© Hindawi Publishing Corp.
ON ALMOST PRECONTINUOUS FUNCTIONS
SAEID JAFARI and TAKASHI NOIRI
(Received 19 December 1997 and in revised form 28 September 1998)
Abstract.Nasef and Noiri (1997) introduced and investigated the class of almost precon-tinuous functions. In this paper, we further investigate some properties of these functions.
Keywords and phrases. Preopen, preclosed, preboundary, strongly compact, almost pre-continuous, strongly prenormal.
2000 Mathematics Subject Classification. Primary 54C10; Secondary 54C08.
1. Introduction. Singal and Singal [24] introduced the notion of almost continu-ity. Feeble continuity was introduced by Maheshwari et al. [8]. As a generalization of almost continuity and feeble continuity, Maheshwari et al. [7] introduced the notion of almost feeble continuity. Nasef and Noiri [12] introduced a new class of functions called almost precontinuous functions. They showed that almost precontinuity is a generalization of each of almost feeble continuity and almostα-continuity [17].
The purpose of this paper is to investigate some more properties of almost precon-tinuous functions. It turns out that almost precontinuity is stronger than almost weak continuity introduced by Jankovi˘c [5].
2. Preliminaries. Throughout this paper,(X,τ)and(Y ,σ )(orXandY) are always topological spaces. A set A in a spaceX is calledpreopen[11] (respectively, semi-open[6] andα-open[13]) ifA⊂A¯◦(respectively,A⊂A¯◦andA⊂A¯◦◦). The complement
of a preopen set is calledpreclosed.
The intersection of all preclosed sets containing a subsetAis called thepreclosure[2] ofAand is denoted by Pcl(A). ThepreinteriorofAis the union of all preopen sets of
Xcontained inA. The family of all preopen sets ofXwill be denoted by PO(X). For a pointxofX, we put PO(X,x)= {U|x∈U∈PO(X)}. A setAis calledregular open (respectively,regular closed) ifA=A¯◦ (respectively,A=A¯◦).
Definition2.1. A functionf:X→Y is calledalmost continuous[24] (in the sense of Singal) atx∈Xif for every open setV inY containingf (x), there is an open set
UinXcontainingxsuch thatf (U)⊂V¯◦. Iffis almost continuous at every point of X, then it is called almost continuous.
Definition2.2. A functionf:X→Yis calledalmost weakly continuous[5] (briefly a.w.c.) iff−1(V)⊂f−1(V)¯ ◦for every open setVofY.
only if for each point x ∈ X and every open set V in Y containing f (x), there existsU∈PO(X,x)such thatf (U)⊂V¯. The referee has given a global description as follows: a functionf :X→Y is a.w.c. if and only if for each open setV ofY, there existsU∈PO(X)such thatf−1(V)⊂U⊂f−1(V )¯ .
Definition2.4. A functionf:X→Y is calledalmost precontinuous[12] (briefly a.p.c.) atx∈Xif for each regular open setV⊂Y containingf (x), there existsU∈ PO(X,x)such thatf (U)⊂V. Iff is almost precontinuous at every point ofX, then it is called almost precontinuous.
Definition2.5. A functionf:X→Yis said to beweaklyα-continuous[16] (briefly w.α.c.) if for eachx∈Xand each open setV⊂Y containingf (x), there exists anα -open setUcontainingxsuch thatf (U)⊂V¯.
Definition2.6. A functionf:X→Y is said to beprecontinuous[11] if for every open setV ofY, the inverse image ofV is preopen inX.
Remark2.7. Between almost precontinuity and precontinuity, we have the follow-ing relationship: a functionf:X→Y is a.p.c. if and only iff:X→Ysis precontinuous, whereYs denotes the semi-regularization ofY.
Remark2.8. It easily follows from [20, Theorem 3.1] that precontinuity implies almost precontinuity and almost precontinuity implies almost weak continuity. How-ever, the converses are not true as the following examples show.
Example2.9. LetX= {a,b,c}, τ= {X,∅,{a},{c},{a,c}}andσ= {X,∅,{a},{b}, {a,b},{b,c}}. Define a functionf:(X,τ)→(X,σ )as follows:f (a)=f (b)=band
f (c)=c. Then f is an almost continuous and hence a.p.c. function which is not precontinuous. Because, there exists{b} ∈σ such thatf−1({b})∉PO(X,τ).
