ON SPACES VIA DENSE SETS AND SMPC FUNCTIONS

Kyungpook Malhemalical Joum꾀

Vol.34

,

No.l

,

^109-116

,

June 1994

ON SPACES VIA DENSE SETS AND SMPC FUNCTIONS

D. A. Rose and R. A. Mabmoud

Density is one of the basic propcrties in topological spaces. So

,

^some

spac않 sucb as hyperconnected space

,

submaximal spacc and all resolvabil- ity spaces have been defined via lhis properly. Therefore

,

we devoled this paper for the study and investigated of new properties of these types of spaces. Also

,

a strong M-precontinuous function (abbrivaled as: SMPC) was characterized by using some previolls spaces. Finally, some effects of SMPC on olher spaces are studied.

1.

Defin itions and preliminaries

Topological spaces used here will not include any separation properties which are assumed unless they arc otherwise needed in which case lhey will be explicilly staled. Given a lopological space (X

,

T)

,

for any A 드 X

,

we denote the interior and the c10sure of A with respecl lo T

,

by IntA and CIA

,

respectively. A is called preopen [1] if A IntCIA

,

and PO(X

,

T) means the collection of all preopen sets in (X

,

r). For any space (X

,

r) let

감 be the smallest tO]lolgy on X containing PO(X

,

T). The topolgy r^O [2]

is PO(X

,

r)

n

SO(X

,

T) where A E SO(X

,

T) iff A is semi-open [3]. i.e.

A 드 CllntA. Thus

,

for any space (X

,

T)

,

T 드 ^TO

드

^PO(X

,

T) 드 Tp ' II is also known thal PO(X

,

T''')'= PO(X

,

r) (Corollary 1 of [4]).

For any topology T on X

,

the semil'egulal'ization of r is the topol ogy 감 having for a basis lhe set of regular open subsets of

(X ^,

^{r). The}

semiregular c1ass of T is the set [T], of all topologies on X having the same semiregularizalion as T. A topological property P is a semiregular property if il is shared by all members of [r]s when possessed by any one

Received June 9, 1992 Revised March 20, 1994

109

member. This is equivalent to saying that (X

,

T) and (X

,

T.) both have P whenever either does [5]. An a-topological property is any topologi- cal property shared by aU members of the a-class when possessed by any one member of the a-class. In particular

,

it is any topological property possessed by both (X

,

T) and (X

,

T") when possessed by either [5].

A space (X

,

T) is called hyperconnected [6] if each nonempty open set is dense. (X

,

T) is resolvable [7] if there is a dense subset D 드 X for which X - D is also dense. A space which is not resolvable is called irresolvable A subset of X is resolvable (irresolvable) if it is resolvable (irresolvable) as a subspace. A space is hereditarily irresolvable if each of its nonempty subsets is irresolvahle. (X

,

T) is submaximal [8] if each of its dense subsets are open. Also

,

(X

,

T) is strongly ∞mpact (strongly Lindelöf) [9] if for each preopen cover of X

,

there is a finite (countible) subcover. Spaces

pre낀 (i 0

,

1

,

2) are defined likewise the spacés

T;

(i

=

0

,

1

,

2) except that the open sets are replaced by preopen sets [10]

A function

f :

^(X

,

T) • (Y

,

a) is said to be precontinuous [1], preirres-

이ute [11] and strongly M-precontinuous [12] (SMPC) if the inverse image

。feach open

,

preopen and preopen in (Y

,

a) is preopen

,

preopen and open in (X

,

T)

,

respectively

Resolvable spaces

An important result related to the resolvable spaces which would be called the Hewitt-representation of (X

,

T) in [13]

Theorem 1 [13]. Every space (X

,

T) can be represented uniquelly as a disjoint union X = F U G where F is c/osed and πso/vable and G is open and hereditari/ν 11γ'eso/vab/e.

Hence (X

,

T) 상 reso/vable iJJ G

=

0 and (X

,

T) is heπditaπly iπ'eso/v­

able 퍼 F=

0.

