413 Internat. J. Math. & Math. Sci. VOL. 12 NO. 2 (1989) 413-415 A NOTE ON WEAKLY QUASI CONTINUOUS FUNCTIONS TAKASHI NOIRI Department of Mathematics Yatsushiro College of Technology 866 Japan Yatsushiro, Kumamoto, (Received March i0, 1987 and in revised form September 15,1987) -1 Y. f X F of is weakly quasi V f (CI(V))C Cl(Int(f (CI(V)))) for every open set By utilizing this result, the present author [2] showed that a function continuous if and only if of Y X f -1 In [i], it was shown that a function ABSTRACT. is weakly quasi continuous if and only if for every regular closed set -i Y, f (F) is semi-open in X. In this note, the author shows that these Y results are false and corrects the proofs of Theorem 6.1.7 and Lemma 6.4.4 of [2]. KEY WORDS AND PHRASES. weakly quasi continuous. 1980 AMS SUBJECT CLASSIFICATION CODE. 54CI0. The purpose of this note is to point out that Theorem 2 of [i] and Theorem 4.2 of [2] are false, and to correct the proofs of Theorem 6.1.7 and Lemma 6.4.4 of [2]. Let S be a subset of a topological space. Int(S), respectively. denoted by CI(S) [3] (resp. regular-closed) if f Y X A function of Y (resp. Y X f [3] (resp. (resp. f(V)) is said to be semi-open Cl(Int(S))). S A function [4]) if for every semi-open is semi-open in X (resp. Y). [5] if for each x e X and -I f(x), Cl(f (V)) is a neighborhood of x. f Y is said to be weakly quasi continuous [i] X is said to be almost-continuous each open neighborhood DEFINITION. f-l(v) X), S A subset S C Cl(Int(S)) (resp. is said to be semi-continuous V open set and The closure and the interior of S are V of A function x e X, each open set U containing f(x), there exists an open set G of (briefly w.q.c.) if for each X x and each open such that # G CU set V and f(G) C CI(V). Popa and Stan [i] obtained the following characterization of w.q.c, functions. Y is w.q.c, if and only if X THEOREM A ([i, Theorem 2]). A function f containing f-I(cI(V))C Cl(Int(f-l(cl(V)))) V for every open set of Y. In [2], among others, the author established the following three statements. Y is w.q.c, if and only if X THEOREM B ([2, Theorem 4.2]). A function f for every regular closed set F of THEOREM C ([2, Theorem 6.1.7]). function w.q.c. f X Y -i Y, f (F) is semi-open in X go f The composition and a semi-continuous function g Y Z X. Z of a continuous is not necessarily 414 T. NOIRI LEMMA D ([2, Lemma 6.4.4]). and Y g Z If a function. Let Y X f Z X go f be an open continuous surjection is w.q.c., then B was utilized in the proofs of Theorem C and Lemma D. is w.q.c. g The author utilized Theorem A in order to prove Theorem B. Moreover, Theorem it follows from However, Example 2 (below) that the necessity of Theorem A is false and hence so is Theorem B. Thus, it is necessary to revise the proofs of Theorem C and Lemma D. For this purpose, we have the following modification of Theorem A. V of x e f U set U C -1 (V). X of -i For each open set V # U such that (CI(V)))) G Cl(Int(f-l(cl(V)))) Y and f(U) CI(V). # U x, there exists an open Therefore, it follows that Since X, G f(x). containing G X any open set of CI(V). f(U) and . Int(f-l(cl(V))), # Put G x e containing x -i x e f (V) C By hypothesis, we have G(Int(f-l(cl(V))) and hence U, then we obtain containing Y be any open set of V X be any point of x any open set of Let is w.q.c. of # U C G -i -i f (V) C Cl(Int(f (CI(V)))). and hence Let G Int(f-l(cl(V))). U C and Sufficiency. and is w.q.c, if and only if for every open set Y Necessity. f-I(cI(V)) Cl(Int(f X f -1 (V) C Cl(Int(f (CI(V)))). Suppose that f Y, f PROOF. and A function THEOREM i. -1 Int(f-l(cl(V))) f This shows that is w.q.c. The following example =iows that the necessities of Theorems A and B are both false. {, X, {a}, {b, c}} and o {, X, {a}, {a, b, c}, EXAMPLE 2. Let X {b}, {a, b}}. Let f (X, o) be the identity function. Then f is w. (X, ) {a} {a, c} is a regular closed set q.c. by Theorem i. Let V o, then CI(V) {a}. Thus, c} and of (X, o). However, {a, Cl(Int(f-l(cl(V)))) f-I(cI(V)) Cl(Int(f-l(cl(V)))). f-I(cI(V)) f-I(cI(V)) (X, ) and (I) It follows immediately from Theorem 1 that the sufficiencies of is not semi-open in REMARK 3. Theorems A and B are both true. (2) In the proof of [2, Theorem 6.1.7], the set {b, c, d} and Cl(Int((g (Z, ), (g g f by Theorem 1 {b, c, d} V f)-I(cI(V)))) f)-l(v) is clopen in {b}. Therefore, is not w.q.c, and hence it is not semi-continuous. Next, we give the correct proof of Lemma D in the improved form. Let THEOREM 4. g Y Z Let W g f If a function. PROOF. be a semi-open almost continuous surjection and Y X f X be any open set of -i -i Z Since is w.q.c. g is w.q.c., then Z. is w.q.c., by Theorem 1 g f (W) Cl(Int((g f) (CI(W)))). Since f is almost continuous, for every subset A of X, f(Cl(Int(A)))Cl(f(Int(A))) [6, Theorem 6]. Moreover, since f is semi-open, f(Int(A)) .Cl(Int(f(A))) [4, Theorem 9] and hence (g f) we have f(Cl(Int(A))) Cl(Int(f(A))) -i -i g (W) Cl(Int(g (CI(W)))). THEOREM 5. g Y Z subset (g B f)-l(w) f Y X A g Necessity. -i of W Let of X. Therefore, we obtain It follows from Theorem 1 that g is w.q.c. A function be an open continuous surjection. i .q.c. if and only if the composition PROOF. Theorem 1 Let for every subset be any open set of g f Z. X Since Z is w.q.c. g is w.q.c., by -i (W) Cl(Int(g (CI(W)))). Since f is open continuous, for every -i -i Cl(Int(f (B))). Therefore, we obtain Y f (Cl(Int(B))) Cl(Int((g Sufficiency. f)-I(cI(W)))) and hence by Theorem 1 g f is w.q.c. 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