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Smarandache v-Connected spaces.

[section]1. Introduction

After the introduction of semi open sets by Norman Levine various authors have turned their attentions to this concept and it becomes the primary aim of many mathematicians to examine and explore how far the basic concepts and theorems remain true if one replaces open set by semi open set. The concept of semi connectedness and locally semi connectedness are introduced by Das and J. P. Sarkar and H. Dasgupta in their papers. Keeping this in mind we here introduce the concepts of connectedness using v-open sets in topological spaces. Throughout the paper a space X means a topological space (X, [Tau]). The class of v-open sets is denoted by v - D(X, [Tau]) respectively. The interior, closure, v-interior, v-closure are defined by [A.sup.o], [A.sup.-], v[A.sup.o], v[A.sup.-] respectively. In section 2 we discuss the basic definitions and results used in this paper. In section 3 we discuss about Smarandache v-connectedness and v-components and in section 4 we discuss locally Smarandache v-connectedness in the topological space and obtain their basic properties.

[section]2. Preliminaries

A subset A of a topological space (X, [Tau]) is said to be regularly open if A = [([(A).sup.-]).sup.o], semi open(regularly semi open or v-open) if there exists an open(regularly open) set O such that O [subset] A [subset] [(O).sup.-] and v-closed if its complement is v-open. The intersection of all v-closed sets containing A is called v-closure of A, denoted by v[(A).sup.-]. The class of all v-closed sets are denoted by v-CL(X, [Tau]). The union of all v-open sets contained in A is called the v-interior of A, denoted by v[(A).sup.o]. A function f: (X, [Tau]) [right arrow] (Y, [sigma]) is said to be v-continuous if the inverse image of any open set in Y is a v-open set in X; said to be v-irresolute if the inverse image of any v-open set in Y is a v-open set in X and is said to be v-open if the image of every v-open set is v-open. f is said to be v-homeomorphism if f is bijective, v-irresolute and v-open. Let x be a point of (X, [Tau]) and V be a subset of X, then V is said to be v-neighbourhood of x if there exists a v-open set U of X such that x [member of] U [subset] V. x [member of] X is said to be v-limit point of U iff for each v-open set V containing x, V [intersection] (U - {x}) [not equal to] [phi]. The set of all v-limit points of U is called v-derived set of U and is denoted by [D.sub.v](U). union and intersection of v-open sets is not open whereas union of regular and v-open set is v-open.

Note 1. Clearly every regularly open set is v-open and every v-open set is semi-open but the reverse implications do not holds good. that is, RO(X)[subset] v - O(X) [subset]SO(X).

Theorem 2.1. (i)If B [subset]X such that A [subset] B [subset] [(A).sup.-] then B is v-open iff A is v-open. (ii)If A and R are regularly open and S is v-open such that R [subset] S [subset] (R)-. Then A[intersection]R = [phi] [??] A [intersection] S = [phi].

Theorem 2.2. (i) Let A [??] Y [??] X and Y is regularly open subspace of X then A is v-open in X iff A is v-open in [[Tau].sub./Y].

(ii)Let Y [??] X and A [member of] v - O(Y, [[Tau].sub./Y]) then A [member of] v - O(X, [Tau]) iff Y is v-open in X.

Theorem 2.3. An almost continuous and almost open map is v-irresolute.

Example 1. Identity map is v-irresolute.

[section]3.v-Connectedness.

Definition 3.01. A topological space is said to be Smarandache v-connected if it cannot be represented by the union of two non-empty disjoint v-open sets.

Note 2. Every Smarandache v-connected space is connected but the converse is not true in general is shown by the following example.

Example 2. Let X = {a, b, c} and [Tau] = {[phi], {a}, X}; then (X, [Tau]) is connected but not v-connected

Note 3. Every Smarandache v-connected space is r-connected but the converse is not true in general is shown by the following example.

Example 3. Let X = {a, b, c} and [Tau] = {[phi], {a}, {b}, {a, b}, X } then (X, [Tau]) is r-connected but not Smarandache v-connected

similary one can show that every semi connected space is Smarandache v-connected but the converse is not true in general.

