset] B;

(d) for every pair of disjoint regular closed sets, there exist vg-open sets U and V such that A [subset] U, B [subset] V and vg[bar.(U)] [intersection] vg[bar.(V)] = [phi].

This theorem may be proved by using the arguments similar to those of theorem 7.7.

Also, we observe that mild vg-normality is regular open hereditary.

We define the following:

Definition 7.4. A space X is weakly vg-regular if for each point X and a regular open set U containing {x}, there is a vg-open set V such that x [member of] V [subset] [bar.V] [subset] U.

Theorem 7.11. If f : X [right arrow] Y is an M-vg-open rc-continuous and almost vg-irresolute function from a mildly vg-normal space X onto a space Y, then Y is mildly vg-normal.

Proof. Let A be a regular closed set and B be a regular open set containing A. Then by rc-continuity of f, [f.sup.-1](A) is a regular closed set contained in the regular open set [f.sup.-1](B). Since X is mildly vg-normal, there exists a vg-open set V such that [f.sup.-1](A) [subset] V [subset] vg[bar.(V)] [subset] [f.sup.-1](B) by theorem 7.10. As f is M-vg-open and almost vg-irresolute surjection, it follows that f(V) [member of] vGO(Y) and A [subset] f(V) [subset] vg[bar.(f(V))] [subset] B. Hence Y is mildly vg-normal.

Theorem 7.12. If f : X [right arrow] Y is rc-continuous, M-vg-closed map from a mildly vg-normal space X onto a space Y, then Y is mildly vg-normal.

[section]8. vg-US spaces

Definition 8.1. A sequence <[x.sub.n]> is said to vg-converges to x [member of] X, written as < [x.sub.n]> [[right arrow].sup.vg] x if <[x.sub.n]> is eventually in every vg-open set containing x.

Clearly, if a sequence <[x.sub.n]> [right arrow] UX of X, then <[x.sub.n]>[[right arrow].sup.vg] to X.

Definition 8.2. A space X is said to be sg-US if every sequence <[x.sub.n]> in Xvg-converges to a unique point.

Theorem 8.1. Every vg-US space is vg-[T.sub.1].

Proof. Let X be vg-US space. Let x [not equal to] y [member of] X. Consider the sequence <[x.sub.n]> where [x.sub.n] = x for every n. Cleary, <[x.sub.n]>[[right arrow].sup.vg]X. Also, since x [not equal to] y and X is vg-US, <[x.sub.n]>[[??].sup.vg] y, i.e. there exists a vg-open set V containing y but not X. Similarly, if we consider the sequence <[y.sub.n]> where [y.sub.n] = y for all n, and proceeding as above we get a vg-open set U containing X but not y. Thus, the space X is vg-[T.sub.1].

Theorem 8.2. Every vg-[T.sup.2] space is vg-US.

Proof. Let X be vg-[T.sup.2] space and <[x.sub.n]> be a sequence in X. If <[x.sub.n]>vg-converge to two distinct points X and y. That is, <[x.sub.n]> is eventually in every vg-open set containing X and also in every vg-open set containing y. This is contradiction since X is vg-[T.sup.2] space. Hence the space X is v g-US.

Definition 8.3. A set F is sequentially vg-closed if every sequence in Fvg-converges to a point in F.

Theorem 8.3. X is vg-US iff the diagonal set is a sequentially vg-closed subset of X x X.

Proof. Let X be vg-US. Let <[x.sub.n], [x.sub.n]> be a sequence in [DELTA]. Then <[x.sub.n]> is a sequence in X. As X is vg-US, <[x.sub.n]>[[right arrow].sup.vg]x for a unique x [member of] X, i.e. if <[x.sub.n]>[[right arrow].sup.vg]x and y. Thus, x = y. Hence [DELTA] is sequentially vg-closed set.

Conversely, let [DELTA] be sequentially vg-closed. Let a sequence <[x.sub.n]>[[right arrow].sup.vg]x and y. Hence <[x.sub.n], [x.sub.n]>[[right arrow].sup.vg](x, y). Since [DELTA] is sequentially vg-closed, (x, y) [member of] [DELTA] which means that x = y implies space X is vg-US. Definition 8.4. A subset G of a space X is said to be sequentially vg-compact if every sequence in G has a subsequence which vg-converges to a point in G.

Theorem 8.4. In a sg-US space every sequentially vg-compact set is sequentially vg-closed.

Proof. Let Y be a sequentially vg-compact subset of sg-US space X. Let <[x.sub.n]> be a sequence in Y. Suppose that <[x.sub.n]>vg-converges to a point in X-Y. Let <[x.sub.np]> be subsequence of <[x.sub.n]> that vg-converges to a point y [member of] Y, since Y is sequentially vg-compact. Also, let a subsequence <[x.sub.np]> of <[x.sub.n]>[[right arrow].sup.vg]x [member of] X-Y. Since <[x.sub.np]> is a sequence in the sg-US space X, x = y. Thus, Y is sequentially vg-closed set.

