Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 52–63

A note on some forms of continuity

Una nota sobre algunas formas de continuidad

Zanyar A. Ameen (zanyar@uod.ac)

Deptartment of Mathematics, College of Science, University of Duhok

Duhok, Kurdistan Region, IRAQ.

Abstract

New connections and characterizations of some classes of continuous functions are ob-

tained. In particular, we characterize quasicontinuity and almost quasicontinuity in terms

of weak types of open sets.

Key words and phrases: quasicontinuous; almost quasicontinuous; semicontinuous;

precontinuous; α-continuous; b-continuous; β-continuous; somewhat continuous; somewhat

nearly continuous.

Resumen

Se obtienen nuevas conexiones y caracterizaciones de algunas clases de funciones conti-

nuas. En particular, caracterizamos la cuasi continuidad y casi cuasicontinuidad en t´erminos

de tipos d´ebiles de conjuntos abiertos.

Palabras y frases clave: cuasicontinuo; casi cuasicontinuo; semicontinuo; precontinuo;

α-continuous; b-continuo; β-continuo; algo continuo; algo casi continuo.

1 Introduction

The role of continuity of functions is essential in developing theory in all branches of (pure)

mathematics, especially in topology and analysis, for decades. Then various generalizations of

continuity have been introduced. In 1932, Kempisty [13] deﬁned the notion of quasicontinuity

which has been of interest to many analysts and topologists, and there is a rich literature on these

functions, see the survey article [18]. In 1958, Ptak [26] introduced nearly continuous functions

to generalize the BanachSchauder Theorem. Levine [14] deﬁned the concept of semicontinuity, in

1963, in terms of semiopen sets. Ten years later, Neubrunnova [19] showed that quasicontinuity

and semicontinuity are similar. A big part of this work is motivated by that result. Gentry

[12] has given a weaker class of quasicontinuity called somewhat continuity while studying the

invariance of Baire spaces under mappings. Then α-continuity was given by Njastad [20], which

implies both nearly continuity and quasicontinuity. Mashhour et al. [16] have introduced the

class of precontinuous functions which is equivalent to the class of nearly continuous functions.

Abd-El-Monsef [1] studied β-continuous functions by using the notion of β-openness of sets.

In 1990, Borsik [10] introduced an equivalent notion to β-continuity under the name of almost

Received 18/01/2021. Revised 15/03/2021. Accepted 14/07/2021.

MSC (2010): Primary 54C08; 54C10.

Corresponding author: Zanyar Ameen

A note on some forms of continuity 53

quasicontinuous functions. Almost quasicontinuity is weaker than both nearly continuity and

quasicontinuity. In 1987, Piotrowski [23] deﬁned a weak version of somewhat continuity called

somewhat nearly continuity to generalize problems in separate versus joint continuity and in

the Closed Graph Theorem. In 2009, Ameen [5] deﬁned a subclass of quasicontinuous functions

called sc-continuous. He showed that quasicontinuity and sc-continuity are identical on T1-

spaces. All such classes of functions mentioned earlier are weaker than the class of continuous

functions except sc-continuity which is incomparable. Due to the importance of these classes of

continuous functions, we present some more connections between these functions and give further

characterizations.

2 Preliminaries and Auxiliary Materials

Throughout this paper, the letters N,Qand R, respectively, stand for the set of natural, rational

and real numbers. The word ”space” mean an arbitrary topological space. For a subset Aof a

space (X, τ ), the closure and interior of Awith respect to Xrespectively are denoted by ClX(A)

and IntX(A) (or simply Cl(A) and Int(A)).

Deﬁnition 2.1. A subset Aof a space X is said to be

(1) regular open if A= Int(Cl(A)),

(2) preopen [16] if A⊆Int(Cl(A)),

(3) semiopen [14] if A⊆Cl(Int(A)),

(4) sc-open [5] if Ais semiopen and union of closed sets,

(5) α-open [20] if A⊆Int(Cl(Int(A))),

(6) γ-open [8] if A⊆Int(Cl(A)) ∪Cl(Int(A)),

(7) β-open [1] or semipreopen [7] if A⊆Cl(Int(Cl(A))),

(8) somewhat open (brieﬂy sw-open) [23] if Int(A)6=∅or A=∅,

(9) somewhat nearly open (brieﬂy swn-open) [23] (for more details, see [4]) if Int(Cl(A)) 6=∅or

A=∅. The class of somewhat nearly open sets (except ∅) were studied under the name of

somewhere dense sets in [2].

The complement of a regular open (resp. preopen, semiopen, sc-open, α-open, β-open, γ-

open, sw-open, swn-open) set is regular closed (resp. preclosed, semi-closed, sc-closed, α-closed,

β-closed, γ-closed, sw-closed, swn-closed).

The intersection of all preclosed (resp. semiclosed, α-closed, β-closed, γ-closed) sets in X

containing Ais called the preclosure (resp. semi-closure, α-closure, β-closure, γ-closure) of A,

and is denoted by Clp(A) (resp. Cls(A), Clα(A), Clβ(A), Clγ(A)).

The union of all preopen (resp. semiopen, α-open, β-open, γ-open) sets in Xcontained in A

is called the preinterior (resp. semi-interior, α-interior, β-interior, γ-interior) of A, and is denoted

by Intp(A) (resp. Ints(A), Intα(A), Intβ(A), Intγ(A)).

The family of all preopen (resp. semiopen, α-open, γ-open, β-open) subsets of Xis denoted

by P O(X) (resp. SO(X), αO(X), γO(X), βO(X)).

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54 Zanyar A. Ameen

Remark 2.1.It is well-known that for a space X,τ⊆αO(X)⊆P O(X)∪SO(X)⊆γO(X)⊆

βO(X).

Deﬁnition 2.2. Let Xbe a space and let A⊆X. A point x∈Xis said to be in the preclosure

(resp. semi-closure, α-closure, β-closure, γ-closure) of Aif U∩A6=φfor each preopen (resp.

semiopen, α-open, β-open, γ-open) set Ucontaining x.

Lemma 2.1. Let Abe a subset of a space X.

(i) Ais semiopen if and only if Cl(A) = Cl(Int(A)).

(ii) Ais β-open if and only if Cl(A) = Cl(Int(Cl(A))).

Proof. (i) If Ais semiopen, then A⊆Cl(Int(A)) and so Cl(A)⊆Cl(Int(A)). For other side of

inclusion, we always have Int(A)⊆A. Therefore Cl(Int(A)) ⊆Cl(A). Thus Cl(A) = Cl(Int(A)).

Conversely, assume that Cl(A) = Cl(Int(A)), but A⊆Cl(A) always, so A⊆Cl(Int(A)).

Hence Ais semiopen.

(ii) Theorem 2.4 in [7].

Lemma 2.2. Let Abe a nonempty subset of a space X.

(i) If Ais semiopen, then Int(A)6=∅.

(ii) If Ais β-open, then Int(Cl(A)) 6=∅.

Proof. (i) Suppose otherwise that if Ais a semiopen set such that Int(A) = ∅, by Lemma 2.1 (i),

Cl(A) = ∅which implies that A=∅. Contradiction!

(ii) Similar to (i).

At this place, perhaps a connection among the classes of open sets (deﬁned above excluding

γ-open as we have only used in Theorem 4.3) is needed.

open set sc-open set regular closed set

α-open set semiopen set sw-open set

preopen set β-open set swn-open set

\

Diagram I

In general, none of these implications can be replaced by equivalence as shown below:

Example 2.1. Consider Rwith the usual topology. Let A=R\{ 1

n}n∈N. Obviously Ais α-open

but not open. If B= [0,1],Bis semiopen but not α-open. If C=Q,Cis preopen but not α-open.

