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@ Appl. Gen. Topol. 19, no. 2 (2018), 203-216doi:10.4995/agt.2018.7667
c© AGT, UPV, 2018

On rings of real valued clopen continuous

functions

Susan Afrooz a, Fariborz Azarpanah b and Masoomeh Etebar b

a
Khoramshahr University of Marine Science and Technology, Khoramshahr, Iran (afrooz@yahoo.com)

b
Department of Mathematics, Shahid Chamran University of Ahvaz, Ahvaz, Iran (Azarpanah@ipm.ir,

m.etebar@scu.ac.ir)

Communicated by D. Georgiou

Abstract

Among variant kinds of strong continuity in the literature, the clopen
continuity or cl-supercontinuity (i.e., inverse image of every open set is
a union of clopen sets) is considered in this paper. We investigate and
study the ring Cs(X) of all real valued clopen continuous functions on
a topological space X. It is shown that every f ∈ Cs(X) is constant on
each quasi-component in X and using this fact we show that Cs(X) ∼=
C(Y ), where Y is a zero-dimensional s-quotient space of X. Whenever
X is locally connected, we observe that Cs(X) ∼= C(Y ), where Y is
a discrete space. Maximal ideals of Cs(X) are characterized in terms
of quasi-components in X and it turns out that X is mildly compact
(every clopen cover has a finite subcover) if and only if every maximal
ideal of Cs(X) is fixed. It is shown that the socle of Cs(X) is an
essential ideal if and only if the union of all open quasi-components in
X is s-dense. Finally the counterparts of some familiar spaces, such as
Ps-spaces, almost Ps-spaces, s-basically and s-extremally disconnected
spaces are defined and some algebraic characterizations of them are
given via the ring Cs(X).

2010 MSC: 54C40.

Keywords: clopen continuous (cl-supercontinuous); zero-dimensional; Ps-
space; almost Ps-space; Baer ring; p.p. ring; quasi-component;
socle; mildly compact; s-basically and s-extremally disconnected
space.

Received 11 May 2017 – Accepted 01 July 2017

http://dx.doi.org/10.4995/agt.2018.7667


S. Afrooz, F. Azarpanah and M. Etebar

1. Introduction

If X and Y are topological spaces, then a function f : X → Y is said to be
clopen continuous [9] or cl-supercontinuous [10] if for every x ∈ X and for each
open set V in Y containing f(x), there exists a clopen (closed and open) set U
in X containing x such that f(U) ⊆ V . Since this is a strong form of continuity,
let us rename “clopen continuous” as strongly continuous and for brevity write
s-continuous. If A is a subset of X, then s-interior of A denotes the set of all
x ∈ A such that A contains a clopen set containing x and we denote it by intsA.
A subset G of X is said to be s-open if G = intsG. In fact a set is s-open if
and only if it is a union of clopen sets. The set of all x ∈ X such that every
clopen set containing x intersects A is called s-closure of A and it is denoted
by clsA. Similarly a set H is called s-closed if H = clsH and a set is s-closed if
and only if it is an intersection of clopen sets. A bijection function θ : X → Y
is called s-homeomorphism [10, under the name of cl-homeomorphism] if both
θ and θ−1 are s-continuous. If such function from X onto Y exists, we say that
X and Y are s-homeomorphic and we write X ∼=s Y .

We denote by C(X) the ring of all real-valued continuous functions on a
space X and by Cs(X) the set of all real valued s-continuous functions on X.
It is easy to see that Cs(X) is a ring and in fact it is a subring of C(X). For
each f ∈ C(X), the zero-set of f, denoted by Z(f), is the set of zeros of f
and X \ Z(f) is the cozero-set of f. The set of all zero-sets in X is denoted
by Z(X) and we also denote by Zs(X) the set of all zero-sets Z(f), where
f ∈ Cs(X). Clearly Zs(X) ⊆ Z(X). If Z ∈ Zs(X), then it is the inverse image
of the closed subset {0} of R under an element of Cs(X) and this implies that
every zero-set in Zs(X) is s-closed. Hence every cozero-set X \ Z(f), where
f ∈ Cs(X), is s-open. The converse is not necessarily true. For instance let S
be an uncountable space in which all points are isolated except for a distinguish
point s. Neighborhoods of s are considered to be those sets containing s with
countable complement, see Problem 4N in [6]. Since {s} =

⋂
s6=a∈S(S \ {a}),

the singleton {s} is s-closed but it is not a zero-set. It is well-known that a
space X is a completely regular Hausdorff space if and only if Z(X) is a base
for closed subsets of X, if and only if the set of all cozero-sets is a base for open
subsets of X, see Theorem 3.2 in [6]. Whenever X is zero-dimensional (i.e., a
T1 space with a base consisting of clopen sets), then clearly C(X) and Cs(X)
coincide, see also Remark 2.3 in [10]. If X is a completely regular Hausdorff
space, the converse is also true, we cite this fact as a lemma for later use.

Lemma 1.1. Whenever X is zero-dimensional, then C(X) = Cs(X) and if X
is a completely regular Hausdorff space, the converse is also true.

