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 AInt(Cl(A)),
(3) semiopen [14] if ACl(Int(A)),
(4) sc-open [5] if Ais semiopen and union of closed sets,
(5) α-open [20] if AInt(Cl(Int(A))),
(6) γ-open [8] if AInt(Cl(A)) Cl(Int(A)),
(7) β-open [1] or semipreopen [7] if ACl(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)).
Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 52–63
54 Zanyar A. Ameen
Remark 2.1.It is well-known that for a space X,ταO(X)P O(X)SO(X)γO(X)
βO(X).
Definition 2.2. Let Xbe a space and let AX. A point xXis said to be in the preclosure
(resp. semi-closure, α-closure, β-closure, γ-closure) of Aif UA6=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 ACl(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 ACl(A) always, so ACl(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}nN. 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 ABis 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 ABis semiopen in B, [17, Lemma 1.1].
(ii) If Ais semiopen and Bis preopen, then ABis preopen in A, [17, Lemma 2.1].
Lemma 2.6. Let Ybe a subspace of a space Xand let AY.
(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 UACl(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
ABis 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 ABis β-open for
each semiopen subset Bof X.
Proof. Given subsets A, B Xsuch that Ais preopen and Bis semiopen. Let xABand
let Ube an open set containing x. Since xInt(Cl(A)), then UInt(Cl(A)) is also an open set
containing x. Set V=UInt(Cl(A)). But xCl(Int(B)), so
UInt(Cl(A)) Int(B) = VInt(B)6=.
This implies that
ABCl [Int(Cl(A)) Int(B)] = Cl [Int [Cl(A)Int(B)]] ,
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56 Zanyar A. Ameen
and therefore,
ABCl [Int [Cl(A)Int(B)]] Cl(Int(Cl(AB))).
Hence ABis β-open.
Conversely, assume that ABis β-open for each semiopen set Bin X. We need to show that
AInt(Cl(A)). Suppose contrary that there exists xXsuch that xAand x /Int(Cl(A)).
Then xCl(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 xInt(Cl(A)), which contradicts our assumption. Therefore, if xA,
then xInt(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
AUis swn-open for each open set Uin X.
Lemma 2.10. Let Xbe a space. A subset Aof Xis semiopen if and only if AUis 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 xAand assume that AUis sw-open for each open set Uin X. That
is Int(AU)6=. But 6= Int(AU) = Int(A)Int(U) = Int(A)U, which implies that
xCl(Int(A)) and so ACl(Int(A)). This proves that Ais semiopen.
Lemma 2.11. Let Xbe a space. A subset Aof Xis α-open if and only if AUis 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 xAand assume that AUis sw-open for each α-open set Uin X. That
is Int(AU)6=. But 6= Int(AU) = 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 xClβ(Int(A)U)
Clβ(Int(A)). By Lemma 2.3 (iii), Clβ(Int(A)) = Int(Cl(Int(A))) and so AInt(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.
Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 52–63
A note on some forms of continuity 57
Proof. (i) (ii): Let Abe a β-open set in X. Then ACl(Int(Cl(A))). By (i), Int(Cl(A))
Int(Cl(Int(A))). Therefore, ACl(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[AInt(X\A)]. This implies that AInt(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[AInt(X\A)] and Int[Int(A)X\A]
are (open) dense. Now,
Int[AInt(X\A)] \Int[Int(A)(X\A)] = Int[AInt(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 AInt(Cl(A)). By (viii), = Int(Cl((A))) =
Int((A)) = Int(Cl(A))\Cl(Int(A)). It follows that AInt(Cl(A)) Cl(Int(A)) and so Cl(A) =
Cl(Int(A)). Since AInt[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 xX, each open set Gcontaining f(x) and each open U
containing x, there exists a nonempty open set Vwith VUsuch that f(V)G,
(10) almost quasicontinuous [10], if for each xX, each open set Gcontaining f(x) and each
open Ucontaining x,f1(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:XYis 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:XYis 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:XYis 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:XYis 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 f1|U(H) = f1(H)Uis sw-open in U. Since Uis an open subset of X,
clearly, f1(H)Uis sw-open in Xand so, by Lemma 2.10, f1(H) is semiopen in X. Thus f
is quasicontinuous.
Theorem 3.5. Let X, Y be spaces. A function f:XYis 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, ABis
α-open in A.
Proof. Given the sets A, B, then
ABAInt(Cl(Int(B)))
IntA[AInt(Cl(Int(B)))]
IntA[Cl(Int(A)) Int(Cl(Int(B)))]
IntA[Cl[Int(A)Int(Cl(Int(B)))]]
IntA(Cl(Int(AB)))
IntA(Cl(IntA(AB))).
Since IntA(Cl(IntA(AB))) is an open set in A, so IntA(Cl(IntA(AB)))A= IntA(Cl(IntA(A
B)) A), and hence ABIntA(Cl(IntA(AB)) A) = IntA(ClA(IntA(AB))). This shows
that ABis α-open in A.
Theorem 3.6. Let X, Y be spaces. A function f:XYis α-continuous if and only if f|Uis
quasicontinuous for each semiopen subset UX.
Proof. Assume that fis α-continuous. Let Hbe an open subset of Yand let Ube a semiopen
subset of X. By assumption f1(H) is α-open in X. By Lemma 3.1, f1(H)Uis α-open in U
and thus, by Diagram I, f1(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 f1|U(H) = f1(H)Uis semiopen in U. Since Uis semiopen in X,
by Lemma 2.6 (i), f1(H)Uis semiopen in Xfor each semiopen Uand thus, by Lemma 2.7,
f1(H) is α-open in X. Thus fis α-continuous.