Let D= [0,1]∪((1,2) ∩Q). Then Dis both β-open and sw-open but neither preopen nor semiopen

([8, Example 1]). If E= [0,1) ∩Q, then Eis swn-open but not sw-open. Let F=C ∪ [2,3],

where Cis the Cantor set. Then Fis swn-open but not β-open. Let G= (0,1] = S∞

n=1[1

n,1]. So

Gis sc-open but neither open nor regular closed.

Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 52–63

A note on some forms of continuity 55

Example 2.2. [5, Example 2.2.3] Consider X={a, b, c}with the topology τ={φ, X,{a},{b},

{a, b}}. The set {a}is semiopen but not sc-open.

Lemma 2.3. [7, Theorems 3.13, 3.14 & 3.22][8, Proposition 2.6] For a subset Aof a space X,

we have

(i) Cl(Int(A)) = Cl(Ints(A)) = Clα(Intα(A)) = Cl(Intα(A)) = Clα(Int(A)),

(ii) Cl(Int(Cl(A))) = Cl(Intp(A)) = Cl(Intγ(A)) = Cl(Intβ(A)),

(iii) Int(Cl(Int(A))) = Cls(Int(A)) = Clγ(Int(A)) = Clβ(Int(A)),

Lemma 2.4. Let A, B be subsets of X. If Ais open and Bis α-open (resp. preopen, semiopen,

β-open), then A∩Bis α-open (resp. preopen, semiopen, β-open) in X.

Proof. Proposition 2 in [20] (resp. Lemma 4.1 in [24], Lemma 1 in [21], Theorem 2.7 in [1]).

Lemma 2.5. Let A, B be subsets of a space X.

(i) If Ais semiopen and Bis preopen, then A∩Bis semiopen in B, [17, Lemma 1.1].

(ii) If Ais semiopen and Bis preopen, then A∩Bis preopen in A, [17, Lemma 2.1].

Lemma 2.6. Let Ybe a subspace of a space Xand let A⊂Y.

(i) If Yis semiopen in X, then Ais semiopen in Yif and only if Ais semiopen in X.

(ii) If Yis β-open in X, then Ais β-open in Yif and only if Ais β-open in X.

(iii) If Yis semiopen in X, then Ais swn-open in Yif and only if it is swn-open in X.

Proof. (i) [15, Theorem 2.4].

(ii) The ﬁst direction is proved in [1, Theorem 2.7]. The converse is can be followed from [14,

Theorem 6] and from the fact that Ais β-open if and only if there exist a preopen open Usuch

that U⊆A⊆Cl(U).

(iii) [4, Theorem 3.14].

Lemma 2.7. [20, Proposition 1] Let Xbe a space. A subset Aof Xis α-open if and only if

A∩Bis semiopen for each semiopen subset Bof X.

In a similar way, we prove the following:

Lemma 2.8. Let Xbe a space. A subset Aof Xis preopen if and only if A∩Bis β-open for

each semiopen subset Bof X.

Proof. Given subsets A, B ⊆Xsuch that Ais preopen and Bis semiopen. Let x∈A∩Band

let Ube an open set containing x. Since x∈Int(Cl(A)), then U∩Int(Cl(A)) is also an open set

containing x. Set V=U∩Int(Cl(A)). But x∈Cl(Int(B)), so

U∩Int(Cl(A)) ∩Int(B) = V∩Int(B)6=∅.

This implies that

A∩B⊆Cl [Int(Cl(A)) ∩Int(B)] = Cl [Int [Cl(A)∩Int(B)]] ,

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56 Zanyar A. Ameen

and therefore,

A∩B⊆Cl [Int [Cl(A)∩Int(B)]] ⊆Cl(Int(Cl(A∩B))).

Hence A∩Bis β-open.

Conversely, assume that A∩Bis β-open for each semiopen set Bin X. We need to show that

A⊆Int(Cl(A)). Suppose contrary that there exists x∈Xsuch that x∈Aand x /∈Int(Cl(A)).

