[disjunction] and [conjunction]sets of generalized topologies.1. Introduction In 199[gamma], Professor A. Csaszar [1] nicely presented the open sets and all weak forms of open sets in a topological space Noun 1. topological space  (mathematics) any set of points that satisfy a set of postulates of some kind; "assume that the topological space is finite dimensional" mathematical space X in terms of monotonic monotonic  In domain theory, a function f : D > C is monotonic (or monotone) if for all x,y in D, x <= y => f(x) <= f(y). ("<=" is written in LaTeX as \sqsubseteq). functions defined on [??](X), the collection of all subsets of X. For each such function [gamma], he defined a collection [mu] of subsets of X, called the collection of [gamma]open sets. A is said to be [gamma]open if A [subset A group of commands or functions that do not include all the capabilities of the original specification. Software or hardware components designed for the subset will also work with the original. ] [gamma](A). B is said to be [gamma]closed if its complement is [gamma]open. With respect to this collection of subsets of X, for A [subset] X, the [gamma]interior of A, denoted by [i.sub.[gamma]](A), is defined as the largest [gamma]open set contained in A and the [gamma]closure of A, denoted by [i.sub.[gamma]](A), is the smallest [gamma]closed set containing A. It is established that [mu] is a generalized gen·er·al·ized adj. 1. Involving an entire organ, as when an epileptic seizure involves all parts of the brain. 2. Not specifically adapted to a particular environment or function; not specialized. 3. topology topology, branch of mathematics, formerly known as analysis situs, that studies patterns of geometric figures involving position and relative position without regard to size. [3]. In [5], [gamma]semiopen sets are defined and discussed. In [7], [gamma][alpha]open sets, [gamma]preopen sets and [gamma]open sets are defined and discussed. [gamma][beta]open sets are defined in [6]. If [alpha] is the family of [gamma][alpha]open sets, is the family of all [gamma]semiopen sets, is the family of all [gamma]preopen sets, [pi] is the family of all [gamma][beta]open sets and [beta] is the family of all [gamma]open sets, then each collection is a generalized topology. Since every topological space is a generalized topological space, we prove that some of the results established for topological to·pol·o·gy n. pl. to·pol·o·gies 1. Topographic study of a given place, especially the history of a region as indicated by its topography. 2. spaces are also true for the generalized topologies [Omega] = {[mu], [alpha], [sigma], [pi], b, [beta]}. In section 2, we list all the required definitions and results. In section 3, we define the [[conjunction conjunction, in astronomy conjunction, in astronomy, alignment of two celestial bodies as seen from the earth. Conjunction of the moon and the planets is often determined by reference to the sun. ].sub.k] and [[disjunction disjunction /dis·junc·tion/ (junk´shun) 1. the act or state of being disjoined. 2. in genetics, the moving apart of bivalent chromosomes at the first anaphase of meiosis. ].sub.k], operators for each k [member of] [Omega] and discuss its properties. Then, we define [[conjunction].sub.k]sets, [[disjunction].sub.k]sets, g. [[conjunction].sub.k]sets and g. [[disjunction].sub.k]sets and characterize these sets. In section 4, for each k [member of] [Omega], we define and characterize the separation axioms This is a list of axioms as that term is understood in mathematics, by Wikipedia page. In epistemology, the word axiom is understood differently; see axiom and selfevidence. Individual axioms are almost always part of a larger axiomatic system. k  [T.sub.i], i = 0, 1, 2 and k  [R.sub.i], i = 0, 1. 2. Preliminaries Let X be a nonempty set and [Gamma] = {[gamma]: [??](X) [right arrow] [??](X)[gamma](A) [subset] [gamma](B) whenever A [subset] B}. For [gamma] [member of] [Gamma], a subset A [subset] X is said to be [gamma]open [1] if A [subset] [gamma](A). The complement of a [gamma]open set is said to be a [gamma]closed set. A family [xi] [subset] [??](X) is said to be a generalized topology [3] if 0 [member of] [xi] and [xi] is closed under arbitrary Irrational; capricious. The term arbitrary describes a course of action or a decision that is not based on reason or judgment but on personal will or discretion without regard to rules or standards. union. The family of all [gamma]open sets, denoted by [mu], is a generalized topology [4]. A [subset] X, is said to be [gamma]semiopen [5] if there is a [gamma]open set G such that G [subset] A [subset] [c.sub.[gamma]](G) or equivalently, A [subset] [c.sub.[gamma]][i.sub.[gamma]]. (A) [8, Theorem theorem, in mathematics and logic, statement in words or symbols that can be established by means of deductive logic; it differs from an axiom in that a proof is required for its acceptance. 2.4]. A is said to be [gamma]preopen [7] if A [subset] [i.sub.[gamma]][c.sub.[gamma]] (A). A [subset] X, is said to be [gamma][alpha]open [7] if A [subset] [i.sub.[gamma]][c.sub.[gamma]][i.sub.[gamma]] (A). A is said to be [gamma][beta]open [7] if A [subset] [c.sub.[gamma]][i.sub.[gamma]][c.sub.[gamma]] (A). A is said to be [gamma][beta]open [6] if A [subset] [i.sub.[gamma]][c.sub.[gamma]](A) [union] [c.sub.[gamma]][i.sub.[gamma]] (A). In [4], [5], [6] and [7], it is established that each [member of] [Omega] is a generalized topology and so [c.sub.k] and [i.sub.k], can be defined, similar to the the definition of [c.sub.[gamma]] and [i.sub.[gamma]]. In this paper, for k [member of] [Omega], the pair (X, k) is called a generalized topological space or simply a space. For each [gamma] [member of] [Gamma], a mapping [gamma]*: [??](X) [right arrow] [??](X) [1] is defined by [gamma]*(A) = X  [gamma](X A). Clearly, [gamma]* [member of] [Gamma]. The following lemmas This following is a list of lemmas (or, "lemmata", i.e. minor theorems, or sometimes intermediate technical results factored out of proofs). See also list of axioms, list of theorems and list of conjectures. will be useful in the sequel. Moreover, one can easily prove the following already established results of Lemma lemma (lĕm`ə): see theorem. (logic) lemma  A result already proved, which is needed in the proof of some further result. 