Proof. The first implication is obvious, see also Remark 2.3 in [10]. For the
converse, as we have already mentioned the collection C = {X \ Z(f) : f ∈
C(X)} is a base for open sets in X. Now let f ∈ C(X) and x ∈ X \Z(f). Since
f(x) 6= 0 and f ∈ C(X) = Cs(X), there exists a clopen set U in X containing
x such that f(y) 6= 0 , for each y ∈ U. Hence U ⊆ X \ Z(f) which means that
X has a base with clopen sets. Since X is T1, it is zero-dimensional. �

c© AGT, UPV, 2018 Appl. Gen. Topol. 19, no. 2 204



On rings of real valued clopen continuous functions

We recall that a completely regular Hausdorff space X is a P-space if ev-
ery Gδ-set or equivalently every zero-set in X is open and it is an almost
P-space if every non-empty Gδ-set or equivalently every nonempty zero-set in
X has a non-empty interior. Hence every P-space is an almost P-space but the
converse fails, for instance, the one-point compactification of an uncountable
discrete space is an almost P-space which is not a P-space, see Example 2 in [8]
and problem 4K(1) in [6]. Basically (extremally) disconnected spaces are those
spaces in which every cozero-set (open set) has an open closure. Clearly ev-
ery extremally disconnected space is basically disconnected but not conversely,
see Problem 4N in [6]. Several algebraic and topological characterizations of
aforementioned spaces are given in [3], [4], [6] and [8].

An ideal I in a commutative ring is called a z-ideal if Ma ⊆ I for each a ∈ I,
where Ma is the intersection of all maximal ideals in the ring containing a. It is
easy to see that an ideal I in C(X) (Cs(X)) is a z-ideal if and only if whenever
f ∈ I, g ∈ C(X) (g ∈ Cs(X)) and Z(f) ⊆ Z(g), then g ∈ I, see Problem
4A in [6]. A non-zero ideal in a ring is said to be essential if it intersects
every non-zero ideal non-trivially. Intersection of all essential ideals in a ring is
called the socle of the ring. A topological characterization of essential ideals in
C(X) and the socle of C(X) are given in [2] and [7] respectively. An ideal I in
C(X) or Cs(X) is said to be fixed if

⋂
f∈I Z(f) 6= ∅, otherwise it is called free.

The space βX is the Stone-C̆ech compactification of X and for any p ∈ βX,
Mp (resp., Op) is the set of all f ∈ C(X) for which p ∈ clβXZ(f) (resp.,
p ∈ intβXclβXZ(f)).

The component Cx of a point x in a topological space X is the union of all
connected subspaces of X which contain x. The quasi-component Qx of x is the
intersection of all clopen subsets of X which contain x. Clearly Cx ⊆ Qx for
each x ∈ X and the inclusion may be strict, see Example 6.1.24 in [5]. It is well-
known that for any two distinct points x and y in a space X, either Qx = Qy
or Qx ∩ Qy = ∅, see [5]. Components and quasi-components in a space X
are closed and whenever X is locally connected, then they are also open, see
Corollary 27.10 in [12]. The converse of this fact is not true in general, see the
example which is given preceding Lemma 1.1. Whenever the components in a
space X are the points, then X is called totally disconnected. Equivalently, X
is totally disconnected if and only if the only non-empty connected subsets of
X are the singleton sets.

2. Some properties of clopen continuous functions

Behaviour of s-continuous functions on quasi-components is investigated in
this section. The results of this section play an important role in the next
sections.

Proposition 2.1. Let X and Y be topological spaces and f : X → Y be a
s-continuous function. Then the following statements hold.

(1) If x ∈ X, y ∈ Y and f(x) = y, then f(Qx) ⊆ Qy.
(2) If Y is a T1-space, then f is constant on each quasi-component in X.

c© AGT, UPV, 2018 Appl. Gen. Topol. 19, no. 2 205



S. Afrooz, F. Azarpanah and M. Etebar

(3) Whenever f is one-to-one and Y is a T1-space, then X is totally dis-
connected.

(4) If f is s-homeomorphism and f(x) = y, then f(Qx) = Qy. Moreover,
if X or Y is T1, then both X and Y are totally disconnected.

Proof. (1) Let z ∈ Qx but f(z) /∈ Qy. Then there is a clopen set U in Y such
that f(z) ∈ U and U ∩ Qy = ∅ which implies y /∈ U. Since f is s-continuous,
there exists a clopen set G in X containing z such that f(G) ⊆ U. But Qx ⊆ G
implies that f(Qx) ⊆ U and this yields that f(x) = y ∈ U, a contradiction.

(2) Let y ∈ Qx and f(x) 6= f(y). Then there exists an open set H in Y such
that f(x) ∈ H but f(y) /∈ H. Since f is s-continuous, there exists a clopen set
G in X containing x such that f(G) ⊆ H. But any clopen set containing x
contains y as well, for y ∈ Qx, so f(y) ∈ H, a contradiction.

(3) Using part (2), f is constant on Qx for every x ∈ X. But f is one-to-one,
so Qx is singleton for each x ∈ X. This implies that every quasi-component in
X, and thus every component of X is singleton, i.e., X is totally disconnected.

(4) It is evident by parts (1) and (3). �

Corollary 2.2. For every f ∈ Cs(X), the following statements hold.

(1) f is constant on each quasi-component in X.
(2) The zero-set Z(f) is a union of quasi-components in X. In fact Z(f) =⋃

x∈Z(f) Qx.

A space X is called ultra Hausdorff [11] if for each pair of distinct points
in X, there exists a clopen set in X containing one but not the other. Two
disjoint subsets of a space are called s-completely separated if there is a function
f ∈ Cs(X) which separates them. Similar to the proof of Theorem 1.15 in [6], it
is easy to see that two disjoint sets are s-completely separated if and only if they
are contained in two disjoint members of Zs(X). By the following proposition,
ultra Hausdorff spaces are characterized by quasi-components in the spaces
and every s-closed set is s-completely separated from every quasi-component
disjoint from it.

Proposition 2.3. Let X be a topological space. Then the following statements
hold.

