Download FAINT PRECONTINUOUS FUNCTIONS 1. Introduction

SOOCHOW JOURNAL OF MATHEMATICS
Volume 21, No. 3, pp. 273-289, July 1995
FAINT PRECONTINUOUS FUNCTIONS
BY
MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
Abstract. In this paper the concept of faint precontinuity is introduced
with the aid of preopen sets some basic properties of this mapping are
obtained its relationships with other generalised forms of continuity are
studied. Within the framework of this mapping, some properties of a few
spaces that exist in the literature along with those of some new spaces
introduced here, are investigated.
1. Introduction
In 1982, Mashhour et al10] introduced the notions of preopen sets and precontinuity in a topological space and obtained their various properties. Quite
recently, R: Paul and the second author of this paper introduced, with the
aid of preopen sets, a weakened form of continuity called quasi-precontinuity
16]. The object of this paper is to introduce, with the help of preopen sets,
a new class of functions called faint precontinuous functions, as a generalisation of quasi-precontinuous functions 16] and to investigate the relationhships
between this class of functions and the other generalised forms of continuity
such as almost continuity 20], slight continuity 5], and pre-irresoluteness 19].
Faint precontinuous functions turn out to be the natural and eective tool for
studying 0-dimensional spaces 22], the weak separation axioms introduced by
Kar and the second author 8], covering axioms of Mashhour et al 11] and
those of Jain 5]. In addition, these functions are also used to study a few
properties of some other spaces such as p-normal 15], pre-connected 15] and
Received February 1, 1993 revised December 10, 1993 revised December 30, 1994.
AMS Subject Classication. 54C.
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
P -closed spaces. Characterisations of faint precontinuous functions are given
in Section 2. Section 3 deals with the relationships between this class and other
generalised classes of continuous functions. Section 4 is concerned with some
basic properties of this class of functions while the last section deals with the
results on separation axioms, pre-connected, slightly compact, P -closed and
other spaces.
Throughout the paper, (X ) and (Y ) etc. (or simply X Y etc.) always
denote topological spaces. The closure and interior of a subset A of a space X
are denoted by ClX (A) and IntX (A) respectively or simply by ClA and IntA
if there is no possibility of confusion. A subset A of a space is called preopen
10] (resp. regular open 22], semi-open 9]) briey p.o. (resp. r.o., s.o.) if
A Int(Cl(A)) (resp. A = Int(Cl(A)) A Cl(Int(A)): The complement of a
p.o. set is called preclosed 10]. The family of all preopen (resp. regular open,
semi-open, preclosed) sets in X is denoted by PO(X ) (resp. RO(X ), SO(X ),
PC(X )). A subset A of a space is called clopen 1] (resp. pre-clopen 8]) if
A is both open and closed (resp. preopen and preclosed). The family of all
clopen (resp. pre-clopen) sets in X is denoted by CO(X ) (resp: PCO(X )). A
set G X is called -open 21] if for each x 2 G there exists V 2 CO(X )
such that x 2 V G. A set F X is called -closed 21] i X ; F is
-open. A subset NX of X is said to be pre-neighbourhood 2] of a point
x 2 X if there exists a p.o. set G such that x 2 G Nx. A is p.o. in X 2] i
it is a pre-neighbourhood of each of its points. A X is said to be quasi-H
closed relative to X , if for every open cover fV : 2 g of A, there exists a
subfamily of such that A fClV : 2 g 18].
2. Characterisations
Denition 2.1. A function f : X ! Y will be termed faint precontinuous
at x 2 X i for every V 2 CO(Y ) containing f (x) there is a U 2 PO(X ) such
that x 2 U and f U ] V .
Denition 2.2. A function f : X ! Y will be termed faint precontinuous
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275
(briey f.p.c.) i f is faint precontinuous at each point of X .
