Almost contra v-continuity.

[section] 1. Introduction

In 1980, Joseph and Kwack introduced the notion of ([theta], s)-continuous functions. In 1982, Jankovic introduced the notion of almost weakly continuous functions. In 1996, Dontchev introduced contra-continuous functions. C. W. Baker defined Subcontra-continuous functions in 1998 and contra almost [beta]--continuous functions in 2006. Dontchev, Ganster and Reilly introduced regular set-connected functions and Dontchev and T. Noiri introduced Contra-semicontinuous functions in 1999. S. Jafari and T. Noiri introduced and studied Contra-super-continuous functions and Jafari introduced the notion of (p, s)-continuous functions in 1999; Contra-[alpha]--continuous functions in 2001 and contra-precontinuous functions in 2002. M. Caldas and S. Jafari studied Some Properties of Contra-[beta]--Continuous Functions in 2001. T. Noiri and V. Popa studied unified theory of contra-continuity in 2002, Some properties of almost contra-precontinuity in 2005 and unified theory of almost contra-continuity in 2008. E. Ekici introduced Almost contra-precontinuous functions in 2004 and recently have been investigated further by Noiri and Popa. A. A. Nasef studied some properties of contra-[gamma]--continuous functions in 2005. M. K. R. S. Veera Kumar introduced Contra-Pre-Semi-Continuous Functions in 2005. Ekici E., introduced Almost contra-precontinuous functions and studied another form of contra-continuity in 2006. During 2007, N. Rajesh studied total [omega]--Continuity, Strong [omega]--Continuity and contra [omega]--Continuity. Recently, Ahmad Al-Omari and Mohd. Salmi Md. Noorani studied Some Properties of Contra-b-Continuous and Almost Contra-b-Continuous Functions in 2009 and Jamal M. Mustafa introduced Contra Semi-I-Continuous functions in 2010. Inspired with these developments, we introduce a new class of functions called contra v--continuous functions. Moreover, we obtain basic properties, preservation theorem of contra v--continuous function and relationship with other types of functions are verified.

[section] 2. Preliminaries

Definition 2.1. A [subset] X is said to be

(i) regular open [pre-open; semi-open; [alpha]-open; [beta]-open] if A = [([bar.A]).sup.o][A [subset or equal to] [([bar.A]).sup.o]; A [subset or equal to] [bar.([A.sup.o])]; A [subset or equal to][bar.(([A.sup.o])).sup.o]; A [subset or equal to] (([[bar.A].sub.o])] and regular closed [pre-closed; semi-closed; [alpha]-closed; [beta]-closed] if A = [bar.[A.sup.0]][bar.([A.sup.o])] [subset or equal to] A; [([bar.A]).sup.o] [subset or equal to] A; (([bar.A]).sup.o]) [subset or equal to] A; [[bar.(([A.sup.o]))].sup.o] [subset or equal to] A].

(ii) v--open [r[alpha]--open] if there exists a regular open set O such that O [subset] A [subset] [bar.O][O [subset] A [subset] [alpha]([bar.O])].

(iii) [theta]-semi-closed if A = [sCl.sub.[theta]](A) = {x [member of] X : [bar.V] [intersection] A [not equal to] [phi]; [for all] V [member of] SO(X, x)}; s[Cl.sub.[theta]](A) is [theta]-semi-closure of A. The complement of a [theta]-semi-closed set is said to be [theta]-semi-open.

(iv) v-dense in X if v([bar.A]) = X.

(v) [theta]-closed if A = [Cl.sub.[theta]](A). The complement of a [theta]-closed set is said to be [theta]-open. Remark 1. We have the following implication diagrams for closed sets.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Definition 2.2. A cover [SIGMA] = {[U.sub.[alpha]] : [alpha] [member of] I} of subsets of X is called a v--cover if [U.sub.[alpha]] is v--open for each [alpha] [member of] I.

Definition 2.3. A filter base [LAMBDA] is said to be v--convergent (resp. rc-convergent) to a point x in X if for any U [member of] vO(X, x)(resp. U [member of] RC(X, x)), there exists a B [member of] [LAMBDA] such that B [subset] U.

Definition 2.4. A function f: X [right arrow] Y is called

(i) almost-[resp: almost-semi-; almost-pre-;almost-r[alpha]--; almost-[alpha]--; almost-[beta]--; almost-[omega]--; almost-pre-semi-; almost-[gamma]--]continuous if [f.sup.-1](V) is open [resp: semi-open; pre-open; r[alpha]--open; [alpha]--open; [beta]--open; [omega]--open; pre-semi-open; [lambda]--open] in X [for all] V [member of] RO(Y).

(ii) contra-[resp: contra-semi-; contra-pre-;contra-r-;contra-r[alpha]--; contra-[alpha]-- ; contra-[beta]--; contra-[omega]--; contra-pre-semi-; contra-v--]continuuos if [f.sup.-1] (V) is closed [resp: semi-closed; preclosed; regular-closed; r[alpha]--closed; [alpha]--closed; [beta]--closed; [omega]--closed; pre-semi-closed; v--closed] in X [for all] V [member of] [sigma].

(iii) regular set-connected if inverse image of every regular open set V in Y is clopen in X.

(iv) perfectly continuous if inverse image of every open set V in Y is clopen in X.

(v) almost s-continuous if for each x [member of] X and each V [member of] SO(Y) with f(x) [member of] V, there exists an open set U in X containing x such that f(U) [subset] scl(V).

(vi) M-v--open if image of each v--open set is v-- open.

(vii) (v, s)-continuous if for each x [member of] X and each V [member of] SO(Y, f (x)), [there exists] U [member of] vO(X, x) [contains as member] f(U) [subset] [bar.V].

(viii) weakly continuous if for each x [member of] X and each V [member of] ([sigma](Y), f(x)), [there exists] U [member of] ([tau](X), x) [contains as member] f(U) [subset] [bar.V].

(ix) ([theta], s)-continuous iff for each [theta]-semi-open set V of Y , [f.sup.-1](V) is open in X.

(x) faintly v--continuous if for each x [member of] X and each [theta]-open set V of Y containing f(x), [there exists]U [member of] vO(X, x)] [contains as member] f(U) [subset] V.

Definition 2.5. A space X is said to be

(i) strongly compact [resp:strongly Lindelof] if every preopen cover of X has a finite [resp: countable] subcover and P-closed [resp: P-Lindelof] if every preclosed cover of X has a finite [resp:countable] subcover.

(ii) strongly countably compact if every countable cover of X by preopen sets has a finite subcover and countably S-closed [resp:countably P-closed] if every countable cover of X by regular closed [resp: preclosed] sets has a finite subcover.

(iii) mildly compact (mildly countably compact, mildly Lindelof) if every clopen cover (respectively, clopen countable cover, clopen cover) of X has a finite (respectively, a finite, a countable) subcover.