(1) If A is a s-closed subset of X and x ∈ X\A, then there exists g ∈ Cs(X)
such that g(A) = {1} and g(Qx) = {0}.

(2) X is ultra Hausdorff if and only if every quasi-component in X is sin-
gleton.

Proof. (1) Since X \ A is s-open, there exists a clopen set U containing x such
that U ∩ A = ∅. Now define an idempotent e with Z(e) = U, i.e., e(U) = {0}
and e(X \ U) = {1}. Clearly e ∈ Cs(X), e(Qx) = 0, since Qx ⊆ U and
e(A) = 1.

(2) Whenever X is ultra Hausdorff and x, y ∈ X are distinct, then there
exists a clopen set U containing x but not y. This implies that y /∈ Qx, i.e. Qx

c© AGT, UPV, 2018 Appl. Gen. Topol. 19, no. 2 206



On rings of real valued clopen continuous functions

is singleton. Conversely, let x, y ∈ X be distinct points. Since Qx = {x}, there
is a clopen set U such that x ∈ U and y /∈ U, i.e., X is ultra Hausdorff. �

3. Cs(X) is a C(Y )

As an equivalent definition, a space X is zero-dimensional if and only if it is
T1 and for each point x ∈ X and each closed subset A of X not containing x,
there exists a clopen set G in X containing x such that G∩A = ∅. Clearly every
zero-dimensional space is Tychonoff. Whenever we consider the collection of
all clopen subsets of (X, τ) as a base for a topology τ∗ on X, then Cs(X, τ) =
C(X, τ∗), by Theorem 5.1 in [10]. But the space (X, τ∗) is not necessarily
T1 and so it may not be zero-dimensional. In the following theorem we show
that Cs(X) is in fact a C(Y ) for a zero-dimensional space Y which is also a
s-quotient space of X.

Theorem 3.1. For each topological space X, there exists a zero-dimensional
space Xz such that Cs(X) ∼= C(Xz).

Proof. For each x ∈ X, let Qx be the quasi-component of x and consider the
decomposition Xz = {Qx : x ∈ X}. Take a topology τ on Xz so that G ∈ τ
if and only if

⋃
Qx∈G

Qx is s-open in X. To see that τ is a topology, clearly

X =
⋃

Qx∈Xz
Qx and ∅ =

⋃
Qx∈∅

Qx imply that Xz and ∅ are open. Whenever

H and K are open sets in Xz, then
⋃

Qx∈H∩K
Qx = (

⋃
Qx∈H

Qx)
⋂
(
⋃

Qx∈K
Qx)

imply that H ∩ K is open in Xz. It is also easy to see that
⋃

α Hα is open
in Xz for each open set Hα in Xz. The space Xz is Hausdorff, in fact if Qx
and Qy are two distinct points in Xz, where x, y ∈ X, then x /∈ Qy and since
Qy is s-closed, there is an idempotent e ∈ Cs(X) such that e(Qy) = 0 and
e(Qx) = 1, by part (1) of Proposition 2.3. If we set H = {Qz : z ∈ Z(e)},
then Z(e) =

⋃
Qz∈H

Qz implies that H is a clopen subset of Xz. Moreover

Qy ∈ H but Qx /∈ H, i.e., Xz is ultra Hausdorff and hence it is hausdorff as
well. To show that Xz is zero-dimensional, let H be a closed set in Xz and
Qy /∈ H. Hence T =

⋃
Qx∈H

Qx is a s-closed subset of X and y /∈ T . Therefore
by Proposition 2.3, there exists a clopen set U in X containing T but not
containing y. Now

⋃
z∈U Qz = U implies that K = {Qz : z ∈ U} is a clopen

subset of Xz. Clearly H ⊆ K and Qy /∈ K, hence Xz is zero-dimensional.
Now it remains to show that Cs(X) ∼= C(Xz). To this end, define ϕ :

Cs(X) → C(Xz) by ϕ(f) = fz for each f ∈ Cs(X), where fz is defined by
fz(Qx) = f(x), for each x ∈ X. By Corollary 2.2, clearly ϕ is well-defined.
Moreover fz ∈ C(Xz) for each f ∈ Cs(X). In fact if fz(Qx) = f(x) = c,
then for each ε > 0, there exists a clopen set G in X containing x such that
f(G) ⊆ (c−ε, c+ε), for f ∈ Cs(X). Now we take the open set H = {Qz : z ∈ G}
in Xz containing Qx. Hence fz(H) = f(G) ⊆ (c − ε, c + ε) implies that
fz ∈ C(Xz). Whenever ϕ(f) = ϕ(g), f, g ∈ Cs(X), then fz = gz implies that
f(x) = fz(Qx) = gz(Qx) = g(x), for all x ∈ X. Hence f = g, i.e., ϕ is one-
to-one. ϕ is also homomorphism, for ϕ(f + g) = (f + g)z and (f + g)z(Qx) =
(f+g)(x) = f(x)+g(x) = fz(Qx)+gz(Qx), for each Qx ∈ Xz. This implies that

c© AGT, UPV, 2018 Appl. Gen. Topol. 19, no. 2 207



S. Afrooz, F. Azarpanah and M. Etebar

ϕ(f +g) = ϕ(f)+ϕ(g), for all f, g ∈ Cs(X) and hence ϕ is homomorphism. To
complete the proof, we must show that ϕ is onto. To this end, let g ∈ C(Xz).
The function f : X → R defined by f(x) = g(Qx), for all x ∈ X is s-continuous.
In fact, if x ∈ X, f(x) = g(Qx) = c and ε > 0 is given, then there is an
open set H in Xz containing Qx such that g(H) ⊆ (c − ε, c + ε). Now it is
enough to take the s-open subset G =

⋃
Qz∈H

Qz of X. Clearly x ∈ G and

f(G) ⊆ (c − ε, c + ε) which implies that f ∈ Cs(X). By definitions of f and ϕ,
it is clear that ϕ(f) = g and we have thus shown that ϕ is onto. �

Corollary 3.2. If every quasi-component in X is open, in particular, if X is
locally connected, then Cs(X) ∼= C(Y ) for a discrete space Y .