2.1. Lemma 2.1. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following hold. (a) A [subset] [c.sub.k] (A) for every subset A of X. (b) If A [subset] B, then [c.sub.k](A) [subset] [c.sub.k](B). (c) For every subset A of X, x [member of] [c.sub.k] (A) if and only if there exists a kopen set G such that A [intersection intersection /in·ter·sec·tion/ (sek´shun) a site at which one structure crosses another. intersection a site at which one structure crosses another. ] G [not equal to] 0 (For k = [mu], it is established in Lemma 2.1 of [2]). (d) A is kclosed if and only if A = [c.sub.k](A). (e) [c.sub.k] (A) is the intersection of all kclosed sets containing A. Lemma 2.2. Let X be a nonempty set and [gamma] [member of] [Gamma](X). Then X is [gamma]semiopen [5, Proposition 1.2]. 3. Usets and nsets In this section, for the space (X, k), k [member of] [Omega], we define [disjunction]sets, [conjunction]sets, g. [disjunction]sets and g. [conjunction]sets. For [sigma] [member of] [Omega], [disjunction]sets and [conjunction]sets are defined in [5]. For A [subset] X, we define [[conjunction].sub.k] (A) = [intersection] {U [subset] X A [subset] U and U [member of] k} and [[disjunction].sub.k] (A) = [union] {U [subset] XU [subset] A and U is kclosed}. The following Theorem 3.1 gives the properties of the operator [[conjunction].sub.k]. Example 3.2 below shows that the two sets in 3.1(e) are not equal. Theorem 3.1. Let A, B and {[C.sub.[??]][??] [member of] [Delta]} be subsets of X, [gamma] [member of] [Gamma] and [member of] [Omega]. Then the following hold. (a) If A [subset] B, then [[conjunction].sub.k](A) [subset] [[conjunction].sub.k](B). (b) A [subset] [[conjunction].sub.k](A). (c) [[conjunction].sub.k] ([[conjunction].sub.k](A)) = [[conjunction].sub.k](A). (d) [[conjunction].sub.k] ([union]{[C.sub.[??]][??][member of][Delta]}) = [union]{[[conjunction].sub.k]([C.sub.[??]])[??][member of][Delta]} (e) [[conjunction].sub.k] ([intersection]{[C.sub.[??]][??][member of][Delta]}) [subset] [intersection]{[[conjunction].sub.k]([C.sub.[??]])[??][member of][Delta]}. (f) If A [member of] k, then [[conjunction].sub.k](A) = A. (g) [[conjunction].sub.k](A) = {x[c.sub.k]({x})[intersection] A [not equal to] 0}. (h) y [member of] [[conjunction].sub.k]({x}) if and only if x [member of] [c.sub.k]({y}). (i) [[conjunction].sub.k]({x}) [not equal to] [[conjunction].sub.k]({y}) if and only if [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Proof. (a). Suppose x [??] [[conjunction].sub.k](B). Then there exists G [member of] k such that B [subset] G and x [??] G. Since A [subset] B, there exists G [member of] k such that A [subset] G and x [??] G and so x [??] [[conjunction].sub.k](A) which proves (a). (b). The proof follows from the definition of [[conjunction].sub.k]. (c). By (b), A [subset] [[conjunction].sub.k](A) and so by (a), [[conjunction].sub.k](A) [subset] [[conjunction].sub.k]([[conjunction].sub.k](A)). Let x [??][[conjunction].sub.k](A). then there exists G [member of] such that A [subset] G and x [??] G which implies (logic) implies  (=> or a thin right arrow) A binary Boolean function and logical connective. A => B is true unless A is true and B is false. The truth table is A B  A => B + F F  T F T  T T F  F T T  T It is surprising at first that A => that [[conjunction].sub.k](A) [subset] G and x [??] G. Therefore, x [??] [[conjunction].sub.k] ([[conjunction].sub.k](A)) which implies that [[conjunction].sub.k]([[conjunction].sub.k](A)) [subset] [[conjunction].sub.k](A). This completes the proof. (d) Clearly, by (a), [union]{[[conjunction].sub.k] ([C.sub.[??]]) [??] [member of] [Delta]} [subset] [[conjunction].sub.k]([union]{[C.sub.[??]][??] [member of] [Delta]}). Conversely con·verse^{ 1} intr.v. con·versed, con·vers·ing, con·vers·es 1. To engage in a spoken exchange of thoughts, ideas, or feelings; talk. See Synonyms at speak. 2. , suppose x [??] [union] {[[conjunction].sub.k]([C.sub.[??]])[??] [member of] [Delta]}. Then x [??][[conjunction].sub.k]([C.sub.[??]]) for every [??] [member of] [Delta]. Therefore, for every [??] [member of] [Delta], there exists [G.sub.[??]] [member of] k such that [C.sub.[??]] [subset] [G.sub.[??]] and x [??][G.sub.[??]]. Let G = [union]{[G.sub.[??]][??] [member of] [Delta]}. Then x [??] G and [union]{C [member of] [Delta]} [subset] G which implies that x [??] [[conjunction].sub.k]([union]{[C.sub.[??]][??] [member of] [Delta]}). This completes the proof. (e) The proof follows from (a). (f) The proof follows from the definition of n. (g) Let x [member of] [[conjunction].sub.k](A). If [c.sub.k]({x}) [intersection] A = 0, then X  [c.sub.k]({x}) is a kopen set such that A [subset] X  [c.sub.k]({x}) and x [??] X  [c.sub.k]({x}). Therefore, x [??] [[conjunction].sub.k](A), a contradiction CONTRADICTION. The incompatibility, contrariety, and evident opposition of two ideas, which are the subject of one and the same proposition. 