(iv) S-closed [resp: S-Lindelof] if every regular closed cover of X has a finite [resp:countable] subcover and nearly compact [resp:nearly Lindelof] if every regular open cover of X has a finite [resp: countable] subcover.

(v) v--connected provided that X is not the union of two disjoint nonempty v--open sets.

(vi) v--ultra-connected if every two non-void v--closed subsets of X intersect.

(vii) hyperconnected if every open set is dense.

(viii) weakly Hausdorff if each element of X is an intersection of regular closed sets.

(ix) v--[T.sub.0] if for each pair of distinct points in X, there exists a v--open set of X containing one point but not the other.

(x) v--[T.sub.1] [resp: v--[T.sub.2]] if for each pair of distinct points x and y of X, there exist [resp: disjoint] v--open sets U and V containing x and y respectively such that y [not member of] U and x [not member of] V.

(xi) almost regular if for each regular closed set F of X and each x [member of] X--F, there exists disjoint open sets U and V of X such that x [member of] U and F [subset] V.

(xii) extremally disconnected (briefly E.D.) if the closure of every open set of X is open in X.

Lemma 2.1. If V is an open [r-open] set, then

(i) sCl(V) = Int(Cl(V)).

(ii) [sCl.sub.[theta]](V) = sCl(V).

(iii) If B [subset or equal to] A [subset or equal to] X and A [member of] RO(X), then v[bar.[sub.A](B)] [subset or equal to] v[bar.B].

Lemma 2.2. For V [subset] Y, the following properties hold:

(i) [alpha][bar.V] = [bar.V] for every V [member of] [beta](Y).

(ii) v[bar.V] = [bar.V] for every V [member of] SO(Y).

(iii) s[bar.V] = [([bar.V]).sup.[omicron]] for every V [member of] RO(Y).

Lemma 2.3. For f: X [right arrow] Y, the following properties are equivalent:

(i) f is faintly-v--continuous.

(ii) [f.sup.-1](V) [member of] vO(X) for every [theta]-open set V of Y.

(iii) [f.sup.1](K) is v--closed in X for every [theta]-closed set K of Y.

Lemma 2.4. f is al.v.c. iff [Angstrom] x [member of] X and each V [member of] RO(Y, f(x)), [there exists] U [member of] vO(X, x) [contains as member] f(U) [subset] V.

Definition 2.6. For a function f: X [right arrow] Y,

(i) The subset {(x, f(x)) : x [member of] X} [subset] X x Y is called the graph of f and is denoted by G(f).

(ii) A graph G(f) of a function f is said to be v--regular if for each (x, y) [member of] (X x Y)--G(f), [there exists] U [member of] vC(X, x) and V [member of] RO(Y, y) [contains as member] (U x V) [intersection] G(f) = [phi].

Lemma 2.5. The following properties are equivalent for a graph G(f) of a function:

(i) G(f) is v--regular;

(ii) for each point (x,y) [member of] (X x Y)--G(f), [there exists] U [member of] vC(X, x) and V [member of] RO(Y, y) [contains as member] f(U)[intersection]V = [phi].

[section] 3. Almost contra v-continuous maps

Definition 3.1. A function f: X [right arrow] Y is said to be

(i) almost contra v--continuous at x if for each regular closed set F in Y containing f(x), there exists a v--open set U in X containing x such that f(U) [subset] F.

(ii) almost contra v--continuous if the inverse image of every regular-open set is v--closed.

Note 1. Here after we call almost contra v--continuous function as al.c.v.c function shortly.

Example 1. (i) X = Y = {a, b, c}; [tau] = {[phi], {a}, {b}, {a, b}, X} and [sigma] = {[phi], {a}, {b, c}, Y}. Let f be identity function, then f is al.c.v.c.

(ii) f: on R defined by f(x) = [x], where [x] is the Gaussian symbol; is al.c.v.c; al.c.s.c. but not al.c.c; r-irreslute and c.r.c.

Example 2. (i) X = Y = {a, b, c}; [tau] = {[phi], {a}, {a, b}, X} and [sigma] = {[phi], {a}, {b}, {a, b}, Y}. Let f be identity function, then f is al.c.r[alpha].c. but not al.c.v.c.

(ii) The identity function on K with usual topology is not al.c.v.c and al.c.s.c. but it is al.c.c;c.c; c.r.c. and r-irresolute.

Theorem 3.1. (i) f is al.c.v.c iff [f.sup.-1](U) [member of] vO(X) whenever U [member of] RC(Y).

(ii) If f is c.v.c., then f is al.c.v.c. Converse is true if X is discrete space.

Theorem 3.2. (i) f is al.c.v.c. iff for each x [member of] X and each [U.sub.Y] [member of] vO(Y,f(x)), [there exists]A [member of] vO(X) [contains as member] x [member of] A and f(A) [subset] [U.sub.Y].

(ii) f is al.c.v.c. iff for each x [member of] X and each V [member of] RO(Y, f(x)), [there exists] U [member of] vO(X, x) [contains as member] f(U) [subset] V.

Proof. Let [U.sub.Y] [member of] RC(Y) and let x [member of] [f.sup.-1]([U.sub.Y]). Then f(x) [member of] [U.sub.Y] and thus [there exists] [A.sub.x] [member of] vO(X) [there exists] x [member of] [A.sub.x] and f([A.sub.x]) [subset] [U.sub.Y]. Then x [member of] [A.sub.x] [subset] [f.sup.-1]([U.sub.Y]) and [f.sup.-1]([U.sub.Y]) = [union][A.sub.x]. Hence [f.sup.-1]([U.sub.Y]) [member of] vO(X).

Remark 2. We have the following implication diagram for a function f: (X, [tau]) [right arrow] (Y, [sigma])

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Example 3. Let X = Y = {a, b, c}; [tau] = {[tau], {b}, {a, b}, {b, c}, X} and [sigma] = {[phi], {a}, {b}, {a, b}, {a, c}, Y}. The identity function f is not al.c.c., al.c.s.c., al.c.p.c., al.c.[alpha].c., al.c.r[alpha].c., al.c. [beta].c., al.c.v.c., contra r-irresolute and r-irresolute and f defined as f(a) = f(b) = a; f(c) = c is al.c.c., al.c.s.c., al.c.p.c., al.c.[alpha].c., al.c.[beta].c., but not al.c.v.c., al.c.r[alpha].c., contra r-irresolute and r-irresolute.

Example 4. Let X = Y = {a, b, c}; [tau] = {([phi], {a}, {b}, {a, b}, X} and [sigma] = {[phi], {a}, {b}, {a, b}, {a, c}, Y}. The identity function f is al.c.s.c., al.c.[alpha].c., al.c.r[alpha].c., al.c.[beta].c., al.c.v.c., but not al.c.c.., al.c.p.c., contra r-irresolute and r-irresolute. Under usual topology on R both al.c.c. and r-irresolute are same as well both al.c.s.c. and al.c.v.c are same.