Denition 2.3. A function f : (X ) ! (Y ) is called precontinuous
10] (resp. quasi-precontinuous 16] briey p.c. (resp. q.p.c.) if for each x 2 X
and each V 2 containing f (x) there exists a set U 2 PO(X ) containing x
such that f U ] V (resp. f U ] ClY V ).
Recently Popa and Noiri Th. 3.1, 17] showed that quasi-precontinuity is
equivalent to almost weak continuity due to Jankovic 7]. As a result of this,
faint precontinuity is also weaker than almost weak continuity introduced by
Jankovic.
Remark 2.4. A p.c. function is f.p.c. But the converse is not, in general,
true as shown by
Example 2.5. Let X = fa b c dg Y = fa b c g be two sets equipped
with topologies X = f? X fag fbg fa bgg and Y = f? Y fa gfc g fa c gg
respectively. Let f : X ! Y be dened as f (a) = a f (b) = b f (c) = c and
f (d) = a . Then f is f.p.c. but not p.c. at c and hence not p.c..
Remark 2.6. Every q.p.c. function is f.p.c. But the converse is not, in
general, true as shown by
Example 2.7. Let f : (X X ) ! (Y Y ) be the function of Example 2.5.
Then f is f.p.c. but not q.p.c. at c and hence not q.p.c..
(a)
(b)
(c)
(d)
(e)
(f)
Theorem 2.8. For a function f : X ! Y , the following are equivalent:
f is f.p.c.,
f ;1V ] 2 PO(X ) for every V 2 CO(Y ),
f ;1V ] 2 PC(X ) for every V 2 CO(Y ),
f ;1V ] 2 PCO(X ) for every V 2 CO(Y ),
f ;1V ] 2 PO(X ) for every -open set V in Y ,
f ;1V ] 2 PC(X ) for every -closed set V in Y .
Proof. (a) ! (b): Let V 2 CO(Y ) and let x 2 f ;1V ]: Then f (x) 2 V .
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
The f.p.c. of f at x gives the existence of a U 2 PO(X ) such that x 2 U and
f U ] V whence we observe x 2 U f ;1V ]. This implies, then, that f ;1V ]
is a pre-neighbourhood of x. Hence f ;1V ] 2 PO(X ).
(b) ! (c): Let V 2 CO(Y ): Then Y ; V 2 CO(Y ). By (b), f ;1Y ; V ] 2
PO(X ) and this implies that f ;1V ] 2 PC(X ).
(c) ! (d): It is obvious from (b) and (c).
(d) ! (e): Let G be a -open set in Y and let x 2 f ;1 G]. Then f (x) 2 G.
The -openness of G gives a V 2 CO(Y ) such that f (x) 2 V G. This
implies that x 2 f ;1V ] f ;1 G]. The f.p.c. of f shows that f ;1 V ] 2 PO(X ).
Hence f ;1G] is a pre-neighbourhood of each of its points. Consequently,
f ;1G] 2 PO(X ).
(e) ! (f): It is obvious from the fact that the complement of a -closed set
is -open.
(f)! (a): It readily follows from the fact that every clopen set is -open.
3. Comparisons
Denition 3.1. A function f : (X ) ! (Y ) is called almost continuous 20] (resp. slightly continuous 5]) if for each x 2 X and each V 2 (resp.
V 2 CO(Y )) containing f (x) there exists a set U 2 containing x such taht
f U ] Int(ClV ) (resp. f U ] V ).
Denition 3.2. A function f : X ! Y is said to be pre-irresolute 19] if
f ;1V ] 2 PO(X ) for every V 2 PO(Y ).
Paul and Bhattacharyya 16] showed that the class of quasi-precontinuity
contains the class of precontinuity which clearly includes in it the class of
continuity. But by the Remarks 2.4 and 2.6 it is now clear that the class
of faint precontinuity contains the class of quasi-precontinuity of Paul and
Bhattacharyya 16]. Again the Denition 3.1 indicates that the class of slight
continuity includes in it the class of almost continuity while it is included in our
class of faint precontinuity. On the other hand, the class of pre-irresoluteness
FAINT PRECONTINUOUS FUNCTIONS
277
19] is included in the class of precontinuity 10] and hence is contained in the
class of faint precontinuity.
