On the δ continuous fixed point property

ON

THE

CONTINUOUS FIXED

POINT PROPERTY

F. CAMMAROTO DipartimcntodiMammatica UniversiuldiMessina 98166Sant’Agata 0Vlessina) ITALY

and T. NOIRI

DepartmentofMathematics Yatsushiro College of Technology Yatsushiro,Kumamoto

866JAPAN

(Received July 7, 1988 and in revised form November 6, 1988)

ABSTRACT. In

this

paper,

wedefineand investigate the continuous retraction andthe-continuousfixed pointproperty.Theorem ofConnell

11]

and Theorem3.4 of

Arya

andDeb

[2]

are improved.

KEY WORDS ANDPHRASES.

5-continuous, 0-continuous,

weakly-continuous, semi-regular,

almost-regular, Voted point property.

1980

MATHEMATICS SUBJECT

CLASSIFICATION

CODE.

54

C

10, 54

C

20.

0.

INTRODUCTIQN..

The notionof 0-continuous functions wasflu’stintroducedbyFomin i].Afterthat,thisnodonhas been widely investigatedintheliterature.

By

utilizing

0-continuous functions,

Arya

andDcb

[2]

definedand

investigated

the 0-continuous retraction,

the 0-continuous

fixed

point

property

and

the 0-continuous

homotopy.

On

theotherhand,in

[3]

and

[4]

thepresent authors haveindependentlyintroduced the notionof8--continuous functions.The

purpose

of this

paper

isto

apply

8-continuitytothe retraction and thefixedpointproperty.

In

1.

PRELIMINARIES.

Throughoutthe present

paper, spaces

willalwaysmean topological

spaces

on which noseparation axie:ns areassumedunless

explicitly

stated.

We

shall denotea

topological space by (X, x)

or simply by

X.

Let (X, x)

be a

space

and

A

asubsetofX.The closure of

A

andtheinteriorof

A

aredenoted by

,

and

’

or

simply

,

and

k),

respectively.

A

subset

A

of

X

is saidtoberegular open (resp. regular closed)if

A=(,)

(resp.

A

A"

).

Thefamilyofregular opensetsof

X

will be denotedby

RO(X). A

point x of

X

is saidto be in the

S-

closure[5]of

A,

denoted by

CI(A),

if

A V #

forevery

V

RO(X)

containingx. Asubset A is said to be 8-closed_[5] if

A=CIn(A).

Thecomplement of a-closed setis saidtobe 5-open.The topologyon

X

whichhas

RO(X)

as abasis is called the.semi-regularization of x and is denotedby

:’.

Itis obviousthat

every

element of

’"

is a

5

open setof

(X,

x). A space (X, x)

is saidtobesemi-regular if x

:’.

A space

(X,

x)

is said to be almost-regular

[6]

if for each regular closed set

F

and each x

X

F,

there exist

open

sets

U

and

V

such that

x

U, F C V

and.

U r V

DEFINITION

1.1.

A

functionf"

(X,

x)--- (Y, o)

issaidtobe -continuous 3, 4

(resp.

almost-continuous 7 ],O-continuous and

weakly

continuous

[8]

iffor each

x

X

and each

open

neighborhood

V

of f(x), there exists an

open

neighborhood

U

of x such that f

U C V

(resp. f

U C

V,

f

U C V

and f

(U) C V).

REMARK

1.1.

It

isshownin 2, 3, 9 thatthefollowing implications hold: 8-continuous almost-continuity

=

O-continuous =#weak-continuity, where noneof these implicationsisreversible.

2. _5-

CONTINUOUS

RETRACTIONS.

Arya

and Deb

[2]

defined a subset

A

ofaspace

X

tobe a 0-continuous retractof

X

if there exists a 0-continuous function f"

X

---)

A

such that f/A is the identity on

A.

We

shall similarly define a

5-continuous retract.

DEFINITION

2.1.

A

subset

A

of

space X

is called a -continuousretract of

X

if there exists a 5-continuous function f"

X

---)

A

suchthat fistheidentityon

A,

that is,f(x =x forevery x A.

Andsuch a function f iscalled a 8-continuous retraction.

REMARK

2.1.

It

is obvious thatevery &continuousretract is a 0-continuousretract.However, every i-continuousretractisnotnecessarilyacontinuousretractasthefollowing exampleshows.