Next, we give a hereditary property of sg-US spaces.

Theorem 8.5. Every regular open subset of a sg-US space is sg-US.

Proof. Let X be a vg-US space and Y [subset] X be an regular open set. Let <[x.sub.n]> be a sequence in Y. Suppose that <[x.sub.n]>[[right arrow].sup.vg]x and y in Y. We shall prove that <[x.sub.n]>[[right arrow].sup.vg]x and y in X. Let U be any vg-open subset of X containing X and V be any vg-open set of X containing y. Then, U [intersection] Y and V [intersection] Y are vg-open sets in Y. Therefore, <[x.sub.n]> is eventually in U [intersection] Y and V [intersection] Y and so in U and V. Since X is sg-US, this implies that x = y. Hence the subspace Y is sg-US.

Theorem 8.6. A space X is vg-[T.sup.2] iff it is both vg-[R.sub.1] and vg-US. Proof. Let X be vg-[T.sup.2] space. Then X is vg-[R.sub.1] and vg-US by theorem 8.2. Conversely, let X be both vg-[R.sub.1] and vg-US space. By theorem 8.1, X is both vg-[T.sub.1] and vg-[R.sub.1] and, it follows that space X is vg-[T.sup.2].

Definition 8.5. A point y is a vg-cluster point of sequence <[x.sub.n]>iff<[x.sub.n]> is frequently in every vg-open set containing X.

The set of all vg-cluster points of <[x.sub.n]> will be denoted by vg[bar.(xn)].

Definition 8.6. A point y is vg-side point of a sequence <[x.sub.n]> if y is a vg-cluster point of <[x.sub.n]> but no subsequence of <[x.sub.n]>vg-converges to y. Now, we define the following.

Definition 8.7. A space X is said to be vg-[S.sub.1] if it is sg-US and every sequence <[x.sub.n]> vg-converges with subsequence of <[x.sub.n]>vg-side points.

Definition 8.8. A space X is said to be vg-[S.sub.2] if it is sg-US and every sequence <[x.sub.n]> in X vg-converges which has no vg-side point.

Lemma 8.1. Every vg-[S.sub.2] space is vg-[S.sub.1] and Every vg-[S.sub.1] space is vg-US. Now using the notion of sequentially continuous functions, we define the notion of sequentially vg-continuous functions.

Definition 8.9. A function f is said to be sequentially vg-continuous at x [member of] X if f([x.sub.n])[[right arrow].sup.U]gf(x) whenever <[x.sub.n]>[[right arrow].sup.U]gX. If f is sequentially vg-continuous at all x [member of] X, then f is said to be sequentially vg-continuous.

Theorem 8.7. Let f and g be two sequentially vg-continuous functions. If Y is sg-US, then the set A = {x|f(x) = g(x)} is sequentially vg-closed.

Proof. Let Y be sg-US and suppose that there is a sequence <[x.sub.n]> in A vg-converging to x [member of] X. Since f and g are sequentially vg-continuous functions, f([x.sub.n])[[right arrow].sup.vg]f(x) and g([x.sub.n])[[right arrow].sup.vg] g(x). Hence f(x) = g(x) and x [member of] A. Therefore, A is sequentially vg-closed.

Next, we prove the product theorem for sg-US spaces.

Theorem 8.8. Product of arbitrary family of sg-US spaces is vg-US.

Proof. Let X = [[PI].sub.[lambda] [member of] [LAMBDA]][X.sub.[lambda]] where [X.sub.[lambda]] is v g-US. Let a sequence <[x.sub.n]> in Xvg-converges to x(= [x.sub.[lambda]]) and y(= [y.sub.[lambda]]). Then the sequence <[x.sub.n]>vg-converges to [x.sub.[lambda]] and [y.sub.[lambda]] for all [lambda] [member of] [LAMBDA]. For suppose there exists a [mu] [member of] [LAMBDA] such that <[x.sub.n[mu]]>[[??].sup.vg][x.sub.[mu]]. Then there exists a [[tau].sub.[mu]]-vg-open set U containing [x.sub.[mu]] such that <[x.sub.n[mu]]> is not eventually in U Consider the set U = P[i.sub.[lambda] [member of] [LAMBDA]][X.sub.[lambda]] x [U.sub.[mu]]. Then U [member of] vGO(X, x). Also, <[x.sub.n]> is not eventually in U, which contradicts the fact that <[x.sub.n]>vg-converges to X. Thus we get <[x.sub.n[lambda]]>vg-converges to [x.sub.[lambda]] and [y.sub.[lambda]] for all [lambda] [member of] [LAMBDA]. Since X is vg-US for each [lambda] [member of] [LAMBDA]. Thus x = y. Hence X is vg-US.

[section]9. Sequentially sub-vg-continuity

In this section we introduce and study the concepts of sequentially sub-vg-continuity, sequentially nearly vg-continuity and sequentially vg-compact preserving functions and study their relations and the property of vg-US spaces.