Then x∈Cl(Int(Ac)) and obviously Int(Ac)∪{x}is semiopen. By assumption, A∩(Int(Ac)∪{x})

is β-open. But A∩(Int(Ac)∪ {x}) = {x}. By Lemma 2.2 (ii) and [6, Lemma 2.1], {x}is

preopen. This implies x∈Int(Cl(A)), which contradicts our assumption. Therefore, if x∈A,

then x∈Int(Cl(A)) and so Ais preopen.

Lemma 2.9. [4, Proposition 3.16] Let Xbe a space. A subset Aof Xis β-open if and only if

A∩Uis swn-open for each open set Uin X.

Lemma 2.10. Let Xbe a space. A subset Aof Xis semiopen if and only if A∩Uis sw-open

for each open set Uin X.

Proof. Since each semiopen set is sw-open and the intersection of a semiopen set with an open

set is semiopen, by Lemma 2.4, so the ﬁrst part follows.

Conversely, let x∈Aand assume that A∩Uis sw-open for each open set Uin X. That

is Int(A∩U)6=∅. But ∅ 6= Int(A∩U) = Int(A)∩Int(U) = Int(A)∩U, which implies that

x∈Cl(Int(A)) and so A⊆Cl(Int(A)). This proves that Ais semiopen.

Lemma 2.11. Let Xbe a space. A subset Aof Xis α-open if and only if A∩Uis sw-open for

each α-open set Uin X.

Proof. Since the intersection of two α-open sets is α-open and each α-open set is sw-open, so the

ﬁrst part is proved.

Conversely, let x∈Aand assume that A∩Uis sw-open for each α-open set Uin X. That

is Int(A∩U)6=∅. But ∅ 6= Int(A∩U) = Int(A)∩Int(U)⊆Int(A)∩Int(Cl(U)) = Int(A)∩

Int(Cl(Int(U))), which implies that Int(A)∩Clβ(U)6=∅and therefore x∈Clβ(Int(A)∩U)⊆

Clβ(Int(A)). By Lemma 2.3 (iii), Clβ(Int(A)) = Int(Cl(Int(A))) and so A⊆Int(Cl(Int(A))).

This proves that Ais α-open.

Lemma 2.12. Let Xbe a space. The following are equivalent:

(i) each preopen subset of Xis α-open,

(ii) each β-open subset of Xis semiopen,

(iii) each preopen subset of Xis semiopen,

(iv) each dense subset of Xis semiopen,

(v) each dense subset of Xhas an interior dense,

(vi) each co-dense subset of Xis nowhere dense,

(vii) each swn-open subset of Xis sw-open,

(viii) each subset of Xhas a nowhere dense boundary.

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A note on some forms of continuity 57

Proof. (i) ⇒(ii): Let Abe a β-open set in X. Then A⊆Cl(Int(Cl(A))). By (i), Int(Cl(A)) ⊆

Int(Cl(Int(A))). Therefore, A⊆Cl(Int(Cl(A))) ⊆Cl[Int(Cl(Int(A)))] = Cl(Int(A)). Hence Ais

semiopen.

The implications “(ii) ⇒(iii)” and “(iii) ⇒(iv)” are clear as each dense is preopen and each

preopen is β-open.

(iv) ⇒(v): Let Dbe a dense subset of X. By Lemma 2.1 (i), X= Cl(D) = Cl(Int(D)).

Thus Int(D) is dense.

(v)⇔(vi): Let Abe co-dense. Then Int(A) = ∅ ⇐⇒ Cl(X\A) = X. By (v), Cl(Int(X\A)) =

X⇐⇒ Int(Cl(A)) = ∅. Hence Ais nowhere dense.

(vi) ⇔(vii): Let Abe an swn-open set in X. Suppose Ais not sw-open. That is, Ais

co-dense. By (vi), Int(Cl(A)) = ∅. Contradiction that assumption that Ais swn-open. The

other way is similar.