Proof. Whenever each quasi-component of X is open, then each set {Qx} is
open in the space Xz, defined in the proof of Theorem 3.1. Therefore each Qx
is an isolated point in Xz, so Y = Xz is discrete and Cs(X) ∼= C(Y ). �

The space Xz defined in the proof of Theorem 3.1 is a s-continuous image of
X. In fact, if we regard the natural function τ : X → Xz, with τ(x) = Qx for
each x ∈ X, then τ is continuous and fz ◦τ = f or equivalently, ϕ(f)◦τ = f. In
order to prove that τ is s-continuous at x ∈ X, let H be a clopen subset of Xz
containing Qx (in fact H is an element of a base of the zero-dimensional space
Xz). Now it is enough to take U =

⋃
Q∈H Q which is s-open in X containing

x and clearly τ(U) = H, i.e., τ is continuous at x. In this case we may say
naturally, like the notion of quotient space, that Xz is a s-quotient of X and
the induced map τ is a s-quotient map. The equality fz ◦ τ = f is evident.

We conclude this section by the following proposition.

Proposition 3.3. For two spaces X and Y , if X ∼=s Y , then Xz ∼= Yz.

Proof. let σ : X → Y be a s-homeomorphism. By Proposition 2.1, if σ(x) = y,
then σ(Qx) = Qy. In fact every quasi-component in Y is exactly the image of
a unique quasi-component in X under σ. Define φ : Xz → Yz by φ(Q) = σ(Q)
for each quasi-component Q in X. Clearly φ is a bijection map. Given a quasi-
component Qx in X, we show that φ is continuous at Qx. Let H be an open
subset of Yz containing φ(Qx) = σ(Qx) = Qy and take V =

⋃
Q∈H Q. Hence

by definition of open sets in Yz, V is s-open in Y containing σ(Qx) = Qy. Since
σ is onto, there exists an element of Qx, say x without loss of generality, such
that σ(x) = y. Therefore there exists a clopen set U in X containing x (and
hence containing Qx) such that σ(U) ⊆ V . Now we set G = {Qz : z ∈ U}.
Clearly G is open in Xz containing Qx and φ(G) ⊆ H. Similarly, φ

−1 is also
continuous and we are done. �

The converse of the Proposition 3.3 is not true. If we take the discrete space
X = {a, b} and the space Y = (0, 1) ∪ (1, 2) as a subspace of R, then clearly
X ∼= Xz and Yz is a discrete space containing two elements (0, 1) and (1, 2).
Hence Xz ∼= Yz, but X and Y are not s-homeomorphic.

c© AGT, UPV, 2018 Appl. Gen. Topol. 19, no. 2 208



On rings of real valued clopen continuous functions

4. Maximal ideals of Cs(X)

In this section, for each space X, we consider Xz and the isomorphism
ϕ : Cs(X) → C(Xz) as defined in the proof of Theorem 3.1. First we show
that ϕ takes fixed (free) ideals to fixed (free) ideals and using this, we transfer
some well-known facts in the context of C(X) to that of Cs(X).

Lemma 4.1. An ideal I in Cs(X) is fixed if and only if ϕ(I) is a fixed ideal
in C(Xz). In particular, ϕ takes fixed maximal ideals to fixed maximal ideals.

Proof. If g ∈ ϕ(I), then there exists f ∈ I such that g = ϕ(f) = fz. Now
f(x) = 0 if and only if g(Qx) = fz(Qx) = f(x) = 0. Therefore, x ∈

⋂
f∈I Z(f)

if and only if Qx ∈
⋂

g∈ϕ(I) Z(g) =
⋂

f∈I Z(fz). �

Theorem 4.2. For a topological space X, the fixed maximal ideals in Cs(X)
are precisely the sets

MQx = {f ∈ Cs(X) : Qx ⊆ Z(f)} x ∈ X.

The ideals MQx are distinct for distinct Qx. For each x ∈ X, Cs(X)/MQx is
isomorphic with the real field R.

Proof. Using Lemma 4.1, fixed maximal ideals of Cs(X) are precisely of the
form ϕ−1(My), where y ∈ Xz (i.e., y = Qx for some x ∈ X). Now f ∈ ϕ

−1(My)
if and only if fz ∈ My = MQx, or equivalently Qx ⊆ Z(f), by Theorem 3.1.
Hence ϕ−1(My) = MQx. Whenever Qp 6= Qq for p, q ∈ X, using Proposition
2.3, there exists g ∈ Cs(X) such that g(Qp) = 0 and g(Qq) = 1 and this
means that g ∈ MQp \ MQq . Finally σ : Cs(X) → R with σ(f) = f(x) (note
that f(y) = f(x), for all y ∈ Qx by Corollary 2.2), for all f ∈ Cs(X) is a
homomorphism with kernel MQx, so Cs(X)/MQx

∼= R. �

A space is said to be mildly compact [11] if every clopen cover of X has a finite
subcover. Clearly every compact space is mildly compact but not conversely.
For instance, consider the space X = (0, 1) ∪ (1, 2) as a subspace of R. By the
following proposition, for a space X, the compactness of Xz is equivalent to
mildly compactness of X.

Proposition 4.3. For a space X, the following statements are equivalent.

