On z-θ-open Sets and Strongly θ-z-continuous Functions

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1 Int. J. Contemp. Math. Sciences, Vol. 8, 2013, no. 8, HIKARI Ltd, On z-θ-open Sets and Strongly θ-z-continuous Functions Murad Özkoç Muğla Sıtkı Koçman University Faculty of Science Department of Mathematics Muğla/TURKEY Copyright c 2013 Murad Özkoç. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract A new class of generalized open sets, called z-open sets which are weaker than both θ-semiopen sets and θ-preopen sets defined by Altunöz and Aslım [2], is introduced and some properties are obtained by Özkoç [14]. In this paper in order to investigate some different properties we introduce two strong form of z-open sets called z-regular sets and z-θ-open sets. By means of z-θ-open sets we also introduce a new class of functions called strongly θ-z-continuous functions which is a generalization of strongly θ-precontinuous functions. We obtain some characterizations and several properties of such functions. Moreover, we define strongly z-closed graphs, z-regular spaces and z-t 2 spaces and investigate the relationship between their properties and strongly θ-z-continuous functions. Mathematics Subject Classification: Primary 54C08, 54C10 Keywords: z-open sets, z-θ-open sets, z-regular sets, strongly θ-z-continuous functions 1 Introduction The notion of continuity is an important concept in general topology. Some properties of strongly θ-continuous functions defined via θ-open sets are studied by Long and Herrington [9]. Recently, five generalizations of strong θ- continuity are obtained by Jafari and Noiri [7], Noiri [12], Noiri and Popa [13],

2 356 Murad Özkoç Park [16] and Özkoç [15]. The main goal of this paper is to introduce and investigate some fundamental properties of strongly θ-z-continuous functions defined via z-open sets introduced by Özkoç [14] in a topological space. It turns out that strong θ-z-continuity is weaker than strong θ-precontinuity [12]. 2 Preliminaries Throughout the present paper, spaces X and Y always mean topological spaces. Let X be a topological space and A a subset of X. The closure of A and the interior of A are denoted by cl(a) and int(a), respectively. All open neighborhoods of the point x of X is denoted by U(x). A subset A is said to be regular open (resp. regular closed) if A = int(cl(a)) (resp. A = cl(int(a))). The subset A is called θ-open [18] if A = int θ (A), where int θ (A) ={x ( U U(x))(cl(U) A)}. The complement of a θ-open set is called θ-closed. The family of all θ-open (resp. θ-closed) sets in X is denoted by θo(x) (resp. θc(x)). The z-interior [14] of a subset A of X is the union of all z-open sets of X contained in A and is denoted by z-int(a). The z-closure [14] of a subset A of X is the intersection of all z-closed sets of X containing A and is denoted by z-cl(a). A subset A of X is called α-open [11] (resp., semiopen [8], θ-semiopen [2], preopen [10], θ-preopen [2], b-open [4], e-open [6], z-open [14], semipreopen [3] (or β-open [1])) if A int(cl(int(a))) (resp. A cl(int(a)),a cl(int θ (A)),A int(cl(a)),a int(cl θ (A)),A int(cl(a)) cl(int(a)),a int(cl δ (A)) cl(int δ (A)),A int(cl θ (A)) cl(int θ (A)),A cl(int(cl(a)))) and the complement of an α-open (resp. semiopen, θ-semiopen, preopen, θ- preopen, b-open, e-open, z-open, semi-preopen, (or β-open)) set are called α-closed (resp. semiclosed, θ-semiclosed, preclosed, θ-preclosed, b-closed, e- closed, z-closed, semi-preclosed (or β-closed)). The intersection of all semiclosed (resp. preclosed, θ-semiclosed, θ-preclosed, b-closed, e-closed, z-closed) sets of X containing A is called the semi-closure [8] (resp. pre-closure [10], θ-semi-closure [2], θ-pre-closure [2], b-closure [4], e-closure [6], z-closure [14]) of A and is denoted by scl(a) (resp. pcl(a), θ- scl(a), θ-pcl(a), bcl(a), e-cl(a), z-cl(a)). Dually, the semi-interior (resp. preinterior, θ-semi-interior, θ-pre-interior, b-interior, e-interior, z-interior) of A is defined to be the union of all semiopen (resp. preopen, θ-semiopen, θ-preopen, b-open, e-open, z-open) sets contained in A and is denoted by sint(a) (resp. pint(a), θ-sint(a), θ-pint(a), bint(a), e-int(a), z-int(a)). The family of all θ-semiopen (resp. θ-preopen, b-open, e-open, z-open) sets in X is denoted by θso(x) (resp. θpo(x), BO(X), eo(x), zo(x)). Lemma 2.1. ([2], [14]). Let A and B be any subsets of a space X. Then

3 On z-θ-open sets and strongly θ-z-continuous functions 357 the following hold: (a) θ-scl(a) =A int(cl θ (A)), θ-sint(a) =A cl(int θ (A)), (b) θ-pcl(a) =A cl(int θ (A)), θ-pint(a) =A int(cl θ (A)), (c) z-cl(a) =θ-scl(a) θ-pcl(a). 3 z-regular Sets and z-θ-open Sets Definition 3.1. A subset A of a topological space X is z-regular if it is both z-open and z-closed. The family of all z-regular sets (resp. which involve the point x of X) inx will be denoted by zr(x)(resp. zr(x, x)). Theorem 3.2. The following properties hold for a subset A of a topological space X. (a) A zo(x) if and only if z-cl(a) zr(x). (b) A zc(x) if and only if z-int(a) zr(x). Proof. We will prove only the first statement. The second one can be proved similarly. Necessity. Let A zo(x). Then we have A int(cl θ (A)) cl(int θ (A)) and hence by Lemma 2.1, z-cl(a) z-cl(int(cl θ (A)) cl(int θ (A))) =θ-scl[int(cl θ (A)) cl(int θ (A))] θ-pcl[int(cl θ (A)) cl(int θ (A))] =[{int(cl θ (A)) cl(int θ (A))} int(cl θ {int(cl θ (A)) cl(int θ (A))})] [{int(cl θ (A)) cl(int θ (A))} cl(int θ {int(cl θ (A)) cl(int θ (A))})] [{int(cl θ (A)) cl(int θ (A))} int{cl θ (int(cl θ (A))) cl θ (cl(int θ (A)))}] [{int(cl θ (A)) cl(int θ (A))} cl{int(cl θ (A)) cl(int θ (A))}] [{int(cl θ (A)) cl(int θ (A))} {int(cl θ (A)) cl θ (cl(int θ (A)))}] [{int(cl θ (A)) cl(int θ (A))} {cl(int(cl θ (A))) cl(cl(int θ (A)))}] [{int(cl θ (A)) cl(int θ (A))} (int(cl θ (A)) cl(int θ (A)))] [cl(int(cl θ (A))) cl(int θ (A))] =int(cl θ (A)) cl(int θ (A)) Since A z-cl(a), we have z-cl(a) int(cl θ [z-cl(a)]) cl(int θ [z-cl(a)]). This shows that z-cl(a) is a z-open set. On the other hand, z-cl(a) is always a z-closed set. Therefore z-cl(a) is a z-regular set. Sufficiency. Let z-cl(a) zr(x). Then by Theorem 4.10 in [14] we have A z-cl(a) int(cl θ (z-cl(a))) cl(int θ (z-cl(a))) = int(cl θ (A)) cl(int(cl(int θ (A)))) = int(cl θ (A)) cl(int θ (A)). Hence we have A zo(x).