The above discussions can be summarised in the following diagram.
Continuity
Pre-irresoluteness !
#
Precontinuity
#
! Almost continuity
#
Slight continuity
#
Quasi-Precontinuity ! Faint Precontinuity
None of the above implications can, in general, be reversed. This can be
seen from the Examples 2.5 and 2.7 given earlier and the Examples 3.3 - 3.8
given below.
Example 3.3. (13], Example 5.10). Let X = fa b c dg and =
f? X fbg fcg fb cg fa bg fa b cg fb c dgg = . Let f : (X ) ! (X )
be dened as f (a) = c f (b) = d f (c) = b and f (d) = a. Then f is q.p.c. but
not p.c..
Example 3.4. Let X = Y = fa b cg. Let the topology on X be indiscrete and we topologise Y with f? Y fag fb cgg. Let f : X ! Y be dened
as f (a) = b f (b) = c and f (c) = a. Then f is f.p.c. but not slightly continuous
at c and hence is not slightly continuous.
Example 3.5. Let f be the function of Example 2.5. Then f is not
almost continuous at c and hence is not almost continuous, but it is slightly
continuous.
Example 3.6. Let f : (X ) ! (Y ) be injective with = Indiscrete
topology, = Discrete topology. Then f is p.c. but not continuous.
Example 3.7. (20], Example 2.1). Let R = the set of reals, X = fa bg,
= Cocountable topology on R while = f? X fagg. Dene f : (R ) !
(X ) as f (x) = a if x is rational and f (x) = b if x is irrational. Then f is
almost continuous but not continuous.
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
Example 3.8. (19], Example 3). Let X = fa b c dg and Y = fx y zg
be equipped with topologies =f? X fag fb cg fa b cgg and U = f? Y fxgg.
Dene f : (Y U ) ! (X ) as f (x) = b, f (y) = c and f (z ) = d. Then f is p.c.
but not pre-irresolute.
4. Basic Properties
Denition 4.1. (22]) X is 0-dimensional i each point of X has a
neighbourhood base consisting of clopen sets.
Theorem 4.2. Let f : X ! Y and Y be 0-dimensional. Then f is f.p.c.
i f is p.c..
Proof. Necessity. Let x 2 X and V
Y be any open set containing
f (x). By the 0-dimensionality of Y , there exists a W 2 CO(Y ) such that
f (x) 2 W V . The f.p.c. of f gives the existence of a U 2 PO(X ) such that
x 2 U and f U ] W . This implies that f U ] V . Hence f is p.c..
Suciency. It follows from the fact that every p.c. function is f.p.c..
Remark 4.3. Composite function of two f.p.c. functions need not be, in
general, f.p.c. as shown by
Example 4.4. Let X0 = X1 = X2 = fa b c g be topologized with
0 = f? X0 fagg 1 = f? X1 g 2 = f? X2 fag fb cgg respectively. Let
f0 : X0 ! X1 be the identity mapping and f1 : X1 ! X2 be dened as
f1(a) = b f1(b) = c f1 (c) = a. Then f1 f0 is not f.p.c. though f0 and f1 are
both f.p.c..
However, we do have
Theorem 4.5. Let f : X ! Y and g : Y ! Z . Then:
(1) g f is f.p.c. whenever f is pre-irresolute and g is f.p.c..
(2) g f is f.p.c. whenever f is f.p.c. and g is slightly continuous.
FAINT PRECONTINUOUS FUNCTIONS
279
Proof of (2). The slight continuity of g gives that g;1V ] 2 CO(Y ) for
every V 2 CO(Z ). Again the f.p.c. of f gives f ;1g;1 V ]] 2 PO(X ), i.e.
(g f );1 V ] 2 PO(X ). This shows that g f is f.p.c..