Theorem 3.3. If f is v--open and al.c.v.c, then [f.sup.-1](U) is v--closed if U is v--open in Y.

Proof. Let U be v--open in Y. Then [there exists]V [member of] RO(Y) [contains as member] V [subset or equal to] U [bar.V]. [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Theorem 3.4. Let f be al. c v c and r-open, then

(i) [f.sup.-1](A) [member of] SO(X) [[f.sup.-1](A) [member of] SC(X)] for each A [member of] SC(Y) [A [member of] SO(Y)].

(ii) [f.sup.-1](A) [member of] RO(X) [[f.sup.-1](A) [member of] RC (X)] for each A [member of] RC(Y) [A [member of] RO(Y)].

(iii) If f is r-open and r-irresolute, then [f.sup.-1](U) [member of] vC(X) for each U [member of] vO(Y).

Theorem 3.5. Let [f.sub.i] : [X.sub.i] [right arrow] [Y.sub.i] be al.c.v.c for i = 1, 2. Let f: [X.sub.1] x [X.sub.2] [right arrow] [Y.sub.1] x [Y.sub.2] be defined as follows: f([x.sub.1],[x.sub.2]) = ([f.sub.1]([x.sub.1]), [f.sub.2]([x.sub.2])). Then f: [X.sub.1] x [X.sub.2] [right arrow] [Y.sub.1] x [Y.sub.2] is al.c.v.c.

Proof. Let [U.sub.1] x [U.sub.2] [subset] [Y.sub.1] x [Y.sub.2] where [U.sub.i] be regular open in [Y.sub.i] for i = 1 , 2. Then [f.sup.-1]([U.sub.1] x [U.sub.2]) = [f.sup.-1.sub.1]([U.sub.1]) x [f.sup.-1.sub.2]([U.sub.2]). But [f.sup.- 1.sub.1]([U.sub.1]) and [f-sup.-1.sub.2]([U.sub.2]) are v--closed in [X.sub.1] and [X.sub.2] respectively and thus [f.sup.-1.sub.1]([U.sub.1]) x [f.sup.-1.sub.2]([U.sub.2]) is v--closed in [X.sub.1] x [X.sub.2]. Therefore f is al.c.v.c.

Theorem 3.6. Let h : X [right arrow] [X.sub.1] x [X.sub.2] be al.c.v.c, where h(x) = ([h.sub.1](x), [h.sub.2](x)). Then [h.sub.i] : X [right arrow] [X.sub.i] is al.c.v.c for i =1, 2.

Proof. Let [U.sub.1] is regular open in [X.sub.1]. Then [U.sub.1] x [X.sub.2] is regular open in [X.sub.1] x [X.sub.2], and [h.sup.-1]([U.sub.1] x [X.sub.2]) is v--closed in X. But [h.sup.-1.sub.1]([U.sub.1]) = [h.sup.-1]([U.sub.1] x [X.sub.2]), therefore [h.sub.1] : X [right arrow] [X.sub.1] is al.c.v.c Similar argument gives [h.sub.2] : X [right arrow] [X.sub.2] is al.c.v.c and thus [h.sub.i] : X [right arrow] [X.sub.i] is al.c.v.c for i = 1, 2.

In general we have the following extenstion of Theorem 3.5 and 3.6:

Theorem 3.7. (i) If f : X [right arrow] [PI][Y.sub.[lambda]] is al.c.v.c, then [P.sub.[lambda]] [omicron] f : X [right arrow] [Y.sub.[lambda]] is al.c.v.c for each [lambda] [member of] [LAMBDA], where [P.sub.[lambda]] is the projection of [PI][Y.sub.[lambda]] onto [Y.sub.[lambda]].

(ii) f: [PI][X.sub.[lambda]] [right arrow] [PI][Y.sub.[lambda]] is al.c.v.c, iff [f.sub.[lambda]] : [X.sub.[lambda]] [right arrow] [Y.sub.[lambda]] is al.c.v.c for each [lambda] [member of] [LAMBDA].

Note 2. Converse of Theorem 3.7 is not true in general, as shown by the following example.

Example 5. Let X = [X.sub.1] = [X.sub.2] = [0,1]. Let [f.sub.1] : X [right arrow] [X.sub.1] be defined as follows: [f.sub.1](x) = 1 if 0 [less than or equal to] x [less than or equal to] 1/2 and [f.sub.1](x) = 0 if 1/2 < x [less than or equal to] 1. Let [f.sub.2] : X [right arrow] [X.sub.2] be defined as follows: [f.sub.2](x) = 1 if 0 [less than or equal to] x < 1/2 and [f.sub.2] (x) = 0 if 1/2 < x < 1. Then [f.sub.i] : X [right arrow] [X.sub.i] is clearly al.c.v.c for i = 1, 2, but h(x) = ([f.sub.1]([x.sub.1]), [f.sub.2]([x.sub.2])) : X [right arrow] [X.sub.1] x [X.sub.2] is not al.c.v.c, for [S.sub.1/2] (1,0) is regular open in [X.sub.1] x [X.sub.2], but [h.sup.-1]([S.sub.1/2] (1, 0)) = {1/2} which is not v--closed in X.

Remark 3. In general,

(i) The algebraic sum and product of two al.c.v.c functions is not al.c.v.c However the scalar multiple of a al.c.v.c function is al.c.v.c.

(ii) The pointwise limit of a sequence of al.c.v.c functions is not al.c.v.c as shown by.

Example 6. Let X = [X.sub.1] = [X.sub.2] = [0,1]. Let [f.sub.1] : X [right arrow] [X.sub.1] and [f.sub.2] : X [right arrow] [X.sub.2] are defined as follows: [f.sub.1](x) = x if 0 < x < 1/2 and [f.sub.1](x) = 0 if 1/2 < x < 1; [f.sub.2](x) = 0 if 0 < x < 1/2 and [f.sub.2](x) = 1 if 1/2 < x < 1.

Example 7. Let X = Y = [0,1]. Let [f.sub.n] is defined as follows: [f.sub.n](x) = [x.sub.n] for n [greater than or equal to] 1 then f is the limit of the sequence where f(x) = 0 if 0 [less than or equal to] x < 1 and f(x) = 1 if x = 1. Therefore f is not al.c.v.c For (1/2, 1] is open in Y, [f.sup.-1]((1/2, 1]) = (1) is not v--closed in X.

However we can prove the following theorem.

Theorem 3.8. Uniform Limit of sequence of al.c.v.c. functions is al.c.v.c.

Problem. (i) Are sup{f, g} and inf{f, g} are al.c.v.c if f, g are al.c.v.c.

(ii) Is [C.sub.al.c.v.c](X, R), the set of all al.c.v.c functions.

