Subcontra continuous functions

VOL. 21 NO. (1998) 19-24

SUBCONTRA-CONTINUOUS FUNCTIONS

C.W.BAKER

Department

of Mathematics IndianaUniversitySoutheast

New

Albany, Indiana 47150

(Received December 27, 1996)

ABSTRACT.

A weak formof contra-continuity, called subcontra-continuity, is introduced. It is

shown that subcontra-continuity is strictly weaker than contra-continuity and stronger than both subweakcontinuityandsub-LC-continuity. Subcontra-continuityisused toimprove severalresults in the literatureconcerning compact spaces.

KEY WORDS AND PHRASES:

subcontra-continuity, contra-continuity, subweak continuity,

sub-LC-continuity.

1991AMS

SUBJECT CLASSIFICATION CODE:

54C10

1.

INTRODUCTION

In

Dontchev introduced the notion of a contra-continuous function.

In

thisnote wedevelop

a weak form of contra-continuity, which we call subcontra-continuity. We show that subcontra-continuityimplies both subweak continuityand sub-LC-continuity. We also establish some of the

properties ofsubcontra-continuous functions.

In

particularit is shown that thegraph ofa subcontra-continuous function into a Tl-space is closed. Finally, we showthatmany of the applications of contra-continuous functions to compact spaces established by Dontchev

[1]

hold for

subcontra-continuous functions.

For

example, we establish that the subcontra-continuous, nearly continuous

image ofanalmostcompact

space

iscompactandthatthe subcontra-continuous,

-continuous

image ofanS-closed spaceiscompact.

2.

PRELIMINARIES

The

symbols

X

and

Y

denote topological spaceswith no separationaxiomsassumed unless

explicitly stated. The closureand interiorofasubset

A

ofaspace

X

are signified by

Cl(A)

and

Int(A),

respectively.

A

set

A

is regular open (semi-open, nearly open) provided that

A

Int(Cl(A))(A

C_

Cl(Int(A)),

A

C_

Int(Cl(A)))

and

A

is regular closed

(semi-closed)

if its

complementisregular open (semi-open).

A

set

A

islocally closed providedthat

A

U

N

F,

where

U

isanopenset and

F

is aclosedset.

DEFINITION

1. Dontchev

[1

]. A

function

f

X-,

Y

is said to contra-continuousprovided

that forevery

open

set

V

in

Y,

f-1

(V)

isclosedin

X.

20 C.W. BAKER

DEFINITION

3. Ganster and Reilly

[3].

A

function

f"

X-.

Y

is said to be

sub-LC-continuousprovided thereisanopenbase3for thetopologyon

Y

suchthat

f-(V)

islocallyclosed

for every

V

6/3.

i)EFINITION 4.

A

function

f"

X

Y

is said to be semi-continuous

(Levine

[4])

(nearly

continuous

(Ptak

[5]),

3-continuous

(Abd

EI-Monsefet al.

[6]))

if for every

open

set

V

in

Y,

f-(V)

C_

Ul(It(f-(V))) (f-(V)

C_

It(Ul(f-(V))), f-(V)

C_

Ul(It(Cl(f-(V))))).

DEFINITION

5.

Gentry

andHoyle

[7].

A

function

f"

X--,

Y

is said tobec-continuousif,

forevery :r

X

and every openset

V

in

Y

containing

f(z)

and withcompact complement,there exists anopenset

U

in

X

containingzsuch that

f

(U)

C_

V.

3.

SUBCONTRA-CONTINUOUS FUNCTIONS

Wedefine afunction

f

X-

Y

tobe subcontra-ontinuous provided thereexists an

open

base

B

for the topology on

Y

such that

f-(V)

is closed in

X

for every

V

/3. Obviously

contra-continuity implies subontra-ominuity. The followingexample shows thatthe reverse implication

doesnothold.

EXAMPLE

1. Let

X

be a nondiscrete

Tx-space

and let

Y

be the set

X

with the discrete

topology. Finallylet

f

X-

Y

be the identitymapping. If/3isthe ollection of all singleton subsets

of

Y,

then/3 is anopen base forthe

topology

on

Y.