(v) ⇔(viii): Let Abe a subset of X. Then X= Cl(A)∪(X\Cl(A)) = Cl(A)∪Int(X\A)⊆

Cl[A∪Int(X\A)]. This implies that A∪Int(X\A) is dense in X. By the same, we can conclude

that Int(A)∪X\Ais also dense in X. By (v), both Int[A∪Int(X\A)] and Int[Int(A)∪X\A]

are (open) dense. Now,

Int[A∪Int(X\A)] \Int[Int(A)∪(X\A)] = Int[A∪Int(X\A)\Int(A)∪(X\A)]

= Int[Int(A)[Int(X\A)]

=X\∂(A),

where ∂(A) means the topological boundary of A. Since the intersection of two open dense is

dense, so X\∂(A) is open dense. Thus ∂(A) is nowhere dense.

(viii) ⇔(i): Let Abe preopen. That is A⊆Int(Cl(A)). By (viii), ∅= Int(Cl(∂(A))) =

Int(∂(A)) = Int(Cl(A))\Cl(Int(A)). It follows that A⊆Int(Cl(A)) ⊆Cl(Int(A)) and so Cl(A) =

Cl(Int(A)). Since A⊆Int[Cl(A)] = Int[Cl(Int(A))]. This proves that Ais α-open.

3 Relationships and properties

This section is devoted to some properties of the following classes of continuous functions and

their relationships.

Deﬁnition 3.1. A function ffrom a space Xto a space Yis called

(1) rc-continuous [11], if the inverse image of each open set in Yis regular closed in X,

(2) sc-continuous [5], if the inverse image of each open set in Yis sc-open in X,

(3) semicontinuous [14], if the inverse image of each open set in Yis semiopen in X,

(4) nearly continuous [26], or precontinuous [16], if the inverse image of each open set in Yis

preopen in X,

(5) α-continuous [20], if the inverse image of each open set in Yis α-open in X,

(6) β-continuous [1], if the inverse image of each open set in Yis β-open in X,

(7) somewhat continuous [12] (brieﬂy sw-continuous), if the inverse image of each open set in Y

is sw-open in X,

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58 Zanyar A. Ameen

(8) somewhat nearly continuous [23, 4] (brieﬂy swn-continuous), if the inverse image of each

open set in Yis swn-open in X,

(9) quasicontinuous [13], if for each x∈X, each open set Gcontaining f(x) and each open U

containing x, there exists a nonempty open set Vwith V⊆Usuch that f(V)⊆G,

(10) almost quasicontinuous [10], if for each x∈X, each open set Gcontaining f(x) and each

open Ucontaining x,f−1(G)∩Uis not nowhere dense.

Remark 3.1.(i) It is proved in [25] that almost quasicontinuity and β-continuity are equivalent

(see also [9, Theorem 1]). An easier proof can be followed from the deﬁnition of almost quasi-

continuity and Lemma 2.9.

(ii) The equivalence of semicontinuity and quasicontinuity is given in [19, Theorem 1.1].

(iii) somewhat nearly continuous functions coincide with surjective SD-continuous in [3].

The following diagram shows the relationship between above functions, which is an enlarge-

ment of the Diagram I given in [23]:

continuous sc-continuous rc-continuous

α-continuous quasicontinuous sw-continuous

nearly continuous almost quasicontinuous swn-continuous

\

Diagram II

In general, none of the implications is reversible. Examples 5.2-5.3 in [4] show that the

existence of swn-continuous functions that are not almost quasicontinuous or sw-continuous.

Counterexamples for other cases are available in the literature.

Theorem 3.1. Let X, Y be spaces such that Xis T1. A function f:X→Yis α-continuous if

and only if it is both sc-continuous and nearly continuous.

Proof. Proposition 2.2.10 in [5] and Theorem 3.2 in [22].

Theorem 3.2. [9, Proposition 1] Let X, Y be spaces. A function f:X→Yis almost quasicon-

tinuous if and only if f|Uis swn-continuous for each open subset Uof X.