EXAMPLE

2.1.

LetX

{a,

b,c,

d}

and

{,X, {a}, {a, b}, {a, c}, {a,

b,

c} }. Let A

{a,

b,

c}

and f" (X, x

(A,

x /

A)

beafunction defined asfollows"

f(a)

a,

f(b)

b,

f(c)

candf(d) d. Then

A

is a -continuousretractof

X

butit isnotacontinuousretractof

X

since

f-I

a

})

x for a x /

A.

REMARK

2.2.

In Example

3.1 of

[2],

Arya

and Deb showedthatevery 0-continuousretractsis not necessarily a continuous retract.

However,

this example is false.The 0-continuous functionf"X---)

A

in

[2,Example 3.1]

is necessarilycontinuous since the subspace

A

is discrete and regular. Since every 5-continuous function is 0-continuous,

Example

2.1 also shows thatevery0-continuous retract is not a continuousretract.

We

shall investigate relationships between -continuousretractand continuousretract.

PROPOSITION

2.1. If

X

is a semi-regular

space

and

A

is a continuous retract of

X,

then

A

is a 5-continuousretractof

X.

PROOF.

Thisfollows from the fact thatacontinuousfunctionfromasemi-regular

space

is &continuous

[3,

Prop.

1.5].

LEMMA

2.1. If

A

is either

open

or dense in a

space X

and

Ve

RO(X),then VC3A isregularopen inthe

subspace A.

PROOF.

If

A

isdense in

X,

then thisfollows from[10, p. 175, B)].

Next,

supposethat

A

isopenin

X

and

Ve

RO(X).

Then, we have

Moreover,

we have

V

A

c

A

D

(V

A

)’

A

V

A .On

the other hand,

*__

-VA

AC

(vX)’q

A

VA

WA.

Th.erefore,

we obtain

Vr

(A)=

Vrh

A

and hence

V

A

isregular

open

in

A.

PROPOSITION

2.2.

Let

X

be a semi-regular space and

A

eitheropenor dense in

X.

Then

A

is a continuousretractof

X

ifandonlyif

A

isa i-continuousretractof

X.

PROOF. From

Lemma 2.1, for

A

either

open

dense and

X

semiregular, x x* and

(’/A)

(’*/A)

C__

(’/A)*

C__

(r/A).

Thcrcforc,

A

isscmircgularsothat f"

X

A

is &continuous if andonlyif it is continuous.

PROPOSITION

2.3.

Lct X

be a

spacc

and

A

ascmi-rcgular

(rcsp.

almost-rcgular) subspacc of

X.

If

A

is a 5-continuous

(rcsp.

continuous)retractof

X,

thcn it is a continuous(rcsp. continuous)rctractof

X.

PROOF.

Lct

f’X

A

be a 5-continuous rctraction and

A

be.

scmi-rcgular.

Evcry

5-continuous function into ascmi-rcgular

spacc

is continuous

[3, Prop.

1.4 ]. Thcrcforc,

A

is a continuous rctract of

X.

Evcry

continuousfunction intoanalmostregular spaccis &continuous[3,

Prop.

1.8].Thcrcforc, thc sccond part follows.

THEOREM2.1. IfA is a -continuous retractofX andB is a -continuousrctractofA,thcn B is a &continuous retractof

X.

PROOF. Lct

f"

X.-->

A

and

g

A---B

bc 5-continuous retractions. Thccompositcfunctiongof"

X---

B

is 5-continuous [3,

Prop.

3.2].

Moreover, we have gof)

(x)

g(f(x))

g(x

x for

every

xe

B C A.

Therefore, gof"

X.-o

B

is a 5-continuous retraction and hence

B

is a 5-continuous retractof

X.

THEOREM

2.2.

A

subset

A

ofa

space X

isa 5-continuousretractof

X

ifand onlyif forevery

space Y,

every 5-continuous function f"

A.-o

Y

canbeextendedto a 5-continuous of

X

into

Y.

PROOF.

Necessity.

Let

g:

X

A

be a i- continuous retraction.Let

Y

beany spaceand f"

A--->

Y

be any5-continuous function.Thencompositefunctionfog"

X--->

Y

is i-continuous[3,

Prop.

3.2]. Moreover, wehave(fog) x f (g(x)) f(x for every xe A. Therefore,f ogis an extension of f.