2. In general, when a party accused of a crime contradicts himself, it is presumed he does so because he is guilty for to the assumption and so [c.sub.k]({x}) [intersection] A [not equal to] 0. Hence [[conjunction].sub.k](A) [subset] {x[c.sub.k]({x})[intersection] A [not equal to] 0}. Conversely, suppose for x [member of] X, [c.sub.k]({x}) [intersection] A [not equal to] 0. If x [??][[conjunction].sub.k](A), then there exists a kopen set G such that A [subset] G and x [??] G. Therefore, x [member of] X  G which implies that [c.sub.k]({x}) [subset] [c.sub.k] (X  G) = X  G [subset] X  A and so [c.sub.k]({x}) [intersection] A = 0, a contradiction. Therefore, {x [c.sub.k] ({x}) [intersection] A [not equal to] 0} [subset] [[conjunction].sub.k](A). This completes the proof. (h) Suppose y [member of] [[conjunction].sub.k]({x}). Then y [member of] G whenever G is a kopen set containing x. Suppose x [??] [c.sub.k] ({y}), then there is a kclosed set F such that {y} [subset] F and x [??] F. Since X  F is a kopen set containing x, y [member of] F and so [c.sub.k]({y}) [subset] [c.sub.k](F) = F which implies that [c.sub.k]({y})[intersection]{x} = 0. By (g), y [??][[conjunction].sub.k]({x}), a contradiction. Hence x [member of] [c.sub.k]({y}). Conversely, suppose x [member of] [c.sub.k]({y}). If y [??] [[conjunction].sub.k]({x}), there exists a kopen set G containing x such that y [??] G. Now y [member of] X  G implies that [c.sub.k]({y}) [subset] [c.sub.k](X  G) = X  G [subset] X  {x} and so [c.sub.k]({y}) [intersection] {x} = 0 which implies that x [??] [c.sub.k]({y}), a contradiction to the hypothesis An assumption or theory. During a criminal trial, a hypothesis is a theory set forth by either the prosecution or the defense for the purpose of explaining the facts in evidence. . Therefore, y [member of] [[conjunction].sub.k]({x}). This completes the proof. (i) Suppose [[conjunction].sub.k]({x}) [not equal to] [[conjunction].sub.k]({y}). Assume that z [member of] [[conjunction].sub.k]({x}) and z [??] [[conjunction].sub.k]({y}). Then by (h) and (g), x [member of] [c.sub.k]({z}) and {y} [intersection] [c.sub.k]({z}) = 0 and so [c.sub.k]({x}) [subset] [c.sub.k]({z}) and {y} [intersection] [c.sub.k]({z}) = 0. Therefore, [c.sub.k]({x}) [intersection] {y} = 0 which implies that [c.sub.k]({x}) [c.sub.k]({y}). Conversely, suppose [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Assume that z [member of] [c.sub.k]({x}) and z [??] [c.sub.k]({y}). By (h), x [member of] [[conjunction].sub.k]({z}) and y [??] [[conjunction].sub.k]({z}) and so [[conjunction].sub.k]({x}) [subset] [[conjunction].sub.k]({z}) and {y} [intersection] [[conjunction].sub.k]({z}) = 0 which implies that {y} [intersection] n({x}) = 0. Therefore, [[conjunction].sub.k]({y}) [[conjunction].sub.k]({x}) which completes the proof. Example 3.2. Let X = {a, b} and [gamma]: [??](X) [right arrow] [??](X) be defined by [gamma](0) = 0, [gamma]({a}) = {b}, [gamma]({b}) = {b}, [gamma](X) = X. Then [mu] = {0, {b}, X}. If A = {a}, B = {b}, then [[conjunction].sub.[mu](A) = X, [[conjunction].sub.[mu]](B) = B and [[conjunction].sub.[mu] (A [intersection] B) = [[conjunction].sub.[mu]] (0) = 0. Since [[conjunction].sub.[mu]](A) [intersection] [[conjunction].sub.[mu]](B) = B, [[conjunction].sub.[mu]](A) [intersection] [[conjunction].sub.[mu]](B) (A [intersection] B). The proof of the following Theorem 3.3 is similar to that of Theorem 3.1 and hence the proof is omitted. Example 3.4 shows that the two sets in 3.3(d) are not equal. Theorem 3.5 below gives the relation between the operators [[conjunction].sub.k] and [[disjunction].sub.k]. Theorem 3.3. Let A, B and {[C.sub.[??]][??] [subset] [member of] [Delta]} be subsets of X, [gamma] [member of] [Gamma] and k [member of] [Omega]. Then the following hold. (a) If A [subset] B, then [[disjunction].sub.k](A) [subset] [[disjunction].sub.k](B). (b) [[disjunction].sub.k] (A) [subset] A. (c) [[disjunction].sub.k] ([[disjunction].sub.k](A)) = [[disjunction].sub.k](A). (d) [[disjunction].sub.k] ([union]{C.sub.[??][??] [member of] [Delta]}) [contains] [union] {[[disjunction].sub.k](C.sub.[??])[??] [member of] [Delta]}. (e) [[disjunction].sub.k] ([intersection]{C.sub.[??][??] [member of] [Delta]}) = [intersection] {[[disjunction].sub.k](C.sub.[??])[??] [member of] [Delta]}. (f) If A is kclosed, then [[disjunction].sub.k] (A) = A. Example 3.4. Let X = {a, b, c, d} and [gamma]: [??](X) [right arrow] [??](X) be defined by [gamma](0) [gamma]({a}) = {b}, [gamma]({b}) = {c}, [gamma]({c}) = {b}, [gamma]({d}) = {d}, [gamma]({a, b}) = {b, c}, [gamma]({a, c}) = {b}, [gamma]({a, d}) = {b, d}, [gamma]({b, c}) = {b, c}, [gamma]({b, d}) = {[c, d}, [gamma]({[c, d}) = {b, d}, [gamma]({a, b, c}) = {b, c}, [gamma]({b, c, d}) = {b, c, d}, [gamma]({a, c, d}) = X, [gamma]({a, b, d}) {b, c, d} and y(X) = X. Then [mu] = {0, {d}, {b, c}, {a, c, d}, {b, c, d}, X}. If A = {a}, B = {d}, then [V.sub.[mu]] (A) {a}, [V.sub.[mu]] (B) = 0 and [V.sub.[mu]] (AU B) = [V.sub.[mu]] ({a, d}) {a, d}. Since [V.sub.[mu]] (A) [union] [V.sub.[mu]] (B) = A, [V.sub.[mu]] (A) [union] [V.sub.[mu]] (B) [V.sub.[mu]] (A [union] B). If A = {a, c, d}, then [V.sub.[mu]] (A) = A but A is not [mu]closed. This shows that the reverse direction of Theorem 3.3 (f) is not true. If A = {[c, d}, then [[conjunction].sub.k](A) = A but A is not [mu]open. This shows that the reverse direction of Theorem 3.1(f) is not true. Theorem 3.5. Let A be a subset of X, [gamma] [member of] [Gamma] and [member of] [Omega]. Then the following hold. (a) [[conjunction].sub.k](X  A) = X  [[conjunction].sub.k](A). (b) [[disjunction].sub.k](X  A) = X  [[disjunction].sub.k](A). (c) [[conjunction].sub.k]* = [[disjunction].sub.k]. (d) [[disjunction].sub.k]* = [[conjunction].sub.k]. Proof. (a) and (b) follow from the definitions of [[conjunction].sub.k] and [[disjunction].sub.k]. (c) If A [subset] X, then ([[conjunction].sub.k])* (A) = X  [[conjunction].sub.k] (X A) = X  (X  [[disjunction].sub.k](A)) = [[disjunction].sub.k](A) and so ([[conjunction].sub.k])* = [[disjunction].sub.k]. (d) The proof is similar to the proof of (c). If X is a nonempty set, [gamma] [member of] [Gamma] (X) and k [member of] [Omega], a subset A of X is said to be a [[disjunction].sub.k]set if A = [[disjunction].sub.k](A) and A is said to be a [[conjunction].sub.k]set if A = [[conjunction].sub.k](A). In any space (X, k), the following Theorem 3.6 lists out the [[disjunction].sub.k]sets and the [[conjunction].sub.k]sets. Theorem 3.6. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then the following hold. (a) 0 is a [[conjunction].sub.k]set. (b) A is a [[conjunction].sub.k]set if and only if X  A is a [[disjunction].sub.k]set. (c) X is a [[disjunction].sub.k]set. (d) The union of [[conjunction].sub.k]sets is again a [[conjunction].sub.k]set. (e) The union of [[disjunction].sub.k]sets is again a [[disjunction].sub.k]set. (f) The intersection of [[conjunction].sub.k]sets is again a [[conjunction].sub.k]set. (g) The intersection of [[disjunction].sub.k]sets is again a [[disjunction].sub.k]set. (h) If k [member of] [[Omega].sub.1] = [Omega]  {[mu], [alpha], [pi]}, then X is a [[conjunction].sub.k]set and so 0 is a [[disjunction].sub.k]set. Proof. (a) follows from Theorem 3.1(f) since 0 [member of] k for every [member of] [Omega]. (b) Suppose A is a [[conjunction].sub.k]set. Then A = [[conjunction].sub.k](A). Now X  A = X  [[conjunction].sub.k](A) = [[disjunction].sub.k](X  A), by Theorem 3.5(b). Therefore, X  A is a [[disjunction].sub.k]set. The proof of the converse (logic) converse  The truth of a proposition of the form A => B and its converse B => A are shown in the following truth table: A B  A => B B => A + f f  t t f t  t f t f  f t t t  t t is similar with follows from Theorem 3.5(a). (c) follows from (a) and (b). (d) Let {[A.sub.[??]][??] [member of] [Delta]} be a family of [[conjunction].sub.k]sets. Therefore, [A.sub.[??]] = [[conjunction].sub.k]([A.sub.[??]]) for every [member of] [Delta]. Now [[conjunction].sub.k]([union]{[A.sub.[??]][??] [member of] [Delta]}) = [union] {[[conjunction].sub.k] ([A.sub.[??]])[??] [member of] [Delta]}, by Theorem 3.1(d) and so [[conjunction].sub.k]([union]{[A.sub.[??]][??] [member of] [Delta]}) = [union]{[A.sub.[??]][??] [member of] [Delta]}. (e) Let {[A.sub.[??]][??] [member of] [Delta]} be a family of [[disjunction].sub.k]sets. Therefore, [A.sub.[??]] = [[disjunction].sub.k]([A.sub.[??]]) for every [member of] [Delta]. Now [union]{[A.sub.[??]][??] [member of] [Delta]} = [union] {[[disjunction].sub.k] ([A.sub.[??]])[??] [member of] [Delta]} [subset] [[disjunction].sub.k]([union]{[A.sub.[??]][??] [member of] [Delta]}) by Theorem 3.3(d) and so [[disjunction].sub.k]([union]{[A.sub.[??]][??] [member of] [Delta]}) = ([union]{[A.sub.[??]][??] [member of] [Delta]}) = U{A [member of] O} by Theorem 3.3(b). (f) Let {{[A.sub.[??]][??] [member of] [Delta]} be a family of [[conjunction].sub.k]sets. Therefore, [A.sub.[??]] = [[conjunction].sub.k]([A.sub.[??]]) for every [??] [member of] [Delta]. Now [intersection]{[A.sub.[??]][??] [member of] [Delta]} = [intersection]{[[conjunction].sub.k] [A.sub.[??]][??] [member of] [Delta]} [contains] [[conjunction].sub.k] [intersection]{[A.sub.[??]][??] [member of] [Delta]}) by Theorem 3.1 (e) and so [[conjunction].sub.k] [intersection]{[A.sub.[??]][??] [member of] [Delta]}) = [intersection]{[A.sub.[??]][??] [member of] [Delta]} by Theorem 3.1(b). (g) Let {[A.sub.[??]][??] [member of] [Delta]} be a family of [[disjunction].sub.k]sets. Therefore, [A.sub.[??]] = [[disjunction].sub.k]([A.sub.[??]]) for every [??] [member of] [Delta]. Now [[disjunction].sub.k] [intersection]{[A.sub.[??]][??] [member of] [Delta]}) = [intersection] {[[disjunction].sub.k], ([A.sub.[??]])[??] [member of] [Delta]}, by Theorem 3.3(e) and so [[disjunction].sub.k] [intersection]{[A.sub.[??]][??] [member of] [Delta]}) = [intersection] {[A.sub.[??]][??] [member of] [Delta]}. (h) Since X [member of] [sigma] by Lemma 2.2, X [member of] k for every [member of] [[Omega].sub.1] and so the proof follows from (a) and (b). Remark 3.7. Let [[tau].sub.k] = {A [subset] XA = [[conjunction].sub.k](A)} and [[tau].sup.k] = {A [subset] XA = [[disjunction].sub.k] (A)}. Then [[tau].sub.k] and [[tau].sup.k] are topologies by Theorem 3.6, such that arbitrary intersection of [[tau].sub.k]open sets is a [[tau].sub.k]open set and an arbitrary intersection of [[tau].sup.k]open sets is a [[tau].sup.k]open set. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. A subset A of X is called a generalized [[conjunction].sub.k]set (in short, g. [[conjunction].sub.k]set) if [[conjunction].sub.k](A) [subset] F whenever A [subset] F and F is kclosed. B is called a generalized [[disjunction].sub.k]set (in short, g. [[disjunction].sub.k]set) if X  B is a g. [[conjunction].sub.k]set. We will denote de·note tr.v. de·not·ed, de·not·ing, de·notes 1. To mark; indicate: a frown that denoted increasing impatience. 2. the family of all g. [[conjunction].sub.k]sets by [D.sup.[[conjunction].sub.k]] and the family of all g. [[disjunction].sub.k]sets by [D.sup.[[disjunction].sub.k]]. The following Theorem 3.8 shows that [D.sup.[[conjunction].sub.k]] is closed under arbitrary union and D is closed under arbitrary intersection. Theorem 3.9 below gives a characterization A rather long and fancy word for analyzing a system or process and measuring its "characteristics." For example, a Web characterization would yield the number of current sites on the Web, types of sites, annual growth, etc. of g. [[disjunction].sub.k]sets. Theorem 3.8. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then the following hold. (a) If [B.sub.[??]] [member of] [D.sup.[[conjunction].sub.k]] for every [??] [member of] [Delta], then [union]{[B.sub.[??]][??] [member of] [Delta]} [member of] [D.sup.[[conjunction].sub.k]]. (b) If [B.sub.[??]] [member of] [D.sup.[[disjunction].sub.k]] for every [??] [member of] [Delta], then [intersection]{[B.sub.[??]][??] [member of] [Delta]} [member of] [D.sup.[[disjunction].sub.k]]. Proof. (a) Let [B.sub.[??]] [member of] [D.sup.[[conjunction].sub.k]] for every [??] [member of] [Delta]. Then each [B.sub.[??]] is a g. [[conjunction].sub.k]set. Suppose F is kclosed and [union] {[B.sub.[??]][??] [member of] [Delta]} [subset] F. Then for every [??] [subset] [member of] [Delta], [B.sub.[??]] [subset] F and F is kclosed. By hypothesis, for every [??] [member of] [Delta], [[conjunction].sub.k]([B.sub.[??]]) [subset] F and so [union]{[[conjunction].sub.k] ([B.sub.[??]]) [??] [member of] [Delta]} [subset] F. By Theorem 3.1(d), [[conjunction].sub.k] ([union]{[B.sub.[??]][??] [member of] [Delta]}) [subset] F and so [union]{[B.sub.[??]][??] [member of] [Delta]} [member of] [D.sup.