Theorem 3.3. [4, Theorem 5.7] Let X, Y be spaces. A function f:X→Yis nearly continuous

if and only if f|Uis swn-continuous for each α-open subset Uof X.

Similar to the above results, we prove the following:

Theorem 3.4. Let X, Y be spaces. A function f:X→Yis quasicontinuous if and only if f|U

is sw-continuous for each open subset Uof X.

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A note on some forms of continuity 59

Proof. The ﬁrst part follows from [17, Theorem 1.3], which implies that each quasicontinuous

restricted to an open set is again quasicontinuous and hence sw-continuous.

Conversely, suppose that f|Uis sw-continuous for each open subset Uof X. Let Hbe an

open in Y. Then f−1|U(H) = f−1(H)∩Uis sw-open in U. Since Uis an open subset of X,

clearly, f−1(H)∩Uis sw-open in Xand so, by Lemma 2.10, f−1(H) is semiopen in X. Thus f

is quasicontinuous.

Theorem 3.5. Let X, Y be spaces. A function f:X→Yis α-continuous if and only if f|Uis

sw-continuous for each α-open subset Uof X.

Proof. Similar steps given in the proof of the above theorem and Lemma 2.11.

Lemma 3.1. Let A, B be subsets of a space X. If Ais semiopen and Bis α-open, A∩Bis

α-open in A.

Proof. Given the sets A, B, then

A∩B⊆A∩Int(Cl(Int(B)))

⊆IntA[A∩Int(Cl(Int(B)))]

⊆IntA[Cl(Int(A)) ∩Int(Cl(Int(B)))]

⊆IntA[Cl[Int(A)∩Int(Cl(Int(B)))]]

⊆IntA(Cl(Int(A∩B)))

⊆IntA(Cl(IntA(A∩B))).

Since IntA(Cl(IntA(A∩B))) is an open set in A, so IntA(Cl(IntA(A∩B)))∩A= IntA(Cl(IntA(A∩

B)) ∩A), and hence A∩B⊆IntA(Cl(IntA(A∩B)) ∩A) = IntA(ClA(IntA(A∩B))). This shows

that A∩Bis α-open in A.

Theorem 3.6. Let X, Y be spaces. A function f:X→Yis α-continuous if and only if f|Uis

quasicontinuous for each semiopen subset U⊆X.

Proof. Assume that fis α-continuous. Let Hbe an open subset of Yand let Ube a semiopen

subset of X. By assumption f−1(H) is α-open in X. By Lemma 3.1, f−1(H)∩Uis α-open in U

and thus, by Diagram I, f−1(H)∩Uis a semiopen subset of U. Hence, f|Uis quasicontinuous.

Conversely, suppose that f|Uis quasicontinuous for each semiopen subset Uof X. Let Hbe

an open set in Y. Then f−1|U(H) = f−1(H)∩Uis semiopen in U. Since Uis semiopen in X,

by Lemma 2.6 (i), f−1(H)∩Uis semiopen in Xfor each semiopen Uand thus, by Lemma 2.7,

f−1(H) is α-open in X. Thus fis α-continuous.

Theorem 3.7. Let X, Y be spaces. A function f:X→Yis nearly continuous if and only if f|U

is almost quasicontinuous for each semiopen subset U⊆X.

Proof. Suppose that fis nearly continuous. Let Hbe an open subset of Yand let Ube a

semiopen subset of X. By hypothesis f−1(H) is preopen in X. By Lemma 2.5 (ii), f−1(H)∩U

is preopen in Uand thus, by Diagram I, f−1(H)∩Uis β-open in U. Therefore, f|Uis almost

quasicontinuous.

Conversely, suppose that f|Uis almost quasicontinuous for each semiopen subset Uof X.

Let Hbe an open set in Y. Then f−1|U(H) = f−1(H)∩Uis a β-open subset of U. Since U

is semiopen in Xand each semiopen is β-open, by Lemma 2.6 (ii), f−1(H)∩Uis β-open in

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60 Zanyar A. Ameen

Xfor each semiopen Uand thus, by Lemma 2.8, f−1(H) is preopen in X. Thus fis nearly

continuous.

