CONTENTS

J. Duhok Univ., Vol.14, No.1 (Pure and Eng. Sciences), Pp 9-16, 2011
Theorem 3.2. Let f :X �Y be any function. If
f is a preirresolute function, then it is cpcontinuous
function.
Proof. For each x�X and each cp-open set V of
Y containing f (x). Then V is preopen set
containing f (x). Since f is preirresolute
function, there exists a preopen set U of X
containing x such that f (U) � V. Hence f is
cp-continuous.
The converse of Theorem 3.2 is not true
as it is shown in the following example:
Example 3.3. Let f from R with usual
topology onto R with co-finite topology be the
identity function. Then f is cp-continuous but
it is not preirresolute because (0, 1] is preopen
set in R with co-finite topology and inverse
image (0, 1] is (0, 1] which is not preopen set in
R with usual topology.
Theorem 3.4.If the co-domain of cp-continuous
function f :X �Y is strongly compact, then it is
preirresolute.
Proof. For each x�X and each preopen set V of
Y containing f (x), so Y\V is preclosed and Y\V
is a subset of strongly compact Y, then by
Lemma 1.10, Y\V is strongly compact. Since f
is cp-continuous there exists a preopen set U of
X containing x such that f (U) � V it follows
that f is preirresolute.
Theorem 3.5. For a function f : X �Y, the
following statements are equivalent:
1. f is cp-continuous,
2. f
cp-open set V in Y,
3. f
� 1
(V) is preopen set in X, for each
� 1
(E) is preclosed set in X, for each
cp-closed set E in Y,
4. f (pcl(A)) � cp-cl( f (A)), for each
subset A of X,
� 1
(B)) � f
� 1
(cp-cl(B)), for
5. pcl( f
each subset B of Y,
6. f
� 1
(cp-int(B)) � pint( f
� 1
(B)), for
each subset B of Y,
7. cp-int( f (A)) � f (pint(A)) , for each
subset A of X.
Proof. (1) �(2). Let V be any cp-open set in Y.
We have to show that f
X. Let x� f
� 1
(V) is preopen set in
� 1
(V), then f (x)�V. By (i), there
exists a preopen set U of X containing x such
that f (U) � V which implies that
x�U � f
� 1
(V) then f
� 1
(V) } is preopen set in X.
� 1
(V) = � { U :
x� f
(2)�(3). Let E be any cp-closed set in Y, then
Y\E is a cp-open set of Y. By(ii),
f
� 1
(Y\E)=X\ f
� 1
(E) is preopen set in X and
� 1
(E) is preclosed set in X.
hence f
(3)�(4). Let A be any subset of X, then
� 1
(cp-
f (A) � cp-cl( f (A)) and A � f
cl( f (A))). Since cp-cl( f (A)) is cp-closed set
in Y. By(iii), we have f
� 1
(cp-cl( f (A)) is
preclosed set in X. So pcl(A) � f
� 1
(cp-
cl( f (A)))
cl( f (A)).
and hence f (pcl(A)) � cp-
(4)�(5). Let B be any subset of Y, then
f
� 1
(B) is a subset of X. By (iv), we have
f (pcl( f
� 1
(B))) � cp-cl( f ( f
� 1
(B)))= cp-
� 1
cl(B). It follows that pcl( f (B)) � f
cl(B)).
� 1
(cp-
(5) � (6). Let B be any subset of Y, then apply
(v)to Y\B, then pcl( f
cl(Y\B)) � pcl(X\ f
� 1
(Y\B)) � f
� 1
(B)) � f
int(B)) � X\pint((B)) � X\ f
� 1
(cp-intB) � pint f
� 1
(B)).
� 1
(cp-
� 1
(Y\cp-
� 1
(cp-
int(B)) � f
(6)�(7). Let A be any subset of X, then f (A)
is a subset of Y. By (vi), f
� 1
(cp-int( f (A)) �
� 1
( f (A)))=pint(A). It follows that cp-
pint( f
int( f (A)) � f ( pint(A)).
(7)�(1). Let x�X and lef V be any cp-open set
of Y containing f (x), then x� f
and f
( f ( f
� 1
(V)
� 1
(V) is a subset of X. By(vii), cp-int
� 1
(V))) � f ( pint( f
int(V) � f ( pint( f
open set. Then V � f ( pint f
that f
f
� 1
(V) � pint( f
� 1
(V))). So cp-
� 1
(V))), since V is a cp-
� 1
(V))) implies
� 1
(V)) and hence
� 1
(V) is preopen set in X which contains x
and clearly f ( f
� 1
(V)) � V.
Definition 3.6. A function f : X � Y is said to
be a cp-open function if for each x�X and for
each preopen set U containing x, there exists a
cp-open set V containing f (x) such that
V � f (U).
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