(1) a Group.

(2) a Ring.

(3) a Vector space.

(4) a Lattice.

(iii) Suppose [f.sub.i] : X [right arrow] [X.sub.i] (i = 1, 2) are al.c.v.c If f : X [right arrow] [X.sub.1] x [X.sub.2] defined by f(x) = ([f.sub.1](x), [f.sub.2](x)), then f is al.c.v.c.

Solution. No.

Note 3. In general al.c.c., al.c.[alpha].c. and al.c.p.c. are independent of al.c.v.c as shown by Example 1 and 3.

Theorem 3.9. (i) If f is v--irresolute and g is al.c.v.c, then g [omicron] f is al.c.v.c.

(ii) If f is al.c.v.c and g is continuous [resp: r-continuous] then g [omicron] f is al.c.v.c.

(iii) If f and g are r-irresolute then g [omicron] f is v--continuous.

(iv) If f is al.c.v.c and g is r-irresolute, then g [omicron] f is al.c.v.c; al.c.s.c and al.c.[beta].c.

(v) If f is al.c.v.c[contra r-irresolute] g is al.g.c.[al.rg.c] and GO(Y) = RGO(Y) = RO(Y), then g [omicron] f is al.c.v.c.

Theorem 3.10. If f is v--irresolute, v--open and vO(X) = [tau] and g be any function, then g [omicron] f: X [right arrow] Z is c.v.c iff g is al.c.v.c.

Proof. If part: Theorem 3.9 only if part: Let A [member of] RC(Z). Then [(g [omicron] f).sup.-1](A) is a v--open and hence open in X [by assumption]. Since f is v--open f[(g [omicron] f).sup.-1](A) = [g.sup.-1](A) is v--open in Y. Thus g : Y [right arrow] Z is al.c.v.c.

Corollary 3.1: (i) If f is a surjective M-v--open [resp: M-v--closed] and g is a function such that g [omicron] f is al.c.v.c., then g is al.c.v.c.

(ii) If f is v--irresolute, M-v--open and bijective, g is a function. Then g is al.c.v.c. iff g [omicron] f is al.c.v.c.

Theorem 3.11. If g : X [right arrow] X x Y, defined by g(x) = (x, f(x))[for all]x [member of] X be the graph function of f. Then g is al.c.v.c iff f is al.c.v.c.

Proof. Let V [member of] RC(Y), then X x V = X x [bar.[V.sup.0]] = [bar.[X.sup.0]] x [bar.[V.sup.0]] = [bar.[(X x V).sup.0]] [member of] RC(X x Y). Since g is al.c.v.c., then [f.sup.-1](V) = [g.sup.-1](X x V) [member of] vO(X). Thus, f is al.c.v.c.

Conversely, let x [member of] X and F [member of] RC(X x Y) containing g(x). Then F[intersection] ({x} x Y) is r-closed in {x} x Y containing g(x). Also {x} x Y is homeomorphic to Y. Hence {y [member of] Y : (x, y) [member of] F} is r-closed subset of Y. Since f is al.c.v.c. [universal]{[f.sup.-1](y) : (x,y) [member of] F} is v--open in X. Further x [member of] [universal]{[f.sup.-1](y) : (x,y) [member of] F} [subset or equal to] [g.sup.-1](F). Hence [g.sup.-1](F) is v--open. Thus g is al.c.v.c.

Remark 4. In general, composition of two al.c.v.c functions is not al.c.v.c. However we have the following example:

Example 8. Let X = Y = Z = {a, b, c} and [tau] = {[phi], {a}, {b}, {a, b}, X}; [sigma] = {[phi], {a}, {b, c}, Y}, and [eta] = {[phi], {a}, {b}, {a, b}, Z}. Let f be identity map and g be be defined as g(a) = a = g(b); g(c) = c; are al.c.v.c and g [omicron] f is also al.c.v.c.

Theorem 3.12. Let X, Y, Z be spaces and every v--closed set be r-open in Y, then the composition of two al.c.v.c maps is al.c.v.c.

Corollary 3.2. If f is al.c.v.c [r-irresolute],

(i) g is al.c [r-continuous], then g [omicron] f is al.c.s.c. and hence al.c.[beta].c.

(ii) g is al.g.c.{al.rg.c.} and Y is r - [T.sub.1/2], then g [omicron] f is al.c.s.c. and hence al.c.[beta].c.

Theorem 3.13. (i) If R[alpha]C(X) = RC(X) then f is al.c.r[alpha].c. iff f is contra r-irresolute.

(ii) If R[alpha]C(X) = vC(X) then f is al.c.r[alpha].c. iff f is al.c.v.c.

(iii) If vC(X) = RC(X) then f is r-irresolute iff f is al.c.v.c.

(iv) If vC(X) = [alpha]C(X) then f is al.c.a.c. iff f is al.c.v.c.

(v) If vC(X) = SC(X) then f is al.c.s.c. iff f is al.c.v.c.

(vi) If vC(X) = [beta]C(X) then f is al.c.[beta].c. iff f is al.c.v.c.

Note 4. Pasting Lemma is not true with respect to al.c.v.c functions. However we have the following weaker versions.

Theorem 3.14. Let X and Y be such that X = A [union] B. Let [f.sub./A] : A [right arrow] Y and [g.sub./B] : B [right arrow] Y are r-irresolute functions such that f(x) = g(x)[for all]x [member of] A [intersection] B. Suppose A and B are r-closed sets in X and RC(X) is closed under finite unions, then the combination [alpha] : X [right arrow] Y is al.c.v.c.

Theorem 3.15. Pasting Lemma. Let X and Y be such that X = A [union] B. Let [f.sub./A] : A [right arrow] Y and [g.sub./B] : B [right arrow] Y are al.c.v.c maps such that f(x) = g(x) [for all]x [member of] A [intersection] B. Suppose A, B are r-closed sets in X and vC(X) is closed under finite unions, then the combination [alpha] : X [right arrow] Y is al.c.v.c.

Proof. Let F be r-open set in Y, then [[alpha].sub.-1](F) = [f.sup.-1](F)[union][g.sup.-1](F) where [f.sup.-1](F) is v--closed in A and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence a is al.c.v.c.

Theorem 3.16. The following statements are equivalent for a function f:

(i) f is al.c.v.c.;

(ii) [f.sup.-1](F) [member of] vO(X) for every F [member of] RC(Y);

(iii) for each x [member of] X and each regular closed set F in Y containing f(x), there exists a v--open set U in X containing x such that f(U) [subset] F;

(iv) for each x [member of] X and each regular open set V in Y non-containing f(x), there exists a v--closed set K in X non-containing x such that [f.sup.-1] (V) [subset] K;

(v) [f.sup.-1]([([bar.G]).sup.[omicron]]) [member of] vC(X) for every open subset G of Y;

(vi) [f.sup.-1]([[bar.D].sup.[omicron]]) [member of] vO(X) for every closed subset F of Y.