(1) X is mildly compact.
(2) Xz is compact.
(3) Every ideal of Cs(X) is fixed.
(4) Every maximal ideal of Cs(X) is fixed.

Proof. A collection {Hα : α ∈ S} is an open cover of Xz if and only if the
collection {Gα : α ∈ S}, where Gα =

⋃
Q∈Hα

Q, is a s-open cover of X. This

implies the equivalence of parts (1) and (2). The equivalence of third and fourth
parts with part (1) is an immediate consequence of Lemma 4.1 and Theorem
4.11 in [6]. �

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S. Afrooz, F. Azarpanah and M. Etebar

Using Theorem 3.1, and in view of the fact that there is a correspondence
between elements of the space βXz and the set of all maximal ideals of C(Xz)
by Theorem 7.3 in [6], all maximal ideals of Cs(X), fixed or free, will be char-
acterize by the following theorem for each space X.

Theorem 4.4. For every space X, the maximal ideals of Cs(X) are precisely
of the form

Mp = {f ∈ Cs(X) : p ∈ clβXz Z(fz)} p ∈ βXz.

Remark 4.5. As in C(X), for each maximal ideal M of Cs(X), we define OM =
{f ∈ Cs(X) : fg = 0 for some g /∈ M}, see 7.12(b) in [6] and the argument
preceding Theorem 2.12 in [1]. Whenever M is fixed, then M = MQx for some
x ∈ X by Theorem 4.2, and therefore we have OM = OQx = {f ∈ Cs(X) :
Qx ⊆ intsZ(f)}. In fact, if f ∈ OM , then fg = 0 for some g /∈ MQx. Hence
Qx ⊆ X \ Z(g) ⊆ Z(f). But X \ Z(g) is s-open, hence Qx ⊆ intsZ(f), so
f ∈ OQx. Whenever f ∈ OQx, then Qx ⊆ intsZ(f). Since intsZ(f) is s-open,
there exists a clopen set U containing Qx contained in intsZ(f). Now if we take
an idempotent e such that Z(e) = U, then Qx ⊆ Z(e) ⊆ intsZ(f). Therefore
1 − e /∈ MQx and (1 − e)f = 0 which implies that f ∈ OQx = OM .

By Theorem 2.4 in [4], X is zero-dimensional if and only if for each x ∈ X,
the ideal Ox is generated by a set of idempotents. Hence for each y ∈ Xz
(y = Qx for some x ∈ X), the ideal Oy is generated by a set of idempotents.
Now in view of Theorem 2.4 in [4] and using the isomorphism ϕ defined in the
proof of Theorem 3.1, the following corollary is evident.

Corollary 4.6. For each x ∈ X, the ideal OQx in Cs(X) is generated by a set
of idempotents.

Remark 4.7. By Theorem 4.9 in [6], two compact spaces X and Y are home-
omorphic if and only if C(X) and C(Y ) are isomorphic. It is clear that if
X ∼=s Y , then Cs(X) ∼= Cs(Y ). In fact, if σ : Y → X is a s-homeomorphism,
then f → f ◦ σ is a s-isomorphism between Cs(X) and Cs(Y ). But in contrast
to Theorem 4.9 in [6], we observe that Cs(X) ∼= Cs(Y ) does not necessarily
imply X ∼=s Y even if X and Y are mildly compact. To see this, consider
spaces X = { 1

n
: n ∈ N} ∪ {0} and Y = (

⋃∞
n=1(

1
n+1

, 1
n
)) ∪ {0} as subspaces

of R. Clearly, X and Y are mildly compact. Also Cs(X) ∼= Cs(Y ), in fact,
every g ∈ Cs(Y ) is constant on each interval (

1
n+1

, 1
n
), by Proposition 2.1,

say g(( 1
n+1

, 1
n
)) = {an}. Now if we define fg : X → R by fg(

1
n
) = an and

fg(0) = g(0), then fg ∈ Cs(X) and θ : Cs(Y ) → Cs(X), θ(g) = fg for each
g ∈ Cs(Y ) is an isomorphism. Since there is no bijection function between X
and Y , these two spaces are not s-homeomorphic.

Proposition 4.8. If Xz ∼= Yz, then Cs(X) ∼= Cs(Y ) and whenever X and Y are
mildly compact, then the converse is also true.

Proof. Xz ∼= Yz implies that C(Xz) ∼= C(Yz) and hence by Theorem 3.1,
Cs(X) ∼= Cs(Y ). For the converse, Cs(X) ∼= Cs(Y ) implies C(Xz) ∼= C(Yz)

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On rings of real valued clopen continuous functions

by Theorem 3.1. Now using Proposition 4.3, Xz and Yz are compact, hence
Xz ∼= Yz, by Theorem 4.9 in [6]. �

5. Some relations between algebraic properties of Cs(X) and
topological properties of X

We call a space X a Ps-space if every zero-set in Zs(X) is open. Clearly,
every P-space is a Ps-space but not conversely. For instance, X = (0, 1)∪(1, 2)
as a subspace of R is not a P-space whereas it is a Ps-space, for Zs(X) =
{∅, X, (0, 1), (1, 2)}, by Corollary 2.2. Every Ps-space is not necessarily a com-
pletely regular space. it is enough to consider a non-completely regular space
with two components.

Whenever f ∈ Cs(X), then Z(fz) = {Qx : f(x) = 0} and Z(f) =
⋃

Qx∈Z(fz)
Qx.

These imply, by definition of open sets in Xz, that Z(f) is s-open in X if and
only if Z(fz) is open in Xz. On the oder hand, since Cs(X) ∼= C(Xz), by
Theorem 3.1, the ring C(Xz) is regular if and only if Cs(X) is a regular ring.
In view of these points, the following result is an immediate consequence of
Problem 4J in [6].