Example2.10. Let X = {a,b,c,d} and τ = {X,∅,{b},{c},{a,b},{b,c},{a,b,c}, {b,c,d}}. Define a functionf:(X,τ)→(X,τ)as follows:f (a)=c, f (b)=d, f (c)= b, andf (d)=a. Thenfis a.w.c. However,fis not a.p.c. because there exists a regular open set{c}of(X,τ)such thatf−1({c})∉PO(X,τ).
Recall that a filter baseᏲis calledδ-convergent[25] (respectively,p-convergent[4]) to a pointxinXif for any open setUcontainingx(respectively, anyU∈PO(X,x)), there existsB∈Ᏺsuch thatB⊂U¯◦(respectively,B⊂U).
3. Some properties. In [9], Mashhour et al. introduced the following notion.
Definition3.1. A functionf :X→Y is called M-preopenif the image of each preopen set is preopen.
We have the following result.
Theorem3.2. Iff:X→Y isM-preopen a.w.c., thenfis a.p.c.
M-preopen,f (U)is preopen inY and hencef (U)⊂f (U)◦⊂V¯◦=V¯◦. It follows that f (U)⊂V¯◦. Hencef is a.p.c.
Recall that a spaceXis calledsubmaximalif every dense subset ofXis open inX. It is shown in [22, Theorem 4] that a spaceXis submaximal if and only if every preopen set ofXis open inX.
Theorem3.3. If a functionf:X→Y is a.p.c., then for each pointx∈Xand each filter baseᏲinX p-converging tox, the filter basef (Ᏺ)isδ-convergent tof (x). IfX
is submaximal, then the converse also holds.
Proof. Suppose thatxbelongs toXandᏲis any filter base inX p-converging to
x. By the almost precontinuity off, for any regular open setVinY containingf (x), there existsU∈PO(X,x)such thatf (U)⊂V. ButᏲisp-convergent toxinX, then there existsB∈Ᏺsuch thatB⊂U. It follows thatf (B)⊂V. This means thatf (Ᏺ)is
δ-convergent tof (x).
Now suppose thatXis submaximal. Letxbe a point inXandV any regular open set containingf (x). SinceXis submaximal, every preopen set ofXis open [22, The-orem 4]. If we setᏲ=PO(X,x), thenᏲwill be a filter base whichp-converges tox. So there existsUinᏲsuch thatf (U)⊂V. This completes the proof.
The following corollary is suggested by the referee.
Corollary3.4. LetXbe a submaximal space. Then a functionf:X→Y is a.p.c. if and only iff:X→Ysis continuous.
Definition3.5. A spaceXis calledpre-T2[18] if for every pair of distinct points
xandyinX, there exist preopen setsUandVcontainingxandy, respectively, such thatU∩V= ∅.
Theorem3.6. Iff:X→Y is an a.p.c. injection andY is Hausdorff, thenXis pre-T2. Proof. Sincef:X→Y is a.p.c. injective,f :X→Ys is a precontinuous injection and Ys is Hausdorff. Let x and y be any distinct points ofX. Since f is injective, f (x)≠f (y)and hence there exist disjoint open setsVandWofYssuch thatf (x)∈V and f (y)∈W. Therefore, we obtainf−1(V)∈PO(X,x), f−1(W )∈PO(X,y), and
f−1(V)∩f−1(W )= ∅. This shows thatXis pre-T2.
Recall that a spaceX is called a door space if every subset of X is either open or closed. Reilly and Vamanamurthy proved the following result in [22, Theo-rem 2].
Lemma3.7. IfXis a door space, then every preopen set inXis open.
Theorem3.8. Letf ,g:X→Y be functions,Y Hausdorff andXa door space. Iff
andgare a.p.c. functions, then the setE= {x∈X|f (x)=g(x)}is closed inX. Proof. Letx∈X−E. It follows thatf (x)≠g(x). SinceYis Hausdorff, then there exist open setsV1andV2inY such thatf (x)∈V1, g(x)∈V2, andV1∩V2= ∅. Since
U=U1∩U2, so, by Lemma 3.7, U is an open set inX containing x. Thus we have
f (U)∩g(U)= ∅. It follows thatx∉E¯. Hence ¯E⊂EandEis closed inX.