Sufficiency. Let _A

A

..-->

A

be the identity function on

A.

Then

iA

is

-

continuous and henceby the hypothesis there exists a

&

continuous function g"

X--o

A

such that

g/A

A.Therefore,

A

is a

5-

continuous retract of

X.

THEOREM

2.3. If

A

isa

5-

continuousretractofaHausdorff

space X,

then

A

is 5-closed in

X.

PROOF. Let

f:

X

A

be a

5-

continuous retraction.

Suppose

that

A

isnoti-closed in

X.

Thereexists apoint xe

C1,

(A)

A.

Since

x

e

A, f(x

:

x

and hencethere exist

open

sets

U

and

V

suchthat

x

e

U,

f(x

e

V

and

U

V

hence

U

V

.Let W

be

any

regular

open

set

containin.g

x.Then

U W

is a regular open set containing x.Sincexe

CI

(A),

U

W

A

:

q).Let a e

U

W

A, then

f(a)

ae

U

and hence

f(a)

e

V.

Thisshows that

f(W)

q

V

for any regular openset

W

containing x. This contradicts thefactthat f is 5-continuous.

3.

THE

5-CONTINUOI,/S

FIXED

PQINT pRQPERTY,

Arya

and Deb

[2]

defined a space

X

to have the 0-continuous fixed point property if,forevery 0-continuous function f"

X--

X,there exists anxe

X

suchthatf(x x.Weshall define the 5-continuous (resp. weakly continuous)fixedpointproperty asfollows

DEFINITION

3.1.

A

space

X

is saidtohave the5-continuous(resp. weakly continuous fixed point property, briefly denotedby i

cFPP

(resp. wcFPP),ifforevery5-continuous(resp. weakly continuous) function f"

X---> X,

there existsan xe

X

such that f(x x.

We

giveanexamplethat aspace

,,,

lththefixedpoint property nednothave the 8cFPP.

EXAMPLE

3.1.Let X a,b,c and z

q,

X, a}, a, b}, a,c }.]’henthespace (X, ,)hasthe fix pointproperty[2, Example 3.2].

Now,

let f:(X, ) + (X, z)beafunctiondefinedby

f(a)

f(c)=bandf(b). Then f is S-continuousbut doesnotafixedpoint. Therefore, (X,) doesnothave theFPP.

RERK

3.2.

We

needthefollowingtwo

spaces

which we wereunabletoobtain:

(1)

aspacewhichhas

FPP

but dsnothavethe fixedpointproperty. (2)aspacewhichhas the 0cFPP but doesnothavethewcFPP.

TOM

3.1.

t

A eitheropenordensein aspace X. "if

X

has the

FPP

and

A

isa &continuous rewactofX,then

A

has the

FPP.

PROOF. t

f:

A

₊

A

any S-continuous function. Since

A

is a 6-continuous rcmct of X, by Theorem 2.2 fcan extendedto a &continuous function F

X

A.Let

A+

X

the inclusion. IfV isaregulopensetof

X,

thenj;(V)

A

V isregular openinthe

sdbspace A

by

Lemma

2.1.Therefore,

F

-

₀

(V))=

0oF)(V)

is

&open

in

X

and hencejoF:X

+X

is &continuous. Since

X

has the

FPP,

x

0oF) (x)

=j

(F(x))

=j

(f(x))

f(x) for somexe

AC

X.

Thisshowsthat

A

has the

FPP.

Thefollowing theorem is aslightmificationof Theorem of 11].

TOREM

3.2.

t

(X,) an almost-regular space with the 8cFPP. If is a topology for

X

onger

than and

(*)=

()

tbr

eve

G

e o, then

X, o)

has the fixpointproperty.

PROOF. Suppose

that f: (X, ) + (X, ) is any continuous function. t g (X, )+ (X, z)and h (X,g)_{+ (X, z)} the functions defined by g(x h(x f(x

tr

eve

xeX. Let (X, z)_)(X,)

theidentityfunction.

en,

since C

,

is anopen

b0ection.

Moreoversince f ogiscontinuous,g is

contuous. Next,we

shall show that h is &continuous.Let xe

X

andh(x )e

Ve

RO(X, ).Since(X,) isNmost-regul,there exists

G

e such that h(x)e

G C

()

C

V.