Theorem 3.01. Let (X, [Tau]) and (Y, [sigma]) be two topological spaces. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is a v-open and v-continuous mapping, then the inverse image of each v-open set in Y is v-open in X

Corollary 3. Let (X, [Tau]) and (Y, [sigma]) be two topological spaces. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is an r-open and r-continuous mapping, then the inverse image of each v-open set in Y is v-open in X

Theorem 3.02. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is a v-continuous mapping, and (X, [Tau]) is Smarandache v-connected space, then (Y, [sigma]) is also v-connected.

Corollary 4. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is a r-continuous mapping, and (X, [Tau]) is Smarandache v-connected space, then (Y, [sigma]) is also Smarandache v-connected.

Theorem 3.03. Let (X, [Tau]) be a topological space and

(i) A be v-open. Then A is Smarandache v-connected if and only if (A, T/A) is Smarandache v-connected

(ii) A be r-open. Then A is Smarandache v-connected if and only if (A, TlA) is Smarandache v-connected

Lemma 3.01.If A and B are two subsets of a topological space (X, [Tau]) such that A [subset] B then v[(A).sup.-] [subset] v[(B).sup.-]

Lemma 3.02. If A is v-connected and A [subset] C [union] D where C and D are v-separated, then either A C C or A C D.

Proof. Write A = (A [intersection] C) [union] (A [intersection] D). Then by lemma 3.01, we have (A [intersection] C) [intersection] (v[(A).sup.-] [intersection] v[(D).sup.-]) [subset] C [intersection] v[(D).sup.-]. Since C and D are v-separated, C [intersection] v[(D).sup.-] = [phi] and so (A [intersection] C) [intersection] (v[(A).sup.-] [intersection] v[(D).sup.-]) = [phi]. Similar argument shows that (v[(A).sup.-] [intersection] v[(C).sup.-]) [intersection] (A [intersection] D) [phi]. So if both (A [intersection] C) [not equal to] [phi] and (A [intersection] D) [not equal to] [phi], then A is not Smarandache v-connected, which is a contradiction for A is Smarandache v-connected. Therefore either (A [intersection] C) = [phi] or (A [intersection] D) = [phi], which in turn implies that either A [subset] C or A [subset] D.

Lemma 3.03. The union of any family of Smarandache v-connected sets having nonempty intersection is a Smarandache v-connected set.

Proof. If E = [union][E.sub.[alpha] is not v-connected where each [E.sub.[alpha]] is Smarandache v-connected, then E = A [union] B, where A and B are v-separated sets. Let x [member of] [intersection][E[alpha]] be any point, then x [member of] [E.sub.[alpha]] for each [E.sub.[alpha]] and so x [member of] E which implies that x [member of] A [union] B in turn implies that either x [member of] A or x [member of] B.

Without loss of generality assume x [member of] A. Since x [member of] [E.sub.[alpha]], A [intersection] [E.sub.[alpha]] [not equal to] [phi] for every [alpha]. By lemma 3.02, either each [E.sub.[alpha]] [subset] A or each [E.sub.[alpha]] [subset] B. Since A and B are disjoint we must have each [E.sub.[alpha]] [subset] A and hence each E [subset] A which gives that B = [phi].

Lemma 3.4. If A is Smarandache v-connected and A [subset] B [subset] v[(A).sup.-], then B is Smarandache v-connected set.

Proof. If E is not v-connected, then E = A [union] B, where A and B are v-separated sets. By lemma 3.02 either E [subset] A or E [subset] B. If E [subset] A, then v[(E).sup.-] [subset] v[(A).sup.-] and so v[(E).sup.-] [intersection] B [subset] v[(A).sup.-] [intersection] B = [phi]. On the other hand B [subset] E [subset] v[(E).sup.-] and so v[(E).sup.-] [intersection] B. Thus we have B = [phi], which is a contradiction. Hence the Lemma.

Corollary 5. If A is Smarandache v-connected and A [subset] B [subset] (A)-, then B is Smarandache v-connected set.

Lemma 3.5. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is v-open and v-continuous, A [subset] X is v-open. Then if A is v-connected, f (A) is also Smarandache v-connected.