Lemma3.9(Popa and Noiri [20]). IfAis anα-open set of a spaceXandB∈PO(X), thenA∩B∈PO(X).
Theorem3.10. Letf ,g:X→Y be functions andY Hausdorff. Iff is w.α.c. andg
is a.p.c., then the setE= {x∈X|f (x)=g(x)}is preclosed inX.
Proof. Suppose thatx∉E. Thenf (x)≠g(x). SinceY is Hausdorff, there exist open setsVandWofYsuch thatf (x)∈V , g(x)∈W, andV∩W= ∅; hence ¯V∩W¯◦= ∅. Sincef is w.α.c., there exists anα-open setU containingx such thatf (U)⊂V¯. Sincegis a.p.c., there existsG∈PO(X,x)such thatg(G)⊂W¯◦. PutO=U∩G, then O∈PO(X,x)by Lemma 3.9 andO∩E= ∅. Therefore, we obtain x∉Pcl(E). This shows thatEis preclosed inX.
Corollary3.11(Popa [19]). Letf ,g:X→Y be functions andY Hausdorff. Iffis continuous andgis precontinuous, then the setE= {x∈X|f (x)=g(x)}is preclosed inX.
Theorem3.12. Letf :X1→Y andg:X2→Y be two a.p.c. functions. If Y is a Hausdorff space, then the set{(x1×x2)∈X1×X2|f (x1)=g(x2)}is preclosed in
X1×X2.
Proof. Let (x1,x2)∉ E. Then f (x1)≠g(x2). Since Y is Hausdorff, there exist disjoint open neighborhoodsVandWoff (x1)andg(x2), respectively. SinceVand
Ware disjoint, we have ¯V◦∩W¯◦= ∅. Sincefandgare a.p.c., there existU∈PO(X
1,x1) andG∈PO(X2,x2)such thatf (U)⊂V¯◦ and g(G)⊂W¯◦, respectively. PutO=U×
G, then(x1,x2)∈O,O is preopen inX1×X2 andO∩E= ∅. Therefore, we obtain
(x1,x2)∈Pcl(E). This shows thatEis preclosed inX1×X2.
Corollary3.13. IfY is Hausdorff andf:X→Y is an a.p.c. function, then the set
E= {(x,y)|f (x)=f (y)}is preclosed inX×X.
Proof. By settingX=X1=X2andg=fin Theorem 3.12, the result follows. Corollary3.14(Mashhour et al. [11]). If f :X →Y is a precontinuous function andY is Hausdorff, then the set{(x,y)|f (x)=f (y)}is preclosed inX×X.
Corollary3.15(Popa [19]). Letf:X1→Y andg:X2→Y be two precontinuous functions. IfY is a Hausdorff space, then the set{(x,y)|f (x)=g(y)}is preclosed in
X1×X2.
We introduce the following concept.
Definition3.16. For a functionf:X→Y, the graphG(f )= {(x,f (x))|x∈X} is calledstrongly almost preclosedif for each(x,y)∈X×Y−G(f ), there existU∈ PO(X,x)and a regular open setV containingysuch that(U×V )∩G(f )= ∅.
Proof. It is an immediate consequence of the above definition.
Theorem3.18. If f :X →Y is a.w.c. and Y is Hausdorff, thenG(f ) is strongly almost preclosed.
Proof. Suppose that(x,y)is any point ofX×Y−G(f ). Theny≠f (x). ButY is Hausdorff and hence there exist open setsG1andG2inYsuch thaty∈G1, f (x)∈G2, andG1∩G2= ∅. SinceG1andG2are disjoint, we obtain ¯G◦1∩G¯2= ∅. And sincef is a.w.c., then there existsU∈PO(X,x)such thatf (U)⊂G¯2. Hence,f (U)∩G¯◦
1= ∅. It follows from Lemma 3.17 thatG(f )is strongly almost preclosed.