Since

X

is

T,

f

is subcontra-ontinuous with

respectto/3. Obviously

f

is not contra-continuous.

Subcontra-ontinuity is independentof continuity. The function inExample is subcontra-continuousbutnot continuous. The nextexample showsthat ontinuity does notimply subcontra-continuity.

EXAMPLE

2

Let

X

{a, b}

bethe Sierpinski

space

withthe topology

T

{X,

,

{

a

} }

and let

f

X--,

X

betheidentity mapping. Obviously

f

is continuous.

However,

any openbase for the

topology

on

X

must contain

{a}

and

f-({a})

is not closed. It follows that

f

is not subcontra-continuous.

Sinceclosedsetsare locally

closed,

subcontra-continuity impliessub-LC-continuity.

We

see

from the

following

theorem that subcontra-cominuity also implies subweak continuity.

THEOREM

1.

Every

subcontra-continuous function issubweaklycontinuous.

PROOF. Assume

f

X--,

Y

is subcontra-continuous. Let/ be an

open

baseforthetopology

on

Y

for which

f-’(V)

is closed in

X

for every

V

/. Then for

V

/,

Cl(f-X(V))

f-(V)

C_

f-(l(v))

andhence

f

issubweaklycontinuous. I-!

Since asubweaklycontinuousfunctioninto aHausdorff

space

hasaclosed

graph (Baker [8]),

a subcontra-continuous function into aHausdorff space has aclosed

graph. However,

the following stronger result holds forsubcontra-continuous functions.

THEOREM

2. If

f

X--,

Y

isa subcontra-continuous function and

Y

is

T,

thenthegraphof

f, G

(f),

isclosed.

PROOF.

Let

(x,y)

6

X

x

Y

G(f).

Theny

# f(x).

LetB

bean

open

basefor the topology

on

Y

forwhich

f-x

(V)

isclosed in

X

for every

V

6

B.

Since

Y

is

T,

thereexists

V

6

B

suchthat

y

e

v

and

f(x)

:

V.

Thenwesee that

(x,

y)

6

(X

-/-a(v))

x

V

c_

x

x

Y

G(I).

t

follows

that

G(/’)

isclosed, l’-!

COROLLARY

1. If

f"

X

Y

is contra-continuous and

Y

is

T,,

then the graph of

f

is

closed.

Long

and Hendrix

[9]

proved

that the closed

graph

property

implies

c-continuity. Thereforewe have thefollowing corollary.

Thenext tworesultsarealsoimplied bytheclosed graph property

(Fuller 10]).

COROLLARY

3. If

f

X- Y

issubcontra-continuous and

Y

is

T1,

thenforevery compact

subset

C

of

Y,

f-l(u)

isclosedin

X.

COROLLARY

4. If

f

X

Y

issubcontra-eontinuous andI/is

T1,

thenfor every compact subset

C

of

X,

f

(C’)

isclosed.

Forafunction

f"

X--,I/, the graphfunction of

f

is the functiong"X-,X xI/given by

g(x)

(z,f(x)).

We

shall see in the following example that the graph functionofa

subeontra-continuousfunction is notnecessarilysubcontra-continuous.

EXAMPLE

3.

Let

X

{a,b}

bethe Sierpinski spacewiththe topology

T

{X,0,

and let

f"

X--,

X

begivenby

f(a)

b and

f(b)

a. Obviously

f

is subeontra-continuous, in fact eontra-continuous. Let/3 be anyopenbasefortheproduct topology on

X

x I,’. Then there exists

V

E/3forwhich

(a,b)

V

C_

{(a,a),(a,b)}.

We

seethat

V=

{(a,a), (a,b))

andthat, ifg"X-,

X

x

X

isthe

graph

functionfor

f,

then

g-l(V)

{a}

which is notclosed. Thusthegraphfunction

of

f

is not subeontra-continuous.

However,

thefollowing result does hold forthegraphfunction.

THEOREM3. Thegraphfunctionof a subcontra-continuous function is sub-LC-continuous.

PROOF.

Assume

f"

X- Y

is subcontra-continuous and letg"

X

-

X

x I/bethe graph

function

of

f.