Proof. Let f: (X, [Tau]) [right arrow] (Y, [sigma]) is open and v-continuous, A [subset] X be open. Since A is v-connected and open in (X, [Tau]), then (A, [[Tau].sub./A]) is also v-connected (by Th. 3.03). Now [f.sub./A]: (A,[[Tau].sub./A]) [right arrow] (f(A),[[sigma].sub.f(A)]) is onto and v-continuous and so by theorem 3.02 f(A) is also v-connected in (f(A), [[sigma].sub.f(A>]). Now for f is open, f(A) is open in (Y, [sigma]) and so by theorem 3.03, f (A) is Smarandache v-connected in (Y, [sigma])

We have the following corollaries from the above theorem

Corollary 6. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is r-open and r-continuous, A [subset] X is r-open. Then if A is v-connected, then f (A) is also Smarandache v-connected.

Corollary 7. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is v-open and v-continuous, A [subset] X is r-open. Then if A is connected, then f (A) is also Smarandache v-connected.

Definition 3.02. Let (X, T) be a topological space and x [member of] X. The v-component of x, denoted by v C(x), is the union of Smarandache v-connected subsets of X containing x.

Further if E [subset] X and if x [member of] E, then the union of all v-connected set containing x and contained in E is called the v-component of E corresponding to x. By the term that C is a v-component of E, we mean that C is v-component of E corresponding to some point of E.

Lemma 3.06. Show that vC(x) is Smarandache v-connected for any x [member of] X

Proof. As the union of any family of Smarandache v-connected sets having a nonempty intersection is a Smarandache v-connected set, it follows that vC(x) is Smarandache v-connected

Theorem 3.04. In a Topological space (X, [Tau]),

(i) Each v-component v(x) is a maximal Smarandache v-connected set in X.

(ii) The set of all distinct v-components of points of X form a partition of X and (iii) Each v(x) is v-closed in X

Proof. (i) follows from the definition 3.02

(ii) Let x and y be any two distinct points and vC(x) and vC(x) be two r-components of x and y respectively. If vC(x) [intersection] vC(y) [not equal to] [phi], then by lemma 3.03, vC(x) [union] vC(y) is Smarandache v-connected. But vC(x) [subset] vC(x) [union] vC(y) which contradicts the maximality of vC(x).

Let x [member of] X be any point, then x [member of] vC(x) implies [union]{x} [subset] [union]vC(x) for all x [member of] X which implies X [subset] [union]vC(x) [subset] X. Therefore [union]vC(x) = X

(iii) Let x [member of] X be any point, then (vC(x))- is a v-connected set containing x. But vC(x) is the maximal Smarandache v-connected set containing x, therefore (vC(x))- C vC(x). Hence vC(x) is v-closed in X.

[section]4. Locally v-connectedness

Definition 4.01. A topological space (X, [Tau]) is called

(i) Smarandache locally v-connected at x [member of] X iff for every v-open set U containing x, there exists a Smarandache v-connected open set C such that x [member of] C [subset] U.

(ii) Smarandache locally v-connected iff it is Smarandache locally v-connected at each x [member of] X.

Remark 3. Every Smarandache locally v-connected topological space is Smarandache locally connected but converse is not true in general.

Remark 4. Smarandache local v-connectedness does not imply Smarandache v-connectedness as shown by the following example.

Example 4. X = {a. b, c} and [Tau] = {[phi], {a}, {c}, {a, b}, {a, c}, X } then (X, [Tau]) is Smarandache locally v-connected but not Smarandache v-connected.

Remark 5. Smarandache v-connectedness does not imply Smarandache local v-connectedness in general.

Theorem 4.01. A topological space (X, [Tau]) is Smarandache locally v-connected iff the v-components of v-open sets are open sets.

Theorem 4.02. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is a v-continuous open and onto mapping, and (X, [Tau]) is Smarandache locally v-connected space, then (Y, [sigma]) is also locally Smarandache v-connected.

