International Mathematical Forum, Vol. 6, 2011, no. 22, 1059 - 1064 Sh - PDF document

International Mathematical Forum, Vol. 6, 2011, no. 22, 1059 - 1064 Shyamapada ModakDepartment of Mathematics 1060 S. ModakA topological space (X, ) is said to be resolvable if there is a subset D of X such that both D and X – D are dense in X, otherwise it is said to be irresolvable. The space of reals with usual topology provides an example of a resolvable space while any topological space with an isolated point furnishes for an irresolvable one. It follows that a space is irresolvable if and only if every dense subset of it has a nonempty interior. A space is called hereditarily irresolvable if every subspace of it is irresolvable. Hewitt proved the following very remarkable theorem: Theorem 1.1. Every topological space (X, ) can be represented as a disjoint union X = FG where F is closed and resolvable and G is open and hereditarily irresolvable. (X, ) is resolvable if and only if G = and (X, ) is hereditarily irresolvable if and only if F = The above representation will henceforth be called Hewitt representation of X. Later on in eighties of last century, Ganster et. al[7] and in nineties Bandyopadhyay and Chattopadhyay[2], [3] had made extensive study on resolvability and irresolvability using the above Hewitt representation. For further reference we refer to [4],[5],[6]. The aim of the paper is to define when a subset A of a topological space is resolvable or irresolvable relative to X. Following Arhangrl’skii we wish to define relativization in the manner when A = X, X has the same property. We shall denote closure and interior with respect to (X, ) as cl and int respectively and shall use suffix to denote the closure and interior with respect to other topologies. 2. Relative resolvability and irresolvability Definition 2.1. A nonempty subset A of a topological space (X, ) is called resolvable relative to X or resolvable in X if there are two dense subsets D and of (X, ) with D A , D A such that D A = ; otherwise irresolvable relative to X or irresolvable in X. Thus A is irresolvable in X if for any two dense subsets D and D of Xsuch that D, D, we have D A Note that if X, resolvability (respectively irresolvability ) of in X does not necessarily imply resolvability (respectively irresolvability ) of with respect to 1062 S. ModakTheorem 2.7. Let A X such that A = {x}. Then A is irresolvable in X . Proof. Let D(X, ) such that D A and D. Hence D= {x} = D A, so D{x} = {x} , so that A is irresolvable in X . 2.8. If x is an isolated point of X, from Theorem2.4., it follows that {x} is irresolvable in X. Now we discuss resolvability in X. Theorem2.9. Let (X, ) be resolvable. Then A X is resolvable in X if intAProof. Given that (X, ) is resolvable, so there exists D , D(X, ) such that X = D and D. Let intA . Then D, D , whereas A = and hence A is resolvable in X. Theorem2.10. Let A X with cl intA = X. If A is resolvable in X, (X, ) is resolvable. Proof. Since A is resolvable in X, there exit dense sets D in (X, ) such that . Since cl intA= X, we get DA and DA both are dense in X. Since (D. We derive that (X, ) is resolvable. For next Theorem we use following result: 2.11 [4] . Let X = FG be the Hewitt representation of a space (X, ). If D is dense in (X, ) then G cl intD. Theorem2.12. Let (X, ) be irresolvable. Then A X with intA is irresolvable in X if and only if int(A, where X = FG is the Hewitt representation of X. Proof. Suppose int(A. We proof that A is irresolvable in X. If not, let A be resolvable in X. Then there exists dense subsets D and D of (X, ) such that , D, but D A= . Let = int(AG). If D is dense in X then G cl intD [by Result 2.11]. Hence G cl intD . Similarly G cl is nonempty open set in G we get intD. Let = . Again since G cl intD , . Let = intD . Thus = int(AG) when and . Hence int(D so that D, a contradiction to that D A = . Hence A is irresolvable in X. Conversely let A be irresolvable in X we are to show that . If possible supposed that int(AG) = . Since intA , intAG= F [by Theorem1.1.]. But F is resolvable implies that any nonempty open subset of F is also resolvable. Hence intA = E , E = , intA cl E, intA cl . 1064 S. Modak[4] C. Chattopadhyay, A study on resolvability and irresolvability in topological spaces and bitopological spaces with relevant structures and mappings, Thesis 1995 Burdwan Univ., W.B., India. [5] A.G. El’kin, Ultrafilters and undecomposable spaces,Vesnik Mosk. Uni. Mat. 24(5)(1969), 51-56. [6] M. Ganster, Preopen sets and resolvable spaces, Kyungpook Math. J. 27(2) (1987), 135-143. [7] M. Ganster, I.L. Reilly and M.K. Vamanamurthy, Dense sets and irresolvable spaces, Ricerche de Mathematica, XXXVI (1987), 163-170. [8] L. Genglei and W. Huidong, The properties of relative regularity and compactness, Modern Applied Science, Vol.2, no.4(2008). [9] L. Genglei ,W. Huidong and W.Yunxia, Some properties of relative regularity and compactness, Journal of Mathematics Research, Vol.1, no.1(2009). [10] E. Hewitt, A problem of set theoretic topology, Duke Math. J. (1943), 309-333. [11] M.S.Sarsak and H.Z.Hdeib, Relatively strongly normal and relatively normal subspaces, International Mathematical Forum, 5, 2010, no. 12, 579 – 586. Received: October, 2010