The straightforward proof of (1) is almost similar and hence is omitted.
Corollary 4.6. Let f : X ! Y and g : Y ! Z . Then:
(1) g f is f.p.c. whenever f is pre-irresolute and g is slightly continuous.
(2) g f is f.p.c. whenever f is p.c. and g is slightly continuous.
(3) g f is f.p.c. whenever f is f.p.c. and g is continuous.
Proof. These are immediate consequences of Theorem 4.5 and the impli-
cations of the Diagram in Section 3.
Lemma 4.7. (3], Theorem 1.1) If f : X ! Y is continuous and open
then f ;1ClY (A)] = ClX (f ;1 A]) for every A Y .
Denition 4.8. (6]) A function f : X ! Y is called p-open if f A] 2
PO(Y ) for all A 2 PO(X ).
Lemma 4.9. If f : X ! Y is continuous and open then f is p-open.
Proof. Let A 2 PO(X ). Then A IntX ClX (A): By openness and con-
tinuity of f , it then follows that f A] f IntX (ClX (A))] IntY (f ClX (A)])
IntY (ClY (f A])). This shows that f A] 2 PO(Y ). Hence f is p-open.
Lemma 4.10. If f : X ! Y is open and continuous then f is pre-
irresolute.
Proof. Let A 2 PO(Y ). Then A
IntY (ClY (A)). This and continuity of f together imply f A] f IntY (ClY (A))] IntX (f ;1ClY (A)]) =
IntX (ClX (f ;1 A])) by Lemma 4.7. This , therefore, shows that f ;1 A] 2
PO(X ). Hence f is pre-irresolute.
;1
;1
Theorem 4.11. Let f : X ! Y be a pre-irresolute p-open surjection
and g : Y ! Z be any function, then g f is f.p.c. i g is f.p.c..
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
Proof. Necessity. Suppose g f is f.p.c. Then (g f );1V ] 2 PO(X )
for every V 2 CO(Z ), i.e. f ;1g;1 V ]] 2 PO(X ): Since f is surjective and
p-open, we then observe that f f ;1g;1 V ]]] 2 PO(Y ), i.e. g;1 V ] 2 PO(Y ).
This implies that g is f.p.c..
Suciency. Suppose now g is f.p.c. Then g;1 V ] 2 PO(Y ) for every V 2
CO(Z ). Since f is pre-irresolute, f ;1g;1 V ]] 2 PO(X ), i.e., (g f );1 V ] 2
PO(X ) for every V 2 PO(Z ). This indicates that g f is f.p.c..
Corollary 4.12. Let f : X ! Y be a continuous open surjection and g
is any function, then g f is f.p.c. i g is f.p.c..
Proof. This is an immediate consequence of Lemma 4.10 and Theorem
4.11.
Lemma 4.13. (12], Lemma 2.10 ) If V 2 PO(X ) and X0 2 SO(X ),
then X0 \ V 2 PO(X0 ).
Lemma 4.14. (8], Lemma 1) Let A X0 X and X0 2 PO(X ): Then
A 2 PO(X ) i A 2 PO(X0).
Theorem 4.15. For a function f : X ! Y , the following are true:
(1) If f is f.p.c. and A 2 SO(X ) then f jA : A ! Y is f.p.c..
(2) If B 2 PO(X ) and x 2 B such that f jB : B ! Y is f.p.c. at x, then f is
f.p.c. at x.
(3) If fC : 2 g is a cover of X such that C 2 PO(X ) for every 2 ,
then f is f.p.c. if f jC : C ! Y is f.p.c. for every 2 .
Proof of (1). Let U 2 CO(Y ). Since f is f.p.c. f ;1U ] 2 PO(X ): Also
(f jA);1 U ] = A \ f ;1U ]: Since A 2 SO(X ), f ;1U ] 2 PO(X ), by Lemma
4.13, A \ f ;1 U ] 2 PO(A) and hence f jA is f.p.c..