Let/3 be anopenbasefor thetopologyon

Y

forwhich

f-(V)

isclosedin

X

for every

V

/3. Then

{U

x

V

U

isopenin

X,

V

/3}

isanopenbasefortheproducttopologyon

X

Y.

Since

g-(U

x

V)

U

n

f-l(V),

weseethat gis sub-LC-continuous.

El

Thegraph function ofasubweaklycontinuous function issubweaklycontinuous

(Baker

[8])

and thegraphfunctionofasub-LC-continuous function is sub-LC-continuous

(Ganster

and Reilly

[3]).

It follows that thegraphfunction inExample 3 is subweaklycontinuousand sub-LC-continuous but not subcontra-continuous. Therefore subcontra-continuity is strictlystrongerthan sub-LC-continuity and subweak continuity.

THEOREM

4. If

Y

is aTl-spaceand

f

X-,

Y

is a subcontra-continuousinjection,then

X

is

T.

PROOF.

Let

x

_{and x2}bedistinctpointsin

X.

Let/3beanopen base for the topologyon

for which

f-(V)

is closedin

X

for every

V

/3. SinceI/is

T1

and

f(:rl)

4

f(:r2),

there exists

V

/3 such

thatf(x)

V

and

f(x)

E

V.

Then aq

.

X-

f-(V)

which is open and

2

X-/-I(v).

D

THEOREM

5.

Let

A

C_

X

and

f-X-

X

be a subcontra-continuous function such that

f(X)

A

and

flA

isthe identityon

A.

Then,if

X

is

TI,

A

isclosedin

X.

PROOF.

Suppose

A

is not closed.

Let

z

Ul(A)

A.

Let/3 be an open base for the

topology

on

Y

forwhich

f-I(V)

isclosed for every

V

/3. Since:r

A,

wehavethata:

f(:r).

Since

X

is

T

1,there exists

V

/3suchthat x E

V

and

f(x)

V. Let U"

be an

open

setcontainingz.

Then:r

U

f

V

which isopen. Since z

UI(A), (U

f

V)

q

A

.

Let

_y

(U

_f

V)

q

A.

Since

y

A,

f(v)

V

V. So

V

f-l(V)

ThusV

U

f

f-I(V)

and hence

U

f-l(V)

:/:

.

We

see that z

C’[(f-I(v))

f-I(V)

which is a contradiction. Therefore

A

isclosed. Ill

Thenextresult follows easily forthe definition.

THEOREM

6.

If

f

X- Y

is subcontra-continuous, then for every

open

set

V

in

Y,

f-l(v)

is aunionof closedsetsin

X.

22 C.W. BAKER

EXAMPLE

4. Let

X

R

with theusual topologyand let

f

X

X

betheidentity mapping.

Since

X

isconnected,

f

isnotsubcontra-continuous. However,since

X

is

Tx,

f

hasthepropertythat the inverseimage of every

(open)

setis a unionof closedsets.

4.

APPLICATIONS TO COMPACT SPACES

In

[1]

Dontchev establishes that the image of an almost compact

space

under a

contra-continuous,nearlycontinuousmappingiscompactand that the contra-continuousimage ofastrongly

S-closedspace iscompact.

In

thissection,westrengthen bothof theseresultsby replacing

contra-continuitywithsubcontra-continuity. The

proofs

mostly follow Dontchev’s.

DEFINITION

6. Dontchev

].

A

space

X

isalmost compact providedthatevery

open

cover

of

X

hasa finitesubfamily the closures of whose memberscover

X.

THEOREM

7. Theimageofanalmost compact

space

under a subcontra-continuous,nearly

continuousmappingiscompact.

PROOF. Let

f

X

Y

be subcontra-continuous and nearlycontinuous and assume that

X

is almostcompact.

Let

B

beanopen base forthetopologyon

Y

forwhich

f-l(v)

closedin

X

for every

V

6

B.

Let

C

be anopencover of

f(X).

For

eachz6

X,

let

C

C

suchthat

f(z)

C.

Then let

Vz

B

for which

f(z)

Vx c_ C’x.