4 358 Murad Özkoç Theorem 3.3. For a subset A of a topological space X, the following are equivalent: (a) A zr(x), (b) A = z-cl(z-int(a)), (c) A = z-int(z-cl(a)). Proof. (a) (b) : Let A zr(x). Then A = z-int(a) =z-cl(a) and hence A = z-cl(z-int(a)). (b) (c) : Let A = z-cl(z-int(a)). Then z-cl(a) =z-cl(z-cl(z-int(a))) z-cl(a) =z-cl(z-int(a)) z-int(z-cl(a)) = z-int(z-cl(z-int(a))) z-int(z-cl(a)) = z-int(a)...(1) z-cl(a) zc(x) z-int(z-cl(a)) zr(x) z-cl(z-int(z-cl(a))) = z-int(z-cl(a)) z-cl(a) =z-int(z-cl(a)...(2) (1), (2) z-int(a) = z-cl(a) = z-int(z-cl(a)) A = z-int(z-cl(a). (c) (a) : Let A = z-cl(z-int(a)). Since z-int(a) is z-open, by Theorem 3.1(a) we have z-cl(z-int(a)) zr(x). Hence from hypothesis A zr(x). Definition 3.4. A point x of X is called a z-θ-cluster point of A if z- cl(u) A for every U zo(x, x). The set of all z-θ-cluster points of A is called z-θ-closure of A and is denoted by z-cl θ (A). A subset A is said to be z-θ-closed if A = z-cl θ (A). The complement of a z-θ-closed set is said to be z-θ-open. The family of all z-θ-open (resp. z-θ-closed) sets in X will be denoted by zθo(x) (resp. zθc(x)). Theorem 3.5. For a subset A of a space X, the following properties hold: (a) If A zo(x), then z-cl(a) =z-cl θ (A), (b) A zr(x) if and only if A is both z-θ-open and z-θ-closed. Proof. (a) Generally we have z-cl(a) z-cl θ (A) for every subset A of X. Let A zo(x) and suppose that x/ z-cl(a). Then there exists U zo(x, x) such that U A =. Since A zo(x), we have z-cl(u) A =. This shows that x/ z-cl θ (A). Hence we obtain z-cl(a) =z-cl θ (A). (b) Let A zr(x). Then A zo(x) and by (a), A = z-cl(a) =z-cl θ (A). Therefore, A is z-θ-closed. Since X\A zr(x), by the argument above, X\A is z-θ-closed and hence A is z-θ-open. The converse is obvious. Remark 3.6. It can be easily shown that z-regular z-θ-open z-open. But the converses are not necessarily true as shown by the following examples.

5 On z-θ-open sets and strongly θ-z-continuous functions 359 Example 3.7. Let τ be the usual topology for R and R\N R, where N denotes the set of natural numbers. Then the subset R\N of R is z-θ-open in R but not z-regular. Theorem 3.8. Let X be a topological space and A X. A is z-θ-open in X if and only if for each x A there exists U zo(x, x) such that z-cl(u) A. Proof. Necessity. Let A zθo(x) and x A. Then X\A zθc(x) and X\A = z-cl θ (X\A). Hence x / z-cl θ (X\A). Therefore there exists U zo(x, x) such that z-cl(u) (X\A) = and so z-cl(u) A. Sufficiency. Let A X and x A. From hypothesis there exists U zo(x, x) such that z-cl(u) A. Therefore z-cl(u) (X\A) =. Hence X\A = z-cl θ (X\A) and A zθo(x). Theorem 3.9. For any subset A of a space X, we have z-cl θ (A) = {V (A V )(V zθc(x))} = {V (A V )(V zr(x))}. Proof. We prove only the first equality since the other is similarly proved. First, suppose that x/ z-cl θ (A). Then there exists V zo(x, x) such that z-cl(v ) A =. By Theorem 3.1, X\z-cl(V )isz-regular and hence X\zcl(V ) isaz-θ-closed set containing A and x/ X\z-cl(V ). Therefore, we have x/ {V (A V )(V zθc(x))}. Conversely, suppose that x/ {V (A V )(V zθc(x))}. There exists a z-θ-closed set V such that A V and x / V. There exists U zo(x) such that x U z-cl(u) X\V. Therefore, we have z-cl(u) A z-cl(u) V =. This shows that x/ z- cl θ (A). Theorem Let A and B be any subsets of a space X. Then the following properties hold: (a) x z-cl θ (A) if and only if U A for each U zr(x, x), (b) If A B, then z-cl θ (A) z-cl θ (B), (c) z-cl θ (z-cl θ (A)) = z-cl θ (A), (d) If A α is z-θ-closed in X for each α Λ, then α Λ A α is z-θ-closed in X. Proof. Clear Corollary Let A and A α (α Λ) be any subsets of a space X. Then the following properties hold: (a) A is z-θ-open in X if and only if for each x A there exists U zr(x, x) such that x U A, (b) z-cl θ (A) is z-θ-closed, (c) If A α is z-θ-open in X for each α Λ, then α Λ A α is z-θ-open in X.