Lemma 1. (Corollary 5 of [7]) If (X

,

T) is resolvable then Tp = 2x. Proof Let D) and D2 be disjoint dense subsets of X and let x E X.

Then D) U {x} and D₂ U {x} are dense and hence preopen. Thus {x} = (C) U {x})

n

(D2 U {x}) E 강·

Other properties of resolvable spaces have studied in [6] by using an anti-operation due to Bankston

Now

,

we give the condition under which that is property of being a resolvable space is hereditrary.

On spaces via dense sets and SMPC functions 111

Theorem 2. Each semi-open subset of a resolvable spaée is resolvable.

Proof Let A E SO(X

,

T)

,

i.e. A 드 ClIntA 드 X and X be resolvable Then IntA is resolvable and A-IntA is nowhere dense in (A

,

TjA). Thus

,

if Dl U Dz is a disjoint union of dense subsets of IntA then Do = D1 ^U (A-IntA) and Dz are disjoint and also are dense in A.

It is obvious that every subspace of a hereditarily irresolvable space is hereditarily irresolvable

,

whereas the converse follows nextly.

Theorem 3. If(X

,

T) is a space

,

X

=

^X

,

^{UXz and X}

,

^nxz

= 0

and X1 is

c/osed then ifboth (X1

,

_TjX1) and (XZ

,

TjXZ) ^are hereditarily irresolvable then (X

,

T) is hereditaπIy i^1"1'esolvable.

Proof Suppose that

ø

^{￥ A 드 X and (A}

^,

^{T / A) is} resolvahle. Then there exist disjoint

,

dense in A

,

subsets Dl and Dz with A ⁼ D

,

^{U D2 . Suppose}

that D1

n

Xz

#

0 and Dz

n x

2

#

0. Then since Xz is open in X

,

_D1

n

Xz and Dz

n

Xz are disjoint and dense in A

n

X2 . For if x E D2

n

Xz and V is open with x E V

,

since D1 is dense in A

,

V

n

A

n

D

,

^{폼 0. lf}

U is open in X and x E U then

,

for V

=

U

n

Xz

,

V E T and x E V so that U

n

A

n

(D

, ⁿ

^{Xz) ￥ 0. Thus}

^,

^D

, ⁿ

^{Xz and similarly Dz}

ⁿ

^X2

are dense in A

n

X₂ and disjoint. Thus

,

A

n

X₂ is a resolvable subspace of X2 which contradicts X2 being hereditarily irresolvable. Apparently

,

either Dl

n

X2

= ø

^{or D}2

n

Xz

=

0. But in either case A

n x

1 contains a dense set in A. Thus

,

CIA(A

n

X

t}

= A 드 An X

,

^{드 X}

,

^{since X! is}

closed. Thus A is a resolvable subspace of X₁ which cannot be since X1

is hereditarily irresolvable. This final contradiction proves that (X

,

T) is hereditarily irresolvable.

Theorem 4 [5J. Let f : (X

,

T) • (Y

,

0") be a bijection. Then f is a semihomeomorphism i

fJ

f is an a-homeomorphism

Where a bijection

f

^{(X ,}

T)

• (Y,O") is said to be a semihomeo morphism [14J if both

f

and

f - '

preserve semiopen sets. Any property transmitted by semihomeomorphisms is called semitopological [14J

By a previous result

,

a property P is semitopological if and only if P is an a-topological property and it is clear that the laUer holds if and only if X and X" both have P whenever either does [5J.

Since it is obviously that spaces (X

,

T) and (X

,

T"') share the same family of dense subsets

,

also resolvability is one of the a-topological and hence semitopological properties. This illustrates our belief that gener ally the best way to demonstrate that a property P is semitopological

is to show that it is a-topological. Whereas serniregular properties are a-topological

[15J.

Hence as the following example shows that semitopo logical properties are not semiregular. In particular

,

resolvability space.

Example 1. Let (X

,

T) be the two-point Sierpinski space. Then (X

,

T) is not resolvable whereas the indiscrete semiregularization (X

,

T s ) is resolv able.