Recall that a subsetAof a spaceXis said to bestrongly compact relative toX[9] (respectively,N-closed relative toX[1]) if every cover ofAby preopen (respectively, regular open) sets ofXhas a finite subcover.
Definition3.19. A spaceX is calledstrongly compact[10] (respectively,nearly compact [23]) if every preopen (respectively, regular open) cover of X has a finite subcover.
Theorem3.20. Iff:X→Y is a.p.c. andKis a strongly compact relative toX, then
f (K )isN-closed relative toY.
Proof. Let{Gα|α∈A}be any cover off (K )by regular open sets ofY. Then, {f−1(G
α)|α∈A} is a cover of K by preopen sets of X [12, Theorem 3.1]. Since K is strongly compact relative to X, there exists a finite subset A◦ ofA such that
K⊂ ∪{f−1(G
α)|α∈A◦}. Therefore, we obtainf (K )⊂ ∪{Gα|α∈A◦}. This shows thatf (K )isN-closed relative toY.
Corollary3.21. Iff:X→Y is an a.p.c. surjection andXis strongly compact, then
Y is nearly compact.
Definition3.22. A functionf:X→Y is said to beδ-continuous[14] if for each
x∈Xand each open set V ofY containing f (x), there exists an open setU in X
containingxsuch thatf (U¯◦)⊂V¯◦.
Theorem3.23. Iff:X→Yis a.p.c. andg:Y →Zisδ-continuous, theng◦f:X→Z
is a.p.c.
Proof. The proof is obvious and is omitted.
Theorem3.24. Iff:X→Y is anM-preopen surjection andg:Y→Zis a function such thatg◦f:X→Zis a.p.c., thengis a.p.c.
Proof. Lety ∈Y and x∈X such that f (x)=y. LetG be a regular open set containing(g◦f )(x). Then there existsU∈PO(X,x)such thatg(f (U))⊂G. Sincef
isM-preopen,f (U)∈PO(Y ,y)such thatg(f (U))⊂G. This shows thatgis a.p.c. aty.
Theorem3.25. Iff:X→Y is a.p.c. andAis a semi-open set ofX, then the restric-tionf|A:A→Y is a.p.c.
semi-open inX, it follows from [11, Lemma 2.1] thatA∩f−1(V)∈PO(A). Therefore,
f|Ais a.p.c.
Remark3.26. It should be noted that every restriction of an a.p.c. function is not necessarily a.p.c. In [15, proof of Theorem 6.2.5], it is pointed out that there is a precontinuous function whose restriction to a not semi-open set is not even a.w.c. It might also be noted that neither is almost precontinuity for a functionf:X→Y
preserved by restriction of the codomain tof (X). The following example is due to referee.
Example3.27. Letf:Q→Rbe the inclusion map of the rationals into the reals. Let the domain have the usual subspace topology and let the nonempty open sets in the codomain have the formP∪A, whereP=R−Qis the set of irrationals and where
A⊆Q. ThenRsis indiscrete so thatfis a.p.c. Yet,f (Q)is a discrete subspace ofRso thatf:Q→f (Q)is not a.p.c. since not every subset of the domain space is preopen. Theorem3.28. Letf:X→Y be a function andx∈X. If there existsU∈PO(X,x)
such that the restriction off toUis a.p.c. atx, thenf is a.p.c. atx.
Proof. Suppose thatV2 is any regular open set containingf (x). Since f |U is a.p.c. atx, there existsV1∈PO(U,x)such thatf (V1)=(f|U)(V1)⊂V2. SinceU∈ PO(X,x), it follows from [11, Lemma 2.2] thatV1∈PO(X,x). This shows clearly that
fis a.p.c. atx.
Definition3.29. LetA⊂X. ThepreboundarypFr(A)ofAis defined by pFr(A)= Pcl(A)∩Pcl(X−A).
Theorem3.30. The set of all pointsxofXat whichf:X→Yis not a.p.c. is identical with the union of the preboundaries of the inverse images of regular open subsets ofY
containingf (x).