6 360 Murad Özkoç 4 Strongly θ-z-continuous Functions and Some Properties In this section, we introduce a new type of continuous functions and look into some relations with some other types. Definition 4.1. A function f : X Y is said to be strongly θ-z-continuous (briefly, st.θ.z.c.) if for each x X and each open set V of Y containing f(x), there exists a z-open set U of X containing x such that f[z-cl(u)] V. Definition 4.2. A function f : X Y is said to be strongly θ-continuous (briefly, st.θ.c.) [9] (resp. strongly θ-semicontinuous (briefly, st.θ.s.c.) [7], strongly θ-precontinuous (briefly, st.θ.p.c.) [12], strongly θ-b-continuous (briefly, st.θ.b.c.) [16], strongly θ-β-continuous (briefly, st.θ.β.c.) [13], strongly θ-e-continuous (briefly, st.θ.e.c.) [15]) if for each x X and each open set V of Y containing f(x), there exists an open (resp. semi-open, preopen, b-open, semi-preopen, e-open) set U of X containing x such that f[cl(u)] V (resp.f[scl(u)] V,f[pcl(U)] V,f[bcl(U)] V,f[spcl(U)] V,f[e-cl(U)] V ). Remark 4.3. From Definitions 4.1 and 4.2, we have the following diagram: st.θ.p.c st.θ.e.c st.θ.c st.θ.z.c st.θ.b.c st.θ.β.c st.θ.s.c However, none of these implications is reversible as shown by the following examples. Example 4.4. Let X = {a, b, c, d, e} and τ = {,X,{a}, {c}, {a, c}, {c, d}, {a, c, d}} and σ = {,X,{d}}. Then the identity function f :(X, τ) (X, σ) is st.θ.z.c. but neither st.θ.b.c. nor st.θ.β.c. Example 4.5. Let X = {a, b, c, d, e} and τ = {,X,{a}, {c}, {a, c}, {c, d}, {a, c, d}} and σ = {,X,{b, c, d}}. Then the identity function f :(X, τ) (X,σ) is st.θ.z.c. but not st.θ.e.c. Example 4.6. Let τ be the usual topology for R and σ = {[0, 1] ((1, 2) Q),, R}, where Q denotes the set of rational numbers. Then the identity function f :(R,τ) (R,σ) is st.θ.z.c. but neither st.θ.p.c. nor st.θ.s.c. Example 4.7. Let τ be the usual topology for R and σ = {, R, [0, 1) Q}. Then the identity function f :(R,τ) (R,σ) is st.θ.β.c. but not st.θ.z.c.