Proof. (i) [??] (ii): Let F [member of] RC(Y). Then Y - F [member of] RO(Y). By (i), [f.sup.-1](Y - F) = X - [f.sup.-1](F) [member of] vC(X). We have [f.sup.-1](F) [member of] vO(X). Reverse can be obtained similarly.

(ii) [??] (iii): Let F [member of] RC(Y, f(x)). By (ii), x [member of] [f.sup.-1](F) [member of] vO(X). Take U = [f.sup.-1](F). Then f(U) [subset] F.

(iii) [??] (ii): Let F [member of] RC(Y) and x [member of] [f.sup.-1](F). From (iii), [there exists] [U.sub.x] [member of] vO(X, x) [contains as member] U [subset] [f.sup.-1](F). We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Thus [f.sup.-1](F) is v--open.

(iii) [??] (iv): Let V [member of] RO(Y) not containing f(x). Then, Y - V [member of] RC(Y, f). By (3), 3 U [member of] vO(X, x) [contains as member] f(U) [subset] Y - V. Hence, U C [f.sup.-1] (Y - V) [subset] X - [f.sup.-1] (V) and then [f.sup.-1] (V) [subset] X - U. Take H = X - U, then H is v--closed in X non-containing x. The converse can be shown easily.

(i) [??] (v): Let G [member of] [sigma](Y). Since [([bar.G]).sup.[omicron]] [member of] RO(Y), by (i), [f.sup.-1]([([bar.G]).sup.[omicron]]) [subset] vC(X). The converse can be shown easily.

(ii) [??] (vi): It can be obtained smilar as (i)[??](v).

Example 9. Let X = {a, b, c}, [tau] = {[phi], {a}, {b}, {a, b}, X} = [sigma]. Then the identity function f: X [right arrow] X is al.c.v.c. But it is not regular set-connected.

Example 10. Let X = {a,b,c}, [tau] = {[phi], {a},X} and [sigma] = {[phi], {a}, {a,b},X}. Then the identity function f on X is al.c.v.c. but not c. v. c. and v.c.

Theorem 3.17. Let f be al.c.v.c and A [member of] RC(X), then [f.sub./A] : A [right arrow] Y is al.c.v.c.

Proof. Let V [member of] RO(Y) [??] [f.sup.-1.sub./A](V) = [f.sup.-1](V) [intersection] A is v--closed in A. Hence [f.sub./A] is al.c.v.c.

Remark 5. Every restriction of an al.c.v.c. function is not necessarily al.c.v.c.

Example 11. Let X = {a, b, c, d}, [tau] = {[phi], {a, b}, X} and [sigma] = {[phi], {a}, {b, c, d}, X}. The identity function f: X [right arrow] X is al.c.v.c., but, if A = {a,c,d} is not regular-open in (X, [sigma]) and [[sigma].sub.A] is the relative topology on A induced by [sigma], then [f.sub.|A] : (A, [[sigma].sub.A]) [right arrow] (X, [tau]) is not al.c.v.c.

Note that {b, c, d} is regular closed in (X, [tau]), but that [f.sup.-.sub.|A] ({b, c, d}) = {c, d} is not v--open in (A, [[sigma].sub.A]).

Theorem 3.18. Let f be a function and [summation] = {[U.sub.[alpha]] : [alpha] [member of] I} be a v--cover of X. If for each [alpha] [member of] I, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is al.c.v.c., then f is an al.c.v.c. function.

Proof. Let F [member of] RC(Y). [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is al.c.v.c. for each [alpha] [member of] I, (F) [member of] [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since [U.sub.[alpha]] [member of] vO(X), by the previous lemma, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for each [alpha] [member of] I. Then [f.sup.-1](F) = [[universal].sub.[alpha][member of]I] [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. This gives f is an al.c.v.c.

Theorem 3.19. Let f be a function and x [member of] X. If there exists U [member of] vO(X) [contains as member] x [member of] U and the restriction of [f.sub.|U] is al.c.v.c. at x, then f is al.c.v.c. at x.

Proof. Suppose that F [member of] RC(Y) containing f (x). Since [f.sub.|U] is al.c.v.c. at x, [there exists] V [member of] vO(U,x) [contains as member] f(V) = ([f.sub.|U])(V) [subset] F. Since U [member of] vO(X,x); then V [member of] vO(X,x). Thus f is al.c.v.c. at x.

Theorem 3.20. For f and g, the following properties hold:

(i) If f is al.c.v.c. and g is regular set-connected, then g [omicron] f is al.c.v.c. and al.v.c.

(ii) If f is al.c.v.c. and g is perfectly continuous, then g [omicron] f is v--continuous and c.v.c.

Proof. (i) Let V [member of] RO(Z). Since g is regular set-connected, [g.sup.-1] (V) is clopen. Since f is al.c.v.c., [f.sup.-1]([g.sup.-1](V)) = [(g [omicron] f).sup.-1](V) is v--clopen. Therefore, g [omicron] f is al.c.v.c. and al.v.c.

(ii) can be obtained similarly.

Theorem 3.21. If f is r-irresolute and al.c.c., then f is regular set-connected.

Theorem 3.22.If f is al.c.v.c., then for each point x [member of] X and each filter base [LAMBDA] in X v--converging to x, the filter base f([LAMBDA]) is rc-convergent to f (x).

Theorem 3.23. For f, the following properties are equivalent:

(i) f is (v, s)-continuous;

(ii) f is al.c.v.c.;

(iii) [f.sup.-1] (V) is v--open in X for each [theta]-semi-open set V of Y;

(iv) [f.sup.-1] (F) is v--closed in X for each [theta]-semi-closed set F of Y.

Proof. (i)[??](ii): Let F [member of] RC(Y) and x [member of] [f.sup.-1] (F). Then f(x) [member of] F and F is semi-open. Since f is (v, s)-continuous, [there exists]U [member of] vO(X,x) [contains as member] f(U) [subset] [bar.F] = F. Hence x [member of] U C [f.sup.-1](F) which implies that x [member of] v[([f.sup.-1] (F)).sup.0]. Therefore, [f.sup.-1] (F) [subset] v[([f.sup.-1] (F)).sup.0] and hence [f.sup.-1](F) = v[([f.sup.-1](F)).sup.0]. This shows that [f.sup.-1](F) [member of] vO(X). Hence f is al.c.v.c.

(ii) [??] (iii): Follows from the fact that every [theta]-semi-open set is the union of regular closed sets.

(iii) [??] (iv): This is obvious.

(iv) [??] (i): Let x [member of] X and V [member of] SO(Y,f(x)). Since [bar.X] is regular closed, it is [theta]-semi- open. Now, put U = [f.sup.-1]([bar.V]). Then U [member of] vO(X, x) and f(U) [subset] [bar.V]. This shows that f is (v, s)- continuous.