Proposition 5.1. A space X is a Ps-space if and only if Cs(X) is a regular
ring.

The counterparts of the other conditions of Problem 4J in [6] can be obtained
more or less for regularity of Cs(X). For example, Cs(X) is regular if and only
if MQx = OQx, for each x ∈ X if and only if every ideal in Cs(X) is a z-ideal
and so on. Note that for each f, g ∈ Cs(X), it is easy to see that Z(f) ⊆ Z(g)
if and only if Z(fz) ⊆ Z(gz) and this implies that an ideal I in Cs(X) is a
z-ideal if and only if ϕ(I) is a z-ideal in C(Xz).

We already observed that every Ps-space is not necessarily a P-space. By
the following result, this happens if and only if X is zero-dimensional.

Proposition 5.2. A space X is a P-space if and only if X is a zero-dimensional
Ps-space.

Proof. Clearly, every P-space is basically disconnected, hence using Problem
16O in [6], every P-space is zero-dimensional. Every P-space is a Ps-space as
well. Conversely, since X is zero-dimensional, C(X) = Cs(X) by Lemma 1.1
and since X is a Ps-space, Cs(X) is a regular ring, by Proposition 5.1. This
implies that C(X) is also a regular ring. Now using Problem 4J in [6], X is a
P-space. �

We call a space X an almost Ps-space if every non-empty zero-set in Zs(X)
has a non-empty s-interior. However the notion of almost Ps-space is the
counterpart of that of almost P-space but the class of almost P-spaces and
the class of almost Ps-spaces are dissimilar. The following example shows that
these classes are not comparable and non of them is larger than the other.

Example 5.3. Whenever every quasi-component in a space X is open, in
particular, if X is locally connected, then X is a Ps-space. In fact if Z(f) 6= ∅,

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S. Afrooz, F. Azarpanah and M. Etebar

f ∈ Cs(X), then Z(f) is a union of quasi-components in X, by Corollary 2.2.
Since each quasi-component in X is clopen, Z(f) is s-open. This implies that
Y = (0, 1) ∪ (1, 2) as a subspace of R is a Ps-space. Hence Y is an almost
Ps-space but clearly, it is not an almost P-space. Also every almost P-space
need not be an almost Ps-space. To see this let X be a (completely regular
Hausdorff) connected almost P-space, see Proposition 2.3 in [8] for existence
of such a space. Take a point σ ∈ X and let Y = X ∪ N with the topology
as follows: elements of N are considered to be isolated points, neighborhoods
of all points of X, except σ will be the same as in the space X while each
neighborhood of σ in Y will be of the form G ∪ A, where G is an open set in
X containing σ and A is a subset of N such that N \ A is finite. If we define
f : Y → R with f(n) = 1

n
and f(X) = {0}, then f ∈ Cs(Y ). Since X is

connected and X =
⋂∞

n=1(Y \ {n}), it is a quasi-component in Y . Also it does
not contain any clopen subset of Y (note that X itself is not open in Y , for σ is
the cluster point of the subset N of Y and hence N is not closed in Y ). Hence
intsZ(f) = ∅, i.e., Y is not an almost Ps-space.

It remains to show that Y is an almost P-space. Let f ∈ C(Y ). If Z(f)∩N 6=
∅, then clearly intY Z(f) 6= ∅. Now suppose that Z(f) ∩ N = ∅. Whenever
σ /∈ Z(f), then intY Z(f) = intXZ(f|X) 6= ∅, for X is an almost P-space.
Finally, suppose that σ ∈ Z(f), then Z(f) 6= {σ}, for otherwise intXZ(f|X) =
intX{σ} 6= ∅ implies that σ is an isolated point of X which contradicts the
connectedness of X. Therefore there exists x 6= σ such that x ∈ Z(f). Since
Y is completely regular Hausdorff, define h ∈ C(Y ) so that h(x) = 0 and
h(σ) = 1. Now take g = f2 + h2, then σ /∈ Z(g) ⊆ X implies that ∅ 6=
intXZ(g) = intY Z(g) ⊆ intY Z(f). Hence Y is an almost P-space.

For the proof of the following proposition, we need the following lemma.

Lemma 5.4. Let f, g ∈ Cs(X).

(1) If Z(g) ⊆ intsZ(f), then f is a multiple of g.
(2) Zs(X) is closed under countable intersection.

Proof. (1) Let Xz, ϕ and fz for each f ∈ Cs(X) be as in the proof of Theorem
3.1. Let Qx ∈ Z(gz), where x ∈ X. Hence x ∈ Z(g) and so x ∈ intsZ(f), by
our hypothesis. Since intsZ(f) is s-open, there exists a clopen set U such that
x ∈ U ⊆ Z(f). Now H = {Qy : y ∈ U} is clopen in Xz and Qx ∈ H ⊆ Z(fz).
This implies that Z(gz) ⊆ intXz Z(fz) and using Problem 1D in [6], there is
kz ∈ C(Xz), where k ∈ Cs(X) such that fz = kzgz. Now it is clear that f = kg.

(2) It is easy to see that whenever {Sn} is a sequence in Cs(X) converges
uniformly to a function f, then f ∈ Cs(X). Now, as in 1.14(a) in [6], if for
each n ∈ N, we consider Zn = Z(fn), where fn ∈ Cs(X) and |fn| ≤ 1 (note

that if f ∈ Cs(X), then
f

1+|f|
∈ Cs(X), Z(

f
1+|f|

) = Z(f) and | f
1+|f|

| ≤ 1), then

the sequence Sn =
∑n

i=1 fi/2
i converges uniformly to a function f ∈ Cs(X).