7 On z-θ-open sets and strongly θ-z-continuous functions 361 Theorem 4.8. For a function f : X Y, the following are equivalent: (a) f is strongly θ-z-continuous, (b) For each x X and each open set V of Y containing f(x), there exists U zr(x, x) such that f[u] V, (c) f 1 [V ] is z-θ-open in X for each open set V of Y, (d) f 1 [F ] is z-θ-closed in X for each closed set F of Y, (e) f[z-cl θ (A)] cl(f[a]) for each subset A of X, (f) z-cl θ (f 1 [B]) f 1 [cl(b)] for each subset B of Y. Proof. (a) (b): It follows from Theorem 3.1(a). (b) (c): Let V be any open set of Y and x f 1 [V ]. There exists U zr(x, x) such that f[u] V. Therefore, we have x U f 1 [V ]. Hence by Corollary 3.1(a), f 1 [V ]isz-θ-open in X. (c) (d): This is obvious. (d) (e): Let A be any subset of X. Since cl(f[a]) is closed in Y, by (c) f 1 [cl(f[a])] is z-θ-closed and we have z-cl θ (A) z-cl θ (f 1 [f[a]]) z-cl θ (f 1 [cl(f[a])]) = f 1 [cl(f[a])]. Therefore, we obtain f[z-cl θ (A)] cl(f[a]). (e) (f): Let B be any subset of Y. By (e), we obtain f[z-cl θ (f 1 [B])] cl(f[f 1 [B]]) cl(b) and hence z-cl θ (f 1 [B]) f 1 [cl(b)]. (f) (a): Let x X and V be any open set of Y containing f(x). Since Y \V is closed in Y, we have z-cl θ (f 1 [Y \V ]) f 1 [cl(y \V )] = f 1 [Y \V ]. Therefore, f 1 [Y \V ]isz-θ-closed in X and f 1 [V ]isaz-θ-open set containing x. There exists U zo(x, x) such that z-cl(u) f 1 [V ] and hence f[zcl(u)] V. This shows that f is st.θ.z.c. Theorem 4.9. Let Y be a regular space. Then f : X Y is st.θ.z.c. if and only if f is z-continuous. Proof. Let x X and V an open set of Y containing f(x). Since Y is regular, there exists an open set W such that f(x) W cl(w ) V. If f is z- continuous, there exists U zo(x, x) such that f[u] W. We shall show that f[z-cl(u)] cl(w ). Suppose that y/ cl(w ). There exists an open set G containing y such that G W =. Since f is z-continuous, f 1 [G] zo(x) and f 1 [G] U = and hence f 1 [G] z-cl(u) =. Therefore, we obtain G f[z-cl(u)] = and y / f[z-cl(u)]. Consequently, we have f[z-cl(u)] cl(w ) V. The converse is obvious. Definition AspaceX is said to be z-regular if for each closed set F and each point x X\F, there exist disjoint z-open sets U and V such that x U and F V.

8 362 Murad Özkoç Lemma For a space X the following are equivalent: (a) X is z-regular, (b) For each point x X and for each open set U of X containing x, there exists V zo(x) such that x V z-cl(v ) U. Theorem A continuous function f : X Y is st.θ.z.c. if and only if X is z-regular. Proof. Necessity. Let f : X X be the identity function. Then f is continuous and st.θ.z.c. by our hypothesis. For any open set U of X and any point x U, we have f(x) =x U and there exists V zo(x, x) such that f[z-cl(v )] U. Therefore, we have x V z-cl(v ) U. It follows from Lemma 4.8 that X is z-regular. Sufficiency. Suppose that f : X Y is continuous and X is z-regular. For any x X and open set V containing f(x), f 1 [V ] is an open set containing x. Since X is z-regular, there exists U zo(x) such that x U z- cl(u) f 1 [V ]. Therefore, we have f[z-cl(u)] V. This shows that f is st.θ.z.c. Theorem Let f : X Y be a function and g : X X Y be the graph function of f. Ifg is st.θ.z.c., then f is st.θ.z.c. and X is z-regular. Proof. First, we show that f is st.θ.z.c. Let x X and V an open set of Y containing f(x). Then X V is an open set of X Y containing g(x). Since g is st.θ.z.c., there exists U zo(x, x) such that g[z-cl(u)] X V. Therefore, we obtain f[z-cl(u)] V. Next, we show that X is z-regular. Let U be any open set of X and x U. Since g(x) U Y and U Y is open in X Y, there exists G zo(x, x) such that g[z-cl(g)] U Y. Therefore, we obtain x G z-cl(g) U and hence X is z-regular. Theorem Let f : X Y and g : Y Z be functions. If f is st.θ.z.c. and g is continuous, then the composition g f : X Z is st.θ.z.c. Proof. It is clear from Theorem 4.6. We recall that a space X is said to be submaximal [17] if each dense subset of X is open in X. It is shown in [17] that a space X is submaximal if and only if every preopen set of X is open. A space X is said to be extremally disconnected [5] if the closure of each open set of X is open. A space X is said to be regular [5] if for each closed set F and each point x X\F, there exist disjoint open sets U and V such that x U and F V. Note that an extremally disconnected space is exactly the space where every semiopen set is α-open.