Theorem 3.8. For a function f:X→Y, the following are equivalent:

(i) each nearly continuous function is α-continuous,

(ii) each almost quasicontinuous function is quasicontinuous,

(iii) each nearly continuous function is quasicontinuous,

(iv) each swn-continuous function is sw-continuous.

Proof. Apply Lemma 2.12

4 Characterizations

Theorem 4.1. Let X, Y be spaces. For a function f:X→Y, the following are equivalent:

(1) fis quasicontinuous;

(2) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x,f−1(G)∩U

is sw-open;

(3) For each x∈X, each open set Gcontaining f(x)and each α-open Ucontaining x, there

exists a nonempty open set Vwith V⊆Usuch that f(V)⊆G;

(4) For each x∈X, each open set Gcontaining f(x)and each α-open Ucontaining x, there

exists a nonempty α-open set Vwith V⊆Usuch that f(V)⊆G;

(5) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exists

a nonempty α-open set Vwith V⊆Usuch that f(V)⊆G;

(6) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exists

a nonempty semiopen set Vwith V⊆Usuch that f(V)⊆G.

Proof. (1)⇒(2): Let Gbe an open set containing f(x) and let Ube any open containing x. By

(1), there is a nonempty open set Vwith V⊆Usuch that f(V)⊆G. Therefore V⊆f−1(G)

and so V⊆Int(f−1(G)). Thus ∅ 6=V=V∩U⊆Int(f−1(G)) ∩U= Int(f−1(G)∩U), which

implies that f−1(G)∩Uis sw-open.

(2)⇒(3): Let Gbe an open set in Ycontaining f(x) and let Ube an α-open set in X

containing x. Since each α-open set is semiopen, by Lemma 2.2 (i), Int(U) is a nonempty open

set. By (2) Int(f−1(G)∩Int(U)) = Int(f−1(G)) ∩Int(U)6=∅. Set V= Int(f−1(G)) ∩Int(U).

Clearly, Vis a nonempty open set Uand

f(V)⊆f(Int(f−1(G)) ∩Int(U)) ⊆f(f−1(G)) ⊆G.

This proves (3).

The implications “(3)⇒(4)”, “(4)⇒(5)” and “(5)⇒(6)” are clear from the Diagram I.

(6)⇒(1): Let Gbe an open set containing f(x). By (6), for each open set Ucontaining x,

there is a nonempty semiopen set Vwith V⊆Usuch that f(V)⊆G. Therefore V⊆f−1(G)

and so V⊆Ints(f−1(G)). Thus ∅ 6=V=V∩U⊆Ints(f−1(G)) ∩U, which implies that

Ints(f−1(G)) ∩U6=∅for each open Ucontaining x. Hence x∈Cl(Ints(f−1(G))). By Lemma

2.3 (i), x∈f−1(G)⊆Cl(Int(f−1(G))). As xwas taken arbitrarily, so fis quasicontinuous.

Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 52–63

A note on some forms of continuity 61

The proofs of the following theorems are quite similar to the proof of Theorem 4.1. But for

the sake of completeness, we provide them.

Theorem 4.2. Let X, Y be spaces. For a function f:X→Y, the following are equivalent:

(1) fis α-continuous;

(2) For each x∈X, each open set Gcontaining f(x)and each semiopen Ucontaining x, there

exist a nonempty open set Vwith V⊆Usuch that f(V)⊆G.

Proof. (1)⇒(2): Let Gbe an open set containing f(x) and let Ube a semiopen containing x. By

(1) and Lemma 2.3 (iii), x∈f−1(G)⊆Int(Cl(Int(f−1(G)))) = Cls(Int(f−1(G))). This implies

that Int(f−1(G)) ∩B6=∅for each semiopen Bcontaining xand so Int(f−1(G)) ∩U6=∅. If

W= Int(f−1(G)) ∩U, by Lemma 2.4, Wis a nonempty semiopen set. Set V= Int(W). By

Lemma 2.2 (i), Vis nonempty open and

f(V)⊆f(Int(f−1(G)) ∩U)⊆f(f−1(G)) ⊆G.