Clearly
⋂∞

n=1 Z(fn) = Z(f). �

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On rings of real valued clopen continuous functions

Proposition 5.5. For a topological space X, the following statements are
equivalent.

(1) X is an almost Ps-space.
(2) For each non-unit f ∈ Cs(X), f = ef for some idempotent e 6= 1 in

Cs(X).
(3) For each non-unit f ∈ Cs(X), fe = 0 for some idempotent e 6= 0 in

Cs(X).
(4) Every non-empty countable intersection of s-open sets has a non-empty

s-interior.

Proof. (1)⇒(2) If f ∈ Cs(X) is not unit, then intsZ(f) 6= ∅. Since intsZ(f)
is s-open, there exists a non-empty clopen set U contained in Z(f). Take the
idempotent e with Z(e) = U. Clearly e 6= 1, for U 6= ∅. Since Z(e) ⊆ intsZ(f),
f is a multiple of e, by Lemma 5.4. Hence f = eg for some g ∈ Cs(X). But
f = g on X \ Z(e), so f = ef.

(2)⇒(3) f = ef implies f(1 − e) = 0, where 1 − e is a non-zero idempotent.
(3)⇒(4) Let A =

⋂∞
n=1 An 6= ∅, where each An is s-open. Let x ∈ A. Hence

there is an idempotent en ∈ Cs(X) such that x ∈ Z(en) ⊆ An, for each n ∈ N.
Now by Lemma 5.4,

⋂∞
n=1 Z(en) is a zero-set, say Z(g), where g ∈ Cs(X).

Since g is non-unit (x ∈ Z(g)), there exists an idempotent 0 6= e ∈ Cs(X) such
that ge = 0, by our hypothesis. Therefore ∅ 6= Z(1 − e) ⊆ Z(g) ⊆ A which
means that A has a non-empty s-interior.

(4)⇒(1) Since every zero-set in Zs(X) is a countable intersection of s-open
sets, the proof is evident. We note that whenever f ∈ Cs(X), then Z(f) =⋂∞

n=1 f
−1((− 1

n
, 1
n
)) and each f−1((− 1

n
, 1
n
)) is s-open, by Theorem 2.2 in [10].

�

We call a space X s-basically (s-extremally) disconnected if for every zero-
set Z(f) ∈ Zs(X) (s-closed subset H of X), intsZ(f) (intsH) is s-closed.
Equivalently, X is a s-basically (s-extremally) disconnected space if and only
if for each X \ Z(f), f ∈ Cs(X) (s-open subset G of X), cls(X \ Z(f)) (clsG)
is s-open. We show the counterparts of Theorems 3.3 and 3.5 in [4] that the
s-basically (s-extremally) disconnectedness of X is equivalent to saying that
Cs(X) is a p.p. ring (Baer ring). Recall that a ring R is said to be p.p. ring (Baer
ring) if for each a ∈ R ( S ⊆ R), Ann(a) (AnnS) is generated by an idempotent,
where Ann(a) := {r ∈ R : ar = 0} (AnnS := {r ∈ R : rs = 0, ∀s ∈ S}). First
we need the following lemma.

Lemma 5.6. Let X be a topological space and Xz be the space mentioned in
the proof of Theorem 3.1.

(1) If f ∈ Cs(X), then
⋂

g∈Ann(f) Z(g) = cls(X \ Z(f)).

(2) X is s-extremally (s-basically) disconnected if and only if Xz is ex-
tremally (basically) disconnected.

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S. Afrooz, F. Azarpanah and M. Etebar

Proof. (1) Since
⋂

g∈Ann(f) Z(g) =
⋂

X\Z(f)⊆Z(g) Z(g), we have

X \ Z(f) ⊆
⋂

g∈Ann(f)

Z(g).

But
⋂

g∈Ann(f) Z(g) is s-closed, hence cls(X \ Z(f)) ⊆
⋂

g∈Ann(f) Z(g). Con-

versely, let x ∈
⋂

g∈Ann(f) Z(g) but x /∈ cls(X \ Z(f)). Hence there exists

an idempotent e ∈ Cs(X) such that e(x) = 1 and e(cls(X \ Z(f))) = 0, by
Proposition 2.3. This implies that e ∈ Ann(f), but e(x) = 1 yields that
x /∈

⋂
g∈Ann(f) Z(g), a contradiction.

(2) Let X be s-extremally disconnected and H be an open set in Xz. Then
G =

⋃
Qx∈H

Qx = {x ∈ X : Qx ∈ H} is a s-open set in X. Hence clsG is
s-open, by our hypothesis. Now let Qx ∈ clXz H. Then every clopen set in
Xz containing Qx intersects H and this implies that every clopen set in X
containing x intersects G as well. Therefore x ∈ clsG. But clsG is s-open,
then there exists a clopen set U in X containing x contained in clsG. Now
V = {Qy : y ∈ U} is a clopen set in Xz containing Qx contained in clXz H, i.e.,
clXz H is open in Xz and hence Xz is extremally disconnected. The proof of
the converse is similar. In case of s-basically disconnectedness, the proof goes
along the lines of the above arguments, so it is left to the reader. �

Proposition 5.7. Let X be a topological space.

(1) Cs(X) is a p.p. ring if and only if X is a s-basically disconnected space.
(2) Cs(X) is a Baer ring if and only if X is a s-extremally disconnected

space.