9 On z-θ-open sets and strongly θ-z-continuous functions 363 Theorem Let X be a regular submaximal extremally disconnected space. Then the following properties are equivalent for a function f : X Y : (a) f is st.θ.c. (b) f is st.θ.s.c. (c) f is st.θ.p.c. (d) f is st.θ.b.c. (e) f is st.θ.e.c. (f) f is st.θ.z.c. (g) f is st.θ.β.c. Proof. It follows from the fact that if X is regular submaximal extremally disconnected, then open set, preopen set, semiopen set, b-open set, e-open set, z-open set and β-open set are equivalent. 5 Separation Axioms Definition 5.1. AspaceX is said to be z-t 2 if for each pair of distinct points x and y in X, there exist U zo(x, x) and V zo(x, y) such that U V =. Lemma 5.2. AspaceX is z-t 2 if and only if for each pair of distinct points x and y in X, there exist U zo(x, x) and V zo(x, y) such that z-cl(u) z-cl(v ) =. Theorem 5.3. If f : X Y is a st.θ.z.c. injection and Y is T 0, then X is z-t 2. Proof. For any distinct points x and y of X, by hypothesis f(x) f(y) and there exists either an open set V containing f(x) not containing f(y)oranopen set W containing f(y) not containing f(x). If the first case holds, then there exists U zo(x, x) such that f[z-cl(u)] V. Thus, we obtain f(y) / f[zcl(u)] and hence X\z-cl(U) zo(x, y). If the second case holds, then we obtain a similar result. Thus, X is z-t 2. Theorem 5.4. If f : X Y is a st.θ.z.c. function and Y is Hausdorff, then the subset A = {(x, y) f(x) =f(y)} is z-θ-closed in X X. Proof. It is clear that f(x) f(y) for each (x, y) / A. Since Y is Hausdorff, there exist open sets V and W of Y containing f(x) and f(y), respectively, such that V W =. Since f is st.θ.z.c., there exist U zo(x, x) and G zo(x, y) such that f[z-cl(u)] V and f[z-cl(g)] W. Put D = f[zcl(u)] f[z-cl(g)]. It follows that (x, y) D zr(x X) and D A =. This means that z-cl θ (A) A and thus, A is z-θ-closed in X X. We recall that for a function f : X Y, the subset {(x, f(x)) x X} of X Y is called the graph of f and is denoted by G(f). Definition 5.5. The graph G(f) of a function f : X Y is said to be strongly z-closed if for each (x, y) (X Y )\G(f), there exist U zo(x, x) and an open set V in Y containing y such that (z-cl(u) V ) G(f) =.