Theorem 3.7. Let X, Y be spaces. A function f:XYis nearly continuous if and only if f|U
is almost quasicontinuous for each semiopen subset UX.
Proof. Suppose that fis nearly continuous. Let Hbe an open subset of Yand let Ube a
semiopen subset of X. By hypothesis f1(H) is preopen in X. By Lemma 2.5 (ii), f1(H)U
is preopen in Uand thus, by Diagram I, f1(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 f1|U(H) = f1(H)Uis a β-open subset of U. Since U
is semiopen in Xand each semiopen is β-open, by Lemma 2.6 (ii), f1(H)Uis β-open in
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60 Zanyar A. Ameen
Xfor each semiopen Uand thus, by Lemma 2.8, f1(H) is preopen in X. Thus fis nearly
continuous.
Theorem 3.8. For a function f:XY, 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:XY, the following are equivalent:
(1) fis quasicontinuous;
(2) For each xX, each open set Gcontaining f(x)and each open Ucontaining x,f1(G)U
is sw-open;
(3) For each xX, each open set Gcontaining f(x)and each α-open Ucontaining x, there
exists a nonempty open set Vwith VUsuch that f(V)G;
(4) For each xX, each open set Gcontaining f(x)and each α-open Ucontaining x, there
exists a nonempty α-open set Vwith VUsuch that f(V)G;
(5) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exists
a nonempty α-open set Vwith VUsuch that f(V)G;
(6) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exists
a nonempty semiopen set Vwith VUsuch 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 VUsuch that f(V)G. Therefore Vf1(G)
and so VInt(f1(G)). Thus 6=V=VUInt(f1(G)) U= Int(f1(G)U), which
implies that f1(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(f1(G)Int(U)) = Int(f1(G)) Int(U)6=. Set V= Int(f1(G)) Int(U).
Clearly, Vis a nonempty open set Uand
f(V)f(Int(f1(G)) Int(U)) f(f1(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 VUsuch that f(V)G. Therefore Vf1(G)
and so VInts(f1(G)). Thus 6=V=VUInts(f1(G)) U, which implies that
Ints(f1(G)) U6=for each open Ucontaining x. Hence xCl(Ints(f1(G))). By Lemma
2.3 (i), xf1(G)Cl(Int(f1(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:XY, the following are equivalent:
(1) fis α-continuous;
(2) For each xX, each open set Gcontaining f(x)and each semiopen Ucontaining x, there
exist a nonempty open set Vwith VUsuch 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), xf1(G)Int(Cl(Int(f1(G)))) = Cls(Int(f1(G))). This implies
that Int(f1(G)) B6=for each semiopen Bcontaining xand so Int(f1(G)) U6=. If
W= Int(f1(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(f1(G)) U)f(f1(G)) G.
This completes the proof of (2).
(2)(1): Let xXand let Gbe an open set containing f(x). By (2), for each semiopen
set Ucontaining x, there is a nonempty open set Vwith VUsuch that f(V)G. Then
Vf1(G) and so VInt(f1(G)). Thus 6=V=VUInt(f1(G)) U, which implies
that Int(f1(G)) U6=for each semiopen Ucontaining x. Therefore xCls(Int(f1(G))).
By Lemma 2.3 (iii), xf1(G)Int(Cl(Int(f1(G)))). Hence fis α-continuous.
Theorem 4.3. Let X, Y be topological spaces. For a function f:XY, the following are
equivalent:
(1) fis almost quasicontinuous;
(2) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exist
a nonempty preopen set Vwith VUsuch that f(V)G;
(3) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exist
a nonempty γ-open set Vwith VUsuch that f(V)G;
(4) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exist
a nonempty β-open set Vwith VUsuch 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), xf1(G)Cl(Int(Cl(f1(G)))) = Cl(Intp(f1(G))). This implies
that Intp(f1(G)) O6=for each open Ocontaining xand so Intp(f1(G)) U6=. Set
V= Intp(f1(G)) U. By Lemma 2.4, Vis a nonempty preopen set with VUand
f(V)f(Intp(f1(G)) U)f(f1(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 VUsuch that f(V)G.
Then Vf1(G) and so VIntβ(f1(G)). Thus 6=V=VUIntβ(f1(G)) U, which
implies that Intβ(f1(G))U6=for each open Ucontaining x. Therefore xCl(Intβ(f1(G))).
By Lemma 2.3 (ii), xf1(G)Cl(Intβ(f1(G))) = Cl(Int(Cl(f1(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:XY, the following are
equivalent:
(1) fis almost quasicontinuous;
(2) For each xX, each open set Gcontaining f(x)and each open Ucontaining x,f1(G)U
is swn-open;
(3) For each xX, each open set Gcontaining f(x)and each α-open Ucontaining x, there
exists a nonempty open set Vwith VUsuch that VCl(f1(G));
(4) For each xX, each open set Gcontaining f(x)and each α-open Ucontaining x, there
exists a nonempty α-open set Vwith VUsuch that VCl(f1(G));
(5) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exists
a nonempty α-open set Vwith VUsuch that VCl(f1(G));
(6) For each xX, each open set Gcontaining f(x)and each open Ucontaining x, there exists
a nonempty semiopen set Vwith VUsuch that VCl(f1(G)).
Proof. Follows from Theorem 4.1 and Lemma 2.9.
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