Proof of (2). Let U 2 CO(Y ) such that f (x) 2 U . Then the f.p.c. of f jB
at x gives the existence of a V 2 PO(B ) containing x such that (f jB )V ] U
i.e. f V ] U: Now V 2 PO(B ) and B 2 PO(X ). So by Lemma 4.14,
FAINT PRECONTINUOUS FUNCTIONS
281
V 2 PO(X ). Thus there exists a V 2 PO(X ) such that f V ] U . Hence f is
f.p.c. at x.
Proof of (3). Let V 2 CO(Y ): Then by the f.p.c. of f jC (f jC );1V ] 2
PO(C ) for every 2 . Since C 2 PO(X ), by Lemma 4.14, (f jC );1 V ] 2
S
PO(X ) for each 2 . But f ;1 V ] = (f jC );1 V ]): Since the union of
2
p.o. sets is a p.o. set 10], the above indicates f ;1 V ] 2 PO(X ). Hence f is
f.p.c..
Remark 4.16. The condition \semi-openness" on A in (1) of Theorem
4.15 cannot be dropped as shown by
Example 4.17. Let X and Y of Example 2.5 be equipped with topologies
X = the same topology of Example 3.3, Y = f? Y fa g fb c gg: If f be
the same functin of Example 2.5 then f jA is not f.p.c. when A = fa b dg 62
SO(X ).
Theorem 4.18. A function f : X ! Y is f.p.c. if the graph function
g : X ! X Y dened by g(x) = (x f (x)) is f.p.c..
Proof. Suppose g is f.p.c. Let V 2 CO(Y ). Then (X V ) 2 CO(X Y ):
Since g is f.p.c., g;1 X V ] 2 PO(X ). It can easily be veried that g;1 X V ] = f ;1V ]. Hence f ;1V ] 2 PO(X ). This indicates that f is f.p.c..
Theorem 4.19. Let f : X ! Y be a function for each 2 and
f : Q X ! Q Y a function dened by f fx g] = ff (x )g for each point
fx g 2 Q X . If f is f.p.c., then f is f.p.c. for each 2 .
Proof. Suppose f : Q X ! Q Y is f.p.c.. Let 2 be xed and
consider the point x 2 X . Let V 2 CO(Y ) such that f (x ) 2 V . Now
Q
consider any point fx g 2 X whose -th coordinate is x . Then the set
V = Q Y V 2 CO(Q Y ) and ff (x )g 2 V . The f.p.c. of f indicates
6=
f ;1V ] 2 PO(Q X ). Now -th projection function p is open and continuous
and hence by Lemma 4.9, p is p-open. So, p f ;1V ]] 2 PO(Y ) i.e. f;1V ] 2
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
PO(Y ). This indicates that f is f.p.c.. It follows then that f is f.p.c. for
each 2 .
5. Further Properties of f.p.c. Functions
I. f.p.c. and some separation axioms
Denition 5.1. X is called
(i) pre-T0 8] (resp. pre-T1 8]) i for x y 2 X such that x 6= y, there exists a
p.o. set containing x but not y, or (resp. and ) a p.o. set containing y
but not x
(ii) pre-T2 8] i for x y 2 X such that x 6= y, there exist disjoint p.o. sets U
and V such that x 2 U and y 2 V (iii) p-regular 4] i for each closed set F of X and each point x 62 F , there
exist disjoint p.o. sets U and V such that F U and x 2 V (iv) p-normal 15] if for every pair of disjoint closed sets F1 and F2 of X , there
exist disjoint p.o. sets U and V such that F1 U and F2 V (v) extremally disconnected 22] if the closure of every open set in X is open.
Remark 5.2. Kar and Bhattacharyya showed (8], Th. 3) that every
topological space is pre-T0 .
Remark 5.3. When p.o. sets are replaced by r.o. sets in (i) and (ii) of
Denition 5.1, one obtains the denitions of rT0 rT1 and rT2 spaces introduced
by Jain 5].