Proof. Let U be any v-open subset of Y and C be any v-component of U, then [f.sup.-1](U) is v-open in X. Let A be any v-component of [f.sup.-1(U). Since X is locally v-connected and [f.sup.-1](U) is v-open, A is open by theorem 4.01. Also f (A) is v-connected subset of Y and since C is v-component of U, it follows that either f(A) [subset] C or f(U) [intersection] C = [phi]. Thus [f.sup.-1](C) is the union of collection of v-components of [f.sup.-1](U) and so [f.sup.-1](C) is open. As f is open and onto, C = f [??] [f.sup.-1](C) is open in Y. Thus any v-component of v-open set in Y is open in Y and hence by above theorem Y is Smarandache locally v-connected.

Corollary 8. If f: (X, [Tau]) [right arrow] (Y, [sigma]) is a r-continuous r-open and onto mapping, and (X, [Tau]) is Smarandache locally v-connected space, then (Y, [sigma]) is also Smarandache locally v-connected.

Proof. Immediate consequence of the above theorem.

Note 4. semi connectedness need not imply and implied by locally semi connectedness. Similarly a Smarandache v-connected space need not imply and implied by Smarandache locally v-connected in general.

Theorem 4.03. A topological space (X, [Tau]) is Smarandache locally v-connected iff given any x [member of] X and a v-open set U containing x, there exists an open set C containing x such that C is contained in a single v-component of U.

Proof. Let X be Smarandache locally v-connected, x [member of] X and U be a v-open set containing x. Let A be a v- component of U containing x. Since X is Smarandache locally v-connected and U is v-open, there is a Smarandache v-connected set C such that x [member of] C [subset] U. By theorem 3.01, A is the maximal Smarandache v-connected set containing x and so x [member of] C [subset] A [subset] U. Since v-components are disjoint sets, it follows that C is not contained in any other v-component of U.

Conversely, suppose that given any point x [member of] X and any v-open set U containing x, there exists an open set C containing x which is contained in a single v-component F of U. Then x [member of] C [subset] F [subset] U. Let y [member of] F, then y [member of] U. Thus there is an open set O such that y [member of] O and O is contained in a single v-component of U. As the v-components are disjoint sets and y [member of] F, y [member of] O [subset] F. Thus F is open. Thus for every x [member of] X and for every v-open set U containing x, there exists a Smarandache v-connected open set F such that x [member of] F [subset] U. Thus (X, [Tau]) is Smarandache locally v-connected at x. Since x [member of] X is arbitrary, (X, [Tau]) is Smarandache locally v-connected.

Remark 6.

Connected [??] semi-connected

r-Connected [??] v-Connected.

none is reversible

Example 5. FOR X = {a, b, c, d}; [[Tau].sub.1] = {[phi], {b}, {a, b}, {b, c}, {a, b, c}, X} [[Tau].sub.2] = {[phi], {a}, {b}, {a, b}, X } and [[Tau].sub.3] = {[phi], {a}, {b}, {d}, {a, b}, {a, d}, {b, d}, {a, b, c}, {a, b, d}, X }

(X, [[Tau].sub.1]) is both r-connected and Smarandache v-connected; (X, [[Tau].sub.2]) is r-connected but not Smarandache v-connected and (X, [[Tau].sub.3]) is neither r-connected and nor Smarandache v-connected

Conclusion.

In this paper we defined new type of connectedness using v-open sets and studied their interrelations with other connectedness.

Acknowlegement.

The Authors are thankful for the referees for their critical comments and suggestions for

S. Balasubramanian ([dagger]), C. Sandhya ([double dagger]) and P. Aruna Swathi Vyjayanthi (#)

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

([double dagger]) Department of Mathematics, C.S.R. Sharma College, Ongole (A.P.), India

(#) Department of Mathematics, C.R. College, Chilakaluripet(A.P.), India

E-mail: mani55682@rediffmail.com sandhya_karavadi@yahoo.co.uk. vyju_9285@rediffmail.com
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Author:Balasubramanian, S.; Sandhya, C.; Vyjayanthi, P. Aruna Swathi
Publication:Scientia Magna
Article Type:Report
Geographic Code:9INDI
Date:Jun 1, 2009
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