[[conjunction].sub.k]]. (b) Let [B.sub.[??]] [member of] [D.sup.[[disjunction].sub.k]] for every [??] [member of] [Delta]. Then each [B.sub.[??]] is a g. [[disjunction].sub.k]set and so X  [B.sub.[??]] [member of] [D.sup.[[conjunction].sub.k]] for every [??] [member of] [Delta]. Now X  ([intersection]{[B.sub.[??]] [??][member of] [Delta]}) = [union]({X  [B.sub.[??]][??][member of] [Delta]) [member of] [D.sup.[[conjunction].sub.k]], by (a). Therefore, [intersection]{[B.sub.[??]] [??][member of] [Delta]} [member of] [D.sup.[[disjunction].sub.k]]. Theorem 3.9. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then a subset A of X is a g. [[disjunction].sub.k]set if and only if U [subset] [[disjunction].sub.k](A) whenever U [subset] A and U is kopen. Proof. Suppose A is a g. [[disjunction].sub.k]set. Let U be a kopen set such that U [subset] A. Then X  U is a kclosed set such that X  U [contains] X  A and so [[conjunction].sub.k](X  U) [contains] [[conjunction].sub.k](X  A). Therefore, X  U [contains] [[conjunction].sub.k] (X  A) = X  [[disjunction].sub.k](A) and so U [subset] [[disjunction].sub.k](A). Conversely, suppose the condition holds. Let A be a subset of X. Let F be a kclosed subset of X such that X  A [subset] F. Then X  F [subset] A and so by hypothesis, X  F [subset] [[disjunction].sub.k](A). Then X  [[disjunction].sub.k](A) [subset] F and so [[conjunction].sub.k](X  A) [subset] F which implies that X  A is a g. [[conjunction].sub.k]set. Therefore, A is a g. [[disjunction].sub.k]set. The remaining theorems This is a list of theorems, by Wikipedia page. See also
Theorem 3.10. Let x [member of] X, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then the following hold. (a) {x} is either a kopen set or X  {x} is a g. [[conjunction].sub.k]set. (b) {x} is either a kopen set or a g. [[disjunction].sub.k]set. Proof. (a) Suppose {x} is not a kopen set. Then X is the only kclosed set containing X  {x} and so [[conjunction].sub.k](X  {x}) [subset] X. Therefore, X  {x} is a g. [[disjunction].sub.k]set. (b) Suppose {x} is not a kopen set. By (a), X  {x} is a g. [[conjunction].sub.k]set and so {x} is a g. [[disjunction].sub.k]set. Theorem 3.11. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. If B is a g. [[disjunction].sub.k]set and F is a kclosed set such that [[disjunction].sub.k](B) [union] (X  B) [subset] F, then F = X. Proof. Since B is a g. [[disjunction].sub.k]set, X  B is a g. [[conjunction].sub.k]set such that X  B [subset] F. Therefore, [[conjunction].sub.k](X  B) [subset] F which implies that X  F [subset] [[disjunction].sub.k] (B). Also, [[disjunction].sub.k](B) [subset] F and so X  F [subset] X  [[disjunction].sub.k](B). Hence X  F [subset] [[disjunction].sub.k](B) [intersection] (X  [[disjunction].sub.k](B)) . Therefore, F = X. Corollary corollary: see theorem. 3.12. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. If B is a g. [[disjunction].sub.k]set such that [[disjunction].sub.k](B) U (X  B) is kclosed, then B is a [[disjunction].sub.k]set. Proof. By Theorem 3.11, [[disjunction].sub.k](B) U (X  B) = X and so X  ([[disjunction].sub.k](B) [union] (X  B)) which implies that (X  [[disjunction].sub.k](B)) [intersection] B = 0. Therefore, B [subset] [[disjunction].sub.k](B) and so by Theorem 3.3(b), B = [[disjunction].sub.k](B). Theorem 3.13. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. If A and B are subsets of X such that A [subset] B [subset] [[conjunction].sub.k](A) and A is a g. [[conjunction].sub.k]set, then B is a g. [[conjunction].sub.k]set. In particular, if A is a g. [[conjunction].sub.k]set, then [[conjunction].sub.k](A) is a g. [[conjunction].sub.k]set. Proof. Since A [subset] B [subset] [[conjunction].sub.k](A), [[conjunction].sub.k](A) [subset] [[conjunction].sub.k](B) [subset] [[conjunction].sub.k]([[conjunction].sub.k] (A)) = [[conjunction].sub.k](A) and so [[conjunction].sub.k](A) [[conjunction].sub.k](B). If F is any kclosed set such that B [subset] F, then A [subset] F and so [[conjunction].sub.k](B) = [[conjunction].sub.k](A) [subset] F. Therefore, B is a g. [[conjunction].sub.k]set. Corollary 3.14. Let X be a nonempty set, [gamma][member of] [Gamma](X) and [member of] [Omega]. If A and B are subsets of X such that [[disjunction].sub.k](A) [subset] B [subset] A and A is a g. [[disjunction].sub.k]set, then B is a g. [[disjunction].sub.k]set. In particular, if A is a g. [[disjunction].sub.k]set, then [[disjunction].sub.k](A) is a g. [[disjunction].sub.k]set. 4. Some separation axioms in generalized topological spaces Let X be a nonempty set, [gamma] [member of] [Gamma](X) and k [member of] [Omega]. Then (X, k) is called a  [T.sub.1] space if for all x, y [member of] X, x [not equal to] y, there exists a kopen set G such that x [member of] G and y [??] G and there exists a kopen set H such that x [??] H and y [member of] H. [sigma] [T.sub.1] space is defined in [5]. The following Theorem 4.1 gives a characterization of k  [T.sub.1] spaces in terms of kclosed sets and Theorem 4.2 gives a characterization of k  [T.sub.1] spaces in terms of [[conjunction].sub.k]sets. Theorem 4.1. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k  [T.sub.1] space if and only if every singleton set is a kclosed set. Proof. Suppose (X, k) is a k  [T.sub.1] space. Let x [member of] X. If g [member of] X  {x}, then x g. By hypothesis, there exists a kopen set [H.sub.y] such that y [member of] [H.sub.y] and x [??][H.sub.y] and so y [member of] [H.sub.y] [subset] X {x}. Hence X  {x} = [union] {[H.sub.y]y [member of] X  {x}} is kopen and so {x} is closed. Conversely, suppose each singleton set is a kclosed set. Let x, y [member of] X such that x [not equal to] y. Then X  {x} and X  {y} are kopen sets such that y [member of] X  {x} and x [member of] X  {y}. Therefore, (X, k) is a k  [T.sub.1] space. Theorem 4.2. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a  [T.sub.1] space if and only if every subset of X is a [[conjunction].sub.k]set. Proof. Suppose (X, k) is a k  [T.sub.1] space. Let A be subset of X. By Theorem 3.1(b), A [subset] [[conjunction].sub.k](A). Suppose x [??] A. Then X  {x} is a kopen set such that A [subset] X  {x} and so [[conjunction].sub.k](A) [subset] X  {x}. Hence every subset of X is a [[conjunction].sub.k]set. Conversely, suppose every subset of X is a [[conjunction].sub.k]set and so [[conjunction].sub.k]({x}) = {x} for every x [member of] X. Let x, y [member of] X such that x [not equal to] y. Then y [??] [[conjunction].sub.k]({x}) and x [??][[conjunction].sub.k]({y}). Since y [??] [[conjunction].sub.k]({x}), there is a kopen set U such that x [member of] U and y [??] U. Similarly, since x [??] [[conjunction].sub.k]({y}), there is a kopen set V such that y [member of] V and x [??] V. Therefore, (X, k) is a k  [T.sub.1] space. Corollary 4.3. Let X be a nonempty set, [member of] [Gamma](X) and k [member of] [Omega]. Then the following are equivalent. (a) (X, k) is a  [T.sub.1] space. (b) Every subset of X is a [[conjunction].sub.k]set. (c) Every subset of X is a [[disjunction].sub.k]set. Proof. (a) and (b) are equivalent by Theorem 4.2. (b) and (c) are equivalent by Theorem 3.5(b). Theorem 4.4. Let X be a nonempty set, [gamma] [member of] [Gamma](X), [member of] [Omega] and A [subset] X. Then the following hold. (a) If A is a [[conjunction].sub.k]set, then A is a g. [[conjunction].sub.k]set. (b) If A is a [[disjunction].sub.k]set, then A is a g. [[disjunction].sub.k]set. The reverse directions are true if (X, k) is a k  [T.sub.1] space. Proof. (a) Suppose A is a [[conjunction].sub.k]set. Then A = [[conjunction].sub.k](A). If A [subset] F where F is kclosed, then A = [[conjunction].sub.k](A) [subset] F. Therefore, A is a g. [[conjunction].sub.k]set. (b) The proof is similar to the proof of (a). The reverse directions follow from Corollary 4.3. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. (X, k) is said to be a k  [T.sub.0] space if for distinct points x and y in X, there exists a kopen set G containing one but not the other. Clearly, every  [T.sub.1] space is a  [T.sub.0] space. Since every topology is a generalized topology and in a topological space, [T.sub.0] spaces need not be [T.sub.1] spaces,  [T.sub.0] spaces need not be  [T.sub.0] spaces. Theorem 4.5 gives a characterization of  [T.sub.0] spaces. The easy proof of Corollary 4.6 is omitted. Theorem 4.5. Let (X, k) be a kspace where [gamma][member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k  [T.sub.0] space if and only if distinct points of X have distinct kclosures. Proof. Suppose (X, k) is a k  [T.sub.0] space. Let x and y be points of X such that x [not equal to] y. Then there exists a kopen set G containing one but not the other, say x [member of] G and y [??] G. Then y [??] [c.sub.k]({x}) and so [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Conversely, suppose distinct points of X have distinct kclosures. Let x and y be points of X such that x [not equal to] y. Then [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Suppose z [member of] [c.sub.k]({x}) and z [??] [c.sub.k]({y}). If x [subset] [c.sub.k]({y}), then [c.sub.k]({x}) [subset] [c.sub.k]({y}) and so z [member of] [c.sub.k]({y}), a contradiction. Therefore, x [??] [c.sub.k]({y}) which implies that x [member of] X  [c.sub.k]({y}) and X  [c.sub.k]({y}) is kopen. Hence (X, k) is a k  [T.sub.0] space. Corollary 4.6. Let (X, k) be a kspace where [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k  [T.sub.0] space if and only if for distinct points x and y of X, either x [??] [c.sub.k]({y}) or y [??] [c.sub.k]({x}). Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. (X, k) is said to be a k  [T.sub.2] space if for distinct points x and y in X, there exist disjoint dis·joint v. To put out of joint; dislocate. kopen sets G and H such that x [member of] G and y [member of] H. Clearly, every k  [T.sub.2] space is a k  [T.sub.1] space and the converse is not true. Theorem 4.7 below gives characterizations of k  [T.sub.2] spaces. Theorem 4.7. In a kspace (X, k), where [gamma] [member of] [Gamma](X) and [member of] [Omega], the following statements are equivalent. (a) (X, k) is a k  [T.sub.2] space. (b) For each x [member of] X and y [not equal to] x, there exists a kopen set U such that x [member of] U and y [??] [c.sub.k](U). (c) For every x [member of] X, {x} = [intersection]{[c.sub.k](U) x [member of] U and U is k  open}. Proof. (a)[??](b). Let x, y [member of] X such that y [not equal to] x. Then, there exists disjoint kopen sets U and H such that x [member of] U and y [member of] H. Then X  His a kclosed set such that U [subset] X  H and so [c.sub.k](U) [subset] X  H. U is the required kopen set such that x [member of] U and y [??] [c.sub.k](U). (b)[??](c). Let x [member of] X. If y [member of] X such that x [not equal to] y, by (b), there exists a kopen set U such that x [member of] U and y [c.sub.k](U). Clearly, {x} = [intersection]{[c.sub.k](U) x [member of] U and U is k  open}. (c)[??](a). Let x, y [member of] X such that y [not equal to] x. Then y [??] {x} = [intersection]{[c.sub.k](U) x [member of] U and U is kopen}, by (c). Therefore, y [??] [c.sub.k](U) for some kopen set containing x. U and X  [c.sub.k](U) are the required disjoint kopen sets containing x and y respectively. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is said to be a k  [R.sub.0] space if every kopen subset of X contains the kclosure of its singletons. (X, k) is said to be a k  [R.sub.1] space if for x, y [member of] X with [c.sub.k]({x}) [c.sub.k]({y}), there exist disjoint kopen sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H. Clearly, every k  [R.sub.1] space is a k  [R.sub.0] space but the converse is not true. The following Theorem 4.8 follows from Theorem 4.1. Theorem 4.9 below gives a characterization of k  [R.sub.0] spaces. Theorem 4.8. Let X be a nonempty set, [gamma] [member of] [Gamma] (X) and [member of] [Omega]. Then every  [T.sub.1] space is a  [R.sub.0] space. Theorem 4.9. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k  [R.sub.0] space if and only if every kopen subset of X is the union of kclosed sets. Proof. Suppose (X, k) is a k  [R.sub.0] space. If A is kopen, then for each x [member of] A, [c.sub.k]({x}) [subset] A and so [union]{[cl.sub.k],{x} x [member of] A} [subset] A. It follows that A = [union]{[cl.sub.k],{x} x [member of] A}. Conversely, suppose A is kopen and x [member of] A. Then by hypothesis, A = [union]{[B.sub.[??]][??] [subset] [member of] [Delta]} where each [B.sub.[??]] is kclosed. Now x [member of] A, implies that x [member of] [B.sub.[??]] for some [??] [member of] [Delta]. Therefore, [c.sub.k]({x}) [subset] [B.sub.[??]] [subset] A and so (X, k) is a k  [R.sub.0] space. Theorem 4.10. Let X be a nonempty set, [gamma] [member of] [gamma] (X) and [member of] [Omega]. Then (X, k) is a k  [R.sub.0] space if and only if for x, y [member of] X, [c.sub.k]({x}) [c.sub.k]({y}) implies that [c.sub.k]({x}) [intersection] [c.sub.k]({y}) = 0. Proof. Suppose (X, k) is a k  [R.sub.0] space. Let x,y [member of] X, such that [c.sub.k]({x}) [??][c.sub.k]({y}). Suppose z [member of] [c.sub.k]({x}) and z [c.sub.k]({y}). Since z [c.sub.k]({y}), there exists a kopen set G containing z such that y G. Since z [member of] [c.sub.k]({x}), x [member of] G. Since y G, it follows that x [c.sub.k]({y}) and so x [member of] X[c.sub.k]({y}). By hypothesis, [c.sub.k]({x}) [subset] X[c.sub.k]({y}) and so [c.sub.k]({x}) [intersection][c.sub.k]({y}) = 0. Conversely, suppose the condition holds. Let G be a kopen set such that x [member of] G. If y [??] G, then x [not equal to]y and so x [c.sub.k]({y}) which implies that [c.sub.k]({x}) [c.sub.k]({y}). By hypothesis, [c.sub.k]({x}) [intersection] [c.sub.k]({y}) and so y [c.sub.k]({x}). Hence [c.sub.k]({x}) [subset] G which implies that (X, k) is a k  [R.sub.0] space. Theorem 4.11. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then (X, k) is a k  [R.sub.0] space if and only if for x, y [member of] X, [[conjunction].sub.k] ({x}) [not equal to] [[conjunction].sub.k]({y}) implies that [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({y}) = 0. Proof. Suppose (X, k) is a k  [R.sub.0] space. Let x, y [member of] X such that [[conjunction].sub.k]({x}) [not equal to] [[conjunction].sub.k]({y}). Suppose that z [member of] [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({y}). Then z [member of] [[conjunction].sub.k]({x}) and z [member of] [[conjunction].sub.k]({y}). By Theorem 3.1(h), x [member of] [c.sub.k]({z}) and y [member of] [c.sub.k]({z}) and so [c.sub.k]({x}) [intersection][c.sub.k]({z}) [not equal to] 0 and [c.sub.k]({y}) [intersection][c.sub.k]({z}) [not equal to] 0. By Theorem 4.10, [c.sub.k]({x}) = [c.sub.k]({z}) and [c.sub.k]({y}) = [c.sub.k]({z}) and so [c.sub.k]({x}) = [c.sub.k]({y}). By Theorem 3.1(i), [[conjunction].sub.k]({x}) = [[conjunction].sub.k]({y}), a contradiction. Therefore, [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({y}) = 0. Conversely, suppose the condition holds. Let x, y [member of] X such that [c.sub.k]({x}) [not equal to] [c.sub.k]({y}). Suppose that z [member of] [c.sub.k]({x}) [intersection] [c.sub.k]({y}). Then z [member of] [c.sub.k]({x}) and z [member of] [c.sub.k]({y}). By Theorem 3.1(h), x [member of] [[conjunction].sub.k]({z}) and y [member of] [[conjunction].sub.k]({z}) and so [[conjunction].sub.k]({x}) [intersection] [[conjunction].sub.k]({z}) [not equal to] 0 and [[conjunction].sub.k]({y}) [intersection] [[conjunction].sub.k]({z}) [not equal to] 0. By hypothesis, [[conjunction].sub.k]({x}) = [[conjunction].sub.k]({z}) and [[conjunction].sub.k]({x}) = [conjunction].sub.k]({z}) and so [[conjunction].sub.k] ({x}) = [[conjunction].sub.k],({y}) By Theorem 3.1(i), [c.sub.k]({x}) = [c.sub.k]({y}), a contradiction. Therefore, [c.sub.k]({x}) [intersection] [c.sub.k]({y}) = 0. By Theorem 4.10, (X, k) is a k  [R.sub.0] space. Theorem 4.12. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equal. (a) (X, k) is a k  [R.sub.0] space. (b) For any nonempty set A and a kopen set G such that A [intersection] G [not equal to] 0, there exists a kclosed set F such that A [intersection] F [not equal to] 0 and F [subset] G. (c) If G is kopen, then G = [union]{FF [subset] G and F is kclosed}. (d) If F is kclosed, then F = [intersection]{GF [subset] G and G is kopen}. (e) For every x [member of] X, [c.sub.k]({x}) [subset] [[conjunction].sub.k]({x}). Proof. (a)[??](b). Suppose (X, k) is a k  [R.sub.0] space. Let A be a nonempty set and G be a kopen set such that A [intersection] G [not equal to] 0. If x [member of] A [intersection] G, then x [member of] G and so by hypothesis, [c.sub.k]({x}) [subset] G. If F = [c.sub.k]({x}), then F is the required kclosed set such that A [intersection] F [not equal to] 0 and F [subset] G. (b)[??](c). Let G be kopen. Clearly, G [contains][intersection] [union]{FF [subset] G and F is kclosed}. If x [member of] G, then {x} [intersection] G [not equal to] 0 and so by (b), there is a kclosed set F such that {x} [intersection] F [not equal to] 0 and F [subset] G which implies that x [member of] {FF [subset] G and F is kclosed}. Therefore, G [subset] {FF [subset] G and F is kclosed}. This completes the proof. (c)[??](d). Let F be kclosed. By (c), X  F = [union]{KK [subset] X  F and K is kclosed} and so F = [intersection]{X  KF [subset] X  K and X  K is kopen}. (d)[??](e). Let x [member of] X. If y [??] [[conjunction].sub.k]({x}), then by Theorem 3.1(g), {x} [intersection] [c.sub.k]({y}) = 0. By (d), [c.sub.k]({y}) = [intersection]{G [c.sub.k]({y}) [subset] G and G is kopen}. Therefore, there is a kopen G such that [c.sub.k]({y}) [subset] G and x [??] G which implies that y [??] [c.sub.k]({x}). Therefore, [c.sub.k]({x}) [subset] [[conjunction].sub.k]({x}). (e)[??](a). Let G be a kopen set such that x [member of] G. If y [member of] [c.sub.k]({x}), then by (e), y [member of] n({x}). Since [[conjunction].sub.k]({x}) [subset] [[conjunction].sub.k](G) = G, y [member of] G. Hence (X, k) is a k  [R.sub.0] space. Corollary 4.13. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent. (a) (X, k) is a k  [R.sub.0] space. (b) For every x [member of] X, [c.sub.k]({x}) = [[conjunction].sub.k]({x}). Proof. (a)(b). Let x [member of] X. By Theorem 4.12, [c.sub.k]({x}) [subset] [[conjunction].sub.k]({x}). To prove the direction, assume that y [member of] [[conjunction].sub.k]({x}). By Theorem 3.1(h), x [member of] [c.sub.k]({y}) and so [c.sub.k]({x}) [subset] [c.sub.k]({y}) which implies that [c.sub.k]({x}) [intersection] [c.sub.k]({y}) [not equal to] 0. By Theorem 4.10, [c.sub.k]({x}) = [c.sub.k]({y}) and so y [member of] [c.sub.k]({x}). Hence [c.sub.k]({x}) [[conjunction].sub.k]({x}). (b)(a). The proof follows from Theorem 4.12. Theorem 4.14. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent. (a) (X, k) is a k  [R.sub.0] space. (b) For all x, y [member of] X, x [member of] [c.sub.k]({y}) if and only if y [member of] [c.sub.k]({x}). Proof. (a)[??](b). Let x, y [member of] X such that x [member of] [c.sub.k]({y}). By Corollary 4.13, x [member of] [[conjunction].sub.k]({y}) and so by Theorem 3.1(h), y [member of] [c.sub.k]({x}). Thus x [member of] [c.sub.k]({y}) implies that y [member of] [c.sub.k]({x}). Similarly, we can prove that y [member of] [c.sub.k]({x}) implies that x [member of] [c.sub.k]({y}). (b)[??](a). Conversely, suppose the condition holds. Let x [member of] X. If y [member of] [c.sub.k]({x}), then by hypothesis, x [member of] [c.sub.k]({y}) and so by Theorem 3.1(h), y [member of] n({x}) which implies that [c.sub.k]({x}) [subset] n({x}). By Theorem 4.12(e), (X, k) is a k  [R.sub.0] space. The following Theorem 4.15 gives a characterization of k  [R.sub.1] space and Theorem 4.16 gives a characterization of k  [T.sub.2] space. Theorem 4.15. Let X be a nonempty set, [gamma] [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent. (a) (X, k) is a k  [R.sub.1] space. (b) For x, y [member of] X such that [[conjunction].sub.k]({x}) [[conjunction].sub.k]({y}), there exist disjoint kopen sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H. Proof. (a)[??](b). Let x, y [member of] X such that [[conjunction].sub.k]({x}) [[conjunction].sub.k]({y}). Then by Corollary 4.13, since (X, k) is a k  [R.sub.0] space, [c.sub.k]({x}) [c.sub.k]({y}) and so there exists disjoint kopen sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H. (b)[??](a). Let x, y [member of] X such that [c.sub.k]({x}) [c.sub.k]({y}). By Theorem 3.1(i), [[conjunction].sub.k]({x}) [[conjunction].sub.k]({y}). By hypothesis, there exist disjoint kopen sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H and so (X, k) is a k  [R.sub.1] space. Theorem 4.16. Let X be a nonempty set, y [member of] [Gamma](X) and [member of] [Omega]. Then the following are equivalent. (a)(X, k) is a k  [T.sub.2] space. (b) (X, k) is both a  [R.sub.1] space and a k  [T.sub.1] space. (c)(X, k) is both a  [R.sub.1] space and a k  [T.sub.0] space. Proof. (a)[??](b). Suppose (X, k) is a k  [T.sub.2] space. Clearly, (X, k) is a k  [T.sub.1] space and so singletons are closed sets, by Theorem 4.1. If x, y [member of] X with [c.sub.k]({x}) [c.sub.k]({y}), then x [not equal to] y and so there exist disjoint kopen sets G and H such that x [member of] G and y [member of] H. Therefore, [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H which implies that (X, k) is a k  [R.sub.1] space. (b)[??](c). The proof is clear. (c)[??](a). Let x, y [member of] X such that x [not equal to] y. Since (X, r) is a k  [T.sub.0] space, by Theorem 4.5, [c.sub.k]({x}) [c.sub.k]({y}). Since (X, k) is a k  [R.sub.1] space, there exist disjoint open sets G and H such that [c.sub.k]({x}) [subset] G and [c.sub.k]({y}) [subset] H. Therefore, (X, k) is a k  [T.sub.2] space. References [1] A. Csaszar, Generalized Open Sets, Acta acta (ăk`tə), official texts of ancient Rome, written or carved on stone or metal. Usually acta were texts made public, although publication was sometimes restricted. Acta were first posted or carved for general reading c.131 B.C. Math. Hungar., 75(1997), No.l2, 6587. [2] A. Csaszar, On the [gamma]interior and [gamma]closure of a set, Acta Math. Hungar., 80(1998), 8993. [3] A. Csaszar, Generalized topology, generalized continuity, Acta Math. Hungar., 96(2002), 351357. [4] A. Csaszar, Generalized open sets in generalized topologies, Acta Math. Hungar., 106(2005), 5366. [5] A. Guldurdek and O.B. Ozbakir, On [gamma]semiopen sets, Acta Math. Hungar., 109(2005), No.4, 34[gamma]355. [6] P. Sivagami, Remarks on [gamma]interior, Acta Math. Hungar., 119(2008), 8194. [7] P. Sivagami and D. Sivaraj, Properties of some generalized topologies, Intern intern /in·tern/ (in´tern) a medical graduate serving in a hospital preparatory to being licensed to practice medicine. in·tern or in·terne n. . J. General Topology In mathematics, general topology or pointset topology is the branch of topology which studies properties of topological spaces and structures defined on them. It is distinct from other branches of topology in that the topological spaces may be very general, and do not have ., 1(2008), No.2, 119124. [8] P. Sivagami and D. Sivaraj, On [gamma]semiopen sets, Tamkang J. Math., 39(4)(2008), 303308. P. Sivagami ([dagger]) and D. Sivaraj ([double dagger double dagger n. A reference mark () used in printing and writing. Also called diesis. Noun 1. ]) ([dagger]) Department of Mathematics, Kamaraj College, Thoothukudi628 003, Tamil Nadu Tamil Nadu (tăm`əl nä`d), formerly Madras (mədrăs`, mədräs`), state (2001 provisional pop. , India India, officially Republic of India, republic (2005 est pop. 1,080,264,000), 1,261,810 sq mi (3,268,090 sq km), S Asia. The second most populous country in the world, it is also sometimes called Bharat, its ancient name. India's land frontier (c. ([double dagger]) Department of Computer Science, D.J. Academy for Managerial Excellence, Coimbatore641 032, Tamil Nadu, India Email: ttn_sivagamiyahoo.co.in 

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