This completes the proof of (2).

(2)⇒(1): Let x∈Xand let Gbe an open set containing f(x). By (2), for each semiopen

set Ucontaining x, there is a nonempty open set Vwith V⊆Usuch that f(V)⊆G. Then

V⊆f−1(G) and so V⊆Int(f−1(G)). Thus ∅ 6=V=V∩U⊆Int(f−1(G)) ∩U, which implies

that Int(f−1(G)) ∩U6=∅for each semiopen Ucontaining x. Therefore x∈Cls(Int(f−1(G))).

By Lemma 2.3 (iii), x∈f−1(G)⊆Int(Cl(Int(f−1(G)))). Hence fis α-continuous.

Theorem 4.3. Let X, Y be topological spaces. For a function f:X→Y, the following are

equivalent:

(1) fis almost quasicontinuous;

(2) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exist

a nonempty preopen set Vwith V⊆Usuch that f(V)⊆G;

(3) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exist

a nonempty γ-open set Vwith V⊆Usuch that f(V)⊆G;

(4) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exist

a nonempty β-open set Vwith V⊆Usuch that f(V)⊆G.

Proof. (1)⇒(2): Let Gbe an open set containing f(x) and let Ube an open set containing x. By

(1) and Lemma 2.3 (ii), x∈f−1(G)⊆Cl(Int(Cl(f−1(G)))) = Cl(Intp(f−1(G))). This implies

that Intp(f−1(G)) ∩O6=∅for each open Ocontaining xand so Intp(f−1(G)) ∩U6=∅. Set

V= Intp(f−1(G)) ∩U. By Lemma 2.4, Vis a nonempty preopen set with V⊆Uand

f(V)⊆f(Intp(f−1(G)) ∩U)⊆f(f−1(G)) ⊆G.

This proves (2).

The implications “(2)⇒(3)” and “(3)⇒(4)” are follow from Remark 2.1.

(4)⇒(1): Let xbe any point in Xand let Gbe an open set containing f(x). By (4), for each

open set Ucontaining x, there is a nonempty β-open set Vwith V⊆Usuch that f(V)⊆G.

Then V⊆f−1(G) and so V⊆Intβ(f−1(G)). Thus ∅ 6=V=V∩U⊆Intβ(f−1(G)) ∩U, which

implies that Intβ(f−1(G))∩U6=∅for each open Ucontaining x. Therefore x∈Cl(Intβ(f−1(G))).

By Lemma 2.3 (ii), x∈f−1(G)⊆Cl(Intβ(f−1(G))) = Cl(Int(Cl(f−1(G)))). This prove that f

is almost quasicontinuous.

Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 52–63

62 Zanyar A. Ameen

Corollary 4.1. Let X, Y be topological spaces. For a function f:X→Y, the following are

equivalent:

(1) fis almost quasicontinuous;

(2) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x,f−1(G)∩U

is swn-open;

(3) For each x∈X, each open set Gcontaining f(x)and each α-open Ucontaining x, there

exists a nonempty open set Vwith V⊆Usuch that V⊆Cl(f−1(G));

(4) For each x∈X, each open set Gcontaining f(x)and each α-open Ucontaining x, there

exists a nonempty α-open set Vwith V⊆Usuch that V⊆Cl(f−1(G));

(5) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exists

a nonempty α-open set Vwith V⊆Usuch that V⊆Cl(f−1(G));

(6) For each x∈X, each open set Gcontaining f(x)and each open Ucontaining x, there exists

a nonempty semiopen set Vwith V⊆Usuch that V⊆Cl(f−1(G)).

Proof. Follows from Theorem 4.1 and Lemma 2.9.

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