3.

Hyperconnected and submaximal spaces

Lemma 2. (Proposition 1 of [7]) A E PO(X

,

T) iff A = U n D f01" some U E T and deηse D 드 X

Proof A E PO(X

,

T) • A 드 IntCIA

=

U E T. L.et D

=

X - (U - A) =

(X -U)UA. Then D is dense since X = CIAU(X - CIA) 드 CIAU(X-U) = ClD. Also. A = U n D. Conversely

,

if A = U n D with U E T and D dense

,

A 드 U

ç

IntClU = IntCl(A) so that A E PO(X

,

T)

Theorem 5. A space (X

,

T) is hyperconnecled iff the class of dense sets Of(X

,T)

is

T

p - {Ø}.

Proof Follow directly by using previous lemma and the fact that ^T 드 _{Tp •}

Lemma 3. If(X， T꺼) is submaxiη7πmí

Proof Clearly T 드 PO(X，T끼). N‘o아、w A E PO(X， T서) • A

=

^U n D for some U E T and dense D 드 X. Therefore

,

if (X

,

T) is submaximal

,

D E T • A E T.

Theorem 6. For a space (X

,

T) the fol/owi때 are equivalent

(i) (X

,

T) is submaximal

(ii) Theπ exists aη open

,

dense and heredilarily irresolνable subspace D

c

X and ^T

=

T p

(iii) PO(X

,

^T) 드 SO(X

,

^T) a뼈 T= 감，

(iv) A 드 X is η0ψhere deηse iff InlA

= ø

^{and T}

⁼

^Tp

(v) (X

,

^Tp ) is s띠 maximal

Proof By (i) and Lemma 3 we get T = PO(X

,

T) = T_p and hence (X

,

T_{p )} is submaximal. Therefore, the equivalents of (i) with each of other statments follow from Theorems 2 and 4 of [7J

Since submaximal spaces are hereditarily irresolvable. Which sub spaces of submaximal spaces are submaximal? Since each open and hence

On spaces via dense sets and SMPC functions 113

preopen subsets of a submaximal space are submaximal and we show a bit more.

We first note the following useful known lemma Lemma 4. If A E SO(X

,

r) then r"'/A = (r/A)"

Theorem 7. If(X

,

r) is submaximal and A E SO(X

,

r) then (A

,

r/A) is submaximal

Proof Since (X

,

r) is submaximal and A E SO(X

,

r). Then r r'"

and there is an open

,

dense

,

hereditarily irresolvable subset D 드 X. lf A ￥

0 ,

^{then Dn IntA is a dense}

,

^open

,

herediiarily irresolvable su bspace of (A

,

^r/A)

,

^and also CIA(Dn IntA)

=

^{An Cl(Dn IntA)}

=

^{An ClIntA}

=

^A

5ince r/A = r"/A = (r/A)"'

,

we have that (A

,

r/A) is s배maximal‘

Theorem 8. Sμ bmaximalitν is preserved bν open surjeclions

Proof If f: (X

,

^r) • (Y

,

σ) is an open surjection and (X

,

^{r) is submaxi-}

mal and if D 드 Y is dense

,

f-l(D) is dense and hence open in X so that D = f(J-l(D)) is open.

Corollary 1. [f fI X" is submaximal then each X" is submaximal

4. SMPC functions

Now

,

we characterize the SMPC-function based on the concepts : r.

and some previously spaces via dense sets

Theorem 9. f: (X

,

r) • (Y

, O" )

is SMPC iff f : (X

,

r) • (Y

,O".)

is continuous.

Proof A basic open set in

0".

^{has the form V = n}

k

⁼¹ Bk where each

Bk E PO(Y

, 0")

^‘ 50 if f ‘ (X

,

r) • (Y

,O")

is SMPC

,

and V is a basic

open set in

0".,

f-l(V) = n

k

^{=lf-1(Bk) E r so that f: (X}

^,

^{r) • (Y}

^,O".)