10 364 Murad Özkoç Lemma 5.6. The graph G(f) of a function f : X Y is strongly z-closed if and only if for each (x, y) (X Y )\G(f), there exist U zo(x, x) and an open set V in Y containing y such that f[z-cl(u)] V =. Theorem 5.7. If f : X Y is st.θ.z.c. and Y is Hausdorff, then G(f) is strongly z-closed in X Y. Proof. It is clear that f(x) y for each (x, y) (X Y )\G(f). Since Y is Hausdorff, there exist open sets V and W in Y containing f(x) and y, respectively, such that V W =. Since f is st.θ.z.c., there exists U zo(x, x) such that f[z-cl(u)] V. Thus, f[z-cl(u)] W = and then by Lemma 5.4, G(f) is strongly z-closed in X Y. 6 Covering Properties Definition 6.1. AspaceX is said to be (a) z-closed if for every z-open cover {V α α A} of X, there exists a finite subset A 0 of A such that X = {z-cl(v α ) α A 0 }; (b) countably z-closed if for every countable z-open cover {V α α A} of X, there exists a finite subset A 0 of A such that X = {z-cl(v α ) α A 0 }. A subset A of a space X is said to be z-closed relative to X if for every cover {V α : α Λ} of A by z-open sets of X, there exists a finite subset Λ 0 of Λ such that A {z-cl(v α ):α Λ 0 }. Theorem 6.2. If f : X Y is a st.θ.z.c. relative to X, then f[a] is a compact set of Y. function and A is z-closed Proof. Let {V α : α Λ} be a cover of f[a] by open sets of Y. For each point x A, there exists α(x) Λ such that f(x) V α(x). Since f is st.θ.z.c., there exists U x zo(x, x) such that f[z-cl(u x )] V α(x). The family {U x : x A} is a cover of A by z-open sets of X and hence there exists a finite subset A 0 of A such that A x A0 z-cl(u x ). Therefore, we obtain f[a] x A0 V α(x). This shows that f[a] is compact. Corollary 6.3. Let f : X Y be a st.θ.z.c. surjection. Then the following properties hold: (a) If X is z-closed, then Y is compact. (b) If X is countably z-closed, then Y is countably compact. Theorem 6.4. If a function f : X Y has a strongly z-closed graph, then f[a] is closed in Y for each subset A which is z-closed relative to X.

11 On z-θ-open sets and strongly θ-z-continuous functions 365 Proof. Let A be z-closed relative to X and y Y \f[a]. Then for each x A we have (x, y) / G(f) and by Lemma 5.4 there exist U x zo(x, x) and an open set V x of Y containing y such that f[z-cl(u x )] V x =. The family {U x : x A} is a cover of A by z-open sets of X. Since A is z-closed relative to X, there exists a finite subset A 0 of A such that A {z-cl(u x ):x A 0 }. Put V = {V x : x A 0 }. Then V is an open set containing y and f[a] V [ x A0 f[z-cl(u x )]] V x A0 [f[z-cl(u x )] V x ]=. Therefore, we have y/ cl(f[a]) and hence f[a] is closed in Y. Theorem 6.5. Let X be a submaximal extremally disconnected space. If a function f : X Y has a strongly z-closed graph, then f 1 [A] is θ-closed in X for each compact set A of Y. Proof. Let A be a compact set of Y and x / f 1 [A]. Then for each y A we have (x, y) / G(f) and by Lemma 5.4 there exists U y zo(x, x) and an open set V y of Y containing y such that f[z-cl(u y )] V y =. The family {V y : y A} is an open cover of A and there exists a finite subset A 0 of A such that A y A0 V y. Since X is submaximal extremally disconnected, each U y is open in X and z-cl(u y )=cl(u y ). Set U = y A0 U y, then U is an open set containing x and f[cl(u)] A x A0 [f[cl(u)] V y ] x A0 [f[z-cl(u y )] V y ]=. Therefore, we have cl(u) f 1 [A] = and hence x/ cl θ (f 1 [A]). This shows that f 1 [A] isθ-closed in X. Corollary 6.6. Let X be a submaximal extremally disconnected space and Y be a compact Hausdorff space. For a function f : X Y, the following properties are equivalent: (a) f is st.θ.z.c., (b) G(f) is strongly z-closed in X Y, (c) f is strongly θ-continuous, (d) f is continuous, (e) f is z-continuous. Proof. (a) (b). It follows from Theorem 5.5. (b) (c). It follows from Theorem 6.4. (c) (d) (e). These are obvious. (e) (a). Since Y is regular, it follows from Theorem 4.7. This study is dedicated to Professor Zekeriya his 65th birthday. GÜNEY on the occasion of

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