Definition 9.1. A function f is said to be sequentially nearly vg-continuous if for each x [member of] X and each sequence <[x.sub.n]>[[right arrow].sup.U]gx [member of] X, there exists a subsequence <[x.sub.nk]> of <[x.sub.n]> such that <f([x.sub.nk])>[[right arrow].sup.U]gf(x).

Definition 9.2. A function f is said to be sequentially sub-vg-continuous if for each x [member of] X and each sequence <[x.sub.n]>[[right arrow].sup.U]gx [member of] X, there exists a subsequence <[x.sub.nk]> of <[x.sub.n]> and a point y [member of] Y such that <f([x.sub.nk])>[[right arrow].sup.U]gy.

Definition 9.3. A function f is said to be sequentially vg-compact preserving if f(K) is sequentially vg-compact in Y for every sequentially vg-compact set K of X.

Lemma 9.1. Every function f is sequentially sub-vg-continuous if Y is a sequentially vg-compact.

Proof. Let <[x.sub.n]> be a sequence in Xvg-converging to a point x of X. Then {f([x.sub.n])} is a sequence in Y and as Y is sequentially vg-compact, there exists a subsequence {f([x.sub.nk])} of {f([x.sub.n])}vg-converging to a point y [member of] Y. Hence f is sequentially sub-vg-continuous.

Theorem 9.1. Every sequentially nearly vg-continuous function is sequentially vg-compact preserving.

Proof. Let f be sequentially nearly vg-continuous function and let K be any sequentially vg-compact subset of X. Let <[y.sub.n]> be any sequence in f(K). Then for each positive integer n, there exists a point [x.sub.n] [member of] K such that f([x.sub.n]) = [y.sub.n]. Since <[x.sub.n]> is a sequence in the sequentially vg-compact set K, there exists a subsequence <[x.sub.nk]> of <[x.sub.n]>[[right arrow].sup.vg]x [member of] K. By hypothesis, f is sequentially nearly vg-continuous and hence there exists a subsequence <[x.sub.j]> of <[x.sub.nk]> such that f([x.sub.j])[[right arrow].sup.vg]f(x). Thus, there exists a subsequence <[y.sub.j]> of <[y.sub.n]>vg- converging to f(x) [member of] f(K). This shows that f(K) is sequentially vg-compact set in Y.

Theorem 9.2. Every sequentially vg-compact preserving function is sequentially sub-vg-continuous.

Proof. Suppose f is a sequentially vg-compact preserving function. Let x be any point of X and <[x.sub.n]> any sequence in Xvg-converging to x. We shall denote the set {[x.sub.n]|n = 1,2,3,...} by A and K = [Aunion]{x}. Then K is sequentially vg-compact since [x.sub.n][[right arrow].sup.U]gX. By hypothesis, f is sequentially vg-compact preserving and hence f(K) is a sequentially vgcompact set of Y. Since {f([x.sub.n])} is a sequence in f(K), there exists a subsequence {f([x.sub.nk])} of {f([x.sub.n])}vg-converging to a point y [member of] f(K). This implies that f is sequentially sub-vg-continuous. Theorem 9.3. A function f : X [right arrow] Y is sequentially vg-compact preserving iff [f.sub./K] : K [right arrow] f(K) is sequentially sub-vg-continuous for each sequentially vg-compact subset K of X.

Proof. Suppose f is a sequentially vg-compact preserving function. Then f(K) is sequentially vg-compact set in Y for each sequentially vg-compact set K of X. Therefore, by lemma 9.1 above, [f.sub./K]:K [right arrow] f(K) is sequentially vg-continuous function.

Conversely, let K be any sequentially vg-compact set of X. Let <[y.sub.n]> be any sequence in f(K). Then for each positive integer n, there exists a point [x.sub.n] [member of] K such that f([x.sub.n]) = [y.sub.n]. Since <[x.sub.n]> is a sequence in the sequentially vg-compact set K, there exists a subsequence <[x.sub.nk]> of <[x.sub.n]>vg-converging to a point x [member of] K. By hypothesis, [f.sub./K] : K [right arrow] f(K) is sequentially sub-vg-continuous and hence there exists a subsequence <[y.sub.nk]> of <[y.sub.n]>vg-converging to a point y [member of] f(K). This implies that f(K) is sequentially vg-compact set in Y. Thus, f is sequentially vg-compact preserving function.

The following corollary gives a sufficient condition for a sequentially sub-vg-continuous function to be sequentially vg-compact preserving.

Corollary 9.1. If f is sequentially sub-vg-continuous and f(K) is sequentially vg-closed set in Y for each sequentially vg-compact set K of X, then f is sequentially vg-compact preserving function.

Proof. omitted.

Acknowledgment. The author is thankful to the referees for their comments and suggestions for the development of the paper.

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S. Balasubramanian

Department of Mathematics, Government Arts College

(Autonomous), Karur-639005, Tamilnadu, India

E-mail: mani55682@rediffmail.com

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