Lemma 5.4. (4], Th. 3.2(2)) X is p-regular i for each x 2 X and each
open set U of X containing x there exists V 2 PO(X ) such that x 2 V
pclV
U.
Lemma 5.5. A 0-dimensional space is always regular, while a regular
extremally disconnected space is 0-dimensional.
Proof. First part of the Lemma follows from the denitions of regular
FAINT PRECONTINUOUS FUNCTIONS
283
and 0-dimensional spaces.
For the second part, suppose X is a regular space which is also extremally
disconnected. Let x 2 X and G be any open set cotaining x. The regularity
of X gives the existence of an open set V such that x 2 V ClV G. On
the other hand, extremally disconnectedness of X implies that ClV is open
and hence ClV is a clopen set such that x 2 ClV G. This indicates that the
family of clopen sets from a neighbourhood base. Hence X is 0-dimensional.
Lemma 5.6. In an extremally disconnected space X , RO(X ) = CO(X ):
Proof. The proof is straightforward and hence is omitted.
Theorem 5.7.
(a) Let Y be 0-dimensional and f : X ! Y be a f.p.c. injection. Then X is
pre-T1 (resp. pre-T2 ) if Y is T1 (resp. T2 ).
(b) Under the hypotheses of (a), if f is either open or closed, then X is
p-regular.
(c) Under the hypotheses of (a), if f is closed and Y is normal, then X is
p-normal.
Proof of (a). We prove only one part of (a) where X is shown to be
pre-T2 .
Since f is injective, for any pair of distinct points x1 x2 2 X , f (x1 ) 6=
f (x2). Since Y is T2, there exist open sets V1 V2 in Y such that f (x1) 2 V1
f (x2) 2 V2 and V1 \ V2 = ?. Since Y is 0-dimensional, there exists U1 2 CO(Y )
such that f (xi ) 2 Ui Vi i = 1 2: Consequently, xi 2 f ;1 Ui ] f ;1Vi ]
i = 1 2 and f ;1U1 ] \ f ;1U2 ] = ?. Since f is f.p.c., f ;1Ui ] 2 PO(X ), i = 1 2
and this implies that X is pre-T2 .
Proof of (b). First suppose f is open. Let x 2 X and U be an open
set containing x. Therefore, f (x) 2 f U ] which is open in Y because of the
openness of f . On the other hand, 0-dimensionality of Y gives the existence
of a V 2 CO(Y ) such that f (x) 2 V f U ]: So, x 2 f ;1V ] U (since
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
f is injective). Again the f.p.c. of f implies that f ;1V ] 2 PCO(X ) by
Theorem 2.8(d) and hence x 2 f ;1 V ] pcl(f ;1 V ]) U . By Lemma 5.4
this, therefore, indicates that X is p-regular.
Next suppose f is closed. Let x 2 X and F be a closed subset of X
such that x 62 F . So, f (x) 62 f F ] and therefore f (x) 2 Y ; f F ] which is
an open set in Y because of the closedness of f . Since Y is 0-dimensional,
there exists V 2 CO(Y ) such that f (x) 2 V Y ; f F ]. Since f is f.p.c., we
have x 2 f ;1 V ] 2 PO(X ) and F X ; f ;1V ] 2 PO(X ). Therefore, X is
p-regular.
