TEXT CAFEARTICLE DISTRIBUTION AND SYNDICATION

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The concepts of 0-closure, 6-closure, 0-interior and 6-interior operators
were first introduced by Velickho. These operators have since then been studied
intensively by many authors. Although 0-interior and 0-closure operators are
not idempotents, the collection of all 5-open sets in a topological space (X, I)
forms a topology ro on X, called the semiregularization topology of F, weaker
than F and the class of all regular open sets in Tforms an open basis for Fs.
Similarly, the collection of all 0-open sets in a topological space (X,/) forms a
topology To on X, weaker than T. So far, numerous applications of such
operators have been found in studying different types of continuous like maps,
separation of axioms, and above all, to many important types of compact like
properties. In 1961, [6] introduced the concept of weak continuity as a
generalization of continuity, later in 1966, Husain introduced almost continuity
as another generalization, and Andrew and Whittlesy [2], the concept of closure
continuity which is stronger than weak continuity. In 1968, Singal and Singal
introduced a new almost continuity which is different from that of Husain. A few
years later, P. E. Long and Carnahan [8] studied similarities and dissimilarities
between the two concepts of almost continuity. The purpose of this paper is to
further the study of the concepts of closure and strong continuity. We get similar
results to those in [8], [11] applied to closure and strong continuity. Among
other results we prove that the graph mapping of f is closure continuous iff f
is closure continuous. In Theorem 3, we show that if the graph mapping off is
strongly continuous then f is strongly continuous but not conversely. Theorem
12 is a stronger result of Theorem 5 in [11]. Theorem 8 shows that a strong
retraction of a Hausdorff space is 0-closed. Several decomposition theorems of
closure and strong continuity are given in this paper. Example 2 shows that [9,
Corollary to Theorem 6] is not true.
For a set A in a space X, let us denote by Int(A) and cls(A) for the
interior and the closure of A in X, respectively. Following Velickho, a point x of
An-Rajah Univ. J. Res., Vol. 12, (1998).
Mohammad Saleh 9
a space X is called a 0-adherent point of a subset A of X iff cls(U)nA 0, for
every open set U containing x. The set of all 0 -adherent points of A is called
the 0-closure of A, denoted by else A. A subset A of a space X is called 8-
closed iff A = else A. The complement of a 0 -closed set is called 0-open.
Similarly, the 0 -interior of a set A in X, written Int,9 A, consists of those points
x of A such that for some open set U containing x, cls(U)c – A. A set A is
8-open iff A Into A, or equivalently, X-A is 0-closed. Clearly every 0-closed
(8-open) is closed (open). It is well-known that one of the most weaker forms of
compactness is closure compactness (QHC). A closure compact Hausdoiff
space is called H-closed, first defined by Alexandroff and Urysohn.
A function f: X—>Y is weakly continuous at xe X if given any open set
V in Y containing f(x), there exists an open set U in X containing x such that
f(U) c cls( V). If this condition is satisfied at each XE X, then f is said to be
weakly continuous. A function f X—>Y is closure continuous (8 -continuous) at
xe- X if given any open set V c Y containing f(x), there exists an open set U in
X containing x such that f(cls(U)) cls(V). If this condition is satisfied at each
XE X, then f is said to be closure continuous (0-continuous). A function f: X–>Y
is strongly continuous (strongly 0-continuous) at xe X if given any open set V
Y containing f(x), there exists an open set U CX containing x such that
f(cls(U))c V. If this condition is satisfied at each xe X, then f is said to be
strongly continuous (strongly 0-continuous). A function f X—> Y is said to be
almost continuous in the since of Singal and Singal (briefly a. c_ S) if for each
point XE X and each open set V c Y containing f (x), there exists an open set
U X containing x such that f.(U)c Int(cls(V)). A function f X–>Y is said to be
almost continuous in the since of Husain (briefly a.c.H) if for each xEX and each
open set V c Y containing f(x), cls(f AV)) is a neighborhood of x E X. A
space X is called completely Hausdorff or Urysohn if for every x ye X, there
An-Najah Univ. J. Res.. Vol. 12, (1998).
I 0 Some Remarks on Closure and Strong Continuity
exist an open set U containing x and an open set V containing y such that
cls(U)n cls(V)=4
2. The Results
Clearly cls(A) c clse A, but not equal as it is shown in the next example.
Over a regular space, it is clear that cls(A)=c1s e A.
Example 1. Let R be the reals with the cofinite topology. Then every finite
subset of R is closed, but the 0-closure of every none empty set is R.
Theorem 1. Let f:X–> Y.Then the following are equivalent:
a) f(clse A)cclsf(A), for every AcX;
b) The inverse image of every closed is 0-closed;
c) The inverse image of every open is 0-open;
d) f is strongly continuous.
Proof (a) (b). Let B be a closed set and let A=f -1(B). Let xe clse A. Then
f(x) E f(clseA) ccls(f(A))c cls(B)=B. Therefore, x E f -1 (B)=A.
Thus cis() A=A.
(b) (c) . Let V be an open subset of Y and thus YV is closed. Let
A= f (YV). Then f ()AV) = X 1 f -1 (V) is 0 – closed and thus
f -1 (V) is 0-open.
(c). (d). Let x EX and let V be an open set containing f(x). By the
hypothesis, it follows that f -1 (V) is 0 -open and thus there exists
U an open set containing x such that cls(U) c r i(V). Thus
f(cls(U)) c V, proving that f is strongly continuous.