Theorem 3.24. For f, the following properties are equivalent:

(i) f is al.c.v.c.;

(ii) f(v([bar.A])) [subset] s[Cl.sub.[theta]](f(A)) for every subset A of X;

(iii) v[bar.([f.sup.-1](B))] [subset] [f.sup.-1](s[Cl.sub.[theta]](B)) for every subset B of Y.

Proof. (i)[??] (ii): Let A be any subset of X. Suppose that x [member of] v[bar.(A)] and G [member of] SO(Y, f(x)). Since f is al.c.v.c., by Theorem 3.23, [there exists]U [member of] vO(X, x) [contains as member] f(U) [subset] [bar.G]. Since x [member of] v[bar.(A)], U [intersection] A [not equal to] [phi]; and hence [phi] [not equal to] f(U) [intersection] f(A) [subset] [bar.G][intersection]f(A). Therefore, f(x) [member of] s[Cl.sub.[theta]](f(A)) and hence f(v[bar.(A)]) [subset] s[Cl.sub.[theta]](f(A)).

(ii) [??] (iii): Let B be any subset of Y. Then f(v[bar.([f.sup.-1](B))]) [subset] s[Cl.sub.[theta]](f([f.sup.-1](B))) [subset] s[Cl.sub.[theta]](B) and hence v[bar.([f.sup.-1](B)]) [subset] [f.sup.-1] (s[Cl.sub.[theta]](B)).

(iii) [??] (i): Let V [member of] SO(Y, f(x)). Since [bar.V] [intersection] (Y - [bar.V]) = [phi], we have f(x) [not member of] [sCl.sub.[theta]] (Y - V) and hence x [not member of] [f.sup.-1](s[Cl.sub.[theta]](Y - [bar.V])). By (3), x [not member of] v[bar.x([f.sup.-1] (Y - [bar.V]))], then [there exists] U [member of] vO(X,x) [contains as member] U [intersection] [f.sup.-1] (Y - [bar.V]) = [phi]; hence f(U) [intersection] (Y - [bar.V]) = [phi]. This shows that f(U) [subset] [bar.V]. Therefore f is al.c.v.c.

Theorem 3.25. For f, the following properties are equivalent:

(i) f is al.c.v.c.;

(ii) [f.sup.-1]([bar.V]) is v--open in X for every V [member of] [beta](Y);

(iii) [f.sup.-1]([bar.V]) is v--open in X for every V [member of] SO(Y);

(iv) [f.sup.-1]([([bar.V]).sup.[omicron]]) is v--closed in X for every V [member of] RO(Y).

Proof. (i) [??] (ii): Let V be any [beta]-open set of Y. It follows from Theorem 2.4 of [2] that [bar.V] is regular closed. Then by Theorem 3.16, [f.sup.-1]([bar.V]) [member of] vO(X).

(ii) [??] (iii): This is obvious since SO(Y) [subset] [beta](Y).

(iii) [??] (iv): Let V [member of] RO(Y). Then Y - ([[bar.V]).sup.[omicron]] is regular closed and hence semi-open. Then, X - [f.sup.-1]([([bar.V]).sup.[omicron]]) = [f.sup.-1](Y - [([bar.V]).sup.[omicron]]) = [f.sup.-1] [bar.((Y - ([bar.V])[omicron]))] [member of] vO(X). Hence [f.sup.-1]([([bar.V]).sup.[omicron]]) [member of] vC(X).

(iv) = (i): Let V [member of] RO(Y). Then V [member of] vO(Y) and hence [f.sup.-1](V) = [f.sup.-1] ([([bar.V]).sup.[omicron]]) [member of] vC(X).

Corollary 3.3. For f, the following properties are equivalent:

(i) f is al.c.v.c.;

(ii) [f.sup.-1] ([alpha][bar.V]) is v--open in X for every V [member of] [beta](Y);

(iii) [f.sup.-1](v[bar.V]) is v--open in X for every V [member of] SO(Y);

(iv) [f.sup.-1](s[bar.V]) is v--closed in X for every V [member of] RO(Y).

Proof. This is an immediate consequence of Theorem 3.25 and Lemma 2.2.

The v--frontier of A [member of] X; is defined by vFr(A) = v[bar.(A)] - v[bar.(X - A)] = v[bar.(A)] - v[(A).sup.0].

Theorem 3.26. For f, the following conditions are equivalent:

(i) f is al.c.v.c.;

(ii) v[bar.([f.sup.-1](V))] [subset or equal to] [f.sup.-1](s[Cl.sub.[theta]](V)) for every open subset V of Y;

(iii) v[bar.([f.sup.-1](V))] [subset or equal to] [f.sup.-1](s[bar.(V)]) for every open subset V of Y;

(iv) v[bar.([f.sup.-1](V))] [subset or equal to] [f.sup.-1][(([bar.V]).sup.o]) for every open subset V of Y;

(v) [bar.[([f.sup.-1](V)).sup.[omicron]]] [subset or equal to] [f.sup.-1]([([bar.V]).sup.o]) for every open subset V of Y.

Proof. (i) [??] (ii) follows from Theorem 3.24(c).

(ii) [??] (iii) follows from Lemma 2.1(ii).

(iii) [??] (iv) follows from Lemma 2.1(iii).

(iv) [??] (v). Since v[bar.([f.sup.-1](V))] = [f.sup.-1](V) [union] [bar.[([f.sup.-1](V)).sup.o])], it follows from (iv) that [bar.[([f.sup.-1](V)).sup.[omicron]])] [subset or equal to] [f.sup.-1]([([bar.V]).sup.o]).

(v) [??] (i). Let V [member of] RO(Y). Then by (v), [bar.[([f.sup.-1](V)).sup.o])] [subset or equal to] [f.sup.-1]([([bar.V]).sup.o]) = [f.sup.-1](V). Therefore [f.sup.-1](V) is v--closed, which proves that f is al.c.v.c.

The next result is an immediate consequence of Theorems 3.24 and 3.26.

Theorem 3.27. Let f be a function and let S be any collection of subsets of Y containing the open sets. Then f is al.c.v.c. iff v[bar.([f.sup.-1](S))] [subset] [f.sup.-1](s[Cl.sub.[theta]](S)) for every S [member of] S.

Corollary 3.4. For f, the following properties are equivalent:

(i) f is al.c.v.c.;

(ii) v[bar.([f.sup.-1](V))] [subset or equal to] [f.sup.-1](s[Cl.sub.[theta]](V)) for every V [member of] SO(Y);

(iii) v[bar.([f.sup.-1](V))] [subset or equal to] [f.sup.-1](s[Cl.sub.[theta]](V)) for every V [member of] PO(Y);

(iv) v[bar.([f.sup.-1](V))] [subset or equal to] [f.sup.-1](s[Cl.sub.[theta]](V)) for every V [member of] [beta]O(Y).