Proof of (c). Let F1 and F2 be any two closed sets in X such that
F1 \ F2 = ?. Since f is closed and injective we observe from this that f F1]
and f F2 ] are two closed sets in Y with f F1 ] \ f F2 ] = ?: By normality of Y ,
there exist, then, two open sets U and V in Y such that f F1 ] U , f F2 ] V
and U \ V = ?. Let y 2 f F1 ], then y 2 U . Since Y is 0-dimensional and U is
open in Y , there exists a Uy 2 CO(Y ) such that y 2 Uy U . It, then, follows
that f F1 ] fUy : Uy 2 CO(Y ) y 2 f F1 ]g U whence F1 f ;1fUy :
Uy 2 CO(Y ) y 2 f F1]g] f ;1U ] i.e. F1 ff ;1Uy ] : Uy 2 CO(Y )
y 2 f F1]g f ;1U ]: Since f is f.p.c., f ;1Uy ] 2 PO(X ) for each Uy 2 CO(Y )
S f ;1U ] 2 PO(X ) and F G f ;1U ]. In like manner, 0so that G =
y
1
y2f F1 ]
dimensionality of Y and the f.p.c. of f give the existence of a set H 2 PO(X )
such that F2 H f ;1 V ]: Also G \ H f ;1U \ V ] = ?. This, therefore,
indicates that X is p-normal.
Corollary 5.8. If Y is regular extremally disconnected space and f :
X ! Y is a f.p.c. injection then X is p-regular if f is either open or closed.
Proof. This is an immediate consequence of the second part of Lemma
5.5 and the Theorem 5.7.
Theorem 5.9. Let Y be extremally disconnected and f : X ! Y be a
f.p.c. injection. Then X is pre-T1 (resp. pre-T2 ) if Y is rT1 (resp. rT2 ).
FAINT PRECONTINUOUS FUNCTIONS
285
Proof. With the aid of Lemma 5.6, the proof follows in like manner as
in (a) of Theorem 5.7.
II f.p.c. and preconnected spaces
Denition 5.10. X is said to be preconnected 15] if it is not the union
of two non-empty disjoint p.o. sets. A space which is not preconnected is
called predisconnected.
It is not dicult to check that every preconnected space is connected and
the converse of it may not hold.
That preconnectedness is not preserved under q.p.c. and hence not under
f.p.c. is evident from the following two examples.
Example 5.11. Let X = fa b cg and Y = fa b c g be two sets.
Let the topology on Y be indiscrete and we topologize X with f? X fag
fcg fa cgg. Let f : X ! Y be dened as f (x) = x for all x 2 X . Then Y
is not preconnected though f is q.p.c. and hence f.p.c. surjection and X is
preconnected.
However, the following is true.
Theorem 5.12. If X is preconnected and f : X ! Y is a f.p.c. surjec-
tion then Y is connected.
Proof. Suppose Y is not connected. Then there exist ? 6= V1 V2 2
CO(Y ), V1 \ V2 = ? such that Y = V1 V2 . Since f is f.p.c. we observe that
?=
6 f ;1V1], f ;1V2] 2 PO(X ). Also f ;1V1 ] f ;1V2] = X and f ;1V1] \
f ;1V2] = ?. This indicates that X is not preconnected, a contradiction.
Again the inverse image of a disconnected space under any f.p.c. may not
be disconnected, as is seen from
Example 5.13. Let X = f? X fag fa bg fa cgg and Y = f? Y fa g,
fb c gg be the topologies on X and Y of Example 5.11. Let f : X ! Y be
dened as f (a) = f (b) = f (c) = b : Then f is continuous and f.p.c. but
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
not surjective. One the other hand Y is disconnected while f ;1Y ] = X is
connected.
However, the following is valid.
Theorem 5.14. Inverse image of a disconnected space under a f.p.c.
surjection is predisconnected.
Proof. Let f : X ! Y be a f.p.c. surjection where Y is disconnected.
Now the disconnectedness of Y gives the existence of a proper subset ? =
6 V2
;
1
CO(Y ). Since f is f.p.c. and surjective, ? =
6 f V ] is a proper preclopen set
in X by Theorem 2.8(d). This indicates that X is predisconnected.
Remark 5.15. That the condition \surjection" cannot be dropped from
Theorem 5.14 is seen from Example 5.13.
Corollary 5.16. Inverse image of a non-indiscrete 0-dimensional space
under a f.p.c. surjection is predisconnected.
Proof. Follows from the fact that a non-indiscrete 0-dimensional space
is always disconnected.