^is

continuous. The converse is clear since PO(Y

, 0" )

ζ

0" •.

We offer the following consequences

Theorem 10. If (Y

, 0")

^{is resolvable}

,

the following are equivalent

(i )

f: (X

,

r) • (Y

,O")

is SMPC.

(ii) f-l(B) is clopen (closed and open) for each B 드 Y (iii) f-l(y) is clopen for each y E Y.

(iv) f-l(y) is opeη for each y E Y.

(v)

f: (X

,

r) • (Y

,

2^Y) is coηti때

Proof By Lemma 1 and Theorem 9

Corollary 2. If (X

,

T) is connected and (Y

,

a) is resolvable then f (X

, T)

^• (Y

,

a) 성 SMPC iff f is a constant function.

For example if R is the usual space of real numbers

,

every non-constant function

f :

R ^• R is not SMPC.

Corollary 3.^’ If (X ，πT샤) is d값ense- 'l1따7

(Y， a이) is a nonempty resolvable space

,

then there is πo SMPC injection f:(X

,

T) • (Y

,

^a).

Lemma 5. (Theroem 5 of [7]) For a space (X, 'T) let X = F u G deη ote

the Hewitt-representation of (X

, T).

Then PO(X

, T )

is a topology on X iff CI G is open and {x} E PO(X,'T) for each x EIπt F.

Corollary 4. If a space (Y,a) as (X,'T) in Lemma 5, theη f: (X ,'T) • (Y,a) 엉 SMPC iff f : (X, T) • (Y, PO(Y, (7)) is continuous

Theorem 11. For f : (X,'T) • (Y, (7), the following holds

(i) If (X

,

T) is submaximal

,

then f is SMPC 펴 it is preirresolute.

(ii) SMPC coincides with continu때 if (Y

,

(7) is submaximal

(i i페l

prπ'Ccontinu“i“ty and continuity m'C equivalent.

Theorem 12. Let f : (Xπ) • (Y

,

(7) be a surjective SMPC function and (X

,

T) is compact. Then (Y

,

a) is st^1'Ongly compact.

Proof Let {끼

:

i E I} be a preopen cover of Y. By SMPC of f , f-l(

V; )

E 'T for each i E 1. Hence {f-l(κ) : i E I} is open cover of X which is compact‘ then there is 10 (finite) of 1 such that X = _{U {f-l(κ):iE 1o}.}

So

,

Y = u{ 끼 : i E Io}. Hence (Y

,

a) is strongly compact.

Corollary 5. St^1'Ongly compactness is preseπed uπder a SMPC surjec- tíon.

Theorem 13. Jf f : (X, T) • (Y,a) is SMPC suηective and (X, T) is LindelöJ, then (Y

,

(7) is st^1'Ongly Lindelöf

Proof Let {κ : i E I} be a preopen cover of Y

,

then {f-l(κ) : i E I} is open cover of X . Since X is Lindelöf, there exists a countable subcover with X = U

{f

-I(

V;) :

i E 10 (countable) }. Hence Y = u{κ ^‘ i E 10}, this gives the result.

Theorem 14. The inverse image under' an SMPC injection of a pr^'C-T

,

On spaces via dense sets and SMPC functions 115

space, is T; for i = (0,1,2)

Proof We prove this result in one case (say i = 0). So

,

let f : (X

,

T) • (Y, (7) be SMPC injective, (Y, (7) _{be pre-To} and Xl, X2 be two distinct points of X. Then f(Xl) 켜 f(X2) io Y. Hence for each preopen set V 드 Y containing one of f(xj),(j = 1,2). So, f-'(V) is open aod contains a corresponding point Xj, (j = 1,2). Then (X, T) is a 낀o-space. The proof of the others are similarly

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’‘ ’

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DEPARTMENT OF MATHEMATICS, EAST CENTRAL UNIVERSJTY, ODA, OKLAHOMA, 74820, U.S.A

DEPARTMENT OF MATHEMATtCS, FACULTY OF SCIENCE, MENOUFIA UNIVERSITY,

EGYPT