Theorem 3.28. {x [member of] X : f : X [right arrow] Y is not al.c.v.c.} is identical with the union of the v--frontier of the inverse images of regular closed sets of Y containing f(x).

Proof. Suppose that f is not al.c.v.c. at x [member of] X. By Theorem 3.16, [there exists] F [member of] RC(Y, f(x) [contains as member] f(U) [intersection] (Y - F) [not equal to] [phi] for every U [member of] vO(X, x). Then, x [member of] v[bar.([f.sup.- 1](Y - F))] = v[bar.(X - [f.sup.-1](F)). On the other hand, we get x [member of] [f.sup.-1](F) [subset] v[bar.([f.sup.-1](F))] and hence x [member of] vFr([f.sup.- 1](F)).

Conversely, suppose that f is al.c.v.c. at x and let F [member of] RC(Y, f(x)). By Theorem 3.16, there exists U [member of] vO(X,x) [contains as member] x [member of] U [subset] [f.sup.-1](F). Therefore, x [member of] v[([f.sup.-1](F)).sup.o]. This contradicts that x [member of] vFr([f.sup.-1](F)). Thus f is not al .c.v.c.

Theorem 3.29. If f is al.c.v.c. and Y is [T.sub.2], then G (f) is v--regular graph in X x Y.

Proof. Assume Y is [T.sub.2]. Let (x, y) [member of] (X x Y) - G(f). It follows that f(x) [not equal to] y. Since Y is [T.sub.2], there exist disjoint open sets V and W containing f (x) and y, respectively. We have ([([bar.V]).sup.o]) [intersection] ([([bar.W]).sup.o]) = [phi]. Since f is al.c.v.c., [f.sup.-1] ([([bar.V]).sup.o]) is v--closed in X containing x. Take U = [f.sup.-1]([([bar.V]).sup.o]). Then f(U) [subset] ([([bar.V]).sup.o]). Therefore, f(U) [intersection] ([([bar.W]).sup.o]) = [phi] and G(f is v--regular in X x Y.

Remark 6. al.v.c. and al.c.v.c. are independent to each other.

It is shown that [Cl.sub.[theta]](V) = [bar.V] for every open set V and [Cl.sub.[theta]](S) is closed for every subset S of X.

Theorem 3.30. Let (Y, [sigma]) be E.D. Then, a function f is al.c.v.c. iff it is al.v.c.

Proof. Necessity. Let x [member of] X and V [member of] RO(Y, f(x)). Since Y is E. D., by Lemma 5.6 of [26] V is clopen and hence V is regular closed. By Theorem 3.16, there exists U [member of] vO(X, x) [contains as member] f(U) [subset] V. By Lemma 2.4, f is al.v.c.

Sufficiency. Let F be any regular closed set of Y. Since (Y, [sigma]) is E. D., F is also regular open and [f.sup.-1](F) is v--open in X. This shows that f is al.c.v.c.

[section] 4. The preservation theorems and some other properties

Theorem 4.1. If f is al.c.v.c.[resp: al.c.r.c] surjection and X is v--compact, then Y is nearly closed compact.

Proof. Let {[G.sub.i] : i [member of] I} be any regular-closed cover for Y. Since f is al.c.v.c., {[f.sup.-1] ([G.sub.i])} forms a v--open cover for X and hence have a finite subcover, since X is v--compact. Since f is surjection, Y = f(X) = [[universal].sup.n.sub.i=1] [G.sub.i]. Therefore Y is nearly closed compact.

Corollary 4.1. If f is al.c.v.c.[r-irresolute], surjection, then the following statements hold:

(i) If X is locally v--compact, then Y is locally nearly closed compact; locally mildly compact.

(ii) If X is v--Lindeloff [locally v--lindeloff], then Y is nearly closed Lindeloff [resp: locally nearly closed Lindeloff; locally mildly lindeloff].

(iii) If X is v--compact [countably v--compact], then Y is S-closed [countably S-closed].

(iv) If X is v--Lindelof, then Y is S-Lindelof [nearly Lindelof].

(v) If is v--closed [countably v--closed], then Y is nearly compact [nearly countably compact].

Theorem 4.2. If f is contra r-iresolute and al.c., surjection and is mildly compact (resp. mildly countably compact, mildly Lindelof), then Y is nearly compact (resp. nearly countably compact, nearly Lindelof) and S-closed (resp. countably S-closed, S-Lindelof).

Proof. Since f is contra r-iresolute and al.c., for {[V.sub.[alpha]] : [alpha] [member of] I} be any regular closed (resp: regular open) cover of Y, {[f.sup.-1]([V.sub.[alpha]] : [alpha] [member of] I} is a clopen cover of X and since X is mildly compact, [there exists] a finite subset [I.sub.0] of I [contains as member] X = [universal]{[f.sup.-1]([V.sub.[alpha]]) : [alpha] [member of] [I.sunb.0]}. Since f is surjective, Y = [universal] {[V.sub.[alpha]] : [alpha] [member of] [I.sub.0]}. Hence Y is S-closed (resp: nearly compact). The other proofs can be obtained similarly.

Theorem 4.3. If f is al.c.v.c., surjection and

(i) X is v--compact [v--lindeloff] then Y is mildly closed compact [mildly closed lindeloff].

(ii) X is s-closed then Y is mildly compact [mildly lindeloff].

Theorem 4.4. If X is v--ultra-connected and f is al.c.v.c. and surjective, then Y is hyperconnected.

Proof. If Y is not hyperconnected. Then [there exists] V [member of] [sigma] [contains as member] [bar.V] [not equal to] Y. Then 3 disjoint nonempty regular open subsets [B.sub.1] and [B.sub.2] in Y. Since f is al.c.v.c. and onto, [A.sub.1] = [f.sup.-1]([B.sub.1]) and [A.sub.2] = [f.sup.-1]([B.sub.2]) are disjoint non-empty v--closed subsets of X. By assumption, the v-- ultraconnectedness of X implies that [A.sub.1] and [A.sub.2] must intersect, which is a contradiction. Therefore Y is hyperconnected.

Theorem 4.5. If f is al.c.v.c.[contra v--irreolute] surjection and X is v--connected, then Y is connected [v--connected].

Proof. If Y is disconnected. Then Y = [V.sub.1] [union] [V.sub.2] and [V.sub.1] [intersection] [V.sub.2] = [phi]. Since f is al.c.v.c., [f.sup.-1] ([V.sub.1]) and [f.sup.-1]([V.sub.2]) are disjoint v--open sets in X and X = [f.sup.-1]([V.sub.1]) [union] [f.sup.-1]([V.sub.2]), which is a contradiction for v--connectedness of X. Hence Y is connected.