Proof. Iffis not a.p.c. atx∈X, then there exists a regular open setVcontaining
f (x) such that for every U ∈PO(X,x), f (U)∩(Y−V )≠∅. This means that for everyU∈PO(X,x), we must haveU∩(X−f−1(V ))≠∅. Hence, it follows from [2, Lemma 2.2] thatx∈Pcl(X−f−1(V)). Butx∈f−1(V )and hencex∈Pcl(f−1(V )). This means thatxbelongs to the preboundary off−1(V). Suppose thatxbelongs to the preboundary off−1(V
1)for some regular open subsetV1ofY such thatf (x)∈V1. Suppose thatf is a.p.c. atx. Then there exists U∈PO(X,x)such thatf (U)⊂V1. Then, we have:x∈U⊂f−1(f (U))⊂f−1(V
1). This shows thatxis a preinterior point off−1(V
1). Therefore, we havex∉Pcl(X−f−1(V1))andx∈pFr(f−1(V1)). But this is a contradiction. This means thatf is not a.p.c.
Recall that a subsetAof a spaceXis said to beH-set[25] orquasiH-closed relative toX [21] if for every cover{Ui|i∈I}ofAby open sets ofX, there exists a finite subsetI0ofIsuch thatA⊂ ∪{U¯i|i∈I0}.
Theorem3.31. Iff:X→Y is a.w.c. andKis strongly compact relative toX, then
f (K )is quasiH-closed relative toY.
Recall that a functionf:X→Y is calledr-preopen[3] if the image of a preopen set inXis open inY.
Theorem3.32. Letf:X→Y be an a.w.c. bijection. IfXis strongly compact andY
is Hausdorff, thenf isr-preopen.
Proof. Suppose thatUis a preopen subset ofX. ThenX−Uis preclosed subset of the strongly compact spaceX. This means thatX−Uis strongly compact relative toX. By Theorem 3.31,f (X−U)is quasiH-closed relative toY. Sincef is bijective, we havef (X−U)=Y−f (U), whereY−f (U)is quasiH-closed relative toY. SinceY
is Hausdorff, thereforeY−f (U)is closed inY. Hencef (U)is open inY.
Corollary3.33. Letf:X→Y be an a.p.c. bijection. IfXis strongly compact and
Y is Hausdorff, thenf isr-preopen.
Proof. Since every a.p.c. function is a.w.c., hence the proof follows from Theo-rem 3.32.
Definition3.34. LetEandFbe any two subsets ofX.EandF are calledstrongly
p-separatedif there exist disjoint preopen setsUandVsuch thatE⊂UandF⊂V. Definition3.35. A functionf:X→Y is said to bestrongly preclosed[18] if the image of a preclosed set inXis preclosed inY.
Definition3.36. A spaceXis calledstrongly prenormal[18] if for disjoint pre-closed subsetsEandFofX, there exist disjoint preopen setsUandVsuch thatE⊂U
andF⊂V.
Theorem3.37. Iffis an a.p.c., strongly preclosed function of strongly pre-normal spaceXonto a spaceY, then any two disjoint regular closed subsets ofYcan be strongly
p-separated.
Proof. Let F and D be two disjoint regular closed subsets of Y. Then f−1(F) and f−1(D)are disjoint, preclosed subsets of the strongly prenormal spaceX and therefore there exist preopen setsU andW such thatU∩W = ∅, f−1(F)⊂U, and
f−1(D)⊂W. Suppose that
P1=y|f−1(y)⊂U, P2=y|f−1(y)⊂W. (3.1) Sincef is strongly preclosed, thenP1andP2are preopen sets. Then we have
F⊂P1, D⊂P2, P1∩P2= ∅. (3.2)
Now we obtain the following results whose proofs are omitted since they are straight-forward.
Recall that a spaceXis said to beextremally disconnectedif the closure of each open set ofXis open inX.
Acknowledgement. The authors are very grateful to the referee for his careful work and suggestions that improved this paper very much.
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Saeid Jafari: Department of Mathematics and Physics, Roskilde University, P.O. Box 260 4000Roskilde, Denmark
E-mail address:[email protected]
Takashi Noiri: Department of Mathematics, Yatsushiro College of Technology, Yatsushiro-shi, Kumamoto-ken,866, Japan