Smcc

g iscontinuous,

ga

(G)e oand h

-

(G)

P

(G)

g

[G).

Therefore, h

(G)

f

(G)

e o and hence,utilizing continuity of f we obtNn

{*)

e

Ue RO

X, andh

C V

Ths shows that h s 8continuous

Now, wesetU=h-(G)

,thenwehave

x

)"

(U)

".

Since(X, ,) has

e

FPP,

there existsxe

X

suchthatx h(x f(x ). Thisshowsthat (X, o)has the fix pointproperty.

COROLLARY

3.1 (Connell 11 ).

Suppose

(X,,) is aregularspacewiththe fixpointproperty. If o is atopologyforX,

zC

g_d

()

G

(*)_{for each}

G

e

,

then(X, o)has the fixpointproperty.

PROOF. It

isshown in[3,

Corollau

1.8]flat if

Y

isregular,then f

X

Y

is 8-continuous if andonly if is continuous. Since

eveu

regular spaceisahnost regular,this is an immediateconsequence oftheorem3.2.

We

shlgivealemmawhich will used in theproofof

tte

finaltheorem.

LEMMA

3.1.

t

f

X

+

Y

and g

Y

Z

be functions:

(1)

f iswetlycontinuousfandonlyif

f- (V) C f (V)

foreachopenset

V

of

Y.

(2)

If compositego f:

X Z

isweaklycontinuousandg

Z

is anopen bijection,then fis

wey

continuous.

PROOF. Statement (1)

is

eorcm

7 of[

121.

We

shall showStatement (2) by utilizing Statement (1).

Let

V any opensetof Y.Since g isopen,gCV)isopenin

Z

and

(go

₀

(g(V))

C

(go

-

(g(V)). Since g s bjecnve, (g o

f) (g(V))

f (V). Moreover,since g isopen,

(g

o

0

(g(V))

f

(g (g (V))) C

f

gi-

’g(9))

f (V).Consequently,

we obtain ";

(V) C

q

CV)

and hence fiswe&lycontinuous.

e

following theoremisanimprovementof[2.Tlcorem3.4]and 11, Theorem

!.

THEOM

3.3.Let(X,z)bearegulspacewith the fixed point property. If isa topology for

X

stronger than z

and()

(tbr

eveu

Ge

,

en

(X, o)has thewcFPP.

h:(X, ’r)

(X,

"r)

and i: (X,

’r)

(X, o)

be the same functions as in Proof ofTheorem 3.2. Since :f=i o

g

is weaklycontinuous and is an

open

bijection,

g

is

weakly

continuous

by Lemma

3.1.Since (X, :)isregular, giscontinuous

[8]. Next,

weshall showthat h iscontinuous.

Let x

e

X

and

V

be anopen setof (X,

")

containingh(x). Since

(X, x)

isregular, thereexists

G

e a: suchthat h(x

G C

()C

V.

Since g is continuous,

g-I

(G)

o and

ha(G)

fl(G)

g

(G). Therefore,we haveh

(G)

fl (G)

o.

Sind6"f

is

weakly

continuous,

by

Lemma 3.1

f

-t)

C f

(0o

(05).

It

follows fromthe sameargument asin Proof ofTheorem3.2that h is continuous.Since (X, a:) hasthefixedpoint property,thereexists apoint x

X

suchthat x

h(x

f(x ).

Thisshows that f hasthe fixedpoint property.

COROLLARY

3.2

(Arya

andDeb

[2]).

If

(X,

a:)is aregular spacewiththe fixedpointproperty and if o

((o)___

isatopologyfor

X

stronger than x such that

(

()for each

G

e

,

then(X, a)hasthe 0-continuous fixedpointproperty.

ACKNOWLEDGEMENT

This

paper

waswrittenduringthe second authorstayedin MessinaUniversity for

May

and

June

1988.

He

wouldliketothanktoC.N.’tl.and Messina University for itshospitality.Thesecond author’sresearch wassupported by

M.P.I.

"fondi40%"

(ITALY).

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Math.

44

(1943),

471-480. 2.

S.P.

Arya

and

Mamata

Deb,

On

0-continuous

mappings,

Math.

Student 42

(1974),

81-89. 3.

F. Cammaroto, On

5-continuous

and/5-open

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J. (submitted).

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Math.

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16

(1980),

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M.K.

Singaland

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Levine,

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