Corollary 4.2. The inverse image of a disconnected [v--disconnected] space under a al.c.v.c.[contra v--irreolute] surjection is v--disconnected.

Theorem 4.6. If fis al.c.v.c., injection and

(i) Y is U[T.sub.i], then X is v--[T.sub.i] [hence semi-[T.sub.i] and [beta] - [T.sub.i]], i = 0, 1, 2.

(ii) Y is U[R.sub.i], then X is v--[R.sub.i] [hence semi-[R.sub.i] and [beta] - [R.sub.i]], i = 0, 1.

(iii) If f is closed and Y is U[T.sub.i], then X is v - [T.sub.i] [hence semi-[T.sub.i] and [beta] - [T.sub.i]], i = 3, 4.

(iv) Y is U[C.sub.i][resp : UD] then X is v - [T.sub.i][resp: v - [D.sub.i]], hence X is semi-[T.sub.i] [resp: semi-[D.sub.i]] and [beta] - [T.sub.i] [resp: [beta] - [D.sub.i]], i = 0, 1, 2.

Theorem 4.7. If f is al.c.v.c.[resp: al.c.r.c] and Y is U[T.sub.2], then the graph G(f) of f is v--closed in the product space X x Y.

Proof. Let ([x.sub.1], [x.sub.2]) [not member of] G(f) [??] y [not equal to] f(x) [??] [there exists] disjoint clopen sets V and W [contains as member] f(x) [member of] V and y[member of] W. Since f is al.c.v.c., [there exists]U [member of] vO(X) [contains as member] x [member of] U and f(U) [subset] W. Therefore (x, y) [member of] U x V [subset] X x Y - G(f). Hence G(f) is v--closed in X x Y.

Corollary 4.3. If f is al.c.v.c. and Y is U[T.sub.2], then the graph G(f) of f is semi-closed [resp: [beta]--closed and semi-[theta]-closed] in the product space X x Y.

Theorem 4.8. If f is al.c.v.c.[al.c.r.c] and Y is U[T.sub.2], then A = {([x.sub.1], [x.sub.2])|f([x.sub.1]) = f([x.sub.2])} is v--closed[and hence semi-closed and [beta]--closed] in the product space X x X.

Proof. If ([x.sub.1], [x.sub.2]) [member of] X x X - A, then f([x.sub.1]) [not equal to] f([x.sub.2]) [??] [there exists] disjoint [V.sub.j] [member of] CO ([sigma]) [contains as member] f([x.sub.j]) [member of] [V.sub.j], and since f is al.c.v.c., [f.sup.-1]([V.sub.j]) [member of] vO(X, [x.sub.j]) for each j = 1, 2. Thus ([x.sub.1], [x.sub.2]) [member of] [f.sup.-1]([V.sub.1]) x [f.sup.-1]([V.sub.2]) [member of] vO(X x X) and [f.sup.-1]([V.sub.1]) x [f.sup.-1]([V.sub.2]) [subset] X x X - A. Hence A is v--closed.

Theorem 4.9. If f is r-irresolute{al.c.c.}; g : X [right arrow] Y is c.v.c; and Y is U[T.sub.2], then E = {x [member of] X : f(x) = g(x)} is v--closed [and hence semi-closed and [beta]--closed] in X.

Theorem 4.10. If f is an al.c.v.c. injection and Y is weakly Hausdorff, then X is v - [T.sub.1].

Proof. Suppose that Y is weakly Hausdorff. For any x [not equal to] y [member of] X, [there exists]V, W [member of] RC(Y) [contains as member] f(x) [member of] V, f(y) [not member of] V, f(x) [not member of] W and f(y) [member of] W. Since f is al.c.v.c., [f.sup.-1](V) and [f.sup.-1](W) are v--open subsets of X such that x [member of] [f.sup.-1](V),y [not member of] [f.sup.-1](V), x [not member of] [f.sup.-1](W) and y [member of] [f.sup.-1](W). This shows that X is v - [T.sub.1].

Theorem 4.11. If for each pair [x.sub.1] [not equal to] [x.sub.2] [member of] X there exists a function f of X into a Urysohn space Y such that f([x.sub.1]) [not equal to] f([x.sub.2]) and f is al.c.v.c., at [x.sub.1] and [x.sub.2], then X is v-- [T.sub.2].

Proof. Let [x.sub.1] [not equal to] [x.sub.2]. Then by the hypothesis [there exists] a function f which satisfies the condition of this theorem. Since Y is Urysohn and f([x.sub.1]) [not equal to] f([x.sub.2]), [there exists] open sets [V.sub.1] and [V.sub.2] containing f([x.sub.1]) and f([x.sub.2]), respectively, such that [bar.[V.sub.1]] [intersection] [bar.[V.sub.2]] = [phi]. Since f is al.v.c., at [x.sub.i], [there exists][U.sub.i] [member of] vO(X, [x.sub.i]) 3 f([U.sub.i]) [subset] [bar.[V.sub.i]] for i = 1, 2. Hence, we obtain [U.sub.1] [intersection] [U.sub.2] = [phi]. Therefore, X is v - [T.sub.2].

Corollary 4.4. If f is r-irresolute injection and Y is Urysohn, then X is v - [T.sub.2].

Definition 4.1. A function f is said to have a strongly contra-v--closed graph if for each (x, y) [member of] (X x Y)--G(f) there exists U [member of] vO(X, x) and a regular closed set V of Y containing y such that (U x V) [there exists] G(f) = [phi].

Lemma 4.1. f has a strongly contra-v--closed graph iff for each (x,y) [member of] (X x Y) - G(f [there exists]U [member of] vO(X, x) and V [member of] RC(Y, y) [contains as member] f(U) [intersection] V = [phi].

Theorem 4.12. If fis injective al.c.v.c. with the strongly contra-v--closed graph, then X is v - [T.sub.2].

Proof. Let x [not equal to] y [member of] X. Since f is injective, we have f(x) [not equal to] f(y) and (x, f(y)) [member of] (X x Y) - G(f). Since G(f) is strongly contra-v--closed, by Lemma 5.1, [there exists] U [member of] vO(X, x) and a V [member of] RC(Y,f(y)) [contains as member] f(U) [intersection] V = [phi]. Since f is al.c.v.c., by Theorem 3.16, [contains as member] G [member of] vO(X, y) [contains as member] f(G) [subset] V. Therefore, we have f(U) [intersection] f(G) = [phi]; hence U [intersection] G = [phi]. Hence X is v - [T.sub.2].

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S. Balasubramanian ([dagger]) and P. Aruna Swathi Vyjayanthi ([double dagger])

([dagger]) Department of Mathematics, Government Arts College (Autonomous), Karur-639 005 (T.N.)

([double dagger]) Research Scholar, Dravidian University, Kuppam (A.P.)

E-mail: [e-mail] [e-mail]