Now

f-X(Vz)

is closed andnearlyopen. It followsthat

f-l(V)

is

clopen and hence that

{f-X(Vx)

:c6

X}

is aclopen cover of

X.

Since

X

isalmost compact, thereis afinitesubfamily

{f-X(Vx,):

i=

1,...,n}

forwhich

X

U

Cl(f-l(vz,))

U

f-x(Vx,)

c_

i=I i=I

[3

f-1(Cz,).

Thuswehave tha

f(X)

c_

[3 C,

and therefore tha

f(X)

iscompact [3

i=1 i=1

DEFINITION

7. Dontchev

[1].

A space

X

isstrongly S-closed providedthatevery closed

coverof

X

has a finitesubcover.

THEOREM

$. Thesubeontra-ontinuousimage ofastrongly S-closed spaceiscompact.

PROOF.

Let

f

X

-

Y

be subcontra-continuous and assume that

X

isstrongly S-closed.

Let

B

beanopen base for the topologyon

Y

forwhich

f-(V)

is closed in

X

forevery

V

B.

Let C

be anopencoverof

f(X).

For

eachz

X,

let

6’

C

with

f(z)

6

C.

Thenlet

V

6

B

forwhich

f(z)

V

C_

C.

Since

{f-X(Vx)

z

X}

is aclosed cover of

X

and

X

isstrongly S-closed, thereis afinitesubcover

{f-X(V,)

l

n}

of

X.

Thenwe seethat

f(X)

f

(

i=1

I,J f

(f-1

(Vz,))

c_

U

Vx,

C_

U

Cx,

andhencethat

f

(X)

iscompact.

E!

i=1 i=1 i=1

In [1]

Dontchevalso shows that the contra-continuous,/-continuous image ofan S-closed

space

is compact.

We

extendthis result by replacing contra-continuity withsubcontra-continuity.

The

proof

parallelsthatof Dontchev’s.

DEFINITION

8. Mukherjee and

Basu [11 ].

A space

X

isS-closedprovidedthat every

semi-open

cover of

X

has a finitesubfamilythe closuresofwhosememberscovers

X.

From Herrmann

[12],

aspace

X

isS-closedifandonlyifevery regular closedcover of

X

has afinitesubcover.

THEOREM

9. The subcontra-continuous,

/-continuous

image of an S-closed

space

is

compact.

PROOF. Assume

that

f

X--,

Y

issubcontra-continuous

and/-continuous

and that

X

is

S-closed.

Let

B

be an

open

base forthe

topology

on

Y

forwhich

f-X(V)

is closedin

X

for every

V

B.

Let

C

be an

open

coverof

f(X).

Then for each z

X

there exists

C

6 for which

f(z)

E

Cz.

For

eachz

e

X,

let

Vx

B

suchthat

f(z)

6

V

C_

Cx.

Since

f

issubcontra-continuous,

{f-(Vx):xX}

is a closed cover of

X.

The ]-continuity of

f

implies that

SUBCONTRA-CONTINUOUS FUNCTIONS 23

f-X(Vx)

is

regular

closed. Since

X

is S-closed, theregularclosed cover

{f-X(Vx)

:z E

X}

has a

finitesubcover

{f-l(V,)

1

n}.

Thenwehave

I(X)

I

U

J’-l(Vx,)

-

U

V:,

_

i=1 i=1

U

C,

andtherefore

J’(X)

is compact.

KI

i=1

In

the aboveproofweshowedthat, if

J’

X

Y

is subcontra-continuous and-continuous,

then there exists anopen base

B

forthetopologyon

Y

suchthatfor every

V

E

B,

f-l(v)

isregular closedandhence semi-open. Since unionsof semi-opensetsaresemi-open (Arya andBhamini

[13]),

itfollowsthat inverseimages of

open

setsare semi-open. Thereforewehave thefollowing theorem

whichstrengthensthe correspondingresult for contra-continuous functions establishedbyDontchev

[11.

TI-IEOREM

10.

Every

subcontra-continuous,/-continuousfunction is semi-continuous.

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J.

Contra-continuousfunctionsand strongly S-closed spaces,

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&

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Rose, D.

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I.

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N.

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C. W.

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