III f.p.c. and some covering axioms
Denition 5.17. X is said to be
(i) precompact if every preopen cover of X admits a nite subcover (Mashhour et al 11] dened this space as strongly compact)
(ii) P-closed (resp. strongly P-closed) if every preopen cover of X has a nite
subfamily such that the union of their closures (resp. preclosures) covers
X (Noiri 14], recently, showed that \P-closedness" is equivalent to quasi
H-closedness 18])
(iii) slightly compact 5] if every clopen cover of X has a nite subcover.
The following Lemma is useful in the sequel.
Lemma 5.18. If X is strongly P-closed, then every preclopen cover of
FAINT PRECONTINUOUS FUNCTIONS
287
X has a nite subcover.
Proof. Suppose C = fC : 2 g be a preclopen cover of the strongly
P-closed space X . Then the strongly P-closedness of X and preopenness of
the cover C give the existence of a nite set 0 such that X = fpcl(C ) :
2 0g = fC : 2 0g (since each C is preclopen) which shows that
C has a nite subcover. C being arbitrary, this implies that every preclopen
cover has a nite subcover.
Theorem 5.19. If X is strongly P-closed and f : X ! Y is a f.p.c.
surjection, then Y is slightly compact.
Proof. Let fC : 2 g be a cover of Y by clopen sets. Since f is f.p.c.,
f ;1C ] 2 PCO(X ) for every 2 , by Theorem 2.8(d). So, ff ;1C ] : 2 g
is a preclopen cover of X . Since X is a strongly P-closed space, by Lemma
5.18, this preclopen cover has a nite subcover. It means that there exists a
nite set 0 such that X = ff ;1 C ] : 2 0 g = f ;1 fC : 2 0 g]:
Consequently the surjection of f gives Y = fC : 2 0 g which, in its turn,
implies that Y is slightly compact.
Corollary 5.20. If f : X ! Y is a f.p.c. surjection and X is precom-
pact, then Y is slightly compact.
Proof. Let fC : 2 g be a clopen cover of Y . Since f is f.p.c.
f ;1C ] 2 PO(X ) for every 2 . Consequently, ff ;1C ] : 2 g is a
preopen cover of X . By precompactness of X , this cover then, admits of a
nite subcover. This means that there exists a nite subst 0 such that
X = ff ;1C ] : 2 0g. The surjection of f , then, gives Y = fC : 2
0 g, which shows that Y is slightly compact.
Theorem 5.21. If X is strongly P-closed, Y is extremally disconnected
and f : X ! Y is a f.p.c. surjection, then Y is P-closed.
Proof. Let fC : 2 g be a preopen cover of Y . Since C 2 PO(Y ), for
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MANINDRA CHANDRA PAL AND P. BHATTACHARYYA
each 2 , C Int(Cl(C )) = V , say. From this we observe that fCl(V ) :
2 g is a cover of Y . The extremally disconnectedness of Y gives that
Cl(V ) 2 CO(Y ) for every 2 . Consequently, fCl(V ) : 2 g is a clopen
cover of Y . Again the f.p.c. of f gives that ff ;1 Cl(V )] : 2 g is a preclopen
cover of X . Since X is strongly P-closed, by Lemma 5.18, this preclopen cover
admits of a nite subcover, which means that there exists a nite set 0 such that X = ff ;1Cl(V )] : 2 0 g = f ;1fCl(V ) : 2 0 g]: The
surjection of f then gives Y = fCl(V ) : 2 0 g fCl(C ) : 2 0 g
(since Cl(V ) Cl(C )). So, Y = fCl(C ) : 2 0 g and thus Y is P-closed.
Acknowledgement
The authors are deeply grateful to the referee for his valuable suggestions
and remarks.
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Department of Mathematics, Maulana Azad College, 8, Ra Ahmed Kidwai Road Calcutta
- 700013, West Bengal, India.
Department of Mathematics, University of Kalyani, Kalyani - 741235, West Bengal, India.