Proof. (1) We may apply each part of Lemma 5.6, we prefer to use part (1).
If Cs(X) is a p.p. ring, then for each f ∈ Cs(X), Ann(f) = (e) for some
idempotent e. Now by Lemma 5.6, Z(e) =

⋂
g∈Ann(f) Z(g) = cls(X \ Z(f))

which implies that cls(X \ Z(f)) is clopen and hence it is s-open. Therefore X
is a s-basically disconnected space.

Conversely, suppose that X is s-basically disconnected. Hence
⋂

g∈Ann(f) Z(g) =

cls(X \ Z(f)) is s-open and hence it is clopen. Now take an idempotent e
with Z(e) = cls(X \ Z(f)). Since X \ Z(f) ⊆ cls(X \ Z(f)) = Z(e), we
have ef = 0, i.e., e ∈ Ann(f). On the other hand if g ∈ Ann(f), then
Z(e) = cls(X \ Z(f)) ⊆ Z(g) implies that Z(e) ⊆ intsZ(g) and by Lemma
5.4, g ∈ (e), i.e., Ann(f) ⊆ (e).

(2) If Cs(X) is a Baer ring, then C(Xz) is also a Baer ring, for Cs(X) ∼=
C(Xz), by Theorem 3.1. Now by Theorem 2.5 in [4], Xz is extremally discon-
nected. Thus using Lemma 5.6, X is s-extremally disconnected. The proof of
the converse is similar. �

It is manifest that every basically (extremally) disconnected space is a s-
basically (s-extremally) disconnected space. The converse is not true in general.
For example let X = (0, 1) ∪ (1, 2) be as a subspace of R. In fact X is a Ps-
space which is not basically disconnected and it is not extremally disconnected

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On rings of real valued clopen continuous functions

as well. It is not hard to see that every s-basically disconnected almost Ps-space
is a Ps-space. The following result states that the zero-dimensionality and s-
basically (s-extremally) disconnectedness is equivalent to basically (extremally)
disconnectedness.

Proposition 5.8. A space is basically (extremally) disconnected if and only if
it is s-basically (s-extremally) disconnected zero-dimensional.

Proof. By Problem 16O in [6], every basically (extremally) disconnected space
is zero-dimensional and since every basically (extremally) disconnected space
is also s-basically (s-extremally) disconnected, the left-to-right implication is
immediate. For the converse, whenever X is zero-dimensional, then by Lemma
1.1, C(X) = Cs(X). Now if X is s-basically (s-extremally) disconnected, then
by Proposition 5.7, C(X) = Cs(X) is a p.p. (Baer) ring. Now by Theorems 3.3
and 3.5 in [4], X is basically (extremally) disconnected. �

The socle CF (X) of C(X) which is the intersection of all essential ideals
in C(X) is the set of all functions which vanish everywhere except on a finite
number of isolated points of X, see Proposition 3.3 in [7]. Corollary 2.3 in [2]
and Proposition 2.1 in [7] show that the socle of C(X) is essential if and only
if the set of isolated points of X is dense in X.

Proposition 5.9. Let X be a topological space and Xz be the space defined in
the proof of Theorem 3.1.

(1) The socle Ss(X) of Cs(X) is free if and only if every quasi-component
in X is open, if and only if Xz is discrete.

(2) The socle of Cs(X) is essential if and only if the union of open quasi-
components in X is s-dense in X (A subset D of X is called s-dense
in X if every non-empty s-open subset of X intersects D).

Proof. We remind the reader that a subset H of Xz is open if and only if⋃
Qx∈H

Qx is s-open in X. This implies that for each x ∈ X, the quasi-
component Qx is an isolated point of Xz if and only if Qx is s-open in X and
hence it should be clopen. Since CF (Xz) is the set of all functions in C(Xz)
which vanish everywhere except on a finite set of isolated points of Xz, Ss(X)
will be the set of all functions in Cs(X) which vanish everywhere except on a
finite union of open quasi-components in X. Therefore,

⋂
f∈Ss(X)

Z(f) is the

union of all non-open quasi-components in X. Now it is clear that Ss(X) is
free if and only if every quasi-components in X is open and this is equivalent
to saying that Xz is discrete. For the proof of part (2), it is easy to see that
the density of isolated points of Xz is equivalent to the density of the union
of open quasi-components in X. Since Cs(X) ∼= C(Xz), the socle of C(Xz) is
essential if and only if the socle of Cs(X) is. Now using Corollary 2.3 in [2], we
are done. �

It is known that every extremally disconnected P-space of non-measurable
cardinal is discrete, see Problem 12H in [6]. We conclude the paper by the
counterpart of this fact.

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S. Afrooz, F. Azarpanah and M. Etebar

Proposition 5.10. Every quasi-component in a s-extremally disconnected Ps-
space of non-measurable cardinal is open.

Proof. Let X be s-extremally disconnected Ps-space of non-measurable cardi-
nal. Then Cs(X) is a Baer ring by Proposition 5.7. But using Proposition 3.1,
Cs(X) ∼= C(Xz), hence C(Xz) is also a Baer ring. Therefore Xz is extremally
disconnected, by Theorem 3.5 in [4]. On the other hand, since X is Ps-space,
Cs(X) will be regular by Proposition 5.1, whence C(Xz) is also regular and
hence Xz is a P-space by Problem 4J in [6]. Finally, |Xz| ≤ |X| implies that
the cardinal of Xz is non-measurable, see part (i) in the proof of Theorem 12.5
in [6]. Now Xz is extremally disconnected P-space of non-measurable cardinal
which means that Xz is discrete, by Problem 12H in [6]. Therefore each Qx is
an isolated point in Xz and hence each Qx should be open in X. �

Acknowledgements. The authors would like to thank the referee for a careful

reading of this article.

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