(d) (a). Let f:X—>Y be strongly continuous and let x E cl se A. Let V
be an open set containing f(x). By strong continuity of f there
exists an open set U containing x such that f(cls(U))c V.
Therefore, cls(U) meets A and thus V meets f(A). Hence
f(x)Ecls(f(A)) as we claim.
An-Najah Univ. J. Res., Vol. 12, (1998).
Alohammad Saleh 11
The proofs of the following Lemmas are straightforward from the
definitions.
Lemma 1. Let f:X -*Y be strongly continuous and let g:Y–* Z be continuous.
Then gof is strongly continuous.
Lemma 2. Let f:X—>Y be closure continuous and let g:Y -Z be closure
continuous. Then go f is closure continuous.
Lemma 3. Let f:X —>Y be closure continuous and let g:Y –>Z be strongly
continuous. Then gof is strongly continuous.
Lemma 4. Let X or Y be regular. Then f:X—> Y is continuous iff f is strongly
continuous.
Remark. We conclude from Lemmas 1 & 3 that the composite of two strongly
continuous functions is strongly continuous.
In [11] it is shown that a function f is weakly continuous iff its graph
mapping g is weakly continuous. This is still true for the case of closure
continuity as it is shown in the next Theorem but it is not the case for strong
continuity as it is shown in Example 2.
Theorem 2. Let f:X –>Y be a mapping and let g:X —>XxY be the graph
mapping of f given by g(x)=(x,f(x)) for every point xG X. Then
g:X -*Xx Y is closure continuous iff f:X—> Y is closure
continuous.
Proof. If g is closure continuous. Then it follows from Lemma 2 that f is closure
continuous, since the projection map 7c: X x Y —> Y is continuous and
f=it og. Conversely, assume f is closure continuous and Let x E X and let
W be an open set in X xY containing g(x). Then there exist an open set
A c X and an open set V cY such that g(x) —(x,f(x)) E Ax V c W.
Since f is closure continuous there exists an open set U containing x such
An-Najah Univ. J. Res., Vol. 12, (1998).
1 2 Some Remarks on Closure and Strong Continuity
that f (cls (U)) c cls (V). Let K = tin A.
Then g (cls (K)) c cls (A) x cls (V) = cls (Ax V) c cls (W), proving
that g is closure continuous.
Theorem 3. Let f:X—> Y be a mapping and let g:X—> X xY be the graph
mapping off given by g(x)=(x,f(x)) for every point x EX. If g:X
X xY is strongly continuous then f:X —>Y is strongly continuous.
Moreover, if the graph mapping g off is strongly continuous then
X is regular.
Proof. It follows directly from Lemma 1 that f is strongly continuous, since the
projection map ir:X xY—> Y is continuous and f=7 og. To prove the
regularity of X. Let x EX and let U be an open set containing x. Then
U nY is an open set containing (x,f(x)). The strong continuity of the
graph mapping of f guarantees the existence of an open set W
containing x such that g(cls(W))=cls(W) x f(cls(W)) c U xY. Thus
xEC1S(W) cU, proving that X is regular.
In [9, Corollary to Theorem 6] it is claimed that the converse of
Theorem 3 is also true which is not as it is shown in the next example.
Example 2. Let X=Y= { 1,2,3} with topologies Fx = {4), { 1}, {2}, { 1,2},X1,
Fy = {C{3},Y}; f(x)=3, for all x. Then f is strongly continuous but
the graph mapping g of f, where g(x)=(x,f(x)) is not strongly
continuous at 1 and 2.
If the domain of f is a regular space then the converse of Theorem 3 is
also true .
Theorem 4. Let f: X –>Y be a mapping with X a regular space, and let g:
X —> X x Y be the graph mapping of f given by g(x)=(x, f(x)) for
every point X E X. If f:X-3Y is strongly continuous then g: Xx
Y is strongly continuous.
An-.Vajah Umv. J. Res., I of 12, (1998).
.11oharnmad Saleh 13
Proof Assume f is strongly continuous and Let x EX and let W be an open set
in Xx Y containing g(x). Then there exist an open set A cX and an open
set V c Y such that g(x)=(x, f(x))e A xV cW. Since f is strongly
continuous, there exists an open set U containing x such that
f(cls(U)) cV. By the regularity of X, there exists an open set K
containing x such that cls(K)c U nA. Therefore, g(cls(K)) c A xV cW,
proving that g is strongly continuous.
By a closure retraction we mean a closure continuous function f.X—> A,
where Ac X and flA is the identity function on A. In this case, A is said to be a
closure retraction of X
Theorem 5. Let Ac X and let f:X—> A be a closure retraction of X onto A. If
X is a completely Hausdorff space, then A is a 0 -closed subset of
X.
Proof Suppose not, then there exists a point xe els e A A. Since f is a closure
retraction we have f(x) # x. Since X is completely Hausdorff, there
exist open sets U and V of x and f(x), respectively, such that
cls(U)n cls(V)=4. Now let W be any open set in X containing x. Then
Un W is an open set containing x and hence cls(Un W)n AAti,
since xe cls o A. Therefore, there exists a point ye cls(Un W)n A. Since
yE A, f(y)=yE cis(U) and hence f(y) ,z cls(V). This shows that f(cls(W))
is not contained in cls(V). This contradicts the hypothesis that f is closure
continuous. Thus A is 0-closed as claimed.
Recall that an almost retraction is an almost continuous function X—>
A, where AcX and fl A is the identity function on A. In this case, A is said to be
an almost retraction of X.

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